SSC Placement-Paper

 S.S.C. Combined Graduate Level Exam(Tier-II) 2011 
Arithmetical Ability

1.   The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –  

a.   74

b.   78

c.   80

d.   90

Ans : Let the numbers be x and y .

∴  xy = 1575

And  x/y = 9/7

∴  xy/x/y = 1575/9/7

∴   y2  = 1225

∴ y = 35 and x = 45

∴  The sum of the numbers = 45+35

  = 80

 

2.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

 a.   3

b.   6

c.   9

d.   8.2

 

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

 

3.   √6+√6+√6+… is equal to –

 

a.   2

b.   5

c.   4

d.   3

 

Ans : Let √6+√6+√6+…. be x.

∴   x =√6+x

∴   x2 = 6+x

∴   x2 + x – 6 = 0

∴   (x-3) (x+2) = 0

∴   x = 3

 

4.   The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

 

a.   24

b.   32

c.   40

d.   28

 

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴  x2 + x2 + 4x + 4 = 394

∴  2 x2 +4x – 390 = 0

∴  x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴  x = 13

∴   Required sum = 13 +15 =28

 

 

5.   When (6767 +67) is divided by 68, the remainder is-

 

a.   1

b.   63

c.   66

d.   67

 

Ans:  (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴  6766 is an even number

∴  (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴  The remainder = 67

 

 6.   In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

 

a.   4b-a2 /a  =3

b.   4b-2a /a2 =2

c.   (a+1)2 = 4b

d.   A(a+2)/b = 4

 

Ans : As  divisor is a, and dividend is b.

∴   Quotient = a/4

And Remainder = a/2

∴   b = a * a/4 + a/2

∴   4b = a2 + 2a

∴   a(a+2)/b = 4

 

7.   If 738 A6A is divisible by 11, then the value of A is-

 

a.   6

b.   3

c.   9

d.   1

Ans :  As 738A6 A is divisible by 11.

∴   A + A + 3 = 6 + 8 + 7

∴   A = 9

 

8.   The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

 

a.   3009

b.   3010

c.   3011

d.   3012

Ans :  ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴   Required least number = 3011

 

9.   P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

 

a.   2850

b.   3750

c.   2750

d.   1000

Ans :  Work done by Q = 19/23 + 8/23 – 1

4/23

∴  Wage of Q = 4/23 * 5750

= Rs. 1000

 

10.  A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

 

a.   15

b.   20

c.   18

d.   30

 

Ans :  Work done by A = 6/24 = ¼

 

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵  ¼ + x – 6/32 + x/64 = 1

∴  16 + 24 – 12 + x /64 = 1

∴  3x + 4 = 64

∴  x = 60/3 = 20 days

 

11.  The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

 

a.   1

b.   2

c.   3

d.   4

 

Ans :  The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

 12.  Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

 

a.   70.4

b.   70.464

c.   71.104

d.   71.4

 

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

 

13.  By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

 

a.   5 : 3

b.   3 : 5

c.   3 : 4

d.   4 : 3

 

Ans :  Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

 14.  A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

 

a.   Rs.7,000

b.   Rs.14,000

c.   Rs.5,000

d.   Rs.10,000

 

Ans :  Let the monthly income of B be Rs. x.

∴  Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒  6x – 12000 = 5x – 5000

∴  x = Rs.7000

 

15.  The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

 

a.   2 : 3

b.   3 : 2

c.   5 : 1

d.   1 : 5

 

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒  5x – 5y = x + y

⇒  4x = 6y

⇒  x/y = 6/4

x : y = 3 : 2

 

16.  A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a.   7.12 p.m.

b.   7.15 p.m.

c.   7.10 p.m.

d.   7.18 p.m.

 

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x – 5/1

⇒  4x – 12 + 3x – 12/12 = x – 5/1

⇒  7x – 24 = 12x – 60

⇒  5x = 36

∴  x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

 

 

17.  If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

 

a.   10 days

b.   12 days

c.   6 days

d.   8 days

 

Let x days are taken when they work together.

∴  Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴  1/x = x + 16 + x + 4/(x + 4) (x +16)

∴  2×2 + 20x = x2 + 20x +64

∴  x2 = 64 (8)2

∴  x = 8 days

 

18.  250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

 

a.   310

b.   300

c.   313

d.   312

 

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

    313

 

19.  2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

 

a.   36 days

b.   21 days

c.   30 days

d.   42 days

 

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒  24m + 60w = 45m + 18w

⇒  21m = 42w

⇒  1m = 2w

∴  2m + 5w = 9w

∴  required no. of days = 9 * 12/3

= 36

 20.  While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

 

a.   10%

b.   20%

c.   16 2/3%

d.   16 1/3%

 

Ans : Let the M.P. be Rs.100.

∴  S.P = (100-40) = Rs.60

and  C. P. = 60 * 100/(100-30) = Rs.600/7

∴  Reqd.  % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

 

21.  Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

 

a.   36%

b.   64%

c.   80%

d.   56%

 

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 –  90*80*50/10000]%

= [100-36]%

= 64%

 

22.  A retailer offers the following discount schemes for buyers on an article-

 

I.            Two successive discounts of 10%

II.          A discount of 12% followed by a discount of 8%.

III.       Successive discounts of 15% and 5%

IV.        A discount of 20%

 

The selling price will be minimum under the scheme-

 

a.   I

b.   II

c.   III

d.   IV

 

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

 

23.  Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

 

a.   72

b.   216

c.   354

d.   726

 

Ans : Let the first number be x.

∴  Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒  19x/4 = 342

∴  x = 342*4/19 = 72

∴  The largest number = 3*72

= 216   

 

 

24.  A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

 

a.   12.625 km/hr

b.   13.625 km/hr

c.   14.625 km/hr

d.   15.625 km/hr

 

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴  Average speed = x/64x/1000

= 15.625 km/hr

 

 

25.  The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

 

a.   264

b.   132

c.   88

d.   66

 

Ans :  Let the first number be x.

∴  Second number = x/2  and  Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴  x = 462*4/7

=264

 

 

26.  The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

 

a.   10

b.   15

c.   25

d.   50

 

Ans :  Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴  2500000 =14000x + 2000000 – 4000x

∴  x = 500000/10000

= 50

 

27.  The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

 

a.   70

b.   72

c.   71.9

d.   72.1

 

Ans :  Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

 

28.  A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

 

a.   5 : 7

b.   5 : 9

c.   7 : 5

d.   9 : 5

 

Ans :  Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴  Required ratio = 7/6 : 5/6 =  7 : 5

 

29.  In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

 

a.   4 : 15

b.   9 : 8

c.   21 : 8

d.   1 : 2

 

Ans :  Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴  x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

 

30.  A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

 

a.   1/3 rd

b.   3/7 th

c.   2/3 rd

d.   2/7 th

 

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴  x = 1.6/5.6 = 2/7th part

 

31.  An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15.  If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

 

a.   20 : 21

b.   21 : 20

c.   20 : 19

d.   19 : 21

 

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴  xy = 150 and  The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴  Required ratio = 18000 : 18900

= 20 : 21

 

32.  The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

 

a.   165

b.   170

c.   172

d.   174

 

Ans : Let the max. number of runs be x.

∴  The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴  x= 174

 

33.  Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

 

a.   1 ½ Km

b.   2 Km

c.   2 ½ Km

d.   5 Km

 

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

 

34.  Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

 

a.   48 dozen

b.   60 dozen

c.   72 dozen

d.   84 dozen

 

Ans : Let the number of oranges bought be x.

∴  45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

 

35.  A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

 

a.   Rs.360

b.   Rs.400

c.   Rs.420

d.   Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴  9x/20 = 135

∴  x = 135*20/9

= Rs. 300

∴  C.P. of the lower priced chair is Rs. 300.

 

36.  By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

 

a.   20

b.   25

c.   30

d.   32

 

Ans :  Let the S.P. of 100 Oranges be Rs. x.

