NCERT Exemplar Solutions: Ratio & Proportion

# NCERT Exemplar Solutions: Ratio & Proportion | Mathematics (Maths) Class 6 PDF Download

## Exercise Page: 120

In questions 1 to 10, only one of the four options is correct. Write the correct one.
Q.1. The ratio of 8 books to 20 books is
(a) 2 : 5
(b) 5 : 2
(c) 4 : 5
(d) 5 : 4

Correct Answer is Option (a)
The comparison of two numbers or quantities by division is known as the ratio. Symbol ‘:’ is used to denote ratio.
Ratio of 8 books to 20 books = 8/20
Divide both numerator and denominator by 4.
= 2/5
Therefore, ratio of 8 books to 20 books = 2 : 5

Q.2. The ratio of the number of sides of a square to the number of edges of a cube is
(a) 1 : 2
(b) 3 : 2
(c) 4 : 1
(d) 1 : 3

Correct Answer is Option (d)
We know that, number of sides in a square = 4 and number of edges in a cube = 12
So, ratio of sides to edges = 4/12
Divide both numerator and denominator by 4.
= 1/3
Therefore, ratio of sides to edges = 1 : 3

Q.3. A picture is 60cm wide and 1.8m long. The ratio of its width to its perimeter in lowest form is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : 8

Correct Answer is Option (d)
From the question it is given that,
Width of a picture = 60 cm
Length of a picture = 1.8 m
We know that, 1 m = 100 cm
So, 1.8 m = 180 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (180 + 60)
= 2 (240)
= 480
Therefore, The ratio of its width to its perimeter in lowest form = 60/480
Divide both numerator and denominator by 20.
= 3/24
Again, divide both numerator and denominator by 3.
= 1/8
= 1 : 8

Q.4. Neelam’s annual income is Rs. 288000. Her annual savings amount to Rs. 36000. The ratio of her savings to her expenditure is
(a) 1 : 8
(b) 1 : 7
(c) 1 : 6
(d) 1 : 5

Correct Answer is Option (b)
From the question it is given that,
Neelam’s annual income is ₹ 288000
Her annual savings amount to ₹ 36000
So, Neelam’s expenditure = 288000 – 36000
= ₹ 252000
Then, ratio of her savings to her expenditure = 36000/252000
= 36/252
Divide both numerator and denominator by 12.
= 3/21
Again, divide both numerator and denominator by 3.
= 1/7
Therefore, ratio of her savings to her expenditure = 1 : 7

Q.5. Mathematics textbook for Class VI has 320 pages. The chapter ‘symmetry’ runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is
(a) 11 : 320
(b) 3 : 40
(c) 3 : 80
(d) 272 : 320

Correct Answer is Option (c)

From the question it is given that,
Total number of pages in the Mathematics textbook for Class VI = 320 pages
The chapter ‘symmetry’ runs from page 261 to page 272
Number of pages contains symmetry chapter = 12
So, the ratio of the number of pages of symmetry chapter to the total number of pages of the book is,
= 12/320
Divide both numerator and denominator by 2.
= 6/160
Again, divide both numerator and denominator by 2.
= 3/80
Therefore, the ratio of the number of pages of this chapter to the total number of pages of the book is 3: 80.

Q.6. In a box, the ratio of red marbles to blue marbles is 7:4. Which of the following could be the total number of marbles in the box?
(a) 18
(b) 19
(c) 21
(d) 22

Correct Answer is Option (d)
From the question it is given that, the ratio of red marbles to blue marbles is 7:4.
Now, let us assume the common factor of 7 and 4 be x.
So, the total number of marbles in the box = 7x + 4x = 11x
Hence number of marbles in the box is a multiple of 11.
Therefore, 11 × 2 = 22

Q.7. On a shelf, books with green cover and that with brown cover are in the ratio 2 : 3. If there are 18 books with green cover, then the number of books with brown cover is
(a) 12
(b) 24
(c) 27
(d) 36

Correct Answer is Option (c)
From the question it is given that,
On a shelf, books with green cover and that with brown cover are in the ratio 2:3
There are 18 books with green cover
So, let us assume the common factor of 2 and 3 be x.
Then, 2x = 18
x = 18/2
Divide both numerator and denominator by 2.
x = 9
Therefore, the number of books with brown cover is = 3x = 3 × 9
= 27

Q.8. The greatest ratio among the ratios 2 : 3, 5 : 8, 75 : 121 and 40 : 25 is
(a) 2 : 3
(b) 5 : 8
(c) 75 : 121
(d) 40 : 25

