2. If P^{2} - 11P -42 = 0, where P is a natural number, find the value of P.

A. 12

B. 13

C. 14

D. 3

Sol : Option C
P^{2} - 14P + 3P - 42 = 0 ⇒ P (P – 14) +3(P – 14) = 0
⇒ (P + 3) (P – 14) = 0 ⇒ P = (-3) or P = 14.
Hence answer is C option.

Q.3. If a positive number is decreased by 10, then it is 75 times of its reciprocal. Find the number.

A. 30

B. 15

C. 20

D. 25

Sol : Option B
Let number = x. So as per question, (x – 10) = 75 * (1/x)
⇒ x^{2} - 10x - 75 = 0
⇒ (x – 15) (x + 10) = 0
⇒ x = 15. Hence option 2.

Q.4. If a and b are roots of P^{2} - 4P - 21 = 0, where a > b, find the equation whose roots are a/b and b/a.

A. 21P^{2} + 58P + 21 = 0

B. 25P^{2} + 58P + 21 = 0

C. 21P^{2} + 40P + 21 = 0

D. 21P^{2} + 56P + 25 = 0

Sol : Option A
Given the equation as P^{2} - 4P - 21 = 0.
As a and b are its roots, so a + b = 4 and ab = -21.
Given new roots are (a/b) and (b/a)
Hence sum of roots = (a/b) + (b/a) = (a^{2} + b^{2}) / ab

Q.5. Find the value of ‘a’ in a^{2} – 2a -15 = 0, if a is a natural number.

A. 7

B. 6

C. 9

D. 14

Sol : Option D
a^{2} – 2a -15 = 0
a^{2} + 3a – 5a – 15 = 0
a (a + 3) -5 ( a + 3) = 0
(a -5) (a + 3) = 0
a= 5 or a= -3
But, it should be a natural number, Therefore a = 5.

Q.6. If one root of the equation P^{2} - 5P – 36 = 0 is same as P^{2} - 25P + Q = 0, then find the value of Q.

A. 16 or 9

B. 144 or -116

C. -144 or 116

D. -16 or -9

Sol : Option B
1^{st} equation is P^{2} - 5P - 36 = 0 ⇒ (P - 9) (P + 4) = 0 ⇒ P = 9 or P = -4
If P = 9 then P^{2} - 25P + Q = 0 ⇒ (9)^{2} - 25(9) + Q = 0 ⇒ Q = 144
If P = -4 then P^{2} -25P + Q = 0 ⇒ (-4)^{2} - 25 (-4) + Q = 0 ⇒ Q = -116.
Hence answer is 2^{nd} option.

Q.7. Find the value of K such that P^{nd} - (3 + k)P + 3K + 4 = 0 has equal roots.

A. 7

B. -1

C. Either 1 or 2

D. None of these

Sol : Option C
Equation P^{2} - (3 + K) P + 3K + 4 = 0 have equal roots.
So (3 + K) ^{2} - 4(3K + 4) = 0 ⇒ 9 + K^{2} + 6K - 12K - 16 = 0 ⇒ K^{2} - 6K - 7 = 0
⇒ (K - 7) (K + 1) = 0 ⇒ K = 7, or K = -1.
Hence answer is 3^{rd} option.

Q8. Find the larger of the two positive numbers, such that sum of the numbers is 18 and difference of their squares is 9 times the larger number.

A. 20

B. 16

C. 12

D. None of these

Sol : Option C
Let Ankit's age be x years. Then Nikita's age = 240/x years.
Therefore 2 * (240/x) - x = 4 ⇒ 480 - x^{2} = 4x ⇒ X^{2} + 4x – 480 = 0
⇒ (x + 24) (x – 20) = 0 ⇒ x = 20. (Because x ≠ - 24).
Hence, Nikita's age = (240/20) years = 12 years.
Hence 3^{rd} option.

Q9. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit’s age by 4 years, what is Nikita’s age?

A. 20

B. 16

C. 12

D. None of these

Sol : Option C
Let Ankit's age be x years. Then Nikita's age = 240/x years.
Therefore 2*(240/x) - x = 4 ⇒ 480 - x^{2} = 4x ⇒ x2 + 4x – 480 = 0
⇒ (x + 24) (x – 20) = 0 ⇒ x = 20. (Because x ≠ - 24).
Hence, Nikita's age =(240/20) years = 12 years.
Hence C option.

Q10. A rectangular garden has an area of 735m^{2}. If length is decreased by 7m and breadth is increased by 7m, then it became a square. Find the area of the square (in m^{2}).

A. 625

B. 784

C. 484

D. 676

Sol : Option B
Let length = l, breadth = b, Area =lb = 735.
Given l - 7 = b + 7 = side of square.
So l = b + 14 ⇒ So b (b + 14) = 735 b^{2} + 14b - 735 = 0 ⇒ (b + 35) (b - 21) = 0
⇒ b = 21. So side of square = 21 + 7 = 28 ⇒ Area = (28) ^{2} = 784. Hence 2^{nd} option.