# Flipkart Aptitude Placement Papers - Flipkart Aptitude Interview Questions and Answers updated on 08.Dec.2023

Ratio of the present age of Kiran and Syam =5:4=5:4

Let present age of Kiran =5x=5x

Present age of Syam =4x=4x

After 33 years, ratio of their ages =11:9=11:9

⇒(5x+3):(4x+3)=11:9

⇒9(5x+3)=11(4x+3)

⇒45x+27=44x+33

⇒x=33−27=6

⇒(5x+3):(4x+3)=11:9

⇒9(5x+3)=11(4x+3)

⇒45x+27=44x+33

⇒x=33−27=6

Syam's present age =4x=4×6=24

Flower-nectar contains 50% of non-water part.

In honey this non-water part constitutes 85% (100-15).

Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg

Therefore amount of flower-nectar needed = (0.85/0.5) * 1kg = 1.7 kg.

capacity of beakers x,2x,3x

2/3(x)+1/4(2x)=7x/6

wine proportion is =7x/(6*3x)=7/18

Given that time taken for riding both ways will be 22 hours lesser than the time needed for waking one way and riding back.

Therefore,

time needed for riding one way = time needed for waking one way - 22 hours

Given that time taken in walking one way and riding back =5=5 hours 45 min

Hence, the time he would take to walk both ways

=5=5 hours 45 min + 22 hours

=7=7 hours 45 min

Given two liquids proportion as 3:2

from the mixtures:

Suppose second liquid cost = \$x, then first liquid = \$(x+2)

(x-10)/(10-x-2) = 3/2

2x-20 = 24-3x

5x = 44

x=8.8

so first liquid cost is x+2 = 8.80+2 = \$10.80

a:b:c,then b ratio is b/(a+b+c)*35

Let present age of the son =x=x years

Then, present age the man =(x+24)=(x+24) years

Given that, in 22 years, man's age will be twice the age of his son

⇒(x+24)+2=2(x+2)⇒x=22

Let the present age the son = x.

Then present age of the father = 3x + 3

Given that ,three years hence, father's age will be 10 yearsmore than twice the ageof the son

=> (3x+3+3) = 2(x + 3) +10

=> x = 10

Father's present age = 3x + 3 = 3×10+ 3 = 33

25%=5kg

100%=20kg

60% of 20 kg= 12 kg

Let present age of P and Q be 3x3x and 4x4x respectively.

Ten years ago, P was half of Q's age

⇒(3x−10)=12(4x−10)

⇒6x−20=4x−10⇒2x=10

⇒x=5⇒(3x−10)=12(4x−10)

⇒6x−20=4x−10⇒2x=10⇒x=5

Total of their present ages

=3x+4x=7x=7×5=35

Profit on 1st part Profit on 2nd part:

8% 18%

Mean Profi 14%

4 6

Ratio of 1st and 2nd parts = 4 : 6 = 2 : 3

Quantity of 2nd kind = (3/5 ) * 1000 kg = 600 kg

10 lit milk cost 20 Rs For 1Rs, Milk = 1/2 lit

For 16Rs, Milk = (1/2)*16 = 8 lit (Milk)

Total quantity should be 10 litres.

so 10 - 8 = 2 lit ( water have to add)

Now ratio = 2(water) : 8(Milk) = 1:4

A=20/kg,B=24/kg ,C =30/kg

lets assume x kg be rice A,

y kg be rice B

and 2 kgs is rice C

Given final CP = 25

So, 20x+24y+60=25(x+y+2) ------(1)

By Solving above eq. y=10-5x

Since , y cannot be zero or negative

Hence, x can only be 1 giving y = 5kg

Two liters were taken away So we have the only 8L of the mixture.

Amount of milk in 8 L of mixture = 8 * 70% = 5.6 liters

Amount of water in 8 L of mix = (8 - 5.6) = 2.4 L.

Half of milk i.e half of 5.6 = 2.8 L.

