**Question 1. The Present Age Of A Father Is 3 Years More Than Three Times The Age Of His Son. Three Years Hence,father’s Age Will Be 10 Years More Than Twice The Age Of The Son. What Is Father’s Present Age?****Answer :**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

**Question 2. Kamal Was 4 Times As Old As His Son 8 Years Ago. After 8 Years, Kamal Will Be Twice As Old As Hisson. Find Out The Present Age Of Kamal ?****Answer :**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

**Question 3. If 6 Years Are Subtracted From The Present Age Of Ajay And The Remainder Is Divided By 18, Then The Present Age Of Rahul Is Obtained. If Rahul Is 2 Years Younger To Denis Whose Age Is 5 Years, Then What Isajay ‘s Present Age ?****Answer :**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

**Question 4. Ten Years Ago, P Was Half Of Q’s Age. If The Ratio Of Their Present Ages Is 3:4 What Will Be The Total Of Their Present Ages?****Answer :**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

**Question 5. A Man’s Age Is 125%125% Of What It Was 1010 Years Ago, But 8313%8313% Of What It Will Be After 1010years. What Is His Present Age ?****Answer :**Let the age before 1010 years =x=x. Then,

125×100=x+10

⇒125x=100x+1000

⇒x=100025=40125×100=x+10

⇒125x=100x+1000⇒x=100025=40

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

**Question 6. A Man Is 2424 Years Older Than His Son. In Two Years, His Age Will Be Twice The Age Of His Son. What Is The Present Age Of His Son ?****Answer :**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

**Question 7. Present Ages Of Kiran And Syam Are In The Ratio Of 5:45:4 Respectively. Three Years Hence, The Ratio Of Their Ages Will Become 11:911:9 Respectively. What Is Syam’s Present Age In Years ?****Answer :**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

**Question 8. A Man Takes 55 Hours 4545 Min In Walking To A Certain Place And Riding Back. He Would Have Gained 22 Hours By Riding Both Ways. The Time He Would Take To Walk Both Ways Is****Answer :**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

**Question 9. A Person Crosses A 600600 Metre Long Street In 55 Minutes. What Is His Speed In Km Per Hour ?****Answer :**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

**Question 10. Excluding Stoppages, The Speed Of A Bus Is 5454 Kmph And Including Stoppages, It Is 4545kmph. For How Many Minutes Does The Bus Stop Per Hour ?****Answer :**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

**Question 11. Tea Worth Rs. 126 Per Kg And Rs. 135 Per Kg Are Mixed With A Third Variety In The Ratio 1 : 1 : 2. If The Mixture Is Worth Rs. 153 Per Kg, The Price Of The Third Variety Per Kg Will Be:****Answer :**126*1 135*1 2*x/1 1 2=153

126 135 2x=153*4

612-261=2x

351=2x

x=175.50

**Question 12. A Merchant Mixes Three Varieties Of Rice Costing Rs.20/kg, Rs.24/kg And Rs.30/kg And Sells The Mixture At A Profit Of 20% At Rs.30 / Kg. How Many Kgs Of The Second Variety Will Be In The Mixture If 2 Kgs Of The Third Variety Is There In The Mixture ?****Answer :**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

**Question 13. When Processing Flower-nectar Into Honeybees’ Extract, A Considerable Amount Of Water Gets Reduced. How Much Flower-nectar Must Be Processed To Yield 1kg Of Honey, If Nectar Contains 50% Water, And The Honey Obtained From This Nectar Contains 15% Water ?****Answer :**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.

**Question 14. There Are Two Bottles A And B, Each Filled With Milk And Water In The Ratio 5:3 And 1:2 Respectively. A New Mixture Is Formed By Mixing The Contents Of A And B In The Ratio 4:3.what Is The Ratio Of Composition Of Milk And Water In The New Mixture ?****Answer :**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

**Question 15. A Merchant Has 1000 Kg Of Sugar, Part Of Which He Sells At 8% Profit And The Rest At 18% Profit. He Gains 14% On The Whole. The Quantity Sold At 18% Profit Is:****Answer :**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

**Question 16. Two Vessels A And B Contain Spirit And Water Mixed In The Ratio 5:2 And 7:6 Respectively. Find The Ratio In Which These Mixture Be Mixed To Obtain A New Mixture In Vessel C Containing Spirit And Water In The Ratio 8:5 ?****Answer :**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.

