# Numbers-practice test

1.

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A 1677

B 1683

C 2523

D 3363

Answer: Option B

Explanation:

L.C.M. of 5, 6, 7, 8 = 840. Required number is in the of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k= 2.

Required number = (840*2+3) = 1683.

2.

There is a certain four digit number whose fourth digit is twice the first digit. Third digit is three more than second digit. Sum of the first and fourth digits is twice the third number. What was that number?

A 3036

B 4368

C 4148

D 3146

Answer: Option B

Explanation:

From the options, only 4368 is satisfying the conditions

As per First condition; 4*2 = 8

As per second condition 3+3 = 6

As per third condition 4+8 = 2*6

3.

A worker is paid Rs.20/- for a full days work. He works 1,1/3,2/3,1/8,3/4 days in a week. What is the total amount paid for that worker ?

A 57.50

B 57

C 57.9

D 58

Answer: Option A

4.

If the number 42573 * is exactly divisible by 72, then the minimum value of * is:

A 4

B 5

C 6

D 7

Answer: Option C

Explanation:

72 = 9 x8, where 9 and 8 are co-prime.

The minimum value of x for which 73x for which 73x is divisible by 8 is, x = 6.

Sum of digits in 425736 = (4 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 9.

Required value of * is 6.

5.

A number which when divided by 3, 4, 5, 6, 7, leaves the remainder 2, 3, 4, 5 and 6 respectively. Such largest 5 digit number is :

A 99960

B 999579

C 99539

D 99959

Answer: Option D

Explanation:

Observe that 3-2.4-3…7-6 leave 1 as the common difference

LCM (3, 4, 5,6, 7) = 3*4*5*7 = 420

Largest 5 digits number = 99999, Which on being divided by 420 will leave 238 and 39 as remainder

Hence, the largest such number = 99999-(remainder + common difference) = 99999-40 =99959.

6.

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A 4

B 5

C 6

D 8

Answer: Option A

Explanation:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305) = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

7.

Sanal has 60 red marbles, 156 blue marbles and 204 green marbles. He distributes them amongst a group of kids, such that each kid gets equal marbles and no kid has marbles of more than one colour. Find the number of kids.

A 12

B 21

C 35

D 36

Answer: Option C

Explanation:

HCF of 60, 156 and 204 is 12

60/12 = 5, 156/12 = 13, 204/12 = 17

Each kid gets 12 marbles, 5 get red, 13 get blue and 17 get green

:. Total number of kids = 5 +13+17 = 35

8.

Two numbers are in the ratio of 11 : 13 . If the H.C.F of these numbers is 19, determine the numbers

A 304, 369

B 209, 247

C 182, 199

D 182,122

Answer: Option B

Explanation:

Let the numbers be 11x and 13x

Since the H.C.F of given numbers is 19 which indicates that 19 is the common factor of these two numbers. Hence, it is obvious that value of x is 19. Therefore these numbers are 209 and 247 respectively.

9.

I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?

A 40

B 39

C 41

D 45

Answer: Option C

Explanation:

x + x+6 = rs 68

2x + 6 = 68

2x = 68-6

2x = 62

x=31

x is the amt lost in I race

x+ 6 = 31+6=37 is lost in second race

then my friend lost 37 + 4 = 41 Rs

10.

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A 279

B 283

C 308

D 318

Answer: Option C

Explanation:

Other number =[(11*7700)/275]=308