Class 6

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.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.4 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.4

Question 1.
Answer the following :
(а) Take Sarita’s present age to be y years:
(i) What will be her age 5 years from now ?
iii) What was her age 3years back? (lit) Sarita’s grandfather’s age is 6
times her age. What is grand-father’s age ?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(iv) Sarita’s father’s age is 5years more than 3 times Sarita’s age. What is her father’s age ?
(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length if the breadth is ‘6’ metres?
(c) A rectangular box has a height ‘h’ cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step ‘s’, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Leena ? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using ‘s’.
(e) A bus travels at V km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur ? Express it using v.
Solution:
(a) Present age of Sarita = y years.
(i) Her age 5 years from now = (y + 5) years
(ii) Her age 3 years back – (y – 3) years
(iii) Her grandfather’s age = 6y years
(iv) Her grandmother’s age = (6y – 2) years
(v) Her father’s age = (3y + 5) years

(b) Breadth of the hall = b metres
Length of the hall = (3b – 4) metres

(c) Height of the box = h cm
Length of the box = 5h cm
Breadth of the box = (5h – 10) cm

(d) Meena is at step = s
∵ Beena is 8 steps ahead
∴ Beena is at steps =. s + 8.
∵ Leena is 7 steps behind
∴ Leena is at steps = s – 7
Total number of steps = 4s – 10

(e) Speed of the bus = v km per hour
∴ Distance covered in 5 hours = 5 v km.
Hence, distance from Daspur to Beespur = (5 v + 20) km.

Question 2.
Change the following statements into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match. Nalin scores (r + 15) runs. In ordinary language, Nalin scores 15 runs more than Salim.)
(a) A notebook costs Rs. ‘p’. A book costs Rs. 3p.
(b) Tony puts ‘q’ marbles on the table. He has 8q marbles in his box.
(c) Our class has ‘n’ students. The school has 20 n students.
(d) Jaggu is ‘z’ years old. His uncle is 4z years old and his aunt is (4z- 3) years old.
(e) In an array of dots, there are Y rows. Each row contains 5r dots.
Solution:
(a) A book costs t hree times the cost of a notebook.
(b) Tony’s box contains 8 times the marbles on the table.
(c) Total number of students in the school is 20 times that of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and his aunt is 3 years younger than his uncle.
(e) The number of dots in a row is 5 times the number of row.

Question 3.
(a) Given, Munnu’s age to be x years. Can you guess what (x – 2) may show ?
[Hint : Think of Munnu’s younger brother.]
Can you guess what (x + 4) may show ? What (3x + 7) may show ?
(b) Given Sara’s age today to be y years. Think of her age in the future or in past. What will the following expression indicate ?
y+1, y-3, y + 4\(\frac{1}{2}\), y-2\(\frac{1}{2}\)
(c) Given, n students in the class like football, what may 2n show ? What may \(\frac{n}{2}\) show ?
[Hint : Think of games other than football.]
Solution:
(a) Munnu’s younger brother’s age is 2 years less than Munnu’s age.
His elder sister’s age is 4 years more than his age.
His father’s age is 7 years more than thrice his age.
(b) y + 7 shows her age after 7 years.
y – 3 shows her age 3 years back.
y + 4\(\frac{1}{2}\) shows her age after 4\(\frac{1}{2}\) years.
y – 2\(\frac{1}{2}\) shows her age 2\(\frac{1}{2}\) years back.
(c) n students in the class like football.
2n shows that twice the students in the class like cricket.
\(\frac{n}{2}\) shows that half the students in the class like hockey.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.5 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.5

