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) 2.25 hr
(b) 1 hr
(c) 1.5 h
(d) 0.4 hr
e) 0.9 hr
The correct answers to the above question in:
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
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Question : 1
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 : 2
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
Directions:
What will come in the place of (?) mark in following question.
Question : 3
280 ÷ 4 ÷ 2 = 170 - ?
a) 105
b) 115
c) 125
d) 135
e) 145
Answer »Answer: (d)
? = 170 – 35
? = 135
Question : 4
Rs 6000 when invested at a certain rate at SI for 2 years, it fetches Rs 1200. If the same sum is invested at the same rate for a year compounded half-yearly then find compound interest.
a) Rs 615
b) Rs 600
c) Rs 1200
d) Rs 585
e) Rs 1260
Answer »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
Question : 5
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 : 6
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
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