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Maths 3 Mark Question

Data Handling II (Central Values) · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 33 questions.

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Question 13 Marks
Find the median of the following data:
133, 73, 89, 108, 94,104, 94, 85, 100, 120
Answer
Arranging the data in ascending order, we have:
73, 85, 89, 94, 100, 104, 108, 120, 133
Here, the number of observations, n = 10 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{94+100}{2}=97$
Hence, the median of the given data is 49.5
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Question 23 Marks
Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?
Answer
Arranging the given data in ascending order, we have:
$41,43,57,58,61,71,92,99,127$
Here, the number of observations n is 9 (odd).
Median $=$ value of $\frac{ n +1}{2}$ th observation $=$ value of the $5^{\text {th }}$ observation $=61$
Hence, the median $=61$
If 58 is replaced by 85 , and then the new observations arranged in ascending order are:
$41,43,57,61,71,85,92,99,12$
$\therefore$ New median $=$ value of the $5^{\text {th }}$ observation $=71$
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Question 33 Marks
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer
Arranging the given data in ascending order, we have:
41, 43, 57, 58, 61, 71, 92, 99, 127
Here, the number of observations n is 10. Since n is even,
$\Rightarrow \text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of}\ 5^\text{th}\ \text{observation}+\text{Value of}\ 6^\text{th}\ \text{observation}}{2}$
$\Rightarrow63=\frac{\text{x}+(\text{x+2})}{2}$
$\Rightarrow63=\frac{2\text{x}+2}{2}$
$\Rightarrow63=\frac{2(\text{x}+1)}{2}$
$\Rightarrow63 = \text{x} + 1$
$\Rightarrow\text{x} = 63 - 1$
$\Rightarrow\text{x} = 62$
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Question 43 Marks
The mean weight of 8 numbers is 15kg. If each number is multiplied by 2, what will be the new mean?
Answer
Let $x _1, x _2, x _3 \ldots x _8$ be the eight numbers whose mean is 15 kg . Then,
$15=\frac{x_1+x_2+x_3+\ldots+x_8}{8}$
$x_1+x_2+x_3+\ldots+x_8=15 \times 8$
$x_1+x_2+x_3+\ldots+x_8=120$
Let the new numbers be $2 x_1, 2 x_2, 2 x_3 \ldots 2 x_8$. Let $M$ be the arithmetic mean of the new numbers.
Then,
$M=\frac{2 x_1+2 x_2+2 x_3+\ldots+2 x_8}{8}$
$\Rightarrow M=\frac{2\left(x_1+x_2+x_3+\ldots+x_8\right)}{8}$
$\Rightarrow M=\frac{2 \times 120}{8}$
$\Rightarrow M=30$
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Question 53 Marks
Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Answer
Arranging the given data in ascending order, we have:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here, the number of observations n is 11 (odd).
Since the number of observations is odd, therefore.
Median = value of $\frac{\text{n}+1}{2}\text{th}$ observation = value of the $6^{th}$ observation = 58
Hence, median = 58
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
$\therefore$ New median = value of the $6^{th}$ observation = 58
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Question 63 Marks
Find the median of the following data:
31, 38, 27, 28, 36, 25, 35, 40
Answer
Arranging the data in ascending order, we have:
25, 27, 28, 31, 35, 36, 38, 40
Here, the number of observations, n = 8 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 4}^\text{th}\ \text{observation}+\text{Value of 5}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{31+35}{2}=33$
Hence, the median of the given data is 33
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Question 73 Marks
The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the mean number of children per family.
Answer
We have,
The mean number of children per family $=\frac{\text{Sum of the total number of children}}{\text{Total Number of families}}$
$\text{Mean}= \frac{2+4+3+4+2+3+5+1+1+5}{10}$
$=\frac{30}{10}$
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Question 83 Marks
Calculate the mean for the following distribution:
x:
 5
6
7
8
9
y:
4
8
14
11
3
Answer
Calculation of mean:
$x _{ i }$ $f _{ i }$ $x _{ i } f _{ i }$
5
4
20
6
 8
48
7 14
98
8
11
88
9 3
27
Total
$\sum\text{f}_\text{i}=40$
$\sum\text{f}_\text{i}\text{x}_\text{i}=281$
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{281}{40}=7.025.$
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Question 93 Marks
The following table shows the weights (in kg) of 15 workers in a factory:
Weight (in kg):
60
63
66
72
75
Numbers of workers:
4
5
3
1
2
Calculate the mean weight.
Answer
Calculation of mean:
$x _{ i }$ $f _{ i }$ $x _{ i } f _{ i }$
60
4
240
63
5
315
66
3
198
72
1
72
75
2
150
Total
$\sum\text{f}_\text{i}=15$
$\sum\text{f}_\text{i}\text{x}_\text{i}=975$
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{975}{15}=65\text{kg.}$
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Question 103 Marks
Calculate the mean and median for the following data:
Marks: 10 11 12 13 14 16 19 20
Number of Students: 3 5 4 5 2 3 2 1
Using empirical formula, find its mode.
Answer
Calculation of Mean:

