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Number Line Prime Class 7 Solutions Chapter 7 Ratio, Proportion and Unitary Method
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 7, Ratio, Proportion and Unitary Method. Here students can easily find step by step solutions of all the problems for Ratio, Proportion and Unitary Method, Exercise 7A, 7B and 7C Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions. Here in this post all the solutions are based on latest Syllabus.
Ratio, Proportion and Unitary Method Exercise 7A Solution :
Question no – (1)
Solution :
(a) 3 hours and 45 minutes
3 hour = 3 × 60 min = 180 min
∴ Ratio 160/45
= 4 : 1
(b) 12 kg and 18 kg
= 12/18
= 2 : 3
(c) 15 m and 750 cm
= 15/750
= 1 : 50
(d) ₹15 and 55 paise
= 15 × 1000/2000
= 15 : 2
Question no – (2)
Solution :
(a) 5 : 6 and 3 : 5
Given, 5/6
= 5 × 5/6 × 5
= 25/30,
Given, 3/5
= 3 × 6/5 × 6
= 18/30
∴ 25/30 > 18/30 > = 5/6 > 3/5
= 5:6 > 3:5
(b) 4:9 and 5:7
Given, 4/9
= 4 × 7/9 ×7
=28/63,
Given, 5/7
= 5 × 9/7 × 9
= 45/63
∴ 4/9 < 5/7 = 4 : 9 < 5 : 7
(c) 10:15 and 20 : 15
10 : 15 < 20 : 15
(d) 12 : 18 and 16 : 20
Given, 12/18
= 2/3
= 2 × 5/3 × 5
= 10/15,
16/20
= 4 × 3/5 × 3
= 12/15
∴ 12/18 < 16/20
= 12 : 18 < 16 : 20
Question no – (3)
Solution :
Sum of the ferms of the ration
= ( 3 + 4 + 5 )
= 12
∴ 1st = 4800 × 3/12
= 1200
2nd = 4800 × 4/12
= 1600
3 rd = 4800 × 5/12
= 2000
Question no – (4)
Solution :
Sum of the terms of the ratio,
= 7 + 11
= 18
∴ Mangoes in the basket,
= 720 × 7/18
= 280
Question no – (5)
Solution :
Let, x Lit oil was taken out.
∴ Oil in the tank
= ( 150 – x 0) lit
Decreased ration = 6 : 4
∴ 4x = 150 – x
= 4x + 1
∴ 6x – 4x = 150 – x
2x + x = 150
3x = 150
x = 50
∴ Remaining oil,
= (150 – 50)
= 100 L
Question no – (6)
Solution :
Let, we add x
∴ 3/5 + x
∴ x = 4/5 – 3/5
∴ x = 1/5
Question no – (7)
Solution :
(a) 3: 2, 5:3, 6:5, 4:3, 5:4
3/2 , 5/3, 6/5, 4/3, 5/4
Given, 3/2
= 3 × 30/2 × 30
= 90/60,
Given, 5/3
= 5 × 20/3 × 20
= 100/60,
Given, 6/5
= 6 ×12/5 × 12
= 72/60,
Given, 4/3
= 4 × 20/3 × 20
= 80/60,
Given,5/4
= 5 × 15/4 × 15
= 75/60
∴ Ascending order,
6/5, 5/4, 4/3, 3/2, 5/3
= 6:5, 5:4, 4:3, 3:2, 5:3
(b) 3:4, 4:7, 7:9, 3:7
LCM of 4, 3, 5, 3, 4 – 60
Given, 3/4
= 3 × 15/4 × 15
= 45/60,
Given, 5/3
= 5 × 20/3 × 20
= 100/60,
Given, 6/5
= 6 × 12/5 × 12
= 72/60,
Given, 4/3
= 4 × 20/3 × 20
= 80/60,
Given, 5/4
= 5 × 15/4 × 15
= 75/60
∴ Ascending order
3:4, 6:5, 5:4, 4:3, 5:3
Question no – (8)
Solution :
(a) First angel : second angle
1st angel – 60°,
2nd angel – 76°
= 60°/76°
= 15:19
(b) Second angle : Third angle
Third angel = 180 – (60°+76°)
= 180° – 36°
= 44°
∴ 76/44
= 19 : 11
(c) Third angel : Sum of the first two angles
= 44/(60 + 76)
= 44/136
= 11:34
Question no – (9)
Solution :
= 1 1/2 : 4/3
= 3/2 : 4/3
= 9 : 8
∴ 85 × 9/17
= 45,
85 × 8/17 = 40
Question no – (10)
Solution :
∴ 3x + 4x + 5x = 384
