# NCERT Solutions for Class 6 Maths Chapter 12

## NCERT Solutions for Class 6 Maths Chapter 12 – Ratio and Proportion

Ratio and Proportion is an important chapter in Class 6. Students will learn the basics of each concept of Ratio and Proportion and their application. NCERT Solutions for Class 6 Maths Chapter 12 by Extramarks have answers to all the questions given at the end of Chapter 12. Students can refer to these to solve their questions and to cross-check answers.

The NCERT Solutions Chapter 12 Ratio and Proportion for Class 6 are prepared by subject-matter experts while keeping the guidelines by CBSE in mind. These are excellent reference material to help students in completing their homework as well as preparing for the exams

## NCERT Solutions for Class 6 Maths

Students need to practise daily in order to understand the concepts in a better way. By referring to the NCERT Solutions for Class 6 Maths students will also understand the right way of answering questions and scoring higher marks in examinations.

NCERT Solutions for Class 6 Maths of Extramarks covers all other chapters sequentially:

### Chapter 12 – Ratio and Proportion

The first section of NCERT Class 6 Maths Chapter 12 focuses on the basic concepts of Ratio and Proportion and their applications. There are questions at the end to help students revise the concepts they have learned and practise the chapter in a better way.

The NCERT Solutions for Class 6 Maths Chapter 12 by Extramarks will help students in answering the questions given at the end of the chapter. Let’s look at what all is incorporated in the practise exercise of the chapter.

#### Ratio

Ratio is used to compare two or more numbers. It also indicates how big or small a quantity is when compared to another. In simple words, “ratio is the comparison of two quantities that have the same units, and it indicates how much of one quantity is present in another quantity”.

#### Proportion

If two ratios are equal, they are said to be in proportion. Therefore, four different entities are required to describe a proportion. A proportion is written as A:B C:D. A and D are known as extreme terms and B and C are known as middle terms.

#### Exercise 12.1 Questions and Answers

Exercise 12.1 has 16 questions in total. Out of the 16 questions, 10 are short questions and 6 are long questions.

#### Exercise 12.2 Questions and Answers

Exercise 12.2 has 5 short questions in total.

#### Exercise 12.3 Questions and Answers

Exercise 12.3 has 11 questions in total, of these 1 question is short answer and 10 are long answer questions.

NCERT Solutions for Class 6 Maths Chapter 12 Exercises are given in the table below:

 Chapter 12 – Ratio and Proportion Exercises Exercise 12.1 16 Questions & Solutions Exercise 12.2 4 Questions & Solutions Exercise 12.3 11 Questions & Solutions

### Key Features of NCERT Solutions for Class 6 Maths Chapter 12

The NCERT solutions of Class 6 Maths will help the students in the following manner:

• Students can refer to NCERT Solutions to solve exercise questions accurately
• These solutions answer all chapter related questions
• Solutions help students in understanding the right way to answer a question, which helps in scoring higher marks in examinations
• NCERT Solutions are helpful in last minute exam preparation

Q.1 There are 20 girls and 15 boys in a class.
(a) What is the ratio of number of girls to the number of boys?
(b) What is the ratio of number of girls to the total number of students in the class?

Ans-

(a) Ratio of number of girls to the number of boys
= 20: 15
= 4:3
(b) Total number of students = 20 + 15
= 35
Ratio of number of girls to the total number of
students = 20:35
= 4:7

Q.2 Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of
(a) Number of students liking football to number of students liking tennis.
(b) Number of students liking cricket to total number of students.

Ans-

Total students in the class = 30
Number of students like football = 6
Number of students like cricket = 12
Number of students like tennis = 30 – (6 + 12)
= 12

(a) Ratio of the number of students liking football to number of students liking tennis = 6: 12
= 1:2
(b) Ratio of the number of students liking cricket to total number of students = 12: 30
= 2:5

Q.3 See the figure and find the ratio of
(a) Number of triangles to the number of circles inside the rectangle.
(b) Number of squares to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.

Ans-

(a) Number of triangles inside the rectangle = 3
Number of circles inside the rectangle = 2
Ratio of number of triangles to the number of
circles inside the rectangle = 3:2

(b) Number of squares inside the rectangle = 2
Number of all figures inside the rectangle = 7
Ratio of the number of squares to all the figures
inside the rectangle = 2:7

(c) Ratio of the number of circles to all the
figures inside the rectangle = 2:7

Q.4 Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.

