Measures of Central Tendency Test

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Measures of Central Tendency Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The number of trees in different parks of a city are $$33, 38, 48, 33, 34, 33$$ and $$24$$. The mode of this data is _______.

A
$$24$$

B
$$34$$

C
$$33$$

D
$$48$$

Solution

We have, $$33,\,38,\,48,\,33,\,34,\,34,\,33$$ and $$24$$.
On arranging the data in ascending order, we get $$24,\,33,\,33,\,33,\,34,\,34,\,38$$ and $$48$$.
Here, $$33$$ occurs more frequently, i.e. $$3$$ times.
$$\therefore$$ Mode of data is $$33$$.
Question 2
1 / -0

The mean age of $$25$$ students in a class is $$12$$. If the teacher's age is included, the mean age increases by $$1$$. Then teacher's age (in years) is:

A
$$24$$

B
$$38$$

C
$$30$$

D
$$48$$

Solution

Sum of $$25$$ students ages be $$x$$

$$\Rightarrow $$ Mean $$=\dfrac {x}{25}=12\Rightarrow x=300$$ equation - (1)

Let the teacher's age be $$y$$

$$\Rightarrow $$ New mean $$=\dfrac {x+y} {26}=13$$

$$\Rightarrow 300+y=338$$

$$\Rightarrow y=38$$

Teacher's age $$=y=38$$
Question 3
1 / -0

The average life expectancy in three countries $$A,B $$ and $$C$$ was $$75,90$$ and $$80$$ years respectively. The ratio of population in the three countries wa $$4:2:3$$ respectively. What was the average life expectancy of the three countries combined?

A
$$75$$

B
$$80$$

C
$$81$$

D
$$92$$

Solution

given avergare life expectency are $$75,90,80$$ respectively

ratio of population $$=4:2:3$$

let the populations be $$4x,2x,3x$$

therefore the total population combined$$=9x$$

average life expectency $$=\dfrac{75.4x+90.2x+80.3x}{9x}=80$$

average life expectency combined is $$80$$
Question 4
1 / -0

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

A
$$\alpha -\cfrac{1}{4}$$

B
$$\alpha -\cfrac{1}{2}$$

C
$$\alpha -2$$

D
$$\alpha +\cfrac{5}{4}$$

Solution

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

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

$$=\alpha-\dfrac{1}{4}$$
Question 5
1 / -0

Find the mode of the following data:
$$90, 40, 68, 90, 50, 60$$

A
$$40$$

B
$$68$$

C
$$90$$

D
$$50$$

Solution

$$\Rightarrow$$ 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.
$$\therefore$$ $$Mode=90$$
Question 6
1 / -0

Most frequently occuring values of a set of observation

A
Mean

B
Mode

C
Median

D
none

Solution

mode is the observation that occurs most frequently in the data. Each of the values in a data are occurring one time (or equal no of time) then all are mode.
Question 7
1 / -0

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 -

A
$$53$$

B
$$43$$

C
$$55.3$$

D
$$52.25$$

Solution

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

$$=\dfrac{8465}{162}$$

$$=52.25$$
Question 8
1 / -0

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:

A
Rs. 255.80

B
Rs. 257.50

C
Rs. 267.20

D
Rs. 267.50

Solution

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

$$=\dfrac{4120}{16}$$

$$=Rs.257.50$$
Question 9
1 / -0

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?

A
Rs. $$7.98$$

B
Rs. $$8$$

C
Rs. $$8.50$$

D
Rs. $$9$$

Solution

Avg cost = $$ \dfrac{7.5+8+8.5}{3}$$
$$ = \dfrac{24}{3}$$
$$ = 8 $$
Hence, average cost per liter is '8' rupees
Question 10
1 / -0

Find the median of the data $$a,ar,ar^{2}$$,......$$ar^{n}$$ where $$n$$ is an odd number.

A
$$ar^{\cfrac{n-1}{2}}$$

B
$$\dfrac{a}{2}(1-r)r^{\dfrac{n+1}{2}}$$

C
$$\dfrac{a}{2}(1+r)r^{\dfrac{n-1}{2}}$$

D
$$\dfrac{a}{2}(1-r)r^{\dfrac{n-1}{2}}$$

Solution

n is odd
Median is $$(\cfrac{n+1}{2})th$$ term
First term $$=a$$
Common ratio $$=r$$
$$\therefore t_{\cfrac{n+1}{2}}=ar^{\cfrac{n+1}{2}-1}\\ \quad=ar^{\cfrac{n-1}{2}}$$

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Measures of Central Tendency Test - 22

Result

Measures of Central Tendency Test - 22

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Question 1

$$24$$

$$34$$

$$33$$

$$48$$

Solution

Question 2

$$24$$

$$38$$

$$30$$

$$48$$

Solution

Question 3

$$75$$

$$80$$

$$81$$

$$92$$

Solution

Question 4

$$\alpha -\cfrac{1}{4}$$

$$\alpha -\cfrac{1}{2}$$

$$\alpha -2$$

$$\alpha +\cfrac{5}{4}$$

Solution

Question 5

$$40$$

$$68$$

$$90$$

$$50$$

Solution

Question 6

Mean

Mode

Median

none

Solution

Question 7

$$53$$

$$43$$

$$55.3$$

$$52.25$$

Solution

Question 8

Rs. 255.80

Rs. 257.50

Rs. 267.20

Rs. 267.50

Solution

Question 9

Rs. $$7.98$$

Rs. $$8$$

Rs. $$8.50$$

Rs. $$9$$

Solution

Question 10

$$ar^{\cfrac{n-1}{2}}$$

$$\dfrac{a}{2}(1-r)r^{\dfrac{n+1}{2}}$$

$$\dfrac{a}{2}(1+r)r^{\dfrac{n-1}{2}}$$

$$\dfrac{a}{2}(1-r)r^{\dfrac{n-1}{2}}$$

Solution

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