∴  S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x – x/5

= Rs. 4x/5

∴  Reqd. Profit % = x/5 * 100*5/4x %

= 25%

 

37.  60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

 

a.   20% loss

b.   16 2/3% profit

c.   20% profit

d.   10% loss

 

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴  S.P = 60*100/50

= Rs. 120

∴  Reqd. % profit = (120 – 100)% = 20%

 

38.  Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

 

a.   Rs.30,000

b.   Rs.30,200

c.   Rs.30,300

d.   Rs.30,400

 

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and  S.P. of two horse = 80000 – 3600 = Rs. 76400

∴  S.P. of the other horse = 76400 – 46000 = 30400

 

39.  A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

 

a.   x2 – y2 / xy % loss

b.   x2 – y2 / xy % gain

c.   x2 – y2 / y2 % loss

d.   x2 – y2 / y2 * 100% gain

 

Ans :  C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴  Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 – y2 /y2 *100%

 

40.  A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

 

a.   25

b.   30

c.   35

d.   40

 

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

 

41.  If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

 

a.   12

b.   16

c.   35

d.   24

Ans : Let  the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2  – 12x = 0

⇒ x – 12 = 0

∴  x = 12

 

42.  The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

 

a.   Rs.1250

b.   Rs.1000

c.   Rs.1125

d.   Rs.1200

 

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴  x = Rs. 1000

 

43.  The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

 

a.   50%

b.   33 1/3%

c.   33%

d.   30%

 

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

 

44.  A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

 

a.   Rs.800 from C

b.   Rs.625 from C

c.   Rs.400 from B

d.   Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

 

45.  If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

 

a.   3

b.   5

c.   6

d.   8

 

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴  n = 5

 

46.  A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

 

a.   200√2 sq.m

b.   400√2 sq.m

c.   400 sq.m

d.   800 sq.m

 

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴  Area of the new square = ½*(20.√2)2 = 400 sq. m

 

47.  The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

 

a.   2 : 9 : 2

b.   4 : 9 : 1

c.   4 : 27 : 2

d.   2 : 3 : 1

 

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

 

48.  A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

 

a.   3 km/hr

b.   4 km/hr

c.   2 km/hr

d.   5 km/hr

 

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒  x2 = 9 = (3)2

∴  x = 3 km/hr

 

49.  A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

 

a.   45 minutes

b.   60 minutes

c.   75 minutes

d.   120 minutes

 

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴   4y = 3y + 1

∴  y = 1 hr = 60 minutes

 

50.  A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

 

a.   2 km/hr

b.   3 km/hr

c.   4 km/hr

d.   5 km/hr

 

Ans :  Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8               …(1)

and  52/x + y + 35/x – y = 11          …(2)

Solving equations (1) and (2), we get-

∴  100 – 105/x – y = 32 – 33

∴  x – y = 5

and  x + y = 13

∴ y = 4 km/hr

 

51.  If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

 

a.   25 years

b.   20 years

c.   15 years

d.   10 years

 

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

 

52.  A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

 

a.   Rs.48,800

b.   Rs.57,600

c.   Rs.62,400

d.   Rs.84,400

 

Ans : Let the present share of the younger son be Rs. x.

∴  The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴  x = Rs. 57600

 

 

53.  If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

 

a.   m : a

b.   am : 1

c.   1/m : 1/a

d.   1/am : 1

 

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴  x : y = am : 1

 

54.  Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

 

a.   9 cm

b.   10 cm

c.   11 cm

d.   12 cm

 

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

 

55.  If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

 

a.   0

b.   -2

c.   1

d.   2

 

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒  (a + 1)2 = 0

⇒   a +1 = 0

∴  a = -1

∴  a37 -1/a100 = (-1) – (1) = -2

 

56.  The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

 

a.   ½

b.   -1/2

c.   2

d.   -2

Ans : (k – 1) x + y – 2 = 0

∴  y = (1 – k) x + 2                ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3                 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 – k

∴  k = 1/2

 

57.  If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

 

a.   -1

b.   3

c.   1

d.   -2

 

Ans : a2 + b2  + c2 = 2(a – b – c) – 3

⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴  a + b – c = 1 – 1 + 1 =1

 

58.  If  x2 + 3x + 1 = 0, then the value of x3 + 1/x3  is-

 

a.   -18

b.   18

c.   36

d.   -36

 

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴  x3 + 1/ x3 = -18

 

59.  If xa, xb, xc = 1, then the value of a3 + b3 + c3  is –

 

a.   9

b.   abc

c.   a + b + c

d.   3abc

 

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c  = x0

⇒ a +b + c =0

∴  a3 + b3 + c3 = 3abc

 

60.  Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

 

a.   7 m

b.   8 m

c.   9 m

d.   10 m

 

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

 

61.  The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking  = 22-7) is-

 

a.   3√12 cm

b.   3√13 cm

c.   3√6 cm

d.   3√20 cm

 

Ans : Let the radius of the upper circular part of the frustum be r cm.

 

 

(Picture)

 

 

Then r/3 = x/9 = AC/AD

∴  x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx – r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 – 9r2 – 3r3 – 27r = 42

⇒ 3r3 = 39

∴  r = 3√13 cm

 

62.  The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

 

a.   5 : 6

b.   3 : 4

c.   4 : 5

d.   2 : 3

 

Ans : Required ratio = 2*2r*5h/2*3r*4h  = 5 : 6

 

63.  A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a.   530 cm3

b.   536 cm3

c.   539 cm3

d.   545 cm3

 

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴  r + h/h = 462/154 =3

∴  r + h = 3h

∴  r = 2h

∴  2*2h2 = 154

∴  h2 = 154*7/22*4 = 49/4

= (7/2)2

∴  h = 7/2 cm 

and  r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

 

64.  A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

 

a.   1 : 2

b.   1 : 3

c.   2 : 3

d.   3 : 4

 

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴  10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴  h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴  r : h = 3 : 4

 

65.  A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

 

a.   1 : 3

b.   1 :1

c.   √3 : 1

d.   1 : √3

 

Ans :                

 

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴  r : h = r/h

= 1 : √3

 

66.  If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

 

a.   550

b.   1100

c.   700

d.   350

 

Ans : ∴  ∠BOC = 1800 – (350 + 350) = 1100

 

 

    

 

 

 ∴  ∠BAC = 1/2 * 1100 = 550

 

67.  If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

 

a.   Acute angled

b.   Equilateral

c.   Right angled

d.   Obtuse angled

 

Ans :                    

 

 

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴   ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

 

 68.  If  sin2 (∝ + β/2) is-

 

a.   1

b.   -1

c.   0

d.   0.5

 

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 – cos2 β = 2

⇒  cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and  cos β = 0

⇒ ∝ = /2 and β = /2

∴  cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos  /2 = 0

 

69.  The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

 

a.   300

b.   450

c.   600

d.   900

 

Ans :                 

 

 

∵ tan θ = h/1√3 h = √3 = tan 600

 

70.  The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

 

a.   8/3 units

b.   4/3 units

c.   1 unit

d.   2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x – 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

 

71.  If x+1/16x = 1, then the value of 64×3 + 1/64×3 is-

 

a.   4

b.   52

c.   64

d.   76

∵ x + 1/16x = 1

⇒ 16×2 – 16x + 1 =0

⇒ 16×2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64×3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴   64×3 + 1/64×3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 – 15√3/676 -675

=52

 

72.  If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca  is-

 

a.   3

b.   2

c.   0

d.   1

 

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒  a2 + b2 + c2 = 2b2 – 2ac

∴  a2 + b2 + c2/ b2 ac = 2

 

 73.  If  a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

 

a.   -1

b.   0

c.   2

d.   1

 

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab)  = 2

⇒ a2 – ab + b2 = 2                          ….(1)

and  a2 + ab + b2 = 4                      …..(2)

⇒ 2ab = 2

⇒ ab = 1

 

74.  If a = 25, b = 15, c = -10, then the value of  a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

 

a.   30

b.   -15

c.   -30

d.   15

 

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a – b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴  Required value = 29250/1950 =15

 

75.  A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

 

a.   840

b.   920

c.   960

d.   1040

 

Ans :                  (Picture)       

 

 

∠ACB = 400 + 440 = 840

∴  ∠ACO = 900 – 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB – ∠ACO

= 840 – 460 = 380 = ∠OBC

∴  ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

 

76.  If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

 

a.   13 cm

b.   15 cm

c.   26 cm

d.   30 cm

 

Ans :     (Picture)

 

 

 

 

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴  Diameter of the circle = 2*OA

= 2*13 = 26cm

 

77.  In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

 

a.   4 cm

b.   2 cm

c.   12 cm

d.   6 cm

 

∵ PR/RQ = ½

But,   PR = 2cm

RQ  = 2*PR

= 4cm

 

(Picture)

 

∴  PQ = PR + RQ

= 2 + 4 = 6 cm

∴  BC = 2*PQ = 12CM

 

78.  If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is  equal to –

 

a.   ¾

b.   10/3

c.   3 ¾

d.   3

 

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 – tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴  θ = 300

∴  sin2  2θ + tan2 2θ =  sin2 600 +  tan2  600

=  ¾ + 3

=3 ¾

 

79.  The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

 

a.   -1

b.   ½

c.   0

d.   1

 

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

=  cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

=  cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

 

80.  If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

 

a.   5/17

b.   3/19

c.   7/10

d.   7/13

∵   sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴   2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴   sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒  (sin θ – cos θ)2 = (7/13)2

∴   sin θ – cos θ = 7/13

 

   

S.S.C. Combined Graduate Level Exam
(Tier-II) 2011
Solved Paper – I
Arithmetical Ability

1.   The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –  

a.   74

b.   78

c.   80

d.   90

Ans : Let the numbers be x and y .

