Numerical Applications Test 48

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Numerical Applications Test 48

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Weekly Quiz Competition

Question 1
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The average of statistical data is called :

A
Arithmetic mean

B
Median

C
Mode

D
Frequency

Solution

The average of statistical data is called $$Mean$$ or $$Arithmatic\ mean$$

Hence, $$Op-A$$ is correct.
Question 2
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Two men and $$5$$ boys together can finish a piece of work in $$4$$ days, while $$3 $$ men and $$ 6$$ boys can finish in $$3$$ days. Find the time taken :
(i) by one man alone to finish the work
(ii) by one boy alone to finish the work.

A
$$17$$ days and $$ 19$$ days

B
$$14$$ days and $$22$$ days

C
$$18$$ days and $$36$$ days

D
$$17$$ days and $$13$$ days

Solution

Let the amount of work done by a Man alone in a day be $$ M $$ and by a Boy alone be $$ B $$

Given, $$2$$ men and $$5$$ boys together can finish a piece of work in $$ 4 $$ days

Amount of work they can do in one day $$2M + 5B $$$$ = \dfrac {1}{4} $$ -- (1)

Similarly, $$ 3M + 6 B = \dfrac {1}{3} $$ --- (2)

Solving equations $$ (1) $$ and $$ (2) $$, we get $$ B = \dfrac {1}{36} $$

Substituting $$ B $$ in equation $$ 2 $$, we get $$ M = \dfrac {1}{18} $$

Hence, one man alone will take $$ 18 $$ days to do the work and one boy will take $$ 36 $$ days.
Question 3
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A contractor undertakes to dig a canal, $$6$$ kilometers long in $$35$$ days and employed $$90$$ men. He finds that after $$20$$ days only $$2$$ km of canal have been completed . How many more men must be employed to finish the work in time?

A
$$120$$ men

B
$$170$$ men

C
$$180$$ men

D
$$150$$ men

Solution

Total time given $$=35$$ days.

To dig a canal of $$2$$ km in $$20$$ days, $$90$$ men were required.

So, to dig a canal of $$1$$ km in $$1$$ day, $$\dfrac{90\times20}{2}$$ men will be required.

So, to dig a canal of $$4$$ km in $$15$$ days $$\dfrac{900\times4}{15}$$ $$=240$$ men will be required.

As $$90$$ men were already employed, so $$150$$ more men are required.
Question 4
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Pooja and Ritu can do a piece of work in $$17\displaystyle \frac{1}{7}$$ days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each alone will do the work.

A
Pooja in $$44$$ days and Ritu in $$8$$ days

B
Pooja in $$80$$ days and Ritu in $$ 500$$ days

C
Pooja in $$20$$ days and Ritu in $$120 $$ days

D
Pooja in $$40$$ days and Ritu in $$30$$ days

Solution

Let Pooja's one day work be $$x$$
And Ritu's one day work be $$y$$
Accprding to first condition,
Pooja and Ritu can do work in $$17 \dfrac{1}{7}$$ days i.e $$\dfrac{120}{7}$$ days
$$\therefore x + y = \dfrac {1}{\dfrac{120}{7}}$$
$$\therefore x + y = \dfrac{7}{120}$$ .........Equation $$1$$
According to second condition,
$$x = \dfrac{3}{4} y$$
$$\therefore 4x - 3y = 0$$ .......... Equation $$2$$
Multiplying eqution $$1$$ by $$3$$ and solving equation $$1$$ and $$2$$ simultaneosly, we get
$$3x + 3y = \dfrac{7}{40}$$
$$4x - 3y = 0$$
$$\therefore 7x = \dfrac{7}{40}$$
$$\therefore x = \dfrac{1}{40}$$
Putting the value of $$x$$ in equation $$1$$ , we get
$$\dfrac{1}{40} + y = \dfrac{7}{120}$$
$$\therefore y = \dfrac{4}{120} = \dfrac{1}{30}$$
$$\therefore$$ Pooja's one day work is $$x = \dfrac{1}{40}$$
i.e. her whole work will be completed in $$40$$ days
Similarly Pooja's work will be completed in $$30$$ days
Answer : Pooja can do the work alone in $$40$$ days and Ritu in $$30$$ days
Question 5
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A cabinet of ministers consists of 11 ministers, one minister being the chief minister. A meeting is to be held in a room having a round table and 11 chairsround it, one of them being meant for the chairman. The number of ways in which the ministers can take their chairs, the chief minister occupying the chairman's place, is

A
$$\displaystyle \frac{1}{2} (10!)$$

B
$$\displaystyle 9!$$

C
$$\displaystyle 10!$$

D
none of these

Solution

These are circular permutations
so,
$$=(n-1)!$$ = $$10!$$
Question 6
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$$16$$ men can complete a work in $$12$$ days, $$24$$ children can complete the same work in $$18$$ days, $$12$$ men and $$8$$ children started working and after $$8$$ days, $$3$$ more children joined them. How many days will they now take to complete the remaining work ?

A
$$2$$ days

B
$$4$$ days

C
$$6$$ days

D
$$8$$ days

Solution

One day work of 16 men is $$\dfrac { 1 }{ 12 } $$.

One day work of 24 children is $$\dfrac { 1 }{ 18 } $$.

One men's one day work is $$\dfrac { 1 }{ 12\times 16 } $$.

One child's one day work is $$\dfrac { 1 }{ 24\times 18 } $$.

One day work of 12 men and 8 children is $$\dfrac { 12 }{ 12\times 16 } +\dfrac { 8 }{ 24\times 18 } $$.

One day work = $$\dfrac { 35 }{ 54\times 8 } $$.

Work done in 8 days = $$8\times \dfrac { 35 }{ 54\times 8 } =\dfrac { 35 }{ 54 } $$.

Remaining work = $$1-\dfrac { 35 }{ 54 } =\dfrac { 19 }{ 54 } $$.

The remaining work is 12$$\times$$ one men's one day work + 11$$\times$$ one child's one day work.

Hence, $$12\times \dfrac { 1 }{ 12\times 16 } +11\times 24\times 18$$

One day work multiplied by number of days to complete remaining work = remaining work

Therefore, the work is done in 4 days.

$$\left( \dfrac { 1 }{ 16 } +\dfrac { 11 }{ 24 } \times 18 \right) x=\dfrac { 19 }{ 54 } $$

$$\Rightarrow $$$$x=4$$

Thus they take 4 days to complete the remaining work.

Hence option B is correct.
Question 7
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A particular work can be completed by $$6$$ men and $$6$$ women in $$24$$ days, whereas the same work can be completed by $$8$$ men and $$12$$ women in $$15$$ days. Find the time taken by $$4$$ men and $$6$$ women to complete the same work .

A
$$20$$ days

B
$$30$$ days

C
$$50$$ days

D
$$40$$ days

Solution

From the question, we have
($$6$$ men and $$6$$ women) $$1$$ day work $$=\dfrac{1}{24}$$

($$12$$ men and $$12$$ women)$$1$$ day work$$=\dfrac{1}{12}$$ ... (1)

($$8$$ men and $$12$$ women)$$1$$ day work$$=\dfrac{1}{15}$$ ... (2)
Subtracting (1) and (2), we get
$$4$$ men work $$=\dfrac{1}{12}$$ $$-\dfrac{1}{15}$$ $$=\dfrac{1}{60}$$

$$2$$ men work $$=\dfrac{1}{120}$$

$$4$$ men and $$6$$ women work $$=1$$ day work of $$6$$ men and $$6$$ women $$-1$$ day work of $$2$$ men.
$$=\dfrac{1}{24}$$ $$-\dfrac{1}{120}$$ $$=\dfrac{4}{120}$$  $$=\dfrac{1}{30}$$

The time taken by $$4$$ men and $$6$$ women to complete the same work $$= 30$$ days.

Hence, option B.
Question 8
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$$A$$ and $$B$$ can do a piece of work in $$5$$ days, $$B$$ and $$C$$ can do it in $$7$$ days. $$A$$ and $$C$$ can do it in $$4$$ days. Who among these will take the least time if put to do it alone ?

A
$$A$$

B
$$B$$

C
$$C$$

D
Data inadequate

Solution

$$Let \, \,the \, \,work \, \,be \, \,W.$$
$$A \, \, and\, \, B\, \, can\, \, do\, \, a \, \, piece \, \,of\, \, work\, \, in \, \, 5 \, \,days$$$$\therefore \, \,5A+5B=W \, \,..........(1)$$

$$B \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,7 \, \, days$$$$\therefore, \, \,7B+7C=W \, \,......(2)$$

$$A \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,4 \, \, days$$$$\therefore, \, \,4A+4C=W \, \,.......(3)$$

$$Solving \, \, (1) \, \,(2) \, \,and \, \,(3) \, \,We \, \,get,$$

$$Speed \, \, of \, \, A=\dfrac{43}{280} \, \, work/day$$

$$Speed \, \, of \, \, B=\dfrac{13}{280} \, \, work/day$$

$$Speed \, \, of \, \, C=\dfrac{27}{280} \, \, work/day$$

$$Since \, \,speed \, \,of \, \,A \, \,is \, \,maximum \, \,among \, \,all \, \,three, \, $$
$$Hence \, \,A \, \, will \, \,take \, \,least \, \,time \, \,to \, \,complete \, \,the \, \,work.$$
Question 9
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A and B can do a job in $$12$$ days, B and C can do it in $$16$$ days. After A has worked for $$5$$ days and B has worked for $$7$$ days, C can finish the rest in $$13$$ days. In how many days, can C do the work alone?

A
$$16$$ days

B
$$24$$ days

C
$$36$$ days

D
$$48$$ days

Solution

$$A$$ and $$B$$ can do a job in $$12$$ days.
$$B$$ and $$C$$ can do a job in $$16$$ days.
$$A$$ worked for $$5$$ days and $$B$$ worked for $$7$$ days.
Part of work done by $$A$$ and $$B$$ in $$5$$ days=$$\dfrac{5}{12}$$
$$\therefore$$ Part of work left after $$A$$ left=$$1-\dfrac{5}{12}=\dfrac{12-5}{12}=\dfrac{7}{12}$$
After $$A$$ left work $$B$$ worked for $$2$$ more days.
Part of work done by $$B$$ and $$C$$ in $$2$$ days=$$\dfrac{2}{16}=\dfrac{1}{8}$$
$$\therefore$$ part of work left after $$B$$ left=$$\dfrac{ 7 }{ 12 } -\dfrac{ 1 }{ 8 } =\dfrac{ 14-3 }{ 24 } =\dfrac{ 11 }{ 24 } $$
Now $$\dfrac{ 11 }{ 24 }$$ part o work $$C$$ has to complete in $$11$$ days
Since $$C$$ already had worked for $$2$$ dayswith $$B$$
(part)complete work is done by $$C$$ in $$\dfrac { 11 }{ \dfrac { 11 }{ 24 } } $$
=$$24$$days.
Question 10
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$$3$$ men and $$4$$ boys can do a piece of work in $$14$$ days, while $$4$$ men and $$6$$ boys do the same work in $$10$$ days. How long would it take one man to finish the same work ?

A
$$40$$ days

B
$$20$$ days

C
$$35$$ days

D
$$70$$ days

Solution

Let one man's one day's work be $$x$$
Let one boy's one day's work be $$y$$
According to first condition,
$$3x+4y = \dfrac{1}{14}$$ .......Equation $$1$$
According to second condition,
$$4x+6Y = \dfrac{1}{10}$$ .......Equation $$2$$
Multiplying Equation $$1$$ by $$4$$ and Equation $$2$$ by $$3$$,
$$12x + 16y = \dfrac{4}{14}$$ .....Equation $$3$$
$$12x + 18y = \dfrac{3}{10}$$ .....Equation $$4$$
Now solving Equation $$3$$ and $$4$$ simultaneously by substracting Equation $$3$$ by Equation $$4$$,
we get,
$$\therefore -2y = \dfrac{4}{14} -\dfrac{3}{10}$$

$$\therefore -2y = \dfrac{40-42}{140}$$

$$\therefore y = \dfrac{1}{140}$$
Putting the value of y in equation $$1$$, we get,
$$3x + \dfrac{4}{140} = \dfrac{1}{14}$$

$$\therefore x = \dfrac{1}{70}$$

Hence one man's one day work is $$\dfrac{1}{70}$$

$$\therefore$$ one man completes the whole work in $$70$$ days.

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Numerical Applications Test 48

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Numerical Applications Test 48

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SHARING IS CARING

Question 1

Arithmetic mean

Median

Mode

Frequency

Solution

Question 2

$$17$$ days and $$ 19$$ days

$$14$$ days and $$22$$ days

$$18$$ days and $$36$$ days

$$17$$ days and $$13$$ days

Solution

Question 3

$$120$$ men

$$170$$ men

$$180$$ men

$$150$$ men

Solution

Question 4

Pooja in $$44$$ days and Ritu in $$8$$ days

Pooja in $$80$$ days and Ritu in $$ 500$$ days

Pooja in $$20$$ days and Ritu in $$120 $$ days

Pooja in $$40$$ days and Ritu in $$30$$ days

Solution

Question 5

$$\displaystyle \frac{1}{2} (10!)$$

$$\displaystyle 9!$$

$$\displaystyle 10!$$

none of these

Solution

Question 6

$$2$$ days

$$4$$ days

$$6$$ days

$$8$$ days

Solution

Question 7

$$20$$ days

$$30$$ days

$$50$$ days

$$40$$ days

Solution

Question 8

$$A$$

$$B$$

$$C$$

Data inadequate

Solution

Question 9

$$16$$ days

$$24$$ days

$$36$$ days

$$48$$ days

Solution

Question 10

$$40$$ days

$$20$$ days

$$35$$ days

$$70$$ days

Solution

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