Explore topic-wise InterviewSolutions in .

(A) numerical equation

4/2 = 2

By cross multiplication we get,

4 = 4

(B) 2

Consider the given equation x – 4 = -2

Transform – 4 from left hand side to right hand side it becomes 4.

x = – 2 + 4

x = 2

z multiplied by -3 = -3z

– 3z subtracted from 13 = 13 – (-3z) = 13 + 3z

(A) 6

Consider the given equation q/2 = 3

By cross multiplication we get, q = 6

(C) 3 – 2x

From the question it is given that,

X is multiplied by 2 = x × 2 = 2x

Then, x is multiplied by 2 and then subtracted from 3 = 3 – 2x

(A) x × x

We know that, area of square = side × side

Given, square having a side x.

So, area of triangle = x × x

= x²

(A) x – 1700

From the question it is given that,

Savitri has a sum of Rs x

She spent money on grocery = ₹ 1000

She spent money on clothes = ₹ 500

She spent money on education = ₹ 400

She received gift = ₹ 200

Total money spent by Savitri = 1000 + 500 + 400 = ₹ 1900

Then,

Total money left with her after deducting = ₹ (x – 1900)

Therefore, money left with her after adding gift money = (x – 1900) + 200

= x – 1700

(i) 114²⁰ + 115²¹ + 116²²

In 114²⁰ unit digit of base 114 is 4 and power is 20 (even power).

∴ Unit digit of 114²⁰ is 6.

In 115²¹ unit digit of base 115 is 5 and power is 21 (Positive Integer).

∴ Unit digit of 115²¹ is 5.

In 116²² unit digit of base 116 is 6 and power is 22 (Positive Integer).

∴ Unit digit of 116²² is 6.

∴ Unit digit of 114²⁰ + 115²¹ + 116²² can be obtained by adding 6 + 5 + 6 = 17.

Unit digit of 114²⁰ + 115²¹ + 116²² is 7.

(ii) 10000¹⁰⁰⁰⁰ + 11111¹¹¹¹¹

In 10000¹⁰⁰⁰⁰ the unit digit of base 10000 is 0 and power is 10000.

Unit digit of 10000¹⁰⁰⁰⁰ is 0.

In 11111¹¹¹¹¹ the unit digit of base 11111 is 1 and power is 11111.

Unit digit of 11111¹¹¹¹¹ is 1.

Unit digit of 10000¹⁰⁰⁰⁰⁰ + 11111¹¹¹¹¹ is 0 + 1 = 1

Answer is (iv) 1

(i) 3⁵ x 3⁸ 

[since a^m x aⁿ = a^{m + n}]

= 3^{5 + 8} 

= 3¹³ 

(ii) a⁴ x a¹⁰

= a^{4 + 10} 

= a¹⁴

(iii) 7^x x 7² = 7^{x + 2}

(iv) 2⁵ ÷ 2³ 

[since \(\frac{a^m}{a^n}\) = a^{m - n}]

= 2^{5 - 3}

= 2²

(v) 18⁸ ÷ 18⁴

= 18^{8 - 4} 

= 18⁴

(vi) (6⁴)³

[since (a^m)ⁿ = a^{m x n}]

= 6^{4 x 3} 

= 6¹²

(vii) (x^m)⁰

[since (a^m)ⁿ = a^{m x n}; a⁰ = 1]

= x^{m x 0} 

= x⁰ 

= 1

(viii) 9⁵ x 3⁵

[since (a^m x b^m) = (a  x b)^m]

= (9 x 3)⁵ 

= 27⁵

(ix) 3^y x 12^y

= (3 x 12)^y 

= 36^y

(x) 25⁶ x 5⁶

= (25 x 5)⁶ 

= 125⁶

1. 6 x 6 x 6 x 6 = 6^1+1+1+1 

= 64 

[Since a^m x aⁿ = a^m+n]

2. t x t = t¹⁺¹ 

= t²

3. 5 x 5 x 7 x 7 x 7 = 5¹⁺¹ x 7¹⁺¹⁺¹ 

= 5² x 7³

4. 2 x 2 x a x a = 2¹⁺¹ x a¹⁺¹ 

= 2² x a² 

= (2a)²

(i) a^b + b^a

a = 3 and b = 2

We get 3² + 2³ = (3 x 3) + (2 x 2 x 2) 