∴  xy = 1575

And  x/y = 9/7

∴  xy/x/y = 1575/9/7

∴   y2  = 1225

∴ y = 35 and x = 45

∴  The sum of the numbers = 45+35

  = 80

 

2.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

 a.   3

b.   6

c.   9

d.   8.2

 

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

 

3.   √6+√6+√6+… is equal to –

 

a.   2

b.   5

c.   4

d.   3

 

 

 

Ans : Let √6+√6+√6+…. be x.

∴   x =√6+x

∴   x2 = 6+x

∴   x2 + x – 6 = 0

∴   (x-3) (x+2) = 0

∴   x = 3

 

4.   The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

 

a.   24

b.   32

c.   40

d.   28

 

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴  x2 + x2 + 4x + 4 = 394

∴  2 x2 +4x – 390 = 0

∴  x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴  x = 13

∴   Required sum = 13 +15 =28

 

 

5.   When (6767 +67) is divided by 68, the remainder is-

 

a.   1

b.   63

c.   66

d.   67

 

Ans:  (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴  6766 is an even number

∴  (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴  The remainder = 67

 

 6.   In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

 

a.   4b-a2 /a  =3

b.   4b-2a /a2 =2

c.   (a+1)2 = 4b

d.   A(a+2)/b = 4

 

Ans : As  divisor is a, and dividend is b.

∴   Quotient = a/4

And Remainder = a/2

∴   b = a * a/4 + a/2

∴   4b = a2 + 2a

∴   a(a+2)/b = 4

 

7.   If 738 A6A is divisible by 11, then the value of A is-

 

a.   6

b.   3

c.   9

d.   1

Ans :  As 738A6 A is divisible by 11.

∴   A + A + 3 = 6 + 8 + 7

∴   A = 9

 

8.   The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

 

a.   3009

b.   3010

c.   3011

d.   3012

Ans :  ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴   Required least number = 3011

 

9.   P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

 

a.   2850

b.   3750

c.   2750

d.   1000

Ans :  Work done by Q = 19/23 + 8/23 – 1

4/23

∴  Wage of Q = 4/23 * 5750

= Rs. 1000

 

10.  A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

 

a.   15

b.   20

c.   18

d.   30

 

Ans :  Work done by A = 6/24 = ¼

 

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵  ¼ + x – 6/32 + x/64 = 1

∴  16 + 24 – 12 + x /64 = 1

∴  3x + 4 = 64

∴  x = 60/3 = 20 days

 

11.  The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

 

a.   1

b.   2

c.   3

d.   4

 

Ans :  The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

 12.  Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

 

a.   70.4

b.   70.464

c.   71.104

d.   71.4

 

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

 

13.  By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

 

a.   5 : 3

b.   3 : 5

c.   3 : 4

d.   4 : 3

 

Ans :  Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

 14.  A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

 

a.   Rs.7,000

b.   Rs.14,000

c.   Rs.5,000

d.   Rs.10,000

 

Ans :  Let the monthly income of B be Rs. x.

∴  Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒  6x – 12000 = 5x – 5000

∴  x = Rs.7000

 

15.  The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

 

a.   2 : 3

b.   3 : 2

c.   5 : 1

d.   1 : 5

 

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒  5x – 5y = x + y

⇒  4x = 6y

⇒  x/y = 6/4

x : y = 3 : 2

 

16.  A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a.   7.12 p.m.

b.   7.15 p.m.

c.   7.10 p.m.

d.   7.18 p.m.

 

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x – 5/1

⇒  4x – 12 + 3x – 12/12 = x – 5/1

⇒  7x – 24 = 12x – 60

⇒  5x = 36

∴  x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

 

 

17.  If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

 

a.   10 days

b.   12 days

c.   6 days

d.   8 days

 

Let x days are taken when they work together.

∴  Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴  1/x = x + 16 + x + 4/(x + 4) (x +16)

∴  2×2 + 20x = x2 + 20x +64

∴  x2 = 64 (8)2

∴  x = 8 days

 

18.  250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

 

a.   310

b.   300

c.   313

d.   312

 

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

    313

 

19.  2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

 

a.   36 days

b.   21 days

c.   30 days

d.   42 days

 

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒  24m + 60w = 45m + 18w

⇒  21m = 42w

⇒  1m = 2w

∴  2m + 5w = 9w

∴  required no. of days = 9 * 12/3

= 36

 20.  While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

 

a.   10%

b.   20%

c.   16 2/3%

d.   16 1/3%

 

Ans : Let the M.P. be Rs.100.

∴  S.P = (100-40) = Rs.60

and  C. P. = 60 * 100/(100-30) = Rs.600/7

∴  Reqd.  % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

 

21.  Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

 

a.   36%

b.   64%

c.   80%

d.   56%

 

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 –  90*80*50/10000]%

= [100-36]%

= 64%

 

22.  A retailer offers the following discount schemes for buyers on an article-

 

I.            Two successive discounts of 10%

II.          A discount of 12% followed by a discount of 8%.

III.       Successive discounts of 15% and 5%

IV.        A discount of 20%

 

The selling price will be minimum under the scheme-

 

a.   I

b.   II

c.   III

d.   IV

 

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

 

23.  Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

 

a.   72

b.   216

c.   354

d.   726

 

Ans : Let the first number be x.

∴  Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒  19x/4 = 342

∴  x = 342*4/19 = 72

∴  The largest number = 3*72

= 216   

 

 

24.  A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

 

a.   12.625 km/hr

b.   13.625 km/hr

c.   14.625 km/hr

d.   15.625 km/hr

 

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴  Average speed = x/64x/1000

= 15.625 km/hr

 

 

25.  The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

 

a.   264

b.   132

c.   88

d.   66

 

Ans :  Let the first number be x.

∴  Second number = x/2  and  Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴  x = 462*4/7

=264

 

 

26.  The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

 

a.   10

b.   15

c.   25

d.   50

 

Ans :  Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴  2500000 =14000x + 2000000 – 4000x

∴  x = 500000/10000

= 50

 

27.  The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

 

a.   70

b.   72

c.   71.9

d.   72.1

 

Ans :  Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

 

28.  A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

 

a.   5 : 7

b.   5 : 9

c.   7 : 5

d.   9 : 5

 

Ans :  Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴  Required ratio = 7/6 : 5/6 =  7 : 5

 

29.  In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

 

a.   4 : 15

b.   9 : 8

c.   21 : 8

d.   1 : 2

 

Ans :  Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴  x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

 

30.  A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

 

a.   1/3 rd

b.   3/7 th

c.   2/3 rd

d.   2/7 th

 

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴  x = 1.6/5.6 = 2/7th part

 

31.  An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15.  If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

 

a.   20 : 21

b.   21 : 20

c.   20 : 19

d.   19 : 21

 

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴  xy = 150 and  The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴  Required ratio = 18000 : 18900

= 20 : 21

 

32.  The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

 

a.   165

b.   170

c.   172

d.   174

 

Ans : Let the max. number of runs be x.

∴  The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴  x= 174

 

33.  Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

 

a.   1 ½ Km

b.   2 Km

c.   2 ½ Km

d.   5 Km

 

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

 

34.  Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

 

a.   48 dozen

b.   60 dozen

c.   72 dozen

d.   84 dozen

 

Ans : Let the number of oranges bought be x.

∴  45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

 

35.  A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

 

a.   Rs.360

b.   Rs.400

c.   Rs.420

d.   Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴  9x/20 = 135

∴  x = 135*20/9

= Rs. 300

∴  C.P. of the lower priced chair is Rs. 300.

 

36.  By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

 

a.   20

b.   25

c.   30

d.   32

 

Ans :  Let the S.P. of 100 Oranges be Rs. x.

