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Haryana State Board HBSE 6th Class Maths Solutions Chapter 13 Symmetry Ex 13.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 13 Symmetry Exercise

13.2

Question 1.
Find the lines of symmetry for each of the following shapes.

Solution:

Question 2.
Copy the triangle in each of the following figures, on squares paper. In each case draw the line(s) of symmetry if any and identify the type of triangle. (Some of you may like to trace the figures and try paper-folding first.)

Solution:

Question 3.
Complete the following table :

Solution:

Question 4.
Can you draw a triangle which has
(a) exactly one line of symmetry ?
(b) exactly two lines of symmetry ?
(c) exactly three lines of symmetry ?
(d) no lines of symmetry ?
Sketch a rough figure in each case.
Solution:
(a) An isosceles triangle has exactly one line of symmetry.
(b) We cannot draw a triangle which has exactly two lines of symmetry.

(c) An equilateral triangle has exactly three lines of symmetry.
(d) A scalene triangle has no lines of symmetry.

Question 5.
On a squared paper-sketch the following :
[Hint: It will help if you first draw the lines of symmetry and then complete the figures.]
(a) A triangle with a horizontal line of symmetry but no vertical line of symmetry.
(b) A quadrilateral with both horizontal and vertical lines of symmetry.
(c) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
(d) A hexagon with exactly two lines of symmetry.
(e) A hexagon with six lines of symmetry.
Solution:
(a) An isosceles triangle has one line of symmetry. [Fig. (i)]
(b) A rectangle has two lines of symmetry. [Fig. (ii)]
(c) A trapezium has one line of symmetry. [Fig. (iii)]
(d) In Fig.(iv), a hexagon has exactly two lines of symmetry.
(e) A regular hexagon has six lines of symmetry. [Fig. (v)]

Question 6.
Trace each figure and draw the lines of symmetry, if any :

Solution:

Question 7.
Consider the English alphabet A to Z. List among them the letters which have
(a) vertical lines of symmetry, (like A)
(b) horizontal lines of symmetry, (like B)
(c) no line of symmetry (like Q)
Solution:
(a) Letters having vertical lines of symmetry are :
A, H, M, 0, T, U, V, W, X.
Cb) Letters having horizontal lines of symmetry are:
B, D, E, H, I, K, 0, X.
(c) Letters having no line of symmetry are :
C, F, G, J, L, N, P, Q, R, S, Y, Z.

Question 8.
Given here are figures of a few folded sheets and designs drawn about the fold. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 13 Symmetry Ex 13.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 13 Symmetry Exercise

13.3

Question 1.
Find the number of lines of symmetry in each of the following shapes : How will you check your answers ?

Solution:

Question 2.
Copy the following drawing on a squared paper. Complete each one of them such that the resulting figure has the two dotted lines as two lines of symmetry :

How did you go about completing the picture?
Solution:

Question 3.
In each figure below, a letter of the alphabet is shown along with a vertical line. Take the mirror image of the letter in the given line. Find which letters A look the same after reflection (i.ewhich letters look i\ the same in the image), and which do not. Can you guess why ?

Try for O, E, M, N, P, H, L, T, S, V, X.

Solution:
A, O, M, H, T, V and X look the same after reflection.
Because these letters are symmetrical.
B, E, N, P, L and S do not look the same after reflection.
Because these letters are not symmetrical.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 13 Symmetry Ex 13.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 13 Symmetry Exercise

13.1

Question 1.
List any four symmetrical object from your home or school.
Solution:
(i) Divider in the mathematical instrument box.
(ii) A bucket in the bathroom.
(iii) An open text book in the middle.
(iv) A closed lock.

Question 2.
For the given figure, which one is the mirror line, l₁, or l₂ ?
Solution:
Line l₂ is the mirror line for the given figure.

Question 3.
Identify the shapes given below.
Check whether they are symmetric or not. Draw the line of symmetry as well.

Solution:
(a) Lock: Symmetrical,
(b) Bucket: Symmetrical.
(c) Hook : Not symmetrical.
(d) Telephone : Symmetrical.
(e) A field : Symmetrical.
(f) Pentagon: Symmetrical.

Question 4.
Copy the following on a squares paper. A square paper is what you would have used in your arithmetic notebook in earlier classes. Then complete them such that the dotted line is the line of symmetry.

Solution:

Question 5.
In the figure, l is the line of symmetry. Complete the diagram to be symmetric.

Solution:

Question 6.
In the figure, l is the line of symmetry. Draw the image of the triangle and complete the diagram. So that it becomes symmetric.

Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 12 Ratio and Proportion Intext Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 12 Ratio and Proportion Intext Questions

TRY THESE (Page 323)

Question 1.
In a class there are 20 boys and 40 girls. What is the ratio of the number of boys to the number of girls ?
Solution:
Number of boys = 20
Number of girls = 40
∴ Ratio between the number of boys and the number of girls = \(\frac{20}{40}=\frac{1}{2}\) = 1 : 2.

Question 2.
Ravi walks 6 km in an hour while Roshan walks 4 km in an houF. What is the ratio of the distance covered by Ravi to the distance covered by Roshan ?
Solution:
Distance covered by Ravi = 6 km
Distance covered by Roshan = 4 km
∴ Ratio of the distance covered by Ravi to the distance covered by Roshan
= \(\frac{6 \mathrm{~km}}{4 \mathrm{~km}}=\frac{3}{2}\) = 3 : 2

TRY THESE (Page 325)

Question 1.
Saurabh takes 15 minutes to reach school from his house and Sachin takes one hour to reach school from his house. Find the ratio of the time taken by Saurabh to the time taken by Sachin.
Solution:
Time taken by Saurabh = 15 minutes Time taken by Sachin
= 1 hour = 60 minutes
∴ Ratio of the time taken by Saurabh to the time taken by Sachin
\(\frac{15 \text { minutes }}{60 \text { minutes }}=\frac{1}{4}\)
= 1:4.

Question 2.
Cost of a toffee is 50 paise and cost of a chocolate is Rs. 10. Find the ratio of the cost of a toffee to the cost of a chocolate.
Solution:
Cost of a toffee = 50 paise
Cost of a chocolate
= Rs. 10 = 10 x 100
= 1000 paise
∴ Ratio of the cost of a toffee to the cost of a chocolate
\(\frac{50 \text { paise }}{1000 \text { paise }}=\frac{1}{20}\)
= 1: 20.

Question 3.
In a school, there were 73 holidays in one year. What is the ratio of the number of holidays to the number of days in one year ?
Solution:
Number of holidays in the year = 73
Number of days in one year = 365
∴ Ratio of the number of holidays to the number of days in one year
= \(\frac{73}{365}=\frac{1}{5}\) = 1:5.

TRY THESE (Page 336)

(Check whether the given ratios are equal i.e., they are in proportion. If yes, then write them in proper form.
1. 1 : 5 and 3 : 15
2. 2:9 and 18 : 81
3. 15 : 45 and 5 : 25
4. 4 : 12 and 9 : 27
5. Rs. 10 to Rs. 15 and 4 to 6.
Solution:
1. 1 : 5 = \(\frac{1}{5}\) and 3 : 15 = \(\frac{3}{15}=\frac{1}{5}\)
Here, the two ratios 1 : 5 and 3 : 15 are equal.
1. e., they are in proportion
∴ 1 : 5 :: 3 : 15.

2. 2:9 = \(\frac{2}{9}\) and 18:81 = \(\frac{18}{81}=\frac{2}{9}\)
Here, the two ratios 2 : 9 and 18 : 81 are equal.
i.e., they are in proportion.
∴  2 : 9 :: 18 : 81.

3. 15:45= \(\frac{15}{45}=\frac{1}{3}\) and 5 : 25 = \(\frac{5}{25}=\frac{1}{5}\)
Here the two ratios 15 : 45 and 5 : 25 are not equal.
i.e., 15 : 45 * 5 : 25.
∴ They are not in proportion.

4. 4 : 12 = \(\frac{4}{12}=\frac{1}{3}\) and 9 : 27 = \(\frac{9}{27}=\frac{1}{3}\)
Here the two ratios 4 : 12 and 9 : 27 are equal.
i.e. they are in proportion.
∴ 4 : 12 :: 9 : 27.

5. Ratio of Rs. 10 to Rs. 15 = \(\frac{10}{15}=\frac{2}{3}\) and 4:6 = \(\frac{4}{6}=\frac{2}{3}\)
Here the two ratios are equal.
i.e. they are in proportion.
∴ Rs. 10 : Rs. 15 :: 4 : 6.
“In a statement of proportion, the four quantities involved are also known as respective terms. First and fourth terms are known as extreme terms. Second and third terms are known as middle terms.”

35, 70, 2, 4 are respective terms. 35 and 4 are extreme terms and 70 and 2 are middle terms.
Remember that, in a proportion
Product of extreme terms
= Product of middle terms
Here, product of extreme terms = 35 x 4 = 140
and product of middle terms
= 70 x 2 = 140.

