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
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
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
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
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=40
Present age =x+10=40+10=50
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
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
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
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
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
Answer :
126*1 135*1 2*x/1 1 2=153
126 135 2x=153*4
612-261=2x
351=2x
x=175.50
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
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.
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
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
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.
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.
Answer :
a:b:c,then b ratio is b/(a+b+c)*35
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
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
Answer :
capacity of beakers x,2x,3x
2/3(x)+1/4(2x)=7x/6
wine proportion is =7x/(6*3x)=7/18
Answer :
25%=5kg
100%=20kg
60% of 20 kg= 12 kg
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
Answer :
Let price of water per liter be Re. 1
((10*1)+(50*16))/60 =13.33
Answer :
(74-64)/(80-74) = 5/3
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
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%.
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
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 %.
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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