Numerical Applications Test 51

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Numerical Applications Test 51

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Question 1
1 / -0

It takes pump (A) $$4$$ hours to empty a swimming pool. It takes pump (B) $$6$$ hours to empty the same swimming pool. If the two pumps are started together, at what time will the two pumps have emptied $$50$$% of the water in the swimming pool?

A
$$1\ hour\ 12\ minutes$$

B
$$1\ hour\ 20\ minutes$$

C
$$2\ hour\ 30\ minutes$$

D
$$3\ hours$$

E
$$5\ hours$$

Solution

Work efficiency of pump A in emptying pool = $$\dfrac{1}{4}$$ per hour
Work efficiency of pump B in emptying pool = $$\dfrac{1}{6}$$ per hour
$$\therefore$$ work done by A and B together = $$\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{10}{24}$$per hour
Hence time taken to fill the tank $$100$$% = $$\dfrac{24}{10}$$ hours
$$\therefore$$ time taken to fill $$50$$% = $$\dfrac{12\times 50}{5\times100}$$ = $$1.2$$ hours
Since, $$1 hour = 60 minutes$$
$$\therefore 1.2$$ hours= $$1.2\times 60 = 72$$ minutes
$$72$$ minutes = $$(60 +12)$$ minutes = $$1 hour$$ $$12 minutes$$
So, the answer is option A
Question 2
1 / -0

The average (mean) number of children per family in the village is approximately.
Number of children $$0$$ $$1$$ $$2$$ $$3$$
Number of families $$13$$ $$24$$ $$36$$ $$27$$
Data were collected on the number of children per family in a certain village and tabulated as shown.

A
$$2.4$$

B
$$2.0$$

C
$$1.8$$

D
$$1.5$$

E
$$1.0$$

Solution
Question 3
1 / -0

Seven person $$P_1,P_2......, P_7$$ initially seated at chairs $$C_1,C_2,.....C_7$$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $$P_1$$, sits on $$C_2$$ and $$P_2$$ sits on $$C_3$$ ?

A
$$52$$

B
$$53$$

C
$$54$$

D
$$55$$

Solution
Question 4
1 / -0

If $$m$$ denotes the number of $$5$$ digit numbers if each successive digits are in their descending order of magnitude and $$n$$ is the corresponding figure. When the digits and in their ascending order of magnitude then $$(m-n)$$ has the value

A
$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

B
$${ _{ }^{ 9 }{ C } }_{ 5 }$$

C
$${ _{ }^{ 10 }{ C } }_{ 3 }$$

D
$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

$$ {\textbf{Step 1: Find m}} $$

$$ {\text{For m,}} $$

$$ {\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}} $$

$$ {\text{Number of ways = }}{}^{10}{{\text{C}}_5} $$

$$ {\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}} $$

$$ {\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{ 1 = }}{}^{10}{{\text{C}}_5} $$

$$ {\textbf{Step 2: Find n}} $$

$$ {\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}} $$

$$ {\text{We can't select zero as first digit because then the number won't be a 5 - digit number}}{\text{.}} $$

$$ {\text{Therefore number of ways = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{ 1 = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5} $$

$$ {\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r} $$

$$   \Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}} $$

$$   \Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4} $$

$$ {\text{Hence, m - n = }}{}^9{{\text{C}}_4} $$

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

It 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children together can do half of the work in :

A
6 months

B
4 months

C
5 months

D
8 months

Solution

Let total work be $$W$$
Work done by $$1$$ man in $$1$$ month = $$\frac { W }{ 10*7 } = \frac { W }{ 70} $$
Work done by $$1$$ woman in $$1$$ month = $$\frac { W }{ 20*7 } = \frac { W }{ 140} $$
Work done by $$1$$ child in $$1$$ month = $$\frac { W }{ 40*7 } = \frac { W }{ 280} $$
Total work done by 5 men, 5 women and 5 children in 1 month= $$\frac{5W}{70}+\frac{5W}{140}+\frac{5W}{280}= \frac{W}{8}$$
Let number of months be $$n$$
$$\frac{W}{8}*n=\frac{W}{2}$$
n=4
Question 6
1 / -0

Raman can do a work in $$5$$ days, Jatin can do the same work in $$7$$ days and Sachin can do the same work in $$9$$ days. If they do the same work together and they are paid Rs. $$2860$$, then what is the share (in Rs.) of Raman?

A
$$1260$$

B
$$700$$

C
$$900$$

D
$$870$$

Solution

Let the total work be $$315$$ [L.C.M of $$(5,7,9)$$] units. Raman's. Jatin's and Sachin's each day work is $$63,45$$ and $$35$$ units.
Hence, Ratio of share $$63:45:35$$
Raman's share $$=\cfrac { 63 }{ 143 } \times 2860=Rs.1260\quad $$
Question 7
1 / -0

What was the average number of children taking swim lessons from 1990 to 1995?

A
250

B
308

C
385

D
450

E
1,850

Solution
Question 8
1 / -0

If four men working at the same rate can do $$2/3$$ of a job in $$40$$ minutes it take one man working at this rate to do $$2/5$$ of the job takes how many minutes?

A
$$80$$

B
$$88$$

C
$$92$$

D
$$96$$

E
$$112$$

Solution
Question 9
1 / -0

A library has $$'a'$$ copies of one book, $$'b'$$ copies of each of two books, $$'c'$$ copies of each of three books, and single copy each of $$'d'$$. The total number of ways in which these books can be arranged in a row is

A
$$\dfrac{(a+b+c+d)!}{a!b!c!}$$

B
$$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$$

C
$$\dfrac{(a+2b+3c+d)!}{a!b!c!}$$

D
none of these

Solution
Question 10
1 / -0

If $$18$$ pumps can raise $$2170$$ tonnes of water in $$10$$ days working $$7$$ hours a day then in how many days will $$16$$ pumps raise $$1736$$ tonnes of water working $$9$$ hours a day

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

$$18$$ pumps raise $$2170$$ tonnes in $$10$$ days working $$7$$ hours a day.

In $$1$$ day they raise $$\dfrac{2170}{10}=217$$ tonnes

In $$1$$ hour they raise $$\dfrac{217}{7}=31$$ tonnes.

$$1$$ pump raises water $$= \dfrac{31}{18}$$ tonnes.

Now,

$$16$$ pumps will raise $$=\dfrac{31}{18}\times 16$$ tonnes.

Working $$9$$ hrs a day they will raise $$=\dfrac{31}{18}\times 16\times 9$$ tonnes

Let the number of days be $$d$$ days.

Thus total $$=\dfrac{31}{18}\times 16\times 9\times d$$

$$1736=\dfrac{31}{18}\times 16\times 9\times d$$

$$\Rightarrow d=1736 \times \dfrac{18}{9\times 16\times 31}=7$$ days.

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Number of children	$$0$$	$$1$$	$$2$$	$$3$$
Number of families	$$13$$	$$24$$	$$36$$	$$27$$

Numerical Applications Test 51

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Numerical Applications Test 51

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SHARING IS CARING

Question 1

$$1\ hour\ 12\ minutes$$

$$1\ hour\ 20\ minutes$$

$$2\ hour\ 30\ minutes$$

$$3\ hours$$

$$5\ hours$$

Solution

Question 2

$$2.4$$

$$2.0$$

$$1.8$$

$$1.5$$

$$1.0$$

Solution

Question 3

$$52$$

$$53$$

$$54$$

$$55$$

Solution

Question 4

$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

$${ _{ }^{ 9 }{ C } }_{ 5 }$$

$${ _{ }^{ 10 }{ C } }_{ 3 }$$

$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

Question 5

6 months

4 months

5 months

8 months

Solution

Question 6

$$1260$$

$$700$$

$$900$$

$$870$$

Solution

Question 7

250

308

385

450

1,850

Solution

Question 8

$$80$$

$$88$$

$$92$$

$$96$$

$$112$$

Solution

Question 9

$$\dfrac{(a+b+c+d)!}{a!b!c!}$$

$$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$$

$$\dfrac{(a+2b+3c+d)!}{a!b!c!}$$

none of these

Solution

Question 10

$$6$$

$$7$$

$$8$$

$$9$$

Solution

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