∴  S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x – x/5

= Rs. 4x/5

∴  Reqd. Profit % = x/5 * 100*5/4x %

= 25%

 

37.  60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

 

a.   20% loss

b.   16 2/3% profit

c.   20% profit

d.   10% loss

 

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴  S.P = 60*100/50

= Rs. 120

∴  Reqd. % profit = (120 – 100)% = 20%

 

38.  Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

 

a.   Rs.30,000

b.   Rs.30,200

c.   Rs.30,300

d.   Rs.30,400

 

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and  S.P. of two horse = 80000 – 3600 = Rs. 76400

∴  S.P. of the other horse = 76400 – 46000 = 30400

 

39.  A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

 

a.   x2 – y2 / xy % loss

b.   x2 – y2 / xy % gain

c.   x2 – y2 / y2 % loss

d.   x2 – y2 / y2 * 100% gain

 

Ans :  C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴  Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 – y2 /y2 *100%

 

40.  A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

 

a.   25

b.   30

c.   35

d.   40

 

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

 

41.  If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

 

a.   12

b.   16

c.   35

d.   24

Ans : Let  the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2  – 12x = 0

⇒ x – 12 = 0

∴  x = 12

 

42.  The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

 

a.   Rs.1250

b.   Rs.1000

c.   Rs.1125

d.   Rs.1200

 

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴  x = Rs. 1000

 

43.  The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

 

a.   50%

b.   33 1/3%

c.   33%

d.   30%

 

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

 

44.  A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

 

a.   Rs.800 from C

b.   Rs.625 from C

c.   Rs.400 from B

d.   Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

 

45.  If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

 

a.   3

b.   5

c.   6

d.   8

 

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴  n = 5

 

46.  A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

 

a.   200√2 sq.m

b.   400√2 sq.m

c.   400 sq.m

d.   800 sq.m

 

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴  Area of the new square = ½*(20.√2)2 = 400 sq. m

 

47.  The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

 

a.   2 : 9 : 2

b.   4 : 9 : 1

c.   4 : 27 : 2

d.   2 : 3 : 1

 

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

 

48.  A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

 

a.   3 km/hr

b.   4 km/hr

c.   2 km/hr

d.   5 km/hr

 

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒  x2 = 9 = (3)2

∴  x = 3 km/hr

 

49.  A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

 

a.   45 minutes

b.   60 minutes

c.   75 minutes

d.   120 minutes

 

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴   4y = 3y + 1

∴  y = 1 hr = 60 minutes

 

50.  A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

 

a.   2 km/hr

b.   3 km/hr

c.   4 km/hr

d.   5 km/hr

 

Ans :  Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8               …(1)

and  52/x + y + 35/x – y = 11          …(2)

Solving equations (1) and (2), we get-

∴  100 – 105/x – y = 32 – 33

∴  x – y = 5

and  x + y = 13

∴ y = 4 km/hr

 

51.  If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

 

a.   25 years

b.   20 years

c.   15 years

d.   10 years

 

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

 

52.  A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

 

a.   Rs.48,800

b.   Rs.57,600

c.   Rs.62,400

d.   Rs.84,400

 

Ans : Let the present share of the younger son be Rs. x.

∴  The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴  x = Rs. 57600

 

 

53.  If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

 

a.   m : a

b.   am : 1

c.   1/m : 1/a

d.   1/am : 1

 

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴  x : y = am : 1

 

54.  Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

 

a.   9 cm

b.   10 cm

c.   11 cm

d.   12 cm

 

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

 

55.  If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

 

a.   0

b.   -2

c.   1

d.   2

 

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒  (a + 1)2 = 0

⇒   a +1 = 0

∴  a = -1

∴  a37 -1/a100 = (-1) – (1) = -2

 

56.  The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

 

a.   ½

b.   -1/2

c.   2

d.   -2

Ans : (k – 1) x + y – 2 = 0

∴  y = (1 – k) x + 2                ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3                 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 – k

∴  k = 1/2

 

57.  If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

 

a.   -1

b.   3

c.   1

d.   -2

 

Ans : a2 + b2  + c2 = 2(a – b – c) – 3

⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴  a + b – c = 1 – 1 + 1 =1

 

58.  If  x2 + 3x + 1 = 0, then the value of x3 + 1/x3  is-

 

a.   -18

b.   18

c.   36

d.   -36

 

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴  x3 + 1/ x3 = -18

 

59.  If xa, xb, xc = 1, then the value of a3 + b3 + c3  is –

 

a.   9

b.   abc

c.   a + b + c

d.   3abc

 

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c  = x0

⇒ a +b + c =0

∴  a3 + b3 + c3 = 3abc

 

60.  Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

 

a.   7 m

b.   8 m

c.   9 m

d.   10 m

 

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

 

61.  The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking  = 22-7) is-

 

a.   3√12 cm

b.   3√13 cm

c.   3√6 cm

d.   3√20 cm

 

Ans : Let the radius of the upper circular part of the frustum be r cm.

 

 

(Picture)

 

 

Then r/3 = x/9 = AC/AD

∴  x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx – r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 – 9r2 – 3r3 – 27r = 42

⇒ 3r3 = 39

∴  r = 3√13 cm

 

62.  The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

 

a.   5 : 6

b.   3 : 4

c.   4 : 5

d.   2 : 3

 

Ans : Required ratio = 2*2r*5h/2*3r*4h  = 5 : 6

 

63.  A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a.   530 cm3

b.   536 cm3

c.   539 cm3

d.   545 cm3

 

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴  r + h/h = 462/154 =3

∴  r + h = 3h

∴  r = 2h

∴  2*2h2 = 154

∴  h2 = 154*7/22*4 = 49/4

= (7/2)2

∴  h = 7/2 cm 

and  r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

 

64.  A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

 

a.   1 : 2

b.   1 : 3

c.   2 : 3

d.   3 : 4

 

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴  10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴  h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴  r : h = 3 : 4

 

65.  A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

 

a.   1 : 3

b.   1 :1

c.   √3 : 1

d.   1 : √3

 

Ans :                  (Picture)       

 

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴  r : h = r/h

= 1 : √3

 

66.  If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

 

a.   550

b.   1100

c.   700

d.   350

 

Ans : ∴  ∠BOC = 1800 – (350 + 350) = 1100

 

 

(Picture)            

 

 

 ∴  ∠BAC = 1/2 * 1100 = 550

 

67.  If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

 

a.   Acute angled

b.   Equilateral

c.   Right angled

d.   Obtuse angled

 

Ans :                     (Picture)

 

 

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴   ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

 

 68.  If  sin2 (∝ + β/2) is-

 

a.   1

b.   -1

c.   0

d.   0.5

 

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 – cos2 β = 2

⇒  cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and  cos β = 0

⇒ ∝ = /2 and β = /2

∴  cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos  /2 = 0

 

69.  The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

 

a.   300

b.   450

c.   600

d.   900

 

Ans :                  (Picture)       

 

 

∵ tan θ = h/1√3 h = √3 = tan 600

 

70.  The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

 

a.   8/3 units

b.   4/3 units

c.   1 unit

d.   2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x – 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

 

71.  If x+1/16x = 1, then the value of 64×3 + 1/64×3 is-

 

a.   4

b.   52

c.   64

d.   76

∵ x + 1/16x = 1

⇒ 16×2 – 16x + 1 =0

⇒ 16×2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64×3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴   64×3 + 1/64×3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 – 15√3/676 -675

=52

 

72.  If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca  is-

 

a.   3

b.   2

c.   0

d.   1

 

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒  a2 + b2 + c2 = 2b2 – 2ac

∴  a2 + b2 + c2/ b2 ac = 2

 

 73.  If  a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

 

a.   -1

b.   0

c.   2

d.   1

 

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab)  = 2

⇒ a2 – ab + b2 = 2                          ….(1)

and  a2 + ab + b2 = 4                      …..(2)

⇒ 2ab = 2

⇒ ab = 1

 

74.  If a = 25, b = 15, c = -10, then the value of  a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

 

a.   30

b.   -15

c.   -30

d.   15

 

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a – b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴  Required value = 29250/1950 =15

 

75.  A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

 

a.   840

b.   920

c.   960

d.   1040

 

Ans :                  (Picture)       

 

 

∠ACB = 400 + 440 = 840

∴  ∠ACO = 900 – 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB – ∠ACO

= 840 – 460 = 380 = ∠OBC

∴  ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

 

76.  If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

 

a.   13 cm

b.   15 cm

c.   26 cm

d.   30 cm

 

Ans :     (Picture)

 

 

 

 

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴  Diameter of the circle = 2*OA

= 2*13 = 26cm

 

77.  In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

 

a.   4 cm

b.   2 cm

c.   12 cm

d.   6 cm

 

∵ PR/RQ = ½

But,   PR = 2cm

RQ  = 2*PR

= 4cm

 

(Picture)

 

∴  PQ = PR + RQ

= 2 + 4 = 6 cm

∴  BC = 2*PQ = 12CM

 

78.  If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is  equal to –

 

a.   ¾

b.   10/3

c.   3 ¾

d.   3

 

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 – tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴  θ = 300

∴  sin2  2θ + tan2 2θ =  sin2 600 +  tan2  600

=  ¾ + 3

=3 ¾

 

79.  The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

 

a.   -1

b.   ½

c.   0

d.   1

 

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

=  cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

=  cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

 

80.  If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

 

a.   5/17

b.   3/19

c.   7/10

d.   7/13

∵   sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴   2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴   sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒  (sin θ – cos θ)2 = (7/13)2

∴   sin θ – cos θ = 7/13

 

   

S.S.C. Combined Graduate Level Exam
(Tier-II) 2011
Solved Paper – I
Arithmetical Ability

1.   The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –  

a.   74

b.   78

c.   80

d.   90

Ans : Let the numbers be x and y .