Question 1.
State whether the following are equations or not. Give reason for your answer. Further identify which are equations with only numbers and which are equations with variables. Identify the variables from the equations with variables. Give reason for your answer.
(a) 17 = x+7
(b) (t-7) > 5
(c) \(\frac{4}{2}\) = 8
(ci) 7 x 3 – 19 = 8
(e) 5 x 4 – 8 = 2x
(f) x-2 = 0
(g) 2m < 30
(h) 2n+11 = 1
(j) 7 = 11 x 5 – 12 x 4
(j) 7 = 11 x 2 + p
(k) 20= 5y
(l) \(\frac{3 q}{2}\)< 5 (m) z + 12 > 24
(n) 20- (10-5) = 3×5
(o) 7 – x = 5.
Solution:
(6) It is not an equation because there is no sign of (=).
(c) It is an equation with numbers only.
(d) It is an equation with numbers only
(e) It is an equation with variable ‘r’
(f) It is an equation with variable ‘x’.
(g) It is not an equation because there is no sign of (=).
(h) It is an equation with variable ‘n’
(i) It is an equation with numbers only
(j) It is an equation with variable ‘p’
(k) It is an equation with variable ‘y’
(l) It is not an equation because there is no sign of (=).
(m) It is not an equation because there is no sign of (=).
(n) It is an equation with numbers only
(o) It is an equation with variable ‘x’.

Question 2.
Complete the entries in the third column of the table :
Solution:

Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation. .
(a) 5m- 60 (10, 5, 12, 15)
(b) n + 12 = 20 (12, 8, 20, 0)
(c) p – 5 = 5 (O, 10, 5, -5)
(d) \(\frac{q}{2}\) = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, -4, 8, 0).
(f) x + 4 = 2 (-2, 0, 2, 4)
Solution:

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16
Solution:

m = 6 is the solution of the equation m + 10 = 16.

(b) Complete the table and by inspection of the table find the solution to the equation 5t = 35

Solution:
t = 7 is the solution of the equation 5t = 35.

(c) Complete the table and find the table the solution of the equation \(\frac{z}{3}\) = 4

z – 12 is the solution of the equation \(\frac{z}{3}\) = 4.

(d) Complete the table and find the solution to the equation m – 7 = 3

m = 10 is the solution of the equation m – 7 = 3.

Question 5.
Solve:
(a) x + 5 = 12
(b) y – 2 = 10
(c) 7p = 210
(d) \(\frac{q}{2}\) = 5
(e) t + 100 = 125
(f) l – 20 = 30
(g) 9u = 81
(h) \(\frac{k}{8}\) = 20
(i) 3y = 33
(j) x-3 = 0
(k) \(\frac{k}{8}\) = 8
(l) 13y = 65
Solution:
(a) x + 5 = 12
x = 12 – 5 = 7
Thus, x = 7 is the required solution of the given equation.

(b) y- 2 = 10
⇒ y = 10 + 2 = 12
Thus, y = 12 is the required solution of the given equation.

(c) 7p = 210
⇒ P = \(\frac{210}{7}\) = 30
Thus, p = 30 is the required solution of the given equation.

(d) \(\frac{q}{2}\) = 5
⇒ q = 2 x 5 = 10
Thus, q = 10 is the required solution of the given equation.

(e) t + 100 = 125
⇒ t = 125 – 100 = 25
Thus, t = 25 is the required solution of the given equation.

(f) 1 – 20 = 30
⇒ l = 30 + 20 = 50
Thus, 1 = 50 is the required solution of the given equation.

(g) 9 u = 81
⇒ u = \(\frac{81}{9}\) = 9
Thus, u = 9 is the required solution of the given equation.

(h) \(\frac{k}{8}\) = 20
⇒ k = 8 x 20= 160
Thus, ft = 160 is the required solution of the given equation.

(i) 3y = 33
⇒ y = \(\frac{33}{3}\) = 11
Thus, y = 11 is the required solution of the given equation.

(j) x- 3 = 0
⇒ x = 3
Thus, x = 3 is the required solution of the given equation.

(k) \(\frac{k}{8}\) = 8
⇒ k = 8 x 8 = 64
Thus, ft = 64 is the required solution of the given equation.

(l) 13y = 65
y = \(\frac{65}{13}\) = 5
Thus, y = 5 is the required solution of the given equation.

Question 6.
Solve the following riddles, you may yourself construct such riddles. Who am I ?
(i) Go round a square counting every corner thrice and no more. Add the count to me to get exactly thirty four!
(ii) I am a special number. Take away from me a six ! A whole cricket team, you will still be able to fix !
(iii) For each day of the week. Make an up cohnt from me A. If you make no mistake, you will get twenty three !
(iv) Tell “me who I am. I shall give a pretty clue ! You will get me back, if you take me out of twenty two !
Solution:
(i) Let I be V. Since a square has four corners.
Counting every corner thrice, we get 4 x 3 = 12
Now x + 12 = 34
⇒ x = 34 – 12 = 22
Thus, I am 22.

