NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT
SOLUTIONS
CLASS - 7TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-4 Maths

Class : 7th
Subject : Maths
Chapter : 4
Chapter Name : Simple Equations

Exercise 4.1

Q1 Complete the last column of the table

Answer. L.H.S. = x + 3
By putting x = 3,
L.H.S. = 3 + 3 = 6 ≠ R.H.S
No, the equation is not satis ed.

(ii) x + 3 = 0
L.H.S. = x +3
By putting x = 0,
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
No, the equation is not satis ed.

(iii) x + 3 = 0
L.H.S. = x +3
By putting x = -3,
L.H.S. = - 3 + 3 = 0 R.H.S.
Yes, the equation is satis ed.

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

(iv) x - 7 = 0
L.H.S. = x - 7
By putting x = 7,
L.H.S. = 7 − 7 = 0 ≠ R.H.S.
No, the equation is not satis ed.

(v) x - 7 = 1
L.H.S. = x - 7
By putting x = 8,
L.H.S. = 8 - 7 = 1 R.H.S.
Yes, the equation is satis ed.

(vi) 5x = 25
L.H.S. = 5x
By putting
= 5 × x =00,= 0 ≠
L.H.S. R.H.S.
No, the equation is not satis ed.

(vii) 5x = 25
L.H.S. = 5x
By putting x = 5,
L.H.S. = 5 x 5 = 25 R.H.S.
Yes, the equation is satis ed.

(viii) 5x = 25
L.H.S. = 5x
By putting
= 5 × 5, = 25 ≠
x =(−5)
L.H.S. R.H.S.
No, the equation is not satis ed.

(ix) m/3 = 2
L.H.S = m/3
−6
= = −2
3

No, the equation is not satis ed.
By putting m = - 6,

(x) m/3 = 2
L.H.S R.H.S
L.H.S = m/3
By putting m = 0,
L.H.S = R.H.S
−0
= 0
3

No, the equation is not satis ed.

(x) m/3 = 2
L.H.S = m/3
By putting m = 6,

Page 2 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-4 Maths

L.H.S = 6

3
= 2 R.H.S
Yes, the equation is satis ed.

Page : 81 , Block Name : Exercise 4.1

Q2 Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0)

Answer. (a) n + 5 -19 (n = 1)
Putting n = 1 in L.H.S.,
n + 5 = 1 + 5 = 6 ≠ 19

As L.H.S ≠ R. H. S.
Therefore, n = 1 is not a solution of the given equation, n + 5 =19.

(b) 7n + 5 = 19 (n = -2)
Putting n = -2 in L.H.S.,
7n + 5 = 7 × (−2) + 5 = −14 + 5 = −9 ≠ 19

As L.H.S ≠ R. H. S.
Therefore, n = -2 is not a solution of the given equation, 7n + 5 =19.

(c) 7n + 5 = 19 (n = 2)
Putting n = 2 in L.H.S.,
7 n + 5 = 7 x ( 2 ) + 5 = 14 + 5 = 19 = R.H.S
As L.H.S = R.H.S.
Therefore, n = 2 is a solution of the given equation, 7n + 5 =19.

(d) 4p - 3 = 13 (p = 1)
Putting p = 1 in L.H.S.,
4p − 3 = (4 × 1) − 3 = 1 ≠ 13

As L.H.S ≠ R. H. S.
Therefore, p = 1 is not a solution of the given equation, 4p - 3 = 13.

(e) 4p - 3 = 13 (p = -4)
Putting p = -4 in L.H.S.,
4p − 3 = 4 × (−4) − 3 = −16 − 3 = −19 ≠ 13

As L.H.S ≠ R. H. S.
Therefore, p = -4 is not a solution of the given equation, 4p - 3 = 13.

(f) 4p - 3 = 13 (p = 0)

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

Putting p = 0 in L.H.S.,
4p − 3 = (4 × 0) − 3 = −3 ≠ 13

As L.H.S ≠ R. H. S.
Therefore, p = 0 is not a solution of the given equation, 4p - 3 = 13.

Page : 81 , Block Name : Exercise 4.1

Q3 Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4

Answer. (i) 5p + 2 = 17
Putting p = 1 in L.H.S.,
(5 × 1) + 2 = 7 ≠ R. H. S.

Putting p = 2 in L H.S.,
(5 × 2) + 2 = 10 + 2 = 12 ≠ R. H. S.

Putting p = 3 in L H.S.,
(5 × 3) + 2 = 17 = R. H . S

Hence, p = 3 is a solution of the given equation.
(ii) 3m – 14 = 4
Putting m = 4,
(3 × 4) − 14 = −2 ≠ R. H. S.

Putting m =5,
(3 × 5) − 14 = 1 ≠ R. H. S.

Putting m = 6,
(3 × 6) − 14 = 18 − 14 = 4 = R. H . S

Hence, m = 6 is a solution of the given equation.

Page : 81 , Block Name : Exercise 4.1

Q4 Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.

Answer. (i) x + 4 = 9
(ii) y - 2 =8
(iii) 10a = 70
(iv) = 6
b

5

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

(v) 3

4
t = 15

(vi) Seven times of m is 7m.
7m +7 = 77
(vii) One-fourth of a number x is x/4.
x
− 4 = 4
4

(viii) Six times of y is 6y.
6Y -6 = 60
(ix) One-third of z is z/3
z
+ 3 = 30
3

Page : 81 , Block Name : Exercise 4.1

Q5 Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) m 5 = 3
(v) 3 5 m = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) p/2 + 2 = 8

Answer. (i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Twice of a number m is 7.
(iv) One- fth of m is 3.
(v) Three- fth of m is 6.
(vi) Three times of a number p, when added to 4, gives 25.
(vii) When 2 is subtracted from four times of a number p, it gives 18.
(viii) When 2 is added to half of a number p, it gives 8.

Page : 81 , Block Name : Exercise 4.1

Q6 Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than ve times the marbles Parmit has. Irfan has 37
marbles. (Take m to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take
Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is
twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b
in degrees. Remember that the sum of angles of a triangle is 180 degrees).

Answer. (i) Let Parmit has m marbles.

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

5 x Number of marbles Parmit has + 7 = Number of marbles Irfan has
5 x m + 7 = 37
5m + 7 = 37
(ii) Let Laxmi be y years old.
3 x Laxmi's age + 4 = Laxmi's father's age
3 x y + 4 = 49
3 +4 = 49
(iii) Let the lowest marks be l.
2 x Lowest marks + 7 = Highest marks
2 x l +7 = 87
2l +7 = 87
(iv) An isosceles triangle has two of its angles of equal measure.
Let base angle be b.
Vertex angle = 2 x Base angle = 2b
Sum of all interior angles of aΔ = 180 ∘

b + b + 2b = 180∘

4b = 180 ∘

Page : 82 , Block Name : Exercise 4.1

Exercise 4.2

Q1 Give rst the step you will use to separate the variable and then solve the equation: (a) x –
1=0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4

Answer. (a) x + 1 = 0
Subtracting 1 to both sides of the given equation, we obtain
x+1-1=0-1
x=-1
(b) x - 1 = 0
Adding 1 to both sides of the given equation, we obtain
x-1+1=0+1
x=1
(c) x - 1 = 5
Adding 1 to both sides of the given equation, we obtain

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

x-1+1=5+1
x=6
(d) x + 6 = 2
Subtracting 6 to both sides of the given equation, we obtain
x+6-6=2-6
x=-4
(e) y - 4 = -7
Adding 4 to both sides of the given equation, we obtain
y - 4 + 4 = -7 + 4
y = -3
(f) y - 4 = 4
Adding 4 to both sides of the given equation, we obtain v
y-4+4=4+4
y=8
(g) y + 4 = 4
Subtracting 4 to both sides of the given equation, we obtain
y+4-4=4-4
y=0

(g) y + 4 = -4
Subtracting 4 to both sides of the given equation, we obtain
y + 4 - 4 = -4 - 4
y = -8

Page : 86 , Block Name : Exercise 4.2

Q2 Give rst the step you will use to separate the variable and then solve the equation: (a) 3l
= 42
(b) b/2 = 6
(c) p/7 = 4
(d) 4x = 25
(e) 8y = 36
(f) z/3 = 5/4
(g) a/5 = 7/15
(h) 20t = – 10

Answer. (a) 3l = 42
Dividing both sides of the given equation by 3, we obtain
3l 42
=
3 3

l = 14

(b) b/2 = 6
Multiplying both sides Of the given equation by 2, we obtain
b×2
= 6 × 2
2

b = 12

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

(c) p/7 = 4
Multiplying both sides of the given equation by 7, we obtain
p×7
= 4 × 7
7

p = 28

(d) 4x = 25
Dividing both sides Of the given equation by 4, we obtain
4x 25
=
4 4

25

4

(e) 8y = 36
Dividing both sides of the given equation by 8, we obtain
8y 36
=
8 8

9
y =
2

(f) z

3
=
5

4

Multiplying both sides of the given equation by 3, we obtain
z×3 5×3
=
3 4

15
z =
4

(g) a

5
=
7

15

Multiplying both sides of the given equation by 5, we obtain
a×5 7×5
=
5 15

7
a =
3

(h) 20t = -10
Dividing both sides of the given equation by 20, we obtain
20t −10
=
20 20

−1
t =
2

Page : 86 , Block Name : Exercise 4.2

Q3 Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p/3 = 40
(d) 3p/10 = 6

Answer. (a) The given equation is 3n - 2 = 46
Add 2 to both sides,

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

3n − 2 + 2 = 46 + 2 ⇒ 3n = 48

Divide both sides by 3,
3n 48
= ⇒ n = 16
3 3

It is the required solution. For Checking, put n = 16 in the given equation.
L.H.S = 3n − 2 = 3 × 16 − 2 = 48 − 2 = 46 = R. H. S. as required. Hence the solution is
correct.

(b) The given equation is 5m + 7 = 17
Subtract 7 from both the sides
5m + 7 − 7 = 17 − 7 ⇒ 5m = 10

Divide both sides by 5
5m 10
= ⇒ m = 2
5 5

It is the required solution. For checking, put m=2 in the given equation.
L.H.S = 5m + 7 = 5 × 2 + 7 = 10 + 7 = 17 = R. H. S. as required. Hence the solution is
correct.

(c) The given equation is
20p
= 40
3

Multiply both sides by 3,
20p
× 3 = 40 × 3 ⇒ 20p = 120
3

Divide both sides by 20
20p 120
= ⇒ p = 6
20 20

It is the required solution. For checking, put p = 6 in the given equation.
20p
L.H.S = 3
=
20×6

3
= 40 = R. H. S as required. Hence the solution is correct.

(d) 3p/10 = 6
Dividing both sides of the given equation by 3, we obtain
3p 60
=
3 3

p=20

Page : 86 , Block Name : Exercise 4.2

Q4 Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4 = 5
(d) –p/3 = 5
(e) 3p/4 = 6
(f) 3s = –9
(g) 3s + 12 = 0
(h) 3s = 0
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

Answer. (a) 10p = 100
10p 100
=
10 10

p=10
(b) 10p + 10 = 100
10 p + 10 - 10 = 100 - 10
10p = 90
10p 90
=
10 10

p=9
(c) p/4 = 5
p×4
= 5 × 4
4

p = 20
(d) –p/3 = 5
−p×(−3)
= 5 × (−3)
3

p = -15
(e) 3p/4 = 6
3p
= 6
4

3p×4
= 6 × 4
4

3p = 24

3p 24
=
3 3

p=8
(f) 3s = –9
\frac { 3 s } { 3 } = \frac { - 9 } { 3 }
s = -3
(g) 3s + 12 = 0
3 s + 12 - 0 — 12
3 s = -12
3s −12
=
3 3

s = -4
(h) 3 s = 0
3s 0
=
3 3

s=0
(i) 2q = 6
2q 6
=
2 2

q=3
(j) 2q - 6 = 0
2q - 6 + 6 = 0 + 6
2q = 6
2q 6
=
2 2

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

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

2q = -6
2q −6
=
2 2

q = -3
(l) 2q + 6 = 12
2q + 6 - 6 = 12 - 6
2q = 6
2q 6
=
2 2

q=3

Page : 86 , Block Name : Exercise 4.2

Exercise 4.3

Q1 Solve the following equations:

Answer. (a) 2y + 5

2
=
37

2

2y =
37

2
−
5

2
=
32

2
= 16 ( Transposing 5/2 to R.H.S)
Dividing both sides by 2,
y = 16/2 = 8

(b) 5t+28= 10
5t = 10 - 28 = -18( Transposing 28 to R.H.S)
Dividing both sides by 5,
t = -18/5

(c) a

5
+ 3 = 2
a

5
= 2 − 3 = −1 ( Transposing 3 to R.H.S)
Multiplying both sides by 5,
(d) + 7 = 5
q

4

q/4 = -2( Transposing 7 to R.H.S)
Multiplying both sides by 4,
q = -8

(e) 5

2
x = −10

Multiplying both sides by 2,

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

5x = -10 x 2 = -20
Dividing both sides by 5,
−20
x = = −4
5

(f) 5

2
x =
25

4

Multiplying both sides by 2,
25 25
5x = × 2 =
4 2

Multiplying both sides by 5,
25 1 5
x = × =
2 5 2

(g) 7m + 19

2
= 13

(Transposing 19/2 to R.H.S)
19 26−19
7m = 13 − =
2 2

7m = 7/2
Dividing both sides by 7,
m=½

(h) 6z + 10 = -2
6z = - 2 - 10 = -12 (Transposing 10 to R.H.S.)
Dividing both sides by 6,
−12
z = = −2
6

(i) 3l

2
=
2

3

Multiplying both sides by 2,
2 4
3l = × 2 =
3 3

Dividing both sides by 3,
4 1 4
I = × =
3 3 9

(j) 2b

3
− 5 = 3

2b

3
= 3 + 5 = 8 (Transposing -5 to R.H.S.)
Multiplying both sides by 3,
2b = 8 x 3 = 24
Dividing both sides by 2,
b = 24/2 = 12

Page : 89 , Block Name : Exercise 4.3

Q2 Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8

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Answer.
(a) 2(x + 4) = 12

Dividing both sides by 2

12
x + 4 = = 6
2

r = 6 − 4 = 2 (Transposing 4 to R.H.S.)

(b) 3(n − 5) = 21

Dividing both sides by 3

21
n − 5 = = 7
3

n = 7 + 5 = 12 (Transposing − 5 to R.H.S.)

(c) 3(n − 5) = −21

Dividing both sides by 3

−21
n − 5 = = −7
3

n = −7 + 5 = −2 (Transposing − 5 to R.H.S.)

(d) − 4(2 + x) = 8

Dividing both sides by − 4

8
2 + x = = −2
−4

x = −2 − 2 = −4 (Transposing 2 to R.H.S.)

(e) 4(2 − x) = 8

Dividing both sides by 4

2 − x = 2

−x = 2 − 2( Transposing 2 to R.H.S. )

−x = 0

x = 0

Page : 89 , Block Name : Exercise 4.3

Q3 Solve the following equations:
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) =34
(e) 0 = 16 + 4(m – 6)

Answer.
(a) 4 = 5(p − 2)

Dividing both sides by 5

4
= p − 2
5
4

5
+ 2 = p (Transposing—2 to L H.S.)

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

4+10
= p
5
14
= p
5

(b) − 4 = 5(p − 2)

Dividing both sides by 5,

4
− = p − 2
5

4
− + 2 = p ( Transposing − 2 to L. H.S. )
5
−4+10
= p
5

6
= p
5

(c) 16 = 4 + 3(t + 2)

16 − 4 = 3(t + 2) (Transposing 4 to L.H.S.)

12 = 3(t + 2)

Dividing both sides by 3 ,
12
= t + 2
3

4 = t + 2

4 − 2 = t( Transposing 2 to L.H.S. )

2 = t

(d) 4 + 5(p − 1) = 34

5(p − 1) = 34 − 4 = 30 (Transposing 4 to R.H.S.)

Dividing both sides by 5

30
p − 1 = = 6
5

p = 6 + 1 = 7 (Transposing − 1 to R.H.S.)

(e) 0 = 16 +4 (m - 6)
0 = 16 + 4m − 24

0 = −8 + 4m

4m = 8( Transposing − 8 to L.H.S)

Dividing both sides by 4 ,

m = 2

Page : 89 , Block Name : Exercise 4.3

Q4 (a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2

Answer.

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

(a) x = 2

Multiplying both sides by 5,

5x = 10 (i)

Subtracting 3 from both sides,

5x − 3 = 10 − 3

5x − 3 = 7( (ii)

Dividing both sides by 2 ,
5x

2
−
3

2
=
7

2
(iii)
>br/>
(b) x = −2

Subtracting 2 from both sides,

x − 2 = −2 − 2

x − 2 = −4 (i)

Again, x = −2

Multiplying by 6

6 × x = −2 × 6

6x = −12

Subtracting 12 from both sides,

6x − 12 = −12 − 12

6x − 12 = −24 (ii)

Adding 24 to both sides,

6x − 12 + 24 = −24 + 24

6x + 12 = 0 (iii)

Page : 89 , Block Name : Exercise 4.3

Exercise 4.4

Q1 Set up equations and solve them to nd the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One- fth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he nds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 5 2 of the number, the result is 23.

Answer.

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(a) Let the number be x .

8 times of this number = 8x

8x + 4 = 60

8x = 60 − 4( Transposing 4 to R.H.S. )

8x = 56

Dividing both sides by 8

8x 56
=
8 8

x = 7

(b) Let the number be x .
x
One-fifth of this number =
5

x
− 4 = 3
5

x
− 4 = 3
5

x×5
= 7 × 5
5

x = 35

(c) Let the number be x .

