Numerical Applications Test 39

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Numerical Applications Test 39

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Question 1
1 / -0

Three pipes $$A, B$$ and $$C$$ can fill a tank from empty to full in $$30$$ minutes, $$20$$ minutes, and $$10$$ minutes respectively. When the tank is empty, all the three pipes are opened. $$A, B$$ and $$C$$ discharge chemical solutions $$P, Q$$ and $$R$$ respectively. What is the proportion of the solution $$R$$ in the liquid in the tank after $$3$$ minutes?

A
$$\displaystyle\frac{5}{11}$$

B
$$\displaystyle\frac{6}{11}$$

C
$$\displaystyle\frac{7}{11}$$

D
$$\displaystyle\frac{8}{11}$$

Solution

Rate of filling the tank for A: $$\dfrac{1}{30}$$
Rate of filling the tank for B: $$\dfrac{1}{20}$$
Rate of filling the tank for C: $$\dfrac{1}{10}$$

Part filled by $$(A+B+C)$$ in $$3$$ minutes $$3\displaystyle\left(\displaystyle\frac{1}{30}+\frac{1}{20}+\frac{1}{10}\right)$$ $$=\displaystyle\left( 3\times \frac{11}{60}\right)=\frac{11}{20}$$.
Part filled by C in $$3$$ minutes $$=\displaystyle\frac{3}{10}$$.

$$\therefore$$ Required ratio$$=\left(\displaystyle\frac{3}{10}\times \frac{20}{11}\right)=\displaystyle \frac{6}{11}$$.
Question 2
1 / -0

39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?

A
10

B
13

C
14

D
15

Solution

$${\textbf{Step -1: Here, the work is same for both the condition}}{\text{.}}$$

$${\text{Let the required number of days be}}$$ $$x$$ $${\text{when 30 persons working 6 hours a day}}{\text{.}}$$

$${\text{Consider,}}$$ $${M_1} = 39$$ $${\text{persons,}}$$ $$ $$ $${D_1} = 12$$ $${\text{days,}}$$ $${H_1} = 5$$ $${\text{hours a day}}$$
$${M_2} = 30$$ $${\text{persons,}}$$ $${D_2} = x$$ $${\text{days,}}$$ $${H_2} = 6$$ $${\text{hours a day}}$$

$${\text{We know that,}}$$
$${\text{Number of days to complete a task inversely proportional to number of men }}$$
$${\text{and both are directly proportional to work. }}$$
$$\Rightarrow{{M_1}{D_1}{H_1}} = {{M_2}{D_2}{H_2}} \ldots \left( 1 \right)$$

$${\textbf{Step -2: Substitute the known values in equation }}\left(\mathbf 1 \right)$$

$$ \Rightarrow{39 \times 12 \times 5}= {30 \times x \times 6}$$

$$ \Rightarrow \dfrac{{39 \times 12 \times 5}}{{30 \times 6}} = x$$

$$ \Rightarrow x = 13$$
$${\textbf{Hence, option B. is correct.}}$$
Question 3
1 / -0

Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

A
648

B
1800

C
2700

D
10800

Solution

Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)
$$\left.\begin{matrix}Machines & 6 : 10 \\ Time (in minutes) & 1 : 4 \end{matrix}\right\} :: 270 : x$$
$$\therefore 6 \times 1 \times x = 10 \times 4 \times 270$$
$$\Rightarrow x = \dfrac{(10 \times 4 \times 270)}{(6)}$$
$$\Rightarrow x = 1800$$
Question 4
1 / -0

$$A$$ alone can do a piece of work in $$6$$ days and $$B$$ alone in $$8$$ days. $$A$$ and $$B$$ undertook to do it for Rs.$$3200$$. With the help of $$C$$, they completed the work in $$3$$ days. How much is to be paid to $$C$$?

A
Rs.$$375$$

B
Rs.$$400$$

C
Rs.$$600$$

D
Rs.$$800$$

Solution

$$C$$'s $$1$$ day's work $$=\cfrac{1}{3}-\left( \cfrac { 1 }{ 6 } +\cfrac { 1 }{ 8 } \right) =\cfrac { 1 }{ 3 } -\cfrac { 7 }{ 24 } =\cfrac { 1 }{ 24 } $$.
$$A$$'s Wages: $$B$$'s wages: $$C$$'s wages $$=\cfrac{1}{6}:\cfrac{1}{8}:\cfrac{1}{24}=4:3:1$$.
$$\therefore$$ $$C$$'s share (for $$3$$ days)$$=$$ Rs. $$\left( 3\times \cfrac { 1 }{ 24 } \times 3200 \right) =$$ Rs.$$400$$.
Question 5
1 / -0

If $$6$$ men and $$8$$ boys can do a piece of work in $$10$$ days while $$26$$ men and $$48$$ boys can do the same in $$2$$ days, the time taken by $$15$$ men and $$20$$ boys in doing the same type of work will be:

A
$$4$$ days

B
$$5$$ days

C
$$6$$ days

D
$$7$$ days

Solution

Given that
6 men and 8 boys can do a piece of work in 10 days
26 men and 48 boys can do the same in 2 days
As the work done is equal,
$$10(6M+8B)=2(26M+48B)$$
$$60M+80B=52M+96B$$
$$M=2B$$
$$B=\dfrac{M}{2}..........(1)$$
Now put (1) in $$15M+20B$$
$$\Rightarrow 15M+10M=25M$$
Now,
$$(6M+4M)10=100M$$
Then,
$$D(25M)=100M$$
$$D=4 days$$
Therefore 15 men and 20 boys can do the work in 4 days.
Question 6
1 / -0

$$A, B$$ and $$C$$ can do a piece of work in $$20,30$$ and $$60$$ respectively. In how many days can $$A$$ do the work if he is assisted by $$B$$ and $$C$$ on every third day?

A
$$12$$ days

B
$$15$$ days

C
$$16$$ days

D
$$18$$ days

Solution

$$A$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 20 } \times 2 \right) =\cfrac { 1 }{ 10 } $$
$$(A+B+C)$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 20 } +\cfrac { 1 }{ 30 } +\cfrac { 1 }{ 60 } \right) =\cfrac { 6 }{ 60 } =\cfrac { 1 }{ 10 } $$
Work done in $$3$$ days $$=\left( \cfrac { 1 }{ 10 } +\cfrac { 1 }{ 10 } \right) =\cfrac { 1 }{ 5 } $$
Now, $$\cfrac{1}{5}$$ work is done in $$3$$ days.
$$\therefore$$ Whole work will be done in $$(3\times 5)=15$$ days.
Question 7
1 / -0

An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $$1\frac{2}{3}$$ hours, it must travel at a speed of:

A
300 kmph

B
360 kmph

C
600 kmph

D
720 kmph

Solution

Distance = (240 $$\times$$ 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write $$1\frac{2}{3}$$ hours as 5/3 hours]
$$\therefore $$ Required speed = $$\left ( 1200 \times \frac{3}{5} \right )$$ km/hr=720 km/hr.
Question 8
1 / -0

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

A
1

B
$$\dfrac{1}{40}$$

C
40

D
80

Solution

Let the required number of days be x.
Less cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)
$$\left.\begin{matrix}Cow & 1 : 40 \\ Bags & 40 : 1 \end{matrix}\right\} :: 40 : x$$
$$\therefore 1 \times 40 \times x = 40 \times 1 \times 40$$
$$\Rightarrow x = 40$$
Question 9
1 / -0

$$A$$ takes twice as much time as $$B$$ or thrice as much time as $$C$$ to finish a piece of work. Working together, they can finish the work in $$2$$ days. $$B$$ can do the work alone in:

A
$$4$$ days

B
$$6$$ days

C
$$8$$ days

D
$$12$$ days

Solution

Suppose $$A,B$$ and $$C$$ take, $$x,\cfrac{x}{2}$$ and $$\cfrac{x}{3}$$ days respectively finish the work.
Then, $$\left( \cfrac { 1 }{ x } +\cfrac { 2 }{ x } +\cfrac { 3 }{ x } \right) =\cfrac { 1 }{ 2 } $$
$$\Rightarrow$$ $$\cfrac{6}{x}=\cfrac{1}{2}$$
$$\Rightarrow$$ $$x=12$$
So, $$B$$ takes $$(12/2)=6$$ days to finish the work.
Question 10
1 / -0

A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. The number of days for which the remaining food will last, is

A
$$29 \dfrac{1}{5}$$

B
$$37 \dfrac{1}{4}$$

C
$$42$$

D
$$54$$

Solution

After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
$$\therefore 125 : 150 :: 35 : x \Leftrightarrow 125 \times x = 150 \times 35$$
$$\Rightarrow x = \dfrac{150 \times 35}{125}$$
$$\Rightarrow x = 42$$

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 39

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Numerical Applications Test 39

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SHARING IS CARING

Question 1

$$\displaystyle\frac{5}{11}$$

$$\displaystyle\frac{6}{11}$$

$$\displaystyle\frac{7}{11}$$

$$\displaystyle\frac{8}{11}$$

Solution

Question 2

10

13

14

15

Solution

Question 3

648

1800

2700

10800

Solution

Question 4

Rs.$$375$$

Rs.$$400$$

Rs.$$600$$

Rs.$$800$$

Solution

Question 5

$$4$$ days

$$5$$ days

$$6$$ days

$$7$$ days

Solution

Question 6

$$12$$ days

$$15$$ days

$$16$$ days

$$18$$ days

Solution

Question 7

300 kmph

360 kmph

600 kmph

720 kmph

Solution

Question 8

1

$$\dfrac{1}{40}$$

40

80

Solution

Question 9

$$4$$ days

$$6$$ days

$$8$$ days

$$12$$ days

Solution

Question 10

$$29 \dfrac{1}{5}$$

$$37 \dfrac{1}{4}$$

$$42$$

$$54$$

Solution

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