RS Aggarwal Solutions for Class 6 Maths Chapter 10–Ratio, Proportion and Unitary Method

Class 6: Maths Chapter 10 solutions. Complete Class 6 Maths Chapter 10 Notes.

RS Aggarwal Solutions for Class 6 Maths Chapter 10–Ratio, Proportion and Unitary Method

RS Aggarwal 6th Maths Chapter 10, Class 6 Maths Chapter 10 solutions

Ex 10A Solutions

Question 1.
Solution:
(i) 24 to 56
= 2456
= 24÷856÷8

Question 2.
Solution:
(i) 36 : 90
= 3690
= 36÷1890÷18

Question 3.
Solution:
(i) The given ratio = Rs. 6.30 : Rs. 16.80
= Rs.6.30Rs.16.80
= 6301680

Question 4.
Solution:
Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300

Question 5.
Solution:
Rohit monthly earnings = Rs. 15300
and his savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,

Question 6.
Solution:
Let the number of male and female workers in the mill be 5x and 3x respectively. Then,
5x = 115
=> 5×5 = 1155
(Dividing both sides by 5)
=> x = 23
Number of female workers in the mill
= 3x
= 3 x 23 = 69.

Question 7.
Solution:
Let the number of boys and girls in the school be 9x and 5x respectively.
According to the question,
9x + 5x = 44
=> 14x = 448
=> 14×14 = 44814
(Dividing both sides by 14)
=> x = 32.
Number of girls =5x
= 5 x 32
= 160

Question 8.
Solution:
Total amount = Rs. 1575
Ratio in Kamal and Madhu’s share = 7 : 2
Sum of ratios = 7 + 2 = 9

Question 9.
Solution:
Total amount = Rs. 3450
Ratio in A, B and C shares = 3 : 5 : 7
Sum of share = 3 + 5 + 7 = 15

Question 10.
Solution:
Let the numbers be 11x and 12x.
Then. 11x + 12x = 460
=> 23x = 460

Question 11.
Solution:
Length of line segment = 35 cm
Ratio = 4 : 3
Sum of ratio = 4 + 3 = 7

Question 12.
Solution:
Total bulbs produced per day = 630
Out of every 10 bulbs, defective bulb = 1
Out of every 10 bulbs, lighting bulbs = 10 – 1 = 9

Question 13.
Solution:
Price of 20 pencils = Rs. 96
(1 score = 20 pencils)
Price of 1 pencil = Rs. (96 ÷ 20)
= Rs. 4.80
Price of 12 ball pens = Rs. 50.40
(1 dozen = 12)
Price of 1 ball pen = Rs. (50.40 ÷ 12)
= Rs. 4.20.
Ratio of the price of a pencil to that of a ball pen = Rs. 4.80 : Rs. 4.20
= 480 paise : 420 paise
= 480 : 420
= 48 : 42
= 8 : 7.
Required ratio = 8 : 7.

Question 14.
Solution:
It is given that the ratio of the length of a field to its width is 5 : 3.
If the width of the field is 3 metres then length = 5 metres.
If the width of the field is 1 metres than length = 53 metres.
If the width of the field is 42 metres then length
= 53 x 42 metres
= 5 x 14 metres
= 70 metres.

Question 15.
Solution:
Ratio in income and savings of a family = 11 : 2
But Total savings = Rs. 1520
Let income = x
11 : 2 = x : 1520
=> x = 11×15202 = 11 x 760
= Rs 8360
Expenditure = total income – savings
= Rs 8360 – 1520
= Rs 6840

Question 16.
Solution:
Ratio in income and expenditure = 7 : 6
Total income = Rs. 14000
Let expenditure = x, then
7 : 6 :: 14000 : x
=>x = 6×140007 = Rs. 12000
Savings = Total income – Expenditure
= Rs. 14000 – 12000
= Rs. 2000

Question 17.
Solution:
It is given that the ratio of zinc and copper in an alloy is 7 : 9.
If the weight of zinc in the alloy is 7 kg then the weight of copper in the alloy is 9 kg.