Correct Answer is Option (d)
Consider the given ratios, 2 : 3, 5 : 8, 75 : 121 and 40 : 25.
Simplified form of 2: 3 = 2/3 = 0.67
Simplified form of 5 : 8 = 5/8 = 0.625
Simplified form of 75: 121 = 75/121 = 0.61
Simplified form of 40: 25 = 40/25 = 1.6
Therefore, the greatest ratio among the given ratios is 40 : 25

Q.9. There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is:
(a) b/(b + g)
(b) g/(b + g)
(c) b/g
(d) (b + g)/b

Correct Answer is Option (a)
From the question,
Number of boys in the class = b
Number of girls in the class = g
Total number of students in the class = b + g
Therefore, the ratio of the number of boys to the total number of students in the class
= b/(b + g)

Q.10. If a bus travels 160 km in 4 hours and a train travels 320km in 5 hours at uniform speeds, then the ratio of the distances travelled by them in one hour is
(a) 1 : 2
(b) 4 : 5
(c) 5 : 8
(d) 8 : 5

Correct Answer is Option (c)
From the question it is given that,
Bus travels 160 km in 4 hours
Train travels 320 km in 5 hours
So, distance travelled by bus in an hour = 160/4 = 40 km/h
Distance travelled by train in an hour = 320/5 = 64 km/h
Then the ratio of the distances travelled by them in one hour is = 40/64
Divide both numerator and denominator by 8.
= 5/8
Therefore, the ratio of the distances travelled by them in one hour is 5: 8.

### In questions 11 to 15, find the missing number in the box [ ] in each of the proportions:

Q.11. 3/5 = [ ]/20

Let us assume the missing number be y.

Then, (3/5) = (y/20)
By cross multiplication we get,
(3 × 20)/5 = y
y = 60/5
Divide both numerator and denominator by 5.
y = 12
Therefore, 3/5 = [12]/20

Q.12. [ ]/18 = 2/9

Let us assume the missing number be y.

Then, y/18 = 2/9
By cross multiplication we get,
y = (2 × 18)/9
y = 36/9
Divide both numerator and denominator by 9.
y = 4
Therefore, [4]/18 = 2/9

Q.13. 8/[ ] = 3.2/4

Let us assume the missing number be y.

Then, 8/y = 3.2/4
By cross multiplication we get,
y = (8 × 4)/3.2
y = 32/3.2
y = 320/32
Divide both numerator and denominator by 32.
y = 10
Therefore, 8/[10] = 3.2/4

Q.14. [ ]/45 = 16/40 = 24/[ ]

Consider the first two ratios [ ]/45 = 16/40

Let us assume the missing number be P
Then, P/45 = 16/40
By cross multiplication we get,
P = (16 × 45)/40
P = 720/40
P = 72/4
Divide both numerator and denominator by 4.
P = 18
Therefore, [18]/45 = 16/40
Now, consider the last two ratios, 16/40 = 24/[ ]
Let us assume the missing number be Q,
Then, 16/40 = 24/Q
By cross multiplication we get,
Q = (24 × 40)/16
Q = 960/16
Divide both numerator and denominator by 16.
Q = 60
Therefore, 16/40 = 24/[60]

Q.15. 16/36 = [ ]/63 = 36/[ ] = [ ]/117

Consider the first two ratios 16/36 = [ ]/63

Let us assume the missing number be P
Then, 16/36 = P/63
By cross multiplication we get,
P = (16 × 63)/36
P = 1008/36
Divide both numerator and denominator by 36.
P = 28
Therefore, 16/36 = [28]/63
Now, consider the middle two ratios, 28/63 = 36/[ ]
Let us assume the missing number be Q,
Then, 28/63 = 36/Q
By cross multiplication we get,
Q = (36 × 63)/28
Q = 2268/28
Divide both numerator and denominator by 28.
Q = 81
Therefore, 28/63 = 36/[81]
Consider the last two ratios 36/81 = [ ]/117
Let us assume the missing number be R
Then, 36/81 = R/117
By cross multiplication we get,
P = (36 × 117)/81
P = 4212/81
Divide both numerator and denominator by 81.
P = 52
Therefore, 36/81 = [52]/117
So, 16/36 = [28]/63 = 36/[81] = [52]/117

### In questions 16 to 34, state whether the given statements are true (T) or false (F).

Q.16. 3/8 = 15/40

True.