∴  xy = 1575

And  x/y = 9/7

∴  xy/x/y = 1575/9/7

∴   y2  = 1225

∴ y = 35 and x = 45

∴  The sum of the numbers = 45+35

  = 80

 

2.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

 a.   3

b.   6

c.   9

d.   8.2

 

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

 

3.   √6+√6+√6+… is equal to –

 

a.   2

b.   5

c.   4

d.   3

 

 

 

Ans : Let √6+√6+√6+…. be x.

∴   x =√6+x

∴   x2 = 6+x

∴   x2 + x – 6 = 0

∴   (x-3) (x+2) = 0

∴   x = 3

 

4.   The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

 

a.   24

b.   32

c.   40

d.   28

 

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴  x2 + x2 + 4x + 4 = 394

∴  2 x2 +4x – 390 = 0

∴  x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴  x = 13

∴   Required sum = 13 +15 =28

 

 

5.   When (6767 +67) is divided by 68, the remainder is-

 

a.   1

b.   63

c.   66

d.   67

 

Ans:  (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴  6766 is an even number

∴  (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴  The remainder = 67

 

 6.   In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

 

a.   4b-a2 /a  =3

b.   4b-2a /a2 =2

c.   (a+1)2 = 4b

d.   A(a+2)/b = 4

 

Ans : As  divisor is a, and dividend is b.

∴   Quotient = a/4

And Remainder = a/2

∴   b = a * a/4 + a/2

∴   4b = a2 + 2a

∴   a(a+2)/b = 4

 

7.   If 738 A6A is divisible by 11, then the value of A is-

 

a.   6

b.   3

c.   9

d.   1

Ans :  As 738A6 A is divisible by 11.

∴   A + A + 3 = 6 + 8 + 7

∴   A = 9

 

8.   The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

 

a.   3009

b.   3010

c.   3011

d.   3012

Ans :  ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴   Required least number = 3011

 

9.   P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

 

a.   2850

b.   3750

c.   2750

d.   1000

Ans :  Work done by Q = 19/23 + 8/23 – 1

4/23

∴  Wage of Q = 4/23 * 5750

= Rs. 1000

 

10.  A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

 

a.   15

b.   20

c.   18

d.   30

 

Ans :  Work done by A = 6/24 = ¼

 

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵  ¼ + x – 6/32 + x/64 = 1

∴  16 + 24 – 12 + x /64 = 1

∴  3x + 4 = 64

∴  x = 60/3 = 20 days

 

11.  The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

 

a.   1

b.   2

c.   3

d.   4

 

Ans :  The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

 12.  Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

 

a.   70.4

b.   70.464

c.   71.104

d.   71.4

 

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

 

13.  By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

 

a.   5 : 3

b.   3 : 5

c.   3 : 4

d.   4 : 3

 

Ans :  Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

 14.  A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

 

a.   Rs.7,000

b.   Rs.14,000

c.   Rs.5,000

d.   Rs.10,000

 

Ans :  Let the monthly income of B be Rs. x.

∴  Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒  6x – 12000 = 5x – 5000

∴  x = Rs.7000

 

15.  The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

 

a.   2 : 3

b.   3 : 2

c.   5 : 1

d.   1 : 5

 

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒  5x – 5y = x + y

⇒  4x = 6y

⇒  x/y = 6/4

x : y = 3 : 2

 

16.  A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a.   7.12 p.m.

b.   7.15 p.m.

c.   7.10 p.m.

d.   7.18 p.m.

 

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x – 5/1

⇒  4x – 12 + 3x – 12/12 = x – 5/1

⇒  7x – 24 = 12x – 60

⇒  5x = 36

∴  x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

 

 

17.  If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

 

a.   10 days

b.   12 days

c.   6 days

d.   8 days

 

Let x days are taken when they work together.

∴  Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴  1/x = x + 16 + x + 4/(x + 4) (x +16)

∴  2×2 + 20x = x2 + 20x +64

∴  x2 = 64 (8)2

∴  x = 8 days

 

18.  250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

 

a.   310

b.   300

c.   313

d.   312

 

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

    313

 

19.  2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

 

a.   36 days

b.   21 days

c.   30 days

d.   42 days

 

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒  24m + 60w = 45m + 18w

⇒  21m = 42w

⇒  1m = 2w

∴  2m + 5w = 9w

∴  required no. of days = 9 * 12/3

= 36

 20.  While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

 

a.   10%

b.   20%

c.   16 2/3%

d.   16 1/3%

 

Ans : Let the M.P. be Rs.100.

∴  S.P = (100-40) = Rs.60

and  C. P. = 60 * 100/(100-30) = Rs.600/7

∴  Reqd.  % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

 

21.  Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

 

a.   36%

b.   64%

c.   80%

d.   56%

 

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 –  90*80*50/10000]%

= [100-36]%

= 64%

 

22.  A retailer offers the following discount schemes for buyers on an article-

 

I.            Two successive discounts of 10%

II.          A discount of 12% followed by a discount of 8%.

III.       Successive discounts of 15% and 5%

IV.        A discount of 20%

 

The selling price will be minimum under the scheme-

 

a.   I

b.   II

c.   III

d.   IV

 

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

 

23.  Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

 

a.   72

b.   216

c.   354

d.   726

 

Ans : Let the first number be x.

∴  Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒  19x/4 = 342

∴  x = 342*4/19 = 72

∴  The largest number = 3*72

= 216   

 

 

24.  A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

 

a.   12.625 km/hr

b.   13.625 km/hr

c.   14.625 km/hr

d.   15.625 km/hr

 

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴  Average speed = x/64x/1000

= 15.625 km/hr

 

 

25.  The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

 

a.   264

b.   132

c.   88

d.   66

 

Ans :  Let the first number be x.

∴  Second number = x/2  and  Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴  x = 462*4/7

=264

 

 

26.  The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

 

a.   10

b.   15

c.   25

d.   50

 

Ans :  Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴  2500000 =14000x + 2000000 – 4000x

∴  x = 500000/10000

= 50

 

27.  The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

 

a.   70

b.   72

c.   71.9

d.   72.1

 

Ans :  Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

 

28.  A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

 

a.   5 : 7

b.   5 : 9

c.   7 : 5

d.   9 : 5

 

Ans :  Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴  Required ratio = 7/6 : 5/6 =  7 : 5

 

29.  In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

 

a.   4 : 15

b.   9 : 8

c.   21 : 8

d.   1 : 2

 

Ans :  Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴  x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

 

30.  A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

 

a.   1/3 rd

b.   3/7 th

c.   2/3 rd

d.   2/7 th

 

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴  x = 1.6/5.6 = 2/7th part

 

31.  An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15.  If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

 

a.   20 : 21

b.   21 : 20

c.   20 : 19

d.   19 : 21

 

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴  xy = 150 and  The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴  Required ratio = 18000 : 18900

= 20 : 21

 

32.  The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

 

a.   165

b.   170

c.   172

d.   174

 

Ans : Let the max. number of runs be x.

∴  The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴  x= 174

 

33.  Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

 

a.   1 ½ Km

b.   2 Km

c.   2 ½ Km

d.   5 Km

 

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

 

34.  Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

 

a.   48 dozen

b.   60 dozen

c.   72 dozen

d.   84 dozen

 

Ans : Let the number of oranges bought be x.

∴  45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

 

35.  A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

 

a.   Rs.360

b.   Rs.400

c.   Rs.420

d.   Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴  9x/20 = 135

∴  x = 135*20/9

= Rs. 300

∴  C.P. of the lower priced chair is Rs. 300.

 

36.  By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

 

a.   20

b.   25

c.   30

d.   32

 

Ans :  Let the S.P. of 100 Oranges be Rs. x.