**Question 17. A Coffee Seller Has Two Types Of Coffee Brand A Costing 5 Bits Per Pound And Brand B Costing 3 Bits Per Pound. He Mixes Two Brands To Get A 40 Pound Mixture. He Sold This At 6 Bits Per Pound. The Seller Gets A Profit Of 33 1/2 Percent. How Much He Has Used Brand A In The Mixture ?****Answer :**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.

**Question 18. If A And B Are Mixed In 3:5 Ratio And B And C Are Mixed In 8:5 Ratio If The Final Mixture Is 35 Liters, Find The Amount Of B ?****Answer :**a:b:c,then b ratio is b/(a+b+c)*35

**Question 19. 5 Coffee And 4 Tea Costs Rs.96, 5 Badam Milk And 6 Coffee Costs Rs. 32 And 7 Tea And 6 Badam Milk Costs Rs.37. What Is The Combined Price Of 1 Tea, 1 Coffee And 1 Badam Milk ?****Answer :****Acc. to the question:**5c+4t=96 —– (i)

5b+6c=32 —– (ii)

7t+6b=37 —– (iii)

Adding (i), (ii) & (iii)

11c+11b+11t=165

b+c+t= 15Rs

**Question 20. If A Strawberry And A Butterscotch Together Cost Rs. 18.00, A Vanilla And A Strawberry Cost Rs. 9.00 And A Butterscotch Cost Rs.9.00 More Than A Vanilla Or A Strawberry Then Which Of The Following Can Be The Price Of A Butterscotch ?****Answer :**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

**Question 21. Two Beakers Are On The Table. The Capacity Of The First Beaker Is X Liters And That Of The Second Beaker Is 2x Liters. Two Thirds Of The First Beaker And One Fourth Of The Second Beaker Is Filled With Wine. The Remaining Space Is Filled With Water. If The Content In Both The Beakers Are Mixed In A Large Beaker Of Volume 3x Liters, What Is The Proportion Of Wine In The Beaker ?****Answer :**capacity of beakers x,2x,3x

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

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

**Question 22. A Manufacturing Company Has 15% Cobalt ,25% Led And 60% Of Copper. If 5kg Of Led Is Used In A Mixture, How Much Copper We Need To Use:****Answer :**25%=5kg

100%=20kg

60% of 20 kg= 12 kg

**Question 23. In What Ratio Must Water Be Added To 10 Liters Of Milk At Rs.20 Per Liter So That Cost Of Mixture Is Rs.16 Per Liter ?****Answer :**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

**Question 24. A Milkman Mixed 10 Liters Of Water To 50 Litres Of Milk Of Rs.16 Per Liter, Then Cost Price Of Mixture Per Liter Is****Answer :**Let price of water per liter be Re. 1

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

**Question 25. In What Ratio Tea Of Rs.80 Per Kg Be Mixed With 12kg Tea Of Rs.64 Per Kg, So Thai Cost Price Of Mixture Is Rs.74 Per Kg ?****Answer :**(74-64)/(80-74) = 5/3

**Question 26. A Vessel Is Full Of A Mixture Of Spirit And Water In Which There Is Found To Be 17% Of Spirit By Measure. Ten Litres Are Drawn Off And The Vessel Is Filled Up With Water. The Proportion Of Spirit Is Now Found To Be 15 1/9%. How Much Does The Vessel Hold ?****Answer :**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

**Question 27. Thirty Liters Of Water Is Added To 150 Litres Of 20% Solution Of Alcohol In Water.the Resulting Strength Of Alcohol Is****Answer :**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%.

**Question 28. A 10 Liter Mixture Of Milk And Water Contains 30 Percent Water. Two Liters Of This Mixture Is Taken Away. How Many Liters Of Water Should Now Be Added So That The Amount Of Milk In The Mixture Is Double That Of Water ?****Answer :**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

**Question 29. One Type Of Liquid Contains 25% Of Kerosene, The Other Contains 30% Of Kerosene. P Can Is Filled With 6 Parts Of The First Liquid And 4 Parts Of The Second Liquid. Find The Percentage Of The Kerosene In The New Mixture ?****Answer :**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 %.

**Question 30. Two Liquids Are Mixed In The Proportion Of 3:2 And The Mixture Is Sold At $11 Per Liter At A 10% Profit. If The First Liquid Costs $2 More Per Liter Than The Second, What Does It Cost Per Liter ?****Answer :**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

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