CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.3

Question 1.
If the cost of 7 m of cloth is ₹ 294, find the cost of 5 m of cloth.
Answer:
Cost of 7 m of cloth = ₹ 294
∴ Cost of 1 m of cloth = \(\frac{294}{7}\) = ₹ 42
∴ Cost of 5 m of cloth = 42 × 5 = ₹ 210
Thus, the cost of 5 m of cloth is ₹ 210.

Question 2.
Ekta earns ₹ 1500 in 10 days. How much will she earn in 30 days?
Answer:
Earning of 10 days = ₹ 1500
∴ Earning of 1 day = \(\frac{1500}{10}\) = ₹ 150
∴ Earning of 30 days = 150 × 30 = ₹ 4500
Thus, the earning of 30 days is ₹ 9000.

Question 3.
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.
Answer:
Rain in 3 days = 276 mm
Rain in 1 day = \(\frac{276}{3}\) = 92 mm
∴ Rain in 7 days = 92 × 7 = 644 mm
Thus, the rain in 7 days is 644 mm.

Question 4.
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 ₹ 183?
Answer:
(a) Cost of 5 kg of wheat = ₹ 30.50
∴ Cost of 1 kg of wheat = \(\frac{30.50}{5}=\frac{3050}{500}\) = ₹ 6.10
∴ Cost of 8 kg of wheat = 6.10 × 8 = ₹ 48.80

(b) From ₹ 30.50, quantity of wheat can be purchased = 5 kg
∴ From ₹ 1, quantity of wheat can be purchased = \(\frac{5}{30.50}\)
From ₹ 61, quantity of wheat can be purchased = \(\frac{5}{30.50}\) × 61
= \(\frac{5}{3050}\) × 6100 = 10 kg

Question 5.
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?
Answer:
Degree of temperature dropped in last 30 days = 15 degrees
∴ Degree of temperature dropped in last 30 days = \(\frac{15}{30}=\frac{1}{2}\) degree
∴ Degree of temperature dropped in last 10 days
= \(\frac{1}{2}\) × 10 = 5 degree
Thus, 5 degree Celsius temperature dropped in 10 days.

Question 6.
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?
Answer:
Rent paid for 3 months = ₹ 7500
∴ Rent paid for 1 months = \(\frac{7500}{3}\) = ₹ 2500
∴ Rent paid for 12 months = 2500 × 12 = ₹ 30,000
Thus, the total rent of one year is ₹ 30,000.

Question 7.
Cost of 4 dozens bananas is ₹ 60. How many bananas can be purchased for ₹ 12.50?
Answer:
Cost of 4 dozen bananas = ₹ 60
Cost of 48 bananas = ₹ 60
[4 dozen = 4 × 12 = 48]
∴ From ₹ 60, number of bananas can be purchased = 48
∴ From ₹ 1, number of bananas can be purchased = \(\frac{48}{60}=\frac{4}{5}\)
= \(\frac{4}{5}\) × 12.50 = \(\frac{4}{5} \times \frac{1250}{100}=\frac{250}{25}\)
= 10 bananas
∴ From ₹ 12.50, number of bananas can be purchased
Thus, 10 bananas can be purchased for ₹ 12.50.

Question 8.
The weight of 72 books is 9 kg what is the weight of 40 such books?
Answer:
The weight of 72 books = 9 kg
The weight of 1 book = \(\frac{9}{12}=\frac{1}{8}\)
∴ The weight of 40 books
= \(\frac{1}{8}\) × 40 = 5 kg
Thus, the weight of 40 books is 5 kg.

Question 9.
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?
Answer:
For covering 594 km, a truck will be required diesel = 108 litres
For covering 1 km, a truck will be required diesel = \(\frac{108}{594}=\frac{2}{11}\)
∴ For covering 1650 km, a truck will be required diesel
= \(\frac{2}{11}\) × 1650 = 300 litres
Thus, 300 litres diesel required by the truck to cover a distance of 1650 km.

Question 10.
Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pen cheaper?
Answer:
Raju purchase 10 pens for = ₹ 150
∴ Raju purchases 1 pen for = \(\frac{150}{10}\) = ₹ 15
Manish purchases 7 pens for = ₹ 84
∴ Manish purchases 1 pen for = \(\frac{84}{7}\) = ₹ 12
Thus, Manish got the pens cheaper.

Question 11.
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Answer:
Anish made in 6 overs = 42 runs
∴ Anish made in 1 overs = \(\frac{42}{6}\) = 7 runs
∴ Anup made in 1 overs = \(\frac{63}{7}\) = 9 runs.
Thus, Anup made more runs per over.

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.4

Question 1.
What is the measure of
(i) a right angle?
(ii) a straight angle?
Answer:
(i) 90°
(ii) 180°

Question 2.
Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°. (c) The measure of a reflex angle > 180°.
(d) The measure of on complete revolution = 360°.
(e) If m∠A = 53° and m∠B = 35° then m∠A > m∠B.
Answer:
(a) True
(b) False
(c) True
(d) True
(e) True

Question 3.
Write down the measure of:
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each)
Answer:
(a) 35°, 20°
(b) 110°, 135°

Question 4.
Measure the angles give below, using the protractor and write down the measure:


Answer:
(a) 40°
(b) 130°
(c) 90°
(d) 60°

Question 5.
Which angle has a large measure? First estimate and then measure:
Measure of angle A = ?
Measure of angle B = ?

Answer:
∠B has larger measure.
∠A = 40° and ∠B = 65°

Question 6.
From these two angles which has larger measure? Estimate and then confirm by measuring them:

Answer:
Second angle has larger measure.

Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is …………………………
(b) An angle whose measure is greater than that of a right angle is …………………………
(c) An angle whose measure is the sum of the measures of two right angles is …………………………
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is …………………………
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……………………….
Answer:
(a) acute angle
(b) obtuse angle
(c) straight angle
(d) acute angle
(e) obtuse angle

Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Answer:
(i) 30°
(ii) 120°
(iii) 60°
(iv) 150°

Question 9.
Find the angle measure between the hands of the clock in each figure:

Answer:
(i) 90° (Right angle)
(ii) 30° (Acute angle)
(iii) 180° (Straight angle)

Question 10.
Investigate:
In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Answer:
No, the measure of angle will be same

Question 11.
Measure and classify each angle:

Answer:

These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.2

Question 1.
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
Answer:
(a) 15 : 45 = \(\frac{15}{45}=\frac{1}{3}\) = 1 : 3
40 : 120 = \(\frac{40}{120}=\frac{1}{3}\) = 1 : 3
Since, 15 : 45 = 40 : 120
Therefore, 15,45,40,120 are in proportion.

(b) 33 : 121= \(\frac{33}{121}=\frac{3}{11}\) = 3 : 11
9 : 96 = \(\frac{9}{96}=\frac{3}{32}\) = 3 : 32
Since, 33 : 121 ≠ 9 : 96
Therefore, 33,121,9,96 are not in proportion.

(c) 24 : 28 = \(\frac{24}{28}=\frac{6}{7}\) = 6 : 7
36 : 48 = \(\frac{36}{48}=\frac{3}{4}\) = 3 : 4
Since, 24 : 28 ≠ 36 : 48
Therefore, 24, 28,36, 48 are not in proportion.

(d) 32 : 48 = \(\frac{32}{48}=\frac{2}{3}\) = 2 : 3
70 : 210= \(\frac{70}{210}=\frac{1}{3}\) = 1 : 3
Since, 32 : 48 ≠ 70 : 210
Therefore, 32,48, 70, 210 are not in proportion.
(e) 4 : 6 = \(\frac{4}{6}=\frac{2}{3}\) = 2 : 3
8 : 12 = \(\frac{8}{12}=\frac{2}{3}\) = 2 : 3
Since, 4 : 6 = 8 : 12
Therefore, 4,6,8,12 are in proportion.

(f) 33 : 44 = \(\frac{33}{44}=\frac{3}{4}\) = 3 : 4
75 : 100 = \(\frac{75}{100}=\frac{3}{4}\) = 3 : 4
Since, 33 : 44 = 75 : 100
Therefore, 33, 44, 75,100 are in ratio.

Question 2.
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
Answer:
(a) 16 : 24 :: 20 : 30
⇒ \(\frac{16}{24}=\frac{20}{30}\) ⇒ \(\frac{2}{3}=\frac{2}{3}\)
Hence, it is true.

(b) 21 : 6 :: 35 : 10
⇒ \(\frac{21}{6}=\frac{35}{10}\) ⇒ \(\frac{7}{2}=\frac{7}{2}\)
Hence, it is true.

(c) 12 : 18 :: 28 : 12
⇒ \(\frac{12}{18}=\frac{28}{12}\) ⇒ \(\frac{2}{3} \neq \frac{7}{3}\)
Hence, it is false.

(d) 8 : 9 :: 24 : 27
⇒ \(\frac{8}{9}=\frac{24}{27}\) ⇒ \(\frac{8}{9}=\frac{8}{9}\)
Hence, it is true.

(e) 5.2 : 3.9 :: 3 : 4
⇒ \(\frac{5.2}{3.9}=\frac{3}{4}\) ⇒ \(\frac{4}{3} \neq \frac{3}{4}\)
Hence, it is false.

(f) 0.9 : 0.36 :: 10 : 4
⇒ \(\frac{0.9}{0.36}=\frac{10}{4}\) ⇒ \(\frac{5}{2}=\frac{5}{2}\)
Hence, it is true.

Question 3.
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
Answer:
(a) 40 persons : 200 persons
= \(\frac{40}{200}=\frac{1}{5}\) = 1 : 5
₹ 15 : ₹ 75 = \(\frac{15}{75}=\frac{1}{5}\) = 1 : 5
Since, 40 persons : 200 persons = ₹ 15 : ₹ 75
Hence, the statement is true.

(b) 7.5 litres : 15 litres
= \(\frac{7.5}{15}=\frac{75}{150}=\frac{1}{2}\) = 1 : 2
5 kg : 10 kg = \(\frac{5}{10}=\frac{1}{2}\) = 1 : 2
Since, 7.5 litres : 15 litres = 5 kg: 10 kg
Hence, the statement is true.

(c) 99 kg : 45 kg = \(\frac{99}{45}=\frac{11}{5}\) = 11 : 5
₹ 44 : ₹ 20= \(\frac{44}{20}=\frac{11}{5}\) = 11 : 5
Since, 99 kg: 45 kg = ₹ 44 : ₹ 20 Hence, the statement is true.

(d) 32 m : 64 m = \(\frac{32}{64}=\frac{1}{2}\) = 1 : 2
6 sec : 12 sec = \(\frac{6}{12}=\frac{1}{2}\) = 1 : 2
Since, 32 m : 64 m = 6 sec : 12 sec Hence, the statement is true.

(e) 45 km : 60 km = \(\frac{45}{60}=\frac{3}{4}\) = 3 : 4
12 hours : 15 hours
= \(\frac{12}{15}=\frac{4}{5}\) = 4 : 5
Since, 45 km : 60 km ≠ 12 hours : 15 hours
Hence, the statement is not true.

Question 4.
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
Answer:
(a) 25 cm : 1 m = 25 cm : (1 × 100) cm
= 25 cm : 100 cm
= \(\frac{25}{100}=\frac{1}{4}\) = 1 : 4
₹ 40 : ₹ 160 = \(\frac{40}{160}=\frac{1}{4}\) = 1 : 4
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 1 m, ₹ 40 and
Extreme terms = 25 cm, ₹ 160

(b) 39 litres : 65 litres
= \(\frac{39}{65}=\frac{3}{5}\) = 3 : 5
6 bottles : 10 bottles
= \(\frac{6}{10}=\frac{3}{5}\) = 3 : 5
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 65 litres, 6 bottles and Extreme terms = 39 litres, 10 bottles

(c) 2 kg : 80 kg = \(\frac{2}{80}=\frac{1}{40}\) = 1 : 40
25 g : 625 g = \(\frac{25}{625}=\frac{1}{25}\) = 1 : 25
Since, the ratios are not equal, therefore these are not in proportion,

(d) 200 ml : 2.5 litres = 200 ml: (25000) litres = 200 ml : 2500 ml = \(\frac{200}{2500}=\frac{2}{25}\) = 2 : 25
₹ 4: ₹ 50 = \(\frac{4}{50}=\frac{2}{25}\) = 2 : 25
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 2.5 litres, ₹ 4 and Extreme terms = 200 ml, ₹ 50

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation?
Answer:
Comparing the line segments simply by ‘observation’ may not be accurate. For example, the line segments AB and CD (in the following figure) seem to be equal, but actually they are not.

Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Answer:
It is better to use a divider than a ruler, because the thickness of the ruler may cause difficulties in reading off the length. However, divider gives up accurate measurement.

Question 3.
Draw any line segment, say \(\begin{equation}
\overline{\mathbf{A B}}
\end{equation}\). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line, such that AC + CB = AB, then we can be sure that C lies between A and B.]
Answer:
Yes.

\(\begin{equation}
\overline{\mathbf{A C}}
\end{equation}\) = 6.3 cm
\(\begin{equation}
\overline{\mathbf{B C}}
\end{equation}\) = 2.7 cm
\(\begin{equation}
\overline{\mathbf{A B}}
\end{equation}\) – 9.0 cm
∵ AC + BC = 6.3 cm + 2.7 cm = 9.0 cm and AB = 9.0 cm
∴ AC + BC = AB Hence verified.

Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Answer:
∵ AB = 5 cm
∵ BC = 3 cm
∴ AB + BC = 5 cm + 3 cm = 8 cm
But AC = 8 cm
The point B lies between A and C.

Question 5.
Verify whether D is the mid-point of \(\begin{equation}
\overline{\mathbf{A G}}
\end{equation}\).

Answer:
AD = 3 units, DG = 3 units

AD = DG.
Thus, D is the mid-point.

Question 6.
If B is the mid-point of \(\begin{equation} \overline{\mathbf{A C}}\end{equation}\) and C is the mid-point of \(\begin{equation}\overline{\mathbf{B D}}\end{equation}\) where A, B, C, D lie on a straight line, say why AB = CD?
Ans.
B is the mid-point of \(\begin{equation} \overline{\mathbf{A C}}\end{equation}\).

∴ AB = BC … (i)
And C is the mid-point of \(\begin{equation} \overline{\mathbf{BD}}\end{equation}\).
∴ BC = CD … (ii)
From equation (i) and (ii), we get AB = CD

Question 7.
Draw five triangles and measure their sides. Check in each case, of the sum of the lengths of any two sides is always less than the third side.
Answer:
Yes, sum of two sides of a triangle is always greater than the third side.

These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.1

Question 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?
Answer:
(a) The ratio of girls to that of boys = \(\frac{20}{15}=\frac{4}{3}\) = 4 : 3

(b) The ratio of girls to total students = \(\frac{20}{20+15}=\frac{20}{35}=\frac{4}{7}\) = 4 : 7

Question 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.
Answer:
Total number of students = 30
Number of students like football = 6
Number of students like cricket = 12
Thus number of students like tennis = 30 – 6 – 12 = 12

(a) The ratio of students like football that of tennis = \(\frac{6}{12}=\frac{1}{2}\) = 1 : 2
(b) The ratio of students like cricket to that of total students = \(\frac{12}{30}=\frac{2}{5}\) = 2.5

Question 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.
Answer:
(a) Ratio of number of triangle to that of circles = \(\frac{3}{2}\) = 3 : 2
(b) Ratio of number of squares to all figures = \(\frac{2}{7}\) = 2 : 7
(c) Ratio of number of circles to all figures = \(\frac{2}{7}\) = 2 : 7

Question 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.
Answer:
We know that, Speed = \(\frac{\text { Distance }}{\text { Time }}\)
Speed of Hamid = \(\frac{9 \mathrm{~m}}{1 \mathrm{~h}}\) = 9 km/h and Speed of Akhtar \(\frac{12 \mathrm{~m}}{1 \mathrm{~h}}\) = 12 km/h
Ratio of speed of Hamid to that of speed of Akhtar = \(\frac{9}{12}=\frac{3}{4}\) = 3 : 4

Question 5.
Fill in the following blanks:
Answer:

Yes, these are equivalent ratios.

Question 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
Answer:
(a) Ratio of 81 to 108
= \(\frac{81}{108}=\frac{3}{4}\) = 3 : 4

(b) Ratio of 98 to 63
= \(\frac{98}{63}=\frac{14}{9}\) = 14 : 9

(c) Ratio of 33 km to 121 km
= \(\frac{33}{121}=\frac{3}{11}\) = 3 : 11

(d) Ratio of 30 minutes to 45 minutes
= \(\frac{30}{45}=\frac{2}{3}\) = 2 : 3

Question 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
Answer:
(a) 30 minutes to 1.5 hour
1.5 hours = 1.5 × 60 = 90 minutes
[∵ 1 hour = 60 minutes]
Now, ratio of 30 minutes to 1.5 hour = 30 minutes: 1.5 hour
⇒ 30 minutes : 90 minutes
= \(\frac{30}{90}=\frac{1}{3}\) = 1 : 3

(b) 40 cm to 1.5 m
1.5 m = 1.5 × 100 cm = 150 cm
[∵ 1 m = 100 cm]
Now. ratio of 40 cm to 1.5 m
= 40 cm : 1.5 m
⇒ 40 cm : 150 cm
= \(\frac{40}{150}=\frac{4}{15}\) = 4 : 15

(c) 55 paise to ₹ 1
Rs 1 = 100 paise
Now, ratio of 55 paise to ₹ 1
= 55 paise : 100 paise
⇒ \(\frac{55}{100}=\frac{11}{20}\) = 11 : 20

(d) 500 ml to 2 litters
2 litres = 2 × 1000 ml = 2000 ml
[∵ litre = 1000 ml]
Now, ratio of 500 ml to 2 litres
= 500 m 1 : 2 litres
⇒ 500 ml : 2000 ml
= \(\frac{500}{2000}=\frac{1}{4}\) = 1 : 4

Question 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.
Answer:
Total earning
= ₹ 1,50,000 and Saving = ₹ 50,000
∵ Money spent
= ₹ 1,50,000 – ₹ 50,000 = ₹ 1,00,000

(a) Ratio of money earned to money
saved = \(\frac{150000}{50000}=\frac{3}{1}\) = 3 : 1

(b) Ratio of money saved to money
spend = \(\frac{50000}{100000}=\frac{1}{2}\) = 1 : 2

Question 9.
There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Answer:
Ratio of number of teachers to that of studends = \(\frac{102}{3300}=\frac{17}{550}\) = 17 : 550

Question 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.
Answer:
Total number of students in school = 4320
Number of girls = 2300
Therefore, number of boys = 4320 – 2300 – 2020

(a) Ratio of girls to total number of students = \(\frac{2300}{4320}=\frac{115}{216}\) = 115 : 216

(b) Ratio of boys to that of girls = \(\frac{2020}{2300}=\frac{101}{115}\) = 101 : 115

(c) Ratio of boys to total number of students = \(\frac{2020}{4320}=\frac{101}{216}\) = 101 : 216

Question 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.
Answer:
Total number of students = 1800
Number of students opted basketball = 750
Number of students opted cricket = 800
Therefore, number of students opted tennis = 1800 – (750 + 800) – 250

(a) Ratio of students opted basketball to that of opted table tennis
= \(\frac{750}{250}=\frac{3}{1}\) = 3 : 1

(b) Ratio of students opted cricket to students opted basketball
= \(\frac{750}{1800}=\frac{5}{12}\) = 16 : 15

(c) Ratio of students opted basketball to total no. of students
= \(\frac{750}{1800}=\frac{5}{12}\) = 5 : 12

Question 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.
Answer:
Cost of a dozen pens (12 pens) = ₹ 180
∴ Cost of 1 pen = \(\frac{180}{2}\) = ₹ 15
Cost of 8 ball pens = ₹ 56
∴ Cost of 1 ball pen = \(\frac{56}{8}\) = ₹ 7
Ratio of cost of one pen to that of one ball pen = \(\frac{15}{7}\) = 15 : 7

Question 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.

Answer:
Ratio of breadth to length = 2 : 5 = \(\frac{2}{5}\)
∴ Other equivalent ratios are

Question 14.
Divide 20 pens between Sheela and Sangeeta in the ratio of 3 : 2.
Answer:
Ratio between Sheela and Sangeeta = 3 : 2
Total these terms = 3 + 2 = 5
Therefore, the part of Sheela = \(\frac{2}{5}\) of the total pens and the part of Sangeeta = \(\frac{2}{5}\) of total pens
Thus, Sheela gets = \(\frac{3}{5}\) x 20 = 12 pens and Sangeeta gets = \(\frac{2}{5}\) x 20 = 8 pens

Question 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.
Answer:
Ratio of the age of Shreya to that of Bhoomika = \(\frac{15}{12}=\frac{5}{4}\) = 5 : 4
Thus, ₹ 36 divide between Shreya and Bhoomika in the ratio of 5 : 4.
Shreya gets = \(\frac{5}{9}\) of ₹ 36 = \(\frac{5}{9}\) × 36 = ₹ 20 Bhoomika gets = \(\frac{4}{9}\) of ₹ 36 = \(\frac{4}{9}\) × 36 = ₹ 16

Question 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.
Answer:
(a) Ratio of father’s present age to that of son = \(\frac{42}{14}=\frac{3}{1}\) = 3 : 1

(b) When son was 12 years, i.e., 2 years ago, then father was (42 – 2) = 40
Therefore, the ratio of their ages = \(\frac{40}{12}=\frac{10}{3}\) = 10 : 3

(c) Age of father after 10 years
= 42 + 10 = 52 years
Age of son after 10 years
= 14 + 10 = 24 years
Therefore, ratio of their ages = \(\frac{52}{24}=\frac{13}{6}\) = 13 : 6

(d) When father was 30 years old, i.e., 12 years ago, then son was (14 – 12] = 2 years old
Therefore, the ratio of their ages
= \(\frac{30}{2}=\frac{15}{1}\) = 15 : 1

These NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas InText Questions

NCERT In-text Question Page No. 71

Question 17.
Name the line segments in the figure 4.2. Is A, the end point of each line segment?

The line sengments are:
i) \(\begin{equation}
\overline{\mathrm{AB}}(\text { or } \overline{\mathrm{BA}})
\end{equation}\)
(ii) \(\begin{equation}
\overline{\mathrm{AC}}(\text { or } \overline{\mathrm{CA}})
\end{equation}\)
‘Yes, A is the end point of each line segment.

NCERT In-text Question Page No. 74

Question 18.
Name the rays given in this picture.

Answer:
Rays in figure are: TA, TN, TB

Question 19.
Is T a starting point of each of these rays?
Answer:
No. [∵ T is not the starting point of]