Sharing a ratio means something !
Consider the following example :
Raju purchased 3 pens for Rs. 15 and Anu purchased 10 pens for Rs. 50, whose pens are more expensive ?
Ratio of number of pens purchased by Raju to the number of pens purchased by Anu = 3 : 10.
Ratio of their costs = 15 : 50 = 3 : 10
Both the ratios 3 : 10 and 15 : 50 are equal.
Therefore, the pens were purchased for the same price by both.

TRY THIS (Page 339) :

Question 1.
Read the table and fill in the blanks:

Solution:
(i) Distance travelled by Karan in 2 hrs = 8 km.
Distance travelled by Karan in 1 hr = \(\frac{8}{2}\) km = 4 km.
∴ Distance travelled
by Karan in 4 hrk = 4 x 4 = 16 km.
(ii) Distance travelled
by Kriti in 2 hrs = 6 km
Distance travelled
by Kriti in 1 hr = \(\frac{6}{2}\) km = 3 km
∴ Distance travelled
by Kriti in 4 hrs = 3 x 4 = 12 km.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 12 Ratio and Proportion Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions 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 ?
Solution:
Number of girls = 20
Number of boys = 15
(a) ∴ Ratio of number of girls to the number of boys
= \(\frac{20}{15}=\frac{4}{3}\) = 4 : 3 Ans-
(b) Total number of students
= No. of girls + No. of boys
= 20 + 15 = 35
∴ Ratio of number of girls to the total number of students
= \(\frac{20}{35}=\frac{4}{7}\) = 4:7

Question 2.
Out of 30 students in a class, 6 like football, 12 like cricket and the remaining 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.
Solution:
Total number of students = 30 Number of students liking football = 6 Number of students liking cricket = 12 Number of remaining students who like tennis
= 30 – (6 + 12)
= 30 – 18 = 12.
(а) Ratio of the number of students liking football to the number of students liking tennis
= \(\frac{6}{12}=\frac{1}{2}\) = 1 : 2- Ans*
(b) Ratio of the number of students liking cricket to the total number of students = \(\frac{12}{30}=\frac{2}{5}\) =2:5.

Question 3.
See Fig. and find the ratio of
(а) 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.

Solution:
Number of triangles inside the rectangle = 1
Number of circles inside the rectangle = 2
Total number of figures inside the rectangle
= 1 + 1 + 2 = 4

(а) Ratio of number of triangles to the number of circles inside the rectangle
= \(\frac{1}{2}\) = 1 : 2.
(b) Ratio of number of squares to all the figures inside the rectangle
= \(\frac{1}{4}\) =1:4.
(c) Ratio of number of circles to all the figures inside the rectangle
= \(\frac{2}{4}=\frac{1}{2}\) =1:2. Ans.

Question 4.
Distance 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.
Solution:
Speed of Hamid = 9 km/hr
Speed of Akhtar = 12 km/hr
∴ Ratio of speed of Hamid to the speed of Akhtar = \(\frac{9}{12}=\frac{3}{4}\) =3:4.

Question 5.
Fill in the following blanks :

[Are these equivalent ratios ?]
Solution:
In order to get the first missing number, we consider the fact that 18 = 3 x 6. This indicates that the numerator of the first number should be divided by 3.
When we divide we have, 15 ÷ 3 = 5.
Hence, the second ratio is 5/6.
Similarly, to get the third ratio, we multiply both the terms of the second ratio by 2.
Hence, the third ratio is 10/12.
Similarly, to get the fourth ratio, we multiply both the terms of the second ratio by 5.
Hence, the fourth ratio is 25/30.
Yes, these are all 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.
Solution:

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 Re. 1
(d) 500 ml to 2 litres.
Solution:
(a) Ratio of 30 minutes to 1.5 hours
= \(\frac{30 \text { minutes }}{1.5 \text { hours }}=\frac{30}{1.5 \times 60}=\frac{30}{90}=\frac{1}{3}\)
= 1:3.

(b) Ratio of 40 cm to 1.5 cm
= \(\frac{40 \mathrm{~cm}}{1.5 \mathrm{~m}}=\frac{40}{1.5 \times 100}=\frac{40}{150}=\frac{4}{15}\)
= 4: 15.

(c) Ratio of 55 paise to Re. 1
= \(\frac{55 \text { paise }}{\text { Re. } 1}=\frac{55}{100}=\frac{11}{20}\)
= 11 : 20.

(d) Ratio of 500 ml to 2 litres
= \(\frac{500 \mathrm{ml}}{2 \text { litres }}=\frac{500}{2 \times 1000}=\frac{1}{4}\)
= 1:4.

