Numerical Applications Test 22

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Numerical Applications Test 22

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Question 1
1 / -0

If $$15$$ men can complete a piece of work in $$30$$ days, then how many days will it take for $$18$$ men to complete it?

A
$$22$$ days

B
$$25$$ days

C
$$15$$ days

D
$$29$$ days

Solution

More number of people, then less number of days.
Let $$18$$ men complete the work in $$x$$ days, then

$$18:30::15:x$$

$$\Longrightarrow 18\times x=15\times 30$$

$$\Longrightarrow x=\dfrac{15\times 30 }{18}$$

$$\Longrightarrow x=25$$

Hence, $$18$$ men can complete the work in $$25$$ days.
Question 2
1 / -0

A tea party is arranged for $$16$$ people along two sides of a large table with $$8$$ chairs on each side. Four men to sit on one particular side and two on the other side. In how many ways can they be seated?

A
$$\displaystyle\frac{8!\space8!\space10!}{(4!)^2\space6!}$$

B
$$\displaystyle\frac{8!\space8!\space10!}{6!}$$

C
$$\displaystyle\frac{8!\space8!\space10!}{4!!}$$

D
$$\displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$

Solution

There are $$8$$ chairs on each side of the table. Let sides be represented by $$A$$ and $$B$$. Let four persons sit on side $$A$$, then number of ways arranging $$4$$ persons on $$8$$ chairs on side $$A =

\space ^8P_4$$ and then two persons sit on side $$B$$, then the number of ways arranging $$2$$ persons on $$8$$ chairs on side $$B = \space

^8P_2$$ and arranging the remaining $$10$$ persons in remaining $$10$$ chairs in $$10!$$ ways.
Hence the total number of ways in which the persons can be arranged
$$\quad = \space ^8P_4\times\space^8P_2\times\space10!$$
$$\quad = \displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$
Question 3
1 / -0

In order to complete a work in $$28$$ days, $$60$$ men are required. How many men will be required if the same work is to be completed in $$40$$ days?

A
$$42$$

B
$$32$$

C
$$48$$

D
$$38$$

Solution

Less number of days then more number of people.
People and days are in inverse proportion.
Let $$x$$ men are required to complete the job in $$40$$ days.
as this is inverse proportion.
$$x:60::28:40$$

$$\Longrightarrow 40x=60\times 28$$

$$\Longrightarrow x=\dfrac{60\times 28}{40}$$

$$\Longrightarrow x=42$$

Hence, $$42$$ men are required to complete the job in $$40$$ days.
Question 4
1 / -0

$$50$$ labourers can dig a pond in $$16$$ days. How many labourers will be required to dig another pond in $$20$$ days, which is double in size?

A
$$80$$

B
$$30$$

C
$$70$$

D
$$20$$

Solution

$$\textbf{Step 1: Solve using the method of proportions }$$

$$\text{No. of labourers}$$ $$\text{No. of days}$$
$$\text{50}$$ $$\text{16}$$
$$\text{?(x)}$$ $$\text{20}$$

$$\text{There is an indirect relation between no. of days and no. of laborers.}$$

$$\therefore$$ $$\text{50 : x :: 20 : 16}$$

$$\Rightarrow \dfrac{50}{x} = \dfrac{20}{16}$$

$$\Rightarrow x = 40$$

$$\therefore$$ $$\text{Required no. of labourers to make it double in size}$$ $$= 40\times 2 = 80$$

$$\textbf{Hence, the answer is 80}$$
Question 5
1 / -0

Seven girls are to dance in a circle. In how many different ways can they stand on the circumference of the circle?

A
$$120$$

B
$$640$$

C
$$720$$

D
$$840$$

Solution

Since circular permutations of $$n$$ objects is $$(n-1)!$$
Hence, the number of ways is $$\quad (7-1)!=6!=720$$
Question 6
1 / -0

If it was Saturday on 17th December, 2002 then what was the day on 22nd December, 2004?

A
Monday

B
Tuesday

C
Wednesday

D
Sunday

Solution

Clearly, every day repeats itself on the seventh day. Now, 17th Dec. 2002 - 16th Dec.
2003 is a period of 365 days. Dividing by 7, we get 52 weeks and one day. Thus, the 365th day will be the same as the first day, i.e., 16th Dec. 2003 is also Saturday.
Now, 16th Dec. 2003 - 15th Dec. 2004 is a period of 366 days (because 2004, being a leap year, has 29 days in February). Thus, as shown above, 14th Dec. 2004 will be the same as 16th Dec. 2003, i.e., Saturday. So, 21st Dec. 2004 is also Saturday and thus, 22nd Dec. 2004 is a Sunday.
Question 7
1 / -0

If $$3$$ men or $$6$$ boys can finish a work in $$20$$ days , then how long will $$4$$ men and $$12$$ boys take to finish the same work ?

A
$$2$$ days

B
$$4$$ days

C
$$6$$ days

D
$$8$$ days

Solution

$$3$$ men $$ = 6$$ boys
So, $$1$$ man $$=2$$ boys
$$3$$ men or $$6$$ boys can finish a work in $$20$$ days.
$$4$$ men and $$12$$ boys $$= 4 +6$$ men $$=10$$ men.
$$3$$ men can finish a work in $$20$$ days.
$$10$$ men can finish the work in $$\dfrac{20\times3}{10}=6$$ days.
Question 8
1 / -0

The heights (in $$cm$$) of $$8$$ girls of a class are $$140, 142, 135, 133, 137, 150, 148$$ and $$138$$ respectively. Find the mean height of these girls.

A
$$137.375$$

B
$$139.375$$

C
$$140.375$$

D
$$143.375$$

Solution

Mean is the average of the values of the data set.
Given the height of the girls are $$140,142,135,133,137,150,148,138$$
The total number of girls is $$8$$

Therefore, $$\text{mean}=\dfrac{140+142+135+133+137+150+148+138}{8}$$

$$\implies \text{mean}=\dfrac{1123}{8}=140.375$$
Question 9
1 / -0

Six taps can fill an empty cistern in $$8$$ hours. How much more time will be taken, if two taps go out of order? Assume all the taps supply water at the same rate.

A
$$1$$ hour

B
$$2$$ hours

C
$$3$$ hours

D
$$4$$ hours

Solution

Given six taps fill an empty cistern in $$8$$ hours
If two taps go out of order then no. of taps $$= 6-2 = 4$$
Then $$\dfrac{6}{4}=\dfrac{x}{8}$$
$$\Longrightarrow 4x=6\times 8$$
$$\Longrightarrow x=\dfrac{48}{4}=12$$ hours
Then, more time $$= 12-8=4$$ hours.
Question 10
1 / -0

If $$12$$ men or $$18$$ women can complete a piece of work in $$7$$ days , then in how many days can $$4$$ men and $$8$$ women complete the same work ?

A
$$9$$ days

B
$$12$$ days

C
$$6$$ days

D
$$7$$ days

Solution

$$12$$ men $$=18$$ women
$$2$$ men $$=3$$ women
$$12$$ men or $$18$$ women can complete a piece of work in $$7$$ days.
$$4$$ men and $$8$$ women $$=6+8$$ women $$=14$$ women.
If $$18$$ women can complete a piece of work in $$7$$ days.
So, $$1$$ women can complete a piece of work in $$7\times18$$ days.
So, $$14$$ women can complete a piece of work in $$7\times\dfrac{18}{14}$$ days $$=9$$ days.

Hence, option A.

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Numerical Applications Test 22

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Numerical Applications Test 22

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SHARING IS CARING

Question 1

$$22$$ days

$$25$$ days

$$15$$ days

$$29$$ days

Solution

Question 2

$$\displaystyle\frac{8!\space8!\space10!}{(4!)^2\space6!}$$

$$\displaystyle\frac{8!\space8!\space10!}{6!}$$

$$\displaystyle\frac{8!\space8!\space10!}{4!!}$$

$$\displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$

Solution

Question 3

$$42$$

$$32$$

$$48$$

$$38$$

Solution

Question 4

$$80$$

$$30$$

$$70$$

$$20$$

Solution

Question 5

$$120$$

$$640$$

$$720$$

$$840$$

Solution

Question 6

Monday

Tuesday

Wednesday

Sunday

Solution

Question 7

$$2$$ days

$$4$$ days

$$6$$ days

$$8$$ days

Solution

Question 8

$$137.375$$

$$139.375$$

$$140.375$$

$$143.375$$

Solution

Question 9

$$1$$ hour

$$2$$ hours

$$3$$ hours

$$4$$ hours

Solution

Question 10

$$9$$ days

$$12$$ days

$$6$$ days

$$7$$ days

Solution

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