(ii) Let I be ‘x’. By taking away 6 from me, you get x – 6.
∵ There are 11 members in a cricket team.
∴ x – 6 = 11
⇒ x – 11 + 6 = 17.
Thus, I am 17.

(iii) There are 7 days in a week.
∴ A + 7 = 23
⇒ A = 23 – 7 = 16
Thus A = 16.

(iv) Let I be x.
Now, 22 – x = x
⇒ 2x = 22
⇒ x = \(\frac{22}{2}\) = 11
Thus, I am 11.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Intext Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Intext Questions

Question 1.
Measure and write the length of the four sides of the top of your study table. What is the perimeter ?
Solution:

AB = 150 cm
BC = 75 cm
CD = 150 cm
DA = 75 cm
Now, the sum of the L lengths of the four sides
= AB + BC + CD + DA
= 150 cm + 75 cm + 150 cm + 75 cm
= 450 cm
Hence, perimeter = 450 cm.

Question 2.
Measure and write the length of the four sides of a page of your note- book. What is the perimeter of the page?
Solution:
The sum of the lengths of the four sides
= AB + BC + CD + DA
= 20 cm + 15 cm + 20 cm + 15 cm
= 70 cm
Hence, perimeter of the page = 70 cm.

Question 3.
Meera went to a park 150 m long and 80 m wide. She took one complete round of it. What is the distance covered by her ?
Solution:
AB = 150 m,
BC = 80 m
CD = 150 m,
DA = 80 m
∴ Distance covered by Meera
= AB + BC + CD + DA
= 150 m + 80 m + 150 m + 80 m
= 460 m.

Question 4.
Find the perimeter of the following rectangles
Solution:

Question 5.
Biswamitra wants to put coloured tape all around a square picture (Fig. 10.2) of side 1 m as shown. What will be the length of the coloured tape he requires ?
Solution:
Length of one side = 1m
Perimeter of the square = 4 x 1 m = 4 m
Hence, 4 m coloured tape will be required.

Question 6.
Now, look at equilateral triangle (10.3) with each side of 4 cm. Can you find its perimeter ?

Solution:
Length of one side = 4 cm
Perimeter of equilateral triangle = 3 x 4 cm
= 12 cm.

Question 7.
Draw any circle on a graph sheet. Count the squares and use them to estimate the area of the circular region.
Solution:
(i) Ignore the portions of the area that are less than half a square.
(ii) If more than half of a square is in a region, just count it as one square.
(iii) If exactly half the square is counted, take it simply as \(\frac{1}{2}\) sq. cm.

Total area = 5 + 1 + 3 = 9sq. cm.

Question 8.
Trace the shape of a leave on the graph paper and estimate its area.
Solution:

Total area = 1 + 1 = 2 sq. cm.

Question 9.
Find the area of the floor of your classroom.
Solution:
Let length of the classroom = 7 m
and breadth of the classroom = 5 m
∴ Area of the classroom
= 7 m x 5 m = 35 sq. m.

Question 10.
Find the area of any one door in your house.
Solution:
Let length of the door = 3 m
and breadth of the door = 1.5 m
∴ Area of the door = 3 m x 1.5 m
= 4.5 sq. m.

Question 11.
Take a piece of 5 cm square paper and find its area.
Solution:
Side of the square = 5 cm
∴ Area of the square = side x side
= 5 cm x 5 cm = 25 cm².

Question 12.
The Jength of one side of a few squares is giyen, find their areas.
Solution:
Length of one side — Area of the square
3 cm = 3 x 3 = 9 sq. cm
7 cm = 7 x 7 = 49 sq. cm
4 cm = 4 x 4 = 16 sq. cm

Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.2

Question 1.
Find the area of the following figures :

Solution:
(a) This figure is made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares = 9.
∴ Total area = 9 sq. cm.
(b) This figure is also made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares 5
∴ Total area = 5 sq. cm.
(c) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.
(i) Fully filled squares = 2.
(ii) Half-filled squares = 4.
∴ Area covered by full squares
= 2 x 1 = 2 sq. cm
and area covered by half squares
= 4 x \(\frac{1}{2}\) = 2 sq. cm
Hence, total area = 2 + 2 = 4 sq. cm.

