Numerical Applications Test 24

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Numerical Applications Test 24

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Question 1
1 / -0

A clock strikes the number of times of the hour. How many strikes does it make in one day?

A
$$165$$ times

B
$$156$$ times

C
$$78$$ times

D
$$87$$ times

Solution

For the first 12 hours of the day,the clock will strike
$$1+2+3+......+12 = \cfrac{12}{2}(1+12) =78$$ times
For the next $$12$$ hours, there will be another 78 times, so in one day,the clock willl strike $$156$$ times
Question 2
1 / -0

In a Zonal athletic long jump meet the distances jumped by $$10$$ atheletes are: $$205\:cm, 200\:cm, 275\:cm, 260\:cm, 259\:cm, 199\:cm, 252\:cm, 239\:cm, 228\:cm$$ and $$281\:cm$$. Find the arithmetic mean of the jumps.

A
$$212.4\:cm$$

B
$$222.5\:cm$$

C
$$230.8\:cm$$

D
$$239.8\:cm$$

Solution

Distances jumped by $$10$$ athletes (in $$cm$$): $$205, 200, 275, 260, 259, 199, 252, 239, 228$$ and $$281$$
Mean $$= \cfrac{\text{Sum}}{\text{Number of athletes}}$$
Mean $$= \cfrac{205 +200 +275 +260 +259 +199 +252 +239 + 228 + 281}{10}$$
Mean $$= \cfrac{2398}{10}$$
Mean $$= 239.8$$
Question 3
1 / -0

If the mean of $$7$$,$$9$$,$$11$$,$$13$$,$$x$$,$$21$$ is $$13$$, find the value of $$x$$.

A
$$17$$

B
$$71$$

C
$$15$$

D
$$20$$
Question 4
1 / -0

$$17$$ men can complete a piece of work in $$12$$ days. In how many days can $$6$$ men complete the same piece of work?

A
$$\;28\:days$$

B
$$\;34\:days$$

C
$$\;26\:days$$

D
$$\;32\:days$$

Solution

(b)$$\;17$$ men can complete a piece of work in $$12$$ days.
$$\;\;\;\;\;\;\therefore\;1$$ men can complete the work in $$(12\times17)$$ days.
$$\;\;\;\;\;\;\therefore\;6$$ men can complete the work in $$\begin{pmatrix}\displaystyle\frac{12\times17}{6}\end{pmatrix}\:days=34\:days$$.
Question 5
1 / -0

On what day of the week India will celebrate its Republic day on 26th January,2015?

A
Sunday

B
Monday

C
Tuesday

D
Wednesday

Solution

$$26$$ January $$2015$$ $$=$$($$2014$$ years $$+$$ period from $$1.1.2015 to 26.1.2015$$)
Odd days in $$1600$$ years $$= 0$$
Odd days in $$400$$ years $$= 0$$
$$14$$ years =($$11$$ ordinary years +$$3$$ leap years )=$$11\times 1+3\times 2=17$$ odd days
January $$=26$$ days
Therefore $$26$$ days $$=3$$ weeks $$+ 5$$ days$$= 5$$ odd days
Total number of odd days$$= 17+5=22$$
if we divide it by $$7$$ then = $$1$$ odd day
Therefore, given day is Monday.
Question 6
1 / -0

If $$20$$th September in a year is Wednesday, the number of Fridays in the same month is

A
$$6$$

B
$$5$$

C
$$4$$

D
$$3$$

Solution

20th September is Wednesday
Friday will be on 20 + 2 = 22nd September
Friday will be on 22 - 7 = 15th September
Friday will be on 15 - 7 = 8th September
Friday will be on 8 - 7 = 1st September
Friday will be on 22 + 7 = 29th September
So, there will be 5 Fridays

Hence B is right answer
Question 7
1 / -0

A car travelling with $$\displaystyle \dfrac{5}{7} $$ of its usual speed covers $$42$$ km in $$1$$ hour $$40$$ min $$48$$ sec. What is the usual speed of the car?

A
$$\displaystyle 17 \dfrac{6}{7}$$ km/hr

B
$$25$$ km/hr

C
$$30$$ km/hr

D
$$35$$ km/hr

Solution

Let the usual speed of the car be x km/hr.
Time = 1 hour 40 min 48 sec $$= \displaystyle 1 + \dfrac{40}{60} + \dfrac{48}{3600} hrs$$
$$= 1 \displaystyle + \dfrac{2}{3} + \dfrac{1}{75} hrs$$
$$= \displaystyle \dfrac{75 + 50 + 1}{75} hrs$$
$$= \displaystyle \dfrac{126}{75} hrs$$
Given, $$\displaystyle \dfrac{5}{7} x \times \dfrac{126}{75} = 42 \Rightarrow x = \dfrac{42 \times 7 \times 75}{5 \times 126}$$
$$= 35 km/hr$$
Question 8
1 / -0

$$45$$ men working $$8$$ hours a day can finish a project within a given time. Five men are unable to report for the work. How long should the rest work a day to finish the project on time.

A
$$\;10\:hrs$$

B
$$\;12\:hrs$$

C
$$\;9\:hrs$$

D
$$\;8\:hrs$$

Solution

Total Work Hours are
$$\Rightarrow 45\times 8$$

When $$5$$ Do not Report ,then Only $$45-5=40$$ men will remain
So Now, Let us Assume the No of Hours be$$H$$ they now have to work to finish the project

$$\Rightarrow 40\times H=45\times 8$$
$$\Rightarrow H=9$$

Therefoe Answer is $$(C)$$
Question 9
1 / -0

If $$10$$ men and $$15$$ women finish a work in $$6$$ days. One man alone finishes that work in $$100$$ days. In how many days will a women finish the work?

A
$$125$$ days

B
$$150$$ days

C
$$90$$ days

D
$$225$$ days

Solution

1 man can finish the whole work in 100 days
$$\therefore$$ 1 man's 1 days' work $$ = \displaystyle\frac { 1 }{ 100 }$$
$$\Rightarrow$$ 10 men's 1 days' work $$ = 10 \times \displaystyle\frac { 1 }{ 100 } = \displaystyle\frac { 1 }{ 10 }$$
Let the time taken by one woman to finish the work be $$x$$ days.
Then, 1 woman's 1 days' work $$= \displaystyle\frac { 1 }{ x }$$
$$\Rightarrow$$ 15 women's 1 days' work $$ = \displaystyle\frac { 15 }{ x }$$
Given, (10 men's + 15 women's ) 1 days' work $$ = \displaystyle\frac { 1 }{ 6 }$$
$$\Rightarrow \displaystyle\frac { 1 }{ 10 } + \displaystyle\frac { 15 }{ x } = \displaystyle\frac { 1 }{ 6 } \Rightarrow \displaystyle\frac { 15 }{ x } = \displaystyle\frac { 1 }{ 6 } - \displaystyle\frac { 1 }{ 10 } = \displaystyle\frac { 5-3 }{ 30 } = \displaystyle\frac { 2 }{ 30 } = \displaystyle\frac { 1 }{ 15 }$$
$$\Rightarrow x = 225$$ days.
Question 10
1 / -0

$$A, B$$ and $$C$$ can complete a work in $$10, 12\ and\ 15$$ days respectively. They started the work together, but $$A$$ left the work before 5 days of its completion. $$B$$ also left the work $$2$$ days after $$A$$ left. In how many days was the work completed.

A
$$4$$

B
$$5$$

C
$$7$$

D
$$8$$

Solution

Let the work be completed in $$x$$ days.
$$A$$'s 1 days' work $$= \displaystyle\frac { 1 }{ 10 }$$
$$\Rightarrow A$$'s $$\left( x-5 \right)$$ days' work $$ = \displaystyle\frac { \left( x-5 \right) }{ 10 }$$
$$B$$'s 1 days' work $$= \displaystyle\frac { 1 }{ 12 }$$
$$\Rightarrow B$$'s $$\left( x-3 \right)$$ days' work $$ = \displaystyle\frac { \left( x-3 \right) }{ 12 }$$
$$C$$'s 1 days' work $$= \displaystyle\frac { 1 }{ 15 }$$
$$\Rightarrow C$$'s $$x$$ days' work $$= \displaystyle\frac { x }{ 15 }$$
$$\Rightarrow \displaystyle\frac { x-5 }{ 10 } + \displaystyle\frac { x-3 }{ 12 } + \displaystyle\frac { x }{ 15 } = 1$$
$$\Rightarrow \displaystyle\frac { 6\left( x-5 \right) + 5\left( x-3 \right) + 4x }{ 60 } = 1$$
$$\Rightarrow 6x - 30 + 5x - 15 + 4x = 60$$
$$\Rightarrow 15x = 60 + 45 = 105 \Rightarrow x = 7$$
$$\therefore $$ The work was completed in $$7$$ days.

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Numerical Applications Test 24

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Numerical Applications Test 24

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SHARING IS CARING

Question 1

$$165$$ times

$$156$$ times

$$78$$ times

$$87$$ times

Solution

Question 2

$$212.4\:cm$$

$$222.5\:cm$$

$$230.8\:cm$$

$$239.8\:cm$$

Solution

Question 3

$$17$$

$$71$$

$$15$$

$$20$$

Question 4

$$\;28\:days$$

$$\;34\:days$$

$$\;26\:days$$

$$\;32\:days$$

Solution

Question 5

Sunday

Monday

Tuesday

Wednesday

Solution

Question 6

$$6$$

$$5$$

$$4$$

$$3$$

Solution

Question 7

$$\displaystyle 17 \dfrac{6}{7}$$ km/hr

$$25$$ km/hr

$$30$$ km/hr

$$35$$ km/hr

Solution

Question 8

$$\;10\:hrs$$

$$\;12\:hrs$$

$$\;9\:hrs$$

$$\;8\:hrs$$

Solution

Question 9

$$125$$ days

$$150$$ days

$$90$$ days

$$225$$ days

Solution

Question 10

$$4$$

$$5$$

$$7$$

$$8$$

Solution

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