Question 8.
In a year, Seema earns Rs. 1,50,000 and saves Rs. 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.
Solution:
Money earned by Seema = Rs. 1,50,000
Money saved by Seema = Rs. 50,000
Money spent by Seema
= Rs. (1,50,000 – 50,000)
= Rs. 1,00,000

(a) Ratio of the money Seema earns to the money she saves
= \(\frac{\text { Rs. } 1,50,000}{\text { Rs. } 50,000}=\frac{3}{1}\) = 3:1.
(b) Ratio of the money she saves to the money she spends
= \(\frac{\text { Rs. } 50,000}{\text { Rs. } 1,00,000}=\frac{1}{2}\) = 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.
Solution:
Number of teachers = 102
Number of students = 3300
∴ Ratio of the number of teachers to the number of students

Question 10.
In a college put of4320 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.
Solution:
Total number of students = 4320 Number of girls = 2300
∴ Number of boys = 4320 – 2300 = 2020
(a) Ratio of number of girls to the total number of students
= \(\frac{2300}{4320}=\frac{2300 \div 20}{4320 \div 20}=\frac{115}{216}\)
= 115 : 216.
(b) Ratio of number of boys to the number of girls
= \(\frac{2020}{2300}=\frac{2020 \div 20}{2300 \div 20}=\frac{101}{115}\)
= 101 : 115. Ans.
(c) Ratio of number of boys to the total number of students
= \(\frac{2020}{4320}=\frac{2020 \div 20}{4320 \div 20}=\frac{101}{216}\)
= 101 : 216.

Question 11.
Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and the remainingoptedtable tennis. Ifastudent can opt only one game, find the ratio of:
(a) Number of students opting basket¬ball to the number of students opting table tennis.
(b) Number of students opting cricket to the number of students opting basketball.
(c) Number of students opting basketball to the total number of students.
Solution:
Total number of students = 1800
Number of students opting basketball = 750
Number of students opting cricket = 800
Number of students opting table tennis = 1800 – (750 + 800)
= 1800 – 1550 = 250
(a) Ratio of number of students opting basketball to the number of students opting
table tennis = \(\frac{750}{250}=\frac{3}{1}\) = 3 : 1
(b) Ratio of number of students opting cricket to the number of students opting
= \(\frac{800}{750}=\frac{16}{15}\) = 16 : 15.
(c) Ratio of number of students opting basketball to the total number of students
= \(\frac{750}{1800}=\frac{750 \div 150}{1800 \div 150}=\frac{5}{12}\) = 5 : 12

Question 12.
Cost of a dozen pens is Rs. 180 and cost of 8 ball pens is Rs. 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Solution:
Cost of 12 pens = Rs. 180
Cost of 1 pen = Rs. 180 ÷ 12
= Rs. 15

Cost of 8 ball pens = Rs. 56
Cost of 1 ball pen = Rs. 56∴ 8 = Rs. 7
Ratio of the cost of a pen to the cost of a ball pen
= \(\frac{\text { Rs. } 15}{\text { Rs. } 7}=\frac{15}{7}\)
= 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 breadth and length of the hall.

Solution:

Question 14.
Divide 20 pens between Sheela and Sangeeta in the ratio 3 : 2.
Solution:
Given ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
Total no. of pens = 20
Sheela’s share = \(\frac{3}{5}\) x 20 = 12
And Sangeeta’s share = \(\frac{2}{5}\) x 20 = 12

Question 15.
Mother wants to divide Rs. 36 among her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get ?
Solution:
Ratio of their ages
= 15 years : 12 years = 5:4
Sum of ratio = 5 + 4 = 9
Total money = Rs. 36
.’. Shreya’s share = Rs. \(\frac{5}{9}\) x 36 = Rs. 20
And Bhoomika’s share
= Rs. \(\frac{4}{9}\) x 36 = Rs. 16.

Question 16.
Present age of father is 42 years and his son is 14 years. Find the ratio of:
(a) Present age of father to the present age of son.
(b) Age of father to the age of son, when son was 12 yrs 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 yrs old.
Solution:
(a) Present age of father = 42 yrs.
Present age of son = 14 yrs.
∴ Ratio of the present age of father to the present age of son
= 42 : 14 = 3 : 1.

(b) When son was 12 yrs old, then father was 42 – 2 = 40 yrs old
∴ The ratio of their ages
= 40 : 12 = 10 : 3.

(c) Age of father after 10 yrs
= 42 + 10 = 52 yrs.
Age of son after 10 yrs
= 14 + 10 = 24 yrs.
∴ The ratio of their ages
= 52 : 24 = 13 : 6.

