Author name: Prasanna

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.3

Question 1.
Use the bar graph to answer the following questions.
Data Handling
(a) Which is the most popular pet?
(b) How many students have dog as a pet?

Answer:
(a) Since the bar representing 10 is the longest, the most popular pet is cat.
(b) 8 students have dog as a pet.

Question 2.
Read the bar graph which shows the number of books sold by a book store
during five consecutive years and answer the following questions:

(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Answer:
(i) Number of books sold:
In 1989 were about 180
In 1990 were about 475
In 1992 were about 225
(ii) About 475 books were sold in 1990 and about 225 books were sold in 1992.
(iii) Fewer than 250 books were sold in 1989 and 1992.
(iv) The height of the bar for 1989 is slightly less than that for 200 books.
Therefore, about 180 books were sold in 1989.

Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?

(ii) Find the ratio of students of class sixth to the students of class eighth.
Answer:
(a) Start the scale at 0. The greatest value in the data is 135, so end the scale at a
value greater than 135 such as 140.
Use equal divisions along the axes such as increment of 10.
(i) We know that all the bars would lie between 0 and 140.
(ii) We choose the scale such that the length between 0 and 140 is neither too long nor too small.
(iii) Here we take one unit for 10 children.

(b) (i) Fifth class has the maximum number of children. Tenth class has the minimum number of
children.
(ii) Ratio of students of class sixth to
Eighth = 120 : 100 = 12 : 10 = 6 : 5

Question 4.
The performance of a student in 1^st Term and 2^nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:

(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Answer:
The required double bar graph is given below:

This graph shows the performance of students in ^st term and 2^nd term.
(i) The child improved his performance the most in Maths.
(ii) The improvement is the least in the subject of social science.
(iii) Yes, the performance has gone down in the subject of Hindi.

Question 5.
Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Answer:
(i) The required ‘double bar graph’ is given below :
From the bar graph it is inferred that the people of the colony like cricket the most and athletics the least.
(ii) The most popular sport is cricket.
(iii) Watching sports is more preferred than participating in sports.

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this Chapter. Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Temperature of cities as on 20.06.2006

Answer:
Plotting a double bar graph, we get.
City Difference in the minimum and maximum temperature on 20-06-2006

(i) Jammu has the largest difference in the minimum and maximum temperatures on the given data.
(ii) Jammu is the hottest city and Bengaluru is the coldest city.
(iii) Bangaluru and Jaipur or Bengaluru and Ahmedabad.
(iv) Mumbai has the least difference between its maximum temperature and minimum temperature.

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4

Question 1.
Amina thinks of numbers and subtracts \(\frac{5}{2}\) from it. She multiplies the results by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the number be x
By subtracting \(\frac{5}{2}\), we get x – \(\frac{5}{2}\)
According to the given question
\(8\left(x-\frac{5}{2}\right)=3 x\)
8x – \(\frac{8 \times 5}{2}\) = 3x
8x – 20 = 3x
By transposing 3x to L.H.S. and -20 to R.H.S., we get
8x – 3x = 20
5x = 20
Dividing both sides by 5, we get
\(\frac{5 \mathrm{x}}{5}=\frac{20}{5}\)
x = 4
∴ The required number is 4.

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let the number be x
The other positive number is 5x
on adding 21 to both numbers, we get (x + 21) and (5x + 21)
According to the question, we get
2(x + 21) = 5x + 21
2x + 42 = 5x + 21
Transposing 42 to R.H.S and 5x to L.H.S., we get
2x – 5x = 21 – 42
-3x = -21
Dividing both sides by -3, we get
\(\frac{-3 x}{-3}=\frac{-21}{-3}\)
x = 7
the other number 5x = 5 × 7 = 35
Thus, the required numbers are 7 and 35.

