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MATHS [5 marks sum]

Data Handling · ICSE STD 7 (ICSE - English Medium). 8 questions.

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Question 15 Marks
Find the mean, the median, and the mode of 21, 24, 21, 6, 15, 18, 21, 45, 9, 6, 27 and 15.
Answer

$ \begin{aligned} & 21,24,21,6,15,18,21,45,9,6,27 \text { and } 15 \\ & \therefore \text { Mean }=\frac{21+24+21+6+15+18+21+45+9+6+27+15}{12} \\ & =\frac{228}{12}=19 \end{aligned} $
Numbers are $21,24,21,6,15,18,21,45,9,6,27$ and 15
Mostly repeated term $=21$
$\therefore$ Mode $=21$
Now, Arranging the numbers in ascending order $=6,6,9,15,15,18,21,21,21,24,27,45$
Here, number of terms $=12$ which is even
$ \begin{aligned} & \therefore \text { Median }=\frac{1}{2}\left\{\frac{\mathrm{n}}{2} \text { th term }+\left(\frac{\mathrm{n}}{2}+1\right) \text { th term }\right\} \\ & =\frac{1}{2}\left\{\frac{12}{2} \text { th term }+\left(\frac{12}{2}+1\right) \text { th term }\right\} \\ & =\frac{1}{2}\{6 \text { th term }+7 \text { th term }\} \\ & =\frac{1}{2}\{18+21\} \\ & =\frac{1}{2} \times 39 \\ & =19.5 \end{aligned} $
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Question 25 Marks
Find the mean, the median, and the mode of 12, 24, 24, 12, 30 and 12.
Answer
$12,24,24,12,30$ and 12
$ \therefore \text { Mean }=\frac{12+24+24+12+30+12}{6} $
$ =\frac{114}{6}=19 $
Numbers are $12,24,24,12,30$ and 12
Mostly repeated term $=12$
$\therefore$ Mode $=12$
Now, Arranging the numbers in ascending order $=12,12,12,24,24,30$
Here, number of terms 6 which is even
$\therefore$ Median $=\frac{1}{2}\left\{\frac{\mathrm{n}}{2}\right.$ th term $+\left(\frac{\mathrm{n}}{2}+1\right)$ th term $\}$
$=\frac{1}{2}\left\{\frac{6}{2}\right.$ th term $+\left(\frac{6}{2}+1\right)$ th term $\}$
$=\frac{1}{2}\{3$ th term +4 th term $\}$
$=\frac{1}{2}\{12+24\}$
$=\frac{1}{2} \times 36$
$ =18 $
$\therefore$ Median $=18$
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Question 35 Marks
The heights (in cm) of 8 girls of a class are 140, 142, 135, 133, 137, 150, 148 and 138 respectively. Find the mean height of these girls and their median height.
Answer
Arranging in ascending order: $133,135,137,138,140,142,148,150$
Here, the number of girls $=8$ which is even
$ \begin{aligned} & \therefore \text { Median }=\frac{1}{2}\left\{\frac{\mathrm{n}}{2} \text { th term }+\left(\frac{\mathrm{n}}{2}+1\right) \text { th term }\right\} \\ & =\frac{1}{2}\left\{\frac{8}{2} \text { th term }+\left(\frac{8}{2}+1\right) \text { th term }\right\} \\ & =12\{4 \text { th term }+5 \text { th term }\} \\ & =\frac{1}{2}\{138+140\} \mathrm{cm} \\ & =\frac{1}{2} \times 278 \\ & =139 \mathrm{~cm} \\ & \therefore \text { Mean }=\frac{133+135+137+138+140+142+148+150}{8} \\ & =\frac{1123}{8} \mathrm{~cm} \\ & =140.375 \mathrm{~cm} \end{aligned} $
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Question 45 Marks
Find the mean and the mode for the following data:
Term $18$ $22$ $23$ $30$ $34$ $38$
Frequency $3$ $5$ $10$ $2$ $8$ $2$
Answer
We prepare the table given below:
Term
($x_i$)		 Frequency
($f_i$)
($f_ix_i$)
$18$ $3$ $54$
$22$ $5$ $110$
$26$ $10$ $260$
$30$ $2$ $60$
$34$ $8$ $276$
$38$ $2$ $76$
Total $30$ $832$

$\text { Mean }=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{f}_{\mathrm{i}}}=\frac{832}{30}=27.73$
Since the frequency of Number 26 is maximum.
$\therefore \text { Mode }=26$
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Question 55 Marks
The mean of six numbers: x − 5, x − 1, x, x + 2, x + 4 and x + 12 is 15. Find the mean of first four numbers.
Answer
The mean of six numbers are $x-5, x-1, x, x+2, x+4$ and $x+12$ is 15
$ \begin{aligned} & \therefore \text { Mean }=\frac{(\mathrm{x}-5)+(\mathrm{x}-1)+(\mathrm{x})+(\mathrm{x}+2)+(\mathrm{x}+4)+(\mathrm{x}+12)}{6} \\ & =\frac{\mathrm{x}-5+\mathrm{x}-1+\mathrm{x}+\mathrm{x}+2+\mathrm{x}+4+\mathrm{x}+12}{6} \\ & =\frac{12+6 \mathrm{x}}{6}=15 \\ & \Rightarrow 12+6 \mathrm{x}=90 \\ & \Rightarrow 6 x=90-12 \\ & \Rightarrow 6 x=78 \\ & \Rightarrow x=\frac{78}{6}=13 \\ & x=13 \end{aligned} $
$\therefore$ The six numbers are $(13-5),(13-1), 13,(13+2),(13+4),(13+12)$ i.e. $8,12,13,15,17,25$
Now, mean of first four numbers $=\frac{8+12+13+15}{4}=\frac{48}{4}=12$
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Question 65 Marks
The marks obtained by 40 students of a class are given below:
80, 10, 30, 70, 60, 50, 50, 40, 40, 20, 40, 90, 50, 30, 70, 10, 60, 50, 20, 70, 70, 30, 80, 40,20, 80, 90, 50, 80, 60, 70, 40, 50, 60, 90, 60, 40, 40, 60 and 60
(i) Construct a frequency distribution table.
(ii) Find how many students have marks equal to or more than 70?
(iii) How many students obtained marks below 40?
Answer
(i) The frequency distribution table will be shown as below:
Numbers Tally marks Frequency
10 || 2
20 ||| 3
30 ||| 3
40 |||| || 7
50 |||| | 6
60 |||| || 7
70 |||| 5
80 |||| 4
90 ||| 3
Total 40
(ii) Students have marks equal to or more than 70 = 5 + 4 + 3 = 12
(iii) Students obtained marks below 40 = 2 + 3 + 3 = 8 students
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Question 75 Marks
Following data shows the weekly wages (in ₹) of 10 workers in a factory.
3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250 and 4000
(i) Prepare a frequency distribution table.
(ii) What is the range of wages (in ₹)?
(iii) How many workers are getting maximum wages?
Answer
(i) The frequency table for the wages of 10 workers will be as shown below:
Weekly wages in (₹) Tally marks Frequency
3500 | 1
3750 | 1
4000 |||| 4
4250 ||| 3
4750 | 1
Total 10
(ii) Range of wages (₹) = ₹4750 – ₹3500 = ₹1250
(iii) One
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Question 85 Marks
A die was thrown 20 times and the following scores were recorded.
2, 1, 5, 2, 4, 3, 6, 1, 4, 2, 5, 1, 6, 2, 6, 3, 5, 4, 1 and 3.
Prepare a frequency table for the scores.
Answer
The frequency table for the scores will be as shown below:
No. of thrown dies Tally marks Frequency
1 |||| 4
2 |||| 4
3 ||| 3
4 ||| 3
5 ||| 3
6 ||| 3
Total 20
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[5 marks sum] - MATHS STD 7 Questions - Vidyadip