1. A tank has leak which would empty the completely filled tank in 8 hours. If the tank is full of water and a tap is opened which admits 6 liters of water per minute in the tank, the leak takes 12 hours to empty the tank. How many liters of water does the tank hold ?
A. 4800
B. 2400
C. 6400
D. 8640
E. None of these
Solution: A leak can emptied the completely full cistern in 8 hr
In 1 hr leak can empty = 1/8 part of the cistern
When tap is open then leak can empty the in = 12 hr
In 1 hr leak can empty = 1/12 part of cistern
Let tap can fill the cistern in = x hr
So, in 1 hr tap can fill the cistern = 1/x hr
Then, 1/8 - 1/x = 1/12
12x – 96 = 8x
4x = 96
x = 24
Means tap can fill the cistern in 24 hour = 24 × 60 = 1440 mins
It is given that, tap can fill 6 liters of water in 1 min.
In 1440 mins, it can fill i.e. Capacity of cistern = 1440 × 6 = 8640 litres.
2. There are two taps to fill a tank while a third to empty it. When the third tap is closed, they can fill the tank in 10 minutes and 12 minutes, respectively. If all the three taps be opened, the tank is filled in 15 minutes. If the first two taps are closed, in what time can the third tap empty the tank when it is full ?
A. 7 min.
B. 9 min. and 32 sec.
C. 8 min and 34 sec.
D. 6 min.
E. None of these
Sol. Let the time required to empty the third tap to be 'T'
Part of the tank filled by first tap in 1 minute = 1/10
Part of the tank filled by second tap in 1 minute = 1/12
Part of the tank emptied by third tap in 1 minute = 1/T
Part of the tank filled by all three taps in 1 minute = 1/10 + 1/12 - 1/T
All three taps can fill the tank in 15 minutes
Part of the tank filled by all three tap in 1 minute = 1/15
1/10 + 1/12 - 1/T = 1/15
T = 60/7 minutes = 8 minutes and 34 seconds
Time required by third tap to empty the tank = 8 minutes and 34 seconds.
3. Arnub's age is 1/6th of his father's age. Arnub's father, Karan’s age will be twice the age of Rajesh's age after 10 years. If Rajesh's tenth birthday was celebrated three years before, then what is Arnub's present age ?
A. 6
B. 5
C. 7
D. 4
E. None of these
Sol. Let present age of Rajesh = x.
Then we have x = 10 + 3 = 13 years as Rajesh’s 10th birthday was three years ago.
Now, we have Rajesh’s age after 10 years = x + 10 = 13 + 10 = 23 years
Let present age of Arnub and Karan be y and z respectively.
Then we have
z + 10 = 2 × (10 + x)
z + 10 = 2 × (10 + 13)
z + 10 = 20 + 26
z = 46 – 10
z = 36
Hence, age of Arnub = z/6
= 36/6
= 6 years.
4. Sum of the present ages of a father and his son is 48 years. If the product of their ages 5 years back is 165, what is the present age of the father ?
A. 36
B. 38
C. 28
D. 30
E. None of these
Sol. Let us assume that the present ages of father and son are `x’ and `y’ respectively
Given that the sum of the ages of father and the son is `48’
x + y = 48 ………. (1)
Age of father '5' years ago = (x – 5) years
Age of son '5' years ago = (y – 5) years
Given that the product of the ages of father and son five years back is `165’
(x – 5) × (y – 5) = 165
xy – 5(x + y) + 25 = 165
xy – 5(x + y) = 140 ……… (2)
On solving (1) and (2), we get
xy – 5(48) = 140 ( x + y = 48 …..From (1))
x (48 – x) – 240 = 140
48x – x^2 = 140 + 240
x^2 – 48x + 380 = 0
x^2 – 10x – 38x + 380 = 0
(x – 10) (x – 38) = 0
x = 10 or x = 38
But x = 38 is correct (`x’ is father’s age and father’s age should be higher)
Father’s age is `38 years’.
5. When Amit’s age 3 years ago is doubled and subtracted from twice his age 2 years hence, the resultant is exactly half of his present age. Find his present age (in years) ?
A. 24
B. 20
C. 15
D. 10
E. None of these
Sol. Let us assume that Amit’s present age is `x’ years
Amit’s age '3' years ago = (x – 3) years
Amit’s age '2' years hence = (x + 2) years
Given that Amit’s age is doubled `3’ years ago and subtracted from twice his age `2’ years hence is equal to half of the present age
2(x + 2) - 2(x – 3) = x/2
2x + 4 -2x + 6 = x/2
x = 20 years
Amit’s present age is `20 years’
6. 10, 10, 20, ?, 110, 300, 930
A. 40
B. 35
C. 45
D. 50
E. 60
Sol. 10 × 0.5 + 5 = 10
10 × 1 + 10 = 20
20 × 1.5 + 15 = 45
45 × 2 + 20 = 110
110 × 2.5 + 25 = 300
300 × 3 + 30 = 930
7. 1, ?, 22, 188, 2052, 28748
A. 2
B. 3
C. 6
D. 16
E. 10
Sol. 1 × 2 + 4 = 6
6 × 5 - 8 = 22
22 × 8 + 12 = 188
188 × 11 - 16 = 2052
2052 × 14 + 20 = 28748
8. 20, 31, 72, 225, ?
A. 648
B. 668
C. 534
D. 908
E. 968
Sol. 20×1 + 11 = 31
31×2 + 10 = 72
72×3 + 9 = 225
225×4 + 8 = 908
9. 110, _, 42, 20, 6
A.63
B.81
C.72
D.90
E.None of these
Sol. 11^2 – 11 = 110
9^2 – 9 = 72
7^2 – 7 = 42
5^2 – 5 = 20
3^2 – 3 = 6
10. 9, 1009, 928, 1440, 1391, _
A. 1675
B. 1650
C. 1660
D. 1607
E. None of these
Sol. 9 + 10^3 = 1009
1009 - 9^2 = 928
928 + 8^3 = 1440
1440 - 7^2 = 1391
1391 + 6^3 = 1607.
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