Class 7

Haryana State Board HBSE 7th Class Maths Solutions Chapter 3 Data Handling Ex 3.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 3 Data Handling Exercise 3.3

Question 1.
Use the bar-chart to answer the following questions :
(a) Which is the most popular pet ?
(b) How many children have a dog as a pet ?

Solution:
(a) cats
(b) 8.

Question 2.
Read the bar-graph and answer the questions that follows :
Number of books sold by a bookstore during five consecutive years.

(i) About how many books were sold in 1989 ? 1990 ? 1992 ?
(ii) In which year were about 475 books sold ? About 225 books ?
(iiii) In which years were fewer than 250 books sold ?
(iv) Can you explain how you would estimate the number of books sold in 1989 ?
Solution:
(i ) Many books were sold in 1989,1990, 1992
In 1989 = 175. In 1990 = 475, In 1992 = 225 Total = 175 + 475 + 225 = 875 books
(ii) 1990, 1992
(iii) 1989 and 1992.
(iv)Scale : 1 cm = 100 books In 1989 scale is 1.75 cm
∴ Books = 1.75 x 100 = 175 books.

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 eight.
Solution:

(a) Scale : 1 cm = 20 children in y-axis.
(b) (i) Fifth class has the maximum number of children = 135
Tenth class has the minimum number of children = 80.
(ii) Ratio of students of class sixth to eighth = 120 : 100 = \(\frac{120}{100}=\frac{12}{10}=\frac{6}{5}\) = 6:5

Question 4.
The performance of students in 1st Term and 2nd 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 least ?
(iii) Has the performance gone down in any subject ?
Solution:

(i) Maths
(ii) Maths
(iii) 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 sports is most popular ?
(iii) What is more preferred, watching or participating in sports ?
Solution:

(i) Cricket is most popular than other sports, because 1240 watching and 620 participating.
(ii) Cricket
(iii) Watching.

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:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date ?
(ii) Which is the hottest and 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.
Solution:
(i) Jammu,
(ii) Jammu and Bangalore,
(iii) Bangalore, Amritsar
(iv) Mumbai. (See Fig. )

Haryana State Board HBSE 7th Class Maths Solutions Chapter 3 Data Handling Ex 3.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 3 Data Handling Exercise 3.2

Question 1.
The scores in mathematics test (out of 25) of 15 students is 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 ?
Solution:
We arrange the data in ascending order, we get,
5.9.10.12.15.16.19.20.20.20.20. 25, 25.
Mode of this data is 20 because it occurs 4 times more than frequently than other observations.
Now, Median is the middle observation.
Therefore, 20 is the median.
Yes, Mode and Median are same.

Question 2.
The runs scored in a cricket match by 11 players is as follows :
6,15,120,50,100,80,10,15,8,10,10.
Find the mean, mode and median of this data. Are the three same ?
Solution:
Mean = \(\frac{Sum of the scores}{No. of players}\)
= \(\frac{424}{11}\) = 38.5454 = 38.55
We arrange the data in ascending order, we get 6, 8, 10, 10,10, 15, 15, 50, 80, 100, 120.
∴ Mode of this data is 10 because it occurs three times more than other observations.
Now, Median is the middle observation.
Therefore 15 is the median.
∴ Mean, Mode and Median are not 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 ?
Solution:
We arrange the data in ascending order, we get,
32,35,36, 37, 38, 38,38, 40,42,43,43,43, 45, 47, 50.
(i) Mode of this data is 38 and 43 because both occur three times more than- other observations.
Now, Median is the middle observation. Therefore 40 is the median.
(ii) Yes, there are two modes 38 and 43.

Question 4.
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, we get,
12, 12, 13, 13, 14, 14, 14, 16,19.
Therefore, mode of this data is 14 because it occurs three times more than other observations.
Now, median is the middle observation. Therefore 14 is the median.