Ans-

Distance covered by Hamid in one hour = 9 km
Speed of Hamid = 9 km/hr
Distance covered by Akhtar in one hour = 12 km
Speed of Hamid = 12 km/hr
Ratio of speed of Hamid to the speed of Akhtar
= 9:12
= 3:4

$\begin{array}{l}\mathrm{Q.5 Fill}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{blanks}:\\ \frac{15}{18}=\frac{\begin{array}{c}\end{array}}{6}=\frac{10}{\begin{array}{c}\end{array}}=\frac{\begin{array}{c}\end{array}}{30}\left[\mathrm{Are}\mathrm{these}\mathrm{equivalent}\mathrm{ratios}?\right]\end{array}$

Ans-

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{15}{18}=\frac{3×5}{3×6}\\ \text{}=\frac{5}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{6}=\frac{5×2}{6×2}\\ \text{}=\frac{10}{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{6}=\frac{5×5}{6×5}\\ \text{}=\frac{25}{30}\\ \therefore \frac{15}{18}=\frac{\begin{array}{c}5\end{array}}{6}=\frac{10}{\begin{array}{c}12\end{array}}=\frac{\begin{array}{c}25\end{array}}{30}\\ Y\text{es, all these ratios are equivalent ratios}\text{.}\end{array}$

Q.6 Find the ratio of the following :
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes

Ans-

$\begin{array}{c}\left(\text{a}\right)\text{Ratio of 81 and 1}0\text{8}=\frac{\text{81}}{108}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{27}}{4×27}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3:4\\ \left(\text{b}\right)\text{Ratio of 98 and}63=\frac{\text{98}}{63}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{7}×\text{14}}{7×9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{14}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14:9\\ \left(\text{c}\right)\text{\hspace{0.17em}Ratio of 33\hspace{0.17em}km and}121\text{\hspace{0.17em}}\mathrm{km}=\frac{\text{33}}{121}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{11}}{11×11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3:11\\ \left(\mathrm{d}\right)\text{Ratio of}30\text{}\mathrm{minutes}\text{}\mathrm{to}\text{}45\text{}\mathrm{minutes}=\frac{30}{45}\\ =\frac{2×15}{3×15}\\ =\frac{2}{3}\\ =2:3\end{array}$

Q.7 Find the ratio of the following :
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes

Ans-

$\begin{array}{c}\left(\text{a}\right)\text{Ratio of 81 and 1}0\text{8}=\frac{\text{81}}{108}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{27}}{4×27}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3:4\\ \left(\text{b}\right)\text{Ratio of 98 and}63=\frac{\text{98}}{63}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{7}×\text{14}}{7×9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{14}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14:9\\ \left(\text{c}\right)\text{\hspace{0.17em}Ratio of 33\hspace{0.17em}km and}121\text{\hspace{0.17em}}\mathrm{km}=\frac{\text{33}}{121}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{11}}{11×11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3:11\\ \left(\mathrm{d}\right)\text{Ratio of}30\text{}\mathrm{minutes}\text{}\mathrm{to}\text{}45\text{}\mathrm{minutes}=\frac{30}{45}\\ =\frac{2×15}{3×15}\\ =\frac{2}{3}\\ =2:3\end{array}$

Q.8 Find the ratio of the following :
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes

Ans-

$\begin{array}{c}\left(\text{a}\right)\text{Ratio of 81 and 1}0\text{8}=\frac{\text{81}}{108}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{27}}{4×27}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3:4\\ \left(\text{b}\right)\text{Ratio of 98 and}63=\frac{\text{98}}{63}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{7}×\text{14}}{7×9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{14}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14:9\\ \left(\text{c}\right)\text{\hspace{0.17em}Ratio of 33\hspace{0.17em}km and}121\text{\hspace{0.17em}}\mathrm{km}=\frac{\text{33}}{121}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{3}×\text{11}}{11×11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3:11\\ \left(\mathrm{d}\right)\text{Ratio of}30\text{}\mathrm{minutes}\text{}\mathrm{to}\text{}45\text{}\mathrm{minutes}=\frac{30}{45}\\ =\frac{2×15}{3×15}\\ =\frac{2}{3}\\ =2:3\end{array}$

Q.9 Find the ratio of the following:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹ 1
(d) 500 ml to 2 litres

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}}1.5\text{hours}=1.5×60\text{\hspace{0.17em}}\mathrm{minutes}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\text{minutes}\\ \text{Ratio of 30 minutes to 1.5 hours}=\frac{30}{90}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3×10}{3×3×10}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}}=1:3\\ \\ \left(\mathrm{b}\right)1.5\text{m}=1.5×100\text{\hspace{0.17em}}\mathrm{cm}=150\text{cm}\\ \text{Ratio of 40 cm to 1.5 m}=\frac{40}{150}\\ \text{\hspace{0.17em}}=\frac{4×10}{15×10}\\ =\frac{4}{15}\\ =4:15\\ \left(\mathrm{c}\right)\mathrm{Since},\text{\hspace{0.17em}₹ 1}=100\text{paise}\\ \text{So, Ratio of}55\text{}\mathrm{paise}\text{}\mathrm{to}\text{₹}1=\frac{55}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5×11}{5×20}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{11}{20}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11:20\\ \left(\mathrm{d}\right)\text{Since 1 litre}=\text{1000 ml}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2 litres}=\text{2000 ml}\\ \text{So, the ratio of 500 ml to 2 litres}=\frac{500}{2000}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=1:4\end{array}$

Q.10 In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money she spends.