3x
Three-fourth of this number = =
4

3
x + 3 = 21
4

3
x = 18
4

(Transposing 3 to R.H.S.)

Multiplying both sides by 4 ,
3x×4
= 18 × 4
4

3x = 72

Dividing both sides by 3,

3x 72
=
3 3

x = 24

(d) Let the number be x .

Twice of this number = 2x

2x − 11 = 15

2x = 15 + 11 (Transposing − 11 to R.H.S.)

2x = 26

Dividing both sides by 2 ,

2x 26
=
2 2

x = 13

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(e) Let the number of books be x .

Thrice the number of books = 3x

50 − 3x = 8

−3x = 8 − 50 (Transposing 50 to R.H.S.)

−3x = −42

Dividing both sides by − 3 ,

−3x −42
=
−3 −3

x =4
(f) Let the number be x .

x+19
= 8
5

Multiplying both sides by 5

(x+19)×5
= 8 × 5
5

x + 19 = 40

x = 40 − 19( Transposing 19 to R.H.S.)

x = 21

(g) Let the number be x .

5 5x
of this number =
2 2

5x
− 7 = 23
2
5x
= 23 + 7 ( Transposing − 7 to R. H. S)
2

5x
= 30
2

Multiplying both sides by 2

5x×2
= 30 × 2
2

5x = 60

Dividing both sides by 5,

5x 60
=
5 5

x = 12

Page : 91 , Block Name : Exercise 4.4

Q2 Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice
the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the
base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double
century. How many runs did each one score?

Answer. (a) Let the lowest score be l.

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Book : Mathematics Ncert Solutions | Chapter-4 Maths

2 x Lowest marks + 7 = Highest marks
21+7 = 87
21 = 87 - 7 (Transposing 7 to R.H.S.)
21 = 80
Dividing both sides by 2,
2l 80
=
2 2

l = 40

Therefore, the lowest score is 40.
(b) Let the base angles be equal to b.
The sum of all interior angles of a triangle is 180°
b + b + 40° = 180°
2b + 40° = 180°
2b = 180° - 40° = 140° (Transposing 40° to R.H.S.)
Dividing both sides by 2,
∘
2b 140
=
2 2

b = 70°
Therefore, the base angles of the triangle are of 70° measure.
(c) Let Rahul's score be x.
Therefore, Sachin's score = 2x
Rahul's score + Sachin's score = 200 — 2
2x + x = 198
3x = 198
Dividing both sides by 3,
3x 198
=
3 3

x = 66

Rahul's score = 66

sachin's score = 2 × 66 = 132

Page : 91 , Block Name : Exercise 4.4

Q3 Solve the following:
(i) Irfan says that he has 7 marbles more than ve times the marbles Parmit has. Irfan has 37
marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is
Laxmi's age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit
trees. The number of non-fruit trees were two more than three times the number of fruit trees.
What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

Answer. (i) Let Parmit's marbles equal x.
5 times the number of marbles Parmit has = 5x
5x + 7 = 37
5x - 37 - 7 = 30 (Transposing 7 to R.H.S.)
Dividing both sides by 5.

Page 18 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-4 Maths

5x 30
=
5 5

x = 6

(ii) Let Laxmi's age be x years.
3 x Laxmi's age + 4 = Her father's age
3x + 4 = 49
3x = 49 — 4 (Transposing 4 to R.H.S.)
3 x = 4s
Dividing both Sides by 3,
3x 45
=
3 3

x = 15

Therefore, Laxmi's age is 15 years.
(iii) Let the number of fruit trees be x.
3 x Number Of fruit trees + 2 = Number of non-fruit trees
3X+ 2 = 77
3x = 77 — 2 (Transposing 2 to R.H.S.)
3x = 75
Dividing both sides of the equation by 3,
3x 75
=
3 3

x = 25

Therefore, the number of fruit trees was 25.

Page : 91 , Block Name : Exercise 4.4

Q4 Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fty!
To reach a triple century
You still need forty!

Answer. Let the number be x.
(7x + 50) + 40 = 300
7x + 90 = 300
7x = 300 — 90 (Transposing 90 to R.H.S.)
7x = 210
Dividing both sides by 7,
7x 210
=
7 7

x = 30
Therefore, the number is 30.

Page : 91 , Block Name : Exercise 4.4

Page 19 of 19 Aglasem Schools