IBPS RRB PO Prelims 12 sep 2020 Questions and Answers Detailed Explanation
MOST IMPORTANT previous year question answers - 1 EXERCISES
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.
The following question based on IBPS RRB PO topic of previous year question answers
(a) y > x
(b) y < x
(c) y ≥ x
(d) y ≤ x
e) x = y or no relation can be established
The correct answers to the above question in:
Answer: (b)
x2 – 20 + 91 = 0
x2 – 13x – 7x + 91 = 0
x (x –13) – 7 ( x –13) = 0
(x –7) (x –13) = 0
X = 13, 7
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Read more ibps rrb po prelims 12 sep 2020 Based Previous Year Questions and Answers
Question : 1
I. x2 + 13x + 40 = 0 and II. y2 + 7y + 10 = 0
a) y > x
b) y < x
c) y ≥ x
d) y ≤ x
e) x = y or no relation can be established
Answer »Answer: (c)
x2 + 13x + 40 = 0
x2 + 8x + 5x + 40 = 0
x(x + 8) + 5 (x + 8) = 0
(x + 5) (x + 8) = 0
X = –5, – 8
y2 + 7y + 10 = 0
y2 + 2y + 5y + 10 = 0
y (y+2) + 3 (y + 2) = 0
y = – 2, – 5
so, y ≥ x
Question : 2
I. x2 – 11x + 30 = 0 and II. y2 + 12y + 36 = 0
a) y > x
b) y < x
c) y ≥ x
d) y ≤ x
e) x = y or no relation can be established
Answer »Answer: (b)
x2 – 11x + 30 = 0
x2 – 5x – 6x + 30 = 0
x(x – 5) – 6(x – 5) = 0
(x – 5)(x – 6) = 0
x = 6, 5
y2 + 12y + 36 = 0
y2 + 6y + 6y + 36 = 0
y(y + 6) + 6(y + 6) = 0
(y + 6)(y + 6) = 0
y = -6, -6
So, x > y.
Question : 3
I. x2 + 9x + 20 = 0 and II. 8y2 – 15y + 7 = 0
a) y > x
b) y < x
c) y ≥ x
d) y ≤ x
e) x = y or no relation can be established
Answer »Answer: (a)
x2 + 9x + 20 = 0
x2 + 4x + 5x + 20 = 0
x(x + 4) + 5(x + 4) = 0
(x + 5)(x + 4) = 0
x = – 4 , –5
8y2 – 15y + 7 = 0
8y2 – 8y – 7y + 7 = 0
8y(y – 1) – 7(y – 1) = 0
(8y – 7)(y – 1) = 0
y = $7/8$ , 1
So, y > x
Question : 4
I. x2 – x – 12 = 0 and II. y2 + 5y + 6 =0
a) y > x
b) y < x
c) y ≥ x
d) y ≤ x
e) x = y or no relation can be established
Answer »Answer: (e)
x2 – 4x + 3x – 12 = 0
x (x –4) + 3(x –4) = 0
(x +3) (x –4) = 0
X = – 3, 4
y2 + 5y + 6 = 0
y2 + 2y + 3y + 6 = 0
y (y + 2) + 3 (y + 2) = 0
(y+2)(y+3) = 0
y = – 2, – 3
Directions:
Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Calculate quantity I and quantity II on the basis of the given information then compare them and answer the following questions accordingly.
Question : 5
- Quantity I:
- A certain sum is invested for 2 years at a rate of 12% simple interest. If the simple interest is Rs. 1200, then the principal is
- Quantity II:
- : Rs. 6000
a) Quantity I > Quantity I
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or no relation can be established
Answer »Answer: (b)
SI = $\text"PRT"/100$
1200 = $\text"P x 12 x 2"/100$
⇒ P = 5000
Quantity I = Rs. 5000
Quantity II = Rs. 6000
Hence, Quantity I < Quantity II
Question : 6
- Quantity I:
- The selling price of an article is Rs 450. If 20% of profit earned by selling the article, then the profit earned is
- Quantity II:
- The selling price of an article is Rs. 84. If 20% of profit earned by selling the article, then the east price of the article is
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or no relation can be established
Answer »Answer: (a)
Quantity I = $450/1.2$ × 0.2 = Rs. 75
Quantity II = $84/1.2$ = Rs. 70
Hence, Quantity I > Quantity II
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