Numerical Applications Test 10

Result

Numerical Applications Test 10

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$A$$ can finish a work in $$18$$ days and $$B$$ can do the same work in $$15$$ days. $$B$$ worked for $$10$$ days and left the job. In how many days, $$A$$ alone can finish the remaining work?

A
$$5$$

B
$$5\cfrac{1}{2}$$

C
$$6$$

D
$$8$$

Solution

$${\textbf{Step 1 : Write the values given in the question}}{\textbf{.}}$$

$$A{\text{ can finish a work in 18 days}}{\text{.}}$$

$$B{\text{ can do the work in 15 days}}{\text{.}}$$

$$B{\text{ worked for 10 days and left the job}}{\text{.}}$$

$${\textbf{Step 2 : Apply mathematics word problem rule}}{\textbf{.}}$$

$$B's{\text{ 10 day's work = }}\left( {\dfrac{1}{{15}} \times 10} \right) = \dfrac{2}{3}$$

$${\text{Remaining work = }}\left( {1 - \dfrac{2}{3}} \right) = \dfrac{1}{3}$$

$${\text{Now, }}\dfrac{1}{{18}}{\text{ work done by }}A{\text{ in 1 day}}{\text{.}}$$

$$\therefore \dfrac{1}{3}{\text{ work is done by }}A{\text{ in }}\left( {18 \times \dfrac{1}{3}} \right) = 6{\text{ days}}$$

$${\textbf{Hence, }}\mathbf{A}{\textbf{ can remaining work in 6 days}}{\textbf{.}}$$
Question 2
1 / -0

$$A$$ works twice as fast as $$B$$. If $$B$$ can complete a work in $$12$$ days independently, the number of days in which $$A$$ and $$B$$ can together finish the work in:

A
$$4$$ days

B
$$6$$ days

C
$$8$$ days

D
$$18$$ days

Solution

Ratio of rates of working of $$A$$ and $$B$$ $$=2:1$$
So, the ratio of time taken $$=1:2$$
$$B$$'s $$1$$ day's work $$=\cfrac{1}{12}$$
$$\therefore$$ $$A$$'s $$1$$ day's work $$=\cfrac{1}{6}$$; ($$2$$ times of $$B$$'s work)
$$(A+B)$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 6 } +\cfrac { 1 }{ 12 } \right) =\cfrac { 3 }{ 12 } =\cfrac { 1 }{ 4 } $$.
So, $$A$$ and $$B$$ together can finish the work in $$4$$ days.
Question 3
1 / -0

$$A$$ and $$B$$ can complete a work in $$15$$ days and $$10$$ days respectively. They started doing the work together but after $$2$$ days $$B$$ had to leave and $$A$$ alone completed the remaining work. The whole work was completed in:

A
$$8$$ days

B
$$10$$ days

C
$$12$$ days

D
$$15$$ days

Solution

$$(A+B)$$'s $$1$$ day's work $$\left( \cfrac { 1 }{ 15 } +\cfrac { 1 }{ 10 } \right) =\cfrac { 1 }{ 6 } $$
Work done by $$A$$ and $$B$$ in $$2$$ days $$=\left( \cfrac { 1 }{ 6 } \times 2 \right) =\cfrac { 1 }{ 3 } $$
Remaining work $$=\left( 1-\cfrac { 1 }{ 3 } \right) =\cfrac { 2 }{ 3 } $$
Now, $$\cfrac{1}{15}$$ work done by $$A$$ in $$1$$ day.
$$\therefore$$ $$\cfrac{2}{3}$$ work will be done by $$A$$ in $$\left( 15\times \cfrac { 2 }{ 3 } \right) \quad =10$$ days.
Hence, the total time taken $$=(10+2)=12$$ days.
Question 4
1 / -0

$$A$$ does $$80$$% of a work in $$20$$ days. He then calls in $$B$$ and they together finish the remaining work in $$3$$ days. How long $$B$$ alone would take to do the whole work?

A
$$23$$ days

B
$$37$$ days

C
$$37\cfrac{1}{2}$$

D
$$40$$ days

Solution

Whole work is done by $$A$$ in $$\left( 20\times \cfrac { 5 }{ 4 } \right) =25$$ days.
Now, $$\left( 1-\cfrac { 4 }{ 15 } \right) $$ i.e., $$\cfrac{1}{5}$$ work is done by $$A$$ and $$B$$ in $$3$$ days.
Whole work will be done by $$A$$ and $$B$$ in $$(3\times 5)$$ $$=15$$ days.
$$A$$'s $$1$$ day's work $$=\cfrac{1}{25}$$, $$(A+B)$$'s $$1$$ day's work $$=\cfrac{1}{15}$$.
$$\therefore$$ $$B$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 15 } -\cfrac { 3 }{ 25 } \right) =\cfrac { 4 }{ 150 } =\cfrac { 2 }{ 75 } $$
So, $$B$$ alone would do the work in $$\cfrac{75}{2}=37\cfrac{1}{2}$$ days.
Question 5
1 / -0

Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

A
$$3:4$$

B
$$4:3$$

C
$$5:3$$

D
Data inadequate

Solution

$$(20\times 16)$$ women can complete the work in $$1$$ day.
$$\therefore$$ $$1$$ women's $$1$$ day's work $$=\cfrac{1}{320}$$.
$$(16\times 15)$$ men can complete the work in $$1$$ day.
$$\therefore$$ $$1$$ man's $$1$$ day's work $$=\cfrac{1}{240}$$
So, required ratio $$=\cfrac{1}{240}:\cfrac{1}{320}$$
$$=\cfrac{1}{3}:\cfrac{1}{4}$$
$$=4:3$$ (cross multiplied)
Question 6
1 / -0

$$4$$men and $$6$$ women can complete a work in $$8$$ days, while $$3$$ men and $$7$$ women can complete it in $$10$$ days. In how many days will $$10$$ women complete it?

A
$$35$$

B
$$40$$

C
$$45$$

D
$$50$$

Solution

Let $$1$$ man's $$1$$ day's work $$=x$$ and $$1$$ woman's $$1$$ day's work $$=y$$.
Then, $$4x+6y=\cfrac{1}{8}$$ and $$3x+7y=\cfrac{1}{10}$$.
Solving the two equations, we get: $$x=\cfrac{11}{400}, y=\cfrac{1}{400}$$
$$\therefore$$ $$1$$ woman's $$1$$ day's work $$=\cfrac{1}{400}$$.
$$\Rightarrow$$ $$10$$ women's $$1$$ day's work $$=\left( \cfrac { 1 }{ 400 } \times 10 \right) =\cfrac { 1 }{ 40 } $$.
Hence, $$10$$ women will complete the work in $$40$$ days.
Question 7
1 / -0

$$A$$ can finish a work in $$24$$ days, $$B$$ in $$9$$ days and $$C$$ in $$12$$ days. $$B$$ and $$C$$ start the work but are force to leave after $$3$$ days. The remaining work was done by $$A$$ in:

A
$$5$$ days

B
$$6$$ days

C
$$10$$ days

D
$$10\cfrac{1}{2}$$ days

Solution

$$(B+C)$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 9 } +\cfrac { 1 }{ 12 } \right) =\cfrac{7}{36}$$
Work done by $$B$$ and $$C$$ in $$3$$ days $$=\left( \cfrac { 7 }{ 36 } \times 3 \right) =\cfrac { 7 }{ 12 } $$
Remaining work $$=\left( 1-\cfrac { 7 }{ 12 } \right) =\cfrac { 5 }{ 12 } $$
Now, $$\cfrac{1}{24}$$ work is done by $$A$$ in $$1$$ day.
So, $$\cfrac{5}{12}$$ work done by $$A$$ in $$\left( 24\times \cfrac { 5 }{ 12 } \right) =10$$ days.
Question 8
1 / -0

$$A$$ and $$B$$ can do a piece of work in $$30$$ days, while $$B$$ and $$C$$ can do the same work in $$24$$ days and $$C$$ and $$A$$ in $$20$$ days. They all work together for $$10$$ days when $$B$$ and $$C$$ leave. How many days more will $$A$$ take to finish the work?

A
$$18$$ days

B
$$24$$ days

C
$$30$$ days

D
$$36$$ days

Solution

$$2(A+B+C)$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 30 } +\cfrac { 1 }{ 24 } +\cfrac { 1 }{ 20 } \right) =\cfrac { 15 }{ 120 } =\cfrac { 1 }{ 8 } $$.
Therefore, $$(A+B+C)$$'s $$1$$ day's work $$=\cfrac{1}{2\times 8}=\cfrac{1}{6}$$.
Work done by $$A,B,C$$ in $$10$$ days $$=\cfrac{10}{16}=\cfrac{5}{8}$$
Remaining work $$=\left( 1-\cfrac { 5 }{ 8 } \right) =\cfrac { 3 }{ 8 } $$.
$$A$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 16 } -\cfrac { 1 }{ 24 } \right) =\cfrac { 1 }{ 48 } $$.
Now, $$\cfrac{1}{8}$$ work done by $$A$$ in $$1$$ day.
So, $$\cfrac{3}{8}$$ work will be done by $$A$$ in $$\left( 48\times \cfrac { 3 }{ 8 } \right) =18$$ days.
Question 9
1 / -0

$$X$$ can do a piece work in $$40$$ days. He works at it for $$8$$ days and then $$Y$$ finished it in $$16$$ days. How long will they together take to complete the work?

A
$$13\cfrac{1}{3}$$ days

B
$$15$$ days

C
$$20$$ days

D
$$26$$ days

Solution

Work done by $$X$$ in $$8$$ days $$=\left( \cfrac { 1 }{ 40 } \times 8 \right) =\cfrac { 1 }{ 5 } $$.
Remaining work $$=\left( 1-\cfrac { 1 }{ 5 } \right) =\cfrac { 4 }{ 5 } $$.
Now, $$\cfrac{4}{5}$$ work done by $$Y$$ in $$16$$ days.
Whole work will be done by $$Y$$ in $$\left( 16\times \cfrac { 5 }{ 4 } \right) =20$$ days.
$$\therefore$$ $$X$$'s $$1$$ day's work $$=\cfrac{1}{40}$$,
$$Y$$'s $$1$$ day's work $$=\cfrac{1}{20}$$.
$$(X+Y$$)'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 40 } +\cfrac { 1 }{ 20 } \right) =\cfrac { 3 }{ 40 } $$.
Hence $$X$$ and $$Y$$ will together complete the work in $$\left( \cfrac { 40 }{ 3 } \right) =13\cfrac{1}{13}$$ days.
Question 10
1 / -0

$$A$$ and $$B$$ can do a work in $$8$$ days, $$B$$ and $$C$$ can do the same work in $$12$$ days. $$A,B$$ and $$C$$ together can finish it in $$6$$ days. $$A$$ and $$C$$ together will do it in:

A
$$4$$ days

B
$$6$$ days

C
$$8$$ days

D
$$12$$ days

Solution

$$(A+B+C)$$'s $$1$$ day's work $$=\cfrac{1}{6}$$;
$$(A+B)$$'s $$1$$ day's work $$=\cfrac{1}{8}$$;
$$(B+C)$$'s $$1$$ day's work $$=\cfrac{1}{12}$$;
$$\therefore$$ $$(A+C)$$'s $$1$$ day's work $$=\left( 2\times \cfrac { 1 }{ 6 } \right) -\left( \cfrac { 1 }{ 8 } +\cfrac { 1 }{ 12 } \right) $$
$$=\left( \cfrac { 1 }{ 3 } -\cfrac { 5 }{ 24 } \right) =\cfrac { 3 }{ 24 } =\cfrac { 1 }{ 8 } $$.
So, $$A$$ and $$C$$ together will do the work in $$8$$ days.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Numerical Applications Test 10

Result

Numerical Applications Test 10

Score

-

Rank

-

SHARING IS CARING

Question 1

$$5$$

$$5\cfrac{1}{2}$$

$$6$$

$$8$$

Solution

Question 2

$$4$$ days

$$6$$ days

$$8$$ days

$$18$$ days

Solution

Question 3

$$8$$ days

$$10$$ days

$$12$$ days

$$15$$ days

Solution

Question 4

$$23$$ days

$$37$$ days

$$37\cfrac{1}{2}$$

$$40$$ days

Solution

Question 5

$$3:4$$

$$4:3$$

$$5:3$$

Data inadequate

Solution

Question 6

$$35$$

$$40$$

$$45$$

$$50$$

Solution

Question 7

$$5$$ days

$$6$$ days

$$10$$ days

$$10\cfrac{1}{2}$$ days

Solution

Question 8

$$18$$ days

$$24$$ days

$$30$$ days

$$36$$ days

Solution

Question 9

$$13\cfrac{1}{3}$$ days

$$15$$ days

$$20$$ days

$$26$$ days

Solution

Question 10

$$4$$ days

$$6$$ days

$$8$$ days

$$12$$ days

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.