Consider the two fractions, 3/8 = 15/40
15/40 is further simplified by dividing both numerator and denominator by 5 we get,
= 3/8
Therefore, 3/8 = 3/8

Q.17. 4 : 7 = 20 : 35

True.

Consider the two ratio, 4: 7 = 20: 35
= 4/7 = 20/35
20/35 is further simplified by dividing both numerator and denominator by 5 we get,
= 4/7
Therefore, 4/7 = 4/7

Q.18. 0.2 : 5 = 2 : 0.5

False.

Consider the two ratio, 0.2: 5 = 2: 0.5
0.2/5 = 2/0.5
0.04 ≠ 4

Q.19. 3 : 33 = 33 : 333

False.

Consider the two ratios 3: 33 = 33 : 333
3/33 = 33/333
0.0909 ≠ 0.0990

Q.20. 15m : 40m = 35m : 65m

False.

Consider the two ratios 15m : 40m = 35m : 65m
15/40 = 35/65
15/40 is further simplified by dividing both numerator and denominator by 5 we get,
= 3/8
35/65 is further simplified by dividing both numerator and denominator by 5 we get,
= 7/13
Hence, 3/8 ≠ 7/13
Therefore, 15/40 ≠ 35/65

Q.21. 27cm2 : 57cm2 = 18cm : 38cm

True

Consider the two ratios 27cm2 : 57cm2 = 18cm : 38cm
27/57 = 18/38
27/57 is further simplified by dividing both numerator and denominator by 3 we get,
= 9/19
18/38 is further simplified by dividing both numerator and denominator by 2 we get,
= 9/19
Hence, 9/19 = 9/19
Therefore, 27cm2 : 57cm2 = 18cm : 38cm

Q.22. 5kg : 7.5kg = Rs 7.50 : Rs 5

False.

Consider the two ratios 5kg: 7.5kg = Rs 7.50: Rs 5
5/7.5 = 7.50/5
50/75 = 75/50
50/75 is further simplified by dividing both numerator and denominator by 25 we get,
= 2/3
75/50 is further simplified by dividing both numerator and denominator by 25 we get,
= 3/2
Hence, 2/3 ≠ 3/2
Therefore, 5kg: 7.5kg ≠ Rs 7.50: Rs 5

Q.23. 20g : 100g = 1metre : 500cm

True
Consider the given ratios, 20g : 100g = 1metre : 500cm
20/100 = 1/500
We know that, 1 metre = 100 cm
So, 20g: 100g = 100cm : 500 cm
20/100 = 100/500
20/100 is further simplified by dividing both numerator and denominator by 20 we get,
= 1/5
100/500 is further simplified by dividing both numerator and denominator by 100 we get,
= 1/5
Hence, 1/5 = 1/5
Therefore, 20g : 100g = 1metre : 500cm

Q.24. 12 hours : 30 hours = 8km : 20km

True

Consider the given ratios, 12 hours : 30 hours = 8km : 20km
12/30 = 8/20
12/30 is further simplified by dividing both numerator and denominator by 6 we get,
= 2/5
8/20 is further simplified by dividing both numerator and denominator by 4 we get,
= 2/5
Hence, 2/5 = 2/5
Therefore, 12 hours : 30 hours = 8km : 20km

Q.25. The ratio of 10kg to 100kg is 1:10

True.

The ratio of 10kg to 100kg = 10/100
= 1/10
= 1:10

Q.26. The ratio of 150cm to 1metre is 1:1.5.

False

We know that, 1 metre = 100 cm
So, the ratio of 150cm to 1metre is = 150/100
= 15/10
Divide both numerator and denominator by 5 we get,
= 3/2
So, the ratio of 150cm to 1metre is 3: 2

Q.27. 25kg : 20g = 50kg : 40g

True.

We know that, 1 kg = 1000 g
So, 25 × 1000 = 25000 g
50 × 1000 = 50000g
Then, 25000g : 20g = 50000g : 40g
25000/20 = 50000/40
2500/2 = 5000/4
2500/2 is further simplified by dividing both numerator and denominator by 2 we get,
= 1250
5000/4 is further simplified by dividing both numerator and denominator by 4 we get,
= 1250
Hence, 1250 = 1250
So, 25kg : 20g = 50kg : 40g

Q.28. The ratio of 1 hour to one day is 1:1.

