Sol : Option A
L = 2B – 1, Diagonal = 17. D^{2 }= L^{2 }+ B^{2}
⇒ 289 = (2B –1)^{2 }+ B^{2}⇒ B = 8 & L = 15.
Also 8, 15, 17 is a Pythagorean triplet.
So answer can be reached directly.

Q2. The sum of the squares of two consecutive natural numbers is 145. Find those numbers.

Q.3. The sum of the squares of two consecutive even natural numbers is 52. Find those numbers.

A. 2, 4

B. 4, 6

C. 8, 10

D. 7, 9

Sol : Option B
x^{2 }+ (x + 2) ^{2 }= 52 ⇒ x = 4 & x + 2 = 6

Q.4. The length of a rectangle is greater than twice its breadth by 2 cm. The length of its diagonal is 13 cm. Find the length and the breadth of the rectangle.

A. 12 cm, 5 cm

B. 5 cm, 3 cm

C. 6 cm, 15 cm

D. 9 cm, 7 cm

Sol : Option A
L = 2B + 2. Diagonal = 13, L^{2 }+ B^{2}= Diagonal^{2}
⇒ (2B + 2)^{2 }+ B^{2 }= 169 ⇒ B = 5 cm & L = 12 cm
Also 5, 12 and 13 is a Pythagorean triplet.

Q.5. In a right-angled triangle the length of one of the side containing the right angle is greater than twice the length of the other by 4 cm. The area of the triangle is 80 sq. cm. Find the length of the sides containing the right angle.

A. 3 cm, 4 cm

B. 7 cm, 8 cm

C. 20 cm, 8 cm

D. 13 cm, 15 cm

Sol : Option C
B = 2H + 4. Area of triangle = 80 ∴B.H/2 = 80
⇒ (2H + 4) H/2 = 80 ⇒ H = 8 cm and B = 20 cm

Q.6. In a rectangular mango-grove, the number of trees lengthways is 5 more than the number of trees breadthways. If the total number of trees is 1,400, find the number of trees lengthways and breadthways.

A. 35, 40

B. 45, 40

C. 30, 25

D. 40, 35

Sol : Option D
x (x + 5) = 1400 ⇒ x = 35 & x + 5 = 40
∴ No. of trees lengthways = 40 & breadthways = 35.

Q.7. If α and β are the roots of the equation x^{2 }+ ax + b = 0, where b ≠ 0, then the roots of the equation bx ^{2 }+ ax + 1 = 0 are:

A. 1/α, 1/β

B. α^{2}β^{2 }

C. 1/α^{2}, 1/β^{2}

D. α/β, β/α

Sol : Option A
From the given equation, it can be derived that
α + β = - a. Similarly the value of αβ = b.
Now for the second given equation sum of roots = - a / b and product of the roots = 1/b.
As we know the values of a and b, it can be put to calculate the sum of the roots as [α +β] /αβ, which can be simplified as 1/α + 1/β. Similarly, the product of the roots is 1/αβ. Just combine the sum and product and find that the parts have to be 1/α and 1/β only.

Q8. Sum of three numbers in GP is 70. If the two extreme terms are multiplied by 4, and the middle term by 5, the resultants are in AP. What is the highest of the three numbers given?

A. 20

B. 40

C. 10

D. 100

Sol : Option B
Let the numbers be a/r, a, ar.
Their sum is given to be 70 i.e. a/r + a + ar = 70 (i). Solving this, we get a + ar + ar^{2}= 70r.
It also states that 4a/r, 5a and 4ar are in A.P. i.e. 2 × 5a = 4a/r + 4ar.
Canceling 2a from both the sides, we get 5/2 = 1/r + r
⇒ 5r = 2 + 2r^{2 }⇒ 2r^{2}– 5r + 2 = 0.
Solving this, we get r = 2 or ½.
Put the value of r in (i) and get the three terms as either 10, 20 and 40 or 40, 20 and 10.
Thus, option B is the answer.

Q9. There are three terms ; a, b, c between 2 & 18 such that (i) their sum is 25, (ii) 2, a, b are consecutive terms of an A.P. and (iii) b, c, 18 are the consecutive terms of a G.P. Find the value of c.

A. 5

B. 8

C. 15

D. 12

Sol : Option D
As the question states 2, a and b are in A.P. i.e.
2 * a = 2 + b ⇒ 2a = 2 + b (i).
Similarly b, c and 18 are in GP i.e. c^{2}= 18b
⇒ c^{2}/18 = b (ii). As per the first statement of the question a + b + c = 25 i.e. a = 25 – c – b (iii).
Putting the value of (iii) in (i) you get 2(25 – c – b)
= 2 + b i.e. 3b = 48 – 2c, putting the value of b from (ii), you get 3c^{2}/18 = 2(24 – c)
⇒ c^{2 }= 144 – 12c i.e. c^{2}+ 12c – 288 = 0.
Solve the quadratic equation and get c = - 24 or 12.
As – 24 does not come in the range given the value of c becomes 12. Thus b = 8 and a = 5.
Thus the answer is option D

Q10. The sum of three numbers in GP is 21/4 and their product is 1. What are the numbers?

A. 4, 1, 1/4

B. 2, 1, 1/2

C. 3, 1, 1/3

D. None of these

Sol : Option A
Assuming that the numbers are a/r, a, ar and their product is 1, we get a = 1. Now, the sum (a/r) + (ar) = 17/4. Since a = 1, we have (r + 1/r) = 17/4. Solving this as a quadratic equation, we get r = 4 or r = ¼. Alternately, (r + 1/r) = 17/4 = (4 + ¼), giving r = 4 or r = ¼. Therefore, the three numbers are 4, 1 and ¼.