Explore topic-wise InterviewSolutions in .

(a) x – 4

Radha is ‘x’ years of age now 4 years ago, her age was x - 4.

(i) t + 100

(ii) 4q

(iii) 8 – y

(iv) 2x + 56

(v) 9y – 4

(i) x divided by 3

(ii) 12 less to 5 times n

(iii) 11 added to 10 times x

(iv) 70 times s

Manisha’s uncle is five times of Manisha’s age. Her aunt is 4 years younger than her uncle.

6 times q = 6q.

The smallest two digit number = 10

6q subtracted from 10 = 10 – 6q.

(a) We have given,

Perimeter of an equilateral triangle = 3 (the side of an equilateral triangle)

⇒ p = 3a

(b) We have given,

Diameter of a circle

= 2 (the radius of the circle)

⇒ d = 2r

(c) We have given,

Selling price = cost price + profit

⇒ s = c + p

(d) We have given,

Amount = principal + interest

⇒ a = p + i

Perimeter of a triangle is the sum of all its sides.

Number of students in the class = p

Total amount collected from p students = Rs. 50p

Amount paid in advance for transport = Rs. 1800

∴ Amount left with them = Rs. (50p – 1800)

We have given, side of a square = m cm

Area of the square = side × side = m × m sq cm

Number of water tanks = 8

Rain water collected by each tank = x litres

∴ Rain water collected by 8 tanks = 8x litres

But 100 litres of water was already there in one of the tanks.

∴ Total amount of water in the tanks = (8x + 100) litres.

Let number of rows be n

Number of cadets in one row = 5

Total number of cadets = number of cadets in a row x numbers of rows= 5

60 – t = t²/4 – 3

⇒ t² – 12 = 240 – 4t

⇒ t² + 4t – 252 = 0

⇒ t² + 18t – 14t – 252 = 0

⇒ t(t + 18) – 14(t + 18) = 0

⇒ (t + 18) (t – 14) = 0

∴ t = 14 or t = -18 is not possible.

Length = 5 × Height

l = 5h cm

Breadth = 5 × Height – 10

b = (5h – 10)

(3p + 2q)²

Comparing (3p + 2q)² with (a + b)², we get a = 3p and b = 2q.

(a + b)² = a² + 2ab + b²

(3p + 2q)² 

= (3p)²+ 2(3p) (2q) + (2q)²

= 9p² + 12pq + 4q²

Area of square = side x side = (m + n – q) 

We know that, 

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 

[m + n + (-q)]² = m² + n² + (-q)² + 2mn + 2n(-q) + 2(-q)m 

= m² + n² + q² + 2mn – 2nq – 2qm 

Therefore, Area of square = m² + n² + q² + 2mn – 2nq – 2qm 

= [m² + n² + q² + 2mn – 2nq – 2qm] sq. units.

(105)² = (100 + 5)²

Comparing (100 + 5)² with (a + b)², we get a = 100 and b = 5.

(a + b)² = a² + 2ab + b²

(100 + 5)² 

= (100)² + 2(100)(5) + 5² 

= 10000 + 1000 + 25

105² = 11,025

We know that, 

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 

Substituting, a = x, b = 2y and c = 3z 

(x + 2y + 3z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x) 

= x² + 4y² + 9z² + 4xy + 12y² + 6zx

(y – 5)(y + 5)

Comparing (y – 5) (y + 5) with (a – b) (a + b) we get a = y; b = 5

(a – b)(a + b) = a² – b²

(y – 5)(y + 5) 

= y² – 5² 

= y² – 25

Replacing ‘c’ by ‘-c’ in the expansion of 

(a + b + c)² = a² + b² + c² + 2 ab + 2 bc + 2ca 

(a + b + (-c))² = a² + b² + (-c)² + 2ab + 2b(-c) + 2(-c)a 

= a² + b² + c² + 2ab – 2bc – 2ca

(2x – 5d)²

Comparing with (a – b)², we get a = 2x b = 5d.

(a – b)² = a² – 2ab + b²

(2x – 5d)² 

= (2x)² – 2(2x)(5d) + (5d)²

= 2x² – 20 xd + 5²d² 

= 4x² – 20 xd + 25d²

203 × 197 

= (200 + 3)(200 – 3)

Comparing (a + b) (a – b) we have a = 200, b = 3

(a + b)(a – b) = a² – b²

(200 + 3)(200 – 3) = 200² – 3²

203 × 197 = 40000 – 9

203 × 197 = 39991

(3x)² – 5²

Comparing (3x)² – 5² with a² – b² we have a = 3x; b = 5

(a² – b²) = (a + b)(a – b)

(3x)² – 5² 

= (3x + 5)(3x – 5) 

= 3x(3x – 5) + 5(3x – 5)

= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)

= 9x² – 15x + 15x – 25 

= 9x² – 25

(2m + n) (2m + p)

Comparing (2m + n) (2m + p) with (x + a) (x + b) we have x = 2n; a = n ;b = p

(x – a)(x + b) = x² + (a + b)x + ab

(2m +n) (2m +p) 

= (2m²) + (n +p)(2m) + (n) (p)

= 2²m² + n(2m) + p(2m) + np

= 4m² + 2mn + 2mp + np

(i) (7x + 2y)² = (7x)² + 2 (7x) (2y) + (2y)² = 49x² + 28xy + 4y²

(ii) (4m – 3m)² = (4m)² – 2(4m) (3m) + (3m)² = 16m² – 24mn + 9n²

(iii) (4a + 3b) (4a – 3b) [We have (a + b) (a – b) = a² – b²] 

Put [a = 4a, b = 3b] 

(4a + 3b) (4a – 3b) = (4a)² – (3b)² = 16a² – 9b² 

(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x² + (a – b)x – ab]

Put [x = k, a = 2, b = 3] 

(k + 2) (k – 3) = k² + (2 – 3)x – 2 x 3 = k² – x – 6

(98)² 

= (100 – 2)²

Comparing (100 – 2)² with (a – b)² we get
a = 100, b = 2

(a – b)² = a² – 2ab + b²

(100 – 2)² 

= 100² – 2(100)(2) + 2²

= 10000 – 400 + 4 

= 9600 + 4 

= 9604

(i) (3a – 4b)³ We know that 

(x – y)³ = x³ – 3x²y + 3xy² – y³ 

(3a - 4b)³ = (3a)³ – 3(3a)² (4b) + 3(3a)(4b)² – (4b)³ 

= 27a³ – 108a²b + 144 ab² – 64b³

(ii) (x + \(\frac{1}{y}\))³ 

(x + y)³ ≡ x³ + 3x²y + 3xy² + y³

(x + \(\frac{1}{y}\))³ = x³ + \(\frac{3x^2}{y}+\frac{3x}{y^2}+\frac{1}{y^3}\)

(i) 98³ = (100 – 2)³ 

(a – b)³ ≡ a³ – 3a²b + 3ab² – b³

98³ = (100 – 2)³ = 100³ – 3 x 100² x + 3 x 100 x 2² – 2³ 

= 1000000 – 3 x 10000 x 2 + 300 x 4 – 8 

= 1000000 – 60000 + 1200 – 8 = 1001200 – 60008 

= 941192

(ii) 1001³ = (1000 + 1)³ 

(a + b)³ ≡ a³ + 3a²b + 3ab² + b³

(1000 + 1)³ = 1000³ + 3(1000²) x 1 + 3 x 1000 x 1² + 1³ 

= 1000,000,000 + 3,000,000 + 3000 + 1 

= 1,003,003,001

x – y = 5, xy = 14 

x³ – y³ = (x – y)³ + 3xy (x – y) = 5³ + 3 x 14 x 5 

= 125 + 210 = 335

Answer is (3) x³ + y³

Answer is (3) p(3)

Answer is (3) 0

Answer is (1) x² + 2x 

Answer is (2) b, ac

(4) 11 

P(-2) = (-2)³ + 6 (-2)² + k (-2) + 6 = 0 

⇒ – 8 + 24 – 2k +6 = 0

⇒ 22 = 2k 

⇒ k = 11