∴  S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x – x/5

= Rs. 4x/5

∴  Reqd. Profit % = x/5 * 100*5/4x %

= 25%

 

37.  60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

 

a.   20% loss

b.   16 2/3% profit

c.   20% profit

d.   10% loss

 

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴  S.P = 60*100/50

= Rs. 120

∴  Reqd. % profit = (120 – 100)% = 20%

 

38.  Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

 

a.   Rs.30,000

b.   Rs.30,200

c.   Rs.30,300

d.   Rs.30,400

 

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and  S.P. of two horse = 80000 – 3600 = Rs. 76400

∴  S.P. of the other horse = 76400 – 46000 = 30400

 

39.  A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

 

a.   x2 – y2 / xy % loss

b.   x2 – y2 / xy % gain

c.   x2 – y2 / y2 % loss

d.   x2 – y2 / y2 * 100% gain

 

Ans :  C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴  Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 – y2 /y2 *100%

 

40.  A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

 

a.   25

b.   30

c.   35

d.   40

 

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

 

41.  If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

 

a.   12

b.   16

c.   35

d.   24

Ans : Let  the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2  – 12x = 0

⇒ x – 12 = 0

∴  x = 12

 

42.  The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

 

a.   Rs.1250

b.   Rs.1000

c.   Rs.1125

d.   Rs.1200

 

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴  x = Rs. 1000

 

43.  The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

 

a.   50%

b.   33 1/3%

c.   33%

d.   30%

 

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

 

44.  A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

 

a.   Rs.800 from C

b.   Rs.625 from C

c.   Rs.400 from B

d.   Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

 

45.  If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

 

a.   3

b.   5

c.   6

d.   8

 

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴  n = 5

 

46.  A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

 

a.   200√2 sq.m

b.   400√2 sq.m

c.   400 sq.m

d.   800 sq.m

 

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴  Area of the new square = ½*(20.√2)2 = 400 sq. m

 

47.  The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

 

a.   2 : 9 : 2

b.   4 : 9 : 1

c.   4 : 27 : 2

d.   2 : 3 : 1

 

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

 

48.  A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

 

a.   3 km/hr

b.   4 km/hr

c.   2 km/hr

d.   5 km/hr

 

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒  x2 = 9 = (3)2

∴  x = 3 km/hr

 

49.  A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

 

a.   45 minutes

b.   60 minutes

c.   75 minutes

d.   120 minutes

 

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴   4y = 3y + 1

∴  y = 1 hr = 60 minutes

 

50.  A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

 

a.   2 km/hr

b.   3 km/hr

c.   4 km/hr

d.   5 km/hr

 

Ans :  Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8               …(1)

and  52/x + y + 35/x – y = 11          …(2)

Solving equations (1) and (2), we get-

∴  100 – 105/x – y = 32 – 33

∴  x – y = 5

and  x + y = 13

∴ y = 4 km/hr

 

51.  If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

 

a.   25 years

b.   20 years

c.   15 years

d.   10 years

 

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

 

52.  A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

 

a.   Rs.48,800

b.   Rs.57,600

c.   Rs.62,400

d.   Rs.84,400

 

Ans : Let the present share of the younger son be Rs. x.

∴  The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴  x = Rs. 57600

 

 

53.  If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

 

a.   m : a

b.   am : 1

c.   1/m : 1/a

d.   1/am : 1

 

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴  x : y = am : 1

 

54.  Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

 

a.   9 cm

b.   10 cm

c.   11 cm

d.   12 cm

 

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

 

55.  If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

 

a.   0

b.   -2

c.   1

d.   2

 

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒  (a + 1)2 = 0

⇒   a +1 = 0

∴  a = -1

∴  a37 -1/a100 = (-1) – (1) = -2

 

56.  The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

 

a.   ½

b.   -1/2

c.   2

d.   -2

Ans : (k – 1) x + y – 2 = 0

∴  y = (1 – k) x + 2                ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3                 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 – k

∴  k = 1/2

 

57.  If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

 

a.   -1

b.   3

c.   1

d.   -2

 

Ans : a2 + b2  + c2 = 2(a – b – c) – 3

⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴  a + b – c = 1 – 1 + 1 =1

 

58.  If  x2 + 3x + 1 = 0, then the value of x3 + 1/x3  is-

 

a.   -18

b.   18

c.   36

d.   -36

 

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴  x3 + 1/ x3 = -18

 

59.  If xa, xb, xc = 1, then the value of a3 + b3 + c3  is –

 

a.   9

b.   abc

c.   a + b + c

d.   3abc

 

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c  = x0

⇒ a +b + c =0

∴  a3 + b3 + c3 = 3abc

 

60.  Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

 

a.   7 m

b.   8 m

c.   9 m

d.   10 m

 

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

 

61.  The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking  = 22-7) is-

 

a.   3√12 cm

b.   3√13 cm

c.   3√6 cm

d.   3√20 cm

 

Ans : Let the radius of the upper circular part of the frustum be r cm.

 

 

(Picture)

 

 

Then r/3 = x/9 = AC/AD

∴  x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx – r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 – 9r2 – 3r3 – 27r = 42

⇒ 3r3 = 39

∴  r = 3√13 cm

 

62.  The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

 

a.   5 : 6

b.   3 : 4

c.   4 : 5

d.   2 : 3

 

Ans : Required ratio = 2*2r*5h/2*3r*4h  = 5 : 6

 

63.  A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a.   530 cm3

b.   536 cm3

c.   539 cm3

d.   545 cm3

 

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴  r + h/h = 462/154 =3

∴  r + h = 3h

∴  r = 2h

∴  2*2h2 = 154

∴  h2 = 154*7/22*4 = 49/4

= (7/2)2

∴  h = 7/2 cm 

and  r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

 

64.  A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

 

a.   1 : 2

b.   1 : 3

c.   2 : 3

d.   3 : 4

 

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴  10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴  h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴  r : h = 3 : 4

 

65.  A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

 

a.   1 : 3

b.   1 :1

c.   √3 : 1

d.   1 : √3

 

Ans :                  (Picture)       

 

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴  r : h = r/h

= 1 : √3

 

66.  If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

 

a.   550

b.   1100

c.   700

d.   350

 

Ans : ∴  ∠BOC = 1800 – (350 + 350) = 1100

 

 

(Picture)            

 

 

 ∴  ∠BAC = 1/2 * 1100 = 550

 

67.  If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

 

a.   Acute angled

b.   Equilateral

c.   Right angled

d.   Obtuse angled

 

Ans :                     (Picture)

 

 

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴   ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

 

 68.  If  sin2 (∝ + β/2) is-

 

a.   1

b.   -1

c.   0

d.   0.5

 

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 – cos2 β = 2

⇒  cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and  cos β = 0

⇒ ∝ = /2 and β = /2

∴  cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos  /2 = 0

 

69.  The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

 

a.   300

b.   450

c.   600

d.   900

 

Ans :                  (Picture)       

 

 

∵ tan θ = h/1√3 h = √3 = tan 600

 

70.  The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

 

a.   8/3 units

b.   4/3 units

c.   1 unit

d.   2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x – 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

 

71.  If x+1/16x = 1, then the value of 64×3 + 1/64×3 is-

 

a.   4

b.   52

c.   64

d.   76

∵ x + 1/16x = 1

⇒ 16×2 – 16x + 1 =0

⇒ 16×2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64×3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴   64×3 + 1/64×3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 – 15√3/676 -675

=52

 

72.  If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca  is-

 

a.   3

b.   2

c.   0

d.   1

 

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒  a2 + b2 + c2 = 2b2 – 2ac

∴  a2 + b2 + c2/ b2 ac = 2

 

 73.  If  a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

 

a.   -1

b.   0

c.   2

d.   1

 

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab)  = 2

⇒ a2 – ab + b2 = 2                          ….(1)

and  a2 + ab + b2 = 4                      …..(2)

⇒ 2ab = 2

⇒ ab = 1

 

74.  If a = 25, b = 15, c = -10, then the value of  a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

 

a.   30

b.   -15

c.   -30

d.   15

 

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a – b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴  Required value = 29250/1950 =15

 

75.  A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

 

a.   840

b.   920

c.   960

d.   1040

 

Ans :                  (Picture)       

 

 

∠ACB = 400 + 440 = 840

∴  ∠ACO = 900 – 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB – ∠ACO

= 840 – 460 = 380 = ∠OBC

∴  ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

 

76.  If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

 

a.   13 cm

b.   15 cm

c.   26 cm

d.   30 cm

 

Ans :     (Picture)

 

 

 

 

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴  Diameter of the circle = 2*OA

= 2*13 = 26cm

 

77.  In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

 

a.   4 cm

b.   2 cm

c.   12 cm

d.   6 cm

 

∵ PR/RQ = ½

But,   PR = 2cm

RQ  = 2*PR

= 4cm

 

(Picture)

 

∴  PQ = PR + RQ

= 2 + 4 = 6 cm

∴  BC = 2*PQ = 12CM

 

78.  If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is  equal to –

 

a.   ¾

b.   10/3

c.   3 ¾

d.   3

 

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 – tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴  θ = 300

∴  sin2  2θ + tan2 2θ =  sin2 600 +  tan2  600

=  ¾ + 3

=3 ¾

 

79.  The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

 

a.   -1

b.   ½

c.   0

d.   1

 

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

=  cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

=  cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

 

80.  If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

 

a.   5/17

b.   3/19

c.   7/10

d.   7/13

∵   sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴   2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴   sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒  (sin θ – cos θ)2 = (7/13)2

∴   sin θ – cos θ = 7/13

 

   

S.S.C. Combined Graduate Level Exam
(Tier-II) 2011
Solved Paper – I
Arithmetical Ability

1.   The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –  

a.   74

b.   78

c.   80

d.   90

Ans : Let the numbers be x and y .