False
We know that, 1 day = 24 hours
So, 1 hour: 1 day = 1 hour : 24 hours
1/1 ≠ 1/24

Q.29. The ratio 4 :16 is in its lowest form.

False

4 : 16
= 4/16
Divide both numerator and denominator by 4,
= 1/4
Therefore, lowest form of 4: 16 is 1/4

Q.30. The ratio 5 : 4 is different from the ratio 4 : 5.

True.
5: 4 ≠ 4: 5
5/4 ≠ 4/5
1.25 ≠ 0.8

Q.31. A ratio will always be more than 1.

False.

A ratio will be more than or less than 1

Q.32. A ratio can be equal to 1.

True.
Example: 2: 2 = 2/2 = 1

Q.33. If b : a = c : d, then a, b, c, d are in proportion.

False

Four quantities are said to be in proportion, if the ratio of the first and the second quantities is equal to the ratio of the third and the fourth quantities. The symbol ‘::’ or ‘=’ is used to equate the two ratios.

Q.34. The two terms of a ratio can be in two different units.

False.

For a ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit before the ratio is taken.

### In questions 35 to 46, fill in the blanks to make the statements true.

35. A ratio is a form of comparison by ______.

A ratio is a form of comparison by division.

Q.36. 20m : 70m = Rs 8 : Rs ______.

20m : 70m = Rs 8 : Rs 28.

Let us assume the missing number be P.
Then, 20m : 70m = ₹ 8 : ₹ P
20/70 = 8/P
P = (70 × 8)/20
P = 560/20
P = 56/2
P = 28
Therefore, 20m : 70m = Rs 8 : Rs 28.

Q.37. There is a number in the box [ ] such that [ ], 24, 9, 12 are in proportion. The number in the box is _____.

There is a number in the box [ ] such that [ ], 24, 9, 12 are in proportion. The number in the box is 18.

Let us assume the missing number be ‘P’,
Then, P, 24, 9, 12
P: 24 = 9: 12
P/24 = 9/12
9/12 is further simplified by dividing both numerator and denominator by 3.
So, P/24 = 3/4
P = (3 × 24)/4
P = 72/4
P = 18
Therefore, the missing number is 18.

Q.38. If two ratios are equal, then they are in _____.

If two ratios are equal, then they are in proportion.

Use Fig. (In which each square is of unit length) for questions 39 and 40:

Q.39. The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is _______.

The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is 3: 7.
From the figure, perimeter of shaded portion = 1 + 2 + 1 + 2 = 6 units
Perimeter of whole figure = 3 + 4 + 3 + 4 = 14 units
Then, ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure = 6/14
= 3/7
= 3: 7

Q.40. The ratio of the area of the shaded portion to that of the whole figure is ______.

The ratio of the area of the shaded portion to that of the whole figure is 1: 6.
Area of the shaded figure = 2 × 1
= 2 sq. units
Area of whole figure = 3 × 4 = 12 sq. units
The ratio of the area of the shaded portion to that of the whole figure is = 2: 12
= 2/12
= 1/6
= 1: 6

Q.41. Sleeping time of a python in a 24 hour clock is represented by the shaded portion in Fig.
The ratio of sleeping time to awaking time is ______.

The ratio of sleeping time to awaking time is 3: 1.
From the figure, sleeping time = 18 hours
Then, awaking time = 24 – 18 = 6 hours
Therefore, the ratio of sleeping time to awaking time is 18/6
= 3/1
= 3: 1

Q.42. A ratio expressed in lowest form has no common factor other than ______ in its terms.

A ratio expressed in lowest form has no common factor other than one in its terms.

Q.43. To find the ratio of two quantities, they must be expressed in _____units.

To find the ratio of two quantities, they must be expressed in same units.

Q.44. Ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to _____

Ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to 100 paise.
From the question,
5 paise : 25 paise = 20 paise: [ ]
Let us assume the missing number be Q,
5 paise : 25 paise = 20 paise: Q
5/25 = 20/Q
Q = (20 × 25)/5
Q = 500/5
Q = 100
Therefore, Ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to 100 paise

Q.45. Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in lowest form is ______.

Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in lowest form is 149: 160.
From the question,
Saturn takes 9 hours 56 minutes for one spin on their axes
We know that, 1 hour = 60 minutes
So, (9 × 60) + 56 = 540 + 56 = 596 minutes
Jupiter takes 10 hours 40 minutes for one spin on their axes
= (10 × 60) + 40
= 600 + 40
= 640 minutes
The ratio of the time taken by Saturn and Jupiter in lowest form is = 596/640
Divide both numerator and denominator by 2,
= 298/320
Again, divide both numerator and denominator by 2,
= 149/160
Therefore, the ratio of the time taken by Saturn and Jupiter in lowest form is 149 : 160.

