ICSE Class 8 Mathematics · Solutions for Class 8 RS Aggarwal ICSE Foundation Mathematics · 18 solved questions
Arrange the following data in ascending order:
8, 3, 11, 15, 1, 7, 16, 25, 9, 4
Arranging the given data in ascending order, i.e., from the smallest value to the greatest value, we get :
1, 3, 4, 7, 8, 9, 11, 15, 16, 25
Hence, the given data in ascending order is 1, 3, 4, 7, 8, 9, 11, 15, 16, 25.
Arrange the following data in ascending order:
5.7, 9.2, 3.1, 4.6, 1.8, 10.4, 2.5, 0.9, 1.2
Arranging the given data in ascending order, i.e., from the smallest value to the greatest value, we get :
0.9, 1.2, 1.8, 2.5, 3.1, 4.6, 5.7, 9.2, 10.4
Hence, the given data in ascending order is 0.9, 1.2, 1.8, 2.5, 3.1, 4.6, 5.7, 9.2, 10.4.
Arrange the following data in descending order:
15, 3, 0, 11, 7, 2, 23, 14, 29, 5, 1, 17
Arranging the given data in descending order, i.e., from the greatest value to the smallest value, we get :
29, 23, 17, 15, 14, 11, 7, 5, 3, 2, 1, 0
Hence, the given data in descending order is 29, 23, 17, 15, 14, 11, 7, 5, 3, 2, 1, 0.
Arrange the following data in descending order:
2.3, 4.6, 1.2, 0.4, 3.5, 9.7, 6.1, 4.8, 0.8, 1.0
Arranging the given data in descending order, i.e., from the greatest value to the smallest value, we get :
9.7, 6.1, 4.8, 4.6, 3.5, 2.3, 1.2, 1.0, 0.8, 0.4
Hence, the given data in descending order is 9.7, 6.1, 4.8, 4.6, 3.5, 2.3, 1.2, 1.0, 0.8, 0.4.
Construct a frequency table for the following data:
3, 2, 5, 4, 1, 3, 2, 2, 5, 3, 1, 2, 1, 1, 2, 2, 3, 4, 5, 3, 1, 2, 3
Arranging the given data in ascending order, we get :
1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5
Against each variate, we mark a tally mark each time the variate occurs in the given data and then count the tally marks to get the frequency.
The required frequency table is as given below :
| x | Tally marks | Frequency (f) |
|---|---|---|
| 1 | llll | 5 |
| 2 | llll ll | 7 |
| 3 | llll l | 6 |
| 4 | ll | 2 |
| 5 | lll | 3 |
Construct a frequency table for the following data:
7, 8, 6, 5, 6, 7, 7, 9, 8, 10, 7, 6, 7, 8, 8, 9, 10, 5, 7, 8, 7, 6
Arranging the given data in ascending order, we get :
5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 10, 10
Against each variate, we mark a tally mark each time the variate occurs in the given data and then count the tally marks to get the frequency.
The required frequency table is as given below :
| x | Tally marks | Frequency (f) |
|---|---|---|
| 5 | ll | 2 |
| 6 | llll | 4 |
| 7 | llll ll | 7 |
| 8 | llll | 5 |
| 9 | ll | 2 |
| 10 | ll | 2 |
Construct a frequency table for the following data:
152, 165, 172, 144, 135, 156, 175, 140, 132, 150, 153, 147
Arranging the given data in ascending order, we get :
132, 135, 140, 144, 147, 150, 152, 153, 156, 165, 172, 175
Here, the smallest observation is 132 and the greatest observation is 175.
So, we condense the data into class-intervals of class-size 5, i.e., 130 – 135, 135 – 140, 140 – 145, ......, 175 – 180.
The class-intervals are taken in the exclusive form. So, the class-interval 130 – 135 means 130 and more but less than 135, the class-interval 135 – 140 means 135 and more but less than 140, and so on.
The required frequency table is as given below :
| Class-interval | Tally marks | Frequency |
|---|---|---|
| 130 – 135 | l | 1 |
| 135 – 140 | l | 1 |
| 140 – 145 | ll | 2 |
| 145 – 150 | l | 1 |
| 150 – 155 | lll | 3 |
| 155 – 160 | l | 1 |
| 160 – 165 | 0 | |
| 165 – 170 | l | 1 |
| 170 – 175 | l | 1 |
| 175 – 180 | l | 1 |
Construct a frequency table for the following data:
13, 25, 19, 16, 8, 30, 27, 6, 0, 34, 40, 11, 4, 17
Arranging the given data in ascending order, we get :
0, 4, 6, 8, 11, 13, 16, 17, 19, 25, 27, 30, 34, 40
Here, the smallest observation is 0 and the greatest observation is 40.