∴ 12x = 348
∴ x = 384/12
∴ x = 32
∴ Numbers are – 96, 128, 160
∴ Smallest numbers – 96
Question no – (11)
Solution :
Let the length of rectangle – 3x
Breadth of rectangle – 2x
And area is = 216 sqm
∴ 3x × 2x = 216
∴ 6x2 = 216
∴ x2 = 216/6
∴ x = √36
∴ x = 6 m2
∴ Length – 18m, breadth – 12m
∴ Perimeter = 2(18+12)
= 2 × 30
= 60 m
Ratio, Proportion and Unitary Method Exercise 7B Solution :
Question no – (1)
Solution :
(a) 18, 35, 25, 45
∴ 18/35 and 25/45 are not proportional
(b) 16, 20 24, 30
∴ 16/20 = 4/5 and 24 = 4/5 are proportional
(c) 15, 36, 25, 48
∴ 15/36 = 5/12 and 25/48 are not proportional
(d) 12, 18, 19, 24
∴ 12/18 = 2/3 and 19/24 are not proportional
Question no – (2)
Solution :
(a) 9 : x : : x : 16
∴ 9/x = x/16
∴ x2 = 16×9
∴ x = √16 × 9
∴ x = 4 × 3
∴ x = 12
(b) x : 16 : : 16 : 64
∴ x/16 = 16/64
∴ x = 4
(c) 3 : 9 : : 9 : x
∴ 3/9 = 9/x
∴ x = 27
Question no – (3)
Solution :
(a) 4 and 121
∴ 4/x = x/121
∴ x2 = 4 × 121
∴ x = √ 4×121
∴ x = 2 × 11
∴ x = 22
(b) 9 and 25
9 : x : : x : 25
∴ x2 = 9 × 25
∴ x = √9 × 25
∴ x = 3 × 5
∴ x = 15
Question no – (4)
Solution :
Let breadth – x
∴ 11: 5 : : 44 : x
∴ 11/5 = 44/x
∴ x = 20 feet
Question no – (5)
Solution :
(a) 8, 20, 12
∴ 5/20 = 12/x
∴ x = 12 × 20/8
∴ x = 30
(b) 8, 10, 12
∴ 8/10 = 12/x
∴ x = 12 × 10/8
∴ x = 15
Question no – (6)
Solution :
(a) 12 and 18
12 : 18 : : 18 : x
∴ 12/18 = 18/x
∴ x = 18 × 18/122
∴ x = 27
(b) 16 and 36
16 : 36 : : 36 : x
∴ 16/36 = 36/x
∴ x = 36 × 36/16
∴ x = 81
(c) 1.4 and 4.2
1.4 : 4.2 : : 4.2 : x
∴ 1.4/4.2 = 4.2/x × 10
∴ x = 42 × 42/14 × 10
∴ x = 126/10
∴ x = 12.6
(d) 1 1/4 and 5
∴ 5/4 × 1/5 = 5/x
∴ x = 20
Question no – (7)
Solution :
(a) A : B = 2 : 5 and B : C = 4 : 6.
Given, A/B = 2/5
∴ A/B = 2 × 4/5 × 4
∴ A/B = 8/20,
And
Given: B/C = 4/6
∴ B/C = 4 × 5/6 × 5
∴ B/C = 20/30
∴ A : B : C = 8 : 20 : 30 = 4 : 10 : 15
(b) A : B = 5 : 7 and B : C = 2 : 9.
Given, A/B = 5/7
∴ A/B = 5 × 2/7 × 2
∴ A/B= 10/14
And
∴ B/C = 2/9
∴ B/C = 2 × 7/9 × 7
∴ B/C = 14/63
∴ A : B : C = 101 : 14 : 63
Question no – (8)
Solution :
∴ A’s heights = 9x
∴ B’s heights = 10x
∴ 9x = 144
∴ x = 144/9
∴ x = 16
∴ B’s heights
= 10 × 16
= 160 cm
Therefore, B’s height will be 160 cm
Ratio, Proportion and Unitary Method Exercise 7C Solution :
Question no – (1)
Solution :
∴ A total wage,
= 53 × 7576
= 401528
∴ One financial year,
= 401528 × 12
= 4818336
1 worker wage in a month = 7576
∴ Annual wage of 1 worker
= 7575 × 12
= 90912
Question no – (2)
Solution :
∴ 1 employee’s salary per month,
= 710582/157
= 4526
∴ Annual salary of 1 employee,
= 4526 × 12
= 54312
Question no – (3)
Solution :
(a) How much dose he earn in 12 months
∴ Rahul earns in one month = 21000/7
= 3000
∴ In 12 months he earn = 3000 × 12
= 36000
(b) In how many months will he earn ₹ 33,000
∴ 33000/3000
= 11 months.
Therefore, he will earn 33,000 in 11 months.
Next Chapter Solution :