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{Seema earns money in a year}=\text{₹ 1},\text{5}0,000\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Seema saves money in a year}=\text{}₹\text{5}0,\text{}000\\ \text{Ratio of money that Seema earns to the money she saves}\\ \text{}=\frac{1,50,000}{50,000}\\ \text{}=\frac{15}{5}\\ \text{}=\frac{3}{1}\\ \text{}=3:1\\ \left(b\right)Seema\text{spends money in a year}=1,50,000-50,000\\ \text{}=1,00,000\\ \text{Ratio of money that Seema saves to the money she spends}\\ \text{}=\frac{50,000}{1,00,000}\\ \text{}=\frac{1}{2}\\ \text{}=1:2\end{array}$

Q.11 There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.

Ans-

$\begin{array}{l}\text{The ratio of the numberof teachers to the number of students}\\ \text{}=\frac{\text{1}0\text{2}}{3300}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2×3×17}{2×3×550}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{17}{550}=17:550\end{array}$

Q.12 In a college, out of 4320 students, 2300 are girls. Find the ratio of
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.

Ans-

$\begin{array}{l}\mathrm{Number}\text{of students in a college}=4320\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of girls in a college}=2300\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of boys in a college}=\text{4320}-2300\\ \text{\hspace{0.17em}\hspace{0.17em}}=2020\\ \left(\mathrm{a}\right)\mathrm{The}\text{ratio of the number of girls to the total number}\\ \text{of students}=\frac{2300}{4320}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×2×5×115}{2×2×5×216}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{115}{216}\\ \text{\hspace{0.17em}\hspace{0.17em}}=115:216\\ \left(\mathrm{b}\right)\mathrm{The}\text{ratio of the number}\mathrm{of}\text{}\mathrm{boys}\text{}\mathrm{to}\text{}\mathrm{the}\text{}\mathrm{number}\\ \text{}\mathrm{of}\text{}\mathrm{girls}=\frac{2020}{2300}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×2×5×101}{2×2×5×115}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{101}{115}\\ \text{\hspace{0.17em}\hspace{0.17em}}=101:115\\ \left(\mathrm{c}\right)\mathrm{The}\text{ratio of the number of boys to the total number of}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}students}=\frac{2020}{4320}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×2×5×101}{2×2×5×216}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{101}{216}\\ \text{\hspace{0.17em}\hspace{0.17em}}=101:216\end{array}$

Q.12 Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis.
If a student can opt only one game, find the ratio of
(a) Number of students who opted basketball to the number of students who opted table tennis.
(b) Number of students who opted cricket to the number of students opting basketball.
(c) Number of students who opted basketball to the total number of students.

Ans-

$\begin{array}{l}\mathrm{Total}\text{students in a school}=1800\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of students opted for basketball}=\text{750}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of students opted for cricket}=\text{800}\\ \mathrm{Number}\text{of students opted for table tennis}=1800-\left(\text{750+800}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1800-1550\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=250\\ \left(\mathrm{a}\right)\mathrm{The}\text{ratio of students for basketball to}\\ \text{\hspace{0.17em}\hspace{0.17em}the students for table tennis}=\frac{750}{250}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3:1\\ \left(\mathrm{b}\right)\mathrm{The}\text{ratio of students for cricket to the students}\\ \text{\hspace{0.17em}\hspace{0.17em}for basketball}=\frac{800}{750}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5×16}{5×15}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{16}{15}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16:15\\ \left(\mathrm{c}\right)\mathrm{The}\text{ratio of students for basketball to the total number}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of students}=\frac{750}{1800}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{60}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5:12\end{array}$

Q.13 Cost of a dozen pens is ₹ 180 and cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.

Ans-

$\begin{array}{l}\mathrm{Cost}\text{of a dozen pens}=₹\text{180}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Cost}\text{of a pens}=₹\text{}\frac{\text{180}}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=₹\text{}15\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Cost}\text{of 8 ball pens}=₹\text{56}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Cost}\text{of a ball pens}=₹\text{}\frac{\text{56}}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{₹}7\\ \mathrm{Ratio}\text{of the cost of a pen to the cost of a ball pen}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15:7\end{array}$

Q.14 Consider the statement : Ratio of breadth and length of a hall is 2:5. Complete the following table that shows some possible breadths and lengths of the hall.

 Breadth of the hall (in metres) 10 ……. 40 Length of the hall (in metres) 25 50 …….

