Simple Linear Equations ICSE Class-7th Concise Selina mathematics Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 Simple Linear Equations for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-12 A , Exe-12 B Exe-12 C and Exe-12 D to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.
Simple Linear Equations ICSE Class-7th Concise Selina mathematics Solutions Chapter-12
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Exercise – 12 A Simple Linear Equations ICSE Class-7th
Solve the following equations :
Question 1.
x + 5 = 10
Answer
x + 5 = 10
⇒ x = 10 – 5 = 5
Question 2.
2 + y = 7
Answer
2 + y = 7
⇒ y = 7 – 2 = 5
Question 3.
a – 2 = 6
Answer
a – 2 = 6
⇒ a = 6 + 2 = 8
Question 4.
x – 5 = 8
Answer
x – 5 = 8
⇒ x = 8 + 5 = 13
Question 5.
5 – d= 12
Answer
5 – d= 12
⇒ – d = 12 – 5 = 7
⇒ d = – 7
Question 6.
3p = 12
Answer
3p = 12
⇒ P = 12/3 = 4
Question 7.
14 = 7m
Answer
14 = 7m
⇒ m = 14/7=2
Question 8.
2x = 0
Answer
2x = 0
⇒ x = 0/2
=0
Question 9.
= 2
Answer
= 2
⇒ x = 2 × 9 = 18
∴ x = 18
Question 10.
= -4
Answer
= -4
y = (- 4) × (- 12)
∴ y = 48
Question 11.
8x-2 =38
Answer
8x – 2 = 38
8x = 38 + 2 = 40
⇒ x = 40/8
=5
∴ x = 5
Question 12.
2x + 5 = 5
Answer
2x + 5 = 5
⇒ 2x = 5 – 5 = 0
x = 0/2= 0
∴ x = 0
Question 13.
5x – 1 = 74
Answer
5x – 1 = 74
⇒ 5x = 74 + 1 = 75
⇒ x = 75/5
=5
Question 14.
14 = 27-x
Answer
14 = 27 – x
⇒ x = 27- 14
⇒ x = 13
∴ x = 13
Question 15.
10 + 6a = 40
Answer
10 + 6a = 40
10 + 6a = 40
⇒ 6a = 40 -10 = 30
⇒ a =30/6 = 5
∴ a= 5
Question 16.
Answer
Question 17.
Answer
⇒a/15=2
a=2×15=30
∴ a = 30
Question 18.
12 = c – 2
Answer
12 = c – 2
⇒ 12 + 2 = c
⇒ 14 = c
∴ c = 14
Question 19.
4 = x- 2.5
Answer
4 = x – 2.5
⇒4 + 2.5 = x
⇒ 6.5 = x
∴ x = 6.5
Question 20.
Answer
Question 21.
Answer
Question 22.
p + 0.02 = 0.08
Answer
p + 0.02 = 0.08
⇒ p = 0.08 – 0.02 = 0.06
∴ p = 0 06
Question 23.
Answer:
Question 24.
– 3x = 15
Answer
– 3x = 15
⇒x=(15/-3)
=-5
∴ x = -5
Question 25.
1.3 b = 39
Answer
1.3b = 39
⇒b= (39/1.3)
= (39×10)/13
=30
∴ b = 30
Question 26.
Answer
⇒ 5n = 20 × 8 = 160
⇒ n = 160/5
=32
∴ n = 32
Question 27.
Answer
⇒ 3m = 21 × 16 = 336
⇒ m = 336/3
=112
∴ m = 112
Question 28.
2a – 3 =5
Answer
2a – 3 =5
⇒2a = 5 +3
⇒ 2a = 8
⇒ a =8/2
= 4
∴ a = 4
Question 29.
3p – 1 = 8
Answer
3p – 1 = 8
⇒3p = 8 + 1 = 9
⇒ p = 9/3
= 3
∴p = 3
Question 30.
9y -7 = 20
Answer
9y -7 = 20
⇒ 9y = 20 + 7 = 27
⇒ y = 27/9
=3
∴ y = 3
Question 31.
2b – 14 = 8
Answer
2b – 14 = 8
⇒ 2b = 8 + 14 = 22
⇒ b = 22/2
=11
∴ b = 11
Question 32.
Answer
⇒ (7/10)x = 41 – 6 = 35
⇒ 7x = 35 × 10 = 350
⇒ x = 350/7
=50
∴ x = 50
Question 33.
