eqn-8
DESCRIPTION
Ibps eqnTRANSCRIPT
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QUICK TEST QUADRATI EQUATION FORIBPS PO MAIN
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 > y3) if x < y 4) if x < y5) if x = y or no relation can be established between x and y.
1. I. 5x – 7y = –24 II. 13x + 3y = 86
2. I. x2 – 13x + 40 = 0 II. y2 + 3y – 40 = 0
3. I. 8x2 – 26x + 15 = 0 II. 2y2 – 17y + 30 = 0
4. I. x2 = 484 II. y2 – 45y + 506 = 0
5. I. 13x – 21 = 200 – 4x II. y = 3 2197
6. I. (x + y)2 = 3136 II. y + 2513 = 2569
7. I. 4x2 – 16x + 15 = 0 II. 2y2 + 5y – 7 = 0
8. I. x2 = 49 II. y2 + 15y + 56 = 0
9. I. 2x2 + 5x – 12 = 0 II. 2y2 – y – 1 = 0
10. I. x2 – 12x + 35 = 0 II. y2 – 25 = 0
ANSWERS WITH EXPLANATIONS:
1. 3;
530 106x 602 21y91x
–7221y–5x1
x = 5, y = 7 x < y
2. 2; I. x2 – 13x + 40 = 0or x2 – 5x – 8x + 40 = 0or x(x – 5) –8(x – 5) = 0or (x – 5) (x – 8) = 0
x = 5, 8
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II. y2 + 3y – 40 = 0or y2 – 5y + 8y – 40 = 0or y(y – 5) + 8(y – 5) = 0or (y – 5) (y + 8) = 0 y = 5, –8Hence, x > y
3. 4; I. 8x2 – 26x + 15 = 0or 8x2 – 20x – 6x + 15 = 0or 4x(2x – 5) – 3(2x – 5) = 0or (4x – 3) (2x – 5) = 0
x = 43
, 25
II. 2y2 – 17y + 30 = 0or 2y2 – 12y – 5y + 30 = 0or 2y(y – 6) – 5(y – 6) = 0or (2y – 5) (y – 6) = 0
y = 25
, 6
x < y
4. 4; I. x2 = 484 x = +22II. y2 – 45y + 506 = 0or y2 – 22y – 23y + 506 = 0or y(y – 22) – 23(y – 22) = 0or (y – 22) (y – 23) = 0 y = 22, 23 x < y
5. 5; I. 13x – 21 = 200 – 4xor 13x + 4x = 200 + 21
x = 1317221
II. y = 3 2197 y = 13 x = y
6. 3; I. (p + q)2 = 3136 p + q = +56II. q + 2513 = 2569or, q = 2569 – 2513 = 56Putting the value of q in (I) we have,p = 0. –112 p < q
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7. 1; I. 4p2 – 16p + 15 = 0or, 4p2 – 10p – 6p + 15 = 0or, 2p(2p – 5) – 3(2p – 5) = 0or, (2p – 3) (2p – 5) = 0
p = 23
, 25
II. 2q2 + 5q – 7 = 0or, 2q2 + 7q – 2q – 7 = 0or, q(2q + 7) – 1(2q + 7) = 0or, (q – 1) (2q + 7) = 0
q = 1, 27–
p > q
8. 2; I. p2 = 49 p = +7II. q2 + 15q + 56 = 0or, q2 + 8q + 7q + 56 = 0or, q(q + 8) + 7(q + 8) = 0or, (q + 7) (q + 8) = 0 q = –7, –8 p > q
9. 5; I. 2p2 + 5p – 12 = 0or, 2p2 + 8p – 3p – 12 = 0or, 2p(p + 4) – 3(p + 4) = 0or, (2p – 3) (p + 4) = 0
p = 23
, – 4
II. 2q2 – q – 1 = 0or, 2q2 – 2q + q – 1 = 0or, 2q(q – 1) + 1(q – 1) = 0or, (2q + 1) (q – 1) = 0
q = 1, 21–
No reation between ‘p’ and ‘q’.
10. 2; I. p2 – 12p + 35 = 0or, p2 – 5p – 7p + 35 = 0or, p(p – 5) – 7(p – 5) = 0or, (p – 7) (p – 5) = 0 p = 5, 7II. q2 – 25 = 0or, q2 = 25 q = +5 p > q