(d) This figure is made up of line- segments. Moreover, it is covered by full squares. This makes our job simple.
Here, fully filled squares = 8
∴ Total area = 8 x 1 = 8 sq. cm.

(e) This figure is made up of line-segments. Moreover, it is covered by full squares. Here, fully filled squares = 10
∴ Total area = 10 x 1 = 10 sq. cm.

(f) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 2
(ii) half-filled squares = 4
∴ Area covered by full squares
= 2 x 1 = 2 sq. cm
Area covered by half squares = 4 x \(\frac{1}{2}\) = 2 sq. cm
Hence, total area = 2 + 2 = 4 sq. cm.

(g) This figure is also made up of line- segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 4
(ii) half-filled squares = 4

∴ Area covered by full squares
= 4 x 1 = 4 sq. cm
Area covered by half squares
= 4 x \(\frac{1}{2}\) = 2 sq. cm
Hence, total area = 4 + 2 = 6 sq. cm.

(h) This figure is made up of line- segments. Moreover, it is covered by full squares only. This makes our job simple.
Here, fully filled squares = 5
∴ Total area = 5 x 1 sq. cm = 5 sq cm.

(i) This figure is made up of line-segments. Moreover, it is covered by full squares only. This makes our job simple.
Here, fully filled squares = 9
∴ Total area = 9 x 1 sq. cm = 9 sq. cm.

(j) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 2

(ii) half-filled squares = 4
∴ Area covered by full squares = 4 x \(\frac{1}{2}\) = 2 sq. cm
Hence,
total area = (2 + 2) sq. cm :
= 4 sq. cm.

(k) This figure is made up of line segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 4
(ii) half filled squares = 2
∴ Area covered by full squares
= 4 x 1 = 4 sq. cm
Area covered by half squares
= 2 x \(\frac{1}{2}\) = 1 sq. cm ,
Hence,
total area = (4+1) sq. cm
= 5 sq. cm.

(l)

Total area = 4 + 1 + 3 + 0 = 8sq. cm.

(m)

Total area = 8 + 1 + 4 + 0 = 13 sq. cm.

(n)

∴ Total area = 13 + 1 + 3 + 0 = 17 sq. cm.

Question 2.
Use tracing paper and centimeter graph paper to compare the areas of the following pair of figures :

Solution:
(a)

∴ Total area = 3 + 1 + 8 + 0 = 12 sq. cm.

(b)

Total area = 6 + 1\(\frac{1}{2}\) + 7 + 0 = 14\(\frac{1}{2}\) sq. cm.
Thus, area in Fig.(b) is greater than the area in Fig.(a).

Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.3

Question 1.
Find the area of the rectangles whose sides are :
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm.
Solution:
(a) Length of the rectangle (l) = 4 cm
Breadth of the rectangle (b) = 3 cm
.’. Area of the rectangle = l x b
= 4 cm x 3 cm
= 12 sq. cm.

(b) Length of the rectangle (l) = 21 m
Breadth of the rectangle (b) = 12 m
Area of the rectangle = l x b
= 21 m x 12 m
= 252 sq. m.

(c) Length of the rectangle (l) = 3 km
Breadth of the rectangle (b) = 2 km
∴ Area of the rectangle = l x b = 3km x 2km = 6 sq. km.

(d) Length of the rectangle (l)
= 2 m = 200 cm
Breadth of the rectangle (b) = 70 cm
∴ Area of the rectangle = l x b
= 200 x 70 sq. cm
= 14,000 sq. cm.

Question 2.
Find the areas of the squares whose sides are :
(a) 10 cm
(b) 14 cm
(c) 5 m.
Solution:
(a) Side of the square = 10 cm
∴ Area of the square = side x side
= 10 cm x 10 cm
= 100 sq. cm.

