Aptitude model placement papers with solution - set 12

Aptitude model placement papers with solution - set 12


1. Pipes A and B fill a tank in 5 and 6 hours respectively. Pipe C empties the tank in 12 hours, if you open all the three pipes together this tank will be filled in?

A. 3 9/17 hours
B. 4.5 hours
C. 1.25 hours
D. 4 7/17 hours
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ANSWER: A. 3 9/17 hours

Solution: It is given that Pipe A,B and C work simultaneously.

Pipe A and B are filling the tank and Pipe C is emptying it.

In 1 hour volume of tank filled by Pipe A: 1/5th

In 1 hour volume of tank filled by Pipe B: 1/6th

In 1 hour volume of tank emptied by Pipe C: 1/12th

Working together Pipe A B and C fill x volume of tank.

=>1/x = (1/5)+(1/6)-(1/12)

=>x=60/17 hours




2. Two trains each of which is 100 m long moving in opposite direction to one another cross each other taking 8 seconds. If speed of one train is twice the speed of other train find the speed of the faster train.

A. 48 km/hr
B. 30 km/h
C. 60 km/hr
D. 40 km/hr
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ANSWER: C. 60 km/hr

Solution: Let the speed of the slower train be x m/sec.

Then speed of faster train becomes: 2x m/sec

Relative speed =x+2x=3x

(100+100)/8 = 3x

x = 25/3 m/sec

=>Speed of faster train= 2x=2*25/3 =50/3 m/sec
Converting into km/hr

=>(50/3)*(18/5)

=>60 km/hr




3. A boat with speed 15 km/hr in standing water goes 30 km downstream and returns in a total of 4.5 hours. What is the speed of current?

A. 4 kmph
B. 6 kmph
C. 5 kmph
D. 8 kmph
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ANSWER: C. 5 kmph

Solution: Speed of the boat in still water: 15 km/hr

Speed of boat downstream: (15+x)km/hr where x is speed of the current.

The boat travels 30 km downstream and then 30 km upstream and takes 9/2 hours.

Total time= Time taken to travel downstream + Time taken to travel upstream

=> 4/5= (30/(15+x))+(30/(15-x))

=>x= 5




4. Simple interest at x% for x years will come out to be Rs x on a sum of Rs?

A. x
B. 100/x
C. 100/x2
D. 100x
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ANSWER: B. 100/x

Solution: Formula for Simple Interest:

SI = Principal*Rate*Time

SI= x

Rate= x/100

Time= x

=> x= Principal*(x/100)*x

=>Principal Amount= 100/x




5. Presently a mixture in a tub contains substances A and B in the ratio of 7:5. If 9 litres of this solution is drawn out and B is added to the mixture in certain quantity the resulting mixture becomes 7:9. How much litres of liquid A was kept in the tub initially?

A. 32
B. 20
C. 27
D. 21
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ANSWER: D. 21

Solution: Currently the mixture is in ratio of A/B: 7:5

=> This means fraction of A in the solution: 7/12

=> Fraction of B in the solvent: 5/12

=>New concentration of A= 7-(7/12)9

=>New concentration of B= 5-(5/12)9+x

=> (7-(7/12)9)/(5-(5/12)9+x) = 7/9

=>x=21




6. logab x logbc x logcb = ?

A. abc
B. 1
C. 0
D. ab + bc + ca
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ANSWER: B. 1

Solution: To solve this question you must remember that:

Logab= logb/loga (logx= log10x)

=> (logb/loga)*(logc/logb)*(logb/logc)

=>1




7. Volume of a sphere with radius r is obtained by multiplying its surface area by r and a constant. Find the value of constant:

A. 3
B. 1/3
C. 4/3
D. 1
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ANSWER: B. 1/3

Solution: Volume of a sphere: (4/3)* π*r3

Area of a sphere: 4πr2

=>It is mentioned that surface area when multiplied by r and a constant gives Volume and value of constant is to be found.

=> 4πr2*r*k= (4/3)* π*r3

=>k=1/3




8. Find the odd one out:

125, 106, 88, 76, 65, 58, 53

A. 88
B. 106
C. 125
D. 76
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ANSWER: A. 88

Solution: This sequence represents a series in which from the reverse order a prime number is added:

53+5=58

58+7=65

65+11=76

76+13=89

89+17=106

106+19=125

Hence 88 is the answer.




9. A pineapple costs Rs 7 each and a watermelon costs Rs. 5 each. If I spend Rs 38 on total what is the number of pineapple I purchased?

A. 2
B. 3
C. 4
D Data Inadequete
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ANSWER: C. 4

Answer: The equation for this problem can be made as:

5x + 7y =38
Where x is the number of watermelons and y is the number of pineapples.
Now test for 2, 3 and 4:
For y = 2
5x + 14 =38
x is not an integer
For y = 3
5x = 17
X not an integer
For y =4
X = 2
So 4 pineapples and 2 watermelons can be bought by 38 Rs.


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