Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

1. Following data gives total marks (out of 600) obtained by six students of a particular class. Represent the data on a bar graph.

Solution:
(i) To choose an appropriate scale we make equal division taking increments of 100.
Thus, 1 unit represent 100 marks.

(ii) Now represent the data on the bar graph.

2. The following bar graph shows the number of books sold by a bookstore during five consecutive years. Read the bar graph and answer the following questions :

Question (i).
About how many books were sold in 2008, 2009 and 2011 years ?
Solution:
140; 360; 180,

Question (ii).
In which year about 475 books were sold ? And in which year about 225 books were sold ?
Solution:
2012; 2010.

3. Two hundred students of 6th and 7th class were asked to name their favourite colour so as to decide upon what should be the colour of their school building. The results are shown in the table :

Represent the data on a graph.
Answer the following questions with the help of bar graph :

Question (i).
Which is the most preferred colour ?
Answer:
Choose a suitable scale as follows :
Start the scale at 0. The greatest value in data is 55, so end the scale at a value greater than 55, such as 60.

Question (ii).
Which is the least preferred colour ?
Answer:
Use equal divisions along the vertical axis, such as increments = 10. All the bars would between 0 and 60. We choose the scale such that the length between 0 and 60 is neither too long nor too small. Here we take 1 unit for 10 students.

Question (iii).
How many colours are there in all ? What are they ?
Answer:
We then draw and label the graph as shown :

From the graph we conclude that :

• Blue is the most preferred colour Because the bar representing Blue is the tallest).
• Green is the least preferred colour (Because the bar representing Green is the shortest).
• There are five colours. They are Red, Green, Blue, Yellow and Orange.

4. Consider the following data collected from a survey of a colony :

Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph ?
(i) Which sports is the most popular ?
(ii) Which is more preferred, watching or participating in sports ?
Solution:
Take different sports along X = axis and number of persons watching and participating favourite sports Y-axis.
Scale. Take 1 unit height along Y-axis = 200 persons. The double bar graphs representing the given data is shown below.

5. The following table shows the time (in hours) spent by a student of class VII in a day.

Draw a bar graph to represent the above data. What do you infer from the above table ?
Solution:
Choose a suitable scale as follows :
(i) Start the scale at 0. The greatest value in the data is 8, so end the scale at a value greater than 8, such as 9. Take 1 unit of height of bars.
(ii) Use equal division along vertical axis, such as increments is 10. All the bars would be between 0 and 9.
(iii) We then draw and label the graph as below :

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.
If 7 is added to five times a number, the result is 57. Find the number.
Solution:
Let the required number = x
Five times the number = 5x
7 added to five times the number = 5x + 7
According to the problem
5x + 7 = 57
5x = 57 – 7
5x = 50
x = \(\frac {50}{5}\)
So, x = 10
Hence the required number is 10.

Question 2.
9 decreased from four times a number yields 43. Find the number.
Solution:
Let the required number = x
Four times the number = 4x
9 decreased from four times the number = 4x – 9
According to the problem
4x – 9 = 43
4x = 43 + 9
4x = 52
x = \(\frac {52}{4}\)
x = 13
Hence, the required number is 13.

Question 3.
If one-fifth of a number minus 4 gives 3, find the number.
Solution:
Let the required number = x
One fifth of the number = \(\frac {1}{5}\)x
One fifth of the number minus 4 = \(\frac {1}{5}\)x – 4
According to problem
\(\frac {1}{5}\)x – 4 = 3
\(\frac {1}{5}\)x = 3 + 4
\(\frac {1}{5}\)x = 7
x = 35
Hence the required number is 35.

Question 4.
In a class of 35 students, the number of girls is two-fifth the number of boys. Find the number of girls in the class.
Solution:
Let the number of boys = x
∴ number of girls = \(\frac {2}{5}\)x
Total number of students = 35
x + \(\frac {2}{5}\)x =35
\(\frac{5 x+2 x}{5}\) = 35
7x = 5 × 35
x = \(\frac{5 \times 35}{7}\)
x = 25
Therefore number of boys = 25
Number of girls = 35 – 25 = 10.

Question 5.
Sham’s father’s age is 5 years more than three times Sham’s age. Find Sham’s age, if his father is 44 years old.
Solution:
Let Sham’s age = x years
Then Sham’s father age = 3x + 5
But Sham’s fathers age = 44
According to question
3x + 5 = 44
3x = 44 – 5
3x = 39
Dividing both sides by 3
\(\frac{3 x}{3}=\frac{39}{3}\)
or x = 13
Hence Sham’s age is 13 years.

Question 6.
In an isosceles triangle the base angles are equal, the vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°)
Solution:
Let each base angle of an isosceles triangle = x (in degrees)
Vertex angle = 40°
The sum of angles of a triangle = 180°
∴ x + x + 40° = 180°
2x = 180° – 40°
2x = 140°
Divide both sides by 2
\(\frac{2 x}{2}-\frac{140^{\circ}}{2}\)
Or x = 70°
Each equal angle is of 70°

Question 7.
Irfan says that he has 7 marbles more than five times the marbles Pannit has. Irfan has 37 marbles. How many marbles does Pannit have ?
Solution:
Let marbles Parmit has = x
Marbles Irfan has = 5x + 7
But Marbles Irfan has = 37
∴ 5x + 7 = 37
5x = 37 – 7
5x = 30
x = \(\frac {30}{5}\) = 6
Therefore Parmit has 6 marbles.

Question 8.
The length of a rectangle is 3 units more than its breadth and the perimeter is 22 units. Find the breadth and length of a rectangle.
Solution:
Let breadth of rectangle (l)
= x units
∵ length of rectangle (b) = (x + 3) units
∴ Perimeter of rectangle = 2(l + b)
= 2 (x + x + 3) units
= 2(2x + 3) units
According to the question
Perimeter = 22 units
2 (2x + 3) =22
\(\frac{2(2 x+3)}{2}=\frac{22}{2}\)
2x + 3 = 11
2x = 11 – 3
or 2x = 8
Dividing both sides by 2 we get
\(\frac{2 x}{2}=\frac{8}{2}\)
x = 4
∴ breadth = 4 units
Length = (4 + 3) units
= 7 units

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

1. Solve each of the following equation.

Question (i).
6x + 10 = – 2
Answer:
Given equation is 6x + 10 = – 2
Transposing + 10 from L.H.S to R.H.S
we get
6x = -2 – 10
or 6x = -12
Dividing both sides by 6, we get
\(\frac{6 x}{6}=\frac{-12}{6}\)
or x = – 2, which is the required solution.