Ans-

$\begin{array}{l}\text{Ratio of breadth and length of a hall = 2 : 5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{Length}}{\mathrm{Breadth}}=\frac{2}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{Length}}{50}=\frac{2}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}=\frac{2}{5}×50\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}=2×10\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\\ \text{So, required length is 20m.}\\ \text{Again,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{Length}}{\mathrm{Breadth}}=\frac{2}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{40}{\mathrm{Breadth}}=\frac{2}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Breadth}=\frac{5}{2}×40\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Breadth}=5×20\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\\ \text{So, required breadth is 100m.}\end{array}$

Q.15 Divide 20 pens between Sheela and Sangeeta in the ratio of 3: 2.

Ans-

$\begin{array}{l}\text{Let Sheela got pens}=\text{x}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Let Sangeet got pens}=\left(\text{2}0-\text{x}\right)\\ \mathrm{Ratio}\text{of pens got by Sheela and Sangeeta}=\frac{3}{2}\\ ⇒\frac{\mathrm{x}}{20-\mathrm{x}}=\frac{3}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=3\left(20-\mathrm{x}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=60-3\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}=60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{60}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=12\\ \mathrm{Therefore},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sheela got pens}=\text{12}\\ \text{And \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sangeet got pens}=\text{20}-12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\end{array}$

Q.16 Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Ans-

$\begin{array}{l}\mathrm{Let}\text{Shreya got money}=₹\text{x}\\ \text{Let Bhoomika got money}=₹\text{}\left(\text{36}-\text{x}\right)\\ \mathrm{According}\text{to the given condition,}\\ \text{Ratio of money got by Shreya and Bhoomika}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Ratio of the ages of Shreya and Bhoomika}\\ \\ ⇒\frac{\mathrm{x}}{36-\mathrm{x}}=\frac{15}{12}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12\mathrm{x}=15\left(36-\mathrm{x}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12\mathrm{x}=540-15\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}27\mathrm{x}=540\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{540}{27}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\\ \mathrm{Therefore},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Shreya got money}=₹\text{\hspace{0.17em}\hspace{0.17em}}20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Bhoomika got money}=₹\text{\hspace{0.17em}\hspace{0.17em}}\left(36-20\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=₹\text{\hspace{0.17em}}16\end{array}$

Q.17 Present age of father is 42 years and that of his son is 14 years. Find the ratio of
(a) Present age of father to the present age of son.
(b) Age of the father to the age of son, when son was 12 years old.
(c) Age of father after 10 years to the age of son after 10 years.
(d) Age of father to the age of son when father was 30 years old.

Ans-

$\begin{array}{l}\text{Present age of father}=42\text{years}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Present age of son}=14\text{years}\\ \left(\mathrm{a}\right)\text{Ratio of present age of father to the present age of son}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{42}{14}\\ =\frac{3}{1}\\ =3:1\\ \left(\mathrm{b}\right)\mathrm{When}\text{age son was 12 years.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}The age of father}=42\text{\hspace{0.17em}}-2\\ =40\text{\hspace{0.17em}}\mathrm{years}\\ \text{Ratio of present age of father to the present age of son}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{40}{12}\\ =\frac{10}{3}\\ =10:3\\ \left(\mathrm{c}\right)\mathrm{After}\text{10 years,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Age of father}=42+10\\ =52\text{\hspace{0.17em}}\mathrm{years}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Age of son}=14+10\\ =24\text{\hspace{0.17em}}\mathrm{years}\\ \mathrm{Ratio}\text{of age of father after 1}0\text{years to the age of son}\\ \text{after 1}0\text{years}=\frac{52}{24}\\ =\frac{4×13}{4×6}\\ =\frac{13}{6}\\ =13:6\\ \left(\mathrm{d}\right)\mathrm{When}\text{age of father was 30 years.}\\ \text{The difference between present age and age of 30 years}\\ =42-30\\ =12\text{\hspace{0.17em}}\mathrm{years}\\ \mathrm{The}\text{age of son before 12 years}\\ =14-12\\ =2\text{\hspace{0.17em}}\mathrm{years}\\ \mathrm{The}\text{ratio of age of father to the age of son when father}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}was 3}0\text{years old}=\frac{30}{2}\\ =\frac{15}{1}\\ =15:1\end{array}$

Q.18 Determine if the following are in proportion.
(a) 15, 45, 40, 120 (b) 33, 121, 9,96
(c) 24, 28, 36, 48 (d) 32, 48, 70, 210
(e) 4, 6, 8, 12 (f) 33, 44, 75, 100

Ans-

(a) Ratio of 15 and 45 = 15:45
= 1:3
Ratio of 40 and 120 = 40:120
= 1:3
Since, 15:45 = 40:120
Therefore, 15, 45, 40 and 120 are in proportion.

(b) Ratio of 33 and 121= 33:121
= 3:11
Ratio of 9 and 96 = 9:96
= 3:32
since, 33:121 ≠ 9:96
Therefore, 33, 121, 9 and 96 are not in proportion.
(c) Ratio of 24 and 28 = 24:28
= 6:7
Ratio of 36 and 48 = 36:48
= 3:4
since, 24:28 ≠ 36:48
Therefore, 24, 28, 36 and 48 are not in proportion.