Question 18.
Solution:
A bus covers in 2 hours = 128 km
128 It will cover in 1 hour = 1282 = 64 km
A train cover in 3 hours = 240 km
It will cover in 1 hour = 2403
= 80 km
Ratio in their speeds = 64: 80
= 4 : 5
{Dividing by 16, the LCM of 64, 80}

Question 19.
Solution:
(i) (3 : 4) or (9 : 16)
LCM of 4, 16 = 16

Question 20.
Solution:

Ex 10B Solutions

Question 1.
Solution:
(i) 4, 6, 8, 12
If it is in proportion, then
If ad = bc if 4 x 12 = 6 x 8
If 48 = 48
Which is true
∴ 4, 6, 8, 12 are in proportion
(ii) 7, 42, 13, 78
7 : 42 :: 13 : 78
If ad = bc if 7 x 78 = 42 x 13
If 546 = 546
Which is true
∴ 7, 42, 13, 78 are in proportion
(iii) 33, 121, 9, 96 or 33 : 121 :: 9 : 96
are in proportion
If ad = bc
If 33 x 96 = 121 x 9
If 3168 = 1089
Which is not true
∴ 33, 121,9, 96 are not in proportion
(iv) 22, 33, 42, 63 or 22 : 33 :: 42 : 63
are in proportion
If ad = bc
If 22 x 63 = 33 x 42
If 1386 = 1386
Which is true
∴ 22, 33, 42, 63 are in proportion
(v) 32, 48, 70, 210 or 32 : 48 :: 70 : 210
are in proportion
If ad = bc
If 32 x 210 = 48 x 70
If 6720 = 3360
Which is not true
∴ 32, 48, 70, 210 are not in proportion
(vi) 150, 200, 250, 300 or
150 : 200 :: 250 : 300 are in proportion
If ad = bc if 150 x 300 = 200 x 250
If 45000 = 50000
Which is not true
∴ 150, 200,250, 300 are not in proportion

Question 2.
Solution:
(i) We have 60 : 105 :: 84 : 147
Product of means = 105 x 84 = 8820
Product of extremes = 60 x 147 = 8820
∴ Product of means = Product of extremes
Hence 60 : 105 :: 84 : 147 is verified.
(ii) We have 91 : 104 :: 119 : 136
Product of means = 104 x 119 = 12376
Product of extremes = 91 x 136 = 12376
Product of means = Product of extremes
Hence 91 : 104 :: 119 : 136 is verified.
(iii) We have 108 : 72 :: 129 : 86
Product of means = 72 x 129 = 9288
Product of extremes = 108 x 86 = 9288
Product of means = Product of extremes
Hence 108 : 72 :: 129 : 86 is verified.
(iv) We have 39 : 65 :: 141 : 235
Product of means = 65 x 141 = 9165
Product of extremes = 39 x 235 = 9165
∴ Product of means = Product of extremes
Hence 39 : 65 :: 141 : 235 is verified.

Question 3.
Solution:
(i) We have 55 : 11 :: x : 6
Product of means = 11 × x = 11x
Product of extremes = 55 x 6 = 330

Question 4.
Solution:
(i) We have, 51 : 68 = 5168
= 34

Question 5.
Solution:
(i) 25 cm : 1 m and Rs. 40 : Rs. 160
= 25cm1000cm = 14,
Rs.40Rs.160 = 14

Question 6.
Solution:
Let the third term be x.
Then 51 : 68 :: x : 108
Now, product of means = x × 68
And, product of extremes = 51 × 108
x × 68 = 51 × 108
=> x = 51×10868
= 3 × 27 = 81
x = 81
Hence the third term of the given proportion is 81

Question 7.
Solution:
1st term =12, third term = 8 and fourth term = 14
Let 2nd term = x, then

Question 8.
Solution:
(i) The given numbers 48, 60, 75 are in
continued proportion if 48 : 60 :: 60 : 75.
Now, product of means = 60 x 60 = 3600
And, product of extremes = 48 x 75 = 3600
∴ Product of means = Product of extremes
So, 48 : 60 :: 60 : 75
Hence, the numbers 48, 60, 75 are in continued proportion.
(ii) The given numbers 36, 90, 225 are in
continued proportion of 36 : 90 :: 90 : 225
Now, product of means = 90 x 90 = 8100
And, product of extremes = 36 x 225 = 8100
∴ Product of means = Product of extremes
So, 36 : 90 :: 90 : 225
Hence, the numbers 36, 90, 225 are in continued proportion.
(iii)The given numbers 16, 84, 441 are in
continued proportion if 16 : 84 :: 84 : 441.
Now, product of means = 84 x 84 = 7056
And, product of extremes = 16 x 441 = 7056
Product of means = Product of extremes.
So, 16 : 845 :: 84 : 441
Hence 16, 84, 441 are in continued proportion.
(iv) The given numbers 27, 36, 48 are
in continued proportion if 27 : 36 :: 36 : 48
Now, product of means = 36 x 36 = 1296
And, product of extremes = 27 x 48 = 1296
∴ Product of means = Product of extremes.
So, 27 : 36 :: 36 : 48
Hence, the numbers 27, 36, 48 are in continued proportional.