(d) When father was 30 yrs old, then son was 14 – 12 = 2 yrs old
∴ The ratio of their ages
= 30 : 2 = 15 : 1.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions 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.
Solution:
(a) Ratio of 15 to 45 = \(\frac{15}{45}=\frac{1}{3}\) = 1 : 3
Ratio of 40 to 120 = \(\frac{40}{120}=\frac{1}{3}\) =1:3
Since, 15 : 45 = 40 : 120
∴ 15, 45, 40, 120 are in proportion.

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

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

(d) Ratio of 32 to 48 = \(\frac{32}{48}=\frac{2}{3}\) =2:3
Ratio of 70 to 210 = \(\frac{70}{210}=\frac{1}{3}\) =1:3
Since, 32 : 48 ≠ 70 : 210
∴ 32, 48, 70, 210 are not in proportion.

(e) Ratio of 4 to 6 = \(\frac{4}{6}=\frac{2}{3}\) = 2:3
Ratio of 8 to 12 = \(\frac{8}{12}=\frac{2}{3}\) = 2 : 3
Since, 4 : 6 = 8 : 12
∴ 4, 6, 8, 12 are in proportion.

(f) Ratio of 33 to 44 = \(\frac{33}{44}=\frac{3}{4}\) = 3 : 4
Ratio of 75 to 100 = \(\frac{75}{100}=\frac{3}{4}\) = 3:4
Since, 33 : 44 = 75 : 100
33, 44, 75, 100 are in proportion.

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
Solution:
(a)

Product of extreme terms = 16 x 30 = 480
Product of middle terms = 24 x 20 = 480
∵ Product of extreme terms
= Product of middle terms
∴ 16 : 24 :: 20 : 30 is true.

(b)

Product of extreme terms = 21 x 10 = 210
Product of middle terms = 6 x 35 = 210
∵ Product of extreme terms
= Product of middle terms
∴ 21 : 6 :: 35 : 10 is true.

(c) 12 : 18 :: 28 : 12
Product of extreme terms = 12 x 12 = 144 Product of middle terms = 18 x 28 = 504
∵ Product of extreme terms
∴ Product of middle terms 12 : 18 :: 28 : 12 is false.

(d) 8:9 :: 24:27
Product of extreme terms = 8 x 27 = 216
Product of middle terms = 9 x 24 = 216
∵ Product of extreme terms = Product of middle terms
∴ 8:9 :: 24 : 27 is true.

(e) 5.2 : 3.9 :: 3 : 4
Product of extreme terms = 5.2 x 4 = 20.8
Product of middle terms = 3.9 x 3 = 11.7
∵ Product of extreme terms * Product of middle terms
∴ 5.2 : 3.9 :: 3 : 4 is false.

(f) 0.9 : 0.36 :: 10 : 4
Product of extreme terms = 0.9 x 4 = 3.6
Product of middle terms = 0.36 x 10 = 3.6
∵ Product of extreme terms = Product of middle terms
∴ 0.9 : 0.36 :: 10 : 4 is true.

Question 3.
Are the following statements true ?
(a) 40 persons : 200 persons = Rs. 15 : Rs. 75.
(b) 7.5 litres : 15 litres = 5 kg : 10 kg.
(c) 99 kg: 45 kg = Rs. 44: Rs. 20.
(d) 32 m : 64m = 6 sec : 12 sec.
(e) 45 km: 60 km = 12 hours: 15 hours.
Solution:
(a) 40 persons : 200 persons
= \(\frac{40}{200}=\frac{1}{5}\) = 1 : 5
Rs. 15 : Rs. 75
= \(\frac{15}{75}=\frac{1}{5}\) = 1:5
∴ 40 persons : 200 persons = Rs. 15 : Rs. 75 is true.

(b) 7.5 litre : 15 litre
= \(\frac{7.5}{15}=\frac{1}{2}\) = 1:2
5 kg : 10 kg = \(\frac{5}{10}=\frac{1}{2}\) = 1 : 2
∴ 7.5 litre : 15 litre = 5 kg : 10 kg is true.

(c)99 kg : 45 kg = \(\frac{99}{45}=\frac{11}{5}\) =11:5
Rs. 44 : Rs. 20 = \(\frac{44}{20}=\frac{11}{5}\) = 11 : 5
∴ 99 kg : 45 kg = Rs. 44 : Rs. 20 is true.

(d) 32m : 64m = \(\frac{32}{64}=\frac{1}{2}\) = 1 : 2
6 sec : 12 sec = \(\frac{6}{12}=\frac{1}{2}\) = 1 : 2
∴ 32 m : 64 m = 6 sec : 12 sec 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
∵ 3:4 ≠ 4:5
∴ 45 km : 60 km = 12 hours : 15 hours is false.