(b) Side of the square = 14 cm
∴ Area of the square = side x side
= 14 cm x 14 cm – 196 sq. cm.

(c) Side of the square = 5 m
.’. Area of the square = side x side
= 5 m x 5 m
= 25 sq. m.

Question 3.
Three rectangles have the following dimensions :
(a) 9 m and 6 m
(b) 3 m and 17 m
(c) 4 m and 14 m.
Which one has the largest area and which one has the smallest ?
Solution:
(a) Length of the rectangle = 9 m
Breadth of the rectangle = 6 m
.’. Area of the rectangle = length x breadth
= 9 m x 6 m = 54 sq. m.

(b) Length of the rectangle (l) = 17 m
Breadth of the rectangle (b) = 3 m
∴ Area of the rectangle = l x b
= 17 m x 3 m = 51 sq. m.

(c) Length of the rectangle (l) = 14 m
Breadth of the rectangle (b) = 4 m
∴ Area of the rectangle = l x b
= 14 m x 4 m = 56 sq. m.
Rectangle (c) has the largest area and rectangle (b) has the smallest area.

Question 4.
The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the garden.
Solution:
Area of the rectangle = 300 sq. m
Length of the rectangle = 50 m
Width of the rectangle = ?

∴ Area of the rectangle = length x width
∴ Width = \(\frac{\text { Area }}{\text { length }}=\frac{300}{50}\) = 6 cm
Hence, width of the rectangle = 6 m.

Question 5.
What is the cost of tiling a rect¬angular piece of land 500 m long and 200 m wide at the rate of Rs. 8 per hundred sq. m.
Solution:
Total area of the tiles must be equal to the area of the land.
Length of the land (l) = 500 m
Breadth of the land (b) = 200 m

∴ Area of the land = l x b
= 500 m x 200 m
Area of the tiles = 1,00,000 sq.m
Cost of 100 sq. m tiles = Rs. 8
Cost of 1 sq. m tiles = Rs. \(\frac{8}{100}\)
Cost of 1,00,000 sq. m tiles 8
= Rs. \(\frac{8}{100}\) x 1,00,000 = Rs. 8,000.

Question 6.
A table measures 2 m 25 cm by 1 m 50 cm. What is its area in square metres ?
Solution:
Length of the table (l) = 2.25 m
Breadth of the table (b) = 1.50 m

∴ Area of the table = lx b
= 2.25 x 1.50 sq. m
= 3.375 sq. m.

Question 7.
A room is 4 m 20 cm long and 3 m 65 cm wide. How many square metres of carpet is needed to cover the floor of the room ?
Solution:
Length of the room (l) = 4.20 m
Breadth of the room (b) = 3.65 m

∴ Area of the room = l x b
= 4.20 x 3.65 sq. m
= 15.33 sq. m
Hence, area of the carpet needed to cover the floor of the room = 15.33 sq. m.

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:
Length of the floor (l) = 5 m
Breadth of the floor (b) = 4 m

∴ Area of the floor = l x b
= 5 x 4 sq. m
= 20 sq. m
Side of the square carpet = 3 m
∴ Area of the square carpet = 3 m x 3 m
= 9 sq. m
Hence, area of the floor that is not carpeted = 20 – 9 = 11 sq. m.

Question 9.
Five square flower beds each of sides 1.2 m are dug on a piece of land 4.8 m long and 4.2 m wide. What is the area of the remaining part of land ?
Solution:
Length of the land (l) = 4.8 m
Bread of the land (b) = 4.2 m

∴ Area of the land = l x b
= 4.8 m x 4.2 m
= 20.16 sq. m.
Side of one square flower bed = 1.2 m
∴ Area of one square flower bed
= 1.2 m x 1.2 m
= 1.44 sq. m
∴  Area of the square flower beds
= 5 x 1.44 sq. m
= 7.20 sq. m
Hence, area of the remaining part of the land
= 20.16 sq. m – 7.20 sq. m = 12.96 sq. m.

Question 10.
The following figures have been split into rectangles. Find their areas (The measures are given in centimetres).

