NCERT Solutions for Class 6 Science Chapter 9 The Living Organisms and Their Surroundings In Hindi

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

Q.5

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

Ans.

\begin{array}{l}\frac{15}{18}=\frac{3\times 5}{3\times 6}\\ =\frac{5}{6}\\ \frac{5}{6}=\frac{5\times 2}{6\times 2}\\ =\frac{10}{12}\\ \frac{5}{6}=\frac{5\times 5}{6\times 5}\\ =\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.

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

Q.7 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.

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

Q.8 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.

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

Q.9 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}\\ =\frac{\text{1}0\text{2}}{3300}\\ =\frac{2\times 3\times 17}{2\times 3\times 550}\\ =\frac{17}{550}=17:550\end{array}

Q.10 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.

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

Q.11 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.

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

Q.12 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.

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

Q.13 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.

Ans.

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

Q.14 Divide 20 pens between Sheela and Sangeeta in the ratio of 3: 2.

Ans.

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

Q.15 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.

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

Q.16 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}\\ =\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}=42-2\\ =40\mathrm{years}\\ \text{Ratio of present age of father to the present age of son}\\ =\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\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\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\times 13}{4\times 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\mathrm{years}\\ \mathrm{The}\text{age of son before 12 years}\\ =14-12\\ =2\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}$