501. I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
A. If x < y
B. If x > y
C. If x ≤ y
D. If x = y or the relationship between x and y cannot be established.

Explanation:
I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.

502. I. a2 + 8a + 16 = 0,
II. b2 – 4b + 3 = 0 to solve both the equations to find the values of a and b?
A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a > b

Explanation
I. (a + 4)2 = 0 => a = -4
II.(b – 3)(b – 1) = 0
=> b = 1, 3 => a < b

503. (i). a2 + 11a + 30 = 0,
(ii). b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a > b

Explanation:
I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b

504. (i). a2 – 9a + 20 = 0,
(ii). 2b2 – 5b – 12 = 0 to solve both the equations to find the values of a and b
A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a ≥ b

Explanation:
I. (a – 5)(a – 4) = 0
=> a = 5, 4
II. (2b + 3)(b – 4) = 0
=> b = 4, -3/2 => a ≥ b

505. (i). a2 – 7a + 12 = 0,
(ii). b2 – 3b + 2 = 0 to solve both the equations to find the values of a and b?
A. if a < b
B. if a ≤ b
C. if the relationship between a and b cannot be established.
D. if a > b

Explanation:
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b

506. A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
A. 10
B. 8
C. 15
D. 7.50

Explanation:
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10

507. Find the quadratic equations whose roots are the reciprocals of the roots of 2×2 + 5x + 3 = 0?
A. 3×2 + 5x – 2 = 0
B. 3×2 + 5x + 2 = 0
C. 3×2 – 5x + 2 = 0
D. 3×2 – 5x – 2 = 0

Explanation
The quadratic equation whose roots are reciprocal of 2×2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3×2 + 5x + 2 = 0

508. Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
A. 15
B. 14
C. 24
D. 26

Explanation:
a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 – 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 – 2(4)]/4 = 56/4 = 14.

509. If a and b are the roots of the equation x2 – 9x + 20 = 0, find the value of a2 + b2 + ab?
A. -21
B. 1
C. 61
D. 21

Explanation:
a2 + b2 + ab = a2 + b2 + 2ab – ab
i.e., (a + b)2 – ab
from x2 – 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression (9)2 – 20 = 61.

510. The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
A. 18, 20, 22
B. 20, 22, 24
C. 22, 24, 26
D. 24, 26, 28

Explanation:
Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.
(2x – 2)2 + (2x)2 + (2x + 2)2 = 1460
4×2 – 8x + 4 + 4×2 + 8x + 4 = 1460
12×2 = 1452 => x2 = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24.

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