Numerical Applications Test 50

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Numerical Applications Test 50

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Weekly Quiz Competition

Question 1
1 / -0

Each crop duster who works in a firm can dust half an acre of crops in $$25$$ minutes. The company must dust eight $$0.75$$-acre lots and twelve $$1.5$$-acre lots to complete a certain job. Calculate the minimum number of crop dusters needed to complete the job in $$5$$ hours.

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

There are two different size lots:
$$0.75$$ acre and $$1.5$$ acre

If a crop duster can dust $$0.5$$ acre in $$25$$ minutes, then they can dust $$0.75$$ acre in $$\dfrac{0.75\times25}{0.5} = 37.5$$ minutes.
That same crop duster can dust a $$1.5$$ acre lot in $$\dfrac{1.5\times25}{0.5} = 75$$ minutes
Now,
We have $$45$$ hours or $$300$$ minutes.
In $$5$$ hours a crop duster can dust
$$4, 1.5$$-acre lots or $$8 , 0.75$$-acre lots

There are $$12, 1.5$$-acre lots, so $$3$$ dusters will be needed.
There are $$8, 0.75$$-acre lots, so $$1$$ crop duster will be needed here.

$$\therefore 3+1=4$$
Hence, they will need $$4$$ crop dusters to complete the job in $$5$$ hours.
Question 2
1 / -0

A cistern can be filled by one tap in $$5$$ hours and by another in $$4$$ hours. How long will it take to fill if both the taps are opened simultaneously?

A
$$\dfrac{10}{9}\;hrs$$

B
$$9\;hrs$$

C
$$10\;hrs$$

D
$$\dfrac{20}{9}\;hrs$$

Solution

Time taken by first tap to fill the cistern $$=5$$ hours.
Work done by first tap in $$1$$ hour $$=\dfrac{1}{5}$$
Time taken by second tap to fill the cistern $$=4\;hours$$.
Work done by second tap in $$1$$ hour $$=\dfrac{1}{4}$$
Work done by (first tap + second tap) in $$1$$ hour $$=\dfrac{1}{5}+\dfrac{1}{4}=\dfrac{9}{20}$$
Therefore, time taken by (first tap + second tap) to fill the cistern $$=\dfrac{20}{9}\;hours$$.
Question 3
1 / -0

A man can make $$18$$ cakes in $$6$$ days by working $$6$$ hours a day. How many cakes can he make in $$6$$ days if he works $$12$$ hours a day?

A
$$12$$

B
$$24$$

C
$$36$$

D
$$30$$

Solution

No of days is constant.
Hence, no of cakes is directly proportional to no of hours.
$$\dfrac {18}{6} = k = \dfrac {x}{12}$$
$$\Rightarrow x = \dfrac {18}{6} \times 12 = 36$$
Question 4
1 / -0

Wyatt can husk at least $$12$$ dozen ears of corn per hour and at most $$18$$ dozen ears of corn per hour. Based on this information, what is a possible amount of time, in hours, that it could take Wyatt to husk $$72$$ dozen ears of corn?

A
Any number between $$4$$ - $$8$$, inclusive

B
Any number between $$4$$ - $$6$$, inclusive

C
Any number between $$2$$ - $$4$$, inclusive

D
Any number between $$1$$ - $$6$$, inclusive

Solution

Wyatt can husk anywhere from $$12 $$ to $$18$$ dozen ears of corn per hour.
Thus, if we need to find the time taken to husk 72 dozen ears of corn, the maximum time would be $$\cfrac{72}{12} = 6$$ hours and the minimum time would be $$\cfrac{72}{18} = 4$$ hours
Thus, depending on his speed, Wyatt takes anywhere from $$4$$ to $$6$$ hours to husk $$72$$ dozen ears of corn.
Question 5
1 / -0

Thomas can paint the house in $$6$$ days while it takes Joe $$12$$ days to paint the same house. How long would it take Thomas and Joe, working together, to paint the house?

A
$$3$$ days

B
$$4$$ days

C
$$5$$ days

D
$$6$$ days

Solution

If Thomas can paint the house in $$6$$ days then in $$1$$ day, he can paint $$\dfrac {1}{6}$$ of the house.
If Joe can paint the house in $$12$$ days, then in $$1$$ day he can paint $$\dfrac {1}{12}$$ of the house.
Let $$x$$ is the number of days it would take them together to paint the house, then working together, in $$1$$ day they can paint $$\dfrac {1}{x}$$ of the house.
The equation becomes:
$$\dfrac {1}{6}+\dfrac {1}{12}=\dfrac {1}{x}$$
$$\Rightarrow$$ $$\dfrac {3}{12}=\dfrac {1}{x}$$
$$\Rightarrow$$ $$\dfrac {1}{4}=\dfrac {1}{x}$$
$$\Rightarrow$$ $$x=4$$

Therefore, Thomas and Joe takes $$4$$ days to paint the house together.
Question 6
1 / -0

The average bonus per employee was.
The chart shows the amount paid in bonuses to the employees of a certain firm.
Bonus paid to an employee$$(\$)$$ $$50$$ $$100$$ $$150$$ $$200$$
Number of employees $$7$$ $$37$$ $$4$$ $$2$$

A
$$81$$

B
$$91$$

C
$$100$$

D
$$101$$

E
$$105$$

Solution
Question 7
1 / -0

Consider a pyramid with a square base side of $$6$$ inches and with a height of $$12$$ inches, as shown below. If we cut off the top of the pyramid parallel to the base $$3$$ inches from the tip, what is the volume of the remaining solid?

