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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 that a value is a solution of the equation if it satisfies the equation.
i.e,
x1 is a solution if f (x1) = 0
Here,
x1 = 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 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 the 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) = 2a2 + 4a + 2 + 13
2a2 + 5a + 4a + 10 – (2a2 + 4a + 15) = 0
5a + 10 – 15 = 0
5a = 5
We get,
a = 1
(viii) (6p + 9)2 + (8p – 7)2 = (10p + 3)2 – 71
{(36p2 + 81 + 2 (6p) (9)} + {(64p2 + 49 – 2 (8p) (7)} = 100p2 + 9 + 2 (10p) (3) – 71
100p2 + 130 + 2 × 54p – 2 × 56p = 100p2 + 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,
{(9x2 + 1 + 2 (3x) (1)} + {(16x2 + 1 + 2(4x) (1)} = {(25x2 + 1 + 2 (5x) (1)} + 5
25x2 + 2 + 8x + 6x = 25x2 + 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) / (x2 – 1)} = 12 / (x – 1)
(i) {(x + 13) / (x2 – 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) / (x2 – 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) / (x2 – 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,
12x2 + 18x + 14x + 21 = 12x2 + 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 {(3x2 + 2x + 3x + 2)} = (9x2 + 6x + 6x + 4) – (3x2 + 3x – 2x – 2)
6x2 + 4x + 6x + 4 = 9x2 + 6x + 6x + 4 – 3x2 – 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, and 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 is 7x years, and Q’s age is 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 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 m2. Find the length and 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 – x2
As per the statement,
Breadth = (x + 2) m
Length = (40 – x – 2) m
= (38 – x) m
So,
Area = (38 – x) (x + 2)
= 38x + 76 – x2 – 2x
= – x2 + 36x + 76
Now,
According to the given condition,
– x2 + 36x + 76 – (40x – x2) = 36
– x2 + 36x + 76 – 40x + x2 = 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
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