∴  xy = 1575

And  x/y = 9/7

∴  xy/x/y = 1575/9/7

∴   y2  = 1225

∴ y = 35 and x = 45

∴  The sum of the numbers = 45+35

  = 80

 

2.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

 a.   3

b.   6

c.   9

d.   8.2

 

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

 

3.   √6+√6+√6+… is equal to –

 

a.   2

b.   5

c.   4

d.   3

 

 

 

Ans : Let √6+√6+√6+…. be x.

∴   x =√6+x

∴   x2 = 6+x

∴   x2 + x – 6 = 0

∴   (x-3) (x+2) = 0

∴   x = 3

 

4.   The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

 

a.   24

b.   32

c.   40

d.   28

 

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴  x2 + x2 + 4x + 4 = 394

∴  2 x2 +4x – 390 = 0

∴  x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴  x = 13

∴   Required sum = 13 +15 =28

 

 

5.   When (6767 +67) is divided by 68, the remainder is-

 

a.   1

b.   63

c.   66

d.   67

 

Ans:  (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴  6766 is an even number

∴  (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴  The remainder = 67

 

 6.   In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

 

a.   4b-a2 /a  =3

b.   4b-2a /a2 =2

c.   (a+1)2 = 4b

d.   A(a+2)/b = 4

 

Ans : As  divisor is a, and dividend is b.

∴   Quotient = a/4

And Remainder = a/2

∴   b = a * a/4 + a/2

∴   4b = a2 + 2a

∴   a(a+2)/b = 4

 

7.   If 738 A6A is divisible by 11, then the value of A is-

 

a.   6

b.   3

c.   9

d.   1

Ans :  As 738A6 A is divisible by 11.

∴   A + A + 3 = 6 + 8 + 7

∴   A = 9

 

8.   The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

 

a.   3009

b.   3010

c.   3011

d.   3012

Ans :  ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴   Required least number = 3011

 

9.   P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

 

a.   2850

b.   3750

c.   2750

d.   1000

Ans :  Work done by Q = 19/23 + 8/23 – 1

4/23

∴  Wage of Q = 4/23 * 5750

= Rs. 1000

 

10.  A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

 

a.   15

b.   20

c.   18

d.   30

 

Ans :  Work done by A = 6/24 = ¼

 

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵  ¼ + x – 6/32 + x/64 = 1

∴  16 + 24 – 12 + x /64 = 1

∴  3x + 4 = 64

∴  x = 60/3 = 20 days

 

11.  The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

 

a.   1

b.   2

c.   3

d.   4

 

Ans :  The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

 12.  Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

 

a.   70.4

b.   70.464

c.   71.104

d.   71.4

 

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

 

13.  By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

 

a.   5 : 3

b.   3 : 5

c.   3 : 4

d.   4 : 3

 

Ans :  Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

 14.  A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

 

a.   Rs.7,000

b.   Rs.14,000

c.   Rs.5,000

d.   Rs.10,000

 

Ans :  Let the monthly income of B be Rs. x.

∴  Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒  6x – 12000 = 5x – 5000

∴  x = Rs.7000

 

15.  The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

 

a.   2 : 3

b.   3 : 2

c.   5 : 1

d.   1 : 5

 

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒  5x – 5y = x + y

⇒  4x = 6y

⇒  x/y = 6/4

x : y = 3 : 2

 

16.  A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a.   7.12 p.m.

b.   7.15 p.m.

c.   7.10 p.m.

d.   7.18 p.m.

 

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x – 5/1

⇒  4x – 12 + 3x – 12/12 = x – 5/1

⇒  7x – 24 = 12x – 60

⇒  5x = 36

∴  x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

 

 

17.  If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

 

a.   10 days

b.   12 days

c.   6 days

d.   8 days

 

Let x days are taken when they work together.

∴  Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴  1/x = x + 16 + x + 4/(x + 4) (x +16)

∴  2×2 + 20x = x2 + 20x +64

∴  x2 = 64 (8)2

∴  x = 8 days

 

18.  250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

 

a.   310

b.   300

c.   313

d.   312

 

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

    313

 

19.  2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

 

a.   36 days

b.   21 days

c.   30 days

d.   42 days

 

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒  24m + 60w = 45m + 18w

⇒  21m = 42w

⇒  1m = 2w

∴  2m + 5w = 9w

∴  required no. of days = 9 * 12/3

= 36

 20.  While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

 

a.   10%

b.   20%

c.   16 2/3%

d.   16 1/3%

 

Ans : Let the M.P. be Rs.100.

∴  S.P = (100-40) = Rs.60

and  C. P. = 60 * 100/(100-30) = Rs.600/7

∴  Reqd.  % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

 

21.  Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

 

a.   36%

b.   64%

c.   80%

d.   56%

 

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 –  90*80*50/10000]%

= [100-36]%

= 64%

 

22.  A retailer offers the following discount schemes for buyers on an article-

 

I.            Two successive discounts of 10%

II.          A discount of 12% followed by a discount of 8%.

III.       Successive discounts of 15% and 5%

IV.        A discount of 20%

 

The selling price will be minimum under the scheme-

 

a.   I

b.   II

c.   III

d.   IV

 

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

 

23.  Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

 

a.   72

b.   216

c.   354

d.   726

 

Ans : Let the first number be x.

∴  Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒  19x/4 = 342

∴  x = 342*4/19 = 72

∴  The largest number = 3*72

= 216   

 

 

24.  A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

 

a.   12.625 km/hr

b.   13.625 km/hr

c.   14.625 km/hr

d.   15.625 km/hr

 

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴  Average speed = x/64x/1000

= 15.625 km/hr

 

 

25.  The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

 

a.   264

b.   132

c.   88

d.   66

 

Ans :  Let the first number be x.

∴  Second number = x/2  and  Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴  x = 462*4/7

=264

 

 

26.  The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

 

a.   10

b.   15

c.   25

d.   50

 

Ans :  Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴  2500000 =14000x + 2000000 – 4000x

∴  x = 500000/10000

= 50

 

27.  The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

 

a.   70

b.   72

c.   71.9

d.   72.1

 

Ans :  Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

 

28.  A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

 

a.   5 : 7

b.   5 : 9

c.   7 : 5

d.   9 : 5

 

Ans :  Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴  Required ratio = 7/6 : 5/6 =  7 : 5

 

29.  In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

 

a.   4 : 15

b.   9 : 8

c.   21 : 8

d.   1 : 2

 

Ans :  Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴  x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

 

30.  A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

 

a.   1/3 rd

b.   3/7 th

c.   2/3 rd

d.   2/7 th

 

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴  x = 1.6/5.6 = 2/7th part

 

31.  An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15.  If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

 

a.   20 : 21

b.   21 : 20

c.   20 : 19

d.   19 : 21

 

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴  xy = 150 and  The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴  Required ratio = 18000 : 18900

= 20 : 21

 

32.  The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

 

a.   165

b.   170

c.   172

d.   174

 

Ans : Let the max. number of runs be x.

∴  The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴  x= 174

 

33.  Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

 

a.   1 ½ Km

b.   2 Km

c.   2 ½ Km

d.   5 Km

 

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

 

34.  Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

 

a.   48 dozen

b.   60 dozen

c.   72 dozen

d.   84 dozen

 

Ans : Let the number of oranges bought be x.

∴  45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

 

35.  A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

 

a.   Rs.360

b.   Rs.400

c.   Rs.420

d.   Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴  9x/20 = 135

∴  x = 135*20/9

= Rs. 300

∴  C.P. of the lower priced chair is Rs. 300.

 

36.  By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

 

a.   20

b.   25

c.   30

d.   32

 

Ans :  Let the S.P. of 100 Oranges be Rs. x.

