SBI clerk prelims 22 Feb 2020 Questions and Answers Detailed Explanation
MOST IMPORTANT previous year question answers - 2 EXERCISES
The following question based on SBI clerk topic of previous year question answers
(a) Rs 615
(b) Rs 600
(c) Rs 1200
(d) Rs 585
e) Rs 1260
The correct answers to the above question in:
Answer: (a)
Let rate of interest be R%
ATQ, 1200 = $\text"6000×R×2"/100$
R = 10%
Since compounding is done half-yearly, rate of interest = 5%
Effective rate of interest = 5 + 5 + $\text"5×5"/100$ = 10.25%
Required interest = $\text"6000×10.25×1"/100$ = Rs 615
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Question : 1
A boat can cover 28 km downstream in 42 min. ratio of speed of boat in still water to speed of stream is 7 : 3. Find difference between time taken by boat to cover 60 km downstream & 40 km upstream.
a) 2.25 hr
b) 1 hr
c) 1.5 h
d) 0.4 hr
e) 0.9 hr
Answer »Answer: (b)
Let speed of the boat in still water & speed of the stream be 7x & 3x kmph respectively
ATQ, $28/\text"7x+3x" = 42/60$
x = 4
Required difference = $40/\text"7x−3x" − 60/\text"7x+3x"$
= $4/x$ = 1 hour
Question : 2
A & B entered into a business by investing total capital of Rs 17000. B withdraws Rs 1500 after 6 months and gets Rs 8100 as profit out of the total profit of Rs 19500 at the end of the year. Find the capital of B after 6 months from starting.
a) Rs 7000
b) Rs 9500
c) Rs 7500
d) Rs 6000
e)
Rs 6500
Answer »Answer: (d)
Let amount invested by A be Rs x
Profit ratio; A : B
= (x × 12) : (17000 – x) × 6 + (15500 – x) × 6
= 2x : (32500 – 2x)
ATQ, $\text"19500"/\text"32500−2x+2x" × (32500 − 2x)$ = 8100
32500 – 2x = 13500
x = Rs 9500
Required capital of B after 6 months
= 15500 – x = Rs 6000
Question : 3
If the length of a rectangle increase by 40% while keeping breadth constant then the area of rectangle increased by 24 m2 and perimeter of the original rectangle is 32 m. find breadth of the rectangle.
a) 8.4 m
b) 10 m
c) 6 m
d) 14 m
e) 8 m
Answer »Answer : (c)
let length & breadth of rectangle be x & y m respectively
ATQ, 1.4xy – xy = 24
xy = 60………(i)
also, 2(x + y) = 32
x + y = 16….(ii)
from (i) & (ii)
x = 10 m, y = 6 m
Breadth of rectangle = 6 m
Question : 4
A container contains mixture of milk & water in ratio 5 : 3 respectively. If 8 lit milk is added in it then ratio of milk to water becomes 11 : 5. Find difference between initial quantity of milk & that of water.
a) 5 lit
b) 38 lit
c) 18 lit
d) 30 lit
e) 10 lit
Answer »Answer: (e)
Let initial quantity of milk & water be 5x & 3x lit respectively
ATQ, $\text"5x+8"/\text"3x" = 11/5$
25x + 40 = 33x
⇒ x = 5
Required difference = 5x – 3x = 2x
= 10 lit
Directions:
In each question two equations numbered (I) and (II) are given. You have to solve both the equations and mark appropriate answer.
Question : 5
I. x2 + 4x - 45 = 0 and II. y2 - 13y + 40 = 0
a) If x = y or no relation can be established
b) If x > y
c) If x < y
d) If x ≥ y
e) If x ≤ y
Answer »Answer: (e)
I. x2 + 9x – 5x – 45 = 0
x(x + 9) – 5(x + 9) = 0
(x – 5) (x + 9) = 0
x = 5, - 9
II. y2 – 5y – 8y + 40 = 0
y(y – 5) – 8(y – 5) = 0
(y – 5) (y – 8) = 0
Y= 5, 8
So, x ≤ y
Question : 6
I. 17x2 – 14x – 83 = - 80 and II. y2 = 2y + 35
a) If x = y or no relation can be established
b) If x > y
c) If x < y
d) If x ≥ y
e) If x ≤ y
Answer »Answer: (a)
I. 17x2 – 14x – 3 = 0
17x2 – 17x + 3x – 3 = 0
17x (x – 1) + 3(x – 1) = 0
(17x + 3) (x – 1) = 0
x = − 3 17 , 1
II. y2 – 2y – 35 = 0
y2 – 7y + 5y – 35 = 0
y(y −7) + 5(???? − 7) = 0
y = 7, - 5
So, no relation can be established
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