Simplification Questions with Solutions for Bank PO
Simplification Questions with Solutions for Competitive Exams
Q.11. One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
(A) Rs. 50,000
(B) Rs. 30,000
(C) Rs. 90,000
(D) Rs. 60,000
Ans . D
Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then,
$$ {1\over3}x={1\over2}(150000-x) $$
$$ → {x\over3} +{x\over2} = 75000 $$
$$ → {5x\over6} = 75000 $$
$$ → x ={75000×6\over3} = 90000 $$
Q.12. A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?
(A) 60
(B) 45
(C) 90
(D) 75
Ans . C
Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
→ 16x = 480
∴ x = 30.
Hence, total number of notes = 3x = 90.
Q.13. If a - b = 3 and a2 + b2 = 29, find the value of ab.
(A) 12
(B) 10
(C) 18
(D) 15
Ans . B
2ab = (a2 + b2) - (a - b)2
= 29 - 9 = 20
ab = 10.
→ ab = 10.
Q.14. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present?
(A) 35
(B) 10
(C) 58
(D) 62.5
(E) None of these
Ans . D
Let the capacity of 1 bucket = x.
Then, the capacity of tank = 25x.
$$ New \ capacity \ of \ bucket \ = {2\over5}x$$
$$ ∴ \ Required \ number \ of \ buckets = {25x\over{2x\over5}}$$
$$ = \left(25x× {5x\over2} \right) $$
$$ = {125\over2} = 62.5$$
Q.15. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
(A) 80
(B) 20
(C) 200
(D) 100
Ans . D
Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 → x - y = 20 .... (i)
and x + 20 = 2(y - 20) → x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100, y = 80.
∴ The required answer A = 100.