Quant Equations for SBI PO 2017

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Mentor for Bank Exams
Quant Equations for SBI PO 2017
Directions (1 – 5): In the following question two equations are given. You have to solve these equations and determine relation between x and y and give answer
a) x < y
b) x > y
c) x = y OR the relationship cannot be determined
d) x ≥ y
e) x ≤ y
1. I. 20x2 + 48x – 5 = 0 II. 2y2 + 15y + 25 = 0
2. I. 12x2 – 73x + 6 = 0 II. 7y2 – 40y – 12 = 0
3. I. x2 – 22x + 117 = 0 II. y2 – 35y + 300 = 0
4. I. x2 + 23x – 210 = 0 II. y2 – 28y + 147 = 0
5. I. 2x2 – 17x + 35 = 0 II. y2 + 9y – 70 = 0
6. I. 6x2 – 19x + 15 = 0 II. 10y2 – 29y + 21 = 0
7. I. 12x2 + 11x – 56 = 0 II. 4y2 – 15y + 14 = 0
8. I. 3x2 + 13x + 12 = 0 II. y2 + 9y + 20 = 0
9. I. 8x2 – 15x + 7 = 0 II. 2y2 – 7y + 6 = 0
10. I. 7x – 3y = 13 II. 5x + 4y = 40
Solutions:
1. D) I. 20x2 + 48x – 5 = 0
⇒ 20x2 + 50x – 2x – 5 = 0
⇒ 10x(2x + 5) – 1(2x + 5) = 0
⇒ (2x + 5)(10x – 1) = 0
Then, x = - 5/2 or x = 1/10
II. 2y2 + 15y + 25 = 0
⇒ 2y2 + 10y + 5y + 25 = 0
⇒ 2y(y + 5) + 5(y + 5) = 0
⇒ (y + 5)(2y + 5) = 0
Then, y = - 5 or y = - 5/2
So, when x = - 5/2, x > y for y = - 5 and x = y for y = - 5/2
And when x = 1/10, x > y for y = - 5 and x > y for y = - 5/2
∴ So, we can observe that x ≥ y.
2. C) I. 12x2 – 73x + 6 = 0
⇒ 12x2 – 72x –x + 6 = 0
⇒ 12x(x – 6) – 1(x – 6) = 0
⇒ (x – 6)(12x – 1) = 0
Then, x = + 6 or x = + 1/12
II. 7y2 – 40y – 12 = 0
⇒ 7y2 – 42y + 2y – 12 = 0
⇒ 7y(y – 6) + 2(y – 6) = 0
⇒ (y – 6)(7y + 2) = 0
Then, y = + 6 or y = - 2/7
So, when x = + 6, x = y for y = + 6 and x > y for y = - 2/7
And when x = + 1/12, x < y for y = + 6 and x > y for y = - 2/7
∴ So, we can observe that no clear relationship cannot be determined between x and y.
3. A) I. a2 – 22x + 117 = 0
⇒ x2 – 13x – 9x + 117 = 0
⇒ x(x – 13) – 9(x – 13) = 0
⇒ (x – 13)(x – 9) = 0
Then, x = + 13 or x = + 9
II. y2 – 35y + 300 = 0
⇒ y2 – 20y – 15y + 300 = 0
⇒ y(y – 20) – 15(y – 20) = 0
⇒ (y – 20)(y – 15) = 0
Then, y = + 20 or y = +15
So, when x = + 13, x < y for y = + 20 and x < y for y = + 15
And when x = + 9, x < y for y = + 20 and x < y for y = + 15
∴ So, we can observe that x < y.
4. E) I. x2 + 23x – 210 = 0
⇒ x2 + 30x – 7x – 210 = 0
⇒ x(x + 30) – 7(x + 30) = 0
⇒ (x + 30)(x – 7) = 0
Then, x = - 30 or x = + 7
II. y2 – 28y + 147 = 0
⇒ y2 – 21y – 7y + 147 = 0
⇒ y(y – 21) – 7(y – 21) = 0
⇒ (y – 21)(y – 7) = 0
Then, y = + 21 or y = + 7
So, when x = - 30, x < y for y = + 21 and x < y for y = + 7
And when x = + 7, x < y for y = + 21 and x = y for y = +7
∴ So, we can clearly observe that x ≤ y.
5. C) I. 2x2 – 17x + 35 = 0
⇒ 2x2 – 10x – 7x + 35 = 0
⇒ 2x(x – 5) – 7(x – 5) = 0
⇒ (x – 5)(2x – 7) = 0
Then, x = + 5 or x = + 7/2
II. y2 + 9y – 70 = 0
⇒ y2 + 14y – 5y – 70 = 0
⇒ y(y + 14) – 5(y + 14) = 0
⇒ (y + 14)(y – 5) = 0
Then, y = - 14 or y = + 5
So, when x = + 5, x > y for y = - 14 and x = y for y = + 5
And when x = + 7/2, x > y for y = - 14 and x < y for y = + 5
∴ So, we can observe that no clear relationship cannot be determined between x and y.
6. D) I. 6x2– 9x – 10x + 15 = 0
or, 3x(2x – 3) – 5(2x – 3) = 0
or, (3x – 5) (2x – 3) = 0
x = 5/3,3/2
II. 10y2– 15y – 14y + 21 = 0
or, 5y(2y – 3) – 7(2y – 3) = 0
or, (5y – 7) (2y – 3) = 0
y=  7/3 , 5/2
x ≥y
7. E) I. 12X2 + 32x – 21x – 56 = 0
or, 4x(3x + 8) – 7(3x + 8) = 0
or, (4x – 7) (3x + 8) = 0
x= 7/8,4/3
II. 4y2 – 8y – 7y + 14 = 0
or, 4y(y – 2) – 7(y – 2) = 0
or, (4y – 7) (y – 2) = 0
y = 2 , 7/4
x ≤ y
8. B) I. 3X2 + 9x + 4x + 12 = 0
or, 3x(x + 3) + 4(x + 3) = 0
or, (3x + 4) (x + 3) = 0
x= – 4/ 3 , 3
II. y2 + 5y + 4y + 20 = 0
or, y(y + 5) + 4(y + 5) = 0
or, (y + 4) (y + 5) = 0
y = – 4, – 5
x > y
9. A) I. 8X2 – 8x – 7x + 7 = 0
or, 8x(x – 1) -7(x – 1) = 0
or, (8x – 7) (x – 1) = 0
x = 7/8 ,1
II. 2y2– 4y – 3y + 6 = 0
or, 2y(y – 2) -3(y – 2) = 0
or, (y – 2) (2y – 3) = 0
y = 2, 3/2
x < y
10. A) Eqn (I) × 4 + Eqn (II) × 3
28x – 12y = 52
15x + 12y = 120
43x = 172
x = 4 and y = 5
x < y