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, Rs. 40: Rs. 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 Rs. 4: Rs. 50.
Solution:
(a) 25 cm : 1 m = 25 cm : 100 cm
= \(\frac{25}{100}=\frac{1}{4}\) = 1:4
and Rs. 40 : Rs. 160 = \(\frac{40}{160}=\frac{1}{4}\) = 1 : 4
∴ They are in proportion.
i.e. 25 cm : 1 m :: Rs. 40 : Rs. 160
Here, middle terms are 100 cm and Rs. 40.
And extreme terms are 25 cm and Rs. 160.

(b) 39 litres : 65 litres = \(\frac{39}{65}=\frac{3}{5}\) = 3 : 5
and 6 bottles : 10 bottles = \(\frac{6}{10}=\frac{3}{5}\) =3:5
∴ They are in proportion.
i.e. 39 litres: 65 litres:: 6 bottles: 10 bottles.
Here, middle terms are 65 litres and 6 bottles.
And extreme terms are 39 litres and 10 bottles.

(c) 2 kg : 80 kg = \(\frac{2}{80}=\frac{1}{40}\) = 1 : 40
and 25 g . 625 g = \(\frac{25}{625}=\frac{1}{25}\) = 1:25
∴ 2 kg : 80 kg ≠ 25 g : 625
i.e. they are not in proportion.

(d) 200 ml: 2.5 litre = \(\frac{200}{2500}=\frac{2}{25}\) = 2 : 25
Rs. 4 : Rs. 50 = \(\frac{4}{50}=\frac{2}{25}\) = 2 : 25
∴ They are in proportion,
i.e. 200 ml: 2.5 litres :: Rs. 4 : Rs. 50
Here, middle terms are 2500 ml and Rs. 4.
And extreme terms are 200 ml and Rs. 50.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 12 Ratio and Proportion Ex 12.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 12 Ratio and Proportion Exercise

12.3

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

Question 2.
Ekta earns Rs. 1500 in 10 days. How much she will earn in 30 days ?
Solution:
Ekta earns in 10 days = Rs. 1500
She earns in 1 day = Rs. \(\frac{1500}{10}\) = Rs. 150
∴ She will earn in 30 days = Rs. 150 × 30
= Rs. 4500.

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.
Solution:
Rainfall in 3 days = 276 mm
Rainfall in 1 day = \(\frac{276}{3}\) = 92 mm
∴ Rainfall in 7 days
= 92 × 7 = 644 mm
= \(\frac{644}{10}\) cm = 64.4 cm.

Question 4.
Cost of 5 kg of wheat is Rs. 30.50.
(a) What will be the cost of 8 kg of wheat ?
(b) What quantity of wheat can be purchased in Rs. 61 ?
Solution:
(a) Cost of 5 kg of wheat = Rs. 30.50
Cost of 1 kg of wheat
= Rs. \(\frac{30.50}{5}\) = 6.10.
∴ Cost of 8 kg of wheat = Rs. 6.10 × 8
= Rs. 48.80.

(b) In Rs. 30.50 quantity of wheat purchased = 5 kg
In Re 1 quantity of wheat purchased
= \(\frac{5}{30.50}\) kg
∴ In Rs. 61 quantity of wheat purchased
=\(\frac{5}{30.50}\) × 61= 5 × 2 = 10kg.

Question 5.
The temperature dropped 15 degrees 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 ?
Solution:
Temperature dropped in the last 30 days = 15 degrees
Temperature dropped in one day
= \(\frac{15}{30}=\frac{1}{2}\) degree
∴ Temperature will drop in the next 10 days
= \(\frac{1}{2}\) × 10 = 5 degrees.

Question 6.
Shaina pays Rs. 7500 as rent for 3 months. How much does she have to pay for a whole year, if the rent for month remains the same ?
Sol. Rent for 3 months = Rs. 7500
Rent for 1 month = Rs. \(\frac{7500}{3}\) = Rs. 2500
Rent for a whole year
= Rs. 2500 × 12
= Rs. 30,000.

Question 7.
Cost of 4 dozens of banana is Rs. 60. How many bananas can be pur¬chased for Rs. 12.50 ?
Solution:
Number of bananas purchased for Rs. 60 = 4 dozen = 4 x 12 = 48
Number of bananas purchased for Re. 1
= \(\frac{48}{60}=\frac{4}{5}\)
Number of bananas purchased for Rs. 12.50
= \(\frac{4}{5}\) × 12.50 = 4 x 2.50 = 10.