To check Put x = – 2 in the LHS of the equation 6x + 10 = – 2
L.H.S. = 6x + 10
= 6 × -2 + 10
= -12 + 10
= – 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2y – 3 = 2
Answer:
Given equation is 2y – 3 = 2
Transposing – 3 from L.H.S. to R.H.S,
we get
2y = 2 + 3
or 2y = 5
Dividing both sides by 2, we get:
\(\frac{2 y}{2}=\frac{5}{2}\)
or y = \(\frac {5}{2}\), which is the required solution

To check. Put y = \(\frac {5}{2}\) in the L.H.S of the equation 2y – 3 = 2
L.H.S = 2y – 3 = 2 × \(\frac {5}{2}\) – 3
= 5 – 3 = 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iii).
\(\frac{a}{5}\) + 3 = 2
Answer:
Given equation is \(\frac{a}{5}\) + 3 = 2
Transposing + 3 from L.H.S to R.H.S., we get
\(\frac{a}{5}\) = 2 – 3
or \(\frac{a}{5}\) = -1
Multiplying both sides, by 5, we get
5 × \(\frac{a}{5}\) = 5 × – 1
or a = – 5, which is the required solution.

To Check: Put a = – 5 in the L.H.S of the equation
\(\frac{a}{5}\) + 3 = 2,
L.H.S. = \(\frac{a}{5}\) + 3
= \(\frac {-5}{5}\) + 3
= – 1 + 3
= 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
\(\frac{3 x}{2}=\frac{2}{3}\)
Answer:
Given equation is \(\frac{3 x}{2}=\frac{2}{3}\)
Multiplying both sides by 2, we get
2 × \(\frac{3 x}{2}\) = 2 × \(\frac {2}{3}\)
or 3x = \(\frac {4}{3}\)
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{4}{3} \times \frac{1}{3}\)
or x = \(\frac {4}{9}\), which is the required solution.

To Check. Put x = \(\frac {4}{9}\) in the L.H.S. of equation \(\frac{3 x}{2}=\frac{2}{3}\)
L.H.S. = \(\frac{3 x}{2}=\frac{3}{2} \times \frac{4}{9}\) = \(\frac {2}{3}\) = R.H.S.
∴L.H.S. = R.H.S.

Question (v).
\(\frac {5}{2}\)x = -5
Answer:
Given equation is \(\frac {5}{2}\) x = – 5
Multiplying both sides by 2, we get
2 × \(\frac {5}{2}\) x = 2 × – 5
or 5x = – 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{-10}{5}\)
or x = – 2, which is the required solution.

To Check. Put x = – 2 in L.H.S. of the equation \(\frac {5}{2}\)x = – 5
L.H.S. = \(\frac {5}{2}\)x = \(\frac {5}{2}\) × -2
= – 5 = R.H.S.
∴ L.H.S. = R.H.S.

Question (vi).
2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Answer:
Given equation is 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Subtract \(\frac {5}{2}\) from both sides, we get
2x + \(\frac {5}{2}\) – \(\frac {5}{2}\)
= \(\frac {37}{2}\) – \(\frac {5}{2}\)
or 2x = \(\frac{37-5}{2}\)
or 2x = \(\frac {32}{2}\)
or 2x = 16
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{16}{2}\)
or x = 8, which is the required solution.

To Check. Put x = 8 in the L.H.S. of the equation 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
L.H.S. = 2x + \(\frac {5}{2}\)
= 2 × 8 + \(\frac {5}{2}\)
= 16 + \(\frac {5}{2}\)
= \(\frac{32+5}{2}\)
= \(\frac {37}{2}\) = R.H.S.
∴ L.H.S. = R.H.S.

2. Solve the following equation

Question (i).
5 (x + 1) = 25
Answer:
Given equation is 5 (x + 1) = 25
Dividing both sides by 5 we get
\(\frac{5(x+1)}{5}=\frac{25}{5}\)
or x + 1 = 5
Transposing 1 from L.H.S. to R.H.S. we get
x = 5 – 1
or x = 4, which is the required solution.

To Check. Put x = 4 in the L.H.S. of the equation 5 (x + 1) = 25
L.H.S. = 5 (x + 1)
= 5 (4 + 1)
= 5 (5)
= 25 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2 (3x – 1) = 10
Answer:
Given equation is 2 (3x – 1) = 10
Dividing both sides by 2, we get
\(\frac{2(3 x-1)}{2}=\frac{10}{2}\)
or 3x – 1 = 5
Transposing – 1 from L.H.S. to R.H.S we get
3x = 5 + 1
3x = 6
Dividing both sides by 3, we get \(\frac{3 x}{3}=\frac{6}{3}\)
or x = 2, which is the required solution.

To Check. Put x = 2, in the L.H.S. of the equation 2 (3x – 1) = 10
L.H.S. = 2 (3x – 1) = 10
L.H.S = 2 (3x – 1) = 2 (3 × 2 – 1)
= 2 (6 – 1)
= 2 × 5
= 10 = R.H.S.
∴L.H.S. = R.H.S.

Question (iii).
4 (2 – x) = 8
Answer:
Given equation is 4 (2 – x) = 8
Dividing both sides by 4, we get
\(\frac{4(2-x)}{4}=\frac{8}{4}\)
or 2 – x= 2
Transposing 2 from L.H.S. to R.H.S. we get
-x = 2 – 2
or – x = 0
Multiplying both sides by – 1, we get
-x × – 1 = x – 1
or x = 0, which is the required solution.