$\text { mean }=\frac{\sum f_{f} x_1}{\sum f_1} \frac{332}{25}=13.28$

Here, $n =25$, witch is an odd number. Therefore,

Median = value of $\frac{ n + 1 }{2}$ th observation = value of the $13^{\text {th }}$ observation $=13$

Now,

$\Rightarrow \text { Mode }=3 \text { Median }-2 \text { Mean }$

$\Rightarrow \text { Mode }=3(13)-2(13.28)$

$\Rightarrow \text { Mode }=39-26.56$

$\Rightarrow \text { Mode }=12.44$
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Question 113 Marks
Find the mean of the following data:
x:
 19
21
23
25
 27 29
31
f:
 13
15
16
18
 16 15
13
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
19
13
247
21
15
315
23
16
368
25
18
450
27
16
432
29
15
435
31
13
403
Total
$\sum\text{f}_\text{i}=\text{N}=106$
$\sum\text{f}_\text{i}\text{x}_\text{i}=2650$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2650}{106}=25.$
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Question 123 Marks
Find the median of the following data:
$15,6,16,8,22,21,9,18,25$
Answer
Arranging the data in ascending order, we have:
$6,8,9,15,16,18,21,22,25$
Here, the number of observations, $n =9$ (Odd).
$\Rightarrow$ Median $=$ value of $\frac{ n +1}{2}$ th observation i.e., value of $5^{\text {th }}$ observation $=16$
Hence, the median of the given data is 16
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Question 133 Marks
The mean weight per student in a group of 7 students is 55kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Answer
We have,
$\text{Mean}=\frac{\text{Sum of the weights of the students}}{\text{Number of students}}$
Let the weight of the seventh student be x kg.
$\text{Mean}=\frac{52+54+55+53+56+54+\text{x}}{7}$
$55=\frac{52+54+55+53+56+54+\text{x}}{7}$
⇒ 385 = 324 + x
⇒ x = 385 - 324
⇒ x = 61kg.
Thus, the weight of the seventh student is 61kg.
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Question 143 Marks
Find the mode and median of the data: $13,16,12,14,19,12,14,13,14$
By using the empirical relation also find the mean.
Answer
Arranging the data in ascending order such that same numbers are put together, we get:
$12,12,13,13,14,14,14,16,19$
Here, $n =9$
Median $=$ value of $\frac{ n +1}{2}$ th observation $=$ value of the $5^{\text {th }}$ observation $=14$
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.
Now,
$\text { Mode }=3 \text { Median }-2 \text { Mean }$
$\Rightarrow 14=3 \times 14-2 \text { Mean }$
$\Rightarrow 2 \text { Mean }=42-14=28$
$\Rightarrow \text { Mean }=28 \div 2=14$
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Question 153 Marks
The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Answer
We have,$\text{Mean}=\frac{\text{Sum of the marks obtained}}{\text{Total Number of students}}$
$\Rightarrow\text{Mean}= \frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}$
$\Rightarrow\text{Mean}=\frac{474}{12}=39.5\%$
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Question 163 Marks
The weights (in kg ) of 15 students are: $31,35,27,29,32,43,37,41,34,28,36,44,45,42$, and 30 . Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg , find the new median.
Answer
Arranging the given data in ascending order, we have:
$27,28,29,30,31,32,34,35,36,37,41,42,43,44,45$
Here, the number of observations n is 15 (odd).
Since the number of observations is odd, therefore.
Median $=$ value of $\frac{ n +1}{2}$ th observation $=$ value of the $8^{\text {th }}$ observation $=35$
Hence, median $=35 kg$.
If 44 is replaced by 46 and 27 kg by 25 kg , then the new observations arranged in ascending order are:
$25,28,29,30,31,32,34,35,36,37,41,42,43,45,46$
$\therefore$ New median $=$ value of the $8^{\text {th }}$ observation $=35 kg$.
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Question 173 Marks
The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.
Answer
Let $x _1, x _2, x _3 \ldots x _{75}$ be 75 numbers with their mean equal to 35 . Then,
$\Rightarrow 35=\frac{x_1+x_2+x_3+\ldots+x_{75}}{75}$
$x_1+x_2+x_3+\ldots+x_{75}=35 \times 75$
$\Rightarrow x_1+x_2+x_3+\ldots+x_{75}=2625$
The new numbers are $4 \times 1,4 \times 2,4 \times 3 \ldots 4 \times 75$
Let $M$ be the arithmetic mean of the new numbers. Then,
$M=\frac{4 x_1+4 x_2+4 x_3+\ldots+4 x_{75}}{75}$
$\Rightarrow M=\frac{4\left(x_1+x_2+x_3+\ldots+x_{75}\right)}{75}$
$\Rightarrow M=\frac{4 \times 2625}{75}$
$\Rightarrow M=140$
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Question 183 Marks
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Answer
Let $x_1, x_2, x_3, x_4$ and $x_5$ be five numbers whose mean is 18 . Then,
$18=$ Sum of five numbers $\div 5$
$\therefore$ Sum of five numbers $=18 \times 5=90$
Now, if one number is excluded, then their mean is 16.
So,
$16=$ Sum of four numbers $\div 4$
$\therefore$ Sum of four numbers $=16 \times 4=64$
The excluded number = Sum of five observations - Sum of four observations.
$\therefore$ The excluded number $=90-64$
$\therefore$ The excluded number $=26$
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Question 193 Marks
The runs scored in a cricket match by 11 players are as follows:
$6,15,120,50,100,80,10,15,8,10,10$
Find the mean, mode and median of this data.
Answer
Arranging the data in ascending order such that same values are put together, we get:
$6,8,10,10,15,15,50,80,100,120$
Here, $n =11$
Median $=$ value of $\frac{ n +1}{2}$ th observation $=$ value of the $6^{\text {th }}$ observation $=15$
Here, 10 occur three times. Therefore, 10 is the mode of the given data.
$\text { Now, }$
$\text { Mode }=3 \text { Median }-2 \text { Mean. }$
$\Rightarrow 10=3 \times 15-2 \text { Mean. }$
$\Rightarrow 2 \text { Mean }=45-10=35$
$\Rightarrow \text { Mean }=35 \div 2=17.5$
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Question 203 Marks
The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.
Answer
We have,
$\text{Mean}=\frac{\text{Sun of the five numbers}}{5}=27$
So, sum of the five numbers = 5 x 27 = 135
Now,
The mean of the four numbers $=\frac{\text{Sun of the four numbers}}{4}=25$
So, sum of the four numbers = 4 x 25 = 100
Therefore, the excluded number = Sum of the five number - Sum of the four numbers.
The excluded number = 135 - 100 = 35
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Question 213 Marks
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
Arranging the data in ascending order such that same values are put together, we get:
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Here, n = 15
Median = value of $\frac{\text{n}+1}{2}\text{th}$ observation = value of the $8^{th}$ observation = 20
Here, clearly, 20 occurs most number of times, i.e., 4 times. Therefore, the mode of the given data is 20
Yes, the median and mode of the given data are the same.
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Question 223 Marks
The daily wages (in Rs) of 15 workers in a factory are given below:
200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180
Prepare the frequency table and find the mean wage.
Answer
The frequency table for the given data is as follows:
Wages ($x_i$): 130 150 180 200
Number of workers ($f_i$): 2 4 6 3
In order to compute the mean wage, we prepare the following table:
Mean wages of the workers:
$x_i$
$f_i$
$x_if_i$
130
2
260
150
4
600
180
6
1080
200
3
600
Total
$\sum\text{f}_\text{i}=\text{N}=15$
$\sum\text{f}_\text{i}\text{x}_\text{i}=2540$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2540}{15}=169.33$
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Question 233 Marks
Find the median of the following data:
12, 17, 3, 14, 5, 8, 7, 15
Answer
Arranging the data in ascending order, we have:
3, 5, 7, 8, 12, 14, 15, 17
Here, the number of observations, n = 8 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 4}^\text{th}\ \text{observation}+\text{Value of 5}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{8+12}{2}=10$
Hence, the median of the given data is 10
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Question 243 Marks
Find the median of the following data:
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Answer
Arranging the data in ascending order, we have:
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here, the number of observations, n = 10 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{45+54}{2}=49.5$
Hence, the median of the given data is 49.5
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Question 253 Marks
The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.
Answer
Mean = Sum of five numbers ÷ 5
⇒ Sum of the five numbers = 27 × 5 = 135
Now, New mean = 25
25 = Sum of six numbers ÷ 6
⇒ Sum of the six numbers = 25 × 6 = 150
The included number = Sum of the six numbers – Sum of the five numbers.
⇒ The included number = 150 - 135
⇒ The included number = 15
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Question 263 Marks
Following are the weights (in kg) of 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean $\bar{\text{x}}.$
Answer
We have,
$\overline{\text{X}}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow \overline{\text{X}} = \frac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}$
$\Rightarrow\overline{\text{X}}=\frac{40}{10}$
$\Rightarrow\overline{\text{X}}=4\text{kg.}$
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Question 273 Marks
Find the median of the following data:
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Answer
Arranging the data in ascending order, we have:
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Here, the number of observations, n = 10 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{26+29}{2}=27.5$
Hence, the median of the given data is 27.5
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Question 283 Marks
Find the median of the following data:
$41,43,127,99,71,92,71,58,57$
Answer
Arranging the data in ascending order, we have:
$41,43,57,58,71,71,92,99,127$
Here, the number of observations, $n =9$ (Odd).
$\therefore$ Median $=$ value of $\frac{9+1}{2}$ th observation i.e., value of $5^{\text {th }}$ observation $=71$
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Question 293 Marks
Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Answer
Arranging the data in ascending order such that same values are put together, we get:
10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18
Here, clearly, 14 occurs the most number ot times.
Therefore, 14 is the mode of the given data.
Alternate answer:
Arrenging the data in the form of a frequency table, we get:

Clearly, 14 has maximum frequency. So, the mode of the given data is 14
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Question 303 Marks
A die was thrown 20 times and the following scores were recorded:
5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1
Prepare the frequency table of the scores on the upper face of the die and find the mean score.
Answer
The frequency table for the given data is as follows:
x:
1
2
3
4
5
6
f:
2
5
1
4
6
2
In order to compute the arithmetic mean, we prepare the following table:
Computation of Arithmetic Mean:
Scores($x_i$)
Frequency($f_i$)
 $x_if_i$
1
2
2
2
5
10
3
1
3
4
4
16
5
6
30
6
2
12
Total
$\sum\text{f}_\text{i}=20$
 $\sum\text{f}_\text{i}\text{x}_\text{i}=73$
We have, $\sum\text{f}_\text{i}=20\ \text{and}\ \sum\text{f}_\text{i}\text{x}_\text{i}=73$
$\therefore\text{Mean score} = \frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}\frac{73}{20}=3.65$
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Question 313 Marks
Find the median of the following data:
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Answer
Arranging the data in ascending order, we have:
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Here, the number of observations, n = 10 (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{67+69}{2}=68$
Hence, the median of the given data is 68
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Question 323 Marks
The ages (in years) of 50 students of a class in a school are given below:
Age (in years):
14
15
16
17
18
Numbers of students:
15
14
10
8
3
Find the mean age.
Answer
Calculation of mean:
$x_i$
$f_i$
 $x_if_i$
14 15 210
15
14
210
16
10
160
17
8
136
18
3
54
Total
$\sum\text{f}_\text{i}=50$
$\sum\text{f}_\text{i}\text{x}_\text{i}=770$
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{770}{50}=15.4\text{ years}$
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Question 333 Marks
Numbers $50,42,35,2 x+10,2 x-8,12,11,8,6$ are written in descending order and their median is 25 , find $x$.
Answer
Here, the number of observations $n$ is 9 . Since $n$ is odd, the median is the $\frac{ n +1}{2}$ th observation, i.e., the $5^{\text {th }}$ observation. As the numbers are arranged in the descending order, we therefore observe from the last.
Median $=5^{\text {th }}$ observations.
$\Rightarrow 25=2 x-8$
$\Rightarrow 2 x=25+8$
$\Rightarrow 2 x=33$
$\Rightarrow x=\frac{33}{2}$
$\Rightarrow x=16.5$
Hence, $x=16.5$
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