Answer
(5/12)m – 12 = 48
⇒ (5/12)m
= 48 + 12 = 60
⇒ 5m = 60 × 12
= 720
⇒ m = 720/5
=144
∴ m = 144
Exercise -12 B ICSE Class-7th Concise Selina mathematics
Question 1.
8y – 4y = 20
Answer
8y – 4y = 20
⇒ 4y = 20
⇒ y = 20/4
=5
∴ y = 5
Question 2.
9b – 4b + 3b = 16
Answer
9b – 4b + 3b = 16
⇒ (9 – 4 + 3)b = 16
⇒ 8b = 16
⇒ b = 16/8
=2
∴ b = 2
Question 3.
5y + 8 = 8y – 18
Answer
5y + 8 = 8y – 18
⇒ 5y – 8y = – 18 – 8
⇒ – 3y = – 26
⇒ y = (-26/-3)
=26/3
∴ y = 8 (2/3)
Question 4.
6 = 7 + 2p -5
Answer
6 = 7 + 2p -5
⇒ – 2p = 7 – 5 – 6
⇒ – 2p = – 4
⇒ p = (-4/-2)
=2
∴ p = 2
Question 5.
8 – 7x = 13x + 8
Answer
8 – 7x = 13x + 8
⇒ – 7x – 13x = 8 – 8
⇒ – 20x = 0
⇒ x = 0/-20
=0
∴ x = 0
Question 6.
4x – 5x + 2x = 28 + 3x
Answer
4x – 5x + 2x = 28 + 3x
⇒ 4x – 5x + 2x – 3x = 28
⇒ 6x – 8x = 28
⇒ x = (28/-2)
=-14
∴ x = – 14
Question 7.
9 + m = 6m + 8 – m
Answer
9 + m = 6m + 8 – m
⇒ m – 6m + m = 8 – 9
⇒ 2m – 6m = -1
⇒ – 4m = -1
∴ m = (-1/-4)
=1/4
Question 8.
24 = y + 2y + 3 + 4y
Answer
24 = y + 2y + 3 + 4y
⇒ 24 – 3 = y + 2y + 4y
⇒ 21 = 7y
⇒ 7y = 21
⇒ y = 21/7
=3
∴ y = 3
Question 9.
19x -+ 13 -12x + 3 = 23
Answer
19x + 13 -12x + 3 = 23
⇒ 19x – 12x = 23 – 13 – 3
⇒ 7x = 23 – 16 = 7
⇒ x = 7/7
=1
∴ x = 1
Question 10.
6b + 40 = – 100 – b
Answer
6b + 40 = – 100 – b
⇒ 6b + b = – 100 – 40
⇒ 7b = – 140
⇒ b = (-140/7)
=-20
∴ b = – 20
Question 11.
6 – 5m – 1 + 3m = 0
Answer
6 – 5m – 1 + 3m = 0
⇒ – 5m + 3m = -6 + 1
⇒ – 2m = – 5
⇒ m = (-5/-2)=5/2
∴ m = 5/2
=2 (1/2)
Question 12.
0.4x – 1.2 = 0.3x + 0.6
Answer
0.4x – 1.2 = 0.3x + 0.6
⇒ 0.4x – 0.3x = 0.6 + 1.2
⇒ 0.1x = 1.8
⇒ (1/10)x
=18/10
⇒ x = (18/10)×(10/1)=18
∴ x = 18
Question 13.
6(x+4) = 36
Answer
6(x+4) = 36
⇒ 6x + 24 = 36
⇒ 6x = 36 – 24 = 12
⇒ x = 12/6
=2
∴ x = 2
Question 14.
9 ( a+ 5) + 2 = 11
Answer
9 (a+ 5) + 2 = 11
⇒ 9a + 45 + 2 = 11
⇒ 9a = 11 – 45 – 2
⇒ 9a = 11 – 47 = – 36
⇒ a = (-36/9)=-4
∴ a = – 4
Question 15.
4 ( x- 2 ) = 12
Answer
4 (x – 2) = 12
⇒ 4x – 8 = 12
⇒ 4x = 12 + 8 = 20
⇒ x = (20/4)
=5
∴ x = 5
Question 16.
-3 (a- 6 ) = 24
Answer:
-3 (a – 6) = 24
⇒ – 3a + 18 = 24
⇒ -3a = 24 -18 = 6
⇒ a = (6/-3)
=-2
∴ a = -2
Question 17.
7 ( x-2) = 2 (2x -4)
Answer:
7 (x – 2) = 2 (2x – 4)
⇒ 7x – 14 = 4x – 8
⇒ 7x – 4x = – 8 + 14
⇒ 3x = 6
⇒ x = 6/3
=2
∴ x = 2
Question 18.
