Q1. The difference between the roots of a quadratic equation 6x^{2}^{ }+ wx + 1 = 0 is 1/6. If w > 0, then the value of w is

A. 6

B. – 5

C. 5

D. 10

Sol : Option C
Let α,β are the roots of the equation, then
α+β = -w/6 ,αβ = 1/6 , α-β = 1/6 and w>0
Solving α= (-w+1)/12 , β = (-w-1)/12,
Therefore, (1-w)/12 × (-1-w)/12 = 1/6
Solving w= ± 5 as w>0, w= 5
Thus, option C is the answer

Q2. If the roots x_{1 }and x_{2 }of the quadratic equation satisfy the condition 7x_{2 }– 4x_{1 }= 47. The quadratic expression is x^{2 }– 2x + c = 0, find the value of c.

A. – 15

B. 15

C. – 6

D. None of these

Sol : Option A x_{1} + x_{2}= + 2/1 = + 2 ............ (1),
Also – 4x_{1 }+ 7x_{2}= 47 ......... (2)
Solving (1) & (2) we get x _{1 }= -3 and x_{2}= 5
∴ x_{1}x_{2 }= -15 = c

Q.3. If (x^{2 }– y^{2}) = 16 and xy = -15. Which of the following is a possible value of (x + y), if (x + y) is a positive number.

A. 3

B. 2

C. 5

D. None of these

Sol : Option B
x^{2} – y ^{2} = 16 and xy = -15, (x + y) >0 =?
x= -15/y, (-15/y)^{2} – y ^{2 } = 16 , 225/y^{2}- y^{2} = 16
y^{4} + 16y^{2} – 225 = 0
y^{2}= -25, 9 or y = ± 3(avoiding complex roots)
Putting values x = ±5 for y = 3 and x = ±5 for y = -3
Therefore (x+y) = 5+3 = 8, -5 +3 =-2
Or (x+y) = 5-3 =2, -5 -3 = -8
Therefore, x+y = 2 or 8 (because x+y > 0).

Q.4. From any two numbers x and y, we define x * y = x + 0.5 y – xy. Suppose that both x and y are greater than 0.5, then x * x < y ×y if :

A. 1 > x > y

B. x > 1 > y

C. y > 1 > x

D. 1 > y > x

Sol : Option D x * x = x + 0.5 x – x^{2 }= 1.5 x – x^{2} Y * y = y + 0.5 y – y^{2 }= 1.5 y – y^{2}
So if x * x < y * y, then 1.5 x – x^{2 }< 1.5 y – y^{2}
< 1.5 (y - x) or (y – x) x (y + x) < 1.5 (y – x). Now this is valid only if (y – x) is not equal to 0. Also, if(y – x) is a negative number, the equality sign will change.
To maintain the inequality, y – x has to be > 0, i.e. y > x.
And if (y – x) > 0, we can cancel out this factor without changing the equality sign.
Therefore, we have (x + y) < 1.5
Since x and y are both greater than 0.5, (x + y) < 1.5 only if both x and y are less than 1.
Hence we have 1 > y > x.

Q.5. The sum of the squares of two consecutive natural numbers is 85. Find those numbers.

A. 6, 7

B. 5, 8

C. 6, 8

D. -8, 6

Sol : Option A
Two consecutive natural nos. are x and x + 1.
⇒ x^{2 }+ (x + 1)^{2}= 85 ⇒ x = 6. ∴nos. are 6 and 7.

Q.6. The sum of the squares of two consecutive odd natural numbers is 130. Find those numbers.

A. 3, 5

B. 7, 9

C. 5, 7

D. 7, -9

Sol : Option B
Two consecutive odd natural nos. are x and x + 2
⇒ x^{2 }+ (x + 2)^{2 }= 130 ⇒ x = 7 ∴ nos. are 7 & 9.

Q.7. The sum of a natural number and its reciprocal is 50/7. What is the number?

A. 7

B. 10

C. 1/5

D. 5

Sol : Option A
x+ 1/x = 50/7 ⇒ x = 7

Q8. The difference between a natural number and twice its reciprocal is 47/7. What is that number?

A. 8

B. 9

C. 7

D. 1/7

Sol : Option C
x- 2/x = 47/7 ⇒ x = 7

Q9. The length of a rectangular field is less than twice its breadth by 5 metres. Its area is 700 square metres. Find the length and breadth of the field.

A. 35 m, 20 m

B. 25 m, 28 m

C. 70 m, 10 m

D. 35 m, 35 m

Sol : Option A
L = 2B – 5. Area = 700m^{2 }
LB = 700 ⇒(2B – 5) B = 700 ⇒ B = 20m ⇒ L = 35 m.

Q10. The base of a triangle is greater than twice its height by 1 cm. The area of the triangle is 18 sq. cm. Find the base and height of the triangle.

A. 18 cm, 2 cm

B. 9 cm, 4 cm

C. 6 cm, 9 cm

D. 6 cm, 7 cm

Sol : Option B
B = 2H + 1
Area = B * H/2 ⇒ 18 = (2H + 1)/2
H = 4, B = 9