Q.46. 10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda. Amount of caustic soda needed for 1 L of water to make the same type of solution is_________.

Given, 10 g of caustic soda dissolved in 100 mL water.
The ratio of caustic soda to water should be in proportion.
10g: 100 mL : :x g :1 L
[∵ 1 L = 1000 mL]
x g × 100 mL = 10 g × 1000 mL     [by cross multiplication]

x = 100 g

Q.47. The marked price of a table is Rs. 625 and its sale price is Rs. 500. What is the ratio of the sale price to the marked price?

Given, marked price of a table = ₹ 625
Sale price of a table = ₹ 500
Ratio of sale price to marked price = ₹500/625 = 500/625
= 20/25                [on dividing numerator and denominator by 25]
= 4/5                      [on dividing numerator and denominator by 5]
∴ Required ratio  = 4 : 5

Q.48. Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio 7 : 2. How much Khoya did she use?

Given,
Quantity of Burfi = 18 kg and Khoya : Sugar = 7 : 2
Total of ratio = 7 + 2 = 9
Quantity of Khoya = (18/9) x 7 = 14kg
So, Reshma used 14 kg Khoya.

Q.49. A line segment 56 cm long is to be divided into two parts in the ratio of 2 : 5. Find the length of each part.

Given,
Length of the line segment = 56 cm Ratio of two parts = 2 : 5 Sum of ratios = 2 + 5 = 7
Length of first part = (2/7) x 56 = 16 cm
Length of second part = (5/7) x 56 = 40 cm

Q.50. The number of milk teeth in human beings is 20 and the number of permanent teeth is 32. Find the ratio of the number of milk teeth to the number of permanent teeth.

Number of milk teeth in human beings = 20
Number of permanent teeth in human beings = 32
Ratio of the number of milk teeth to the number of permanent teeth = 20/32
= 5/8 [on dividing numerator and denominator by 4]
= 5 : 8

Q.51. A rectangular sheet of paper is of length 1.2 m and width 21 cm. Find the ratio of width of the paper to its length.

Given,
Length of rectangular sheet = 1.2 m [v 1 m = 100 cm]
= 1.2 x 100cm= 120cm Width of rectangular sheet = 21 cm
Ratio of width to length = 21 cm/120 cm
= 7/40 = 7:40 [on dividing numerator and denominator by 3]

Q.52. In a school, the ratio of the number of large classrooms to small classrooms is 3 : 4. If the number of small rooms is 20, then find the number of large rooms.

Given, ratio of number of large classrooms to small classrooms = 3:4 Number of small classrooms = 20 Let the classrooms are multiple of x.

So, large classrooms = 3x Small classrooms = 4x

According to the question, 4x = 20 => x=20/4 = 5 .
Hence, number of large classrooms = 3x
= 3 x 5 = 15

Q.53. Samira sells newspapers at Janpath crossing daily. On a particular day, she had 312 newspapers out of which 216 are in English and remaining in Hindi.
Find the ratio of
(a) the number of English newspapers to the number of Hindi newspapers.
(b) the number of Hindi newspapers to the total number of newspapers.

Given, total newspapers = 312 English newspapers = 216
Hindi newspapers = Total number of newspapers – Newspapers in English = 312 – 216 = 96
(a) Ratio of number of English newspapers to number of Hindi newspapers = 216/96
= 9/4 = 9:4 [on dividing numerator and denominator by 24]
(b) Ratio of number of Hindi newspapers to the total number of newspapers = 96/312
= 4/13 = 4:13.

Q.54. The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 144 and the number of Christian students is 72.
Find the ratio of
(a) the number of Hindu students to the number of Christian students.
(b) the number of Muslim students to the total number of students.

Given, number of Hindu students = 288 Number of Muslim students = 252 Number of Sikh students = 144 Number of Christian students = 72 Total number of students = 288+252+144+72 =756

(a) Ratio of number of Hindu students to the number of Christian students = 288/72

= 4/1 = 4:1 [on dividing numerator and denominator by 72]

(b) Ratio of number of Muslim students to the total number of students = 252/756

= 1/3 =1:3 [on dividing numerator and denominator by 252]

Q.55. When Chinmay visited Chowpati at Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5:4. If the total number of food stalls is 117, find the number of each type of food stalls.