= 9 + 8 

= 17

(ii) (a^a – b^b)

Substituting a = 3 and b = 2

We get 3² – 2² = (3 x 3 x 3) – (2 x 2) 

= 27 – 4 

= 23

(iii) (a + b)^b

Substituting a = 3 and b = 2

We get (3 + 2)² = 5² 

= 5 x 5 

= 25

(iv) (a – b)^a

Substituting a = 3 and b = 2

We get (3 – 2)³ = 1³

= 1 x 1 x 1 

= 1

Answer is (ii) 5

Answer is (i) a⁵

(i) 3² x 3⁴ x 3² = 3²⁺⁴⁺⁸ 

= 3¹⁴ 

[∵ a^m x aⁿ = a^m+n]

So 3² x 3⁴ x 3² = 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰ = 6^15-10 

= 6⁵ 

[∵ a^m x aⁿ = a^m-n]

(iii) a³ x a² = a³⁺² 

= a⁵

(iv) 7^x x 7² = 7^x+2

(v) (5²)³ ÷ 5³ 

= 5^{2 x 3} ÷ 5³ 

= 5⁶ ÷ 5³ 

[∵ (a^m)ⁿ = a^{m x n}]

1. 2³ ÷ 2⁵

= 2^3-5 

= 2^-2

[since \(\frac{a^m}{a^n}\) = a^m-n]

2. 11⁶ ÷ 11³

= 11^6-3 

= 11³

[since \(\frac{a^m}{a^n}\) = a^m-n]

3. (-5)³ ÷ (-5)²

= (-5)^3-2 

= (-5)¹

[since \(\frac{a^m}{a^n}\) = a^m-n]

4. 7³ ÷ 7³

= 7^3-3 

= 7⁰  

= 1

[since \(\frac{a^m}{a^n}\) = a^m-n; a⁰ = 1]

5. 15⁴ ÷ 15

= 15⁴ ÷ 15¹ 

= 15^4-1 

= 15³

[since \(\frac{a^m}{a^n}\) = a^m-n]

1. 2³ x 2⁵ = 2³⁺⁵ 

= 2⁸ 

[since a^m x aⁿ = a^m+n]

2. p² x p⁴ = p²⁺⁴ 

= p⁶ 

[since a^m x aⁿ = a^m+n]

3. x⁶ x x⁴ = x⁶ + 4 

= x¹⁰ 

[since a^m x aⁿ = a^m+n]

4. 3¹ x 3⁵ x 3⁴ = 3¹⁺⁵ x 3⁴ 

[since a^m x aⁿ = a^m+n]

= 3⁶ x 3⁴ 

[since a^m x aⁿ = a^m+n]

= 3¹⁰

5. (-1)² x (-1)³ x (-1)⁵

= (-1)²⁺³ x (-1)⁵ 

[Since a^m x aⁿ = a^m+n]

= (-1)⁵ x (-1)⁵

= (-1)⁵⁺⁵ 

[Since a^m x aⁿ = a^m+n]

= (-1)¹⁰

Answer is (i) a

We have 6x, -5x, x, 16x are like terms

6y, y, 7y, are like terms 

6, –5, 1, 3 are like terms

(iii) -625 

(50 + 20) (-20 – 5) = -(5 + 20)² 

= – (25)² 

= – 625

(i) 64¹¹ Unit digit of base 64 is 4 and the power is 11 (odd power).

∴ Unit digit of 64¹¹ is 4.

(ii) 29¹⁸ Unit digit of base 29 is 9 and the power is 18 (even power).

Therefore, unit digit of 29¹⁸ is 1.

(iii) 79¹⁹ Unit digit of base 79 is 9 and the power is 19 (odd power).

Therefore, unit digit of 79¹⁹ is 9.

(iv) 104³² Unit digit of base 104 is 4 and the power is 32 (even power).

Therefore, unit digit of 104³² is 6.

(i) 106²¹ Unit digit of base 106 is 6 and the power is 21 and is positive.

Thus the unit digit of 106²¹ is 6.

(ii) 25⁸ Unit digit of base 25 is 5 and the power is 8 and is positive.

Thus the unit digit of 25⁸ is 5.

(iii) 31¹⁸ Unit digit of base 31 is 1 and the power 18 and is positive.

Thus the unit digit of 31¹⁸ is 1.

(iv) 20¹⁰ Unit digit of base 20 is 0 and the power 10 and is positive.

Thus the unit digit of 20¹⁰ is 0.

(ii) 6

(a + b)² = 25

13 + 2ab = 25 

2ab = 12 

ab = 6

(i) (p + 2) (p – 2)

Substituting a = p; b = 2 in the identity 

(a + b) (a – b) = a² – b², we get

(p + 2) (p – 2) = p² – 2²

(ii) (1 + 3b) (3b – 1)

(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)

Substituting a = 36 and b = 1 in the identity

(a + b) (a – b) = a² – b², we get

(3b + 1) (3b – 1) = (3b)² – 1² 

= 3² x b² – 1²

(3b + 1) (3b – 1) = 9b² – 1²

(iii) (4 – mn) (mn + 4)

(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)

Substituting a = 4 and b = mn is

(a + b) (a – b) = a² – b², we get

(4 + mn) (4 – mn) = 4² – (mn)² 

= 16 – m² n²

(iv) (6x + 7y) (6x – 7y)

Substituting a = 6x and b = 7y in

(a + b) (a – b) = a² – b², we get

(6x + 7y) (6x – 7y) = (6x)² – (7y)² 

= 6²x² – 7²y²

(6x + 7y) (6x – 7y) = (6x)² – (7y)² 

= 6²x² – 7²y²

(6x + 7y) (6x – 7y) = 36x² – 49y²

(2a² + 3ab – b²) – (3a² – ab – 3b²)

= (2a² + 3ab – b²) + (- 3a² + ab + 3b²)

= 2a² + 3ab – b² – 3a² + ab + 3b²

= 2a² – 3a² + 3ab + ab + 3b² – b²

= 2a² – 3a² + ab (3 + 1) + b²(3 – 1)

= – a² + 4 ab + 2b²

Hence degree of the expression is 2.

Given x + 5 = 12 

(i) Our aim is to find the value of x. Which means we have to eliminate the other values from LHS. 

Since 5 is given with x it should be subtracted. 

(ii) If we add 5 on both sides we cannot eliminate the numbers from LHS and we get x + 10.

Sum of 5x + 7y – 12 and 3x – 5y + 2.

= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)

= 8x + 2y – 10. 

Again Sum of 2x – 7y – 1 and – 6x + 3y + 9 

= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9 

= (2 – 6) x + (- 7 + 3) y + (- 1 + 9) 

= – 4x – 4y + 8 

Now 8x + 2y – 10 – (-4x – 4y + 8) 

= 8x + 2y – 10 + (4x + 4y – 8) 

= 8x + 2y – 10 + 4x + 4y – 8 

= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8)) 

= 12x + 6y – 18

(iii) -2a 

– a – a = – 2a

(i) x²

The exponent in x² is 2. 

∴ Degree of the term is 2.

(ii) 4xyz

In 4xyz the sum of the powers of x, y and z as 3.

(iii) \(\frac{7x^2y^4}{xy}\)

We have \(\frac{7x^2y^4}{xy}\) = 7x^2-1 y^4-1 

= 7x¹y³ 

[Since \(\frac{a^m}{a^n}\) = a^m-n]

In 7 x¹ y³ the sum of the powers of x and y is 4(1 + 3 = 4)

Thus degree of the expression is 4.

(iv) \(\frac{x^2\times\,y^2}{x\times\,y^2}\)

We have \(\frac{x^2\times\,y^2}{x\times\,y^2}\) = x^2-1 y^2-2 

= x¹ y⁰ 

= x¹ 

[∵ y⁰ = 1]

The exponent of the expression is 1.

To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10. 

(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16 

= (2 + 3)m + (8 – 7)n + (10 – 16) 

= 5m + n – 6

(i) (7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p

= (7p + 5p + 16p) + (6q – q + q)

= (7 + 5 + 16)p + (6 – 1 + 1)q

= (12 + 16)p + 6q = 28p + 6q

(ii) (a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c

= a + 2a + 5b + 10b + 7c + 9c

= (1 + 2)a + (5 + 10)b + (7 + 9)c

= 3a + 15b + 16c

(iii) (mn + t) + (2mn – 2t) + (-3t + 3mn)

= mn + t + 2mn – 2t + (-3t) + 3mn

= (mn + 2mn + 3mn) + (t – 2t – 3t)

= (1 + 2 + 3)mn + (1 – 2 – 3)t

= 6mn + (1 – 5)t

= 6mn + (- 4)t

= 6mn – 4t

(iv) (u + v) + (u – v) + (2u + 5v) + (2u – 5v)

= u + v + u – v + 2u + 5v + 2u – 5v

= u + u + 2u + 2u + v – v + 5v – 5v

= (1 + 1 + 2 + 2)u + (1 – 1 + 5 – 5)v = 6u + 0v

= 6u

(v) 5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy

= (5 + 3)xyz + [(-3) + (-5)]xy = 8xyz + (-8)xy

= 8xyz – 8xy

(i) x² – 8x + 16

x² – 8x + 16 = x² – (2 x 4 (x) x) + 4²

This expression is in the form of identity

a² – 2ab + b² = (a – b)²

x² – 2 (x) 4 (x) x + 4² = (x – 4)²

∴ x² – 8x + 16 = (x – 4) (x – 4)

(ii) y² + 20y + 100

y² + 20y + 100 = y² + (2 x (10)) y + (10 x 10)

= y² + (2 x 10 x y) + 10²

This is of the form of identity

a² + 2 ab + b² = (a + b)²

y² + (2 x 10 x y) + 10² = (y + 10)²

y² + 20y + 100 = (y + 10)²

y² + 20y + 100 = (y + 10) (y + 10)

(iii) 36m² + 60m + 25

36m² + 60m + 25 = 62 m² + 2 x 6m x 5 + 5²

This expression is of the form of identity

a² + 2ab + b² = (a + b)²

(6m)² + (2 x 6m x 5) + 5²

= (6m + 5)²

36m² + 60m + 25 = (6m + 5) (6m + 5)

(i) 297 x 303

297 x 303 = (300 – 3) (300 + 3)

Taking a = 300 and b = 3, then

(a + b) (a – b) = a² – b² becomes

(300 + 3) (300 – 3) = 300² – 3²

303 x 297 = 90000 – 9

297 x 303 = 89,991

(ii) 990 x 1010

990 x 1010 = (1000 – 10) (1000 + 10)

Taking a = 1000 and b = 10, then

(a – b) (a + b) = a² – b² becomes

(1000 – 10) (1000 + 10) = 1000² – 10²

990 x 1010 = 1000000 – 100

990 x 1010 = 999900

(iii) 51 x 52

= (50 + 1) (50 + 1)

Taking x = 50, a = 1 and b = 2

Then (x + a) (x + b) = x² + (a + b) x + ab becomes

(50 + 1) (50 + 2) = 50² + (1 + 2) 50 + (1 × 2)

2500 + (3) 50 + 2 = 2500 + 150 + 2

51 x 52 = 2652

1549¹⁰¹ + 6541²⁰

In 1549¹⁰¹, the unit digit of base 1549 is 9 and power is 101 (odd power).

Therefore, unit digit of the 1549¹⁰¹ is 9.

In 6541²⁰, the unit digit of base 6541 is 1 and power is 20 (even power).

Therefore, unit digit of the 6541²⁰ is 1.

∴ Unit digit of 1549¹⁰¹ + 6541²⁰ is 9 + 1 = 10

∴ Unit digit of 1549¹⁰¹ + 6541²⁰ is 0.

(iii) 2 mn

= 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn 

= 2 mn

(i) 4k from 12k

12k – 4k = (12 – 4)k = 8k

(ii) 15q from 25q

25q – 15q = (25 – 15)q = 10q

(iii) 7xyz from 17xyz

17xyz – 7xyz = (17 – 7)xyz = 10xyz

(i) 51²

= (50 + 1)²

Taking a = 50 and b = 1 we get

(a + b)² = a² + 2ab + b²

(50 + 1)² = 50² + 2 (50) (1) + 1² 

= 2500 + 100 + 1

51² = 2601

(ii) 103²

103² = (100 + 3)²

Taking a = 100 and b = 3

(a + b)² = a² + 2ab + b² becomes

(100 + 3)² = 100² + 2 (100) (3) + 3² 

= 10000 + 600 + 9

103² = 10609

(iii) 998²

998² = (1000 – 2)²

Taking a = 1000 and b = 2

(a – b)² = a² + 2ab + b² becomes

(1000 – 2)² = 1000² – 2 (1000) (2) + 2²

= 1000000 – 4000 + 4

998² = 10,04,004

(iv) 47²

47² = (50 – 3)²

Taking a = 50 and b = 3

(a – b)² = a² – 2ab + b² becomes

(50 – 3)² = 50² – 2 (50) (3) + 3²

= 2500 – 300 + 9 = 2200 + 9

47² = 2209

Subtracting -3ab – 8 from 3ab + 8 

= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8) 

= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8) 

= 6ab + 16 

Also subtracting 3ab + 8 from – 3ab – 8 

= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) 

= – 3ab – 8 – 3ab – 8 

= [(-3) + (- 3)]ab + [(-8) + (-8)] = – 6ab + (- 16) 

= -6ab – 16

(i) like terms

Answer is (ii) -7

To get the required expression we must subtract

5a – 3b + 2c from a – 4b – 2c. 

∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c) 

= a – 4b – 2c – 5a + 3b - 2c 

= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c 

= – 4a – b – 4c. 

∴ -4a – b – 4c must be added.

To find the answer we have to find the difference. 

Here greater number 5ab – 3ab + 2c. 

∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) 

= 5ab – 3b + 2c + (- 2ab - 4b + c) 

= 5ab – 3b + 2c – 2ab – 4b + c 

= (5 – 2)ab + (-3 – 4)b + (2 + 1)c = 3ab + (-7)b + 3c 

= 3ab – 7b + 3c 

It is 3ab – 7b + 3c smaller.

Given A = 2a² – 4b – 1; B = 5a² + 3b – 8; C = 2a² – 9b + 3

A – B + C = (2a² – 4b – 1) – (5a² + 3b – 8) + (2a² – 9b + 3)

= 2a² – 4b – 1 + (-5a² – 3b + 8) + 2a² – 9b + 3

= 2a² – 4b – 1 – 5a² – 3b + 8 + 2a² – 9b + 3

= 2a² – 5a² + 2a² – 4b – 3b – 9b – 1 + 8 + 3

= (2 – 5 + 2)a² + (-4 – 3 – 9)6 + (-1 + 8 + 3)

= -a² – 16b + 10

(i) x⁵ – x⁴ + 3

The terms of the given expression are x⁵, -x⁴, 3.

Degree of each of the terms : 5, 4, 0

Term with highest degree: x⁵

Therefore degree of the expression in 5.

(ii) 2 – y⁵ – y³ + 2y⁸

The terms of the given expression are 2, -y⁵ ,-y³, 2y⁸.

Degree of each of the terms : 0, 2, 3, 8.

Term with highest degree: 2y⁸

Therefore degree of the expression in 8.

(iii) 2

Degree of the constant term is 0.

∴ Degree of 2 is 0.

(iv) 5x³ + 4x² + 7x

The terms of the given expression are 5x³, 4x², 7x

Degree of each of the terms : 3, 2, 1

Term with highest degree: 5x³

Therefore degree of the expression in 3.

(v) 4xy + 7x²y + 3xy³

The terms of the given expression are 4xy, 7x²y, 3xy³

Degree of each of the terms : 2, 3, 4

Term with highest degree: 3xy³

Therefore degree of the expression in 4.

(iv) -4x, 7x

Perimeter of a square with side a is 16 cm. 

4 x a = 16