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IBPS PO Mains 5 Nov 2023 Paper

Question Numbers: 147-149

(x - 2)^{2} = (-3x^{2}) + 2^{2} + 25x - P

(10y^{2} - 3^{2}y + 2/3)(3) + 10y = 0

One factor of first equation is 5.

Question : 147 of 155

Marks: +1, -0

Find the value of 7P ÷ 21 × 0.2P - 89.

Solution:

Concept:

First find the value of

P

using the given equation, then substitute it into the required expression.

Explanation:

The given equation is:

(x-2)^2 = (-3x^2) + 2^2 + 25x - P

As it is given that one factor is

5

, substitute

x = 5

(5-2)^2 = -3(5^2) + 2^2 + 25(5) - P

9 = -75 + 4 + 125 - P

9 = 54 - P

P = 54 - 9 = 45

Now substitute

P = 45

7P \div 21 \times 0.2P - 89

7(45) \div 21 \times 0.2(45) - 89

= 315 \div 21 \times 9 - 89

= 15 \times 9 - 89

= 135 - 89 = 46

Since

P + 1 = 45 + 1 = 46

, the expression equals

P + 1

Answer:

Option C:

P + 1

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