Given, ratio of North Indian food stalls to South Indian food stalls = 5:4

Total number of food stalls =117

Total ratio = 5+4 = 9

North Indian food stalls = (5/9) x 117 = 65

South Indian food stalls = (4/9) x 117 = 52

Q.56. At the parking stand of Ramleela ground, Kartik counted that there are 115 cycles, 75 scooters and 45 bikes. Find the ratio of the number of cycles to the total number of vehicles.

Given, at parking stand, number of Cycles = 115

Scooters = 75

Bikes = 45

Total number of vehicles = 115+75+ 45 = 235

Ratio of number of cycles to the total number of vehicles = 115/235

= 23/47 =23:47

[on dividing numerator and denominator by 5]

Q.57. A tea merchant blends two varieties of tea costing it Rs. 234 and Rs. 130 per kg in the ratio of their costs. If the weight of the mixture is 84 kg, then find the weight of each variety of tea.

Given, cost of two varities of tea = Rs. 234 and Rs. 130

Ratio of their costs = 234/130 = 9/5 = 9:5

[on dividing numerator and denominator by 26]

Total weight of mixture = 84 kg Total ratio = 9+5 = 14

Weight of first variety tea = (9/14) x 84

= 54 kg

Weight of second variety tea = (5/14) x 84

= 30 kg

Q.58. An alloy contains only Zinc and Copper and they are in the ratio of 7:9. If the weight of the alloy is 8 kg, then find the weight of Copper in the alloy.

Given, the ratio of Zinc and Copper in alloy = 7:9 and weight of alloy = 8 kg

Let the weight of Zinc and Copper in alloy be 7x and 9x respectively, where x is multiple of weight.

Then, total weight =7x+9x = 6x

16x = 8 kg => x = ½ kg

Weight of copper = 9x = 9 x (1/2) = 4 ½ kg Hence, the weight of copper is 4 ½ kg.

Q.59. Find two numbers, whose sum is 100 and whose ratio is 9:16.

Let the two numbers are 9x and 16x, whose sum is 100.

=> 9x +16x = 100

=> 25x = 100

=> x = 4

Q.60. A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript. What is the ratio of the number of pages typed to the number of pages left?

Total pages of manuscript to type = 40

Typed pages of manuscript = 30

Left pages = 40 – 30 = 10

Ratio of the number of pages to the types pages to the number of left pages = 30/10 = 3/1 = 3:1 .

The document NCERT Exemplar Solutions: Ratio & Proportion | Mathematics (Maths) Class 6 is a part of the Class 6 Course Mathematics (Maths) Class 6.
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## FAQs on NCERT Exemplar Solutions: Ratio & Proportion - Mathematics (Maths) Class 6

 1. What is the concept of ratio and proportion in mathematics?
Ans. Ratio is a comparison of two or more quantities of the same kind, while proportion is an equation that states that two ratios are equal. In mathematics, ratio and proportion are used to compare quantities, solve problems related to scaling, and understand the relationship between different quantities.
 2. How can ratio and proportion be applied in real-life situations?
Ans. Ratio and proportion can be applied in various real-life situations. For example, they can be used to determine the ratio of ingredients in a recipe, calculate the scale factor in map reading, or solve problems related to time and distance. They also help in understanding the concept of similarity and scaling in everyday scenarios.
 3. What are some common types of problems that involve ratio and proportion?
Ans. Some common types of problems involving ratio and proportion are finding the missing value in a given proportion, solving problems related to direct and inverse proportions, and solving problems related to discounts and sales. These types of problems require understanding the concept of ratio and proportion and applying it to real-life scenarios.
 4. How can ratio and proportion be used to solve problems related to scaling?
Ans. Ratio and proportion can be used to solve problems related to scaling by understanding the relationship between different quantities. For example, if the scale factor between two similar figures is known, the dimensions of one figure can be multiplied by the scale factor to find the dimensions of the other figure. This helps in resizing or enlarging objects while maintaining their proportional relationships.
 5. Are there any shortcuts or tricks to solve ratio and proportion problems quickly?
Ans. Yes, there are some shortcuts or tricks to solve ratio and proportion problems quickly. For example, in problems involving direct proportions, multiplying or dividing the quantities by the same factor can help find the missing value. Similarly, in problems involving inverse proportions, multiplying the two quantities together gives a constant value. Understanding and applying these tricks can save time and make solving ratio and proportion problems easier.

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