Numerical Applications Test 4

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Numerical Applications Test 4

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Weekly Quiz Competition

Question 1
1 / -0

Mean of a set of observations is the value which

A
occurs most frequently

B
divides observations into two equal parts

C
is a representative of whole group

D
is the sum of observation

Solution

Mean is the value which is the representative of the whole group. As, this takes into account all the values present in the group and averages them.
Question 2
1 / -0

In a cricket test match the scores of ten players are: $$85$$, $$32$$, $$0$$, $$54$$, $$29$$, $$101$$, $$73$$, $$64$$, $$29$$ and $$36$$. Find the mean of the runs.

A
$$36.4$$ runs

B
$$45.5$$ runs

C
$$50.3$$ runs

D
$$63.7$$ runs

Solution

Runs scored by $$10$$ players $$= 85,32,0,54,29,101,73,64,29,36$$
Mean $$= \cfrac{\text{Sum}}{\text{Number of players}}$$
Mean $$= \cfrac{85 + 32 + 0 + 54 + 29 + 101 + 73 + 64 + 29 + 36}{10}$$
Mean $$= \cfrac{503}{10}$$
Mean $$= 50.3$$
Question 3
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The heights of $$6$$ boys in a group are $$142$$ cm, $$154 $$ cm, $$146$$ cm, $$145$$ cm, $$151$$ cm and $$150$$ cm. Find the mean height per boy

A
$$122$$ cm

B
$$148$$ cm

C
$$156$$ cm

D
$$173$$ cm

Solution

Mean height $$=\cfrac{\text{Sum of the height of the all boys}}{\text{No. of boys}}$$
$$=\cfrac{142+154+146+145+151+150}{6}$$
$$=\cfrac{888}{6}=148$$ cm
Hence, the mean height is $$148$$ cm.
Question 4
1 / -0

Find the mean of first ten odd natural numbers.

A
$$7$$

B
$$10$$

C
$$13$$

D
$$18$$

Solution

$${\textbf{Step - 1: Listing first 10 odd natural numbers}}$$
$${\text{First 10 odd natural numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$$
$${\textbf{Step - 2: Calculating mean}}$$
  $${\text{We know that, mean of n numbers = }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$$
  $$\therefore {\text{ Mean of first 10 odd natural numbers,}}$$
$${\text{M = }}\dfrac{{{\text{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}}}}{{{\text{10}}}}$$
  $$ \Rightarrow {\text{ M = }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{ = 10}}$$
$${\textbf{Thus, the mean of first 10 odd natural numbers is 10}}$$
Question 5
1 / -0

In a monthly test the marks obtained in mathematics by $$16$$ students of a class are as follows
$$0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 $$
The arithmetic mean of the marks obtained is

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Since, Mean $$= \cfrac {0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8}{16} $$
      $$\Rightarrow $$ Mean $$= \cfrac {64}{16} $$
   $$\Rightarrow $$ Mean $$= 4 $$
   $$\therefore $$ Option $$B$$ is correct.
Question 6
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A contractor undertook to do a certain piece of work in $$9$$ days. He employed certain number of men, but $$6$$ of them being absent from the very first day, the rest could finish the work in $$15$$ days. The number of men originally employed were

A
$$12$$

B
$$15$$

C
$$18$$

D
$$24$$

Solution

Let there be $$ x $$ men at the begining less men, more days
$$\therefore$$ $$15 : 9 :: x : (x - 6)$$ $$\Rightarrow$$ $$ 15(x - 6) = 9x$$
$$\Rightarrow$$ $$6x = 90$$ $$\Rightarrow$$ $$ x = 15$$
Question 7
1 / -0

Out of the following four choices which does not show the coinciding of the hour hand and minute hand :

A
$$3 : 16 : 2$$

B
$$6 : 32 : 43$$

C
$$9 : 59 : 05$$

D
$$5 : 27 : 16$$

Solution

Clearly, at 9:59:05 the hour hand and the minute hand does not coincide.
Answer is Option C
Question 8
1 / -0

If in a non leap year it is Tuesday on 28th February then the day which will be on 28th march is ------

A
Monday

B
Tuesday

C
Wednesday

D
Thursday
Question 9
1 / -0

A fort had provision of food for $$150$$ men for $$45$$ days. After $$10$$ days, $$25$$ men left the fort, The number of days for which the remaining food will last, is

A
$$29\displaystyle \frac{1}{5}$$

B
$$37\displaystyle \frac{1}{4}$$

C
$$42$$

D
$$54$$

Solution

After $$10$$ days: $$150$$ men had food for $$35$$ days.
Suppose $$125$$ men had food for $$x$$ days. Less men, More days; More men less days.
$$\therefore$$ $$125 : 150 : : 35 : x$$
$$\Rightarrow\, 125\, \times\, x\, =\, 150\, \times\, 35$$
$$\Rightarrow\, x\, =\, \displaystyle \frac{150\, \times\, 35}{125}\, \Rightarrow\, x\, =\, 42$$
Question 10
1 / -0

If the arithmetic mean of $$6, 8, 5, 7, x$$ and $$4$$ is $$7,$$ then $$x$$ is

A
$$12$$

B
$$6$$

C
$$8$$

D
$$4$$

Solution

$$\textbf{Step - 1: Solving for x.}$$
$$\text{Observation given = }{6, 8, 5, 7, x, 4}$$
$$\text{Arithmetic mean = 7}$$
$$\text{Arithmetic mean = }\dfrac{\text{Sum of observations.}}{\text{Number of observations.}}$$
$$\Rightarrow \dfrac{6 + 8 + 5 + 7 + x + 4}{6} = 7$$
$$30 + x = 7 \times 6$$
$$x = 42 - 30$$
$$x = 12$$
$$\textbf{Hence, the value of x is 12. (Option A)}$$

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 4

Result

Numerical Applications Test 4

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SHARING IS CARING

Question 1

occurs most frequently

divides observations into two equal parts

is a representative of whole group

is the sum of observation

Solution

Question 2

$$36.4$$ runs

$$45.5$$ runs

$$50.3$$ runs

$$63.7$$ runs

Solution

Question 3

$$122$$ cm

$$148$$ cm

$$156$$ cm

$$173$$ cm

Solution

Question 4

$$7$$

$$10$$

$$13$$

$$18$$

Solution

Question 5

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 6

$$12$$

$$15$$

$$18$$

$$24$$

Solution

Question 7

$$3 : 16 : 2$$

$$6 : 32 : 43$$

$$9 : 59 : 05$$

$$5 : 27 : 16$$

Solution

Question 8

Monday

Tuesday

Wednesday

Thursday

Question 9

$$29\displaystyle \frac{1}{5}$$

$$37\displaystyle \frac{1}{4}$$

$$42$$

$$54$$

Solution

Question 10

$$12$$

$$6$$

$$8$$

$$4$$

Solution

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