SBI PO 2019 Questions and Answers Detailed Explanation
MOST IMPORTANT previous year question answers - 1 EXERCISES
Directions:
What approximate value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)
The following question based on SBI PO topic of previous year question answers
(a) 226
(b) 242
(c) 256
(d) 351
e) 248
The correct answers to the above question in:
Answer: (b)
? = 14 × 3 + 200
⇒ ? = 42 + 200
⇒ ? = 242
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Read more sbi po prelims 2019 Based Previous Year Questions and Answers
Question : 1
6 % of [ ( 198.99 ) + ( 15.99 )2 ] = ?
a) 30
b) 17
c) 27
d) 24
e) 3
Answer »Answer: (c)
6 % of [ ( 198.99 ) + ( 15.99 )2 ] = ?
? = $6/100$ × [ 199 + (16)2 ]
? = $6/100$ × [ 199 + 256 ]
? = $6/100$ × 455
? = 27.30
? ≈ 27
Question : 2
$1/14.96$ - $1/3.94$ + $1/?$ = 0
a) 5
b) 8
c) 10
d) 12
e) 3
Answer »Answer: (a)
${1/14.96 - 1/3.94 + 1/? = 0}$
${1/? = 1/4 - 1/15}$
$1/?$ = $\text"15 - 4"/60$
$1/? = 11/60$
? = $60/11$ ≈ 5
Question : 3
(6.02)2 - $√{120.89} + √{144.06}$= ?
a) 35
b) 42
c) 37
d) 40
e) 33
Answer »Answer: (c)
(6.02)2 - $√{120.89} + √{144.06}$= ?
? = (6)2 - $√{121} + √{144}$
? = 36 - 11 + 12
? = 37
Question : 4
8.02 – (? – 9.01) + (10.92)2 = 120
a) 0
b) 18
c) 21
d) 24
e) 23
Answer »Answer: (b)
8.02 – (? – 9.01) + (10.92)2 = 120
⇒ 8 – ? + 9 + (11)2 = 120
⇒ 8 – ? + 9 + 121 = 120
⇒ ? = 138 – 120
⇒ ? = 18
Directions:
In the following question, two equations are given in variables X and Y. You have to solve these equations and determine the relation between X and Y.
Question : 5
- x 2 – 16x + 64 = 0
- Y 2 = 64
a) If X > Y
b) If X < Y
c) If X ≥ Y
d) . If X ≤ Y
e) If X=Y or No relation can be established
Answer »Answer: (c)
x 2 – 16x + 64 = 0
⇒ (x – 8)2 =0
⇒ x = 8
Y 2 = 64
⇒ Y = ± 8
So, X ≥ Y
Question : 6
- 6x2 + 5x + 1 = 0
- 4y2 + 4y + 1 = 0
a) If X > Y
b) If X < Y
c) If X ≥ Y
d) If X ≤ Y
e) If X = Y or no relation can be established
Answer »Answer: (c)
6x2 + 5x + 1 = 0
⇒ 6x2 + (3 + 2)x + 1 = 0
⇒ 3x(2x + 1) + (2x + 1) = 0
⇒ (3x + 1)(2x + 1) = 0
⇒ x = $–1/3$, $–1/2$
4y2 + 4y + 1 = 0
⇒4y2 + (2 + 2)y + 1 = 0
⇒ (2y + 1)2 = 0
⇒ y = $–1/2$
So, X ≥ Y
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