We need (2.8 - 2.4)L water more = 0.4 L

As per question, 20% solution of alcohol in water

=> 30 litres of alcohol & 120 litres of water.

Then 30 litres of water is added, so the solution now contains 150 litres of water and 30 litres of alcohol.

Therefore, the resulting strength of alcohol = 30*100/180 = 16.67%.

Let the age of the son before 8 years = x.

Then age of Kamal before 8 years ago = 4x

After 8 years, Kamal will be twice as old as his son

=> 4x + 16 = 2(x + 16)

=> x = 8 Present age of Kamal = 4x + 8 = 4×8 +8 = 40

Acc. to the question:

5c+4t=96 ----- (i)

5b+6c=32 ----- (ii)

7t+6b=37 ----- (iii)

11c+11b+11t=165

b+c+t= 15Rs

Let price of water per liter be Re. 1

((10*1)+(50*16))/60 =13.33

Let x and y quantities of A and B respectively are mixed.

=> x+y=40 --------- (1)

His total selling price = 40*6=240

His cost price 180 approx(as he makes 33.5 % profit)

5x+3y=180 ------------- (2)

On solving (1) and (2)

x=30 pounds.

Let the age before 1010 years =x=x. Then,

125x100=x+10

⇒125x=100x+1000

⇒x=100025=40125x100=x+10

⇒125x=100x+1000⇒x=100025=40

Present age =x+10=40+10=50

Let's assume C.P. of spirit = Re. 1 per litre.

Spirit in 1 litre mix. of A = 5/7 litre. So C.P of 1 litre mix in A = Re. 5/7.

Spirit in 1 litre mix. of B = 7/13 litre. So C.P of 1 litre mix in B = Re. 7/13.

Spirit in 1-litre mix. of C = 8/13 litre. So C.P. of 1 litre mix in C = Re. 8/13.

By rule of an allegation, we have required ratio X:Y.

(5/7) (7/13)

(8/13)

(1/13) (9/91)

So, required ratio = 1/13 : 9/91 = 7:9.

Bottle A mixture: milk = 5/8 & water=3/8

Bottle B mixture: milk = 1/3 & water=2/3

Given New mixture (Bottle A & B) ratio = 4:3

Now milk ratio = (5/8*4/7) + (1/3*3/7) = 1/2

and water = (3/8*4/7) + (2/3*3/7) = 1/2

So, New mixture ratio milk : water = 1 : 1

Distance =600=600 metre =0.6=0.6 km

Time =5=5 minutes =112=112 hour

Speed=distance time=0.6(112)Speed=distance time=0.6(112) =7.2 km/hr

Present age of Denis = 5 years

Present age of Rahul = 5-2 = 3

Let the present age of Ajay = x

Then (x-6)/18 = present age of Rahul = 3=> x- 6 = 3×18 = 54=> x = 54 + 6= 60

Speed of the bus excluding stoppages =54=54 kmph

Speed of the bus including stoppages =45=45 kmph

Loss in speed when including stoppages =54−45=9 kmph=54−45=9 kmph

⇒⇒ In 11 hour, bus covers 99 km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover 99 km

=distancespeed=954 hour=16 hour =distancespeed=954 hour=16 hour  =606 min=10 min

126*1 135*1 2*x/1 1 2=153

126 135 2x=153*4

612-261=2x

351=2x

x=175.50

In 10L of water drawn amount of spirit present is = (17/100)*10 =1.7 L .

Let's assume x is the capacity of the vessel

from 17% amount of spirit comes down to 15 1/9% .

The difference between this percentage is (17% - 15 1/9%) = 17/9 % x

17/9 % x =1.7 litres .

=> x = 90 L

Acc. to the question

S+B= 18 ---- (1)

V+S= 9 ---- (2)

B-S=9 ---- (3)

Adding (1) and (3) we get

2B= 27 so B= 13.5Rs

Let P be filled by 60 lts of 1st liquid and 40 lts. of 2nd liquid.

Amount of kerosene = (25*60/100) + (30*40/100) = 27 lts.

% of kerosene = 27 %.