Data Handling Test

Result

Data Handling Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

$$2, 10, m, 12, 4$$
A group of $$5$$ integers is shown above. If the average (arithmetic mean) of the numbers is equal to $$m$$, find the value of $$m$$.

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

E
$$11$$

Solution

We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation:
$$\Rightarrow \dfrac { 2+10+m+12+4 }{ 5 } =m$$
$$\Rightarrow 28+m=5m$$
$$\Rightarrow m-5m=-28$$
$$\Rightarrow -4m=-28$$
$$\Rightarrow m=7$$
Question 2
1 / -0

What is the arithmetic mean of the progression $$11, 22, 33, 44, 55, 66, 77?$$

A
$$44$$

B
$$208$$

C
$$308$$

D
$$48$$

Solution

Using the formula for required arithmetic mean $$=\dfrac{\text{sum of the terms}}{\text{number of terms}}$$

After substituing the values we get: $$=\dfrac{11+22+33+44+55+66+77}{7}$$
$$\quad \quad \quad =\dfrac{11(1+2+3+4+5+6+7)}{5}=\dfrac{11\cdot 28}{7}=11\cdot 4=44$$
Question 3
1 / -0

Size 2 3 4 5 6 7 8
Frequency 10 12 25 20 25 15 11
The mode of the following distribution is ______.

A
$$2$$

B
$$8$$

C
Both $$4$$ and $$6$$

D
$$5$$

Solution

Mode is that observation which have highest frequency. Since, both $$4$$ and $$6$$ have highest frequency i.e. $$25$$ and $$25$$, they are the mode of the given distribution.
Hence, option (C) is correct.
Question 4
1 / -0

Arithmetic mean of $$2$$ and $$8$$ is

A
$$5$$

B
$$10$$

C
$$16$$

D
$$3.2$$

Solution

$${\textbf{Step -1: Stating the arithmetic mean of n general numbers.}}$$

$${\text{Let there be }}n{\text{ numbers }}{a_1}{\text{,}}{a_2}{\text{,}}{a_3}.....{a_n}{\text{. }}$$

   $${\text{Arithmetic mean of the numbers = }}\dfrac{{{a_1} + {a_2} + {a_3}{\text{ + }}.....{\text{ + }}{a_n}{\text{.}}}}{n}$$

$${\textbf{Step -2: Finding arithmetic mean of 2 and 8.}}$$

   $${\text{Here we have only two numbers - }}2,8$$

   $${\text{Arithematic mean }}A = \dfrac{{2 + 8}}{2} = \dfrac{{10}}{2} = 5$$

$${\textbf{Hence, }}\ {\textbf{Arithematic mean of 8 annd 2 is 5.}}$$
Question 5
1 / -0

The wickets taken by a bowler in a one-day cricket match are $$4, 5, 6, 3, 4, 0, 3, 2, 3, 5.$$ The mode of the data is ...............

A
$$3$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Mode of the set of data is the observation which occurs the most.
$$4$$ and $$6$$ occurs $$2$$ times each, $$6,~2$$ and $$0$$ occurs $$1$$ time each, whereas $$3$$ occurs $$3$$ times.
Thus, the number $$3$$ occurs the maximum number of times i.e., $$3$$.
Therefore, mode of the data is $$3$$.
Question 6
1 / -0

The given bar graph shows the number of matches played by different teams. Read the bar graph and answer the questions:
How many matches did both India and Bangladesh play?

A
$$34$$

B
$$22$$

C
$$38$$

D
$$40$$

Solution

From the bar graph,
The number of matches played by India $$=26$$.
The number of matches played by Bangladesh $$=8$$.

Therefore, the total number of matches played by both India and Bangladesh $$=26 + 8 = 34$$.

Thus, total number of matches played by both India and Bangladesh $$34$$.

Hence, option $$A$$ is correct.
Question 7
1 / -0

What temperature does the given thermometer shows?

A
$$35^{\circ}C$$

B
$$77^{\circ}F$$

C
$$95^{\circ}C$$

D
$$122^{\circ}F$$

Solution

Thermometer shows $$35^o C$$ or $$95^o F$$
Hence the correct answer is option A.
Question 8
1 / -0

The A.M. of $$a + 2, a, 2-a$$ is

A
$$a$$

B
$$\cfrac { a+4 }{ 3 } $$

C
$$\cfrac { a-4 }{ 3 } $$

D
$$\cfrac { a }{ 2 } $$

Solution

Given values are : $$a+2, a, 2-a$$

Arithmetic Mean, $$AM = \dfrac{a+2+a+2-a}{3}$$

$$AM = \dfrac{a+4}{3}$$

Hence, option (B) is corret
Question 9
1 / -0

Median is based on the...

A
highest $$50$$ $$\%$$ of items

B
lowest $$25$$ $$\%$$ of items

C
highest $$25$$ $$\%$$ of items

D
middle $$50$$ $$\%$$ of items

Solution

Median, by definition is the term that
occurs in the middle. Hence it depends upon
the middle 50% of the given items.
Question 10
1 / -0

Find the mean of $$ 43,54,64,53,36.$$

A
$$50$$

B
$$ 40$$

C
$$60$$

D
$$ 30$$

Solution

Sum of observations $$=$$ $$43+54+64+53+36$$
$$=250$$

No . of observations $$=$$ $$5$$

Mean of data $$=\dfrac{250}{5}$$
$$=50$$

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Data Handling Test - 17

Result

Data Handling Test - 17

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SHARING IS CARING

Question 1

$$7$$

$$8$$

$$9$$

$$10$$

$$11$$

Solution

Question 2

$$44$$

$$208$$

$$308$$

$$48$$

Solution

Question 3

$$2$$

$$8$$

Both $$4$$ and $$6$$

$$5$$

Solution

Question 4

$$5$$

$$10$$

$$16$$

$$3.2$$

Solution

Question 5

$$3$$

$$4$$

$$5$$

$$2$$

Solution

Question 6

$$34$$

$$22$$

$$38$$

$$40$$

Solution

Question 7

$$35^{\circ}C$$

$$77^{\circ}F$$

$$95^{\circ}C$$

$$122^{\circ}F$$

Solution

Question 8

$$a$$

$$\cfrac { a+4 }{ 3 } $$

$$\cfrac { a-4 }{ 3 } $$

$$\cfrac { a }{ 2 } $$

Solution

Question 9

highest $$50$$ $$\%$$ of items

lowest $$25$$ $$\%$$ of items

highest $$25$$ $$\%$$ of items

middle $$50$$ $$\%$$ of items

Solution

Question 10

$$50$$

$$ 40$$

$$60$$

$$ 30$$

Solution

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