(x-4) (2x +3 ) = 2x²
Answer:
(x – 4) (2x + 3 ) = 2x²
⇒ x (2x + 3) – 4 (2x + 3) = 2x²
⇒ 2x² + 3x – 8x – 12 = 2x²
⇒ 2x² + 3x – 8x – 2x² = 12
⇒ – 5x = 12
⇒ x = (12/-5)
=-12/5
∴ x = -2 (2/5)
Question 19.
21 – 3 ( b-7 ) = b+ 20
Answer:
21 – 3 (b – 7) = b+ 20
⇒ 21 – 3b + 21 = b + 20
⇒ – 3b + 42 = b + 20
⇒ – 3b – b = 20 – 42
⇒ – 4b = – 22
⇒ b = (-22/-4)
=11/2
∴ b = 11/2
=5 (1/2)
Question 20.
x (x +5 ) = x² +x + 32
Answer:
x (x + 5) = x² +x + 32
⇒ x² + 5x = x² + x + 32
⇒ x² + 5x – x² – x = 32
⇒ 4x = 32
⇒ x = 32/4
=8
∴ x = 8
Simple Linear Equations Exe- 12 C Selina ICSE Maths Class-7
Question 1.
Answer:
(x/2)+x=9
(x/2)+(x/1)=9
(x + 2x)/2=9
⇒ x + 2x = 2 × 9
⇒ 3x = 18
⇒ x = 18/3
=6
∴ x = 6
Question 2.
Answer:
∴ x = 15
Question 3.
Answer:
⇒ 3x + 16x = 38 × 4
∴ 3x + 16x = 152
⇒ 19x = 152
⇒ x = 152/19
=8
∴ x = 8
Question 4.
Answer:
⇒ (5x + 2x)/10=14
⇒ 5x + 2x = 14 × 10
⇒ 5x + 2x = 140
⇒ 7x = 140
⇒ x = 140/7
=20
∴ x = 20
Question 5.
Answer:
⇒ (4x – 3x)/12
=2
⇒ 4x – 3x = 2 × 12
⇒ x = 24
Question 6.
Answer:
⇒ (y/1)+(y/2)
=(7/4)-(y/4)
⇒ (4y + 2y = 7 – y)/4
⇒ 4y + 2y = 7 – y
⇒ 4y + 2y ⇒ 7y = 7
⇒ y = 7/7
=1
∴ y = 1
Question 7.
Answer:
Question 8.
Answer:
⇒ 2m + 3m – 4m = 10
⇒ 5m – 4m = 10
⇒ m = 10
Question 9.
Answer:
⇒ (8x + 6x – 9x)/12=1
⇒ 8x + 6x – 9x = 12 × 1
⇒ 8x + 6x – 9x = 12
⇒ 14x – 9x = 12
⇒ 5x = 12
⇒ x = (12/5)
=2 (2/5)
Question 10.
Answer:
⇒ (9a + 2a)12=66
⇒ (9a+2a)=66×12
⇒ 9a + 2a = 792
⇒ 11a = 792
⇒ a = 792/11
=72
⇒ a = 72
Question 11.
Answer:
⇒ (10p – 3p)/15=35
⇒ 10p – 3p = 35 × 15
⇒ 10p – 3p = 525
⇒ 7p = 525
⇒ p = 525/7
=75
⇒ p = 75
Question 12.
0.6a +0.2a = 0.4 a +8
Answer:
0.6a +0.2a = 0.4 a + 8
⇒ (6/10)a+(2/10)a
=(4/10)a+(8/1)
⇒(6a + 2a)/10
=4a + 80
⇒ 6a + 2a = 4a + 80
⇒ 6a + 2a – 4a = 80
⇒ 4a = 80
⇒ a = 80/4
=20
∴ a = 20
Question 13.
p + 104p= 48
Answer:
p + 1.4p = 48
⇒ p+(14/10)p=48
⇒ (10p + 14p)/10=48
⇒ 10 P + 14P = 48 × 10
⇒ 10p + 14p = 480
⇒ 24p = 480
⇒ p = 480/24
=20
∴ p = 20
Question 14.