Question 8.
The weight of 72 books is 9 kg. What is the weight of 40 such books ?
Solution:
Weight of 72 books = 9 kg
Weight of 1 book = \(\frac{9}{72}=\frac{1}{8}\) kg
∴ Weight of 40 books = \(\frac{1}{8}\) × 40 = 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 ?
Solution:
To cover a distance of 594 km a truck requires = 108l of diesel
To cover a distance of 1 km a truck requires = \(\frac{108}{594}\) l
To cover a distance of 1650 km a truck requires
= \(\frac{108}{594}\) × 1650 l = 300 l.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.1

Question 1.
Find the rule, which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

Solution:
(a)

To make one T, we use two matchsticks as shown in Fig.(i)

Thus, number of matchsticks required = 2n; n is a variable taking values 1, 2, 3, 4 ………..

(b) To make one Z, we use three matchsticks as shown in Fig. 11.2(f)

Thus, number of matchsticks required = 3n; n is a variable taking values 1, 2, 3, ……………

(c) To make one U, we use three matchsticks as shown in Fig. (i)

Thus, number of matchsticks required = 3n; n is a variable taking values 1, 2, 3, 4,

(d) To make one V, we use two matchsticks as shown in Fig. (i)

Thus, number of matchsticks required = 2n; n is a variable taking values 1, 2, 3, 4, 5, ………….

(e) To make one E, we use five matchsticks as shown in Fig. (i)

Thus, number of matchsticks required = 5n; n is a variable taking values 1, 2, 3, …………

(f) To make one S, we use five matchsticks as shown in Fig. (i)

Thus, number of matchsticks required = 5n; n is a variable taking values 1, 2, 3, 4, 5

(g) To make one A, we use six rnatchsticks as shown in Fig.(i)

Thus, number of matchsticks required = 6n; n is a variable taking values 1, 2, 3, 4, 5, …………….

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Question 1 (given back) give us the same rule as that given by L. Which are these ? Why does this happen ?
Solution:
(a) and (d) give us the same rule as that given by L. Because to make one T and one V, we require two matchsticks.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows ? (Use ‘n’ for the number of rows).
Solution:
The number of cadets will depend on the number of rows. If there is 1 row, there will be 5 cadets. If there are 2 rows, there will be 2 x 5 or 10 cadets and so on. If there are ‘n’ rows, there will be n x 5 or 5n cadets, here n is a variable which stands for the number of rows and takes values 1, 2, 3, 4 ……..

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes ? (Use ‘6’ for the number of boxes.)
Solution:
Number of mangoes in one box = 50
Number of mangoes in two boxes = 50 x 2 = 100
Number of mangoes in three boxes
= 50 x 3 = 150 and so on.
Number of mangoes in ‘b’ boxes = 50 x b = 50b
where ‘b’ is the no. of boxes.

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of
students ? (Use ‘s’ for the number of students.)
Solution:
Number of pencils needed for one student = 5
Number of pencils needed for two students = 5 x 2 = 10
Number of pencils needed for three students
= 5 x 3 = 15 and so on.
∴ Number of pencils needed for ‘s’ students = 5 x s = 50s.
where ‘s’ is the number of students.

Question 6.
A bird flies 1 kilometre in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes ? (Use T for flying time in minutes.)
Sol. Distance covered by the bird in one minute = 1 km
Distance covered by the bird in two minutes = 1 x 2 = 2 km
Distance covered by the bird in three minutes
= 1 x 3 = 3 Ion and so on.
∴ Distance covered by the bird in‘t’ minutes 1 x t = t km.
where ‘t’ is the flying time in minutes.

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder as in (Fig. 11.8). She has 8 dots in a row. How many dots will her Rangoli have for V rows ? How many dots are there if there are 8 rows ? If there are 10 rows ?
Solution:

Number of dots in one row = 8
Number of dots in two rows = 8 x 2 = 16
Number of dots in three rows = 8 x 3 = 24 and so on.
Number of dots in rows = 8 x r = 8r
Now, number of dots in 8 rows = 8 x 8 = 64
and number of dots in 10 rows = 8 x 10 = 80

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age ? Take Radha’s age to be V years.
Solution:
∵ Leela is 4 years younger than Radha.
∴ Leela’s age will be 4 years less than Radha’s age.
If Radha’s age = x years, then Leela’s age = (x – 4) years.

Question 9.
Mother has made laddus. She gives some laddus to guests and family members, still 5 laddus remain. If the number of laddus mother gave away is ‘l’, how many laddus did she make ?
Solution:
Number of laddus given to guests and family members = l
Number of laddus still remain = 5
Total number of laddus she made = l + 5.

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box ?
Solution:
Number of oranges in one small box = x
Number of oranges in two small boxes = 2x
Number of oranges remain outside = 10
∴ Number of oranges in the larger box = 2x + 10.

Question 11.
(a) Look at the following matchstick pattern of squares (Fig. 11.9).

The squares are not separate. Two neighbouring squares have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks in terms of the number of squares.
(Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)
Solution:

Thus, number of matchsticks required = 3x + 1, where x is a variable and represents the number of squares.

(b) Fig. 11.10 gives a matchstick pattern of triangles. As in Exercise 11(a) above, find the general rule that gives the no. of matchsticks in terms of the number of triangles.

Solution:

Thus number of matchsticks required = 2x + 1, where x is a variable and represents the number of triangles.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.2

Question 1.
The side of an equilateral triangle is shown by ‘l’. Express the perimeter of the equilateral triangle using T.

Solution:
Perimeter of the equilateral ΔABC
= AB + BC + AC
= I + l + l
= 3l.

Question 2.
The side of a regular hexagon (Fig.) is denoted by l. Express the peri¬meter of the hexagon using l.
(Hint : A regular hexagon has all its six sides equal in length.)

Solution:
Perimeter of a regular hexagon = l + l + l + l + l + l
= 6l.

Question 3.
A cube is a 1 three-dimensional figure as shown in Fig. It has six faces and all of them are identical squares.
The length of an edge of the cube is given by T. Find the formula for the total length of the edges of a cube.

Solution:
The length of the edges of a cube — l + l + l + l + l + l + l + l + l + l + l + l
= 12 l,

Question 4.
The dia¬meter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining Fig., AB is a diameter of the circle; C is its centre). Express the diameter of the circle (d) in terms of its radius (r).4

Solution:
We know that diameter of the circle = 2 x radius.
∴ d = 2r

Question 5.
Consider the sum of three numbers 14, 27 and 13. We may do the sum in two ways.
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54, or
(b) We may add 27 and 13 to get 40 and then add it to 14 to get the total sum 54. Thus,
(14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on whole numbers, in a general way, by using variables a, b and c.
Solution:
(a + b) + c = a + (b + c).

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.3

Question 1.
Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multi-plication.
[Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), 5 x 8 + 7 make the other expressions.]
Solution:
7 + (8 – 5), 7 – (8 – 5), 7 x 8 + 5, 7 x 8 – 5; 7 – (8 + 5) .
8 + (5 – 7), 8 – (5 – 7), 8 x 5 + 7, 8 x 5 – 7, 8 – (5 + 7) etc.

Question 2.
Which, out of the following, are expressions with numbers only ?
(a) y + 3
(b) 7 x 20 – 82
(c) 5 (21 – 7) + 7 x 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) 7 x 20 – 5 x 10 – 45 + p.
Solution:
(c) and (d) are expressions with numbers only.

Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed :
(a) z +1, z-1, y + 17, y-17
(b) 17 y, ~, 5 z
(c) 2 y + 17, 2y – 17
(d) 7 m, – 7m +3,- 7m – 3
Solution:

Question 4.
Give expressions in the following cases :
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) – p multiplied by 5
(g) – p divided, by 5
(h) p multiplied by – 5.
Solution:
(a) 7 added to p : p + 7
(b) 7 subtracted from p : p—7
(c) p multiplied by 7 : 7p
(d) p divided by 7 : p ÷ 7 or \(\frac{p}{7}\)
(e) 7 subtracted from — m. : — m — 7
(J) —p multiplied by 5 : 5 x (—p) or —5p
(g) —p divided by 5 : (—p) ÷ 5 or –\(\frac{p}{5}\)
(h) p multiplied by —5 : p x (—5) or —5p.

Question 5.
Give expressions in the following cases :
(a) 11 added to 2 m
(b) 11 subtracted from 2 m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by (-8)
(f) y is multiplied by (-8) and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by (-5) and the result is added to 16.
Solution:
(a) 2 m + 11
(b) 2m -11
(c) by + 3
(d) by – 3
(e) -8y
(f) -8y + 5
(g) 16 – 5y
(h) 16 – 5y.

Question 6.
(a) Form expressions using ‘£’ and 4. Use not more than one number operation. Every expression must have ‘t’ in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution:
(a) t + 4, t – 4, 4t, \(\frac{t}{4}\)
(b) 2y + 7, 2y – 7, 7y + 2, 7y – 2.