1. A is 2 years older than B while B is 3 year younger than C. The ratio of age of A, 6 years hence and B, 2 years ago is 5:3. What was age of C, 6 years ago?
(a)12 years
(b)19 years
(c)15 years
(d)14 years
(e)21 years
Sol. Lets assume C is 'x' yrs
Then B will be (x-3) yrs
And A will be (x-3+2) = (x-1) yrs
Age of A 6yrs hence = x-1+6 = (x+5) yrs
Age of B 2 yrs ago = x-3-2 = (x-5) yrs
Now, ratio of these two is 5:3, so,
5/3 = (x+5)/(x-5)
x = 20 yrs
Age of C 6 yrs ago = x-6 = 14 yrs.
2. A, B and C started a business with their investments in the ratio 1 : 2 : 4. After 6 month A invested the half amount more as before and B invested same the amount as before while C withdrew 1/4 th of his investment. Find the ratio of their profits at the end of the year.
(a) 5 : 12 : 13
(b) 5 : 11 : 14
(c) 5 : 12 : 14
(d) 5 : 12 : 10
(e) None of these
Sol. Let their initial investments be x, 2x and 4x.
Their profits for 1st 6months will be,
A = x*6
B = 2x*6
C = 4x*6
Their profits for next 6 months will be,
A = 9x (as increased by half amount)
B = 12x (remained same)
C = 18x (reduced by 1/4th amount)
Thus total profits,
A = 6x+9x = 15x
B = 12x+12x = 24x
C = 24x+18x = 42x
Thus, A:B:C = 5:8:14 .
3. A train makes four stops each 15 min long to cover a distance of 500 km in 3hrs. If the train is 300m long then find the time it takes to cross a pole ?
A. 4.56 seconds
B. 3.24 seconds
C. 4.32 seconds
D. 3.42 seconds
E. None of these
Sol. Lets assume time taken to be 't' seconds
Then speed of the train will be = (300/t) m/sec
Now its also given that the train takes 3hrs to cover a distance of 500 km with the stoppage time being = (15*4) = 60 min = 1 hr
Thus the speed of the train will be,
=500/(3-1) = 250km/hr = (250/3.6)m/sec
Now equating the two,
250/3.6 = 300/t
t = 4.32 seconds
Thus it takes 4.32 seconds to cross a pole.
4. A, B and C started a business with the investment in the ratio 2:1:4. After 4 months A reduces his investment by half, B doubles it and C leaves the partnership. Then what is the ratio of profit of A, B and C after a year ?
A. 3:3:4
B. 4:5:5
C. 3:4:3
D. 4:5:4
E. None of these
Sol. Lets assume initial investment be 2x,x and 4x
For first 4 months,
Profit of A = 2x*4 = 8X
Profit of B = x*4 = 4x
Profit of C = 4x*4 = 16x
For last 8 months,
Profit of A = x*8 = 8x
Profit of B = 2x*8 = 16x
Profit of C = 0*8 = 0
Thus profits fo the year will be,
A = (8x+8x) = 16x
B = (16x+4x) = 20x
C = (16x+0) = 16x
Ratio A:B:C is,
= 16x:20x:16x
= 4:5:4
5. In a batch of 50 students and 8 teachers, gifts were distributed such that each student got 25% of the total teachers and each teacher got 10% of the total students. How many gifts were there ?
A. 150
B. 145
C. 140
D. 135
E. None of these
Sol. Number of gifts each student gets = 25% of 8 = 2
Total gifts to students = 2*50 = 100
Number of gifts to each teacher = 10% of 50 = 5
Total gifts to teachers = 5*8 = 40
Thus total gifts = (100+40) = 140
6. In how many ways 8 men and 5 women can be arranged around a circular table such that no two women are together ?
A. 8!*336
B. 7!*336
C. 5!*336
D. 6!*336
E. None of these
Sol. No. of ways of sitting 8 men around a circular table = (8-1)! = 7!
Now no. of gaps between these men is 8, so 5 women can be seated in these by,
= 8p5 = 8!/5! = 336
Thus total no. of cases,
= 7!*336
7. Ramesh donates 20% of his salary to charity. On the donation day he changed is his mind and donated Rs 15,000 which is 125% of his previous amount. What is his salary ?
A. Rs 60,000
B. Rs 50,000
C. Rs 40,000
D. Rs 30,000
E. None of these
Sol. Lets assume the salary is 'x'.
Previous charity donation = 20% of x = 0.2x
New donation = 125% of 0.2x = 0.25x
Now this is equal to Rs 15,000
0.25x = 15000
x = Rs 60,000
8. Two pipes x and y fill a tank in 10 and 15 hrs respectively. x and y are opened together but after some time x was closed and it took 12 hrs to fill the tank, then for how long x was opened ?
A. 2 hrs
B. 4 hrs
C. 6 hrs
D. 4 hrs
E. None of these
Sol. Lets assume x was opened for = 'a' hrs
Thus work done by x and y together for 'a' hrs,
= a*(1/10 + 1/15) = (a/6)th work
Now work done byy alone,
= (12-a)/15 th work
Adding the two,
a/6 + (12-a)/15 = 1
a = 2hrs
9. A mixture of 15 ltrs contains two liquids A and B in the ratio 2:1. 5 ltrs is drawn and replaced with 5 ltrs of B, if the same process is repeated again, then find the amount of A at the end of it ?
A. 10
B. 12
C. 15
D. 16
E. None of these
Sol. Initial amount of A = (2/3)*15 = 10
Initial amount of B = (1/3)*15 = 5
Amount of A left after 2 withdrawls,
= 10*[(10-5)/5]^2
= 10 ltrs
10. The average age of a couple was 24 years. After their 1st and 2nd children (twins) were born, the average age of the family became 13.5 years. The average age of the family just after 3rd child was born was 13.2 years. The average age of the family after 4th child was born was 16 years. The current average age of the family is 19 years. What is the current age of the twin children?
(a) 14 years
(b) 15 years
(c) 11 years
(d) 12 years
(e) 13 years
Sol. Sum of the ages of the couple = 24 * 2 = 48
After the 1st and 2nd children, sum = 13.5 * 4 = 54
Difference in sum = 54 – 48 = 6 years
Or after 6/ 2 = 3 years, the twins were born to the couple
(Ages of children at the time of birth is 0)
After 3rd child, sum = 13.2 * 5 = 66 yrs
Difference = 66 – 54 = 12
Or after 12/4 = 3 yrs, 3rd child was born (Couple + 2 children were already present. So 4)
After 4th child, sum = 16 * 6 = 96 yrs
Difference = 96 – 66 = 30
Or after 30/5 = 6 yrs, 4th child was born
Current sum = 19 * 6 = 114 yrs
Difference 114 – 96 = 18 yrs
Or after 18/6 = 3 yrs
The gap between the children are as follows: 3 yrs, 6 yrs and 3 yrs
Age of eldest ones = 3 + 6 + 3 = 12yrs
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