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MATHS 1 Marks Question

Stastics · UP Board (UPMSP) STD 10 (UPMSP - English Medium). 30 questions.

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Question 11 Mark
Write the median class of the following distribution:
Classes Frequency
0 - 10 4
10 - 20 4
20 - 30 8
30 - 40 10
40 - 50 12
50 - 60 8
60 - 70 4
Answer
Classes
f
c.f.
0 - 10
4
4
10 - 20
4
8
20 - 30
8
16
30 - 40
10
26
40 - 50
12
38
50 - 60
8
46
60 - 70
4
50
Here n = 50, $\therefore\ \frac{\text{n}}{2}=25$
Now, 30 - 40 is the classes whose c. f 26 is greater than (and nearest to) $\frac{\text{n}}{2}$ i.e., 25
Hence, 30 - 40 is the median class.
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Question 21 Mark
Find the median class of the following data:
Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
Frequency 8 10 12 22 30 18
Answer
30 – 40.
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Question 31 Mark
Which measure of central tendency is given by the x-coordinate of the point of intersection of the “more than ogive” and “less than ogive”?
Answer
Median.
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Question 41 Mark
Two right circular cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1, what is the ratio of their volumes?
Answer
$\text{V}_1=\text{V}_2=\frac{1}{3}\pi(3\text{r})^2\text{h}:\frac{1}{3}\pi\text{r}^2(3\text{h})$
$=3:1$
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Question 51 Mark
Using the empirical formula, find the mode of a distribution whose mean is 8.32 and the median is 8.05.
Answer
3 Median = 2Mean + Mode
Mode = 3(8.05) - 2(8.32)
Mode = 24.15 - 16.64
Mode = 7.51
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Question 61 Mark
Find the class marks of the classes 15–35 and 45–60.
Answer
$\frac{15 +35}{2}= 25 $
$\frac{45+ 60}{2}= 52.5 $
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Question 71 Mark
Find the class-marks of the classes 10-25 and 35-55.
Answer
Class mark = $\frac{\text{Upper limit+Lower limits}}{2}$
$\therefore$ Class mark of $10-25 =\frac{10+25}{2} = 17.5$
And class mark of $35-55 = \frac{35+55}{2} = 45$
class mark = 17.5, 45
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Question 81 Mark
If the mean of first n natural number is 15, then find n.
Answer
The mean of the first n natural number = 15
Mean $=\frac{1+2+3+4....+\text{n}}{\text{n}}$
$\Rightarrow\frac{\big(\frac{\text{n}(\text{n}+1)}{2}\big)}{\text{n}}=15$
$\Rightarrow\frac{\text{n}+1}{2}=15$
$\Rightarrow\text{n}=29.$
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Question 91 Mark
Consider the following frequency distribution:
Class
0-5
6-11
12-17
18-23
24-29
Frequency
13
10
15
8
11
Find the upper limit of the median class.
Answer
The given series is in indusive from. Converting it to exclusive from and preparing cumulative frequency table, we get
Class
Frequency
Cumulative frequency
0.5-5.5
13
13
5.5-11.5
10
23
11.5-17.5
15
38
17.5-23.5
8
46
23.5-29.5
11
57
Here, $\text{N}=57\Rightarrow\frac{\text{N}}{2}=28.5$
The cumulative frequency just greater than 28.5 is 38.
Hence, the median class is 11.5-17.5
$\therefore$ Upper limit of the median class = 17.5
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Question 101 Mark
Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?
Answer
Median is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive.
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MCQ 111 Mark
The percentage of marks obtained by $100$ students in an examination are as follows :
Mark
 $130-135$ $135-140$ $140-145$ $145-150$ $150-155$ $155-160$ $160-165$
Frequency
 $10$ $15$ $18$ $22$ $23$ $8$ $4$
The cumulative frequency of the class interval $150 - 155$ is :
  • A
    $100$
  • B
    $80$
  • C
    $90$
  • ✓
    $88$
Answer
Correct option: D.
$88$
Mark $130-135$ $135-140$ $140-145$ $145-150$ $150-155$ $155-160$ $160-165$
Frequency $10$ $15$ $18$ $22$ $23$ $8$ $4$
Cumulative Frequency $10$ $25$ $43$ $65$ $88$ $96$ $100$
Therefore, the cumulative frequency of the class interval of $150-155 $ is $88.$
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Question 121 Mark
If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.
Answer
Mean = 53.4
Mode = 55.2
Mode = 3Median - Mean
Hence, median = $\frac{\text{Mode}+2\text{Mean}}{3}$
$=\frac{55.2+2(53.4)}{3}$
$=\frac{55.2+106.8}{3}$
$=\frac{162}{3}$
$=54$
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MCQ 131 Mark
An ogive is used to determine :
  • ✓
    Median
  • B
    None of these
  • C
    Mode
  • D
    Mean
Answer
Correct option: A.
Median
An ogive is used to determine how many data values lie above or below a particular value in a data set.
In other words, it is used to determine the Median of a grouped data.
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Question 141 Mark
Which measure of central tendency can be determine graphically?
Answer
Median can be determined graphically.
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Question 151 Mark
Find the mode of the following data:
3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Answer
x
Tally marks
(f)
3
4
5
6
7
8
9
IIII
II
IIII
II
II
I
II
5
2
4
2
2
1
2
we see that 3 occurs in maximum times i.e. 5
$\therefore$ Mode = 3
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Question 161 Mark
Write the modal class for the following frequency distribution:
Class
10-15
15-20
20-25
25-30
30-35
35-40
Frequency
30
35
75
40
30
15
Answer
The modal class is 20-25 as it has the maximum frequency of 75 in the given distribution.
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Question 171 Mark

What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?
Answer
Median = 4, because the coordinates of the point of intersection of two ogives at x-axis is 4.
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Question 181 Mark
Write the empirical relation between mean, mode and median.
Answer
The empirical relation is Mode = 3
Mode = 3 Median - 2 mean
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Question 191 Mark
Define Mean.
Answer
The mean of a set of observations is equal to their sum divided by the total number of observations. Mean is also called an average.
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Question 201 Mark
The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.
Answer
The median of an ungrouped data and the median calculated when the same data is grouped are not always the same because the median for ungrouped data is calculated by arranging the data in increasing or decreasing order. But for calculating the median of a grouped data, the formula used is based on the assumption that the observations are uniformly distributed in the classes.
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Question 211 Mark
What is the algebraic sum of deviations of a frequency distribution about its mean?
Answer
The algebraic sum of deviation of a frequency distribution about its mean is zero.
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Question 221 Mark
The annual profit earned by 30 shops of a shopping complex in a locality are recorded in the table shown below:
Profit (in lakhs Rs.)
Number of shops
More than or equal to 5
30
More than or equal to 10
28
More than or equal to 15
16
More than or equal to 20
14
More than or equal to 25
10
More than or equal to 30
7
More than or equal to 35
3
If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25.
Answer
The frequency table is as follows:
Classes Profit (in lakhs Rs.)
frequency Number of shops
5-10
2
10-15
12
15-20
2
20-25
4
25-30
3
30-35
4
35-40
4
Thus, the frequency corresponding to the class 20-25 is 4.
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Question 231 Mark
Find the mode of the following data:
15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Answer
x
Tally mark
(f)
8
15
18
19
20
24
25
26
I
IIII
I
I
I
II
I
I
1
4
1
1
1
2
1
1
Here we see that 15 occurs in maximum times i.e. 54
$\therefore$ Mode = 15
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Question 241 Mark
Find the mode of the following data:
3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Answer
x
Tally mark
(f)
3
4
5
6
7
8
9
III
II
IIII
II
II
I
II
4
2
5
2
2
1
2
We see that 5 occurs in maximum times which is 5
$\therefore$ Mode = 5
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Question 251 Mark
In the table given below, the times taken by 120 athletes to run a 100 m hurdle race are given:
Class
13.8-14
14-14.2
14.2-14.4
14.4-14.6
14.6-14.8
14.8-15
Frequency
2
4
15
54
25
20
Find the number of athletes who completed the race in less than 14.6 seconds.
Answer
Number of athletes who completed the race in less than 14.6 seconds
= 2 + 4 + 15 + 54
= 75
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Question 261 Mark
Will the median class and modal class of grouped data always be different? Justify your answer.
Answer
The median and modal class may be same if modal class is median class which is not always possible as the number of frequencies may be maximum in any class.
So given statement is not true.
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1 Marks Question - MATHS STD 10 Questions - Vidyadip