# Aptitude model placement papers with solution - set 10

## Aptitude model placement papers with solution - set 10

1. A man sells 45 lemons for Rs 40 and loses 20%. At how much price should he sell 24 lemons to the next customer to make a 20% profit?

A. 32
B. 20
C. 24
D. 16

Solution:

Let cost price of lemons be x.

Selling price of lemons becomes CP - loss.

=>SP=x-(20/100)x

=>40=x(80/100)

=>50

So 45 lemons cost Rs 50
Cost of 1 lemon = 50/45 Rs
Cost of 24 lemons = (50/45)*24 =80/3 Rs
Selling Price of 24 lemons =(80/3)+(20/100)(80/3)=Rs 32

2. If m and n are two whole numbers such than mn=121 then determine the value of (m-1) n+1

A. 1
B. 121
C. 100
D. 1000

Solution: It is mentioned that m and n are 2 whole numbers.

mn=121. Clearly 112 is121.

=>m=11 .and n=2

=>(m+1) n+1 = (10) 3

=>1000

3. If one is added to the numerator of the fraction it becomes one. If one is added to the denominator of the fraction it becomes 1/2. The fraction is?

A. 1/2
B. 2/3
C. 3/5
D. 2/5

Ans: Let the fraction be x/y.

=> (x+1)/y = 1

=> x+1 = y—(1)
x/(y-1) = 1/2

=> 2x = (y+1)—(2)

=>Equating (1) and (2)

=> x+1 = 2x-1

=>x=2

Subsituting the value of x in (1)

=>y=3

=>2/3

4. Ratio of areas of incircle and circumcircle of an equilateral triangle is?

A. 1:4
B. 1:2
C. 2:1
D. 2:3

5. What is the minimum number of years upon which SI on Rs 2600 at 6.67% interest rate will be in whole number?

A. 2
B. 6
C. 3
D. 4

Solution: Simple Interest =Principal * Rate * Time

Principal = 2600 Rs

Interest Rate = 6.67%

=> SI = 2600 * (20/3)*Time

SI of the number will be a whole number for the first time when time is three as denominator of rate is 3.

Hence time should be set to 3 years.

6. Speed of a man with current measures 15 km/hr. Speed of current is 2.5 km/hr. What is the speed of man against the current?

A. 12.5 kmph
B. 10 kmph
C. 9 kmph
D. 8.5 kmph

ANSWER: B. 10 kmph

Solution: Let the speed of man in still water be x km/hr

Now, speed with current=15 km/hr

Speed of current: 2.5 km/hr

Speed of man with current: Speed of man in still water+Speed of current

=>15=x+2.5

=>x=12.5

Speed of man against the current: Speed of man in still water-Speed of current

=>12.5-2.5

=>10 km/hr

7. A jar is filled of liquid which is 3 parts water and 5 parts alcohol. How much of this mixture should be drawn out and replaced such that this mixture may contain half water and half alcohol?

A. 25%
B. 20%
C. 14.29%
D. 33.33%

Solution: Let the jar initially contain 8 litres of mixture:3 litres water and 5 litres alcohol

Let x litres of this mixture is drawn out and is replaced by x litres of water.

Amount of water now in the solution: 3-(3/8)x+x.

Amount of alcohol now in the solution: 5-(5/8)x

Desired ratio: 1/1

=>3-(3/8)x+x=5-(5/8)x

x=8/5 litres

%x=((8/5)/8)*100=20%

8. Find the odd man out:

5, 8, 20, 42, 124, 246

A. 20
B. 42
C. 124
D. 246

Solution: The series is followed on the pattern of 2 alternate sequences with common formula being 2n-2 and 3n-2 respectively.

8 = 2*5-2

22 = 3*8-2

42 = 22*2-2

124 = 42*3-2

246 = 124*2-2

Hence, 20 is the odd no.

9. A and B working separately can complete a piece of work in 9 and 12 days respectively. If they work alternatingly starting from A in how many days will the work be completed?

A. 10.25 days
B. 11.5 days
C. 12.75 days
D. 9 days

ANSWER: A. 10.25 days

Solution: A takes 9 days to complete a piece of work. B takes 12 days to complete a piece of work.

=> In one day A came complete 1/9th of the task(x/9)

=> In one day B can complete 1/12th of the job.(x/12)

First day x/9th work will be done next day x/12th and so on:
Together A and B can do x/9 + x/12 work.

=>(x/9) + (x/12) = 7/36th work done

In 10 days 35/36th of the work will be done.
On 11th day it will be A’s chance. It takes A 1 day to complete 1/9th work.
So, 1/36th work will be done in 1/4 days.

=>10.25 days

10. Determine the value of 51/4*(125) 0.25

A. 5
B. 10
C. 25
D. 125

Solution: 125 can be written as 53.

Substituting this in equation:

=>51/4*(53)0.25

=>51/4*53/4

=>51/4+3/4

=>5