10% of x = 20
Answer:
10% of x = 20
⇒ (10/100)×x=20
⇒ (x/10)=20
⇒ x = 20 × 10 = 200
∴ x = 200
Question 15.
y + 20% of y = 18
Answer:
y + 20% of y = 18
⇒ y + (20/100)×y=18
⇒ (100y+20y)/100=18
⇒ 100y + 20y = 18 × 100
⇒ 100y+ 20y = 1800
⇒ 120y = 1800
⇒ y = 1800/120
=15
∴ y = 15
Question 16.
x – 13% of x = 35
Answer:
x – 30% of x = 35
⇒ x (-30/100)×x=35
⇒ (100x-30x)/100=35
⇒ 100x – 30y = 35 × 100
⇒ 100x – 30x = 3500
⇒ 70x = 3500
⇒ x = 3500/70
=50
∴ x = 50
Question 17.
Answer:
⇒ (3x+12x+2x)/6=7
⇒ 3x + 12 + 2x = 7 × 6
⇒ 3x + 12 + 2x = 42
⇒ 5x = 42 – 12 = 30
⇒ x = 30/5=6
∴ x = 6
Question 18.
Answer:
⇒ (4y+8+3y+15)/12=6
⇒ 4y + 8 + 3y + 15 = 6 × 12
⇒ 7y + 23 = 72
⇒ 7y = 72 – 23 = 49
⇒ y = 49/7=7
∴ y = 7
Question 19.
Answer:
⇒ (12a-8-7a+14)/28=2
⇒ 12a – 8 – 7a + 14 = 2 × 28
⇒ 12a – 8 – 7a + 14 = 56
⇒ 12a – 7a + 14 – 8 = 56
⇒ 5a + 6 = 56
⇒ 5a = 56 – 6 = 50
⇒ a = 50/5=10
∴ a = 10
Question 20.
Answer:
⇒ (3(x+5)-2(x-2))/6=4
⇒ 3x + 15 – 2x + 4 = 4 × 6
⇒ 3x + 15 – 2x + 4 = 24
⇒ 3x – 2x = 24 – 15 – 4
⇒ x = 24 – 19 = 5
∴ x = 5
Question 21.
Answer:
⇒ (6(x-1)-4(x-2)-3(x-3))/12=0
⇒ 6(x – 1) – 4(x – 2) – 3(x – 3) = 0
∴ 6x – 6 – 4x + 8 – 3x + 9 = 0
∴ 6x – 4x – 3x – 6 + 8 + 9 = 0
∴ 6x – 7x = 6 – 8 – 9
⇒ – x = 6 – 17 = – 11
∴ x = 11
Question 22.
Answer
⇒ (35(x+1)+21(x+4)=15(x-4))/105
…(LCM of 3,5,7 = 105)
⇒ 35(x + 1) + 21 (x + 4) = 15(x – 4)
⇒ 35x + 35 + 21x + 84 = 15x – 60
⇒ 35x + 21x – 15x = -60 – 35 – 84
⇒ 56x – 15x = – (60 + 35 + 84)
⇒ 41x = – 179
∴ x = (-179/41)
=-4 (15/21)
Question 23.
15 – 2 (5-3x ) = 4 ( x-3 ) + 13
Answer
15 – 2 (5 – 3x) = 4 (x – 3) + 13
⇒ 15 – 10 + 6x = 4x – 12 + 13
⇒ 6x – 4x = – 12 + 13 – 15 + 10
⇒ 2x = 23 – 27 = – 4
⇒ x =(-4/2)=-2
Hence x = -2
Question 24.
Answer
⇒ (2x+1)/(3x-2)=5/4
By cross multiplication
(3x – 2)× 5 = 4 (2x + 1)
⇒ 15x – 10 = 8x + 4
⇒ 15x – 8x = 4 + 10
⇒ 7x = 14
⇒ x = 14/7
=2
∴ x = 2
Question 25.
21 – 3 (x – 7) = x + 20
Answer
21 – 3 (x – 7) = x + 20
⇒ 21 – 3x + 21 = x + 20
⇒ 42 – 3x = x + 20
⇒ 42 – 20 = x + 3x
⇒ 4x = 22
⇒ x = 22/4
=11/2=5 (1/2)
∴ x = 5 (1/2)
Question 26.
Answer
⇒ (12x-8-7x+14)/28=2
⇒ (5x+6)/28=2
⇒ 5x + 6 = 28 × 2
⇒ 5x + 6 = 56
⇒ 5x = 56 – 6 = 50
⇒ x = 50/5
=10
∴ x = 10
Question 27.
Answer
⇒ (2x-3)/3-(x-5)/1=x/3
⇒ (2x-3-3x+15=x)/3
⇒ – x + 12 = x
⇒ 12 = x + x
⇒ 2x = 12
∴ x = 122=6
Question 28.
Answer
⇒ (5x-20=5x+15+7x+283)/5
⇒ 5x – 5x – 7x = 15 + 28 + 20
⇒ – 7x = 63
⇒ x = 63-7=-9
∴ x = – 9
Question 29.
Answer
⇒ (6x-6-10x=30-15x+30)/30
⇒ – 4x – 6 = 60 – 15x
⇒ 15x – 4x = 60 + 6
⇒ 11x = 66
⇒ x = 66/11=6
∴ x = 6
Question 30.
2x + 20% of x = 12.1
Answer
2x + 20% of x = 12.1
Simple Linear Equations Exercise 12 D – Selina Concise Mathematics Class 7 ICSE Solutions
Question 1.
One-fifth of a number is 5, find the number.
Answer
The number = x
According to the condition
(1/5)x=5
⇒ x = 5 x 5
⇒ x = 25
∴ Number = 25
Question 2.
Six times a number is 72, find the number.
Answer
The number = x
According to the condition
6x = 72
⇒ x = 72/6
⇒ x= 12
∴ Number = 12
Question 3.
If 15 is added to a number, the result is 69, find the number.
Answer
The number = x
According to the condition
x + 15 = 69
⇒ x = 69 – 15 x = 54
∴Number = 54
Question 4.
The sum of twice a number and 4 is 80, find the number.
Answer
The number = x
According to the condition
2x + 4 = 80
⇒ 2x = 80 – 4
⇒ 2x = 76
⇒ x = 76/2
=38
Number = 38
Question 5.
The difference between a number and one- fourth of itself is 24, find the number.
Answer
Let the number = x
According to the condition
x-(1/4)x=24
⇒(4x – x)/x=24
⇒3x/4=24
⇒x=24×(4/3)
⇒x=8×4
⇒ x = 32
∴ Number = 32
Question 6.
Find a number whose one-third part exceeds its one-fifth part by 20.
Answer
Let the number = x
According to the condition
(1/3)x-(1/5)x=20
⇒(5x – 3x)/15=20 …(LCM of 3, and 5 = 15)
⇒2x/15=20
⇒x=(20×15)/2
⇒x=150
∴ Number = 150
Question 7.
A number is as much greater than 35 as is less than 53. Find the number.
Answer
The number = x
According to the condition
x – 35 = 53 – x
⇒ x + x = 53 + 35
⇒ 2x = 88
⇒ x = 88/2
=44
∴ Number = 44
Question 8.
The sum of two numbers is 18. If one is twice the other, find the numbers.
Answer
The first number = x
and the second number = y
According to the condition
x + y= 18 …(i)
and x = 27 ….(ii)
Substitute the eq. (ii) in eq. (i), we get
2y + y= 18
x= 2y = 18
⇒ 3y= 18
⇒ y = 18/3
=6
Now, substitute the value of y in eq. (ii), we get
x = 2 x 6= 12
∴ The two numbers are 12, 6
Question 9.
A number is 15 more than the other. The sum of of the two numbers is 195. Find the numbers.
Answer
The First number = x
and the Second number = y
According to the condition
x = y+ 15 …(i)
x + 7=195 …(ii)
Substitute the eq. (i) in eq. (ii), we get
y + 15 + 7 = 195
⇒ 2y = 195 – 15
⇒ y = 180/2
=90
Now, substitute the value of y in eq. (i), we get
x = 90+ 15 = 105
∴ The two numbers are 105 and 90
Question 10.
The sum of three consecutive even numbers is 54. Find the numbers.
Answer
The first even number = x
second even number = x + 2
and third even number = x + 4
According to the condition,
x + x + 2 + x + 4 = 54
⇒ 3x + 6 = 54
⇒ 3x = 54 – 6
⇒ x = 483=16
∴ First even number = 16
Second even number = 16 + 2 = 18
and third even number = 16 + 4 = 20
Question 11.
The sum of three consecutive odd numbers is 63. Find the numbers.
Answer
The first odd number = x
second odd number = x + 2
and third odd number = x + 4
According to the condition,
x+ x + 2 + x+4 = 63
3x + 6 = 63 ⇒ 3x = 63 – 6
⇒3x = 57 ⇒ x =57/3
=19
∴ First odd number = 19
Second odd number = 19 + 2 = 21
third odd number = 19 + 4 = 23
Question 12.
A man has ₹ x from which he spends ₹6. If twice of the money left with him is ₹86, find x.
Answer
The total amount be x
According to the condition
2x = 86
⇒ x = 86/2
⇒ x = 43
Amount spent by him = 6
∴ Total money he have = ₹43 + ₹6 = ₹49
Question 13.
A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.
Answer
The present age of the son = x years
Present age of the father = 4x years
After 20 years,
Son’s age will be (x + 20) years
and Father’s age will be (4x + 20) years
According to the condition,
4x + 20 = 2 (x + 20)
4x + 20 = 2x + 40
4x – 2x = 40 – 20
2x = 20
⇒ x = 10
∴ Present age of the son = 10 years and Present age of the father = 4×10 years = 40 years
Question 14.
If 5 is subtracted from three times a number, the result is 16. Find the number.
Answer
The number = x
According to the condition,
3x – 5 = 16
⇒ 3x = 16 + 5
⇒ 3x = 21
⇒ x = 213
⇒ x = 7
∴The number = 7
Question 15.
Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.
Answer
The first consecutive number = x,
Second consecutive number = x + 1
and Third consecutive number = x + 2
According to the condition,
x + x + 1 = 15 + x + 2
⇒ 2x + 1 = 17 +x
⇒ 2x -x = 17 – 1
⇒ x= 16
∴ The first consecutive number = 16
Second consecutive number =16+1 = 17
Third consecutive number =16 + 2=18
Question 16.
The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
Answer
The smallest number = x
and the largest number = y
According to the condition,
y – x = 7 …(i)
and 6x + y = 77 ….(ii)
From eq. (i)
y = 7 + x …(iii)
Substitute the eq. (iii) in eq. (ii)
6x + 7 + x = 77
⇒ 7x = 77 – 7
⇒ x = 70/7
=10
Now, substitute the value of x in eq. (iii)
y = 7+ 10= 17
∴ The smallest number 10 and the largest number is 17.
Question 17.
The length of a rectangular plot exceeds its breadth by 5 metre. If the perimeter of the plot is 142 metres, find the length and the breadth of the plot.
Answer
The length of a rectangular plot = x
and the breadth of a rectangular plot = y
According to the condition,
x = y + 5 …(i)
and 2(x + y) = 142
⇒ x + y = 142/2
=71
⇒ x + y = 71 …(ii)
Now, substitute the value of eq. (i) in eq (ii)
y + 5 + y = 71
⇒ 2y = 71 – 5
⇒ y = 66/2=33
Now, put the value of y in eq. (i)
x = 33 + 5 = 38
∴ The length of rectangular plot is 38 m and breadth is 33 m.
Question 18.
The numerator of a fraction is four less than its denominator. If 1 is added to both, is numerator and denominator, the fraction becomes Find the fraction.
Answer
The numerator of a fraction = x
and the denominator of a fraction = y
According to the condition,
x = y – 4 …..(i)
and (x + 1)/(y + 1)=12
⇒ 2(x + 1) = y + 1
⇒ 2x + 2 = y + 1
⇒ 2x – y = -1 …(ii)
Substitute the eq.(i) in eq.(ii)
2(y – 4) – y = -1
2y – 8 – y = – 1
y = – 1 + 8
y = 7
Now, put the value of y in eq.(i), we get
x = 7 – 4 = 3
∴ The numerator of a fraction is 3
and denominator is 7 and the fraction is 3/7
Question 19.
A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time. Find their present ages.
Answer
The present age of the son = x years
and the present age of the father = 3x years
After 12 years,
Son’s age will be (x + 12) years
and father’s age will be (3x + 12) years
According to the condition,
3x + 12 = 2 (x + 12)
3x + 12 = 2x+ 24
3x – 2x = 24 – 12
x = 12
∴Present age of the son = 12 years
and Present age of the father = 3×12 years
= 36 years
Question 20.
A sum of ₹ 500 is in the form of notes of denominations of ₹ 5 and₹ 10. If the total number of notes is 90, find the number of notes of each type.
Answer
The number of ₹ 5 notes = x
∴ The number of ₹10 notes = 90 – x
Value of ₹10 notes = x ×₹ 5 = ₹ 3x
and value of ₹10 notes = (90 – x) x ₹ 10 =₹(900 – 10x)
∴Total value of all the notes = ₹ 500
∴5x+ (900- 10x) = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900
⇒ x = 400/5
⇒ x = 80
∴ The number of ₹5 notes = x = 80
and the number of ₹10 notes = 90 – x
= 90 – 80 = 10
— End of Simple Linear Equations Solutions :–
Return to – Concise Selina Maths Solutions for ICSE Class -7
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