Solution:
(a) Total area of the given figure
= (3×3 + lx2 + 3×3 + 4×2)sq. m = (9 + 2 + 9 + 8) sq. cm
= 28 sq. cm

(b) Total area of the given figure
= (2 x 1 + 5 x 1 + 2 x 1) sq. cm
= (2 + 5 + 2) sq. cm = 9 sq. cm.

Question 11.
Split the following shapes into rectangles and find the area of each. (The measures are given in centimetres).

Solution:
(a) Total area of the given figure
= (10 x 2 + 10 x 2) sq. cm = (20 + 20) sq. cm = 40 sq. cm
(b) Total area of the given figure
= (7 x 7 + 21 x 7 + 7 x 7) Sq. cm = (49 + 147 + 49) sq. cm = 245 sq. cm.
(c) Total area of the given figure
= (4 x 1 + 5 x 1) sq. cm = (4 + 5) sq. cm = 9 sq. cm.

Question 12.
How many tiles with dimensions 5 cm and 12 cm will be needed to fit a region whose length and breadth are respectively :
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm.
Solution:
(a) Length of the region (l) = 144 cm
Breadth of the region (b) = 100 cm

∴ Area of the region = 144 x 100
= 14400 sq. cm
Length of one tile = 12 cm
Breadth of one tile = 5 cm
∴ Area of one tile = 12 x 5
= 60 sq. cm
Number of tiles required = \(\frac{\text { Area of the region }}{\text { Area of one tile }}=\frac{14400}{60}\) = 240 tiles.

(b) Length of the region (l) = 70 cm
Breadth of the region (b) = 36 cm

∴ Area of the region = 70 x 36 sq.cm
= 2520 sq.cm
Length of one tile = 12 cm
Breadth of one tile = 5 cm
∴ Area of one tile = 12 x 5 = 60 sq. cm
Number of tiles required = \(\) = 42 tiles

Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.1

Question 1.
Find the perimeter of each of the following figures :

Solution:
(a) Perimeter = 5 cm + 1 cm + 2 cm + 4 cm = 12 cm.
(b) Perimeter = 40 cm + 35 cm + 23 cm + 35 cm = 133 cm.
(c) Perimeter = 15 cm +15 cm + 15 cm + 15 cm = 60 cm.
(d) Perimeter = 3 cm + 3 cm + 3 cm + 3 cm + 3 cm = 15 cm.
(e) Perimeter = 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm.
(f) Perimeter = 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm
= 52 cm.

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required ?

Solution:
Length of the rectangular box (l) = 40 cm
Breadth of the rectangular box (b) = 10 cm
Perimeter of the rectangular box
= 2 x (l + b)
= 2 x (40 + 10) cm
= 2 x 50 = 100 cm
Hence, length of the tape required = 100 cm or 1m.

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the top of the table ?

Solution:
Length of the table-top (l)
= 2 m 25 cm
= 2.25 m
Breadth of the table-top (b)
= 1 m 50 cm
= 1.50 m
Perimeter of the table-top
= 2 x (l + b)
= 2 x (2.25 + 1.50) m
= 2 x 3.75
= 7.50 m.

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively ?

Solution:
Length of the frame (l)
= 32 cm
Breadth of the frame (b)
= 21 cm
Perimeter of the frame
= 2 x (l + b)
= 2 x (32 cm + 21 cm)
= 2 x 53 cm
= 106 cm.
Hence, length of the wooden strip required to frame a photograph = 106 cm.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of wire needed ?
Solution:
Length (l) = 0.7 km
Breadth (b) = 0.5 km
Perimeter – 2 x (l + b)
= 2 x (0.7 + 0.5) km
= 2 x 1.2 = 2.4 km

Length of wire needed to fence with 4 rows = 4 x 2.4 = 9.6 km.

Question 6.
Find the perimeter of each of the following shapes :
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solution:
(a) Perimeter of the triangle
= 3 cm + 4 cm + 5 cm = 12 cm.

(b) One side of an equilateral
Δ = 9 cm

.’. Perimeter of the triangle
= 3 x length of one side
= 3 x 9 cm = 27 cm.

(c) Perimeter of isosceles triangle
= 8 cm + 8 cm + 6 cm
= 22 cm.

