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## Frank Solutions for Class 9 Maths Chapter 7 Linear Equations Download PDF

## Access Frank Solutions for Class 9 Maths Chapter 7 Linear Equations

**1. In the following equations, verify if the given value is a solution of the equation:**

**(i) 5x – 2 = 18; x = 4**

**(ii) 2x – 5 = 3x; x = 3**

**(iii) 3x + 8 = x – 7; x = 3 **

**(iv) () x + () x = 56 – 2x; x = 7**

**(v) {(3x – 1) / 4} + (3 / 4) = 2; x = 2**

**Solution:**

(i) 5x – 2 = 18; x = 4

We know a value is a solution of the equation if it satisfies the equation.

i.e,

x_{1} is a solution if f (x_{1}) = 0

Here,

x_{1 }= 4

Put in 5x – 2 = 18

We get,

5(4) – 2 = 18

20 – 2 =18

18 = 18

LHS = RHS

Therefore,

x = 4 is a solution of the equation 5x – 2 = 18

(ii) 2x – 5 = 3x; x = 3

This can be written as:

x = – 5

-which is not satisfied by x = 3

Therefore,

x = 3 is not a solution of 2x – 5 = 3x

(iii) 3x + 8 = x – 7; x = 3

This can be written as,

2x + 15 = 0

Now, putting x = 3, we get,

LHS ≠ RHS

Hence,

2(3) + 15 ≠ 0

Therefore, x = 3 is not a solution of the equation 3x + 8 = x – 7

(iv) (

) x + (

) x = 56 – 2x; x = 7

On simplifying, we get,

(5 / 2) x + (7 / 2)x = 56 – 2x

(12 / 2) x + 2x = 56

8x = 56

We get,

x = 56 / 8

x = 7

By substituting x = 7,

(12 / 2) 7 + 2(7) = 56

42 + 14 = 56

56 = 56

We get,

LHS = RHS

Therefore,

x = 7 is a solution of the equation (

) x + (

) x = 56 – 2x

(iv) {(3x – 1) / 4} + (3 / 4) = 2; x = 2

On simplifying, we get,

{(3x – 1 + 3) / 4} = 2

(3x + 2) = 8

3x = 8 – 2

3x = 6

We get,

x = 2

Put x = 2 in above equation,

{(3x – 1) / 4} + (3 / 4) = 2

{3(2) – 1 / 4} + (3 / 4) = 2

{(6 – 1) / 4} + (3 / 4) = 2

(5 / 4) + (3 / 4) = 2

(5 + 3) / 4 = 2

2 = 2

We get,

LHS = RHS

Therefore,

x = 2 is a solution of the equation {(3x – 1) / 4} + (3 / 4) = 2

**2. Solve the following equations for the unknown:**

**(i) 3x + 8 = 35**

**(ii) 8x – 21 = 3x – 11**

**(iii) 2x – (3x – 4) = 3x – 4**

**(iv) 2x + √2 = 3x – 4 – 3√2**

**(v) 15y – 20 = 2y + 6**

**(vi) 5x + 10 – 4x + 6 = 12x + 20 – 3x + 12**

**(vii) (a + 2) (2a + 5) = 2 (a + 1) ^{2} + 13**

**(viii) (6p + 9) ^{2} + (8p – 7)^{2} = (10p + 3)^{2} – 71**

**(ix) (3x – 1) ^{2} + (4x + 1)^{2} = (5x + 1)^{2} + 5**

**(x) 3 (3x – 4) – 2 (4x – 5) = 6**

**Solution:**

(i) 3x + 8 = 35

Collecting like terms,

We get,

3x = 27

x = 9

(ii) 8x – 21 = 3x – 11

Collecting like terms,

We get,

8x – 3x = 21 – 11

5x = 10

x = 2

(iii) 2x – (3x – 4) = 3x – 4

2x – 3x + 4 = 3x – 4

Collecting like terms,

We get,

2x – 3x – 3x = – 8

– 4x = – 8

x = 2

(iv) 2x + √2 = 3x – 4 – 3√2

Collecting like terms,

We get,

x = 4√2 + 4

x = 4 (√2 + 1)