To Check. Put x = 0 in the L.H.S. of the equation 4 (2 – x) = 8
L.H.S. = 4 (2 – x) = 4 (2 – 0)
= 4 × 2
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
– 4 (2 + x) = 8.
Answer:
Given equation is – 4 (2 + x) = 8
Dividing both sides by – 4, we get
\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)
Transposing 2 from L.H.S. to R.H.S. we get :
x = – 2 – 2
or x = – 4, which is the required solution

To Check. Put x = – 4 in the L.H.S. of equation – 4 (2 + x) = 8
L.H.S. = – 4 (2 + x) = – 4 [2 + (- 4)]
= – 4 (2 – 4)
= – 4 (- 2)
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

3. Solve the following equations :

Question (i).
4 = 5 (x – 2)
Answer:
Given equation is 4 = 5 (x – 2)
or 4 = 5x – 10
Transposing 5x to L.H.S. and 4 to R.H.S.,
we get
– 5x = – 4 – 10
or – 5x = – 14
Dividing both sides by – 5, we get
\(\frac{-5 x}{-5}=\frac{-14}{-5}\)
or, x = \(\frac {14}{5}\), which is the required solution.

To Check. Put x = \(\frac {14}{5}\) in the R.H.S. of the equation 4 = 5 (x – 2)
R.H.S. = 5 (x – 2) = 5\(\left(\frac{14}{5}-2\right)\)
= 5\(\left(\frac{14-10}{5}\right)\)
= 5 \(\left(\frac{4}{5}\right)\)
= 4 = L.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
– 4 = 5 (x – 2)
Answer:
Given equation is – 4 = 5 (x – 2)
or – 4 = 5x – 10
Transposing -4 to R.H.S and 5x to L.H.S
we get
-5x = 4 – 10 or -5x = -6
Dividing both sides by – 5 we get
\(\frac{-5 x}{-5}=\frac{-6}{-5}\)
or x = \(\frac {6}{5}\), which is the required solution.

To Check. Put x = \(\frac {6}{5}\) in the R.H.S. of the equation – 4 = 5 (x – 2)
L.H.S. = 5 (x – 2)
= 5\(\left(\frac{6}{5}-2\right)\)
= 5\(\left(\frac{6-10}{5}\right)\)
= 5\(\left(\frac{-4}{5}\right)\)
= -4 = L.H.S.
L.H.S. = R.H.S.

Question (iii).
4 + 5 (p – 1) = 34
Answer:
Given equation is 4 + 5(p – 1) = 34
Transposing 4 to R.H.S. we get
5(p – 1) = 34 – 4
5(p – 1) = 30
Dividing both sides, by 5, we get
\(\frac{5(p-1)}{5}=\frac{30}{5}\)
p – 1=6
Transposing -1 to R.H.S. we get
p = 6 + 1
p = 7 which is the required solution.

To Check : Put p = 7 in L.H.S. of the equation 4 + 5 (p – 1) = 34
L.H.S. = 4 + 5 (p – 1)
= 4 + 5 (7 – 1)
= 4 + 5 (6)
= 4 + 30
= 34 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
6y – 1 = 2y + 1.
Answer:
Given equation is 6y – 1 = 2y + 1
Transposing – 1 to R.H.S. and 2y to L.H.S,
we get
6y – 2y = 1 + 1
or 4y = 2 or y = \(\frac {2}{4}\)
or y = \(\frac {1}{2}\), which is the required solution.

To Check Put y = \(\frac {1}{2}\) in both L.H.S. and R.H.S. of the equation
6y – 1 = 2y + 1
L.H.S. = 6y – 1 = 6 × \(\frac {1}{2}\) – 1 = 3 – 1 = 2
R.H.S. = 2y + 1 = 2 × \(\frac {1}{2}\) + 1 = 1 + 1 = 2.
∴ L.H.S. = R.H.S.

4.

Question (i).
Construct 3 equations starting with x = 2
Answer:
First Equation.
(i) Start with x = 2
Multiplying both sides by 10
10x = 20
Adding 2 to both sides
10x + 2 = 20 + 2
or 10x + 2 = 22
This has resulted in an equation.

Second Equation. Start with x = 2
Divide both sides by 5
∴ \(\frac{x}{5}=\frac{2}{5}\)
This has resulted in an equation.

Third Equation. Start with x = 2
Multiply both sides by 5, we get
5x = 5 × 2
or 5x = 10
Subtracting 3 from both sides, we get
5x – 4 = 10 – 3
or 5x – 3 = 7
This has resulted in an equation.

Question (ii).
Construct 3 equation starting with x = – 2
Answer:
First Equation. Start with x = – 2
Multiplying both sides with 3, we get
3x = – 6
This has resulted in an equation

Second Equation. Start with x = – 2
Multiplying both sides with 3, we get 3x = -6
Adding 7 to both sides, we get 3x + 7
= -6 + 7 or 3x + 7 = 1
This has resulted in an equation.

Third Equation. Start with x = – 2
Multiplying both side with 2 we get 3x = – 6
Adding 10 to both sides we get
3x+ 10 = -6 + 10
or 3x + 10 = 4
This has resulted in an equation.

Multiple Choice Questions :

5. If 7x + 4 = 39, then x is equal to :
(a) 6
(b) -4
(c) 5
(d) 8
Answer:
(c) 5

6. If 8m – 8 = 56 then m is equal to :
(a) -4
(b) -2
(c) -14
(d) 8
Answer:
(d) 8

7. Which of the following number satisfies the equation – 6 + x = -18 ?
(a) 10
(b) – 13
(c) – 12
(d) – 16.
Answer:
(a) 10

8. If \(\frac{x}{2}\) = 14, then the value of 2x + 6 is equal to :
(a) 62
(b) -64
(c) 16
(d) -62.
Answer:
(a) 62

9. If 3 subtracted from twice a number is 5, then the number is :
(a) -4
(b) -2
(c) 2
(d) 4
Answer:
(d) 4

10. If 5 added to thrice an integer is – 7, then the integer is :
(a) – 6
(b) – 5
(c) -4
(d) 4
Answer:
(c) -4

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2

1. Find the median of the following data :
3, 1, 5, 6, 3, 4, 5
Solution:
We arrange the data in ascending order.
1, 3, 3, 4, 5, 5, 6
Here, 3 and 5 both occur twice.
Therefore, both the numbers 3 and 5 are modes of the given data.