Question 5.
Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean can be one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) A data always has a mode.
(v) The data 6, 4, 3, 8, 9,12,13, 9 has mean 9.
Solution:
(i) False, (ii) Yes, (iii) Yes, (it;) Yes,
(v) Mean = \(\frac{6+4+3+8+9+12+13+9}{8}=\frac{64}{8}\) = 8
∴ False.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 3 Data Handling Ex 3.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 3 Data Handling Exercise 3.1

Question 1.
Find the range of heights of 10 students of your class.
Solution:
The height (in feet) of 10 students of my class VII,
= 3.5, 4, 5, 4, 3, 5, 4, 3.5, 4, 6.
∴ Range of the heights (in feet) of the students = 6-3 = 3 Feet.

Question 2.
Organise the following marks in a class assessment, in a tabular form.
(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.

Solution:
Let us put the data in a tabular form :

(i) 9 is the highest grade.
(ii) 1 is the lowest grade.
(wi) Range of the data = 9-1 = 8
(iv) Arithmetic Mean

Question 3.
Find the mean of first five whole numbers.
Solution:
Mean of first 5 whole number
\(\frac{0+1+2+3+4}{5}=\frac{10}{5}=2\)

Question 4.
A cricketer scores the following runs in eight innings: 58, 76,40,35,48,45, 0,100. Find the mean score.
Solution:
Mean score = \(\frac{\text { Sum of scores }}{\text { Total no. of innings }}\)
= \(\frac{58+76+40+35+48+45+0+100}{8}=\frac{402}{8}=\) = 50.25

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

Now answer these 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 1 Why ?
(Hi) B played in all the four games. How would you find the mean ?
(iv) Who is the best performer ?
Solution:
(i) Mean of A
\(\frac{14+16+10+10}{4}=\frac{50}{4}=\frac{25}{2}\) = 12.5

(ii) Mean of C = \(\frac{8+11+13}{3}=\frac{32}{3}\) = 10.66
because three players play.
(iii) Mean of B
= \(\frac{0+8+6+4}{4}=\frac{18}{4}=\frac{9}{2}\) = 4.5

(iv) Mean of A > Mean of C > Mean of B.
∴ A is the best performer.

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 mark obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:
(i) Highest marks of the students = 95
Lowest marks of the students = 39 (ii) Range of the marks = highest marks – lowest marks = 95 – 39 = 56

Question 7.
The enrolment of 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.
Solution:

Question 7.
The rainfall (in mm) in 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) On how many days was the rainfall less
than the mean rainfall. HBSESolutions.com
Solution:
(i) Range of rainfall
= highest rainfall – lowest rainfall.
= 20.5 – 0.0 = 20.5 mm

(ii) Mean rainfall = \(\frac{\text { Sum of rainfall }}{\text { Total no. of days }}\)
= \(\frac{0.0+12.2+2.1+0.0+20.5+5.3+1.0}{7}\)
= \(\frac{41.1}{7}\) = 5.87 mm

(iii) 5 days (Mon, Wed, Thus, Sat, Sun).

Question 8.
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 of height the girls ?
(v) How many girls have heights more than the mean height.
Solution:
(i) 151 cm. is the height of the tallest girl.
(ii) 128 cm. is the height of the shortest girl.
(iii) Range of data = highest height – shortest height = 151 – 128 = 23 cm. ;
(iv) Mean height of the girls

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.7 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.7

Question 1.
Find :
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5
Solution:
(i) 0.4 ÷ 2 = 0.2
(ii) 0.35 ÷ 5 = 0.07
(iii) 2.48 ÷ 4 = \(\frac{218}{400}=\frac{62}{100}\) = 0.62
(iv) 65.6 ÷ 6 = 10.93
(v) 651.2 ÷ 4 = 162.8
(vi) 14.49 ÷ 7 = 2.7
(vii) 3.96 ÷ 4 = 0.99
(viii) 0.80 ÷ 5 = 0.16

