Numerical Applications Test 36

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Numerical Applications Test 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$5$$ people can paint a house in $$16$$ hours, how many people are required to complete the job in $$10$$ hours?

A
$$32$$

B
$$8$$

C
$$24$$

D
$$120$$

Solution

It is the case of inverse proportion.
$$5\times 16 = k = x\times 10$$
$$\Rightarrow x = \dfrac {5\times 16}{10} = 8$$
Question 2
1 / -0

If $$x$$ persons can complete work in $$t$$ hours, in how many hours $$y$$ persons can complete it?

A
$$\dfrac{yt}{x}$$

B
$$\dfrac{yx}{t}$$

C
$$\dfrac{t}{y}$$

D
$$\dfrac{tx}{y}$$

Solution

$$x$$ person can complete in $$t$$ hours
$$1$$ person can complete in $$t\times x$$ hrs
$$y$$ person can complete in $$\dfrac{t\times x}{y}=\frac{t}{y}$$
Question 3
1 / -0

If $$6$$ painters can complete $$9$$ drawing in $$5$$ hours then
How many painters will make $$18$$ drawing in $$5$$ hours?

A
$$11$$

B
$$12$$

C
$$13$$

D
$$14$$

Solution

Let number of painters required for painting 18 drawing is $$x$$
Now since no. of painters is directly proportion to no. of drawing
$$\dfrac {6}{9} = \dfrac {x}{18} \Rightarrow x = \dfrac {6\times 18}{9} = 12$$
Question 4
1 / -0

If A is three times as efficient as B & A can complete a work in $$40$$ days less time than B. Find in how many days both can complete the work

A
$$15$$

B
$$20$$

C
$$40$$

D
$$50$$

Solution

Since $$A$$ is three times more efficient than $$B$$, if $$A$$ completes a work in $$x$$ days, then $$B$$ can complete the same work in $$3x$$ days.

It is also given that $$A$$ can complete a work in $$40$$ days less time than $$B$$, that is,

$$3x-x=40\\ \Rightarrow 2x=40\\ \Rightarrow x=\dfrac { 40 }{ 2 } \\ \Rightarrow x=20$$

And, $$3x=3\times 20=60$$

This means that when $$B$$ can complete the work in $$60$$ days, $$A$$ can complete it in $$20$$ days. So, both can together finish the work in:

$$\dfrac { 20\times 60 }{ 20+60 } =\dfrac { 1200 }{ 80 } =15$$ days.

Hence, both $$A$$ and $$B$$ can complete the work in $$15$$ days.
Question 5
1 / -0

If $$7$$ pipes fill a tank in $$8$$ hours, then $$4$$ pipes will fill it in ......... hours

A
$$14$$

B
$$16$$

C
$$18$$

D
$$20$$

Solution

No of pipes is inversely proportional to the taken to fill the tank
$$\therefore 7\times 8 = k = 4\times x \Rightarrow 56 = 4\times x \Rightarrow x = 14$$
Question 6
1 / -0

If $$4$$ persons do a work in $$8$$ days then how many days it will take if $$8$$ persons do the same work?

A
$$4$$

B
$$16$$

C
$$8$$

D
$$32$$

Solution

The no of persons (p) is inversely proportional to the no. of days (d). Hence $$p\propto \dfrac {1}{d}$$
$$\Rightarrow p_{1}d_{1} = p_{2}d_{2}$$
Here, $$p_{1} = 4, d_{1} = 8, p_{2} = 8, d_{2}$$ is unknown.
$$\Rightarrow 4\times 8 = 8d_{2}$$
$$\Rightarrow d_{2} = 4$$.
Question 7
1 / -0

Find the no. of pipes required to fill the tank in $$16$$ hours.

A
$$\dfrac {5}{2}$$

B
$$\dfrac {7}{2}$$

C
$$7$$

D
$$\dfrac {9}{2}$$

Solution

$$7\times 8 = x \times 16 \Rightarrow x =\dfrac {7\times 8}{16} = \dfrac {7}{2}$$
Question 8
1 / -0

An 11.2 gb image has been taken of the surface of Jupiter by a camera . A tracking station in Earth can receive data from the spacecraft at a data rate of $$3$$ megabits per second for a maximum of $$11$$ hours each day. If $$1$$ gb equals $$1,024$$ mb, find the maximum number of images that the tracking station can receive from the camera each day. (gb: gigabits, mb: megabits)

A
$$3$$

B
$$10$$

C
$$56$$

D
$$144$$

Solution

Given: A tracking station in earth can received data from the spacecraft at a data rate $$3$$ megabits per second for a maximum of $$11$$ hour each day.
Then total data received per day $$=$$ $$3\times 11\times 3600$$ megabits
Then total data received per day $$=$$ $$\dfrac{118800}{1024}=116.01$$ gb
Then tricking station received images by camera $$=$$ $$\dfrac{116.01}{11.02}=10.35$$
Then camera received image $$=10$$.
Question 9
1 / -0

Ram can finish a work in $$45$$ minutes. His son can finish a work in $$3$$ hours. If they work together, how long will it take them to finish the work?

A
$$24$$ minutes

B
$$30$$ minutes

C
$$36$$ minutes

D
$$40$$ minutes

Solution

Ram can finish a work in $$45$$ minute
$$\therefore$$ work done by Ram in $$1$$ minute $$=$$ $$\dfrac{1}{45}$$
Ram's son can finish a work in $$3$$ hr $$=180 $$ minute
$$\therefore $$ Work done by his son in $$1$$ minute $$=\dfrac{1}{180}$$
If they work together then work done by both in $$1$$ minute
$$=\dfrac{1}{45}+\dfrac{1}{180}$$
$$=\dfrac{4+1}{180}=\dfrac{5}{180}=\dfrac{1}{36}$$
Then it will take $$36$$ minute if they work together.
Question 10
1 / -0

Find the arithmetic mean of the progression $$2, 4, 6, 8, 10.$$

A
$$10$$

B
$$20$$

C
$$30$$

D
$$6$$

Solution

Using the formula for Required arithmetic mean $$=\dfrac{\text{sum of the terms}}{\text{number of terms}}$$

After substituting the values we get$$=\dfrac{2+4+6+8+10}{5}=\dfrac{30}{5}=6$$

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 36

Result

Numerical Applications Test 36

Score

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Rank

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SHARING IS CARING

Question 1

$$32$$

$$8$$

$$24$$

$$120$$

Solution

Question 2

$$\dfrac{yt}{x}$$

$$\dfrac{yx}{t}$$

$$\dfrac{t}{y}$$

$$\dfrac{tx}{y}$$

Solution

Question 3

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Question 4

$$15$$

$$20$$

$$40$$

$$50$$

Solution

Question 5

$$14$$

$$16$$

$$18$$

$$20$$

Solution

Question 6

$$4$$

$$16$$

$$8$$

$$32$$

Solution

Question 7

$$\dfrac {5}{2}$$

$$\dfrac {7}{2}$$

$$7$$

$$\dfrac {9}{2}$$

Solution

Question 8

$$3$$

$$10$$

$$56$$

$$144$$

Solution

Question 9

$$24$$ minutes

$$30$$ minutes

$$36$$ minutes

$$40$$ minutes

Solution

Question 10

$$10$$

$$20$$

$$30$$

$$6$$

Solution

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