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?
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
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 ?
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
Let present age of P and Q be 3x3x and 4x4x respectively.
Ten years ago, P was half of Q's age
Total of their present ages
Let the age before 1010 years =x=x. Then,
Present age =x+10=40+10=50
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
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
Syam's present age =4x=4×6=24
Given that time taken for riding both ways will be 22 hours lesser than the time needed for waking one way and riding back.
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
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
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
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 ?
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 ?
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 ?
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
Profit on 1st part Profit on 2nd part:
Mean Profi 14%
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 ?
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.
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 ?
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)
a:b:c,then b ratio is b/(a+b+c)*35
Acc. to the question:
5c+4t=96 ----- (i)
5b+6c=32 ----- (ii)
7t+6b=37 ----- (iii)
Adding (i), (ii) & (iii)
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 ?
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 ?
capacity of beakers x,2x,3x
wine proportion is =7x/(6*3x)=7/18
60% of 20 kg= 12 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
Let price of water per liter be Re. 1
(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 ?
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
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 ?
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 ?
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 ?
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
so first liquid cost is x+2 = 8.80+2 = $10.80
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