**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.