So, we condense the data into class-intervals of class-size 10, i.e., 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50.
The class-intervals are taken in the exclusive form. So, the class-interval 0 – 10 means 0 and more but less than 10, the class-interval 10 – 20 means 10 and more but less than 20, and so on.
The required frequency table is as given below :
| Class-interval | Tally marks | Frequency |
|---|---|---|
| 0 – 10 | llll | 4 |
| 10 – 20 | llll | 5 |
| 20 – 30 | ll | 2 |
| 30 – 40 | ll | 2 |
| 40 – 50 | l | 1 |
Fill in the blanks:
(i) Lower limit of the class interval 14 – 22 is ...............
(ii) Upper limit of the class interval 16 – 25 is ...............
(iii) Class-size of the class interval 20 – 30 is ...............
(iv) If the upper and lower limits of a class interval be 18 and 11 respectively, then the class-interval is ...............
(v) If the lower and upper limits of a class interval be 11.5 and 21.5 respectively, then the class-interval is ...............
(i) Each class is bounded by two figures, which are called class limits. The figure on the left side is called the lower limit of the class.
In the class interval 14 – 22, the figure on the left side is 14.
Hence, the lower limit of the class interval 14 – 22 is 14.
(ii) Each class is bounded by two figures, which are called class limits. The figure on the right side is called the upper limit of the class.
In the class interval 16 – 25, the figure on the right side is 25.
Hence, the upper limit of the class interval 16 – 25 is 25.
(iii) By formula,
Class-size = Upper limit − Lower limit
For the class interval 20 – 30, the lower limit is 20 and the upper limit is 30.
Substituting the values, we get :
Class-size = 30 − 20 = 10
Hence, the class-size of the class interval 20 – 30 is 10.
(iv) In a class-interval, the lower limit is written on the left side and the upper limit is written on the right side.
Here, the lower limit is 11 and the upper limit is 18.
So, the class-interval is 11 – 18.
Hence, the required class-interval is 11 – 18.
(v) In a class-interval, the lower limit is written on the left side and the upper limit is written on the right side.
Here, the lower limit is 11.5 and the upper limit is 21.5.
So, the class-interval is 11.5 – 21.5.
Hence, the required class-interval is 11.5 – 21.5.
The marks out of 50, obtained by 30 students of a class in an examination are given below:
40, 12, 46, 37, 17, 27, 30, 6, 2, 23, 19, 39, 25, 5, 33, 19, 21, 12, 41, 17, 12, 19, 17, 8, 10, 1, 9, 21, 13, 48
Arrange them in ascending order and present it as a grouped data, by taking class-intervals 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50.
Arranging the given marks in ascending order, we get :
1, 2, 5, 6, 8, 9, 10, 12, 12, 12, 13, 17, 17, 17, 19, 19, 19, 21, 21, 23, 25, 27, 30, 33, 37, 39, 40, 41, 46, 48
The class-intervals are taken in the exclusive form. So, the class-interval 0 – 10 means 0 and more but less than 10, the class-interval 10 – 20 means 10 and more but less than 20, and so on.
The required grouped frequency distribution table is as given below :
| Marks | Tally marks | No. of students |
|---|---|---|
| 0 – 10 | llll l | 6 |
| 10 – 20 | llll llll l | 11 |
| 20 – 30 | llll | 5 |
| 30 – 40 | llll | 4 |
| 40 – 50 | llll | 4 |
The fine imposed (in rupees) on 40 pupils for being absent is given below:
18, 8, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 0, 5, 3, 6, 27, 21, 19, 15, 20, 24, 2, 16, 14, 8, 23, 24, 10, 14, 3, 8, 11, 2
Present the data in the form of frequency distribution table by taking class intervals 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25.
Arranging the given fines in ascending order, we get :
0, 2, 2, 2, 2, 3, 3, 5, 5, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 11, 12, 12, 12, 14, 14, 15, 16, 16, 16, 18, 19, 20, 20, 21, 23, 23, 23, 24, 24, 27
Here, the smallest observation is 0 and the greatest observation is 27.
Since 27 does not lie in any of the class-intervals 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, we take one more class-interval 25 – 30 of the same class-size so that all the observations are covered.
The class-intervals are taken in the exclusive form. So, the class-interval 0 – 5 means 0 and more but less than 5, the class-interval 5 – 10 means 5 and more but less than 10, and so on.
The required frequency distribution table is as given below :
| Fine (in ₹) | Tally marks | No. of pupils |
|---|---|---|
| 0 – 5 | llll ll | 7 |
| 5 – 10 | llll llll | 10 |
| 10 – 15 | llll lll | 8 |
| 15 – 20 | llll l | 6 |
| 20 – 25 | llll lll | 8 |
| 25 – 30 | l | 1 |
The weights (in kg) of 30 students are given below:
48, 55, 30, 43, 41, 36, 48, 37, 34, 38, 46, 42, 38, 52, 37, 34, 40, 42, 50, 37, 35, 46, 42, 45, 43, 37, 39, 45, 48, 46
Form the grouped frequency table by taking class-intervals 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55.
Arranging the given weights in ascending order, we get :
30, 34, 34, 35, 36, 37, 37, 37, 37, 38, 38, 39, 40, 41, 42, 42, 42, 43, 43, 45, 45, 46, 46, 46, 48, 48, 48, 50, 52, 55
The class-intervals are taken in the exclusive form. So, the class-interval 30 – 35 means 30 and more but less than 35, the class-interval 35 – 40 means 35 and more but less than 40, and so on.
Here, 55 is the greatest observation and 50 – 55 is the last class-interval. So, 55 is included in the class-interval 50 – 55.
The required grouped frequency table is as given below :
| Weight (in kg) | Tally marks | No. of students |
|---|---|---|
| 30 – 35 | lll | 3 |
| 35 – 40 | llll llll | 9 |
| 40 – 45 | llll ll | 7 |
| 45 – 50 | llll lll | 8 |
| 50 – 55 | lll | 3 |
The daily income (in rupees) of 30 workers is given below:
83, 96, 91, 80, 97, 111, 90, 83, 112, 120, 103, 77, 82, 90, 70, 68, 89, 85, 78, 105, 94, 97, 90, 89, 102, 113, 90, 106, 70, 84
Construct a frequency table using groups 65 – 75, 75 – 85, 85 – 95, 95 – 105, 105 – 115, 115 – 125.
Arranging the given daily incomes in ascending order, we get :
68, 70, 70, 77, 78, 80, 82, 83, 83, 84, 85, 89, 89, 90, 90, 90, 90, 91, 94, 96, 97, 97, 102, 103, 105, 106, 111, 112, 113, 120
The class-intervals are taken in the exclusive form. So, the class-interval 65 – 75 means 65 and more but less than 75, the class-interval 75 – 85 means 75 and more but less than 85, and so on.
The required frequency table is as given below :
| Class-interval | Tally marks | Frequency |
|---|---|---|
| 65 – 75 | lll | 3 |
| 75 – 85 | llll ll | 7 |
| 85 – 95 | llll llll | 9 |
| 95 – 105 | llll | 5 |
| 105 – 115 | llll | 5 |
| 115 – 125 | l | 1 |
The heights of 45 students measured to the nearest centimetre are given below:
155, 166, 159, 161, 158, 154, 153, 159, 157, 152, 160, 157, 158, 170, 152, 151, 162, 164, 171, 154, 165, 161, 150, 168, 162, 154, 160, 170, 163, 164, 165, 169, 165, 155, 157, 158, 160, 161, 156, 153, 150, 165, 167, 169, 172
Prepare a frequency table taking class intervals 150 – 155, 155 – 160, 160 – 165, 165 – 170, 170 – 175.
Arranging the given heights in ascending order, we get :
150, 150, 151, 152, 152, 153, 153, 154, 154, 154, 155, 155, 156, 157, 157, 157, 158, 158, 158, 159, 159, 160, 160, 160, 161, 161, 161, 162, 162, 163, 164, 164, 165, 165, 165, 165, 166, 167, 168, 169, 169, 170, 170, 171, 172
The class-intervals are taken in the exclusive form. So, the class-interval 150 – 155 means 150 and more but less than 155, the class-interval 155 – 160 means 155 and more but less than 160, and so on.
The required frequency table is as given below :
| Class-interval | Tally marks | Frequency |
|---|---|---|
| 150 – 155 | llll llll | 10 |
| 155 – 160 | llll llll l | 11 |
| 160 – 165 | llll llll l | 11 |
| 165 – 170 | llll llll | 9 |
| 170 – 175 | llll | 4 |
Complete the table given below:
| Class-interval | Class-limits (Lower) | Class-limits (Upper) | Class-size | Class-mark | Class boundaries (Lower) | Class boundaries (Upper) |
|---|---|---|---|---|---|---|
| 25 – 30 | ||||||
| 30 – 35 | ||||||
| 35 – 40 | ||||||
| 40 – 45 |
The given class-intervals are in the exclusive form.
Each class is bounded by two figures, called class limits. The figure on the left side is the lower limit and the figure on the right side is the upper limit of the class.
By formula,
Class-size = Upper limit − Lower limit
Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$
In an exclusive form, the class boundaries (or true limits) are the same as the class limits.
For the class-interval 25 – 30 :
Lower limit = 25 and Upper limit = 30
Class-size = 30 − 25 = 5
$$\begin{aligned}\text{Class mark} = \frac{30 + 25}{2}\\[1em] = \frac{55}{2}\\[1em] = 27.5\end{aligned}$$
Class boundaries are 25 and 30.
Proceeding in the same manner for the remaining class-intervals, the completed table is as given below :
| Class-interval | Class-limits (Lower) | Class-limits (Upper) | Class-size | Class-mark | Class boundaries (Lower) | Class boundaries (Upper) |
|---|---|---|---|---|---|---|
| 25 – 30 | 25 | 30 | 5 | 27.5 | 25 | 30 |
| 30 – 35 | 30 | 35 | 5 | 32.5 | 30 | 35 |
| 35 – 40 | 35 | 40 | 5 | 37.5 | 35 | 40 |
| 40 – 45 | 40 | 45 | 5 | 42.5 | 40 | 45 |
Complete the table given below:
| Class-interval | Class-limits (Lower) | Class-limits (Upper) | Class-size | Class-mark | Class boundaries (Lower) | Class boundaries (Upper) |
|---|---|---|---|---|---|---|
| 26 – 30 | ||||||
| 31 – 35 | ||||||
| 36 – 40 | ||||||
| 41 – 45 |
The given class-intervals are in the inclusive form.
Each class is bounded by two figures, called class limits. The figure on the left side is the lower limit and the figure on the right side is the upper limit of the class.
In an inclusive form, the class boundaries (or true limits) are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
By formula,
Class-size = True upper limit − True lower limit
Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$
For the class-interval 26 – 30 :
Lower limit = 26 and Upper limit = 30
Class boundaries are (26 − 0.5) = 25.5 and (30 + 0.5) = 30.5
Class-size = 30.5 − 25.5 = 5
$$\begin{aligned}\text{Class mark} = \frac{30 + 26}{2}\\[1em] = \frac{56}{2}\\[1em] = 28\end{aligned}$$
Proceeding in the same manner for the remaining class-intervals, the completed table is as given below :
| Class-interval | Class-limits (Lower) | Class-limits (Upper) | Class-size | Class-mark | Class boundaries (Lower) | Class boundaries (Upper) |
|---|---|---|---|---|---|---|
| 26 – 30 | 26 | 30 | 5 | 28 | 25.5 | 30.5 |
| 31 – 35 | 31 | 35 | 5 | 33 | 30.5 | 35.5 |
| 36 – 40 | 36 | 40 | 5 | 38 | 35.5 | 40.5 |
| 41 – 45 | 41 | 45 | 5 | 43 | 40.5 | 45.5 |
Assertion (A): The class size of the class interval 40 – 50 is 10.
Reason (R): The difference between the upper and lower limit of a class interval is called the class-size.
The difference between the upper limit and the lower limit of a class interval is called the class-size.
So, reason (R) is true.
For the class interval 40 – 50,
Lower limit = 40 and Upper limit = 50
Class-size = Upper limit − Lower limit
= 50 − 40
= 10
So, assertion (A) is true and reason (R) is the correct explanation of assertion (A).
∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Hence, option 1 is the correct option.
Assertion (A): The class mark of the class interval 100 – 200 is 200.
Reason (R): Class mark = $\frac{\text{upper limit} + \text{lower limit}}{2}$
The class mark (or mid-value) of a class interval is given by :
Class mark = $\frac{\text{upper limit} + \text{lower limit}}{2}$
So, reason (R) is true.
For the class interval 100 – 200,
Lower limit = 100 and Upper limit = 200
$$\begin{aligned}\text{Class mark} = \frac{200 + 100}{2}\\[1em] = \frac{300}{2}\\[1em] = 150\end{aligned}$$
Since the class mark of the class interval 100 – 200 is 150 and not 200, assertion (A) is false.
∴ Assertion (A) is false but Reason (R) is true.
Hence, option 4 is the correct option.
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