Numerical Applications Test 45

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Numerical Applications Test 45

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Weekly Quiz Competition

Question 1
1 / -0

Small tap require $$18\ hrs$$. to fill the tank white big tap require $$12\ hrs$$. to fill the tank. In how many hrs. both taps would fill the tank together?

A
$$7\ hrs$$

B
$$7\dfrac{1}{5}\ hr$$

C
$$8\ hr$$

D
$$9\ hr$$

Solution
Question 2
1 / -0

A and B can do a piece of work in $$8$$ days and
$$12$$ days respectively. How long will they take
to finish it they work together?

A
$$2\frac{3}{5}$$ days

B
$$4\frac{4}{5}$$ days

C
$$5\frac{4}{5}$$ days

D
none of there

Solution

$$ A \rightarrow 8 $$ days $$ ; B\rightarrow 12 $$ days.
speed of $$ A \rightarrow \frac{x}{8} ;$$ speed of $$ B \rightarrow \frac{x}{12}.$$
$$ \therefore A+B \rightarrow \frac{x}{8}+\frac{x}{12} = \frac{3+2}{24}x = \frac{5}{24}x $$
$$ \therefore $$ No. of days required $$ = \frac{24}{5} = 4\frac{4}{5}$$ days
Question 3
1 / -0

$$18$$ men can reap a field in $$35$$ days. For reaping the same field in $$15$$ days, how many men are required?

A
$$42$$

B
$$28$$

C
$$32$$

D
$$40$$

Solution

18 Men $$\rightarrow $$ 35 days.
1 Men $$\rightarrow 35\times 18$$ days.
$$\therefore $$ Reaping in 15 days -
No. of men = $$\frac{35\times 18}{15}= 42$$ men.
Question 4
1 / -0

Working together, pipes $$A$$ and $$B$$ can fill an empty tank in $$10\ hours$$. they worked together for $$4$$ hours and then $$B$$ stopped and $$A$$ continued filling the tank till was full. It took a total of $$13\ hours$$ to fill the tank. How long would it take $$A$$ to fill the empty tank alone?

A
$$13\ hours$$

B
$$15\ hours$$

C
$$17\ hours$$

D
$$18\ hours$$

Solution
Question 5
1 / -0

If $$200$$ students take $$40$$ days to complete the project, how much time will be taken by $$250$$ students?

A
$$31$$ days

B
$$32$$ days

C
$$23$$ days

D
$$21$$ days

Solution

$$200$$ students $$\rightarrow $$ $$40$$ days.
Rate work of 1 student $$\Rightarrow (\frac{x}{40})\times \frac{1}{200}$$
rate of $$250$$ students
$$\Rightarrow \frac{x\times 250}{40\times 200}=\frac{x}{32}$$
$$\therefore $$ Time required = $$32$$ hours.
Question 6
1 / -0

The number of seven letter words that can be formed by using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together is

A
$$24$$

B
$$18$$

C
$$54$$

D
none of these

Solution

using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together
let two C's be 1 unit
$$\therefore $$ no. of ways $$=\frac{6!}{4!}=30$$
We have to put 1 letter between one S
So, No of ways$$=2\times 3!=12$$
No. of ways=30-12=18
Question 7
1 / -0

4 men and 6 woman get Rs 1600 by doing a piece of work in 5 days. 3 men and 7 woman get Rs 1740 by doing the same work in 6 days. In how many days, 7 men and 6 woman can complete the same work getting Rs 3760?

A
$$6$$ days

B
$$8$$ days

C
$$10$$ days

D
$$12$$ days

Solution

In $$5$$ days ,($$4$$ men$$+6$$ women) get $$= $$Rs $$1600$$
$$\therefore$$, In $$1$$ day, ($$4$$ men$$+6$$ women) get $$=\dfrac { 1600 }{ 5 } =$$Rs.$$320$$ ...$$(1)$$
In $$1$$ day, number of persons to get Re.$$1=\dfrac { 320 }{ 4men+6women }$$ ....$$(2)$$
Similarly, in second condition,
In $$1$$ day, number of person to get Re.$$1$$
$$=\dfrac { 1740 }{ 6\times \left( 3men+7women \right) } =\dfrac { 290 }{ 3men+7women }$$ ...$$(3)$$
From Eqs.$$(2)$$ and $$(3)$$ We get
$$\dfrac { 320 }{ 3men+7women } =\dfrac { 290 }{ 3men+7women } $$
$$96$$ men$$+224$$women$$=116$$ men$$+174 $$women
$$\Rightarrow 20men=50women $$
$$\Rightarrow \dfrac { man }{ woman } =\dfrac { 5 }{ 2 } $$
$$\therefore, 1$$ woman$$=\dfrac { 2 }{ 5 }$$ man
from Eq.$$(2)$$ in $$1$$ day,
$$\left( 4men+6women \right) =\left( 4men+6\times \dfrac { 2 }{ 5 } m \right) $$ $$=\dfrac { 32 }{ 5 } men\quad get\quad Rs.320$$
In $$1$$ day , $$1$$ man get$$\dfrac { 320\times 5 }{ 32 } =Rs.50$$
$$\therefore$$ In $$1$$ day, $$1$$ woman get $$\dfrac { 50\times 2 }{ 5 } =Rs.20$$
$$\therefore$$, in $$1$$ day $$\left( 7men+6women \right) get\quad 7\times 50+6\times 20=Rs.470$$
Let, required number of days $$=\dfrac { 3760 }{ 470 } =8$$days
Question 8
1 / -0

Pipe $$A$$ can fill one fourth of a tank in $$5$$ hours and pipe $$B$$ can empty the full tank in $$30$$ hours.If both pipes are opened simultaneously, then in how many hours will the empty tank be filled.

A
$$10$$

B
$$60$$

C
$$25$$

D
$$45$$

Solution

Pipe can do $$\frac{1}{4}$$ work in 5 hours.
$$\Rightarrow $$ pipe can do $$\frac{1}{20}$$ work in 1 hours.
when both are opened simultaneously,
they can do $$ \frac{1}{20} \frac{-1}{30} = \frac{1}{60}$$ work in 1 hours
So, when opened together,
it takes 60 hours to fill the tank.
Question 9
1 / -0

$$6$$ pipes are required to fill a tank in $$1$$ hour $$20$$ minutes. How long will it take if only $$5$$ pipes of the same type are used?

A
$$56$$ minutes

B
$$80$$ minutes

C
$$96$$ minutes

D
$$72$$ minutes

Solution

It is given that $$6$$ pipes are required to fill a tank in $$1$$ hour $$20$$ minutes.
$$6$$ pipes can fill a tank in minutes $$=(60+20)min=80min$$
Let $$5$$ pipes can fill a tank in $$x\,min.$$
Since, it is an inverse proportion,
$$\Rightarrow$$ $$6\times 80=5\times x$$
$$\Rightarrow$$ $$x=\dfrac{6\times 80}{5}$$
$$\Rightarrow$$ $$x=\dfrac{480}{5}$$
$$\therefore$$ $$x=96$$ minutes
$$\therefore$$ $$5$$ pipes can fill a tank in $$96\,min.$$
Question 10
1 / -0

A train 100 meters long takes 6 seconds to cross a man walking at 5km/h in a direction opposite to that of the train. The speed of the train is

A
50 km/h

B
55 km /h

C
57 km/h

D
60 km/h

Solution

$$\begin{array}{l} Given, \\ speed=100 \\ time=6\, \sec \\ speed\, \, of\, \, train=\frac { { 100 } }{ 6 } -speed\, \, \, of\, \, man \\ =16.66-1.38\, \, m/s \\ =15.286\, \, m/s \end{array}$$
$$=15.286 \times \frac{{18}}{5} = 55km/h$$