(d) Ratio of 32 and 48 = 32:48
= 2:3
Ratio of 70 and 210 = 70:210
= 1:3
since, 32:48 ≠ 70:210
Therefore, 32, 48, 70 and 210 are not in proportion.
(e) Ratio of 4 and 6 = 4:6
= 2:3
Ratio of 8 and 12 = 8:12
= 2:3
since, 4:6 = 8:12
Therefore, 4, 6, 8 and 12 are in proportion.
(f) Ratio of 33 and 44 = 33:44
= 3:4
Ratio of 75 and 100 = 75:100
= 3:4
since, 33:44 = 75:100
Therefore, 33, 44, 75 and 100 are in proportion.

Q.19 Write True (T) or False (F) against each of the following statements:
(a) 16 : 24 :: 20 : 30 (b) 21: 6 :: 35 : 10
(c) 12 : 18 :: 28 : 12 (d) 8 : 9 :: 24 : 27
(e) 5.2 : 3.9 :: 3 : 4 (f) 0.9 : 0.36 :: 10 : 4

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 16 and 24}=\frac{16}{24}\\ =\frac{8×2}{8×3}\\ =\frac{2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 20 and 30}=\frac{20}{30}\\ =\frac{10×2}{10×3}\\ =\frac{2}{3}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}16:24::20:30}\\ \mathrm{Therefore},\text{it is true.}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 21 and 6}=\frac{21}{6}\\ =\frac{3×7}{3×2}\\ =\frac{7}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 35 and 10}=\frac{35}{10}\\ =\frac{5×7}{5×2}\\ =\frac{7}{2}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}21:6::35:10}\\ \mathrm{Therefore},\text{it is true.}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 12 and 18}=\frac{12}{18}\\ =\frac{6×2}{6×3}\\ =\frac{2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 28 and 12}=\frac{28}{12}\\ =\frac{4×7}{4×3}\\ =\frac{7}{3}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}12:18}\ne \text{28:12}\\ \mathrm{Therefore},\text{it is False.}\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 8 and 9}=\frac{8}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 24 and 27}=\frac{24}{27}\\ =\frac{3×8}{3×9}\\ =\frac{8}{9}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8:9}=\text{24:27}\\ \mathrm{Therefore},\text{it is True.}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}\mathrm{Ratio}\text{of 5.2 and 3.9}=\frac{5.2}{3.9}\\ =\frac{1.3×4}{1.3×3}\\ =\frac{4}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 3 and 4}=\frac{3}{4}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}5.2:3.9}\ne \text{3:4}\\ \mathrm{Therefore},\text{it is False.}\\ \left(\mathrm{f}\right)\text{\hspace{0.17em}}\mathrm{Ratio}\text{of 0.9 and 0.36}=\frac{0.9}{0.36}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{0.9×1}{0.9×4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of 10 and 4}=\frac{10}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2×5}{2×2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\\ \mathrm{So},\text{​ \hspace{0.17em}0.9:0.36}\ne \text{10:4}\\ \mathrm{Therefore},\text{it is False.}\end{array}$

Q.20 Are the following statements true?
(a) 40 persons: 200 persons = ₹ 15: ₹ 75
(b) 7.5 litres: 15 litres = 5 kg: 10 kg
(c) 99 kg: 45 kg = ₹ 44: ₹ 20
(d) 32 m: 64 m = 6 sec: 12 sec
(e) 45 km: 60 km = 12 hours: 15 hours

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{Ratio of 40 persons and 200 persons}=\frac{40}{200}\\ =\frac{40×1}{40×5}\\ =\frac{1}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of ₹15 and ₹75}=\frac{15}{75}\\ =\frac{15×1}{15×5}\\ =\frac{1}{5}\\ \mathrm{So},\text{}40\text{}\mathrm{persons}:200\text{}\mathrm{persons}=₹\text{}15:₹\text{}75\\ \mathrm{Therefore},\text{the given statement is true.}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 7.5 litres and 15 litres}=\frac{7.5}{15}\\ =\frac{7.5×1}{7.5×2}\\ =\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 5 kg and 10 kg}=\frac{5}{10}\\ =\frac{5×1}{5×2}\\ =\frac{1}{2}\\ \mathrm{So},\text{}7.5\text{}\mathrm{litres}:15\text{}\mathrm{litres}=5\text{}\mathrm{kg}:10\text{\hspace{0.17em}}\mathrm{kg}.\\ \mathrm{Therefore},\text{the given statement is true.}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 99 kg and 45 kg}=\frac{99}{45}\\ =\frac{9×11}{9×5}\\ =\frac{11}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of ₹ 44 and ₹20}=\frac{44}{20}\\ =\frac{4×11}{4×5}\\ =\frac{11}{5}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}99\text{}\mathrm{kg}:45\text{}\mathrm{kg}=₹\text{}44:₹\text{}20.\\ \mathrm{Therefore},\text{the given statement is true.}\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 32 m and 64 m}=\frac{32}{64}\\ =\frac{32×1}{32×2}\\ =\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 6\hspace{0.17em}sec and 12\hspace{0.17em}sec}=\frac{6}{12}\\ =\frac{6×1}{6×2}\\ =\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}32\text{}\mathrm{m}:64\text{}\mathrm{m}=6\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{sec}:12\text{\hspace{0.17em}}\mathrm{sec}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Therefore},\text{the given statement is true.}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 45 km and 60 km}=\frac{45}{60}\\ =\frac{3×15}{4×15}\\ =\frac{3}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of 12 hours and 15 hours}=\frac{12}{15}\\ =\frac{3×4}{3×5}\\ =\frac{4}{5}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}45 km}:\text{60 km}\ne \text{12 hours}:\text{15 hours}.\\ \mathrm{Therefore},\text{the given statement is false.}\end{array}$