Question 2.
Find:
(i) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(v) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Solution:
(i) 4.8 ÷ 10 = 0.48
(ii) 52.5 ÷ 10 = 5.25
(iii) 0.7 ÷ 10 = 0.07
(iv) 33.1 ÷ 10 = 3.31
(v) 272.23 ÷ 10 = 27.223
(vi) 0.56 ÷ 10 = 0.056
(vii) 3.97 ÷ 10 = 0.397

Question 3.
Find :
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Solution:
(i) 2.7 ÷ 100 = \(\frac{2.7}{100}\) = 0.027
(ii) 0.3 ÷ 100 = 0.003
(iii) 0.78 ÷ 100 = 0.0078
(iv) 432.6 ÷ 100 = 4.326
(v) 23.6 ÷ 100 = 0.236
(vi) 98.53 ÷ 100 = 0.9853

Question 4.
Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Solution:
(i) 7.9 ÷ 1000 = \(\frac{79}{1000}\) = 0.0079
(ii) 26.3 ÷ 1000 = \(\frac{263}{1000}\) = 0.0263
(iii) 38.53 ÷ 1000 = \(\frac{38.50}{10000}\) = 0.03850
(iv) 128.9 ÷ 1000 = \(\frac{1289}{10000}\) = 0.1289
(v) 0.5 ÷ 1000 = \(\frac{05}{10000}\) = 0.0005

Question 5.
Find :
(i) 7÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Solution:
(i) 7÷ 3.5 = \(\frac{70}{35}\) = 2
(ii) 36 ÷ 0.2 = \(\frac{360}{2}\) = 180
(iii) 3.25 ÷ 0.5 = \(\frac{325}{50}\) = \(\frac{65}{10}\) = 6.5
(iv) 30.94 ÷ 0.7 = \(\frac{3094}{700}=\frac{442}{100}\) = 4.42
(v) 0.5 ÷ 0.25 = \(\frac{50}{25}\) = 2
(vi) 7.75 ÷ 0.25 = \(\frac{775}{25}\) = 31
(vii) 76.5 ÷ 0.15 = \(\frac{7650}{15}\) = 510
(viii) 37.8 ÷ 1.4 = \(\frac{378}{14}\) = 27
(ix) 2.73 ÷ 1.3 = \(\frac{2.73}{130}\) = 21

Question 6.
A vehicle covers a distance of 43.2 km. in 2.4 litres of petrol. How much distance will it travel in one litre of petrol ?
Solution:
∵ 2.4 litres of petrol a vehicle covers a distance of 43.2 km.
∴ 1 litre of petrol a vehicle covers
\(=\frac{43.2}{2.4} \mathrm{~km}=\frac{432}{24}=\frac{108}{6}=18 \mathrm{~km}\)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.6 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.186
Solution:
(i) 0.2 × 6 = 1.2
(ii) 8 × 4.6 = 36.8
(iii) 2.71 × 5 = 13.55
(iv) 20.1 × 4 = 80.4
(v) 0.05 × 7 = 0.35
(vi) 211.02 × 4 = 844.08
(vii) 2 × 0.86 = 1.72.

Question 2.
Find the area of rectangle whose length is 5.7 cm. and breadth is 3 cm.

Solution:
Area of rectangle = l × b = 5.7× 3 = 17.1cm².

Question 3.
Find :
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 3

Question 4.
A two wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol ?
Solution:
Total distance cover = 55.3 × 10 = 553 km.

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07  × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
(xi) 211.02 × 11.32
(xii) 13.01 × 5.01
Solution:
(i) 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7 – 5.17
(iii) 0.2 × 316.8 = 63.36
(iv) 1.3 × 3.1 = 4.03
(v) 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15 = 1.680
(vii) 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1 = 110.011.
(xi) 211.02 × 11.32 = 2388.7464
(xii) 13.01 × 5.01 = 65.1801

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.5 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.5

Question 1.
What is greater ?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88.
Solution:
(i) 0.5 > 0.05 [Because 5 > 0]
(ii) 0.7 > 0.5 [Because .7 > .5]
(iii) 7 > 0.7
(iv) 1.49 > 1.37 [Because 4 > 3]
(v) 2.30 > 2.03 [Because 3 > 0]
(vi) 0.88 > 0.80. [Because 8 > 0]

