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 existing age the son = x.

Then present age of the father = 3x + three

Given that ,3 years for this reason, father's age will be 10 yearsmore than twice the ageof the son

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

=> x = 10

Father's gift age = 3x + three = three×10+ three = 33

Question 2. Kamal Was four Times As Old As His Son eight 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 earlier than 8 years = x.

Then age of Kamal before 8 years ago = 4x

After 8 years, Kamal can be two times as old as his son

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

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

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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 five Years, Then What Isajay 's Present Age ?

Answer :

Present age of Denis = five years

Present age of Rahul = 5-2 = three

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 four. Ten Years Ago, P Was Half Of Q's Age. If The Ratio Of Their Present Ages Is three:four What Will Be The Total Of Their Present Ages?

Answer :

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

Ten years in the past, P turned into half of 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=five

Total of their present a while

=3x+4x=7x=7×5=35

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Question five. A Man's Age Is a hundred twenty fivepercentone hundred twenty five% Of What It Was 1010 Years Ago, But 8313percent8313% Of What It Will Be After 1010years. What Is His Present Age ?

Answer :

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=forty

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

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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 person =(x+24)=(x+24) years

Given that, in 22 years, man's age may 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 five:forty five: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 prevailing age of Kiran and Syam =5:four=5:4

Let gift age of Kiran =5x=5x

Present age of Syam =4x=4x

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

⇒(5x+3):(4x+three)=eleven:9

⇒9(5x+three)=eleven(4x+three)

⇒45x+27=44x+33

⇒x=33−27=6

⇒(5x+3):(4x+3)=eleven:nine

⇒9(5x+three)=11(4x+three)

⇒45x+27=44x+33

⇒x=33−27=6

Syam's gift age =4x=four×6=24

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Question eight. 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 point taken for riding both approaches may be 22 hours lesser than the time wanted for waking one manner and driving returned.

Therefore,

time wished for driving one way = time wanted for waking one manner - 22 hours

Given that point taken in taking walks one way and using lower back =five=5 hours 45 min

Hence, the time he could take to walk both ways

=5=5 hours forty five min + 22 hours

=7=7 hours 45 min

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

Answer :

Distance =six hundred=six hundred metre =zero.6=0.6 km

Time =5=5 mins =112=112 hour

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

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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 with the exception of stoppages =fifty four=54 kmph

Speed of the bus which includes stoppages =45=forty five kmph

Loss in velocity when such as stoppages =fifty four−forty five=9 kmph=54−45=9 kmph

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

Hence, time in which the bus stops consistent with hour

= Time taken to cover ninety nine km

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

Question eleven. Tea Worth Rs. 126 Per Kg And Rs. A hundred thirty five 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 one hundred thirty five*1 2*x/1 1 2=153

126 a hundred thirty five 2x=153*4

612-261=2x

351=2x

x=one hundred seventy five.50

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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

shall we expect x kg be rice A,

y kg be rice B

and a couple of 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 can't be 0 or negative

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

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Question thirteen. 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 includes 50% of non-water component.

In honey this non-water component constitutes eighty five% (100-15).

Therefore 0.5 X Amount of flower-nectar = zero.Eighty five X Amount of honey = zero.Eighty five X 1 kg

Therefore quantity of flower-nectar wished = (0.85/0.Five) * 1kg = 1.7 kg.

Question 14. There Are Two Bottles A And B, Each Filled With Milk And Water In The Ratio five:3 And 1:2 Respectively. A New Mixture Is Formed By Mixing The Contents Of A And B In The Ratio four:three.What Is The Ratio Of Composition Of Milk And Water In The New Mixture ?

Answer :

Bottle A combination: milk = 5/eight & water=three/8

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

Given New combination (Bottle A & B) ratio = four:3

Now milk ratio = (5/eight*four/7) + (1/three*3/7) = half of

and water = (3/eight*4/7) + (2/three*three/7) = half

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

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

Answer :

Profit on 1st element Profit on 2d component:

eight% 18%

Mean Profi 14%

4 6

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

Quantity of second kind = (3/5 ) * one thousand 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 eight:5 ?

