QUICK TEST QUADRATIC EQUATION FOR IBPS RRB
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Directions (Q. 1 - 10): In each of these questions two equations (I) and (II) are given.
You have to solve both the equations and give answer
(1) if x > y
(2) if x > y
(3) if x < y
(4) if x < y
(5) if x = yor no relation can be established between x and y.
1. I. 2.3x – 20.01 = 0
II. 2.9y – x = 0
2. I. x = 1764
II. y2 = 1764
3. I. x2 – 26x + 168 = 0
II. y2 – 25y + 156 = 0
4. I. x2 – 13x + 42 = 0
II. y2 + y – 42 = 0
5. I. 6x – 5y = –47
II. 5x+ 3y = 11
6. I. 2x2 + 13x – 7 = 0
II. 2y2 –5y + 3 = 0
7. I. 2x2 – 15x + 28 = 0
II. 4y2 – 16y + 15 = 0
8. I. x2 + 8x + 16 = 0
II. y2 = 16
9. I. x2 – 2x – 24 = 0
II. y2 + 8y = 0
10. I. x2 + 4x = 0
II. y2 + 10y + 25 = 0
ANSWERS:
1. 1; I. 2.3x – 20.01 = 0
x=
20 . 01
= 8.7
2 .3
II. 2.9y – x = 0
or, x = 2.9y
y =
8 .7
2 .9
=3
i.e x > y
2. 2; I. x =
1764
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 x = 42
II. y2 = 1764
 y = +42
i.e x > y
3. 5; I. x2 – 26x + 168 = 0
x2 – 12x – 14x + 168 = 0
x(x – 12) – 14(x – 12) = 0
(x – 12) (x – 14) =0
x = 12, 14
II. y2 – 25y + 156 = 0
y2 – 13y – 12y + 156 = 0
y(y – 13) – 12(y – 13) = 0
(y – 12) (y – 13) = 0
y = 12, 13
Hence, no relation can be established between x and y.
4. 2; I. x2 – 13x + 42 = 0
x2 – 6x – 7x + 42 = 0
x(x – 6) – 7(x – 6) = 0
(x – 6) (x – 7) = 0
x = 6, 7
II. y2 + y – 42 = 0
y2 + 7y – 6y – 42 = 0
y(y + 7) – 6(y + 7) = 0
(y – 6) (y + 7) = 0
y = 6, –7
i.e x > y
5. 3; eqn(I) × 3 18x –15y = –141
eqn(II) × 5 25 x  15 y  55
43x
= –86
x =
– 86
43
= –2
5x + 3y = 11
3y = 11 – 5x
3y = 11 + 10
3y = 21
y = 7
i.e x < y
6. 3; I. 2x2 + 13x – 7 = 0
or 2x2 + 14x – x – 7 = 0
or 2x (x + 7) – 1 (x +7) = 0
or (2x – 1) (x+7) = 0
 x = 1 2 , –7
II. 2y2 – 5y + 3 = 0
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or 2y2 – 2y – 3y + 3 = 0
or 2y(y–1) –3(y–1) = 0
or (2y – 3) (y – 1) = 0
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 y = 1, 3 2
Hence x < y
7. 1; I. 2x2 – 8x –7x + 28 = 0
or 2x (x–4) – 7(x–4) = 0
or (2x–7) (x–4) = 0
 x = 4,
7
2
II. 4y2 – 16y + 15 = 0
or 4y2 – 6y – 10y + 15 = 0
or 2y (2y–3)–5(2y–3)= 0
or (2y – 5) (2y –3) = 0
y= 52 , 32
Hence x > y
8. 4; I. x2 + 8x + 16 = 0
or (x+4)2 = 0
or x + 4 = 0  x = –4
II. y2 = 16  y = ± 4
Hence, x < y
9. 5; I. x2 – 2x – 24 = 0
or x2 + 4x – 6x –24 = 0
or x(x+4) –6(x+4) = 0
or (x–6) (x+4) = 0
 x = 6, –4
II. y2 + 8y = 0
or y(y+8) = 0
 y = 0, – 8
i.e No relation can be established between x and y.
10. 1; I. x2 + 4x =0
or x(x + 4) = 0  x = 0, – 4
II. y2 + 10y + 25 = 0
or (y + 5)2 = 0
or y + 5 = 0  y = – 5
x>y
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