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number.
Solution:
Let the units digit be ‘x’.
the tens digits = 9 – x(sum of the digits is 9)
The original two digit number = 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
On interchanging the digits,
the new number = 10x + 9 – x = 9x + 9
According to the given question, we get
New number = Original number + 27
9x + 9 = 90 – 9x + 27
9x + 9 = 117 – 9x
Transposing 9 to R.H.S. and -9x to L.H.S., we get
9x + 9x = 117 – 9
18x = 108
Dividing both sides by 18, we get
\(\frac{18 \mathrm{x}}{18}=\frac{108}{18}\)
x = 6
∴ The original number = 90 – 9x = 90 – 54 = 36.

Question 4.
One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two- digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution:
Let the digit in the unit place be ‘x’
Then, the digit at tens place = 3x
The number = 10(3x) + x
= 30x + x
= 31x
On interchanging the digits,
The new number = 10x + 3x = 13x
According to the question
31x + 13x = 88
44x = 88
Dividing both sides by 44
\(\frac{44 x}{44}=\frac{88}{44}\)
x = 2
∴ The number = 31x = 31 × 2 = 62.

Question 5.
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one-third of his mother’s present age. What are their present ages?
Solution:
Let Shobos present age be ‘x’ years
Mother’s present age = 6x years
After 5 years
Shobos age = (x + 5) years
Shobos mothers age = (6x + 5) years
According to the question, we get
\(\frac{1}{3}\) (mothers present age) = Shobos age after 5 years
\(\frac{1}{3}\) × 6x = x + 5
2x = x + 5
Transposing x to LHS
2x – x = 5
x = 5
∴ Shobo’s present age = 5 years
Mothers present age = (6 × 5) = 30 years

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate of ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the length of the rectangular plot be 11x metres
and the breadth of the rectangular plot be 4x metres
Perimeter of the plot = 2(l + b)
= 2(11x + 4x)
= 2 × 15x
= 30x
Cost of fencing = ₹ 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we get
\(\frac{3000 \mathrm{x}}{3000}=\frac{75000}{3000}\)
x = 25
Length of the plot = 11 × 25 = 275 metres
Breadth of the plot = 4 × 25 = 100 metres

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 metres of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale of ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for shirts be ‘3x’ metres
and the length of cloth for trousers be ‘2x’ metres
Cost of shirts cloth = 3x × 50 = ₹ 150x
Cost of trouser cloth = 2x × 90 = ₹ 180x
S.P of shirts cloth at 12% profit
= ₹ \(\frac{112}{100}\) × 150x
= ₹ 168x
S.P. of trousers cloth at 10% profit
= \(\frac{110}{100}\) × 180x
= ₹ 198x
Total S.P = ₹ 36,600
168x + 198x = ₹ 36,600
366x = 36600
Dividing both sides by 366, we get
\(\frac{366 \mathrm{x}}{366}=\frac{36600}{366}\)
x = 100
Material bought for trousers (2 × 100) = 200 metres

Question 8.
Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let the total number of deer be ‘x’
Number of deer, grazing in the field = \(\frac{\mathrm{x}}{2}\)
Number of deer playing near by = \(\frac{3}{4}\left(x-\frac{x}{2}\right)=\frac{3 x}{4}-\frac{3 x}{8}=\frac{6 x-3 x}{8}=\frac{3 x}{8}\)
Number of deer drinking water = 9
\(\frac{x}{2}+\frac{3 x}{8}\) + 9 = x
Transposing 9 to RHS and x to LHS we get
\(\frac{x}{2}+\frac{3 x}{8}-x=-9\)
\(\frac{4 x+3 x-8 x}{8}=-9\)
\(\frac{-x}{8}=-9\)
Multiplying both sides by -8
\(\frac{-\mathrm{x}}{8}\) × (-8) = -9 × (-8)
x = 72
∴ Number of deer in herd = 72

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let the present age of granddaughter be ‘x’ years
Present age of grandfather = 10x years
According to the given question, we get
10x – x = 54
9x = 54
Dividing both sides by 9, we get
\(\frac{9 \mathrm{x}}{9}=\frac{54}{9}\)
x = 6
Present age of granddaughter = 6 years.
Present age of grandfather = 10 × 6 = 60 years.