2. Find the mode of the following numbers :
2, 2, 2, 3, 4, 5, 5, 5, 6, 6, 8
Solution:
The given data is :
2, 2, 2, 3, 4, 5, 5, 5, 6, 6, 8
Median is the middle observation
∴ 5 is the median

3. The scores in mathematics test (out of 25) of 15 students are as follows :
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20.
Find the mean, mode and median of this data ? Are they same ?
Solution:
Mean =

= \(\frac {263}{15}\)
= 17.53
Now we arrange the data in ascending order
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Here 20 occurs more frequently
∴ Mode = 20
Median is the middle observation
∴ 20 is the median.
Yes, both Mode and Median are same.

4. The weight (in kg) of 15 students of class are :
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47

Question (i).
Find the mode and median of this data.
Solution:
We arrange the data in ascending order
32, 25, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
Mode. Here 38 and 43 occurs more frequently i.e. 3 times
∴ Mode = 38 and 43
Median is the middle observation
∴ 40 is median.

Question (ii).
Is there more than one mode ?
Solution:
Yes, there are two mode i.e. 38 and 43

5. Find the mode and median of the following data :
13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:
We arrange the data in ascending order
12, 12, 13, 13, 14, 14, 14, 16, 19
Here 14 occurs more frequently
∴ Mode = 14
Median is middle observation
∴ 14 is median.

6. Find the mode of the following data :
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14.
Solution:
We arrange the data in ascending order
12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 18, 19.
Here 15 occurs more frequently
∴ Mode =15

7. Multiple Choice Questions :

Question (i).
The mode of the data :
3, 5, 1, 2, 0, 2, 3, 5, 0, 2, 1, 6 is :
(a) 6
(b) 3
(c) 2
(d) 1.
Answer:
(c) 2

Question (ii).
A cricketer scored 38, 79, 25, 52, 0, 8, 100 runs in seven innings, the range of the runs scored is :
(a) 100
(b) 92
(c) 52
(d) 38.
Answer:
(a) 100

Question (iii).
Which of the following is not a central tendency of a data ?
(a) Mean
(b) Median
(c) Mode
(d) Range.
Answer:
(a) Mean

Question (iv).
The mean of 3, 1, 5, 7 and 9 is :
(a) 6
(b) 4
(c) 5
(d) 0.
Answer:
(c) 5

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

1. Find the mean of the following data :

Question (i).
3, 5, 7, 9, 11, 13, 15
Answer:
Mean = \(\frac{3+5+7+9+11+13+15}{7}\)
= \(\frac {63}{7}\)
= 9

Question (ii).
40, 30, 30, 0, 26, 60
Answer:
Mean = \(\frac{40+30+30+0+26+60}{6}\)
= \(\frac {183}{6}\)
= 9

2. Find the mean of the first five whole numbers.
Answer:
The first five whole numbers are : 0, 1, 2, 3, 4
Mean = \(\frac{0+1+2+3+4}{5}\)
= \(\frac {10}{5}\)
= 2
Hence, the mean of first five whole numbers = 2

3. A batsman scored the following number of runs in six innings :
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning.
Answer:
Mean runs
= \(\frac{36+35+50+46+60+55}{6}\)
= \(\frac {282}{6}\)
Hence, the mean runs scored by batsman in an innings = 47

4. The ages in years of 10 teachers of a school are :
32, 41, 28, 54, 35, 26, 23, 33, 38, 40
(i) What is the age of the oldest teacher and that of the youngest teacher ?
(ii) What is the range of the ages of the teachers ?
(iii) What is the mean age of these teachers ?
Answer:
Arranging the ages in ascending order, we get
23, 26, 28, 32, 33, 35, 38, 40, 41, 54
(i) Age of the oldest teacher = 54 years Age of the youngest teacher 23 years
(ii) Range of the ages = 54 – 23 = 31
(iii) Mean age of the teachers

= \(\frac {350}{10}\)
= 35 years.

5. The rain fall (in mm) ip a city on 7 days of a certain week was recorded as follows :

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) How many days had the rainfall less than the mean rainfall ?
Answer:
(i) Arranging the rainfall (in mm) in ascending order, we get
Highest Rainfall = 20.5 mm
Lowest Rainfall = 0.0 mm
Range of the rainfall = 20.5 mm – 0.0 mm
= 20.5 mm.

(ii) Mean Rainfall
= \(\frac{0.01+12.2+2.1+0.0+20.5+5.5+1.0}{7}\)
= \(\frac {41.31}{7}\)
= 5.9 mm.

(iii) Number of days having rainfall less than mean rainfall = 5 days.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

1. Write the first step that you will use to separate the variable and then solve the equation.

Question (i).
x + 1 = 0
Answer:
Given equation x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = -1
or x = – 1

Question (ii).
x – 1 = 5
Answer:
Given equation is x – 1 = 5
Adding 1 to both sides we get
x – 1 + 1 = 5 + 1
or x = 6
Thus x = 6 is the solution of the given equation

Question (iii).
x + 6 = 2
Answer:
Given equation is x + 6 = 2
Subtracting 6 from both sides, we get:
x + 6 – 6 = 2 – 6
or x = – 4
Thus, x = – 4 is the solution of the given equation.

Question (iv).
y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y + 4 – 4 = 4 – 4
or y = 0
Thus, y = 0 is the solution of the given equation.

Question (v).
y – 3 = 3
Answer:
Given equation is y – 3 = 3
Adding 3 to both sides we get
y – 3 + 3 = 3 + 3
or y = 6
Thus, y = 6 is the solution of the given equation.

2. Write the first step that you will use to separate the variable and then sotye the equation :

Question (i).
3x = 15
Answer:
Given equation is 3x = 15
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{15}{3}\)
or x = 5

Question (ii).
\(\frac{P}{7}\) = 4
Answer:
Given equation is \(\frac{P}{7}\) = 4
Multiplying both sides by 7, we get
7 × \(\frac{P}{7}\) = 7 × 4
or p = 28
Thus, p = 28 is the solution of the given equation.

