Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 4 Simple Equations MCQ Questions

Multiple Choice Questions :

Question 1.
Choose simple equation out of the following:
(a) 3x + 11
(b) 2x + 5 < 11
(c) x – 5 = 7x + 6
(d) \(\frac{5 x+6}{6}\)
Answer:
(c) x – 5 = 7x + 6

Question 2.
A quantity which takes a fixed numerical value is called :
(a) Constant
(b) Variable
(c) Equation
(d) Expression
Answer:
(a) Constant

Question 3.
In equation 5x = 25 the value of x is :
(a) 0
(b) 5
(c) -5
(d) 1
Answer:
(b) 5

Question 4.
In equation \(\frac{m}{3}\) = 2 the value of m is :
(a) 1
(b) 0
(c) 6
(d) -6
Answer:
(c) 6

Question 5.
In equation 7x + 5 = 19 the value of n is :
(a) 0
(b) -2
(c) 1
(d) 2
Answer:
(d) 2

Question 6.
In equation 4p – 3 = 13, the value of p is :
(a) 1
(b) 4
(c) 0
(d) -4
Answer:
(b) 4

Question 7.
The equation of the statement, the sum of number x and 4 is 9 is :
(a) x + 4 = 9
(b) x – 4 = 9
(c) x = 4 + 9
(d) x – 9 = 4.
Answer:
(a) x + 4 = 9

Question 8.
The equation of the statement, ‘seven times m plus 7 = gives 77’ is.
(a) 1m × 7 = 77
(b) 7m + 7 = 77
(c) 7m = 77 + 7
(d) m + 7 × 7 = 77
Answer:
(a) 1m × 7 = 77

Fill in the blanks :

Question 1.
A quantity which takes a fixed numerical value is called …………….
Answer:
Constant

Question 2.
The equation for the statement seven time a number is 42 is …………….
Answer:
7x = 42

Question 3.
If x + 4 = 15, then the value of x is …………….
Answer:
x = 11

Question 4.
If 2y – 6 = 4, then y is equal to …………….
Answer:
y = 5

Question 5.
If 8x – 4 = 28, then x is equal to …………….
Answer:
x = 4

Write True or False :

Question 1.
An equation of one variable is called linear equation. (True/False)
Answer:
True

Question 2.
If x – 3 = 1, then value of x is 2. (True/False)
Answer:
False

Question 3.
If 7m + 7 = 77, then value of m is 10. (True/False)
Answer:
True

Question 4.
If 3 subtracted from twice a number is 5, then the number is 4. (True/False)
Answer:
True

Question 5.
If one fourth of a number is 10 then the number is 40. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

1. Complete the following :

Solution:
(i) No
Reason : For x = 5
L.H.S. = x + 5 = 5 + 5 = 10
RHS = 0
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 5

(ii) Yes
Reason : For x = -5
L.H.S. = x + 5
= -5 + 5 = 0
R.H.S. = 0
Since L.H.S. = R.H.S.
Therefore, given equation is satisfied for
x = – 5

(iii) NO
Reason : For x = 3
L.H.S. = x – 3
= 3 – 3 = 0
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 3

(iv) No
Reason : x = – 3
L.H.S. = x – 3
= – 3 – 3 = -6
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = – 3

(v) Yes
Reason : For x = 5
L.H.S. = 2x
= 2 × 5 = 10
R.H.S. = 10
Since L.H.S. = R.H.S.
Therefore given equation is satisfied for
x = 5

(vi) No
Reason : For x = – 6
L.H.S. = \(\frac{x}{3}\)
= \(\frac {-6}{3}\)
= -2
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = – 6

(vii) No,
Reason : For x = 0
L.H.S. = \(\frac{x}{2}\)
= \(\frac {0}{2}\) = 0
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = 0

2. Check whether the value given in the brackets is a solution to the given equation or not.

Question (i).
x + 4 = 11 (x = 7)
Answer:
Yes
Check : For x = 1
L.H.S. = x + 4
= 7 + 4 = 11
R.H.S. = 11
Since L.H.S. = R.H.S.
Therefore x = 7 is the solution to the given equation.

Question (ii).
8x + 4 = 28 (x = 4)
Answer:
No
Check: Forx-4
L.H.S. = 8x + 4
= 8 × 4 + 4
= 32 + 4
= 36
R.H.S. = 28
Since L.H.S. ≠ R.H.S.
Therefore x = 4 is not solution to the given equation.