Question 10.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution:
Let the present age of son be ‘x’ years
Present age of Aman = 3x
Ten years ago Son’s age = (x – 10) years
Aman’s age = (3x – 10) years
According to the given question, we get
3x – 10 = 5(x – 10)
3x – 10 = 5x – 50
Transposing -10 to R.H.S. and 5x to L.H.S
3x – 5x = 10 – 50
-2x = -40
Dividing both sides by -2
\(\frac{-2 x}{2}=\frac{-40}{-2}\)
x = 20
∴ Sons present age = 20 years
Aman’s present age = (3 × 20) = 60 years.

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.2

Question 1.
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 mode and median of this data. Are they same?
Answer:
Arrange the given data in ascending order, we get, 5, 9, 10,12, 15,16, 19, 20, 20, 20, 20, 23, 24, 25, 25
(a) Mode of the data = 20
20 is repeated 4 times

(b) Median of the data = 20
Since the middle value of the data is 20
Obviously, here the mode and median are the same.

Question 2.
The runs scored in a cricket match by 11 players are as follows:
6,15,120, 50, 100, 80, 10,15, 8,10, 15
Find the mean, mode and median of this data. Are the three same?
Answer:
Arrange the given data in ascending order, we get, 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
(i) Mean of the data

(ii) Since 15 is repeated 3 times, the mode of the data = 15
(iii) The value of the middle observation is 15. The Median of the data =15.
Here mean, median and mode are not the same.

Question 3.
The weights (in kg) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38,43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Answer:
Arrange the given data in ascending order, we get,
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Since 38 and 43 are repeated 3 times, the mode of the data = 38 and 43
The value of the middle observation is 40.
∴ Median of the data = 40

(ii) Yes, there are more than one mode.

Question 4.
Find the mode and median of the data: 13,16, 12,14, 19,12,14, 13,14
Answer:
Arrange the given data in ascending order, we get, 12, 12, 13, 13, 14, 14, 14, 16, 19
Since 14 is repeated 3 times, the mode of the data = 14
Again the value of the middle observation is 14, median of the data = 14

Question 5.
Tell whether the following statements is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
Answer:
(i) True
(ii) False
(iii) True
(iv) False

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1

Question 1.
Find the range of heights of any ten students of your class.
Answer:
Do it yourself, one simple solution may be to consider the heights (in cm) of ten students as follows:
111, 118, 110, 120, 114, 113, 118, 117, 115, 119
Average height

= 115.5 cm

Question 2.
Organise the following marks in a class assessment in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6,7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Answer:

(i) The highest number = 9
(ii) The lowest number = 1
(iii) The range = Highest number – Lowest number = 9 – 1 = 8
(iv) Arithmetic Mean
= \(\frac{\text { Sum of the marks }}{\text { Number of students }}\)

= \(\frac{100}{20}\) = 5
Thus, the mean marks = 5.

Question 3.
Find the mean of the first five whole numbers.
Answer:
First five whole numbers are 0, 1, 2, 3 and 4
Sum 0 + 1 + 2 + 3 + 4 = 10
Mean = \(\frac{\text { Sum of the number }}{\text { Number of whole numbers }}\)
= \(\frac{10}{5}\) = 2

Question 4.
A cricketer scores the following runs in eight innings.
58, 76, 40, 35, 46, 45, 0, 100
Find the mean score.
Answer:
Mean score = \(\frac{\text { Sum of the scores }}{\text { Number of innings }}\)
= \(\frac{58+76+40+35+46+45+0+100}{8}\)
= \(\frac{400}{8}\) = 50
Mean score = 50

Question 5.
Following table shows the points of each player scored in four games:

Now answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4?Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Answer:
(i) Mean = \(\frac{\text { Sum of the observations }}{\text { Number of observations }}\)
= \(\frac{14+16+10+10}{4}\) = \(\frac{50}{4}\) = 12.5
A’s average score per game is 12.5

