M.C.Q

Question 11 Mark

The class mark of the class interval 2.4-6.6 is:

Answer

Solution:
Class mark $=\frac{\text{uppar limit+lower limit}}{2}$
$=\frac{2.4+6.6}{2}$
$=\frac{9}{2}$

Question 21 Mark

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The correct mean is:

Answer

39.4

Solution:
Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = (39 × 50) = 1950
Correct sum = (1950 + 43 - 23) = 1970
$\therefore$ Mean = 19750 = 39.4

View full question & answer→

Question 31 Mark

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11 then the value of x is:

Answer

Solution:
Mean of 5 observations = 11
We know:
Mean $=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$\Rightarrow 11=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$
$\Rightarrow11=\frac{\text{5x+20}}{5}$
⇒ 5x + 20 = 55
⇒ 5x = 35
⇒ x = 7

View full question & answer→

Question 41 Mark

Given the class intervals 1-10, 11-20, 21-30, …, then 20 is considered in class:

Answer

11-20

Solution:
In a discontinuous class both lower and upper limits belong to that particular class.

View full question & answer→

Question 51 Mark

In a histogram, each class rectangle is constructed with base as:

Answer

Class interval

Solution:
Class interval is the difference between the upper limit and lower limit of a class, also called as class width.
Hence it forms the base of the rectangle in histogram.

View full question & answer→

Question 61 Mark

If each observation of the data is decreased by 8 then their mean:

Answer

is decreased by 8.

Solution:
If each observation of the data is decreased by 8 then their mean is also decreased by 8.

View full question & answer→

Question 71 Mark

The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8. The mode of the data is:

Answer

5 and 8

Solution:
The given data is 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10 and 8
Make the frequency table.

Value	Tallybars	Frequency
4	\|	1
5	\|\|\|\|\|	4
7	\|\|\|	3
8	\|\|\|\|	4
9	\|\|	2
10	\|	1

Since the value 5 and 8 occurs in the data maximum number of times, that is, 4. Hence, the modal value is 5 and 8. In this case the mode is not unique.

View full question & answer→

Question 81 Mark

Tally are usually marked in a bunch of:

Answer

Solution:
Tally are usually marked in a bunch of 5 : 4 in a vertical line and one is placed diagonally.

View full question & answer→

Question 91 Mark

Write the correct answer in the following: If each observation of the data is increased by $5,$ then their mean:

Answer

Let $x_1, x_2, ...., x_n$ be the $n$ observation,
Then, old mean $\bar{\text{x}}_{\text{old}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}\ \dots(\text{i})$
Now, adding $5$ in each obsevation, the new mean becomes
$\bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+\ ...\ +(\text{x}_\text{n}+5)}{\text{n}}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+\text{x}_2+\ ...\ +\text{x}_\text{n})+5\text{n}}{}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}+5=\bar{\text{x}}_{\text{old}}+5 [$from $Eq. (i)]$
$\Rightarrow\ \bar{\text{x}}_\text{new}=\bar{\text{x}}_{\text{old}}+5$
Hence, the new mean is increased by $5.$

View full question & answer→

Question 101 Mark

Let l be the lower class limit of a class-interval in a frequency distribution and m be the mid point of the class. Then, the upper class limit of the class is:

Answer

$2\text{m}-1$

Solution:
Given that, the lower class limit of a class-interval is l and the mid-point of the class is m. Let u be the upper class limit of the class-interval.
Therefore, we have
$\text{m}=\frac{\text{l+u}}{2}$
⇒ l + u = 2m
⇒ u = 2m - l
Thus the upper class limit of the class is (2m - l).
Hence, the correct choice is (c).

View full question & answer→

Question 111 Mark

The traffic police recorded the speed $\big(\text{ in }\frac{\text{km}}{\text{h}}\big)$ of 10 motorists as 48, 52, 57, 55, 42, 39, 60, 49, 53 and 47. Later an error in recording instrument was found. If the instrument has recorded the speed $\frac{5\text{km}}{\text{h}}$ less in each case, then the correct average speed of the motorists is:

Answer

$\frac{55.2\text{km}}{\text{h}}$

Solution:
Sum of all the recorded speeds is
48 + 52 + 57 + 55 + 42 + 39 + 60 + 49 + 53 + 47 = 502
Because of the error, $\frac{5\text{km}}{\text{h}}$ in each case
The sum increases by 50 i.e., 552
So the average speed of 10 vehicle is $\frac{55.2\text{km}}{\text{h}}$

View full question & answer→

Question 121 Mark

For the frequency distribution given below, the adjusted frequency for the class 25-45 is:

Class Interval	5-10	10-15	15-25	25-45	45-75
Frequency	6	12	10	8	15

Answer

Solution:
Adjusted frequency for the class 25-45 is
10 - 8 = 2

View full question & answer→

Question 131 Mark

The algebraic sum of the deviations of a set of $n$ values from their mean is:

Answer

if is the mean of n observations $x_1, x_2, x_3, x_4 ...x_n.$
then algebraic sum of deviations $=\sum\limits^\text{n}_{\text{i}=0}\Big(\text{x}_\text{i}-{\overline{\text{X}}}\Big)$
$=\sum\limits^\text{n}_{\text{i}=0}\text{x}_\text{i}-\text{n}{\overline{\text{X}}}$
$=\text{n}\bigg(\frac{\sum^\text{n}_{\text{i}=0}\text{x}_\text{i}}{\text{n}}\bigg)-\text{n}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$

View full question & answer→

Question 141 Mark

The mean of the following data is $8.$

$x$	$3$	$5$	$7$	$9$	$11$	$13$
$y$	$6$	$8$	$15$	$p$	$8$	$4$

Then, the value of $p$ is:

Answer

For calculating the mean, we prepare the table below:

$x_i$	$f_i$	$x_i \times f_i$
$3$	$6$	$18$
$5$	$8$	$40$
$7$	$15$	$105$
$9$	$p$	$9p$
$11$	$8$	$88$
$13$	$4$	$52$
	$\sum\text{f}_\text{i}=(41+\text{p})$	$\sum(\text{x}_\text{i}\times\text{f}_\text{i})=(303+9\text{p})$

Mean $=\frac{\sum(\text{x}_\text{i}\times\text{f}_\text{i})}{\sum\text{f}_\text{i}}=\frac{303+9\text{p}}{41+\text{p}}$
But mean $= 8$
$=\frac{303+9\text{p}}{41+\text{p}}=8$
$\Rightarrow303+9\text{p}=8(41+\text{p})$
$\Rightarrow303+9\text{p}=328+8\text{p}$
$\Rightarrow\text{p}=25$

View full question & answer→

Question 151 Mark

Write the correct answer in the following:
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is:

Answer

Solution:
Arranging the data in ascending order, we get
22, 34, 39, 45, 54, 56, 78, 84
Here n = 9, which is an odd number.
$\therefore\ \text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}$
$\text{Value}=\Big(\frac{9+1}{2}\Big)^{\text{th}}$
$\text{value}=5^{\text{th}}\text{value}$
So, median = 54

View full question & answer→

Question 161 Mark

The runs scored by 11 members of a cricket team are.
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0.
The median score is:

Answer

Solution:
Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number. Thus, we have:
Median = Value of $\bigg(\frac{\text{n}+1}{2}\bigg)\text{th}$ observation median score
= Value of $\bigg(\frac{11+1}{2}\bigg)\text{th}$ term
= Value of 6th term
= 29

View full question & answer→

Question 171 Mark

The median of the following data : 0, 2, 2, 2, -3, 5, -1, 5, −3, 6, 6, 5, 6 is:

Answer

Solution:
Data: 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6
Rearranging data in increasing order, we have
-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6
Number of observations = n = 14 (even)
Now,
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$

View full question & answer→

Question 181 Mark

One of the sides of a frequency polygon is:

Answer

The x-axis.

Solution:
In frequency polygon x-axis repesents data value whereas y-axis is used represent the frequencies of the data.
We include one class below lowest value and one class above highest value with zero frequencies. The graph touches the x-axis at these points.
So, one sides of the frequency polygon is x-axis.

View full question & answer→

Question 191 Mark

The class marks of a frequency distribution are given as follows 15, 20, 25 the class corresponding to the class mark 20 is:

Answer

17.5 - 22.5

Solution:
Clearly, lower limit of the class corrseponding to class mark 20 $=\frac{\text{Class mark of precending class + 20}}{2}$
$=\frac{15+20}{2}=17.5$
Uppar limit of the class corresponding to the class mark 20 $=\frac{\text{20 + Class mark of succeending class}}{2}$
$=\frac{20+25}{2}=\frac{45}{2}=22.5$
Hence the required class is 17.5 - 22.5

View full question & answer→

Question 201 Mark

A set of data consists of six numbers: 7, 8, 8, 9, 9 and x. The difference between the modes when x = 9 and x = 8 is:

Answer

Solution:
The mode in a list of numbers refers to the integers that occurs most number of times.
In the given list both 8 and 9 occur two times.
So the value of will decide the mode
If x = 8, then mode will be 8
If x = 9, then mode will be 9
Hence, differnce between two modes is 1.

View full question & answer→

Question 211 Mark

Tallys are usually marked in a bunch of:

Answer

Solution:
Tallies are usually marked in a bunch of 4.
Hence, the correct option is (b).

View full question & answer→

Question 221 Mark

The mean for counting numbers through 100 is:

Answer

50.5

Solution:
Mean of n natural numbers is $\frac{(\text{n+1})}{2}$ So
Mean of 100 numbers $=\frac{100+1}{2}$
Mean of 100 numbers $=\frac{100}{2}=50.5$

View full question & answer→

Question 231 Mark

A histogram is a pictorial representation of the grouped data in which class intervals and frequency are respectively taken along:

Answer

Horizontal axis and vertical axis.

Solution:
In a histogram the class limits are marked on the horizontal axis and the frequency is marked on the vertical axis. Thus, a rectangle is constructed on each class interval.

View full question & answer→

Question 241 Mark

The marks obtained by 10 students in a mathematics test are 75, 90, 70, 50, 70, 50, 75, 90, 70 and 75. Their median mark is:

Answer

72.5

Solution:
The median is the middle value for a set of data that has been arranged in ascending or descending order of magnitude.
Arrange the given observations in ascending order 50, 50, 70, 70, 70, 75, 75, 75, 90, 90
Here two numbers 70 and 75 are in the middle.
So, median is average of 70 and 75
$=\frac{70+75}{2}=72.5\text{ i..e }72.5$

View full question & answer→

Question 251 Mark

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is:

Answer

Solution:
First, we arrange the given observations in ascending order as follows
22, 34, 39, 45, 54, 54, 56, 68, 78 and 84
Here, total number of observation, n = 10
Since, n is even, so we use the formula for median,
Median $=\frac{\big(\frac{\text{n}}{2}\big)^{-1}\text{observation +}\big(\frac{\text{n}}{2}+1\big)\text{th observation}}{2}$
$=\frac{\big(\frac{10}{2}\big)\text{ th observation +}\big(\frac{10}{2}+1\big)\text{th observation}}{2}$ [put n = 10]
$=\frac{5\text{th observation + 6th obsevation}}{2}=\frac{54+54}{2}=\frac{108}{2}=54$
Hence, the median of given data is 54.

View full question & answer→

Question 261 Mark

If $x$ is mean of $x_1, x_2, .....$ then for $\text{a}\neq0,$ the mean of $ax_1, ax_2, ....ax_n, \frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},...\frac{\text{x}_n}{\text{a}},$ is:

Answer

Mean of $ax_1, ax_2, .....ax_n,$ is $ax$
Mean of $\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},...\frac{\text{x}_n}{\text{a}},$is $\frac{1}{\text{a}}\text{x}$
So the their mean is $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\text{x}}{2}$

View full question & answer→

Question 271 Mark

The class mark of the class 90-120 is:

Answer

Solution:
Class mark $=\frac{190+120}{2}=\frac{210}{2}=105$

View full question & answer→

Question 281 Mark

Which of the following is not a measure of central tendency?

Answer

Standard deviation.

Solution:
A measure of central tendency is a single value that attempts to describe a set of data. Mean, median and mode are the measures of central tendency. Standard deviation is not the measure of central tendency.

View full question & answer→

Question 291 Mark

The following observations have been arranged in an ascending order: 18, 20, 25, 26, 30, x, 37, 38, 39, 48. If the median of the data is 35, then the value of x is:

Answer

Solution:
The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
for even number of observations, median is calculated as average of two middle number.
for the given example 30 and x are in the middle and median is 35.
So,
$35=\frac{30+\text{x}}{2}$
$70=30+\text{x}$
$\text{x}=40$

View full question & answer→

Question 301 Mark

Tally marks are used to find:

Answer

Frequency.
Solution:
Tally marks are used to find the frequencies.
Hence, the correct choice is (c).

View full question & answer→

Question 311 Mark

Tally marks are used to find:

Answer

Frequency.

Solution:
When observations are large, it may not be easy to find the frequencies by simple counting.
So, we make use of tally marks.
Thus, Tally marks are used to find frequency.

View full question & answer→

Question 321 Mark

In the class intervals 10-20, 20-30, 20 is taken in:

Answer

The interval 20-30.

Solution:
The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30.Hence, the correct choice is (b).

View full question & answer→

Question 331 Mark

In the given graph, the number of students who scored 60 or more marks is:

Answer

Solution:
Add the values corresponding to the height of the bar from 60 to 100
10 + 5 + 3 + 3 = 21

View full question & answer→

Question 341 Mark

The median of the numbers $84, 78, 54, 56, 68, 22, 34, 45, 39, 54$ is:

Answer

Arranging the points in an ascending order,
We have:
$22, 34, 39, 45, 54, 56, 68, 78, 84$
Here$, n = 10,$ Which is even
$\therefore $median = mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{10}{2}+1\Big)^\text{th}$ terms
$=$ mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{12}{2}\Big)^\text{th}$ term
$=$ mean of $5^{th}$ and $6^{th}$ terms
$=\frac{1}{2}(54+54)$
$\frac{1}{2}\times108$
$=54$

View full question & answer→

Question 351 Mark

In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is:

Answer

Solution:
Mid-value = 10
$\Rightarrow\frac{\text{Upper limit + Lower limit }}{2}=10$
⇒ Upper limit + Lower limit = 20 .....(i)
Also, Class length = 6
⇒ Upper limit - lower limit = 6 .....(ii)
Subtracting (ii) from (i), we get
2 × Lower limit = 14
⇒ Lower limit = 7

View full question & answer→

Question 361 Mark

The class marks of a frequency distribution are as given below: 38, 43, 48, 53, 58. The class corresponding to the class mark 43 is:

Answer

40.5 - 45.5

Solution:
As the class size = 5, Class interval is got by subtracting and adding 2.5 to 43.

View full question & answer→

Question 371 Mark

Find out the mode of the following: 5, 4, 3, 5, 6, 6, 6, 5, 4, 5, 5, 3, 2, 1.

Answer

Solution:
The observation which occurs maximum number of times is called as mode of the given data.

View full question & answer→

Question 381 Mark

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequency of respective classes and abscissa are respectively:

Answer

Class marks of the classes.

Solution:
Frequency polygon is the line graph plotted with class marks on x-axis & frequency of the class on y-axis.

View full question & answer→

Question 391 Mark

In a bar graph if 1cm represents 30km, then the length of bar needed to represent 75km is:

Answer

2.5cm

Solution:
1cm - 30km
So for 75km
$\frac{75}{30}=2.5\text{cm}$

View full question & answer→

Question 401 Mark

The difference between the highest and lowest values of the observations is called:

Answer

Range.
Solution:
The difference between the highest and lowest values of the observations is called the range.

Hence, the correct choice is (c)

View full question & answer→

Question 411 Mark

For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?

Answer

Mean > Mode.

Solution:
Median = 4
Mode = 2
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, (Mean = 5) > (Mode = 2)

View full question & answer→

Question 421 Mark

Write the correct answer in the following:
For drawing a frequency polygon of a continous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abcissae are respectively:

Answer

Class marks of the classes.

Solution:
Abcissac are the class marks of the classes.

View full question & answer→

Question 431 Mark

The class mark of the class 100-120 is:

Answer

Solution:
Class mark $=\frac{\text{Upper limit + lower limit}}{2}=\frac{120+100}{2}=110$

View full question & answer→

Question 441 Mark

The empirical relation between mean, mode and median is:

Answer

Mode = 3 Median - 2 Mean.

Solution:
The empirical Relation between mean, median and mode is:
Mode = 3 Median - 2 mean.

View full question & answer→

Question 451 Mark

In the following distribution:

Wages(in Rs)	No of workers
More than 140	12
More than 130	27
More than 120	60
More than 110	105
More than 100	124
More than 90	141
More than 80	150

The number of workers having wage range (in Rs.) 110-120 is:

Answer

Solution:

Wages(in Rs)	No of workers
140–150	12
130–140	15
120–130	33
110–120	45
100–90	19
90–100	17
80–90	9

Therefore, the number of workers having a wage range (in Rs.) 110-120 is 45

View full question & answer→

Question 461 Mark

If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is:

Answer

Solution:
The given data is 7, 5, 13, x and 9. They are 5 in numbers.
The mean is $\frac{7+5+13+\text{x}+9}{5}=\frac{34+\text{x}}{5}$
But, it is given that the mean is 10. Hence, we have
$\frac{34+\text{x}}{5}=10$
⇒ 34 + x = 50
⇒ x = 50 - 34
⇒ x = 16

View full question & answer→

Question 471 Mark

Write the correct answer in the following:
A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44. The number of classes in the distribution will be:

Answer

Solution:
Minimum value = 14
Maximum value = 112
The classes are
13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102 and 103 - 112.
The number of classes in the distribution will be 10.

View full question & answer→

Question 481 Mark

The difference between the highest and lowest values of the observations is called:

Answer

Range.

Solution:
The difference between the highest and lowest value of observations is called 'Range' of observations.

View full question & answer→

Question 491 Mark

In a frequency distribution, ogives are graphical representation of:

Answer

Cumulative frequency.

Solution:
In a frequency distribution, ogives are graphical representation of cumulative frequency.

View full question & answer→

Question 501 Mark

The graph given below shows the frequency distribution of the age of 22 teachers in a school. The number of teachers whose age is less than 40 years is:

Answer

Solution:
Add the values corresponding to the height of the bar before 40.
6 + 3 + 4 + 2 = 15

View full question & answer→

MATHS M.C.Q

Test yourself on this topic