Question (iii).
3m – 3 = 0 (m = 1)
Answer:
Yes
Check : For m = 1
L.H.S. = 3m – 3
= 3 × 1 – 3 = 3 – 3 = 0
R.H.S.= 0
Since L.H.S. = R.H.S.
Therefore, m = 1 is the solution to the given equation.

Question (iv).
\(\frac{x}{5}\) – 4 = -1 (x = 15)
Answer:
Yes
Check : For x = 15
L.H.S. = \(\frac{x}{5}\) – 4
= \(\frac {15}{5}\) – 4 = 3 – 4 = – 1
R.H.S. = – 1
Since L.H.S. = R.H.S.
Therefore, x = 15 is the solution to the given equation.

Question (v).
4x – 3 = 13 (x = 0)
Answer:
No
Check : For x = 0
L.H.S. = 4x – 3
= 4 × 0 – 3
= 0 – 3
= -3
R.H.S. = 13
Since L.H.S. ≠ R.H.S.
Therefore x = 0 is not the solution to the given equation.

3. Solve the following equations by trial and error method

Question (i).
5x + 2 = 17
Answer:

Value of x	L.H.S	R.H.S
1	5 × 1 + 2 = 5 + 2 = 7	17
2	5 × 2 + 2 = 10 + 2 = 12	17
3	5 × 3 + 2 = 15 + 2 = 17	17

We observe that for x = 3, L.H.S. = R.H.S.
Hence x = 3 is the solution of the given equation.

Question (ii).
3p – 14 = 4
Answer:

Value of p	L.H.S.	R.H.S.
1	3 × 1 – 14 = 3 – 14 = -11	4
2	3 × 2 – 14 = 6 – 14 = -8	4
3	3 × 3- 14 = 9 – 14 = -5	4
4	3 × 4 – 14 = 12 – 14 = -2	4
5	3 × 5 – 14 = 15 – 14 = 1	4
6	3 × 6 – 14 = 18 – 14 = 4	4

We observe that for p = 6, L.H.S. = R.H.S.
Hence p = 6 is the solution of the given equation

4. Write equations for the following statements.

Question (i).
The sum of numbers x and 4 is 9
Answer:
x + 4 = 9

Question (ii).
3 subtracted from y gives 9
Answer:
y – 3 = 9

Question (iii).
Ten times x is 50
Answer:
10x = 50

Question (iv).
Nine times x plus 6 is 87
Answer:
9x + 6 = 87

Question (v).
One fifth of a number y minus 6 gives 3.
Answer:
\(\frac {1}{5}\)x – 6 = 3

5. Write the following equations in statement form :

Question (i).
x – 2 = 6
Answer:
2 substracted from x is 6

Question (ii).
3y – 2 = 10
Answer:
2 subtracted from 3 times a number y is 10.

Question (iii).
\(\frac{x}{6}\) = 6
Answer:
One sixth of a number x is 6

Question (iv).
7x – 15 = 34
Answer:
15 subtracted 7 times a number x is 34.

Question (v).
\(\frac{x}{2}\) + 2 = 8
Answer:
add 2 to half of a number x to get 8.

6. Write an equation for the following statements :

Question (i).
Raju’s father’s age is 4 years more than five times Raju’s age. Raju’s father is 54 years old.
Answer:
Let x years be Raju’s age
Five times Raju’s age is 5x years
His father’s age will be 4 years more than five times
Raju’s age = 5x + 4
But 4 years more han five times Raju’s age = Raju’s father’s age
Therefore 5x + 4 = 54

Question (ii).
A teacher tells that the highest marks obtained by a student in his class is twice the lowest marks plus 6. The highest score is 86. (Take the lowest score to be x).
Answer:
Let the lowest score to be x
Twice the lowest score plus 6 = 2x + 6
The highest score obtained by a student is twice the lowest score plus 6 = 2x + 6
But highest score = 86
Hence 2x + 6 = 86

Question (iii).
In an isosceles triangle, the vertex angle is twice either base angle (Let the base angle be x in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Answer:
Let the base angle be x (in degrees)
Therefore vertex angle is twice the base angle = 2x (in degrees)
Sum of three angles of a triangle = 180°
∴ x + x + 2x = 180° = 4x = 180°

Question (iv).
A shopkeeper sells mangoes in two types of boxes. One small and one large. The large box contains as many as 8 small boxes plus 4 loose mangoes. The number of mangoes in a large box is given to be 100.
Answer:
Let the mangoes in small box be x.
Large box contains mangoes
= 8 small box + 4
= 8x + 4
But mangoes in large box = 100
∴ 8x + 4 = 100

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 3 Data Handling MCQ Questions

Multiple Choice Questions :

Question 1.
Mean of first five natural numbers is :
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

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

Question 3.
The probability of possible event is :
(a) 0
(b) 1
(c) 2
(d) -1
Answer:
(b) 1

Question 4.
The probability of impossible event is :
(a) 1
(b) -1
(c) 0
(d) None of these
Answer:
(c) 0

Question 5.
The probability of selecting letter S from the word ‘STUDENT’ is :
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{7}\)
Answer:
(d) \(\frac {1}{7}\)

Fill in the blanks :

Question 1.
The mean of first five prime numbers is ……………..
Answer:
5.6

Question 2.
The median of the data: 3, 1, 5, 6, 3, 4, 5 is ……………..
Answer:
5

Question 3.
Mode of the data : 1, 0, 1, 2, 3, 1, 2, is ……………..
Answer:
1

Question 4.
The probability of getting a head or tail is ……………..
Answer:
\(\frac {1}{2}\)

Question 5.
When a die is thrown the probability of getting a number 5 is ……………..
Answer:
\(\frac {1}{6}\)

Write True or False :

Question 1.
Mean of the first five whole numbers is 2. (True/False)
Answer:
True

Question 2.
Mode of data : 1, 1, 2,4, 3, 2, 1 is 2. (True/False)
Answer:
False

Question 3.
Median of data : 1, 2, 3, 4, 5 is 3. (True/False)
Answer:
True

Question 4.
An outcome is the result of an experiment. (True/False)
Answer:
True

Question 5.
Events that have many probabilities can have probability between 0 and 1. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

1. State whether the following is certain to happen, impossible to happen, may happen.

Question (i).
Two hundred people sit in a Maruti car.
Answer:
Impossible to happen.

Question (ii).
You are older than yesterday.
Answer:
Certain to happen.

Question (iii).
A tossed coin will land heads up.
Answer:
Can happen but not certain.

Question (iv).
A die when rolled shall land up 8 on top.
Answer:
Impossible to happen.

Question (v).
Tommorrow will be a cloudy day.
Answer:
Can happen but not certain.

Question (vi).
India will win the next test series.
Answer:
Can happen but not certain.

Question (vii).
The next traffic light seen will be green.
Answer:
Can happen but not certain.

2. There are 6 marbles in a box with numbers 1 to 6 marked on them.
(i) What is the probability of drawing a marble with number 5 ?
(ii) What is the probability of drawing a marble with number 2 ?
Solution:
Sample space : {1, 2, 3, 4, 5, 6}
There are 6 equally likely possible outcomes.
Outcomes : 1, 2, 3, 4, 5 or 6.
(i) Probability of drawing a marble with number 5 = \(\frac {1}{6}\)
(ii) Probability of drawing a marble with number 2 = \(\frac {1}{6}\)

3. There are two teams A and B. A coin is flipped to decide which team starts the game. What is the probability that team A will start ?
Solution:
When a coin is tossed once. The number of possible outcomes is head or tail
Sample space : [H, T]
The probability of getting head or tail is equal and is \(\frac {1}{2}\) for each.
∴ The probability that team A will start = \(\frac {1}{2}\)

4. A bag contains 3 red and 7 green balls. One ball is drawn at random from the bag. Find the probability of getting (i) a red ball (ii) a green ball.
Solution:
Total number of balls in the bag = 3 + 7 = 10
(i) The event is getting a red ball.
Probability (getting a red ball) = \(\frac {3}{10}\)

(ii) The event is getting a green ball.
Probability (getting a green ball) = \(\frac {7}{10}\)

5. Multiple Choice Questions :

Question (i).
The probability of an impossible event is : ………………
(a) -1
(b) 0
(c) \(\frac {1}{2}\)
(d) 1.
Answer:
(b) 0

Question (ii).
The probability of selecting letter G from the world ‘GIRL’ is :
(a) 1
(b) \(\frac {1}{2}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{3}\)
Answer:
(c) \(\frac {1}{4}\)

Question (iii).
When a die is thrown, the probability of getting a number 4 is :
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {4}{6}\)
(d) \(\frac {1}{6}\)
Answer:
(d) \(\frac {1}{6}\)

Question (iv).
A bag contains 5 white balls and 10 black balls. The probability of drawing a white ball from the bag is :
(a) \(\frac {5}{10}\)
(b) \(\frac {5}{15}\)
(c) \(\frac {10}{15}\)
(d) 1
Answer:
(b) \(\frac {5}{15}\)

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 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 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.