Question (iii).
8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
or y = \(\frac {9}{2}\)

Question (iv).
20x = – 10
Answer:
Given equation is
20x = – 10
Dividing both sides by 20
\(\frac{20 x}{20}=\frac{-10}{20}\)
or x = \(\frac {-1}{2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

Question (i).
5x + 7 = 17
Answer:
Given equation is 5x + 7 = 17
Subtracting 7 from both sides, we get
5x + 7 – 7 = 17 – 7
or 5x = 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{10}{5}\)
or x = 2

Question (ii).
\(\frac{20 x}{3}\) = 40
Answer:
Given equation is \(\frac{20 x}{3}\) = 40
Multiplying both sides by 3, we get
3 × \(\frac{20 x}{3}\) = 3 × 40
or 20x = 3 × 40
Dividing both sides by 20, we get
\(\frac{20 x}{20}\) = \(\frac{3 \times 40}{20}\)
or x = 6

Question (iii).
3p – 2 = 46
Answer:
Given equation is 3p – 2 = 46
Adding 2 to both sides, we get
3p – 2 + 2 = 46 + 2
or 3 p = 48
Dividing both sides by 3, we get:
\(\frac{3 p}{3}=\frac{48}{3}\)
or p = 16

4. Solve the following equations :

Question (i).
10x + 10 = 100
Answer:
Given equation is 10x + 10 = 100
Subtracting 10 from both sides, we get
10x + 10 – 10 = 100 – 10
or 10x = 90
Dividing both sides by 10, we get
\(\frac{10 x}{10}=\frac{90}{10}\)
or x = 9
Thus x = 9 is the solution of the given equation.

Question (ii).
\(\frac{-p}{3}\) = 5
Answer:
Given equation is \(\frac{-p}{3}\) = 5
Multiplying both sides by – 3, we get
– 3 × \(\frac{-p}{3}\) = -3 × 5
or p = -15
Thus p = – 15 is the solution of the given equation.

Question (iii).
3x + 12 = 0
Answer:
Given equation is 3x + 12 = 0
Subtracting 12 from both sides, we get
3x + 12 – 12 = – 12
or 3x = – 12
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{-12}{3}\)
or x = -4
Thus x = – 4 is the solution of the given equation.

Question (iv).
2q – 6 = 0
Answer:
The given equation is 2q – 6 = 0
Adding 6 to both sides, we get
2q – 6 + 6 = 0 + 6
or 2q = 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{6}{2}\)
or q = 3
Thus, q = 3 is the solution of the given equation.

Question (v).
3p = 0
Answer:
The given equation is 3p = 0
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{0}{3}\)
or p = 0
Thus, p = 0 is the solution of the given equation.

Question (vi).
3s = -9
Answer:
The given equation is
3s = -9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=-\frac{9}{3}\)
or s = – 3
Thus, s = – 3 is the solution of the given equation.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 2 Fractions and Decimals MCQ Questions

Multiple Choice Questions :

Question 1.
Shaded area of given circle represent the fraction.
(a) \(\frac {1}{4}\)
(b) \(\frac {2}{3}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(c) \(\frac {3}{4}\)

Question 2.
2 – \(\frac {3}{5}\) = …………….
(a) 7
(b) -7
(c) \(\frac {7}{5}\)
(d) –\(\frac {7}{5}\)
Answer:
(c) \(\frac {7}{5}\)

Question 3.
Place value of 5 in 17.56 is :
(a) 5
(b) \(\frac {5}{10}\)
(c) \(\frac {5}{100}\)
(d) 50
Answer:
(b) \(\frac {5}{10}\)

Question 4.
1.31 × 10 = ?
(a) 0.131
(b) 131
(c) 13.1
(d) 1.31.
Answer:
(c) 13.1

Question 5.
2.7 ÷ 10 is :
(a) 27
(b) 0.27
(c) 0.027
(d) None of these.
Answer:
(b) 0.27

Fill in the blanks :

Question 1.
Equivalent fraction of \(\frac {2}{5}\) is ………….
Answer:
\(\frac {4}{10}\)

Question 2.
\(\frac {2}{3}\) of 18 is ………….
Answer:
12

Question 3.
Expanded form is 40.38 is :
Answer:
40 + 3 × \(\frac {1}{10}\) + 8 × \(\frac {1}{100}\)

Question 4.
The product of decimal number and zero is always.
Answer:
0

Question 5.
The average of decimal numbers 1.1, 2.1 and 3.1 is ………….
Answer:
2.1

Write True or False :

Question 1.
The place value of 2 in 2.56 is 20 (True/False)
Answer:
False

Question 2.
The value of 15.37 × 100 is 1537. (True/False)
Answer:
True

Question 3.
When a decimal number is multiplied by 100, the decimal point in the product is shifted to the right by two places. (True/False)
Answer:
True

Question 4.
The value of 1.5 × 8 is 12 (True/False)
Answer:
True

Question 5.
On dividing a decimal number by 1000, the decimal point is shifted to the left by three places. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

1. Solve, dividing decimal number by 10, 100 or 1000 in the following :

Question (i).
2.7 ÷ 10
Answer:
2.7 ÷ 10 = \(\frac{27}{10} \times \frac{1}{10}\)
= \(\frac{27}{100}\)
= 0.27

Question (ii).
3.35 ÷ 10
Answer:
3.35 ÷ 10 = \(\frac{335}{100} \times \frac{1}{10}\)
= \(\frac {335}{1000}\)
= 0.335

Question (iii).
0.15 ÷ 10
Answer:
0.15 ÷ 10 = \(\frac{15}{100} \times \frac{1}{10}\)
= \(\frac {15}{1000}\)
= 0.015

Question (iv).
32.7 ÷ 10
Answer:
32.7 ÷ 10 = \(\frac{327}{10} \times \frac{1}{10}\)
= \(\frac {327}{100}\)

Question (v).
5.72 ÷ 100
Answer:
5.72 ÷ 100 = \(\frac{572}{100} \times \frac{1}{100}\)
= \(\frac {572}{10000}\)
= 0.0572

Question (vi).
23.75 ÷ 100
Answer:
23.75 ÷ 100 = \(\frac{2375}{100} \times \frac{1}{100}\)
= \(\frac {2375}{10000}\)
= 0.2375

Question (vii).
532.73 ÷ 100
Answer:
532.73 ÷ 100 = \(\frac{53273}{100} \times \frac{1}{100}\)
= \(\frac {53273}{10000}\)
= 5.3273

Question (viii).
1.321 ÷ 100
Answer:
1.321 ÷ 100 = \(\frac{1321}{1000} \times \frac{1}{100}\)
= \(\frac {1321}{10000}\)
= 0.01321

Question (ix).
2.5 ÷ 1000
Answer:
2.5 ÷ 1000 = \(\frac{25}{10} \times \frac{1}{1000}\)
= \(\frac {25}{10000}\)
= 0.0025

Question (x).
53.83 ÷ 1000
Answer:
53.83 ÷ 1000 = \(\frac{5383}{100} \times \frac{1}{1000}\)
= \(\frac {5383}{100000}\)
= 0.05383

Question (xi).
217.35 ÷ 1000
Answer:
217.35 ÷ 1000 = \(\frac{21735}{100} \times \frac{1}{1000}\)
= \(\frac {21735}{100000}\)
= 0.21735

Question (xii).
0.2 ÷ 1000
Answer:
0.2 ÷ 1000 = \(\frac{2}{10} \times \frac{1}{1000}\)
= \(\frac {2}{10000}\)
= 0.0002

2. Solve, dividing decimal number by whole number.

Question (i).
7.5 ÷ 5
Answer:
7.5 ÷ 5 = \(\frac{75}{10} \times \frac{1}{5}\)
= \(\frac {15}{10}\)
= 1.5

Question (ii).
16.9 ÷ 13
Answer:
16.9 ÷ 13 = \(\frac{169}{10} \times \frac{1}{13}\)
= \(\frac {13}{10}\)
= 1.3

Question (iii).
65.4 ÷ 6
Answer:
65.4 ÷ 6 = \(\frac{654}{10} \times \frac{1}{6}\)
= \(\frac {109}{10}\)
= 10.9

Question (iv).
0.121 ÷ 11
Answer:
0.121 ÷ 11 = \(\frac{121}{1000} \times \frac{1}{11}\)
= \(\frac {11}{1000}\)
= 0.011

Question (v).
11.84 ÷ 4
Answer:
11.84 ÷ 4 = \(\frac{1184}{100} \times \frac{1}{4}\)
= \(\frac {296}{100}\)
= 2.96

Question (vi).
47.6 ÷ 7
Answer:
47.6 ÷ 7 = \(\frac{476}{10} \times \frac{1}{7}\)
= \(\frac {68}{10}\)
= 6.8

3. Solve, dividing the decimal number by decimal number

Question (i).
3.25 ÷ 0.5
Answer:
3.25 ÷ 0.5 = \(\frac{325}{100} \div \frac{5}{10}\)
= \(\frac{325}{100} \times \frac{10}{5}\)
= \(\frac {65}{10}\)
= 6.5

Question (ii).
5.4 ÷ 1.2
Answer:
5.4 ÷ 1.2 = \(\frac{54}{10} \div \frac{12}{10}\)
= \(\frac{54}{10} \times \frac{10}{12}\)
= \(\frac {9}{2}\)
= 4.5

Question (iii).
26.32 ÷ 3.5
Answer:
26.32 ÷ 3.5 = \(\frac{2632}{100} \div \frac{35}{10}\)
= \(\frac{2632}{100} \times \frac{10}{35}\)
= \(\frac {752}{100}\)
= 7.52

Question (iv).
2.73 ÷ 13
Answer:
2.73 ÷ 13 = \(\frac{273}{100} \times \frac{10}{13}\)
= \(\frac {21}{10}\)
= 2.1

Question (v).
12.321 ÷ 11.1
Answer:
12.321 ÷ 11.1 = \(\frac{12321}{1000} \div \frac{111}{10}\)
= \(\frac{12321}{1000} \times \frac{10}{111}\)
= \(\frac {111}{100}\)
= 1.11

Question (vi).
0.0018 ÷ 0.15
Answer:
0.0018 ÷ 0.15 = \(\frac{18}{10000} \div \frac{15}{100}\)
= \(\frac{18}{10000} \times \frac{100}{15}\)
= \(\frac {12}{1000}\)
= 0.012

4. 25 steel chairs were purchased by a school for ₹ 11,883.75. Find the cost of one steel chair.
Answer:
Cost Price of 25 steel chairs = ₹ 11,883.75
Cost Price of 1 steel chair = ₹ 11,883.75 ÷ 15
= ₹ \(\frac{11,88375}{100} \times \frac{1}{15}\)
= ₹ \(\frac {47535}{100}\)
= ₹ 475.35

5. A car covers a distance of 276.75 km in 4.5 hours. What is the average speed of the car ?
Answer:
Total Distance covered = 276.75 km
Time taken = 4.5 hours
Average speed of car = \(\frac{Distance}{Time}\)
= \(\frac{276.75}{4.5}\)
= \(\frac{27675}{100} \times \frac{10}{45}\)
= \(\frac {615}{10}\)
= 61.5 km/hr.

6. Multiple Choice Questions :

Question (i).
27.5 ÷ 10 = ?
(a) 275
(b) 0.275
(c) 2.75
(d) None of these.
Answer:
(c) 2.75

Question (ii).
The value of 1.5 ÷ 3 is :
(a) 5
(b) 0.05
(c) 0.5
(d) 4.5.
Answer:
(c) 0.5

Question (iii).
The average of decimal number 1.1, 2.1 and 3.1 is :
(a) 2.5
(b) 1.1
(c) 2.1
(d) 6.3.
Answer:
(c) 2.1

7. On dividing a decimal number by 100, the decimal point is shifted to the left by one place. (True/False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

1. Find the product of each of the following:

Question (i).
1.31 × 10
Answer:
1.31 × 10
= \(\frac {131}{100}\) × 10
= \(\frac {131}{10}\)
= 13.1

Question (ii).
1.31 × 10
Answer:
25.7 × 10
= \(\frac {257}{10}\) × 10
= 257

Question (iii).
1.01 × 100
Answer:
1.01 × 100
= \(\frac {101}{100}\) × 100
= 101

Question (iv).
0.45 × 100
Answer:
0.45 × 100
= \(\frac {45}{100}\) × 100
= 45

Question (v).
9.7 × 100
Answer:
9.7 × 100
= \(\frac {97}{10}\) × 100
= 970

Question (vi).
3.87 × 10
Answer:
3.87 × 10
= \(\frac {387}{100}\) × 100
= \(\frac {387}{10}\)
= 38.7

Question (vii).
0.07 × 10
Answer:
0.07 × 10
= \(\frac {7}{100}\) × 10
= \(\frac {7}{100}\)
= 0.70

Question (viii).
0.3 × 100
Answer:
0.3 × 100
= \(\frac {3}{10}\) × 100
= 30

Question (ix).
5.37 × 1000
Answer:
5.37 × 1000
= \(\frac {537}{10}\) × 100
= 53700

Question (x).
0.02 × 1000
Answer:
0.02 × 1000
= \(\frac {2}{100}\) × 100
= 20

2. Find the product of each of the following :

Question (i).
1.5 × 3
Answer:
1.5 × 3 = \(\frac {15}{10}\) × 3
= \(\frac {45}{10}\)
= 4.5

Question (ii).
2.71 × 12
Answer:
2.71 × 12 = \(\frac {271}{100}\) × 12
= \(\frac {3252}{100}\)
= 32.52

Question (iii).
7.05 × 4
Answer:
7.05 × 4 = \(\frac {705}{100}\) × 4
= \(\frac {2820}{100}\)
= 28.2

Question (iv).
0.05 × 12
Answer:
0.05 × 12 = \(\frac {5}{100}\) × 12
= \(\frac {60}{100}\)
= 0.6

Question (v).
112.03 × 8
Answer:
112.03 × 8 = \(\frac {89624}{100}\) × 8
= 896.24

Question (vi).
3 × 7.53
Answer:
3 × 7.53 = 3 × \(\frac {753}{100}\)
= \(\frac {2259}{100}\)
= 22.59

3. Evaluate the following :

Question (i).
3.7 × 0.4
Answer:
3.7 × 0.4 = \(\frac{37}{10} \times \frac{4}{10}\)
= \(\frac {148}{100}\)
= 1.48

Question (ii).
2.75 × 1.1
Answer:
2.75 × 1.1 = \(\frac{275}{100} \times \frac{11}{10}\)
= \(\frac {3025}{1000}\)

Question (iii).
0.07 × 1.9
Answer:
0.07 × 1.9 = \(\frac{7}{100} \times \frac{19}{10}\)
= \(\frac {133}{1000}\)
= 0.133

Question (iv).
0.5 × 31.83
Answer:
0.5 × 31.83 = \(\frac{5}{10} \times \frac{3183}{100}\)
= \(\frac {15915}{1000}\)
= 15.915

Question (v).
7.5 × 5.7
Answer:
7.5 × 5.7 = \(\frac{75}{10} \times \frac{57}{10}\)
= \(\frac {4275}{100}\)
= 42.75

Question (vi).
10.02 × 1.02
Answer:
10.02 × 1.02 = \(\frac{1002}{100} \times \frac{102}{100}\)
= \(\frac {102240}{10000}\)
= 10.2204

Question (vii).
0.08 × 0.53
Answer:
0.08 × 0.53 = \(\frac{8}{10} \times \frac{53}{100}\)
= \(\frac {424}{10000}\)
= 0.0424

Question (viii).
21.12 × 1.21
Answer:
21.12 × 1.21 = \(\frac{2112}{100} \times \frac{121}{100}\)
= \(\frac {255552}{10000}\)
= 25.5552

Question (ix).
1.06 × 0.04
Answer:
1.06 × 0.04 = \(\frac{106}{100} \times \frac{4}{100}\)
= \(\frac {424}{1000}\)
= 0.0424

4. A piece of wire is divided into 15 equal parts. If length of one part is 2.03 m, then find the total length of the wire.
Answer:
Length of one part = 2.03 m
Length of 15 parts = 15 × 2.03 m
= 30.45 m

5. The cost of 1 metre cloth is ₹ 75.80. Find the cost of 4.75 metre cloth.
Answer:
Cost of 1 metre cloth = ₹ 75.80
Cost of 4.75 metre cloth = ₹ 75.80 × 4.75
= ₹ 360.05

6. Multiple choice questions :

Question (i).
1.25 × 10 = ?
(a) 0.125
(b) 125
(c) 12.5
(d) 1.25
Answer:
(c) 12.5

Question (ii).
If x × 100 = 135.72 then value of x is equal to
(a) 13.572
(b) 1.3572
(c) 135.72
(d) 13572.
Answer:
(b) 1.3572

Question (iii).
The value of 1.5 × 8 is :
(a) 1.2
(b) 120
(c) 12
(d) 0.12.
Answer:
(c) 12

7.
Question (i).
The product of a decimal number and zero is always zero. (True/False)
Answer:
True

Question (ii).
On multiplying a decimal number by 10, the decimal point is shifted to the left by one place. (True/False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

1. Solve the following :

Question (i).
\(\frac {1}{3}\) of
(a) \(\frac {1}{5}\)
(b) \(\frac {2}{7}\)
(c) \(\frac {3}{2}\)
Answer:

Question (ii).
\(\frac {3}{4}\) of
(a) \(\frac {2}{9}\)
(b) \(\frac {4}{7}\)
(c) \(\frac {8}{3}\)
Answer:

2. Multiply the fractions and reduce it to the lowest form (If possible) :

Question (i).
\(\frac{2}{7} \times \frac{7}{9}\)
Answer:
\(\frac{2}{7} \times \frac{7}{9}\)
= \(\frac {2}{9}\)

Question (ii).
\(\frac{1}{3} \times \frac{15}{8}\)
Answer:
\(\frac{1}{3} \times \frac{15}{8}\)
= \(\frac {5}{8}\)

Question (iii).
\(\frac{12}{27} \times \frac{3}{9}\)
Answer:
\(\frac{12}{27} \times \frac{3}{9}\)
= \(\frac {4}{27}\)

Question (iv).
\(\frac{2}{5} \times \frac{6}{4}\)
Answer:
\(\frac{2}{5} \times \frac{6}{4}\)
= \(\frac {3}{5}\)

Question (v).
\(\frac{81}{100} \times \frac{6}{7}\)
Answer:
\(\frac{81}{100} \times \frac{6}{7}\)
= \(\frac {243}{350}\)

Question (vi).
\(\frac{3}{5} \times \frac{5}{27}\)
Answer:
\(\frac{3}{5} \times \frac{5}{27}\)
= \(\frac {1}{9}\)

3. Multiply the following fractions :

Question (i).
\(\frac{3}{2} \times 5 \frac{1}{3}\)
Answer:
\(\frac{3}{2} \times 5 \frac{1}{3}\)
= \(\frac{3}{2} \times \frac{16}{3}\)
= 8

Question (ii).
\(\frac{1}{7} \times 5 \frac{2}{3}\)
Answer:
\(\frac{1}{7} \times 5 \frac{2}{3}\)
= \(\frac{1}{7} \times \frac{17}{3}\)
= \(\frac {17}{21}\)

Question (iii).
\(2 \frac{5}{6} \times 4\)
Answer:
\(2 \frac{5}{6} \times 4\)
= \(\frac {17}{6}\) × 4
= \(\frac {34}{3}\)
= 11\(\frac {1}{3}\)

Question (iv).
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
Answer:
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
= \(\frac{13}{3} \times \frac{37}{4}\)
= \(\frac {481}{12}\)

Question (v).
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
Answer:
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
= \(\frac{8}{3} \times \frac{29}{8}\)
= 9\(\frac {2}{3}\)

Question (vi).
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
Answer:
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
= \(\frac{16}{5} \times \frac{9}{4}\)
= \(\frac {36}{5}\)
= 7\(\frac {1}{5}\)

4. Which fraction is greater in the following fractions ?

Question (i).
\(\frac {3}{2}\) of \(\frac {2}{7}\) or \(\frac {5}{2}\) of \(\frac {3}{8}\)
Answer:
\(\frac {3}{2}\) of \(\frac {2}{7}\)
= \(\frac{3}{2} \times \frac{2}{7}\)
= \(\frac {3}{7}\)
or
\(\frac {5}{2}\) of \(\frac {3}{8}\)
= \(\frac{5}{2} \times \frac{3}{8}\)
= \(\frac {15}{16}\)
Since, \(\frac {15}{16}\) > \(\frac {3}{7}\)
So, \(\frac {5}{2}\) of \(\frac {3}{8}\) is greater.

Question (ii).
\(\frac {1}{2}\) of \(\frac {6}{5}\) or \(\frac {1}{3}\) of \(\frac {4}{5}\)
Answer:
\(\frac {1}{2}\) of \(\frac {6}{5}\)
= \(\frac {1}{2}\) × \(\frac {6}{5}\)
= \(\frac {3}{5}\)
or
\(\frac {1}{3}\) of \(\frac {4}{5}\)
= \(\frac {1}{3}\) × \(\frac {4}{5}\)
= \(\frac {4}{15}\)
Since, \(\frac {3}{5}\) > \(\frac {4}{15}\)
So, \(\frac {1}{2}\) of \(\frac {6}{5}\) is greater.

5. If the speed of a car is 105 \(\frac {1}{3}\) km/hr, find the distance covered by it m 3\(\frac {2}{3}\) hours.
Answer:
Distance travelled in 1 hour= 105\(\frac {1}{3}\) km
= \(\frac {316}{3}\) km
Distance travelled in 3\(\frac {2}{3}\) hours i.e. in \(\frac {11}{3}\) hours
= \(\frac {11}{3}\) × \(\frac {316}{3}\) km
= \(\frac {3476}{9}\) km
= 386\(\frac {2}{9}\)km

6. The length of a rctangular plot of land is 29\(\frac {3}{7}\) m. If its breadth is 12\(\frac {8}{11}\)m, find its area.
Answer:
Length of rectangular plot (l)
= 29\(\frac {3}{7}\)m = \(\frac {206}{7}\)m
Breadth of rectangular plot (b)
= 12\(\frac {8}{11}\)m = \(\frac {140}{11}\)m
Area of rectangular plot
= l × b = \(\frac{206}{7} \mathrm{~m} \times \frac{140}{11} \mathrm{~m}\)
= \(\frac{4120}{11}\) sq.m
= 374\(\frac{6}{11}\) sq.m

7. If a cloth costs ₹ 120 \(\frac {1}{4}\) per metre, find the cost of 4\(\frac {1}{3}\) metre of this cloth.
Answer:
Per m cost of cloth = ₹ 120\(\frac {1}{4}\)
= ₹ \(\frac {481}{4}\)
Cost of 4\(\frac {1}{3}\) metre of this cloth
= \(\frac {1}{3}\) × ₹ \(\frac {481}{4}\)
= ₹ \(\frac {6253}{12}\)
= ₹ 521\(\frac {1}{12}\)

8. Multiple choice questions :

Question (i).
\(\frac {1}{4}\) of \(\frac {8}{3}\) is :
(a) \(\frac {9}{7}\)
(b) \(\frac {8}{4}\)
(c) \(\frac {2}{3}\)
(d) 1
Answer:
(c) \(\frac {2}{3}\)

Question (ii).
\(\frac {3}{2}\) × \(\frac {2}{3}\) = ?
(a) 1
(b) \(\frac {5}{6}\)
(c) 3
(d) \(\frac {6}{5}\)
Answer:
(a) 1

Question (iii).
The value of the product of two proper fractions is :
(a) greater than both of the proper fractions.
(b) lesser than both of the proper fractions.
(c) lie between the given two fractions.
(d) none of these.
Answer:
(b) lesser than both of the proper fractions.

9. Check if the following equations are ‘True’ or ‘False’ :

Question (i).
\(1 \frac{2}{3} \times 4 \frac{5}{7}=4 \frac{10}{21} ?\) (True/False)
Answer:
False

Question (ii).
\(\frac{3}{4} \times \frac{2}{3}=\frac{2}{3} \times \frac{3}{4} ?\) (True/False)
Answer:
True