Question 7.
Find the peri-meter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of the triangle
= 10 cm + 14 cm + 15 cm
= 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
One side of a regular hexagon = 8 m

Perimeter of regular hexagon = 6 x one side
= 6 x 8 m
= 48 m.

Question 9.
Find the side of a square whose perimeter is 20 m.

Solution:
Let, one side of a square = am
Perimeter of the square = 20 m
4 x a = 20
a = \(\frac{20}{4}\) = 5 cm
Hence, one side of the square = 5 m.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is each side?
Solution:
Let one side of the regular pentagon = a cm
Perimeter of the regular pentagon = 100 cm
5 x a = 100
=> a = \(\frac{100}{5}\) = 20 cm.
Hence, each side of a regular pentagon = 20 cm.

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
(a) a square ?
(b) an equilateral triangle ?
(c) a regular hexagon ?

Solution:
(a) Let one side of a square = a cm
Perimeter of the square = 30 cm
4 x a = 30 cm
=> a = \(\frac{30}{4}\) cm = 7.5 cm
Hence, length of each side of the square = 7.5 cm.

(b) Let each side of the equilateral Δ = a cm
Perimeter of the equilateral Δ = 30 cm
3 x a = 30 cm
a = \(\frac{30}{3}\)
= 10 cm

Hence, each side of the equilateral Δ = 10 cm.

(c) Let each side of the regular hexagon
= a cm

Perimeter of the regular hexagon = 30 cm
6 x a = 30 cm
=> a = \(\frac{30}{6}\) = 5 cm

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side ?
Solution:
Let the third side of a triangle = x cm

Perimeter of the triangle = 36 cm
.-. 14 cm + 12 cm + x cm = 36 cm
=> 26 cm + x cm = 36 cm
x = 36 – 26 = 10 cm
Hence, the third side of the triangle = 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per metre.

Solution:
Each side of the square park = 250 m
Perimeter of the square park
= 4 x 250 m
= 1000 m
Cost of one metre of wire = Rs. 20
Cost of fencing the square park = Rs. 20 x 1000
= Rs. 20,000.

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per metre.
Solution:
Length of the rectangular park (l) = 175 m
Breadth of the rectangular park (b) = 125 m
Perimeter of the rectangular park . = 2 x (l + b)
= 2 x (175 + 125) m
= 2 x 300
= 600 m
Cost of one metre of wire = Rs. 12
.•. Cost of fencing the rectangular park
= Rs. 12 x 600
= Rs. 7200.

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangle with length 60 metre and breadth 45 metre. Who covers less distance ?
Solution:

Each side of the square park = 75 m
Perimeter of the square park = 4 x 75 = 300 m
.’. Distance covered by Sweety = 300 m
Now, length of the rectangular park (l)
= 60 m
breadth of the rectangular park (b) = 45 m
Perimeter of the rectangular park = 2 x (l+ b)
= 2 x (60 + 45)
= 2 x 105 = 210 m
Distance covered by Bulbul = 210 m
Hence, Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures ? What do you infer from the answers ?

Solution:
(a) One side of the square = 25 cm
Perimeter of the square
=4 x 25 = 100 cm.

(b) Length of the rectangle (l) = 40 cm
Breadth of the rectangle (b) = 10 cm
Perimeter of the rectangle = 2 x (l + b)
= 2 x (40 + 10) cm
= 2 x 50 = 100 cm.

(c) Length of the rectangle (l) = 30 cm
Breadth of the rectangle (b) = 20 cm
.’. Perimeter of the rectangle
= 2 x (l + b)
= 2 x (30 + 20) cm
= 2 x 50 = 100 cm.

(d) Sides of the triangle are 40 cm, 30 cm and 30 cm.
Perimeter of the triangle
= 40 cm + 30 cm + 30 cm
= 100 cm.
We infer from the answers that all the figures have the same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each of a side of 1/2 m. He lays them in the form of a square.


(a) What is the perameter of his arrangement [Fig. 10.24(a)]?
(b) Shari does not like his arrangement.
She gets him to lay them out like a cross [Fig. 10.24(b)]. What is the perimeter of her arrangement ?
(c) Which has a greater perimeter ?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this ? (The paving slabs must meet along complete edges, they cannot be broken).
Solution: