Ncert Solutions For Class 6 Maths Ex 12.1

Ncert Solutions For Class 6 Maths Chapter 12 Ex 12.1

Question 1:

The number of girls and boys studying in a class are 25 and 20 respectively.

  1. Calculate the ratio of number of girls to that of boys?
  2. Calculate the ratio of number of girls to that of the total students in the class?

Solution:

Given;

No. of girls = 25

No. of boys = 20

Total number of students studying the class = 25 + 20 = 45

  1. Ratio of no. of girls to boys = \(\frac{25}{20}\) = \(\frac{5}{4}\)
  2. Ratio of girls to total students in the class = \(\frac{25}{45}\) = \(\frac{5}{9}\)

Question 2:

The strength of a class is 40, in which 8 students play football, while 14 others play cricket and all those left over play tennis.

  1. What is the ratio of the students playing football to the students playing tennis?
  2. What is the ratio of the students playing cricket to the strength of the class?

Solution:

Students playing football = 8 students

Students playing cricket = 14 students

Students playing tennis = 40 – (14 + 8) = 18

  1. ratio of the students playing football to the students playing tennis = \(\frac{8}{18}\) = \(\frac{4}{9}\)
  2. ratio of the students playing cricket to the strength of the class = \(\frac{14}{40}\) = \(\frac{7}{20}\)

 

Question 3:

Observe the following figure and answer the below questions

1

i) Calculate the ratio of triangles to the circles within the rectangle?

ii) Calculate the ratio of squares to the figures of all shapes within the rectangle?

iii) Calculate the ratio of circles to the figures of all shapes within the rectangle?

Solution:

No. of triangles within the rectangle = 4

No. of circles within the rectangle = 3

No. of squares within the rectangle = 1

Total figures = 8

i) Ratio of triangles to the circles = \(\frac{4}{3}\)

ii) Ratio of squares to all figures = \(\frac{1}{8}\)

iii) Ratio of circles to all figures = \(\frac{3}{8}\)

Question 4:

Kedhar and Rishab cover a distance of 10 km and 15 km respectively. What is the ratio of Kedhar’s speed to that of Rishab?

Solution:

Distance covered by kedhar in one hour = 10 km

Distance covered by Rishab in one hour = 15 km

Speed of Kedhar = 10 km/h

Speed of Rishab = 15km/h

Ratio of Kedhar’s speed to that of Rishab’s = \(\frac{10}{15}\)= \(\frac{2}{3}\)

Questsion 5:

Fill in the following boxes

\(\frac{18}{21}\) = \(\frac{ \sqcap }{7}\) = \(\frac{12}{ \sqcap }\) = \(\frac{ \sqcap }{35}\)

And tell if these ratios are equivalent.

Solution:

\(\frac{18}{21}\) = \(\frac{18\div 3}{21\div 3}\) = \(\frac{6}{7}\)

\(\frac{6}{7}\) = \(\frac{6\times 2}{7\times 2}\) = \(\frac{12}{14}\)

\(\frac{6}{7}\) = \(\frac{6\times 5}{7\times 5}\) = \(\frac{30}{35}\)

Thus, 6, 14, 30 are the numbers that are to be filled in three boxes respectively.

Yes, these ratios are equivalent.

Questsion 6:

Reduce the following ratios to their lowest terms;

(i) 81 to 108 (ii) 98 to 63

(iii) 33 to 121 (iv) 30 to 45

Solution:

(i) \(\frac{81}{108}=\frac{3\times 3\times 3\times 3}{2\times 2\times 3\times 3\times 3}=\frac{3}{4}\)

(ii) \(\frac{98}{63}=\frac{14\times 7}{9\times 7}=\frac{14}{9}\)

(iii) \(\frac{33}{121}=\frac{3\times 11}{11\times 11}=\frac{3}{11}\)

(iv) \(\frac{30}{45}=\frac{2\times 3\times 5}{3\times 3\times 5}=\frac{2}{3}\)

 

Questsion 7:

Fathima’s salary is Rs. 2,00,000 per annum and out of which she saves 75,000.

i) Calculate the ratio of Seema’s earnings to her savings

ii) Calculate the ratio of Seema’s savings to her expenditure

Solution:

Income = Rs 2,00,000

Savings = Rs 75,000

Expenditure = Rs 1,25,000

i) Ratio of Seema’s earnings to her savings = \(\frac{2,00,000}{75,000}\) = \(\frac{8}{3}\)

ii) ratio of Seema’s savings to her expenditure =\(\frac{75,000}{1,25,000}\) = \(\frac{3}{5}\)

Questsion 8:

3852 pupil studies in a school which has 154 teachers. Calculate the ratio of teachers to the students?

Solution:

Ratio of teachers to students = \(\frac{154}{3850}\) = \(\frac{2\times 11\times 7}{2\times 11\times 175 }\) = \(\frac{7}{175}\)

Questsion 9:

There are 5852 students in a school, out of which 3800 are girls. Calculate the ratio of

i) Girls to total students in school

ii) Boys to girls in school

iii) Boys to total students in school

Solution:

Number of students in school = 5852

Girls in school = 3800

Boys in school = 5852 – 3800 = 2052

i) Required ratio = \(\frac{3800}{5852}\) = \(\frac{1900}{2926}\) = \(\frac{100}{154}\) = \(\frac{50}{77}\)

ii) Required ratio = \(\frac{2052}{3800}\) = \(\frac{1026}{1900}\) = \(\frac{513}{950}\) = \(\frac{27}{50}\)

iii) Required ratio = \(\frac{2052}{5852}\) = \(\frac{27}{77}\)

Questsion 10:

Calculate the ratio for the given

i) 30 seconds to 1.5 minutes ii) 40 cg to 1.5 g

iii) 55 paise to 1 rupee iv) 500 mm to 2 m

Solution:

i) 1.5 minutes = 90 seconds

∴ Required ratio = \(\frac{30}{90}\) = \(\frac{1}{3}\)

ii) 1.5 gram = 150 gram

∴ Required ratio = \(\frac{40}{150}\) = \(\frac{4}{15}\)

iii) 1 rupee = 100 paise

∴ Required ratio = \(\frac{55}{100}\) = \(\frac{11}{20}\)

iv) 2 m = 2000 mm

∴ Required ratio = \(\frac{500}{2000}\) = \(\frac{1}{4}\)

 

Questsion 11:

A college comprises of 1800 pupil. Among those 750 prefer hockey , 800 prefer football and remaining students prefer basketball. If only one game can be preferred by a student, what is the ratio of

i) Students who prefer hockey to students who prefer basketball

ii) Students who prefer football to students who prefer hockey

iii) Students who prefer hockey to the total number of students

Solution:

i) The ratio required is \(\frac{750}{250}\) = \(\frac{3}{1}\)

ii) The ratio required is \(\frac{800}{750}\) = \(\frac{16}{15}\)

iii) The ratio required is \(\frac{750}{1800}\) = \(\frac{5}{12}\)

 

Questsion 12:

The price of a dozen pens is 216 and the price of 8 pencils is 64. Calculate the ratio of the price of a pen to the price of a pencil?

Solution:

Dozen pens cost 216. Therefore, cost of one pen = \(\frac{216}{12}\) = 18

8 pencils cost 64. Therefore, cost of one pencil = \(\frac{64}{8}\) = 8

Ratio of the price of a pen to the price of a pencil = \(\frac{18}{8}\) = \(\frac{9}{4}\)

∴ Required ratio =\(\frac{9}{4}\)

Questsion 13:

The length and breadth of a box are in a ratio of 6:3. Fill in the emply blanks and find the possible lengths and breadths of the box?

Length of the box 30 60
Breadth of the box 15 60

Solution:

Ratio of length to breadth = 6:3 = \(\frac{6}{3}\)

Therefore, their equivalent ratios are =

\(\frac{6}{3}\, \times \, \frac{10}{10}\, =\, \frac{60}{30}\) \(\frac{6}{3}\, \times \, \frac{20}{20}\, =\, \frac{120}{60}\)
Length of the box 30 60 120
Breadth of the box 15 30 60

 

Questsion 14:

How can 20 pens be divided between Sujay and Ram the ratio of 4:1?

Solution:

Ratio between Sujay and ram = 4:1

Total value = 4+1 = 5

Therefore, Sujay’s part of total pencils = \(\frac{4}{5}\)

Ram’s part of total pencils = \(\frac{1}{5}\)

Thus, Pencils with Sujay = \(\frac{4}{5}\) \(\times\) 20 = 16 pencils

Pencils with Ram = \(\frac{1}{5}\) \(\times\) 20 = 4 pencils

Questsion 15:

Sneha and Radhika decide to divide 35 candies between them in the ratio of their ages. If Sneha and Radhika are 16 and 12 years old respectively, find how many candies Sneha and Radhika get.

Solution:

Ratio of Sneha’s age to Radhika’s age = \(\frac{16}{12}\) = \(\frac{4}{3}\)

Total value = 4+3 = 7

Candies received by Sneha = \(\frac{4}{7}\) * 35 = 20

Candies received by Radhika = \(\frac{3}{7}\) * 35 = 15

Questsion 16:

The present age of a son and his father is 15 and 60 years respectively. Calculate the ratio of

i) Current age of father to that of his son

ii) Age of father to that of his son, when son was 9 years old

iii) Age of father to that of his son 10 years later

iv) Age of father to that of his son, when the father was 50 years old.

Solution:

i) Required ratio = \(\frac{60}{15}\) = \(\frac{4}{1}\)

ii) Required ratio = \(\frac{54}{9}\) = \(\frac{6}{1}\)

iii) Required ratio = \(\frac{70}{25}\) = \(\frac{14}{5}\)

iv) Required ratio = \(\frac{50}{5}\) = \(\frac{10}{1}\)

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