Question 2.
Express as rupees using decimals.
(i) 7paise
(ii) 7 rupees 7paise
(iii) 77 rupees 77paise
(iv) 50paise
(v) 235paise.
Solution:
(i) ∵ 100 paise = 1 Rupee
∴ paise = 0.07 Rupees
(ii) 7 rupees 7 paise = 7.07 Rupees.
(iii) 77 rupees 77 paise = 77.77 Rupees.
(iv) 50 paise = 0.50 Rupees
(v) 235 paise = \(\frac{235}{100}\)
Rupees = 2.35 Rupees.

Question 3.
(i) Express 5 cm in metre and kilometre.
(ii) Express 35 mm in cm, m and km.
Solution:
(i) ∵ 100 cm = 1 m and 1000 m
= 1 km
5 cm = 0.05 m and \(\frac{0.05}{1000}\) = 0.00005 km

(ii) 35 mm – 3.5 cm = \(\frac{3.5}{100}\) = 0.035 m = \(\frac{0.035}{1000}\)
= 0.000035 km

Question 4.
Express in kg.
(i) 200g
(ii) 3470g
(iii) 4 kg 8 g
(iv) 2598 mg
Solution:
(i) ∵ 1000 gm = 1 kg
200g = \(\frac{200}{1000}=\frac{2}{10}\) = 0.2kg
(ii) 3470 g = \(\frac{3470}{1000}=\frac{347}{100}\) = 3.47 kg
(iii) 4 kg 8g = 4.008 kg
(iv) 2598 mg = \(\frac{2508}{1000}\)
= \(\frac{2598}{1000 \times 1000} \mathrm{~kg}=\frac{2598}{1000000} \mathrm{~kg}\) = 0.02598 kg

Question 5.
Write the following decimal numbers in the expanded form :
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:
(i) 20.03 = 2 x 10 + 0 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\))
(ii) 2.03 = 2 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\))
(iii) 200.03 = 2 x 100 + 0 x 10 + 0 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\)) + 4 x (\(\frac{1}{1000}\))

Question 6.
Write the place value of 2 in the following decimal numbers :
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352.
Solution:
(i) 2.56 Place value of 2 = Ones
(ii) 21.37 Place value of 2 = Tens
(iii) 10.25 Place value of 2 = Tenths
(iv) 9.42 Place value of 2 = Hundredths
(v) 63.352 Place value of 2 = Thousandths.

Question 7.
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is
12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much ?
Solution:
Total travelled distance of Dinesh = 7.5 km + 12.7 km = 20.2 km
Total travelled distance of Ayub = 9.3 km + 11.8 km = 21.1 km
or, 21.1 km – 20.2 km = 0.9 km
∴ Ayub travelled 0.9 km more.

Question 8.
Shyama bought 5 kg 300g apples and 3 kg 250g mangoes. Sarala bought 4 kg 800g apples and 4 kg 150g bananas. Who bought more fruits ?
Solution:

or, 8.950 kg – 8.550 kg = 0.400 kg
.’. Sarala bought 400 g more fruits.

Question 9.
How much less is 28 km than 42.6 km?
Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.4

Question 1.
Find

Solution:
0

Question 2.


Solution:

Question 3.

Solution:

Question 4.

Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.3

Question 1.
Find

Solution:

Question 2.
Multiply and reduce to lowest form (if possible) :

Solution:

Question 3.
Multiply the following fractions :

Solution:

Question 4.
Which is greater?

Solution:

Question 5.
Saili plants 4 saplings in a row, in her garden. The distance between two adjacent saplings is \(\frac{3}{4}\) m. Find the distance between the first and the last sapling.
Solution:
The distance between two adjacent saplings = \(\frac{3}{4}\) m
There are 4 saplings in a row.
∴ The distance between the first and the last sapling = \(\frac{3}{4}\) x 3 = \(\frac{9}{4}\) m.

Question 6.
Lipika reads a book for 1 \(\frac{3}{4}\) hours every day. She reads the entire book in 6 days. IIow many hours in all were required by her to read the book ?
Solution:
Lipika reads a book for \(\frac{7}{4}\) hours everyday.
∴ Required hours = \(\frac{7}{4} \times 6=\frac{7 \times 3}{2}=\frac{21}{2}=10 \frac{1}{2}\) hours

Question 7.
A car runs 16 km. using 1 litre of petrol. How much distance will it cover using 2\(\frac{3}{4}\) litres of petrol ?
Solution:
A car runs 16 km. using 1 litre of petrol.
∴ Total distance cover =16 x 2\(\frac{3}{4}\)
= 16 x \(\frac{11}{4}\) = 44 km.

Question 8.

Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.2

Question 1.
Which of the drawings (a) to (d) show :


Solution:
(i)—(d), (ii)—(b), (iii)—(a), (iv)—(c).

Question 2.
Some pictures (a) to (c) are given below. Tell which of them shows :

Solution:
(i)—(c), (ii)—(a), (iii)—(b).

Question 3.
Multiply and reduce to lowest form :

Solution:

Question 4.
Shade:
(i) \(\frac{1}{2}\) of the circle in box (a)
(ii) \(\frac{2}{3}\) of the triangles in box (b).
(iii) \(\frac{3}{5}\) of the boaxes in box (c).

Solution:

Question 5.
(i) \(\frac{1}{2}\) of (i) 24 (ii) 46
(ii) \(\frac{2}{3}\) of (i) 18 (ii) 27
(iii) \(\frac{3}{5}\) of (i) 16 (ii) 36
(iv) \(\frac{4}{5}\) of (i) 20 (ii) 35
Solution:

Question 6.
Multiply and express as a mixed fraction :

Solution:

Question 7.
Find:

Solution:

Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bag that contained 5 litres of water. Vidya consumed \(\frac{2}{5}\) the remaining water. Pratap consumed the remaining water.
(i) How much water did Vidya drink ?
(ii) What fraction of the total quantity of water did Pratap drink ?
Solution:
(i) Vidya consumed \(\frac{2}{5}\) of the water
∴ 5 litres of \(\frac{2}{5}\) = 5 x \(\frac{2}{5}\) = 2 litres
∴ Vidya drink = 2 litres.
(ii) Now 5 litres – 2 litres = 3 litres 3
∴ Pratap drink = \(\frac{3}{1}\) litres = 3 litres.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.1

Question 1.
Solve :

Solution:

Question 2.
Arrange the following in descending order:
(i) \(\frac{2}{9}, \frac{2}{3}, \frac{8}{21}\)
(ii) \(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
Solution:

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square ?

Solution:

Question 4.
A rectangular sheet of paper is 12 \(\frac{1}{2}\)cm long and 10\(\frac{2}{3}\)cm wide. Find its perimeter.
Solution:
Perimeter of a rectangular sheet of paper

Question 5.
Find the perimeters (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
Solution:
(i) Perimeter of ΔABE
= AB + BC + EA


Perimeter of ΔABE is greater than the rectangle BCDE.

Question 6.
Salil wants to put a picture-in a frame. The picture is 7\(\frac{3}{5}\) cm wide. To fit in the frame the picture cannot be more than 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
The picture, should be trimmed

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share 7 By how much ?
Solution:
Ritu ate \(\frac{3}{5}\) part of an apple.
Remaining apple = \(1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}\)
Somu ate \(\frac{2}{5}\) part of an apple \(\frac{3}{5}>\frac{2}{5}\)
Ritu had the larger share,
Now \(\frac{3}{5}-\frac{2}{5}=\frac{3-2}{5}=\frac{1}{5}\)

Question 8.
Michael finished colouring a picture \(\frac{7}{12}\) in hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer?
By what fraction was it longer ?
Solution:
Michael finished colouring a pictuie in \(\frac{7}{12}\) hour
Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour.