MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 6

If a variable takes the discrete values

$\alpha +4,\alpha -\cfrac{7}{2},\alpha -\cfrac{5}{2},\alpha -3,\alpha -2,\alpha +\cfrac{1}{2},\alpha -\cfrac{1}{2},\alpha +5(\alpha > 0)$ then the median is

0%
$\alpha -\cfrac{1}{4}$
0%
$\alpha -\cfrac{1}{2}$
0%
$\alpha -2$
0%
$\alpha +\cfrac{5}{4}$

Explanation

Median

$=\dfrac{\alpha +4+\alpha -\cfrac{7}{2}+\alpha -\cfrac{5}{2}+\alpha-3+\alpha -2+\alpha +\cfrac{1}{2}+\alpha -\cfrac{1}{2}+\alpha+5}{8}$

$=\dfrac{8\alpha -2}{8}$

$=\alpha-\dfrac{1}{4}$

Find the mode of the following data:

$90, 40, 68, 90, 50, 60$

0%
$40$
0%
$68$
0%
$90$
0%
$50$

Explanation

$\Rightarrow$ The given numbers are

$90,40,68,90,50,60$

$\Rightarrow$ Arranging given numbers in ascending order,

$40,50,60,68,90,90$

$\Rightarrow$ Mode is a number which occurs maximum times.

$\Rightarrow$ So, here numbers

$90$ occurs

$2$ times.

$\therefore$

$Mode=90$

Most frequently occuring values of a set of observation

0%
Mean
0%
Mode
0%
Median
0%
none

A school has four sections of chemistry in class XII having

$40, 35, 45$ and

$42$ students. The mean marks obtained in Chemistry test are

$50, 60, 55$ and

$45$ respectively for the four sections. The overall average of marks per students is -

0%
$53$
0%
$43$
0%
$55.3$
0%
$52.25$

Explanation

$\Rightarrow$ For

$1st$ section total marks obtain by class

$=40\times 50=2000$

$\Rightarrow$ For

$2nd$ section total marks obtain by class

$=35\times 60=2100$

$\Rightarrow$ For

$3rd$ section total marks obtain by class

$=45\times 55=2475$

$\Rightarrow$ For

$4th$ section total marks obtain by class

$=42\times 45=1890$

$\Rightarrow$ The overall average of marks per students

$=\dfrac{2000+2100+2475+1890}{40+35+45+42}$

$=\dfrac{8465}{162}$

$=52.25$

7 kg of tea costing Rs. 280 per kg is mixed with 9 kg of tea costing Rs. 240 per kg. The average price per kg of the mixed tea is:

0%
Rs. 255.80
0%
Rs. 257.50
0%
Rs. 267.20
0%
Rs. 267.50

Explanation

$\Rightarrow$ Total cost of

$7kg$ of tea

$=Rs.280\times 7=Rs.1960$

$\Rightarrow$ Total cost of

$9kg$ of tea

$=Rs.240\times 9=Rs.2160$

$\Rightarrow$ Average price of mixed tea

$=\dfrac{1960+2160}{7+9}$

$=\dfrac{4120}{16}$

$=Rs.257.50$

Find the median of the data

$a,ar,ar^{2}$ ,......

$ar^{n}$ where

$n$ is an odd number.

0%
$ar^{\cfrac{n-1}{2}}$
0%
$\dfrac{a}{2}(1-r)r^{\dfrac{n+1}{2}}$
0%
$\dfrac{a}{2}(1+r)r^{\dfrac{n-1}{2}}$
0%
$\dfrac{a}{2}(1-r)r^{\dfrac{n-1}{2}}$

Explanation

n is odd

Median is

$(\cfrac{n+1}{2})th$ term

First term

$=a$

Common ratio

$=r$

$\therefore t_{\cfrac{n+1}{2}}=ar^{\cfrac{n+1}{2}-1}\\ \quad=ar^{\cfrac{n-1}{2}}$

A car owner buys petrol at Rs.

$7.50$ , Rs.

$8$ and Rs.

$8.50$ per litre for three consecutive years. What approximately is the average cost per litre if he spends Rs.

$4000$ each year?

0%
Rs. $7.98$
0%
Rs. $8$
0%
Rs. $8.50$
0%
Rs. $9$

Explanation

Avg cost =

$\dfrac{7.5+8+8.5}{3}$

$= \dfrac{24}{3}$

$= 8$

Hence, average cost per liter is '8' rupees

1074842_1183678_ans_528aa25bb4ac4cadad60e0866c87fb88.png

The median of the set of the observation 1,3,5,7,11,13,17 is

0%
$1$
0%
$7$
0%
$9$
0%
$17$

Explanation

The set of the observation

$=1,3,5,7,11,13,17$

$Median=\dfrac{n+1}{2}\ term$

$Median=\dfrac{7+1}{2}\ term$

$Median=\dfrac{8}{2}\ term$

$Median=4\ term$

$Median=7$

Hence, this is the answer.

The median of the observations

$3,18,6,16,12$ and

$10$ is

0%
$11$
0%
$10$
0%
$12$
0%
$20$

Explanation

Arranging given number in ascending order,

$3,6,10,12,16,18$

Number of observations

$=6$

$Median=\dfrac{1}{2}\left[\left(\dfrac{6}{2}\right)^{th}term+\left(\dfrac{6}{2}+1\right)^{th}term\right]$

$Median=\dfrac{1}{2}[3^{rd}term+4^{th}term]$

$3^{rd}term=10$ and

$4^{th}term=12$

$\therefore$

$Median=\dfrac{1}{2}[10+12]$

$=\dfrac{1}{2}\times 22$

$=11$

In a test in Mathematics,

$7$ student scored

$19$ marks,

$11$ student scored

$15$ marks,

$16$ students

$13$ marks and

$12$ student scored

$10\ marks$ . The mode of the data is

0%
$19\ marks$
0%
$15\ marks$
0%
$13\ marks$
0%
$10\ marks$

Explanation

$13$ marks is obtained by

$16$ studends.

(maximum number of students )

Hence,

Mode of data is

$13$ marks.

Option

$C$ is correct answer.

The mean height of 25 male workers in a factory is 61 cm, and the mean height of 35 female workers in the same factory is 58 cm. The combined mean height of 60 workers in the factory is

0%
59.25
0%
59.5
0%
59.75
0%
58.75

Explanation

The mean height of

$25$ male workers in factory is

$61\,cm$

$\therefore$ Total height of

$25$ male worker

$=25\times 61=1525\,cm$

The mean height of

$35$ female workers in factory is

$58\,cm$

$\therefore$ Total height of

$25$ female worker

$=35\times 58=2030\,cm$

$\Rightarrow$ Total height of

$60$ workers

$=1525+2030=3555\,cm$

$\Rightarrow$ Required mean

$=\dfrac{3555}{60}=59.25\,cm$

The mean weight of 9 students is 25 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

0%
25 kg
0%
24 kg
0%
26 kg
0%
23 kg

Explanation

Formula used:

$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$

Apply the above formula, we get the sum of weights of the 9 students =

$25\times 9=225$ kg

If one more student is joined in the group, then total number of students is 10, and the mean isn 25.

The sum of weights of the 10 students is

$25\times 10=250$ kg

The weight of the tenth student is

$250-225=25$ kg

The sum of

$15$ observations is

$434+x.$ If the mean of the data is

$x,$ then find

$x.$

0%
$25$
0%
$27$
0%
$31$
0%
$33$

Explanation

Given, sum of all observations

$=434+x$

Number of observations

$=15$

and

$Mean=x$

$\Rightarrow \dfrac{434+x}{15}=x$

$\Rightarrow 434+x=15x$

$\Rightarrow x-15x=-434$

$\Rightarrow -14x=-434$

$\Rightarrow x=\dfrac{-434}{-14}$

$\Rightarrow x=31$

Hence, the value of

$x$ is

$31$ .

The mode of the unimodular data

$7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 12$ and

$x$ is

$10$ . The value of

$x$ is

0%
$10$
0%
$9$
0%
$8$
0%
$11$

Explanation

Given data-

$7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 12, x$

Mode of the data

$= 10$

Frequency of

$9 = 3$

Frequency of

$10 = 3$

Since frequency of

$9$ is same as frequency of

$10$ but mode of data is

$10$ . Thus the value of

$x$ is

$10$ .

Out of 5 brands of chocolates in a shop, a boy has to purchase the brand which is most liked by children . What measure of central tendency would be most appropriate if the data is provided to him?

0%
Mean
0%
Mode
0%
Median
0%
Any of the three

The mean of first ten odd natural number is

0%
5
0%
10
0%
20
0%
19

Explanation

Since, first ten odd natural numbers are $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$

We know that

$\text{Mean}=\dfrac{\text{sum of observation}}{\text{number of observation}}$

$\therefore$ Mean $=\dfrac{(1 + 19 + 3 + 17 + 5 + 15 + 7 + 13 + 9 + 11) }{ 10} = \dfrac{100}{10} = 10$

From a frequency distribution table if

$N = 100, h = 10 ;c.f. = 38 ;f = 18 ;L = 50$ , then find the median for the distribution. Choose the correct alternative.

0%
$56.67$
0%
$55.76$
0%
$56.76$
0%
$55.87$

Explanation

Medium

$= I + \left(\dfrac{\dfrac{N}{2}- c.f}{f}\right)\times h$

$= 50 + \dfrac{\left(\dfrac{100}{2}-38\right)}{18}\times 10$

$50+\dfrac{(50-38)}{18} \times 10 = 50 + \dfrac{12}{18} \times 10 = 56.67$

The mean of

$100$ observations is

$50$ . If one of the observations which was

$50$ is replaced by

$150$ , the resulting mean will be :

0%
$50.5$
0%
$51$
0%
$51.5$
0%
$52$

Explanation

Mean of

$100$ terms =

$50$
Sum of

$100$ terms =

$50\times 100$ = 5000
After replacing a term by

$150$ , sum =

$5000 - 50 +150 = 5100$
Mean =

$\dfrac{5100}{100}= 51$

A football player scored the following number of goals in the

$10$ matches:

$1, 3, 2, 5, 8, 6, 1, 4, 7, 9$
Since the number of matches is

$10$ (an even number), therefore, the median

$=\dfrac{5^{th} \text{observation} +6^{th} \text{observation}}{2}$

$=\dfrac{8+6}{2}=7$
Is it the correct answer and why?

0%
Correct
0%
Incorrect
0%
Ambiguous
0%
Data insufficient

Explanation

Data have first to be arranged in ascending (or descending) order before finding the median.
Hence, option

$B$ is the correct answer.

If the mean of variate X in distribution 2.6, then the value of k is

Variate X	1	2	3	4	5
Frequency of X	4	5	k	1	2

0%
$8$
0%
$10$
0%
$12$
0%
$18$

Explanation

$\dfrac{1\times 4+2 \times 5+3\times k+4\times 1+5\times 2}{4+5+k+1+2}=2.6$

$\Rightarrow 0.4k=3.2 \Rightarrow k=8$

If the mean of the observations:

$x, x + 3, x + 5, x + 7, x + 10$ is

$9$ , the mean of the last three observations is

0%
$10\dfrac{1}{3}$
0%
$10\dfrac{2}{3}$
0%
$11\dfrac{1}{3}$
0%
$11\dfrac{2}{3}$

Explanation

Mean of

$x, x + 3, x + 5, x + 7, x + 10 = 9$
Sum of the terms

$= 5x + 25$
Mean =

$\dfrac{5x+25}{5} =9$

$x +5 =9$

$x =4$
Last three observations are

$9,11,14$
Mean

$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$

The mean of

$25$ observations is

$36$ . Out of these observations if the mean of first

$13$ observations is

$32$ and that of the last

$13$ observations is

$40$ , the

$13$ th observation is :

0%
$23$
0%
$36$
0%
$38$
0%
$40$

Explanation

Mean of

$25$ observations =

$36$
Sum of

$25$ observations =

$36\times25 = 900$
Mean of first

$13$ observations=

$32$
Sum of first

$13$ observations =

$32\times13 = 416$
Mean of Last

$13$ observation=

$40$
Sum of Last 13 observations =

$40\times13$ =

$520$

Sum of first

$13$ observation+ Sum of last

$13$ obs

$-$

$13$ th observation

$=$ Total sum of

$25$ observations

$416 +520 - 13th\ observation = 900$

$13$ th observation $=36$

The median of the data

$78, 56, 22, 34, 45, 54, 39, 68, 54, 84$ is

0%
$45$
0%
$49.5$
0%
$54$
0%
$56$

Explanation

Terms are:

$78, 56, 22, 34, 45, 54, 39, 68, 54, 84.$
Arranging the terms in ascending order:

$22,34,39,45,54,54,56,68,78,84$ .

Since the number of terms are even, the median will be the mean of

$5^{th}$ and

$6^{th}$ terms:
Median

$= \dfrac{54+54}{2} = 54$

A total of

$25$ patients admitted to a hospital are tested for levels of blood sugar

$(mg/dl)$ and the results obtained were as follows:

$87$

$71$

$83$

$67$

$85$

$77$

$69$

$76$

$65$

$85$

$54$

$70$

$68$

$80$

$73$

$78$

$68$

$85$

$73$

$81$

$78$

$81$

$77$

$75$
Find mean, median and mode

$(mg/dl)$ of the above data.

0%
Mean $= 75.64$ , Median $= 77$ , Mode $= 85$
0%
Mean $= 75.64$ , Median $= 76$ , Mode $= 81$
0%
Mean $= 75.64$ , Median $= 77$ , Mode $= 81$
0%
Mean $= 73.64$ , Median $= 76$ , Mode $= 85$

Explanation

Arrange them in ascending order,

$54,65,67,68,68,69,70,71,73,73,75,76,77,77,78,78,80,81,81,83,85,85,85,85,87$
so total no.

$=25$
sum of total no.

$=1891$

$\therefore$ mean

$=\dfrac{1891}{25}=75.64$
median

$=\left(\dfrac{25+1}{2}\right)^{th}$ element

$= 13^{th}$ element

$=77$
mode

$=$ highest frequency element

$=85$

The points scored by a basket ball team in a series of matches are as follows:

$17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28$
Find the median and mode for the data.

0%
Median $= 12$ , mode $= 10$
0%
Median $= 11$ , mode $= 10$
0%
Median $= 12$ , mode $= 18$
0%
Median $= 10$ , mode $= 18$

Explanation

Arrange in ascending order

$=2,5,7,7,8,10,10,10,14,17,18,24,25,27,28,48$
Total

$16$ elements
So median

$=\left(\dfrac{(8^{th}+9^{th})\: \: \text{element}}{2}\right)$

$=\dfrac{10+14}{2}$

$=12$
mode

$= 10$

$(3$ times repeating

$)$

The mean marks (out of

$100$ ) of boys and girls in an examination are

$70$ and

$73$ , respectively. If the mean marks of all the students in that examination is

$71$ , find the ratio of the number of boys to the number of girls.

0%
$1:1$
0%
$2:1$
0%
$1:2$
0%
$2:3$

Explanation

Let the total number of boys be

$x$ and the total number of girls br

$y$ .

Let the sum of the marks of boys be

$a$ and the sum of the marks of girls be

$b$ .

Mean marks of boys are

$=70$

Mean marks of girls are

$=73$

Therefore,

$\Rightarrow \dfrac{a}{x}=70$

$\Rightarrow a=70x$

and

$\Rightarrow \dfrac{b}{y}=73$

$\Rightarrow b=73y$

Now, mean marks of all the students is

$71$

Therfore,

$\Rightarrow \dfrac{sum\ of\ the\ marks\ of\ all\ the\ students}{x+y}=71$

$\Rightarrow sum\ of\ the\ marks\ of\ all\ the\ students=a+b$

$\Rightarrow a+b=70x+73y$

Therefore,

$\Rightarrow \dfrac{70x+73y}{x+y}=71$

$\Rightarrow 70x+73y=71x+71y$

$\Rightarrow2y=x$

$\Rightarrow \dfrac{x}{y}=\dfrac{2}{1}$

$Hence\ the\ ratio\ is\ 2:1$

Mean of

$50$ observations was found to be

$80.4$ . But later on, it was discovered that

$96$ was misread as

$69$ at one place. Find the correct mean.

0%
$75.9$
0%
$8.4$
0%
$79$
0%
$80.94$

Explanation

$\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 } }{ 50 } =80.4$

Say

${ x }_{ 10 }$ (incorrect value)

$=69$

${ x' }_{ 10 }$ (correct value)

$=96={ x }_{ 10 }+27$

${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }=4020$

Add

$27$ on both sides

${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }+27=4020+27=4047$

${ x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 }=4047$

Actual mean

$=\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 } }{ 50 }$

Therefore,

$\dfrac { 4047 }{ 50 } =80.94$

The median of a set of

$9$ distinct observations is

$20.5$ . If each of the largest

$4$ observation of the set is increased by

$2$ , then the median of the new set

0%
is increased by $2$
0%
is decreased by $2$
0%
is two times the original median
0%
remains the same as that of the original set

Explanation

$Since$

$n= 9,$

$the$

$median$

$will$

$be$

$the$

$5th(\frac{9 + 1}{2}) term.$

$If$

$the$

$largest$

$4$

$terms$

$are$

$increased$

$by$

$2,$

$there$

$is$

$no$

$effect$

$on$

$the$

$5th$

$term,$

$implying$

$median$

$will$

$remain$

$same.$

In a class of 100 students there are 70 boys whose average marks in a subject areIf the average marks of the complete class are 72, then the average marks of the girls :

0%
$73$
0%
$65$
0%
$68$
0%
$74$

Explanation

Average of marks of

$70$ boys =

$75$

Sum of marks of

$70$ boys =

$75\times70$ =

$5250$

Average of marks of whole class =

$72$

Sum of marks of whole class =

$100\times72$ = 7200

Sum of marks of

$30$ girls =

$7200-5250 = 1950$

Average of marks of girls =

$\dfrac{1950}{30}$ =

$65$

If mode of the following data is

$7$ , then the value of

$k$ in the

$2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13$ is:

0%
$3$
0%
$7$
0%
$4$
0%
$2$

Explanation

Mode is the value which occurs most often in the data set of values.

Given data values are

$2,4,6,7,5,6,10,6,7,2k+1,9,7,13$

In the above data set, values

$6,7$ have occured more times i.e., 3 times

But given that mode is

$7$ . So, should occur more times than

$6$ . Hence the variable

$2k+1$ must be 7

$\implies 2k+1=7$

$\implies 2k=6 \implies k=\dfrac 62$

$\implies k=3$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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