Maruti Suzuki India Limited Aptitude Questions and Answers with Explanation,
Aptitude Questions
1. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1 / 3
B. 1 / 4
C. 1 / 5
D. 1 / 7
Ans:C
Ex:Suppose the vessel initially contains 8 litres of liquid. Let x littres of this liquid be replaced with water.
Quantity of water in new mixture = (3 – 3x/8 + x) litres.
Quantity of syrup in new mixture = (5 – 5x/8) litres.
(3 – 3x/8 + x) = (5 – 5x/8) = 5x + 24 = 40 – 5x
=› 10x = 16 =› x = 8/5
So, part of the mixture replaced = (8/5 x 1/8) = 1/5.
2. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?
A. 18750 sq.m
B. 37500 sq.m
C. 40000 sq.m
D. 48000 sq.m
Ans:B
Ex:So length =250 m;breadth=150 m
Area = (250 x 150)m²
= 37500 m²
3. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3.Find his average after 17th inning?
A. 19
B. 29
C. 39
D. 49
Ans:C
Ex:Let the average after 17th inning = x. Then, average after 16th inning = (x – 3)
Average=16 (x-3)+87
= 17x or x=(87-48)
= 39.
4. The banker’s discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs.1872 for the same time at the same rate. Find the time?
A. 3 months
B. 4 months
C. 5 months
D. 6 months
Ans:B
EX:S.I on Rs.1800 = T.D on Rs.1872.
P.W on Rs.1872 is Rs.1800.
Rs.72 is S.I on Rs. 1800 at 12%.
Time =(100×72 / 12×1800)
= 1/3 year
= 4 months.
5. A boat can travel with a speed of 13 km / hr in still water. If the speed of the stream is 4 km / hr. find the time taken by the boat to go 68 km downstream?
A. 2 hours
B. 3 hours
C. 4 hours
D. 5 hours
Ans:C
Ex:Speed Downstream= (13 + 4) km/hr
= 17 km/hr.
Time taken to travel 68 km downstream =(68 / 17)hrs
= 4 hrs.
6. An accurate clock shows 8 o’clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o’clock in the afternoon?
A. 144°
B. 150°
C. 168°
D. 180°
Ans:D
Ex:Angle traced by hour hand in
5 hrs 10 min. = (360/12 x 6)°
= 180°.
7. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is
A. Rs. 1550
B. Rs. 1650
C. Rs. 1750
D. Rs. 2000
Ans:C
Ex:C.I. =Rs[4000x(1+10/100)²-4000]
Rs.(4000×11/10×11/10-4000)= Rs.940.
Sum=Rs. [420×100 /3×8]
= Rs.1750.
8. (7.5×7.5+3.75+2.5×2.5) is equal to
A. 30
B. 60
C. 80
D. 100
Ans:D
Ex:Given expression= (7.5×7.5+2×7.5×2.5+2.5×2.5)²
= (a²+2ab+b²)
=(a+b)²
= (7.5+2.5)²
= 10²
= 100.
9. The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree
A. 30°
B. 45°
C. 60°
D. 90°
Ans:A
EX:Let AB be the tree and AC be its shadow.
Then, < ABC= θ.
Then, AC/AB= √3
cotθ= √3
θ=30°
10. If log 2 = 0.30103, the number of digits in 520 is
A. 14
B. 16
C. 18
D. 25
Ans:A
Ex:Log 520=20 log 5
=20 ×[log(10/2)]
=20 (log 10 – log 2)
=20 (1 – 0.3010)
=20×0.6990
=13.9800.
Characteristics = 13.
Hence, the number of digits in Log 520 is 14.
11. The difference between the place values of 7 and 3 in the prime number 527435 is
A. 4
B. 5
C. 45
D. 6970
Ans:D
Ex:(place value of 7)-(place value of 3)
= (7000 – 30)
= 6970.
12. A started a business with Rs.21,000 and is joined afterwards by B with Rs.36,000. After how many months did B join if the profits at the end of the year are divided equally?
A. 3
B. 4
C. 5
D. 6
Ans:C
Ex:Suppose B joined after x months.
Then, 21000×12=36000×(12 – x)
‹=›36x=180
= x= 5.
13. A woman introduces a man as the son of the brother of her mother. How is the man, related to the woman?
A. Nephew
B. Son
C. Cousin
D. Grandson
Ans:C
Ex:Brother of mother->uncle;
uncle’s son ->cousin
14. The year next to 1990 will have the same calendar as that of the year 1990?
A. 1995
B. 1997
C. 1996
D. 1992
Ans:C
Ex:The year 1990 has 365 days,year 1991 has 365 days ,ie 1 odd day,year 1992 has 366 days ,ie 2 odd days..therefore the sum odd days calculated from year 1990 to 1995 (1+2+1+1+1+1)=7 odd days.
15. If ROSE is coded as 6821, CHAIR is coded as 73456 and PREACH is coded as 961473, what will be the code for SEARCH?
A. 246173
B. 214673
C. 214763
D. 216473
Ans:B
Ex:So,in SEARCH, S is coded as 2, E as 1, A as 4, R as 6, C as 7, H as 3. Thus, the code for SEARCH is 214673.
16. There are five different houses, A to E in a row. A is to the right of B and E is to the left of C and right of A. B is to the right of D.Which of the houses is in the middle?
A. A
B. C
C. D
D. E
E. F
Ans:A
Ex:B is to the right of D. A is to the right of B. E is to the right A and left of C. So the order is D,B,A,E,C. Clearly A is in the middle.
17. A man sitting in a train which is travelling at 50 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.
A. 52 kmph.
B. 62 kmph.
C. 72 kmph.
D. 80 kmph.
Ans:B
Ex:Relative Speed = (280 / 9)m/sec
= (280/9 x 18/5)
= 112 kmph.
Speed of the train = (112 – 50)kmph
= 62 kmph.
18. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are
A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34
Ans:C
Ex:Let their marks be (x+9) and x.
Then, x+9= 56/100(x + 9 +x)
‹=› 25(x+9)
‹=› 14 (2x + 9)
‹=›3x = 99
‹=›x = 33.
SO, their marks are 42 and 33.
19. A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled ?
A. 4.5 hrs
B. 5 hrs
C. 6.5 hrs
D. 7.2 hrs
Ans:D
Ex:Net part filled in 1 hour= (1/4 – 1/9)
= 5/36.
Therefore, the cistern will be filled in 36/5 hrs i.e, 7.2 hrs.
20. The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?
A. 456 kms
B. 482 kms
C. 552 kms
D. 556 kms
Ans:C
Ex:Total distance travelled in 12 hours =(35+37+39+…..upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance= 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.