∴  S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x – x/5

= Rs. 4x/5

∴  Reqd. Profit % = x/5 * 100*5/4x %

= 25%

 

37.  60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

 

a.   20% loss

b.   16 2/3% profit

c.   20% profit

d.   10% loss

 

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴  S.P = 60*100/50

= Rs. 120

∴  Reqd. % profit = (120 – 100)% = 20%

 

38.  Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

 

a.   Rs.30,000

b.   Rs.30,200

c.   Rs.30,300

d.   Rs.30,400

 

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and  S.P. of two horse = 80000 – 3600 = Rs. 76400

∴  S.P. of the other horse = 76400 – 46000 = 30400

 

39.  A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

 

a.   x2 – y2 / xy % loss

b.   x2 – y2 / xy % gain

c.   x2 – y2 / y2 % loss

d.   x2 – y2 / y2 * 100% gain

 

Ans :  C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴  Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 – y2 /y2 *100%

 

40.  A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

 

a.   25

b.   30

c.   35

d.   40

 

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

 

41.  If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

 

a.   12

b.   16

c.   35

d.   24

Ans : Let  the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2  – 12x = 0

⇒ x – 12 = 0

∴  x = 12

 

42.  The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

 

a.   Rs.1250

b.   Rs.1000

c.   Rs.1125

d.   Rs.1200

 

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴  x = Rs. 1000

 

43.  The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

 

a.   50%

b.   33 1/3%

c.   33%

d.   30%

 

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

 

44.  A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

 

a.   Rs.800 from C

b.   Rs.625 from C

c.   Rs.400 from B

d.   Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

 

45.  If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

 

a.   3

b.   5

c.   6

d.   8

 

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴  n = 5

 

46.  A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

 

a.   200√2 sq.m

b.   400√2 sq.m

c.   400 sq.m

d.   800 sq.m

 

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴  Area of the new square = ½*(20.√2)2 = 400 sq. m

 

47.  The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

 

a.   2 : 9 : 2

b.   4 : 9 : 1

c.   4 : 27 : 2

d.   2 : 3 : 1

 

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

 

48.  A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

 

a.   3 km/hr

b.   4 km/hr

c.   2 km/hr

d.   5 km/hr

 

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒  x2 = 9 = (3)2

∴  x = 3 km/hr

 

49.  A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

 

a.   45 minutes

b.   60 minutes

c.   75 minutes

d.   120 minutes

 

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴   4y = 3y + 1

∴  y = 1 hr = 60 minutes

 

50.  A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

 

a.   2 km/hr

b.   3 km/hr

c.   4 km/hr

d.   5 km/hr

 

Ans :  Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8               …(1)

and  52/x + y + 35/x – y = 11          …(2)

Solving equations (1) and (2), we get-

∴  100 – 105/x – y = 32 – 33

∴  x – y = 5

and  x + y = 13

∴ y = 4 km/hr

 

51.  If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

 

a.   25 years

b.   20 years

c.   15 years

d.   10 years

 

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

 

52.  A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

 

a.   Rs.48,800

b.   Rs.57,600

c.   Rs.62,400

d.   Rs.84,400

 

Ans : Let the present share of the younger son be Rs. x.

∴  The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴  x = Rs. 57600

 

 

53.  If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

 

a.   m : a

b.   am : 1

c.   1/m : 1/a

d.   1/am : 1

 

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴  x : y = am : 1

 

54.  Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

 

a.   9 cm

b.   10 cm

c.   11 cm

d.   12 cm

 

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

 

55.  If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

 

a.   0

b.   -2

c.   1

d.   2

 

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒  (a + 1)2 = 0

⇒   a +1 = 0

∴  a = -1

∴  a37 -1/a100 = (-1) – (1) = -2

 

56.  The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

 

a.   ½

b.   -1/2

c.   2

d.   -2

Ans : (k – 1) x + y – 2 = 0

∴  y = (1 – k) x + 2                ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3                 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 – k

∴  k = 1/2

 

57.  If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

 

a.   -1

b.   3

c.   1

d.   -2

 

Ans : a2 + b2  + c2 = 2(a – b – c) – 3

⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴  a + b – c = 1 – 1 + 1 =1

 

58.  If  x2 + 3x + 1 = 0, then the value of x3 + 1/x3  is-

 

a.   -18

b.   18

c.   36

d.   -36

 

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴  x3 + 1/ x3 = -18

 

59.  If xa, xb, xc = 1, then the value of a3 + b3 + c3  is –

 

a.   9

b.   abc

c.   a + b + c

d.   3abc

 

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c  = x0

⇒ a +b + c =0

∴  a3 + b3 + c3 = 3abc

 

60.  Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

 

a.   7 m

b.   8 m

c.   9 m

d.   10 m

 

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

 

61.  The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking  = 22-7) is-

 

a.   3√12 cm

b.   3√13 cm

c.   3√6 cm

d.   3√20 cm

 

Ans : Let the radius of the upper circular part of the frustum be r cm.

 

 

(Picture)

 

 

Then r/3 = x/9 = AC/AD

∴  x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx – r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 – 9r2 – 3r3 – 27r = 42

⇒ 3r3 = 39

∴  r = 3√13 cm

 

62.  The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

 

a.   5 : 6

b.   3 : 4

c.   4 : 5

d.   2 : 3

 

Ans : Required ratio = 2*2r*5h/2*3r*4h  = 5 : 6

 

63.  A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a.   530 cm3

b.   536 cm3

c.   539 cm3

d.   545 cm3

 

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴  r + h/h = 462/154 =3

∴  r + h = 3h

∴  r = 2h

∴  2*2h2 = 154

∴  h2 = 154*7/22*4 = 49/4

= (7/2)2

∴  h = 7/2 cm 

and  r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

 

64.  A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

 

a.   1 : 2

b.   1 : 3

c.   2 : 3

d.   3 : 4

 

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴  10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴  h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴  r : h = 3 : 4

 

65.  A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

 

a.   1 : 3

b.   1 :1

c.   √3 : 1

d.   1 : √3

 

Ans :                  (Picture)       

 

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴  r : h = r/h

= 1 : √3

 

66.  If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

 

a.   550

b.   1100

c.   700

d.   350

 

Ans : ∴  ∠BOC = 1800 – (350 + 350) = 1100

 

 

(Picture)            

 

 

 ∴  ∠BAC = 1/2 * 1100 = 550

 

67.  If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

 

a.   Acute angled

b.   Equilateral

c.   Right angled

d.   Obtuse angled

 

Ans :                     (Picture)

 

 

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴   ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

 

 68.  If  sin2 (∝ + β/2) is-

 

a.   1

b.   -1

c.   0

d.   0.5

 

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 – cos2 β = 2

⇒  cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and  cos β = 0

⇒ ∝ = /2 and β = /2

∴  cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos  /2 = 0

 

69.  The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

 

a.   300

b.   450

c.   600

d.   900

 

Ans :                  (Picture)       

 

 

∵ tan θ = h/1√3 h = √3 = tan 600

 

70.  The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

 

a.   8/3 units

b.   4/3 units

c.   1 unit

d.   2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x – 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

 

71.  If x+1/16x = 1, then the value of 64×3 + 1/64×3 is-

 

a.   4

b.   52

c.   64

d.   76

∵ x + 1/16x = 1

⇒ 16×2 – 16x + 1 =0

⇒ 16×2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64×3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴   64×3 + 1/64×3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 – 15√3/676 -675

=52

 

72.  If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca  is-

 

a.   3

b.   2

c.   0

d.   1

 

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒  a2 + b2 + c2 = 2b2 – 2ac

∴  a2 + b2 + c2/ b2 ac = 2

 

 73.  If  a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

 

a.   -1

b.   0

c.   2

d.   1

 

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab)  = 2

⇒ a2 – ab + b2 = 2                          ….(1)

and  a2 + ab + b2 = 4                      …..(2)

⇒ 2ab = 2

⇒ ab = 1

 

74.  If a = 25, b = 15, c = -10, then the value of  a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

 

a.   30

b.   -15

c.   -30

d.   15

 

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a – b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴  Required value = 29250/1950 =15

 

75.  A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

 

a.   840

b.   920

c.   960

d.   1040

 

Ans :                  (Picture)       

 

 

∠ACB = 400 + 440 = 840

∴  ∠ACO = 900 – 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB – ∠ACO

= 840 – 460 = 380 = ∠OBC

∴  ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

 

76.  If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

 

a.   13 cm

b.   15 cm

c.   26 cm

d.   30 cm

 

Ans :     (Picture)

 

 

 

 

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴  Diameter of the circle = 2*OA

= 2*13 = 26cm

 

77.  In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

 

a.   4 cm

b.   2 cm

c.   12 cm

d.   6 cm

 

∵ PR/RQ = ½

But,   PR = 2cm

RQ  = 2*PR

= 4cm

 

(Picture)

 

∴  PQ = PR + RQ

= 2 + 4 = 6 cm

∴  BC = 2*PQ = 12CM

 

78.  If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is  equal to –

 

a.   ¾

b.   10/3

c.   3 ¾

d.   3

 

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 – tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴  θ = 300

∴  sin2  2θ + tan2 2θ =  sin2 600 +  tan2  600

=  ¾ + 3

=3 ¾

 

79.  The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

 

a.   -1

b.   ½

c.   0

d.   1

 

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

=  cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

=  cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

 

80.  If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

 

a.   5/17

b.   3/19

c.   7/10

d.   7/13

∵   sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴   2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴   sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒  (sin θ – cos θ)2 = (7/13)2

∴   sin θ – cos θ = 7/13

 

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