Answer :

Let's count on C.P. Of spirit = Re. 1 in line with litre.

Spirit in 1 litre mix. Of A = five/7 litre. So C.P of one litre blend in A = Re. Five/7.

Spirit in 1 litre mix. Of B = 7/thirteen litre. So C.P of one litre mix in B = Re. 7/thirteen.

Spirit in 1-litre blend. Of C = 8/thirteen litre. So C.P. Of 1 litre mix in C = Re. 8/13.

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

(five/7) (7/thirteen)

(8/thirteen)

(1/thirteen) (nine/91)

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

Question 17. A Coffee Seller Has Two Types Of Coffee Brand A Costing five Bits Per Pound And Brand B Costing 3 Bits Per Pound. He Mixes Two Brands To Get A forty Pound Mixture. He Sold This At 6 Bits Per Pound. The Seller Gets A Profit Of 33 half Percent. How Much He Has Used Brand A In The Mixture ?

Answer :

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

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

His general selling price = 40*6=240

His value charge one hundred eighty approx(as he makes 33.5 % earnings)

5x+3y=a hundred and eighty ------------- (2)

On fixing (1) and (2)

x=30 kilos.

Question 18. If A And B Are Mixed In 3:5 Ratio And B And C Are Mixed In eight: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

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Question 19. Five Coffee And 4 Tea Costs Rs.Ninety six, 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=a hundred sixty five

b+c+t= 15Rs

Question 20. If A Strawberry And A Butterscotch Together Cost Rs. 18.00, A Vanilla And A Strawberry Cost Rs. Nine.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=nine ---- (three)

Adding (1) and (3) we get

2B= 27 so B= thirteen.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 :

capability of beakers x,2x,3x

2/three(x)+1/four(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

a hundred%=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 price 20 Rs For 1Rs, Milk = 1/2 lit

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

Total amount should be 10 litres.

So 10 - eight = 2 lit ( water need to upload)

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

Question 24. A Milkman Mixed 10 Liters Of Water To 50 Litres Of Milk Of Rs.Sixteen Per Liter, Then Cost Price Of Mixture Per Liter Is

Answer :

Let charge of water in keeping with liter be Re. 1

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

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Question 25. In What Ratio Tea Of Rs.Eighty Per Kg Be Mixed With 12kg Tea Of Rs.64 Per Kg, So Thai Cost Price Of Mixture Is Rs.Seventy four Per Kg ?

Answer :

(74-sixty four)/(eighty-74) = 5/three

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/nine%. How Much Does The Vessel Hold ?

Answer :

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

Let's anticipate x is the capability of the vessel

from 17% quantity of spirit comes down to fifteen 1/9% .

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

17/nine % x =1.7 litres .

=> x = ninety L

Question 27. Thirty Liters Of Water Is Added To one hundred fifty Litres Of 20% Solution Of Alcohol In Water.The Resulting Strength Of Alcohol Is

Answer :

As in line with query, 20% solution of alcohol in water

=> 30 litres of alcohol & one hundred twenty litres of water.

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

Therefore, the resulting electricity of alcohol = 30*one hundred/180 = 16.67%.

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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 had been taken away So we have the most effective 8L of the combination.

Amount of milk in 8 L of aggregate = 8 * 70% = five.6 liters

Amount of water in eight L of blend = (8 - 5.6) = 2.Four L.

Half of milk i.E 1/2 of 5.6 = 2.8 L.

We need (2.8 - 2.Four)L water greater = 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 crammed by way of 60 lts of 1st liquid and 40 lts. Of 2nd liquid.

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

% of kerosene = 27 %.

Question 30. Two Liquids Are Mixed In The Proportion Of three:2 And The Mixture Is Sold At $eleven 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 beverages proportion as three:2

from the combinations:

Suppose 2nd liquid price = $x, then first liquid = $(x+2)

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

2x-20 = 24-3x

5x = forty four

x=eight.8

so first liquid cost is x+2 = eight.Eighty+2 = $10.80