Question 9.
Solution:
It is given that 9, x, x, 49 are in proportion, that is, 9 : x :: x : 49
∴ Product of means = Product of extremes

Question 10.
Solution:
Let the height of the pole be x metres.

Question 11.
Solution:
5 : 3 :: x : 6

Question 1.
Solution:
Cost of 14 m of cloth = Rs. 1890
Cost of 1 m = Rs. 189014
and cost of 6 m = Rs. 1890×614
= Rs. 135 x 6
= Rs. 810

Question 2.
Solution:
Cost of 1 dozen or 12 soaps = Rs. 285.60
Cost of 1 soap = Rs. 285.6012
Cost of 15 soaps = Rs. 285.60×1512
= Rs. 357.00

Question 3.
Solution:
Cost of 9 kg of rice = Rs. 327.60
Cost of 1 kg = Rs. 327.609
and cost of 50 kg = Rs. 327.60×509
= Rs. 36.40 x 50
= Rs. 1820

Question 4.
Solution:
Weight of 22.5 metres of the iron rod: = 85.5 kg
Weight of 1 metre of the iron rod

Question 5.
Solution:
Quantity of oil in 15 tins = 234 kg
Quantity of oil in 1 tin = 23415 kg
Quantity of oil in 10 tins

Question 6.
Solution:
Distance covered by the car in 12 litres of diesel = 222 kms
Distance covered by the car in 1 litre of diesel = 22212 km

Question 7.
Solution:
Charges of 25 tonnes of weight = Rs. 540
charges of 1 ton = Rs.54025

Question 8.
Solution:
Weight of copper in 4.5 g of alloy = 3.5g
Weight of copper in 1 g of alloy

Question 9.
Solution:
In Rs. 87.50, the inland letter are purchased = 35
In Re. 1, letters can be purchased
= 3587.50
and in Rs. 315, letters can be purchased

Question 10.
Solution:
4 dozen = 4 x 12 = 48 bananas
In Rs. 104, banana are purchased = 48

Question 11.
Solution:
In Rs. 22770, chairs are purchased =18
In Re. 1, chairs will be purchased
= 1822770
and in Rs. 10120, chairs will be
purchased = 18×1012022770
= 8

Question 12.
Solution:
(i) A car travels 195 km distance in = 3 hours
It will travel 1 km distance in = 3195 hr.
and it will travel 520 km distance in
= 3×520195
= 8 hr
(ii) A car travels in 3hr = 195 km

Question 13.
Solution:
(i) A laborer earn in 12 days = Rs. 1980
He will earn in 1 day = Rs. 198012
and he will earn in 7 days

Question 14.
Solution:
(i) Weight of 65 books = 13 kg
Then weight of 1 book = 1365 kg

Question 15.
Solution:
Number of boxes needed for 6000 pens = 48
Number of boxes needed for 1 pen 48 = 486000

Question 16.
Solution:
Clearly, less workers will build the wall in more days.
And, more workers will build the wall in less days.
24 workers can build the wall in 15 days
1 worker can build the wall in (15 x 24) days
(less worker, more days)
9 workers will build the wall in

Question 17.
Solution:
Men needed to finish a piece of work in 26 days = 40
Men needed to finish a piece of work in 1 day = 40 x 26 (less days, more men)

Question 18.
Solution:
Clearly, less men will take more days to consume the food.
And, more men will take less days to consume the food.
550 men have provisions for 28 days
1 men has provisions for (28 x 550) days [less men, more days]
700 men will have provisions for