Q.21 Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ₹ 40 : ₹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg: 80 kg and 25 g: 625 g
(d) 200 ml : 2.5 litre and ₹ 4 : ₹ 50

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of}25\text{}\mathrm{cm}\text{and}1\text{}\mathrm{m}=\frac{25}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Ratio}\text{of}₹\text{}40\text{and}₹\text{}160=\frac{40}{160}\\ =\frac{40×1}{40×4}\\ =\frac{1}{4}\\ \mathrm{Here},\text{\hspace{0.17em}}25\text{}\mathrm{cm}:1\text{}\mathrm{m}=₹\text{}40:₹\text{}160.\\ \mathrm{So},\text{}25\text{}\mathrm{cm}:1\text{}\mathrm{m}\text{}\mathrm{and}\text{}₹\text{}40:₹\text{}160\text{are in proportion.}\\ \text{Middle terms of the proportion are: 1m and ₹ 40.}\\ \text{Extreme terms of the proportion are: 25 cm and ₹ 160.}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}Ratio of}39\text{}\mathrm{litres}\text{}\mathrm{and}\text{}65\text{}\mathrm{litres}=\frac{39}{65}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{13×3}{13×5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}Ratio}\mathrm{of}\text{\hspace{0.17em}}6\text{}\mathrm{bottles}\text{}\mathrm{and}\text{}10\text{}\mathrm{bottles}=\frac{6}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2×3}{5×5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{5}\\ \mathrm{Here},\text{\hspace{0.17em}}39\text{}\mathrm{litres}:65\text{}\mathrm{litres}=6\text{}\mathrm{bottles}:10\text{}\mathrm{bottles}.\\ \mathrm{So},\text{}39\text{}\mathrm{litres}:65\text{}\mathrm{litres}::6\text{}\mathrm{bottles}:10\text{}\mathrm{bottles}\text{are in proportion.}\\ \text{Middle terms of the proportion are:}65\text{}\mathrm{litres}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}6\text{}\mathrm{bottles}\text{.}\\ \text{Extreme terms of the proportion are:}39\text{}\mathrm{litres}\text{and}10\text{}\mathrm{bottles}\text{.}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of}2\text{}\mathrm{kg}\text{}\mathrm{and}\text{}80\text{}\mathrm{kg}=\frac{2}{80}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio}\mathrm{of}\text{\hspace{0.17em}}25\text{}\mathrm{g}\text{}\mathrm{and}\text{}625\text{}\mathrm{g}=\frac{25}{625}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25×1}{25×25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{25}\\ \mathrm{Here},\text{\hspace{0.17em}}2\text{}\mathrm{kg}:80\text{}\mathrm{kg}\ne 25\text{}\mathrm{g}:625\text{}\mathrm{g}.\\ \mathrm{So},\text{}2\text{}\mathrm{kg}:80\text{}\mathrm{kg}\text{}\mathrm{and}\text{\hspace{0.17em}}25\text{}\mathrm{g}:625\text{}\mathrm{g}\text{are}\mathrm{not}\text{in proportion.}\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio of}200\text{}\mathrm{ml}\text{}\mathrm{and}\text{}2.5\text{}\mathrm{l}=\frac{200}{2500}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because 1\mathrm{l}=1000\text{\hspace{0.17em}}\mathrm{ml}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ratio}\mathrm{of}\text{\hspace{0.17em}}₹4\text{}\mathrm{and}\text{}₹50=\frac{4}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2×2}{2×25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{25}\\ \mathrm{Here},\text{\hspace{0.17em}}200\text{}\mathrm{ml}:2.5\text{}\mathrm{l}::₹4:₹50.\\ \mathrm{So},\text{}200\text{}\mathrm{ml}:2.5\text{}\mathrm{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}₹4:₹50\text{are in proportion.}\\ \text{Middle terms of the proportion are: 2.5}\mathrm{litres}\text{and ₹}4.\\ \text{Extreme terms of the proportion are: 200 ml and ₹ 50.}\end{array}$

Q.22 If the cost of 7 m of cloth is ₹ 294, find the cost of 5 m of cloth.

