Numerical Applications Test 12

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Numerical Applications Test 12

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

Each of the questions given below consists of a statement and$$/$$ or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is$$/$$are sufficient to answer the given question. Read the both statements and
$$\bullet$$ Give answer(A) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
$$\bullet$$ Give answer(B) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
$$\bullet$$ Give answer(C) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
$$\bullet$$ Give answer(D) if the data even in both Statements I and II together are not sufficient to answer the question.
$$\bullet$$ Give answer(E) if the data in both Statements I and II together are necessary to answer the question.

...view full instructions

How much time will the leak take to empty the full cistern?
I. The cistern is normally filled in $$9$$ hours.
II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.

A
I alone sufficient while II alone not sufficient to answer

B
II alone sufficient while I alone not sufficient to answer

C
Either I or II alone sufficient to answer

D
Both I and II are not sufficient to answer

E
Both I and II are sufficient to answer

Solution

I. Time taken to fill the cistern without leak$$=9$$hours.
Part of cistern filled without leak in $$1$$ hour$$=\displaystyle\frac{1}{9}$$
II. Time taken to fill the cistern in presence of leak$$=10$$hours.
Net filling in $$1$$hour$$=\displaystyle\frac{1}{10}$$
Work done by leak in $$1$$ hour$$=\left(\displaystyle\frac{1}{9}-\frac{1}{10}\right)=\displaystyle\frac{1}{90}$$
$$\therefore$$ Leak will empty the full cistern in $$90$$ hours.
Clearly, both I and II are necessary to answer the question.
$$\therefore$$ Correct answer is (E).
Question 2
1 / -0

A watch which gains $$5$$ seconds in $$3$$ minutes was set right at $$7$$ a.m. In the afternoon of the same day, when the watch indicated quarter past $$4$$ o'clock, the true time is:

A
$$59\cfrac { 7 }{ 12 } $$ min. pase $$3$$

B
$$4$$ p.m.

C
$$58\cfrac { 7 }{ 11 } $$ min. past $$3$$

D
$$2\cfrac { 3 }{ 11 } $$ min. past $$4$$

Solution

Time from $$7$$ a.m. to $$4.15$$ p.m. $$=9hrs $$ $$15min.\cfrac{37}{4}hrs$$.
$$3min.$$ $$5sec.$$ of this clock $$=3min.$$ of the correct clock.
$$\Rightarrow \cfrac { 37 }{ 720 } hrs$$ of this clock $$=\cfrac{1}{20}hrs.$$ of the correct clock.
$$\Rightarrow \cfrac { 37 }{ 4 } hrs.$$ of this clock$$=\left( \cfrac { 1 }{ 20 } \times \cfrac { 720 }{ 37 } \times \cfrac { 37 }{ 4 } \right) hrs.$$ of the correct clock.
$$=9hrs.$$ of the correct clock.
$$\therefore$$ The correct time is $$9hrs.$$ after $$7$$ a.m. i.e., $$4$$p.m.
Question 3
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Three pipes A, B and C can fill a tank in $$6$$ hours. After working at it together for $$2$$ hours, C is closed and A and B can fill the remaining part in $$7$$ hours. The number of hours taken by C alone to fill the tank is.

A
$$10$$

B
$$12$$

C
$$14$$

D
$$16$$

Solution

Part time in $$2$$ hours$$=\displaystyle\frac{2}{6}=\frac{1}{3}$$
Remaining part$$=\left(\displaystyle 1-\frac{1}{3}\right)=\displaystyle\frac{2}{3}$$.
$$\therefore (A+B)$$'s $$7$$ hour's work$$=\displaystyle\frac{2}{3}$$
$$(A+B)$$'s $$1$$ hour's work$$=\displaystyle\frac{2}{21}$$
$$\therefore$$ C's hour's work$$=\{(A+B+C)$$'s $$1$$ hour's work$$\}-\{(A+B)$$'s $$1$$ hour's work$$\}$$
$$=\left(\displaystyle\frac{1}{6}-\frac{2}{21}\right)=\displaystyle\frac{1}{14}$$
$$\therefore$$ C alone can fill the tank in $$14$$ hours.
Question 4
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Two pipes A and B together can fill a cistern in $$4$$ hours. Had they been opened separately, then B would have taken $$6$$ hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

A
$$1$$ hour

B
$$2$$ hours

C
$$6$$ hours

D
$$8$$ hours

Solution

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fit it in $$(x+6)$$ hours.
$$\therefore \displaystyle\frac{1}{x}+\frac{1}{(x+6)}=\frac{1}{4}$$
$$\Rightarrow \displaystyle\frac{x+6+x}{x(x+6)}=\frac{1}{4}$$
$$\Rightarrow x^2-2x-24=0$$
$$\Rightarrow (x-6)(x+4)=0$$
$$\Rightarrow x=6$$. [neglecting the negative value of x]
Question 5
1 / -0

Two pipes A and B can fill a tank in $$20$$ and $$30$$ minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?

A
$$12$$ min

B
$$15$$ min

C
$$25$$ min

D
$$50$$ min

Solution

Part filled by A in $$1$$ min $$=\displaystyle\frac{1}{20}$$.
Part filled by B in $$1$$ min$$=\displaystyle\frac{1}{30}$$.
Part filled by $$(A+B)$$ in $$1$$ min$$=\left(\displaystyle\frac{1}{20}+\frac{1}{30}\right)=\displaystyle\frac{1}{12}$$.
$$\therefore$$ Both pipes can fill the tank in $$12$$ minutes.
Question 6
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One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in $$36$$ minutes, then the slower pipe alone will be able to fill the tank in.

A
$$81$$ min.

B
$$108$$ min.

C
$$144$$ min.

D
$$192$$ min.

Solution

Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in $$\displaystyle\frac{x}{3}$$ minuts.
$$\therefore \displaystyle\frac{1}{x}+\frac{3}{x}=\frac{1}{36}$$
$$\Rightarrow \displaystyle\frac{4}{x}=\frac{1}{36}$$
$$\Rightarrow x=144$$ min.
Question 7
1 / -0

A tank is filled in $$5$$ hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A lone take to fill the tank?

A
$$20$$ hours

B
$$25$$ hours

C
$$35$$ hours

D
Cannot be determined

E
None of these

Solution

Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take $$\displaystyle\frac{x}{2}$$ and $$\displaystyle\frac{x}{4}$$ hours respectively to fill the tank.
$$\therefore \displaystyle\frac{1}{x}+\frac{2}{x}+\frac{4}{x}=\displaystyle\frac{1}{5}$$
$$\Rightarrow \displaystyle\frac{7}{x}=\frac{1}{5}$$
$$\Rightarrow x=35$$ hrs.
Question 8
1 / -0

A tap can fill a tank in $$6$$ hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

A
$$3$$ hrs $$15$$ min

B
$$3$$hrs $$45$$min

C
$$4$$hrs

D
$$4$$hrs $$15$$min

Solution

Time taken by one tap to fill half of the tank$$=3$$ hrs.
Part time by the four taps in $$1$$ hour$$=\left(4\times \displaystyle\frac{1}{6}\right)=\displaystyle\frac{2}{3}$$.
Remaining part$$=\left(\displaystyle 1-\frac{1}{2}\right)=\displaystyle\frac{1}{2}$$.
$$\therefore \displaystyle\frac{2}{3}:\frac{1}{2}::1:x$$
$$\Rightarrow x=\left(\displaystyle\frac{1}{2}\times 1\times\frac{3}{2}\right)=\displaystyle\frac{3}{4}$$ hours i.e., $$45$$ mins.
So, total time taken $$=3$$hrs. $$45$$ mins.
Question 9
1 / -0

A large tanker can be filled by two pipes A and B in $$60$$ minutes and $$40$$ minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?

A
$$15$$ min

B
$$20$$ min

C
$$27.5$$ min

D
$$30$$ min

Solution

Part filled by $$(A+B)$$ in $$1$$ minute $$=\left(\displaystyle\frac{1}{60}+\frac{1}{40}\right)=\displaystyle\frac{1}{24}$$.
Suppose the tank is filled in x minutes.
Then, $$\displaystyle\frac{x}{2}\left(\displaystyle\frac{1}{24}+\frac{1}{40}\right)=1$$
$$\Rightarrow \displaystyle\frac{x}{2}\times \frac{1}{15}=1$$
$$\Rightarrow x=30$$ min.
Question 10
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Three taps A, B and C can fill a tank in $$12, 15$$ and $$20$$ hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in.

A
$$6$$ hours

B
$$6\displaystyle\frac{2}{3}$$hours

C
$$7$$ hours

D
$$7\displaystyle\frac{1}{2}$$ hours

Solution

$$(A+B)$$'s hour's work$$=\left(\displaystyle\frac{1}{12}+\frac{1}{15}\right)=\displaystyle\frac{9}{60}=\frac{3}{20}$$.
$$(A+C)$$'s hour's work $$=\left(\displaystyle\frac{1}{12}+\frac{1}{20}\right)=\displaystyle\frac{8}{60}=\frac{2}{15}$$.
Part filled in $$2$$hrs $$=\left(\displaystyle\frac{3}{20}+\frac{2}{15}\right)=\displaystyle\frac{17}{60}$$.
Part filled in $$6$$hrs$$=\left(\displaystyle 3\times \frac{17}{60}\right)=\displaystyle \frac{17}{20}$$.
Remaining part $$=\left(1-\displaystyle\frac{17}{20}\right)=\displaystyle\frac{3}{20}$$.
Noe, it is turn of A and B and $$\displaystyle\frac{3}{20}$$ part is filled by A and B in $$1$$ hour.
$$\therefore$$ Total time taken to fill the tank $$=(6+1)$$hrs $$=7$$hrs.

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Numerical Applications Test 12

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Question 1

I alone sufficient while II alone not sufficient to answer

II alone sufficient while I alone not sufficient to answer

Either I or II alone sufficient to answer

Both I and II are not sufficient to answer

Both I and II are sufficient to answer

Solution

Question 2

$$59\cfrac { 7 }{ 12 } $$ min. pase $$3$$

$$4$$ p.m.

$$58\cfrac { 7 }{ 11 } $$ min. past $$3$$

$$2\cfrac { 3 }{ 11 } $$ min. past $$4$$

Solution

Question 3

$$10$$

$$12$$

$$14$$

$$16$$

Solution

Question 4

$$1$$ hour

$$2$$ hours

$$6$$ hours

$$8$$ hours

Solution

Question 5

$$12$$ min

$$15$$ min

$$25$$ min

$$50$$ min

Solution

Question 6

$$81$$ min.

$$108$$ min.

$$144$$ min.

$$192$$ min.

Solution

Question 7

$$20$$ hours

$$25$$ hours

$$35$$ hours

Cannot be determined

None of these

Solution

Question 8

$$3$$ hrs $$15$$ min

$$3$$hrs $$45$$min

$$4$$hrs

$$4$$hrs $$15$$min

Solution

Question 9

$$15$$ min

$$20$$ min

$$27.5$$ min

$$30$$ min

Solution

Question 10

$$6$$ hours

$$6\displaystyle\frac{2}{3}$$hours

$$7$$ hours

$$7\displaystyle\frac{1}{2}$$ hours

Solution

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