(ii) Since C played only 3 games (He did not play the 3rd game)
Total will be divided by 3

(iii) Mean score of B
= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)
= \(\frac{0+8+6+4}{4}\) = \(\frac{18}{4}\) = 4.5
Thus, th average’tnumber of points scored by is B is 4.5

(iv) Mean score of C
= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)
= \(\frac{8+11+13}{4}\) = \(\frac{32}{3}\) = 10.67
Thus, mean number of points scored by C is 10.67
A’s average score is = 12.5
B’s average score is = 4.5
C’s average score is = 10.67
The best performer is A.

Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39,48, 56, 95, 81, and 75. Find the:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Answer:
Write the given marks in ascending order, we get 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95, Lowest marks = 39

(ii) Range = Highest marks – Lowest marks = 95 – 39 = 56

(iii) Mean = \(\frac{\text { Sum of all the marks }}{\text { Number of students }}\)

= \(\frac{730}{10}\) = 73
Mean marks obtained by the group is 73.

Question 7.
The enrolment in a school during six consecutive years was as follows:
1555,1670,1750,2013,2540,2820 Find the mean enrolment of the school for this period.
Answer:

∴ The mean enrolment is 2058 per year.

Question 8.
The rainfall (in mm) in a city on 7days 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) On how many days was the rainfall less than the mean rainfall.
Answer:
(i) Range = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm.

= \(\frac{41.3}{7}\) = 5.9mm
Mean rainfall = 5.9 mm

(iii) The rainfall was less than the mean rainfall on 5 days.

Question 9.
The heights of 10 girls were measured in cm and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143,141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height.
Answer:
Arrange the heights of girls in ascending order, we get
128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Height of the tallest girl = 151cm
(ii) Height of the shortest girl = 128cm
(iii) Range= Highest height – Lowest height = 151cm – 128cm = 23cm

The mean height of the girls is 141.4cm
(v) The heights of 5 girls are more than the mean height.

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.

Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
Transposing 2x from RHS to L.H.S, we get
3x – 2x = 18
x = 18

Check:
Put x = 18 in L.H.S. and R.H.S. of the equation.
L.H.S. = 3x = 3 × 18 = 54
R.H.S. = 2x + 18 = 2(18) + 18 = 36 +18 = 54
∴ L.H.S. = R.H.S

Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
Transposing (-3) to R.H.S. and 3t to L.H.S., we get
5t – 3t = -5 + 3
2t = -2
Dividing both sides by 2, we get
\(\frac{2 t}{2}=\frac{-2}{2}\)
t = -1

Check:
Put t = -1 in L.H.S. and RHS of the equation
L.H.S. = 5t – 3 = 5(-1) – 3 = -5 – 3 = -8
R.H.S. = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8
Hence, L.H.S. = R.H.S.

Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
Transposing 9 to R.H.S. and 3x to L.H.S.
5x – 3x = 5 – 9
2x = -4
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{-4}{2}\)
x = -2

Check:
Put x = -2 in L.H.S. and R.H.S. of the equation
L.H.S. = 5x + 9
= 5 (-2) + 9
= -10 + 9
= -1
R.H.S. = 5 + 3x
= 5 + 3(-2)
= 5 – 6 = -1
Hence, L.H.S. = R.H.S.

Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
Transposing 3 to R.H.S. and 2z to LHS
4z – 2z = 6 – 3
2z = 3
Dividing both sides by 2, we get
\(\frac{2 z}{2}=\frac{3}{2}\)
z = \(\frac{3}{2}\)

Check:
Put z = \(\frac{3}{2}\) in L.H.S. and R.H.S of the equation
L.H.S. = 4z + 3
= 4\(\left(\frac{3}{2}\right)\) + 3
= 2(3) + 3
= 6 + 3
= 9
R.H.S = 6 + 2z
= 6 + 2\(\left(\frac{3}{2}\right)\)
= 6 + 3
= 9
Hence, L.H.S. = R.H.S.

Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
Transposing -1 to RHS and -x to L.H.S.
2x + x = 14 + 1
3x = 15
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{15}{3}\)
x = 5

Check:
Put x = 5 in LHS and RHS of the equation
LHS = 2x – 1
= 2(5) – 1
= 10 – 1
= 9
R.H.S = 14 – x
= 14 – 5
= 9
Hence, L.H.S. = R.H.S.

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4
Transposing 4 to R.H.S. and 3x to LHS, we get
8x – 3x = 4 – 4
5x = 0
Dividing both sides by 5
\(\frac{5 x}{5}=\frac{0}{5}\)
x = 0

Check:
Put x = 0 in L.H.S. and R.H.S. of the equation
L.H.S. = 8x + 4
= 8(0) + 4
= 4
R.H.S = 3(x – 1) + 7
= 3(0 – 1) + 7
= -3 + 7
= 4
Hence, L.H.S. = R.H.S.

Question 7.
x = \(\frac{4}{5}\) (x + 10)
Solution:
x = \(\frac{4}{5}\) (x + 10)
Multiplying both sides by 5, we get
5x = 5 × \(\frac{4}{5}\) (x + 10)
5x = 4(x + 10)
5x = 4x + 40
Transposing 4x to L.H.S.
5x – 4x = 40
x = 40

Check:
Put x = 40 in L.H.S. and R.H.S. of the equation
L.H.S. = x = 40
R.H.S. = \(\frac{4}{5}\) (x + 10)
= \(\frac{4}{5}\) (40 + 10)
= \(\frac{4}{5}\) x 50
= 4 × 10
= 40
Hence, L.H.S. = R.H.S

Question 8.
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Solution:
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Transposing 1 to R.H.S. and \(\frac{7 \mathrm{x}}{15}\) to L.H.S.
\(\frac{2 x}{3}-\frac{7 x}{15}=3-1\)
\(\frac{10 x-7 x}{15}\) = 2
\(\frac{3 x}{15}\) = 2
Multiplying both sides by 15, we get
\(\frac{3 x}{15}\) × 15 = 2 × 15
3x = 30
Dividing both sides by 3
\(\frac{3 x}{3}=\frac{30}{3}\)
x = 10

Check:
Put x = 10 in L.H.S. and R.H.S. of the equation

Hence, L.H.S. = R.H.S.

Question 9.
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Transposing \(\frac{5}{3}\) to R.H.S. and -y to L.H.S.
2y + y = \(\frac{26}{3}-\frac{5}{3}\)
3y = \(\frac{26-5}{3}\)
3y = \(\frac{21}{3}\) = 7
Divide both sides by 3, we get
\(\frac{3 y}{3}=\frac{7}{3}\)
y = \(\frac{7}{3}\)

Check:
Put y = \(\frac{7}{3}\) in .L.H.S. and R.H.S of the equation

Hence, L.H.S. = R.H.S

Question 10.
3m = 5m – \(\frac{8}{5}\)
Solution:
3m = 5m – \(\frac{8}{5}\)
Transposing 5m to L.H.S. we get
3m – 5m = \(\frac{-8}{5}\)
-2m = \(\frac{-8}{5}\)
Dividing both sides by -2, we get
\(\frac{-2 m}{-2}=\frac{-8}{5} \div(-2)\)
m = \(\frac{8}{10}=\frac{4}{5}\)

Check:
Put m = \(\frac{4}{5}\) in L.H.S. and R.H.S. of the equation

Hence, L.H.S = R.H.S

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2

Question 1.
If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{8}\). What is the number.
Solution:
Let the number be ‘x’
According to the given condition, we get
\(\left(x-\frac{1}{2}\right) \frac{1}{2}=\frac{1}{8}\)
Multiplying both sides by 2
\(\mathrm{x}-\frac{1}{2}=\frac{1}{8} \times 2=\frac{1}{4}\)
Transposing \(\left(-\frac{1}{2}\right)\) to R.H.S. we get
x = \(\frac{1}{4}+\frac{1}{2}\)
x = \(\frac{1+2}{4}=\frac{3}{4}\)
∴ The required number is \(\frac{3}{4}\).

Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Let the breadth of the pool be x m
∴ Length of the pool = (2 + 2x) m = (2x + 2) m
Perimeter of the pool = 154 m
2(2x + 2 + x) = 154 m
[∵ Perimeter of the rectangle = 2(l + b)]
2(3x + 2) = 154
6x + 4 = 154
Transposing 4 to the R.H.S.
6x = 154 – 4
6x = 150
Dividing both sides by 6,
\(\frac{6 x}{6}=\frac{150}{6}\)
∴ x = 25
∴ Breadth of the pool = 25m
Length of the pool = 2(25) + 2 = 50 + 2 = 52 m

Question 3.
The base of an isosceles triangle is \(\frac{4}{3}\) cm and the perimeter of the triangle is 4\(\frac{2}{15}\) cm. What is the length of the remaining equal sides?
Solution:
Let the length of the equal sides of a triangle be ‘x’
Base of an isosceles triangle = \(\frac{4}{3}\) cm
Perimeter of the triangle = 4\(\frac{2}{15}\) cm

∴ Length of the equal sides is 1\(\frac{2}{5}\) cm.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number be x
then the greater number = x + 15
Sum of two numbers = 95
x + (x + 15) = 95
2x + 15 = 95
Transposing 15 to R.H.S, we get
2x = 95 – 15 = 80
Dividing both sides by 2
x = \(\frac{80}{2}\) = 40
∴ The smaller number = 40
The greater number = (40 + 15) = 55

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two number be 5x and 3x
Difference of two numbers = 18
5x – 3x = 18
2x = 18
Dividing both sides by 2, we get
\(\frac{2 \mathrm{x}}{2}=\frac{18}{2}\)
∴ x = 9
The two numbers are (5 × 9) and (3 × 9) i.e. 45 and 27.
Hence, the required numbers are 45 and 27.

Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
Sum of three integers = 51
x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 to RHS, we get
3x = 51 – 3 = 48
Dividing both sides by 3, we get
\(\frac{3 \mathrm{x}}{3}=\frac{48}{3}\)
x = 16
Now x = 16,
x + 1 = 16 + 1 = 17
x + 2 = 16 + 2 = 18
∴ The required three consecutive integers are 16, 17 and 18.

Question 7.
The sum of three consecutive multiple of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be x, x + 8, x + 16
Sum of three consecutive multiples = 888
x + x + 8 + x + 16 = 888
3x + 24 = 888
Transposing 24 to R.H.S., we get
3x = 888 – 24 = 864
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{864}{3}\)
x = 288
x + 8 = 288 + 8 = 296
x + 16 = 288 + 16 = 304
∴ The required multiples of 8 are 288, 296 and 304.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
According to the given condition
2(x) + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
Transposing 11 to R.H.S., we get
9x = 74 – 11
9x = 63
Dividing both sides by 9, we get
\(\frac{9 x}{9}=\frac{63}{9}\)
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
∴ The consecutive integers are 7, 8 and 9.

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later, the sum of their ages will be 56 years. What are their present ages?
Solution:
Let the present age of Rahul be ‘5x’
and the present age of Harron be 7x;
4 years later
Rahul’s age = (5x + 4) years
Harron’s age = (7x + 4) years
Sum of their ages = 56 years
5x + 4 + 7x + 4 = 56
12x + 8 = 56
Transposing 8 to R.H.S. we get
12x = 56 – 8 = 48
Dividing both sides by 12
\(\frac{12 \mathrm{x}}{12}=\frac{48}{12}\)
x = 4
Present age of Rahul = 5 × 4 = 20 years
Present age of Haroon = 7 × 4 = 28 years

Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength.
Solution:
Let the number of boys in the class be 7x
and the number of girls in the class be 5x
As number of boys is 8 more than the number of girls.
7x = 5x + 8
Transposing 5x to L.H.S.
7x – 5x = 8
2x = 8
Dividing both sides by 2
\(\frac{2 \mathrm{x}}{2}=\frac{8}{2}\)
x = 4
∴ Number of boys = 7 × 4 = 28
Number of girls = 5 × 4 = 20
Total class strength = 28 + 20 = 48 students.

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let Baichung’s present age be x years
Baichung father’s age = (x + 29) years
Baichung grandfather’s age = (x + 29 +26) years = (x + 55) years
Sum of the ages of all the three =135 years
x + x + 29 + x + 55 = 135
3x + 84 = 135
Transposing 84 to RHS, we have
3x = 135 – 84 = 51
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{51}{3}\)
x = 17
Baichung’s age =17 years
Baichung fathers age = 17 + 29 = 46 years
Baichung grandfather’s age = 17 + 55 = 72 years.

Question 12.
Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age be ‘x’ years
4 times his present age = 4x years
15 years from now, Ravi’s age = (x + 15) years
According to the given condition
x + 15 = 4x
Transposing 15 and 4x, we get
-4x + x = -15
-3x = -15
Dividing both sides by (-3) we get
\(\frac{-3 x}{-3}=\frac{-15}{-3}\)
x = 5
∴ Ravi’s present age = 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(\frac{7}{12}\). What is the number?
Solution:
Let the required rational number be ‘x’
According to the given condition

x = \(\frac{-1}{2}\)
∴ The rational number is \(\frac{-1}{2}\).

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10 respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00, 000. How many notes of each denomination does she have?
Solution:
Let the number of ₹ 100 notes be 2x
the number of ₹ 50 notes be 3x
and the number of ₹ 10 notes be 5x
Value of ₹ 100 notes = 2x × 100 = ₹ 200x
Value of ₹ 50 notes = 3x × 50 = ₹ 150x
Value of ₹ 10 notes = 5x × 10 = ₹ 50x
According to the given condition,
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00000
400x = 4,00,000
Dividing both sides by 400, we get
\(\frac{400 \mathrm{x}}{400}=\frac{4,00,000}{400}\)
x = 1000
∴ Number of ₹ 100 notes = 2 × 1000 = 2000
Number of ₹ 50 notes = 3 × 1000 = 3000
Numbers of ₹ 10 notes = 5 × 1000 = 5000

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x
then number of ₹ 2 coins = 3x
and the number of coins of ₹ 1 = 160 – (x + 3x) = 160 – 4x
Now Value of ₹ 5 coins = ₹ 5 × x = 5x
Value of ₹ 2 coins = ₹ 2 × 3x = 6x
Value of ₹ 1 coins = ₹ 1 × (160 – 4x) = (160 – 4x)
According to the given condition,
160 – 4x + 5x + 6x = 300
160 + 7x = 300
Transposing 160 to the R.H.S.
7x = 300 – 160
7x = 140
Dividing both sides by 7
\(\frac{7 \mathrm{x}}{7}=\frac{140}{7}\)
x = 20
∴ Number of ₹ 5 coins = 20
Number of ₹ 2 coins = (3 × 20) = 60
Number of ₹ 1 coin = 160 – (4 × 20) = 80

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be ‘x’
The total number of participants = 63
The number of non-winners = 63 – x
Prize money given to winners = ₹ 100 × x
Prize money given to non-winner participants = ₹ 25(63 – x)
= ₹ 25 × 63 – ₹ 25x
= ₹ 1575 – ₹ 25x
According to the condition given in the question,
100x + 1575 – 25x = 3000
75x + 1575 = 3000
Transposing 1575 to R.H.S. we get
75x = 3000 – 1575
75x = 1425
Dividing both sides by 75
\(\frac{75 \mathrm{x}}{75}=\frac{1425}{75}\)
x = 19
∴ The number of winners = 19.