SBI PO 2019 - Category wise Question and Answer Practice Test with Solutions & PDF
SBI PO PRACTICE TEST [1 - EXERCISES]
SBI PO 2019
Directions:
Read the sentence to find out whether there is an error in it. The error, if any, will be in one part of the sentence. The number corresponding to that part will be your answer. If the given sentence is correct as it is, mark the answer as 'No error'. Ignore the errors of punctuation, if any.
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 : 72 [SBI-PO-2019]
- A) 5x + 2y = 31
- B) 3x + 7y = 36
a) X > Y
b) X ≥ Y
c) X < Y
d) X ≤ Y
e) X = Y or No relation can be established
Answer »Answer: (a)
A) (5x + 2y = 31) × 3
B) (3x + 7y = 36) × 5
15x + 6y = 93 ..... (i)
15x+35y= 180 .... (ii)
On solving both equations,
we get x = 5 and y = 3
so, X > Y
Question : 73 [SBI-PO-2019]
- A) x 2 – 7x + 12 = 0
- B ) y 2 – 9y + 20 = 0
a) X > Y
b) X ≥ Y
c) X < Y
d) X ≤ Y
e) X = Y or No relation can be established
Answer »Answer: (d)
A) x 2 – 7x + 12 = 0
⇒ x(x – 4) – 3(x – 4) = 0
⇒ (x – 4)(x – 3) = 0
So, x = 3, 4
B ) y 2 – 9y + 20 = 0
⇒ y(y – 5) – 4(x – 5) = 0
⇒ (y – 5)(y – 4) = 0
So, y= 4, 5
Hence, X ≤ Y
Question : 74 [SBI-PO-2019]
- A) (x + 1)2 = 122
- B) y 2 + 2y – 143 = 0
a) X > Y
b) X ≥ Y
c) X < Y
d) X ≤ Y
e) X = Y or No relation can be established
Answer »Answer: (e)
A) (x + 1)2 = 122
⇒ (x + 1)2 – 122 = 0
⇒ (x + 1 – 12)(x + 1 + 12) = 0
⇒ (x – 11)(x + 13) = 0
x = 11, –13
B) y 2 + 2y – 143 = 0
⇒ y 2 + 13y – 11y – 143 = 0
⇒ y(y + 13) – 11(y + 13) = 0
⇒ (y + 13)(y – 11) = 0
y = 11, – 13
So, no relation can be established.
Question : 75 [SBI-PO-2019]
- 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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