(v) 15y – 20 = 2y + 6

Collecting like terms,

We get,

13y = 26

y = 2

(vi) 5x + 10 – 4x + 6 = 12x + 20 – 3x + 12

Collecting like terms,

We get,

x + 16 = 9x + 32

8x = – 16

x = – 2

(vii) (a + 2) (2a + 5) = 2 (a + 1)^{2} + 13

Collecting like terms,

We get,

a (2a + 5) + 2 (2a + 5) = 2a^{2} + 4a + 2 + 13

2a^{2} + 5a + 4a + 10 – (2a^{2} + 4a + 15) = 0

5a + 10 – 15 = 0

5a = 5

We get,

a = 1

(viii) (6p + 9)^{2} + (8p – 7)^{2} = (10p + 3)^{2} – 71

{(36p^{2} + 81 + 2 (6p) (9)} + {(64p^{2} + 49 – 2 (8p) (7)} = 100p^{2} + 9 + 2 (10p) (3) – 71

100p^{2} + 130 + 2 × 54p – 2 × 56p = 100p^{2} + 9 – 71 + 60p

Collecting like terms,

We get,

130 + 71 – 9 = 60p + 4p

192 = 64p

We get,

p = 3

(ix**) **(3x – 1)^{2} + (4x + 1)^{2} = (5x + 1)^{2} + 5

On further calculation, we get,

{(9x^{2} + 1 + 2 (3x) (1)} + {(16x^{2} + 1 + 2(4x) (1)} = {(25x^{2} + 1 + 2 (5x) (1)} + 5

25x^{2} + 2 + 8x + 6x = 25x^{2} + 6 + 10x

14x – 10x = 6 – 2

4x = 4

We get,

x = 1

(x) 3 (3x – 4) – 2 (4x – 5) = 6

Simplifying by collecting like terms,

9x – 12 – 8x + 10 = 6

9x – 8x = 6 + 12 – 10

We get,

x = 8

**3. Solve the following equations for the unknown:**

**(a) (4x / 27) = (8 / 9)**

**(b) (1.5y / 3) = (7 / 2)**

**(c) (-3.4m) / (2.7) = (10.2) / 9**

**(d) (1 / 2)p + (3 / 4) p = p – 3**

**(e) (9y / 4) – (5y / 3) = (1 / 5)**

**(f) (x / 2) + (x / 4) + (x / 8) = 7**

**(g) (2m / 3) – (m / 2) = 1**

**(h) {2 (2x – 1) / 9} – {(x – 1) / 2} = 0**

**(i) (4 / 5) x – 21 = (3 / 4) x – 20**

**(j) {(a – 1) / 2} – {(a + 1) / 3} = 5 – a**

**Solution:**

(a) (4x / 27) = (8 / 9)

On cross multiplication, we get,

(4x) (9) = (8) (27)

x = (8 × 27) / (4 × 9)

x = (2 × 3) / (1 × 1)

We get,

x = 6

(b) (1.5y / 3) = (7 / 2)

On cross multiplication, we get,

(1.5y) (2) = (7) (3)

y = (7 × 3) / (1.5 × 2)

y = (70 × 3) / (15 × 2)

We get,

y = 7

(c) (-3.4m) / (2.7) = (10.2) / 9

On cross multiplication, we get,

(-3.4m) (9) = (10.2) (2.7)

m = – (10.2 x 2.7) / (3.4 x 9)

m = – (102 x 27) / (34 x 9 x 10)

m = – (3 x 3) / (1 x 1 x 10)

We get,

m = -0.9

(d) (1 / 2) p + (3 / 4) p = p – 3

On calculating further, we get,

(1 / 2) p + (3 / 4) p – p = – 3

(p / 2) + (3p / 4) – p = – 3

On taking L.C.M we get,

{(2p + 3p – 4p) / 4} = – 3

p / 4 = – 3

p = (- 3) (4)

We get,

p = – 12

(e) (9y / 4) – (5y / 3) = (1 / 5)

On taking L.C.M. we get,

{(27y – 20y) / 12} = (1 / 5)

(7y / 12) = (1 / 5)

(7y) (5) = 12

y = 12 / (7 × 5)

We get,

y = (12 / 35)

(f) (x / 2) + (x / 4) + (x / 8) = 7

On taking L.C.M. we get,

{(4x + 2x + x) / 8} = 7

(7x / 8) = 7

On cross multiplication, we get,

7x = (7) (8)

x = (7 × 8) / 7

We get,

x = 8

(g) (2m / 3) – (m / 2) = 1

On taking L.C.M. we get,

{(4m – 3m) / 6} = 1

4m – 3m = 6

Hence,

m = 6

(h) {2 (2x – 1) / 9} – {(x – 1) / 2} = 0

On further calculation, we get,

{(4x – 2) / 9} – {(x – 1) / 2} = 0

(4x – 2) / 9 = (x – 1) / 2

On cross multiplication, we get,

2 (4x – 2) = 9 (x – 1)

8x – 4 = 9x – 9

We get,

x = 5

(i) (4 / 5) x – 21 = (3 / 4) x – 20

(4 / 5) x – (3 / 4) x = – 20 + 21

Taking L.C.M. we get,

(16x / 20) – (15x / 20) = 1

{(16x – 15x) / 20} = 1

16x – 15x = 20

We get,

x = 20

(j) {(a – 1) / 2} – {(a + 1) / 3} = 5 – a

On further calculation, we get,

{(a – 1) / 2} – {(a + 1) / 3} + a = 5

On taking the L.C.M. we get,

[{3 (a – 1) – 2 (a + 1) + 6a} / 6] = 53a – 3 – 2a – 2 + 6a = 5 × 6

7a – 5 = 30

7a = 30 + 5

7a = 35

We get,

a = 5

**4. Solve the following equations for the unknown:**

**(a) (5 / x) – 11 = (2 / x) + 16, x ≠ 0**

**(b) 11 – (3 / x) = (5 / x) + 3**

**(c) {5 / (3x – 2)} – (1 / 8) = 0, x ≠ 0, x ≠ (2 / 3)**

**(d) {1 / (x – 1)} + (4 / 5) = (2 / 3), x ≠ 1**

**(e) {7 / (x – 2)} – (5 / 3) = 3, x ≠ 2**

**(f) (2x + 3) / (x + 7) = (5 / 8), x ≠ – 7**

**(g) {(3x – 5) / (7x – 5)} = (1 / 9), x ≠ (5 / 7)**

**(h) {3 / (x + 1)} – {(x – 6) / (x ^{2} – 1)} = 12 / (x – 1)**

**(i) {(x + 13) / (x ^{2} – 1)} + {5 / (x + 1)} = 7 / (x + 1)**

**(j) {(6x + 7) / (3x + 2)} = {(4x + 5) / (2x + 3)}**

**(k) – {(x – 2) / 3} = (x – 1) / 3**

**(l) (1 / 2) {y – (1 / 3)} + (1 / 4) {2y + (1 / 5)} = (3 / 4) {y – (1 / 12)}**

**(m) 2 + {(3x – 2) / (3x + 2)} = (3x + 2) / (x + 1)**

**(n) {(7x – 1) / 4} – (1 / 3) {2x – (1 – x) / 2} = **

**Solution:**

(a) (5 / x) – 11 = (2 / x) + 16, x ≠ 0

(5 / x) – (2 / x) = 11 + 16

On further calculation, we get,

{(5 – 2) / x} = 27

(3 / x) = 27

x = (3 / 27)

We get,

x = (1 / 9)

(b) 11 – (3 / x) = (5 / x) + 3

11 – 3 = (5 / x) + (3 / x)

(5 / x) + (3 / x) = 11 – 3

On taking L.C.M. we get,

{(5 + 3) / x} = 11 – 3

(8 / x) = 8

x = (8 / 8)

We get,

x = 1

(c) {5 / (3x – 2)} – (1 / 8) = 0, x ≠ 0, x ≠ (2 / 3)

{5 / (3x – 2)} = (1 / 8)

On cross multiplication, we get,

40 = 3x – 2

3x = 40 + 2

3x = 42

x = (42 / 3)

We get,

x = 14

(d) {1 / (x – 1)} + (4 / 5) = (2 / 3), x ≠ 1

{1 / (x – 1) = (2 / 3) – (4 / 5)

On taking L.C.M. we get,

{1 / (x – 1)} = (10 / 15) – (12 / 15)

{1 / (x – 1)} = (- 2 / 15)

On cross multiplication, we get,

15 = – 2 (x – 1)

15 = – 2x + 2

2x = – 13

x = (- 13 / 2)

We get,

x =

(e) {7 / (x – 2)} – (5 / 3) = 3, x ≠ 2

{7 / (x – 2)} = (5 / 3) + 3

On taking L.C.M. we get,

{7 / (x – 2)} = {(5 + 9) / 3}

{7 / (x – 2)} = (14 / 3)

On cross multiplication, we get,

21 = 14 (x – 2)

21 = 14x – 28

14x = 49

x = (49 / 14)

We get,

x = (7 / 2)

(f) (2x + 3) / (x + 7) = (5 / 8), x ≠ – 7

On cross multiplication, we get,

8 (2x + 3) = 5 (x + 7)

16x + 24 = 5x + 35

16x – 5x = 35 – 24

11x = 11

We get,

x = 1

(g) {(3x – 5) / (7x – 5)} = (1 / 9), x ≠ (5 / 7)

On cross multiplication, we get,

9 (3x – 5) = 7x – 5

27x – 45 = 7x – 5

27x – 7x = – 5 + 45

20x = 40

We get,

x = 2

(h) {3 / (x + 1)} – {(x – 6) / (x^{2} – 1)} = 12 / (x – 1)

{3 / (x + 1)} – {(x – 6) / (x – 1) (x + 1)} = 12 / (x – 1)

Here,

LCM of all the denominators in the equation is (x – 1) (x + 1)

Multiplying throughout by the LCM, we get,

3 (x – 1) – (x – 6) = 12 (x + 1)

3x – 3 – x + 6 = 12x + 12

3x – 12x – x = 12 – 6 + 3

– 10x = 9

We get,

x = (- 9 / 10)

(i) {(x + 13) / (x^{2} – 1)} + {5 / (x + 1)} = 7 / (x + 1)

{(x + 13) / (x – 1) (x + 1)} + {5 / (x + 1)} = 7 / (x + 1)

Here,

LCM of all the denominators in the equation is (x – 1) (x + 1)

Multiplying throughout by the LCM, we get,

x + 13 + 5 (x – 1) = 7 (x – 1)

x + 13 + 5x – 5 = 7x – 7

13 – 5 + 7 = 7x – 5x – x

20 – 5 = 7x – 6x

We get,

x = 15

(j) {(6x + 7) / (3x + 2)} = {(4x + 5) / (2x + 3)}

On cross multiplication, we get,

(6x + 7) (2x + 3) = (4x + 5) (3x + 2)

6x (2x + 3) + 7 (2x + 3) = 4x (3x + 2) + 5 (3x + 2)

On further calculation, we get,

12x^{2} + 18x + 14x + 21 = 12x^{2} + 8x + 15x + 10

18x + 14x + 21 = 8x + 15x + 10

18x – 8x + 14x – 15x = 10 – 21

10x – 1x = – 11

9x = – 11

We get,

x = (- 11 / 9)

(k)

– {(x – 2) / 3} = (x – 1) / 3

This can be written as,

(11 / 5) – {(x – 2) / 3} = (x – 1) / 3

(11 / 5) = {(x – 1) / 3} + {(x – 2) / 3}

On taking LCM, we get,

(11 / 5) = {(x – 1 + x – 2) / 3}

(11 / 5) = (2x – 3) / 3

On cross multiplication, we get,

11 (3) = 5 (2x – 3)

33 = 10x – 15

10x = 33 + 15

10x = 48

We get,

x = 4.8

(l) (1 / 2) {y – (1 / 3)} + (1 / 4) {2y + (1 / 5)} = (3 / 4) {y – (1 / 12)}

(y / 2) – (1 / 6) + (y / 2) + (1 / 20) = (3y / 4) – (1 / 16)

(y / 2) + (y / 2) – (3y / 4) = – (1 / 16) + (1 / 6) – (1 / 20)

y – (3y / 4) = – (1 / 16) + (1 / 6) – (1 / 20)

On taking LCM, we get,

{(4y – 3y) / 4} = {(-15 + 40 – 12) / 240}

(y / 4) = (13 / 240)

y = (13 / 240) × 4

We get,

y = (13 / 60)

(m) 2 + {(3x – 2) / (3x + 2)} = (3x + 2) / (x + 1)

2 = {(3x + 2) / (x + 1)} – {(3x – 2) / (3x + 2)}

On taking LCM, we get,

2 = {(3x + 2) (3x + 2) – (3x – 2) (x + 1)} / (x + 1) (3x + 2)

On cross multiplication, we get,

2{(x + 1) (3x + 2)} = (3x + 2) (3x + 2) – (3x – 2) (x + 1)

2 {(3x^{2} + 2x + 3x + 2)} = (9x^{2} + 6x + 6x + 4) – (3x^{2} + 3x – 2x – 2)

6x^{2} + 4x + 6x + 4 = 9x^{2} + 6x + 6x + 4 – 3x^{2} – 3x + 2x + 2

4x = 6x – 3x + 2x + 2

– x = 2

We get,

x = – 2

(n) {(7x – 1) / 4} – (1 / 3) {2x – (1 – x) / 2} =

{(7x – 1) / 4}- {(2x / 3) + (1 – x) / 6} = (16 / 3)

Here,

LCM of all the denominators is 12

Multiply the equation throughout by 12, we get,

{3 (7x – 1) / 12} – {4 (2x) / 12} + {2 (1 – x) / 12} = 4 (16)

{(21x – 3) / 12} – {(8x / 12)} + {(2 – 2x) / 12} = 64

21x – 3 – 8x + 2 – 2x = 64

11x = 64 + 3 – 2

11x = 65

We get,

x = (65 / 11)

**5. Solve the following equations for the unknown:**

**(a) √x – 5 = 3**

**(b) 7 – (1 / √y) = 0**

**(c) (1 / 5) = (3√x / 3)**

**(d) 2 √(x – 3) / (x + 5) = (1 / 3)**

**Solution:**

(a) √x – 5 = 3

Squaring on both sides, we get,

x – 5 = (3)^{2}

x – 5 = 9

x = 9 + 5

We get,

x = 14

(b) 7 – (1 / √y) = 0

7 = (1 / √y)

Squaring on both sides, we get,

(7)^{2} = (1 / y)

49 = (1 / y)

We get,

y = (1 / 49)

(c) (1 / 5) = (3√x / 3)

(1 / 5) = √x

Squaring on both sides, we get,

(1 / 5)^{2} = x

We get,

x = (1 / 25)

(d) 2 √(x – 3) / (x + 5) = (1 / 3)

On squaring both sides, we get,

{2 √(x – 3) / (x + 5)}^{2} = (1 / 3)^{2}

4 {(x – 3) / (x + 5)} = (1 / 9)

(4x – 12) / (x + 5) = (1 / 9)

On cross multiplication, we get,

9 (4x – 12) = x + 5

36x – 108 = x + 5

36x – x = 5 + 108

35x = 113

We get,

x = (113 / 35)

**6. Find the value of x for which the expression (x / 5) + 2 and (x / 3) – 4 are equal.**

**Solution:**

Given that the two expressions are equal

i.e,

(x / 5) + 2 = (x / 3) – 4

On further calculation, we get,

(x / 5) – (x / 3) = – 4 – 2

On taking LCM, we get,

(3x – 5x) / 15 = – 6

(- 2x / 15) = – 6

On cross multiplication, we get,

2x = 15 × 6

2x = 90

We get,

x = 45

**7. Find the value of x if the difference of one-third of (x + 7) and one-fifth of (3x – 2) is 3.**

**Solution:**

According to the given statement,

(1 / 3) (x + 7) – (1 / 5) (3x – 2) = 3

Taking LCM, we get,

{5 (x + 7) – 3 (3x – 2)} / 15 = 3

On cross multiplication, we get,

5x + 35 – 9x + 6 = 3 × 15

5x – 9x + 35 + 6 = 45

– 4x + 41 = 45

– 4x = 45 – 41

– 4x = 4

We get,

x = – 1

**8. Shweta’s age is six times that of Jayeeta’s age. 15 years hence Shweta will be three times as old as Jayeeta; find their ages.**

**Solution:**

Let Jayeeta’s age = x years and

Shweta’s age = 6x years

After 15 years,

As per the given condition,

6x + 15 = 3 (x + 15)

6x + 15 = 3x + 45

6x – 3x = 45 – 15

3x = 30

We get,

x = 10

Jayeeta’s age = x years

= 10 years

Shweta’s age = 6x

= 6 (10)

= 60 years

Therefore, Jayeeta’s age is 10 years and Shweta’s age is 60 years

**9. The ages of P and Q are in the ratio 7: 5. Ten years hence, the ratio of their ages will be 9: 7. Find their ages.**

**Solution:**

Let the common multiple be x

Hence,

P’s age be 7x years and Q’s age be 5x years

After 10 years,

As per the given condition,

(7x + 10) / (5x + 10) = (9 / 7)

On cross multiplication, we get,

7 (7x + 10) = 9 (5x + 10)

49x + 70 = 45x + 90

49x – 45x = 90 – 70

4x = 20

We get,

x = 5

P’s age = 7x years

7 (5) = 35 years

Q’s age = 5x

5 (5) = 25 years

Therefore, P’s age is 35 years and Q’s age is 25 years

**10. The length of a rectangle is 30 cm more than its breadth. The perimeter of the rectangle is 180 cm. Find the length and the breadth of the rectangle.**

**Solution:**

Let the breadth of the rectangle = x cm

So,

The length of the rectangle = (30 + x) cm

According to the given condition,

Perimeter of the rectangle = 180 cm

2 (l + b) = 180

2 (30 + x + x) = 180

2 (30 + 2x) = 180

60 + 4x = 180

4x = 180 – 60

4x = 120

We get,

x = 30

breadth = x cm

= 30 cm

Length = (30 + x) cm

= (30 + 30) cm

= 60 cm

Therefore, the breadth of the rectangle is 30 cm and the length of the rectangle is 60 cm

**11. The perimeter of a rectangle field is 80 m. If the breadth is increased by 2 m and the length is decreased by 2 m, the area of the field increases by 36 m ^{2}. Find the length and the breadth of the field.**

**Solution:**

Let the breadth of the rectangle = x cm

Perimeter of the rectangle = 80 m

2 (l + x) = 80

l + x = (80 / 2)

l + x = 40

We get,

l = 40 – x

So,

The area = lb = x (40 – x)

= 40x – x^{2}

As per the statement,

Breadth = (x + 2) m

Length = (40 – x – 2) m

= (38 – x) m

So,

Area = (38 – x) (x + 2)

= 38x + 76 – x^{2} – 2x

= – x^{2} + 36x + 76

Now,

According to the given condition,

– x^{2} + 36x + 76 – (40x – x^{2}) = 36

– x^{2} + 36x + 76 – 40x + x^{2} = 36

36x + 76 – 40x = 36

36x – 40x = 36 – 76

-4x = – 40

We get,

x = 10

So, breadth = x = 10 m and

length = 40 – x = 40 – 10 = 30 m

Therefore, the breadth of the field is 10 m and the length of the field is 30 m

**12. A’s age is six times that of B’s age. 15 years hence A will be three times as old as B;** **find their ages.**

**Solution:**

Let the age of B = x years

Then,

The age of A becomes 6x years

After 15 years,

Age of A = 6x + 15

Age of B = x + 15

Given that, after 15 years, A will be three times as old as B

6x + 15 = 3 (x + 15)

6x + 15 = 3x + 45

6x – 3x = 45 – 15

3x = 30

We get,

x = 10 years

Hence,

Age of B = x = 10 years

Age of A = 6x = 6 (10) = 60 years

**13. The present age of a man is double the age of his son. After 8 years, the ratio of their ages will be 7: 4. Find the present ages of the man and his son.**

**Solution:**

Let the present age of the son = x years

Then,

The father’s age = 2x years

After 8 years,

Their ages will b (x + 8) years and (2x + 8) years respectively

Given that, after 8 years, the ratio of their ages will be 7: 4

According to the given condition,

(x + 8) / (2x + 8) = (4 / 7)

On cross multiplication, we get,

7 (x + 8) = 4 (2x + 8)

7x + 56 = 8x + 32

7x – 8x = 32 – 56

-x = – 24

We get,

x = 24 years

Therefore, the present age of the son is 24 years and

Age of father = 2x = 2 (24) = 48 years

**14. What number increased by 8% of itself gives 1620?**

**Solution:**

Let the number be x

Hence,

x + (8 / 100) × x = 1620

On taking LCM, we get,

(100x + 8x) / 100 = 1620

108x = 1620 × 100

x = {(1620 × 100) / 108}

We get,

x = 1500

Therefore, the required number is 1500

**15. What number increased by 15% of itself gives 2921?**

**Solution:**

Let the number be x

Hence,

x + (15 / 100) × x = 2921

{100x + 15x) / 100} = 2921

On cross multiplication, we get,

115x = 2921 × 100

x = {(2921 × 100) / 115}

We get,

x = 2540

Therefore, the required number is 2540