1. The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 yrs and average age of males is 20 yrs. Find the number of males ?
A. 10
B. 12
C. 15
D. 18
E. 20
Sol. Lets assume no. of males to be 'x'.
Thus, sum of ages of males/x = 20
sum of ages of males = 20x
Therefore, (240 + 20x)/(15 + x) = 18
Solving, we get,
x = 15yrs
2. If the sum of the smaller number x and two times the other number is equal to sum of two times the smaller number and 16. The difference between the numbers is 6. Find the smaller number ?
A. 4
B. 5
C. 6
D. 2
E. 3
Sol. Lets assume the larger number be 'y'
Thus, x + 2y = 2x + 16........1st equation
Also, y - x = 6...............2nd eqution
Solving these two, we get,
y = 10
x = 4
3. 16 men and 12 women can complete a work in 8 days, if 20 men can complete the same work in 16 days, in how many days 16 women can complete the same piece of work ?
A. 8
B. 12
C. 10
D. 15
E. 9
Sol. Lets assume it take 'w' days for 1 women to complete the job.
Then, 16/(16*20) + 12/w = 1/8
w = 160 days
Now for 16 women it will be, = 160/16 = 10 days.
4. If A's salary is 10,000 less than B's salary and B's salary is 15,000 less than C's salary and sum of their salaries is 65,000. Find the salary of A.
A. 15,000
B. 10,000
C. 12,000
D. 20,000
E. 5,000
Sol. Lets assume the salary of C be 'x'
Then, salary of B = x - 15,000
salary of A = x - 25,000
Now sum of these salaries is,
x + x - 15,000 + x - 25,000 = 65,000
Solving,
x = 35,000
A's salary = 35,000 - 25,000 = 10,000.
5. The sum of the ages of P and Q is 25 yrs more than the age of R.The present age of Q is 5 yrs more than the age of R. Find the present age of P ?
A. 15
B. 20
C. 12
D. 10
E. 16
Sol. Given, P + Q = 25 + R
Also, given, Q = 5 + R
Putting the value of Q in the 1st eqn, we get,
P + 5 + R = 25 + R
P = 20 yrs.
6. The perimeter of the rectangular field is 240 m. The ratio of length and breadth is 8:7. Find the area of the rectangle ?
A. 3854
B. 3584
C. 3485
D. 3845
E. None of these
Sol. Perimeter of the rectangle = 2*(l + b) = 240
Also given, l/b = 8/7, using this in 1st eqn,
8b/7 + b = 120
b = 56
l = 64
Area of the rectangle = 56*64 = 3584 sq.m.
7. One fourth of two-fifth of 30% of a number x is equal to 15. Find 20% of the same number ?
A. 100
B. 120
C. 105
D. 80
E. None of these
Sol. (1/4)*(2/5)*(30/100)*x = 15
x = 100.
8. The difference between compound interest and simple interest on a sum for 2 yrs at 20% per annum, when the interest is compounded annually is Rs 240. If the interest were compounded half yearly, the difference in two interests over same period would be ?
A. Rs 384.6
B. Rs 344.2
C. Rs 324.8
D. Rs 316.5
E. None of these
Sol. Difference for 2 yrs = P*20*20/100*100 = 240
Solving, we get,
P = 6000
So SI for 2 yrs = 6000*20*2/100 = 2400
Amount at compounded half yearly = 6000 [1 + 10/100]^4
Solving,we get,
Amount at compounded half yearly = 8784.6
So CI at compounded half yearly = 8784.6 – 6000 = 2784.6
So difference = 2784.6 – 2400 = Rs 384.6
Directions(9-10): What should come in place of the question mark(?) in the following number series:
9. 7, 16, 34, 61, 97, ?
A. 142
B. 154
C. 148
D. 164
E. None of these
Sol. 7 + 9*1, 16 + 9*2, 34 + 9*3, 61 + 9*4, ..
10. 8, 2, 2, 4.5, 18, ?
A. 112.5
B. 122.5
C. 134.2
D. 142.5
E. None of these
Sol. *0.5^2, *1^2, *1.5^2, *2^2, *2.5^2,
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