A
$$141.75$$

B
$$140$$

C
$$135.48$$

D
$$144$$

E
$$130$$

Solution

a pyramid of a square base side of 6 inches.
height = $$12$$ inches
for smaller pyramid given :-
height = $$3$$ inches
Base square of side = $$1.5 $$ inches
volume of larger pyramid = $$\dfrac{1}{3}$$ Base area $$\times $$ height
$$= \dfrac{1}{3} (6)^2 \times 12$$
$$= 36 \times 4 = 144 \, inch^3$$
Volume of smaller pyramid $$= \dfrac{1}{3}$$ Base area $$\times$$ height
$$= \dfrac{1}{3} (1.5)^2 \times 3$$
$$= (1.5)^2 = 2.25 inch^3$$
So, volume of remaining solid
= larger volume - smaller volume
$$= 144 - 2.25$$
$$= 141.75 \, inch^3$$
Question 8
1 / -0

Using a hose, it takes $$9$$ hours to fill a swimming pool. Using an irrigation pump, it takes $$3$$ hours.
If you use both the hose and pump together, how long will it take to fill the pool?

A
$$2\ hours$$

B
$$2\ hours\ 15\ minutes$$

C
$$2\ hours\ 45\ minutes$$

D
$$3\ hours$$

E
$$6\ hours$$

Solution

Work efficiency of hose in filling pool = $$\dfrac{1}{9}$$ per hour
Work efficiency of irrigation pump in filling pool= $$\dfrac{1}{3}$$ per hour
$$\therefore$$ work done by both together = $$\dfrac{1}{9} + \dfrac{1}{3} = \dfrac{4}{9}$$per hour
Hence time taken to fill the tank $$100$$% = $$\dfrac{9}{4}$$ hours = $$2.25$$ hours
Since, $$1 hour = 60 minutes$$
$$\therefore 2.25$$ hours = $$2.25\times 60 = 135$$ minutes
now, $$135$$ minutes = $$(60 +60+15)$$ minutes = $$2 hours$$ $$15minutes$$
So, the answer is option B
Question 9
1 / -0

What was the average number of babies that Dr. Jones delivered each year from 1995 to 1998?

A
35

B
40

C
45

D
50

E
55
Question 10
1 / -0

Each side of the base of a square pyramid is reduced by $$20%$$. By what percent must the height be increased so that the volume of the new pyramid is the same as the volume of the original pyramid?

A
20

B
40

C
46.875

D
56.25

E
71.875

Solution

Let $$a$$ be the side of the square.
Length of side of square when reduced by $$20\% = a-\dfrac{20a}{100}=0.8a$$
Let $$a_1=0.8a$$
Volume of pyramid $$V=\dfrac { 1 }{ 3 } \times $$ Area of base $$\times height=\dfrac{1}{3}A\times h$$
Area of base with side $$a = { a }^{ 2 }$$
$${ a }^{ 2 }=0.8a$$

$${V}_{ 1 }=\dfrac { 1 }{ 3 } \times { \left( { a }_{1} \right) }^{ 2 }\times { h }_{ 1 }$$
$${V}_{ 1 }=V$$ ....... [Given]

$$\Rightarrow \dfrac { 1 }{ 3 } { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } { \left( 0.8 \right) }^{ 2 }{ a }^{ 2 }\times { h }_{ 1 }$$

$$\therefore { h }_{ 1 }=1.5625h$$

$$\implies \dfrac { { h }_{ 1 }-h }{ h } =1.5625-1=56.25%$$

$$\therefore$$ 'h' need to be increase by $$56.25$$

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Bonus paid to an employee$$(\$)$$	$$50$$	$$100$$	$$150$$	$$200$$
Number of employees	$$7$$	$$37$$	$$4$$	$$2$$

Numerical Applications Test 50

Result

Numerical Applications Test 50

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Rank

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SHARING IS CARING

Question 1

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 2

$$\dfrac{10}{9}\;hrs$$

$$9\;hrs$$

$$10\;hrs$$

$$\dfrac{20}{9}\;hrs$$

Solution

Question 3

$$12$$

$$24$$

$$36$$

$$30$$

Solution

Question 4

Any number between $$4$$ - $$8$$, inclusive

Any number between $$4$$ - $$6$$, inclusive

Any number between $$2$$ - $$4$$, inclusive

Any number between $$1$$ - $$6$$, inclusive

Solution

Question 5

$$3$$ days

$$4$$ days

$$5$$ days

$$6$$ days

Solution

Question 6

$$81$$

$$91$$

$$100$$

$$101$$

$$105$$

Solution

Question 7

$$141.75$$

$$140$$

$$135.48$$

$$144$$

$$130$$

Solution

Question 8

$$2\ hours$$

$$2\ hours\ 15\ minutes$$

$$2\ hours\ 45\ minutes$$

$$3\ hours$$

$$6\ hours$$

Solution

Question 9

35

40

45

50

55

Question 10

20

40

46.875

56.25

71.875

Solution

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