Question 19.
Solution:
Clearly, less persons will consume the rice in more days.
And more persons will consume the rice in less days.
60 persons consume the bag of rice in 3 days.
1 person will consume the bag of rice in
(3 x 60) days (less persons, more days)
18 persons will consume the bag of rice

Ex 10D Solutions

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1
Solution:
(d) 92115

Question 2.
Solution:
(a) 57x
= 5185

Question 3.
Solution:
(a) 2535
= 45x

Question 4.
Solution:
(c) 45
= x35

Question 5.
Solution:
(b) ab
= cd
=>ad = bc

Question 6.
Solution:
(b) a : b :: b : c

Question 7.
Solution:
(b) 58 < 34 => 4 x 5 < 3 x 8 => 20 < 24

Question 8.
Solution:
Total amount = Rs 760
Ratio A : B = 8 : 11

Question 9.
Solution:
(d) largest = 252×75+7
= 252×712
= 21 x 7
= 147

Question 10.
Solution:
(b) largest = 90×51+3+5
= 90×59
= 50 cm

Question 11.
Solution:
(c) total strength of school
largest = 8405 x (12 + 5)
= 840×175
= 168 x 17
= 2856

Question 12.
Solution:
(b) Cost of 12 pens = Rs 138
Cost of 1 pen = Rs 138×1412
and cost of 14 pens = Rs 161

Question 13.
Solution:
(b) 24×158
= 45 days

Question 14.
Solution:
(a) 26×4020
= 52 men

Question 15.
Solution:
(b) In 6 L of petrol, a car covers = 111 km
In 1 L, it will cover = 1116 km
and in 10 L it will cover = 111×106 km
= 185 km

Question 16.
Solution:
(a) 28×550700
= 22 days

Question 17.
Solution:
Ratio in the angles of triangle
= 3 : 1 : 2

Question 18.
Solution:
(b) Ratio in length and breadth of a rectangle = 5 : 4
Width = 36 m
Length = 54 x 36 m
= 45m

Question 19.
Solution:
Bus covers in 3 hrs = 195 km
It will cover in 1 hr = 1953 = 65 km

Question 20.
Solution:
1 dozen = 12 bars
Cost of 5 bars of soap = Rs.82.50

Question 21.
Solution:
Total pencils in 30 packets of 8 pencils
= 30 x 8= 240
and total pencils of 25 packets of 12
pencils = 25 x 12 = 300
Now cost of 240 pencils = Rs. 600
Then cost of 1 percent = Rs. 60024
and cost of 300 pencils = Rs. 600×300240
= Rs 750 (b)

Question 22.
Solution:
Journey of 75 km costs = Rs 215
Cost of 1 km = Rs 21575

Question 23.
Solution:
1st term = 12
2nd term = 21
fourth term = 14

Question 24.
Solution:
10 boys dig a patch in = 12 hrs
1 boy will dig it in = 12 x 10 hours

RS Aggarwal Solutions for Class 6 Maths Chapter 10: Download PDF

RS Aggarwal Solutions for Class 6 Maths Chapter 10–Ratio, Proportion and Unitary Method

Download PDF: RS Aggarwal Solutions for Class 6 Maths Chapter 10–Ratio, Proportion and Unitary Method PDF

Chapterwise RS Aggarwal Solutions for Class 6 Maths :

• Chapter 1–Number System
• Chapter 2–Factors and Multiples
• Chapter 3–Whole Numbers
• Chapter 4–Integers
• Chapter 5–Fractions
• Chapter 6–Simplification
• Chapter 7–Decimals
• Chapter 8–Algebraic Expressions
• Chapter 9–Linear Equations in One Variable
• Chapter 10–Ratio, Proportion and Unitary Method
• Chapter 11–Line Segment, Ray and Line
• Chapter 12–Parallel Lines
• Chapter 13–Angles and Their Measurement
• Chapter 14–Constructions (Using Ruler and a Pairs of Compasses)
• Chapter 15–Polygons
• Chapter 16–Triangles
• Chapter 17–Quadrilaterals
• Chapter 18–Circles
• Chapter 19–Three-Dimensional Shapes
• Chapter 20–Two-Dimensional Reflection Symmetry (Linear Symmetry)
• Chapter 21–Concept of Perimeter and Area
• Chapter 22–Data Handling
• Chapter 23–Pictograph
• Chapter 24–Bar Graph

About RS Aggarwal Class 6 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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