IBPS PO Prelims 3 oct 2020 - Category wise Question and Answer Practice Test with Solutions & PDF
IBPS PO PRACTICE TEST [1 - EXERCISES]
IBPS PO Prelims 3 oct 2020
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 : 78 [IBPS-PO-Prelims-2020]
I. X2 – 5x + 6 = 0 and II. 2y2 – 7y +3 = 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)
By solving the two equations,
Equation 1 ⇒ X2 -5X + 6 = 0
⇒ X2- 3X -2X + 6 = 0
⇒ X(X - 3) - 2(X - 3) = 0
⇒ (X -3)(X -2) = 0
⇒ X = +3, +2
Equation 2 ⇒ 2y2 – 7y +3 = 0
⇒ 2Y2− 6Y − Y + 3 = 0
⇒ 2Y(Y − 3) −1(Y − 3) = 0
⇒ (2Y − 1)(Y − 3) = 0
⇒ Y = 3, 0.5
Hence, No relation can be established
Question : 79 [IBPS-PO-Prelims-2020]
I. X2 = 92 and II. (Y – 8)2 = 9
a) X > Y
b) X < Y
c) X ≥ Y
d) X ≤ Y
e) X = Y or no relation can be established
Answer »Answer: (e)
I. X2 = 92
⇒ X = +9, –9
II. (Y – 8)2 = 9
⇒ (Y – 8)2 = 32
⇒ Y = 8 + 3 or 8 – 3
⇒ Y = 11 or 5
So, no relation can be established between X and Y
Question : 80 [IBPS-PO-Prelims-2020]
I. X2 – 5x + 6 = 0 and II. Y2 – y – 6 = 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)
I. X2 – 5x + 6 = 0
⇒ X2 – 3x -2x + 6 = 0
⇒ X(X - 3) - 2(X - 3) = 0
⇒ (X -3)(X -2) = 0
∴ X = +3, +2
II. Y2 – y – 6 = 0
⇒ Y2 – 3y +2y – 6 = 0
⇒ (Y − 3)(Y + 2) = 0
∴ Y = 3, -2
∴ no relation can be established
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