Ans-

$\begin{array}{l}\mathrm{The}\text{cost of 7\hspace{0.17em}m of cloth}=₹\text{\hspace{0.17em}}294\\ \mathrm{The}\text{cost of 1\hspace{0.17em}m of cloth}=₹\frac{\text{\hspace{0.17em}}294}{7}\\ \text{\hspace{0.17em}\hspace{0.17em}}=₹\text{\hspace{0.17em}}42\\ \mathrm{The}\text{cost of 5\hspace{0.17em}m of cloth}=₹\text{\hspace{0.17em}}42×5\\ \text{\hspace{0.17em}\hspace{0.17em}}=₹\text{\hspace{0.17em}}210\end{array}$

Q.23 Ekta earns ₹ 1500 in 10 days. How much will she earn in 30 days?

Ans-

$\begin{array}{l}\mathrm{Ekta}\text{earns money in 10 days}=₹\text{\hspace{0.17em}}1500\\ \mathrm{Ekta}\text{earns money in 1 day}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹\text{\hspace{0.17em}}\frac{1500}{10}\\ \mathrm{Ekta}\text{earns money in 30 days}=₹\text{\hspace{0.17em}}150×30\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹\text{\hspace{0.17em}}4500\end{array}$

Q.24 If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.

Ans-

$\begin{array}{l}\mathrm{The}\text{height of rain water in 3 days}=276\text{\hspace{0.17em}}\mathrm{mm}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{The}\text{height of rain water in 1 day}=\frac{276}{3}\text{\hspace{0.17em}}\mathrm{mm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=92\text{\hspace{0.17em}}\mathrm{mm}\\ \mathrm{The}\text{height of rain water in 7 days}=92\text{\hspace{0.17em}}\mathrm{mm}×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=644\text{\hspace{0.17em}}\mathrm{mm}\\ \mathrm{Thus},\text{644\hspace{0.17em}mm of rain will fall in one full week.}\end{array}$

Q.25 Cost of 5 kg of wheat is ₹ 30.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹ 61?

Ans-

$\begin{array}{l}\mathrm{The}\text{cost of 5 kg of wheat}=₹30.50\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{The}\text{cost of 1 kg of wheat}=₹\frac{30.50}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=₹6.10\\ \left(\mathrm{a}\right)\mathrm{The}\text{cost of 8 kg of wheat}=₹6.10×8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=₹48.80\\ \left(\mathrm{b}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Wheat}\text{bought for ₹ 30.50}=5\text{\hspace{0.17em}}\mathrm{kg}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Wheat}\text{bought for ₹1}=\frac{5}{30.10}\text{\hspace{0.17em}}\mathrm{kg}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Wheat}\text{bought for ₹ 61}=\frac{5}{30.10}\text{\hspace{0.17em}}\mathrm{kg}×₹61\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10.13\text{\hspace{0.17em}}\mathrm{kg}~10\text{\hspace{0.17em}}\mathrm{kg}\end{array}$

Q.26 The temperature dropped 15 degree celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?

Ans-

$\begin{array}{l}\mathrm{The}\text{temperature dropped in 30 days}=15\text{\hspace{0.17em}}\mathrm{°}\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{The}\text{temperature dropped in 1 day}=\frac{15}{30}\text{\hspace{0.17em}}\mathrm{°}\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}}=0.5\text{\hspace{0.17em}}\mathrm{°}\mathrm{C}\\ \mathrm{The}\text{temperature dropped in 10 days}=0.5\text{\hspace{0.17em}}\mathrm{°}\mathrm{C}×10\\ \text{\hspace{0.17em}\hspace{0.17em}}=5\text{\hspace{0.17em}}\mathrm{°}\mathrm{C}\\ \mathrm{Therefore},\text{5\hspace{0.17em}°C temperature will fall in next 10 days.}\end{array}$

Q.27 Shaina pays ₹ 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?

Ans-

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Sheena}\text{pays rent for 3 months}=\text{₹7500}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{ }\mathrm{Sheena}\mathrm{pays}\mathrm{rent}\mathrm{for}1\mathrm{month}=\text{₹}\frac{\text{7500}}{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹\text{\hspace{0.17em}}\text{\hspace{0.17em}}2500\\ \mathrm{Sheena}\mathrm{pays}\mathrm{rent}\mathrm{for}12\mathrm{months}=\text{₹2500}×\text{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹\text{\hspace{0.17em}}30,000\end{array}$

Q.28 Cost of 4 dozens bananas is ₹ 60. How many bananas can be purchased for ₹ 12.50?

Ans-

$\begin{array}{l}\mathrm{Bananas}\text{purchased for ₹ 60}=4\text{\hspace{0.17em}}\mathrm{dozens}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}48\left[\because 1\text{\hspace{0.17em}}\mathrm{dozen}=12\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Banana}\text{purchased for ₹1}=\frac{48}{60}\\ \mathrm{Banana}\text{purchased for ₹1}2.50=\frac{4}{5}×12.50\\ \text{\hspace{0.17em}}=10\\ \mathrm{Therefore},\text{10 bananas are purchased for ₹12.50.}\end{array}$

Q.29 The weight of 72 books is 9 kg. What is the weight of 40 such books?

Ans-

$\begin{array}{l}\mathrm{Weight}\mathrm{of}72\mathrm{books}=9\text{\hspace{0.17em}}\mathrm{kg}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Weight}\text{of 1 book}=\frac{9}{72}\text{\hspace{0.17em}}\mathrm{kg}\\ \mathrm{Weight}\text{of 40 such books}=\frac{1}{8}×40\text{\hspace{0.17em}}\mathrm{kg}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\mathrm{kg}\\ \mathrm{Thus},\text{the weight of 40 book is 5 kg.}\end{array}$

Q.30 A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Ans-

$\begin{array}{l}\mathrm{A}\text{truck requires diesel for covering distance of 594 km}\\ =108\text{\hspace{0.17em}}\mathrm{l}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\text{truck requires diesel for covering distance of 1 km}\\ =\frac{108}{594}\text{\hspace{0.17em}}\mathrm{l}\\ \mathrm{A}\text{truck requires diesel for covering distance of 1650 km}\\ =\frac{108}{594}×1650\text{\hspace{0.17em}}\mathrm{l}\\ =300\text{\hspace{0.17em}}\mathrm{l}\\ \mathrm{Thus},\text{300}\mathrm{l}\text{diesel is required for covering 1650 km.}\end{array}$

Q.31 Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper?

Ans-

$\begin{array}{l}\mathrm{Raju}\mathrm{paid}\mathrm{money}\mathrm{for}10\mathrm{pens}=₹150\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Raju}\mathrm{paid}\mathrm{money}\mathrm{for}1\mathrm{pen}=₹\frac{150}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹15\\ \mathrm{Manish}\mathrm{paid}\mathrm{money}\mathrm{for}7\mathrm{pens}=₹84\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Manish}\mathrm{paid}\mathrm{money}\mathrm{for}1\mathrm{pen}=₹\frac{84}{7}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹12\\ \mathrm{Therefore},\mathrm{Manish}\mathrm{got}\mathrm{the}\mathrm{pens}\mathrm{cheaper}\mathrm{.}\end{array}$

Q.32 Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?

Ans-

$\begin{array}{l}\mathrm{Ashish}\mathrm{made}\mathrm{runs}\mathrm{in}6\mathrm{overs}=42\\ \text{\hspace{0.17em}} \mathrm{Ashish}\mathrm{made}\mathrm{runs}\mathrm{in}1\mathrm{over}=\frac{42}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Anup}\mathrm{made}\mathrm{runs}\mathrm{in}7\mathrm{overs}=63\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Anup}\mathrm{made}\mathrm{runs}\mathrm{in}1\mathrm{over}=\frac{63}{7}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=9\\ \mathrm{Therefore},\mathrm{}\mathrm{Anup}\mathrm{made}\mathrm{more}\mathrm{runs}\mathrm{per}\mathrm{over}\mathrm{.}\end{array}$

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### 1. Should students make notes while referring to NCERT Solutions for Maths for Chapter 12?

Students should make notes while referring to NCERT Solutions by Extramarks. By making notes in their own words, they will be able to recall the concepts quickly. Also, writing down the formula will help students in remembering them for a long time.

### 2. Are NCERT Solutions for Class 6 Maths Chapter 12 sufficient to prepare for exams?

The NCERT Solutions for Class 6 Maths Chapter 12 have answers to all the questions from the textbook. However, students should also refer to other learning materials such as sample papers and mock tests available on Extramarks to prepare better for the examinations.

### 3. What are the best ways to grasp concepts in Class 6 Maths of Chapter 12?

The best way for students to grasp the concepts of Class 6 Maths Chapter 12 is by practising the chapter constantly and with due diligence. The students need to build a strong knowledge and have a clear understanding of all the concepts in the chapter to score better marks in exams.

### 4. What are equivalent ratios covered in Chapter 12 of Class 6 Maths?

When the given ratios are equal, they are known as equivalent ratios. They can be obtained by multiplying and dividing the numerator and denominator with the same number, for example, ratios 30:40(=3:4) and 33:44 (=3:4) are equivalent ratios.

### 5. Explain the golden ratio covered in Chapter 12 of Class 6 Maths.

Two quantities are said to be in a golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities.