Comparing Quantities Test

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Comparing Quantities Test - 50

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Question 1
1 / -0

In January the price of a certain item was $$\$120$$. In February the price increased by $$10$$ percent. During a sale in March, the February price was discounted by $$10$$ percent. What was the price of the item during the sale in March?

A
$$\$106.80$$

B
$$\$118.80$$

C
$$\$120.00$$

D
$$\$121.20$$

Solution

C.P(Cost Price) of a certain article is $$ $120$$
In February the cost price increased by 10 percent which means=$$10\%\ of C.P+C.P$$
$$=10\%\ of\ 120+120$$
$$=\dfrac{10}{100}\times 120+120$$
$$=12+120$$
$$=132$$

In March the price was discounted by $$10\%$$ which means we have to subtract $$10\%$$ from the amount we get in February
$$=132-10\%\ of\ 132$$
$$=132-\dfrac{10}{100}\times 132$$
$$=132-13.2$$
$$ =118.8$$$
Question 2
1 / -0

Mr. Ambani purchased a car of $$3,00,000$$ and a bike for his son for $$1,00,000$$. He sold the car at a profit of $$10$$% and bike at a loss of $$20$$%. What is the net gain or loss?

A
$$2$$% gain

B
$$1.5$$% gain

C
$$2.5$$% loss

D
$$2.5$$% gain

Solution

CP of a car = 3,00,000 Rs

Profit on car = 10%

$$\therefore SP\ of\ car=CP+CP\times\dfrac{Profit\ Percentage}{100}=300000+300000\times\dfrac{10}{100}=3,30,000\ Rs$$

CP of a bike = 1,00,000 Rs.

Loss on bike = 20%

$$\therefore SP\ of\ bike=CP-CP\times\dfrac{Loss\ Percentage}{100}=100000-100000\times\dfrac{20}{100}=80,000\ Rs$$

So, $$Total\ CP = 3,00,000 + 1,00,000 = 4,00,000$$ Rs

And $$Total\ SP = 3,30,000+80,000=4,10,000$$ Rs

Here SP > CP.

$$\therefore Gain = SP - CP = 4,10,000-4,00,000=10,000\ Rs$$

$$\Rightarrow Gain\ percentage = \dfrac{gain\times100}{CP}=\dfrac{10,000\times100}{4,00,000}=2.5\%$$
Question 3
1 / -0

A barrel of crude oil is extracted from shale at a cost of $$\$51$$, and then transported to and from the refinery at a cost of $$\$6$$ each direction. Oil is processed three times at the refinery plant, at a cost of $$\$9$$ each time. What is the profit, in dollars per barrel, if one barrel is sold for $$\$93$$? (Profit is equal to revenue minus expenses.)

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Costing in the process of making and transporting oil are:
Cost of extraction$$=\$51$$
Cosr of transportation to and from refinery= $$\$(6+6)=$12$$
Cost of processing $$3$$ times at $$\$9$$ each time $$=\$(3\times 9)=\$27$$
Total Cost$$= \$(51+12+27)=$$ $$ $90$$
Selling Price=$$ $93$$
Profit$$=\$(93-90)=$3$$
Question 4
1 / -0

If an article is sold at $$8\%$$ profit instead of $$8\%$$ loss, it would have brought Rs. $$12$$ more, what is the cost price of the article?

A
Rs. $$60$$

B
Rs. $$70$$

C
Rs. $$72$$

D
Rs. $$75$$

Solution

Let the Cost Price (C.P) be Rs $$x$$ and then at $$ 8\% $$ of profit
Selling Price(S.P) would be :
Gain $$\%$$ = $$ \dfrac{(S.P - C.P)}{C.P} \times 100 $$
$$\Rightarrow 8 \% =\dfrac{S.P- x}{x}\times100$$
$$\Rightarrow \dfrac{8x}{100} = S.P - x $$

$$\Rightarrow S.P = \dfrac{8x}{100} + x$$

$$\Rightarrow S.P = \dfrac{108x}{100}$$ .....(1)
Now, If the loss is $$ 8 \% $$
$$\Rightarrow \dfrac{8x}{100}= x - S.P$$

$$\Rightarrow S.P = \dfrac{92x}{100}$$...(2)

Now we have the condition from 1 and 2 we get :

$$S.P (profit) = S.P(loss) + 12$$

$$\Rightarrow \dfrac{108x}{100} = \dfrac{92x}{100} +12 $$

$$\Rightarrow \dfrac{(108 - 92 )}{100} x = 12 $$

$$\Rightarrow \dfrac{16}{100}x = 12 $$

$$ \Rightarrow 16x = 12 \times 100 $$

$$ \Rightarrow x = \dfrac{12\times 100 }{16}$$

$$\Rightarrow x = 75 $$

$$ \Rightarrow C.P = $$Rs $$75 $$

The Cost Price of the article is $$Rs 75$$ .
Question 5
1 / -0

If a $ $$12,000$$ car loses $$10$$% of its value every year, what is the worth after $$3$$ years?

A
8748

B
8750

C
8752

D
8744

Solution

After $$1^{st}$$ year the value of car is $$12000\left (1- \dfrac{10}{100}\right) = 10800$$
The value of car after $$2^{nd}$$ year is $$10800\left (1-\dfrac{10}{100}\right) = 9720$$
The value of car after $$3^{rd}$$ year is $$9720\left (1-\dfrac{10}{100}\right) = 8748$$
Question 6
1 / -0

In a certain store, the profit is $$320$$% of the cost. If the cost increases by $$25$$% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A
$$30$$%

B
$$70$$%

C
$$100$$%

D
$$250$$%

Solution

Let $$C.P.= Rs. 100$$. Then, $$Profit = Rs. 320, S.P. = Rs. 420$$.
New $$C.P. = 125$$% of $$Rs. 100 = Rs. 125$$
New $$S.P. = Rs. 420$$.
$$Profit = Rs. (420 - 125) = Rs. 295$$.
$$\therefore$$ Required percentage $$=\left (\dfrac {295}{420}\times 100\right )$$% $$= \dfrac {1475}{21}$$% $$= 70$$% (approximately).
Question 7
1 / -0

The question given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.

By selling an article what is the profit percent gained?
I. $$5\%$$ discount is given on list price.
II. If the discount is not given, $$20\%$$ profit is gained.
III. The cost price of the articles is $$\text{Rs. } 5000$$.

A
Only I and II

B
Only II and III

C
Only I and III

D
All I, II and III

E
None of these

Solution

I. Let the list price be $$\text{Rs. } x$$
Then, $$\text{S.P.} = 95\%$$ of $$\text{Rs. } x = \text{Rs. } \left (x\times \dfrac {95}{100}\right ) = \text{Rs. } \dfrac {19x}{20}$$
II. When $$\text{S.P.} = \text{Rs. } x$$ and gain $$= 20\%$$
Then, $$\text{C.P.} =\text{Rs. } \left (\dfrac {100}{120}\times x\right ) =\text{Rs. } \dfrac {5x}{6}$$
$$\therefore \text{Gain} = \left (\dfrac {19x}{20} - \dfrac {5x}{6}\right ) = \left (\dfrac {57x - 50x}{60}\right ) = \dfrac {7x}{60}$$
$$\therefore \text{Gain}\%$$ $$=\left (\dfrac {7x}{60}\times \dfrac {6}{5x} \times 100\right )\%$$ $$= 14\%$$.
Thus, I and II only give the answer.
$$\therefore$$ Correct answer is (A).
Question 8
1 / -0

A trader marks $$10\%$$ higher than the cost price. He gives a discount of $$10\%$$ on the marked price. In this kind of sales how much percent does the trader gain or lose?

A
$$5\%$$ gain

B
$$2\%$$ gain

C
$$1\%$$ loss

D
$$3\%$$ loss

Solution

Let the cost price be $$\text{Rs }100$$
Then the mark-up price is $$10\%$$ above the cost price.
Mark price $$=(100+10\% \ \text{of}\ 100)=$$ $$\text{Rs.}$$ $$110$$
Trader gives $$10\%$$ discount on the marked price, then the
Selling price $$=(110-10\% \ \text{of}\ 110)=$$ $$\text{Rs.}$$ $$99$$
Therefore, Loss $$=$$ cost price $$-$$ selling price
Loss $$=100-99= \text{Rs }1$$ on cost price of $$\text{Rs }100$$
Thus loss percent is $$1\%$$.
Question 9
1 / -0

At the "$$30\%$$ off" book sale, a calculus book costs $$\$24$$. What is the original price of the book?

A
$$\$72.80$$

B
$$\$77$$

C
$$\$80$$

D
$$\$82.60$$

Solution

Let the original price of the book be $$x$$.
After $$30 \%$$ discount on the original price the rate is $$ $24 $$
Therefore, $$30 \%$$ on $$x$$ is $$ $24$$
$$\Rightarrow \dfrac{30}{100}\times$$ $$x$$=$$24$$
$$\Rightarrow \dfrac{3}{10}\times $$ $$x$$=$$24$$
$$\Rightarrow x=\dfrac{24 \times 10}{3}$$
$$\Rightarrow x=\dfrac{240}{3}$$
$$\Rightarrow x=$80 $$
Question 10
1 / -0

In the half yearly exam only $$70\%$$ of the students were passed. Out of these (passed in half yearly) only $$60\%$$ students are passed in annual exam. Out of those who did not pass the half yearly exam, $$80\%$$ passed in annual exam. What per cent of the students passed the annual exam?

A
$$42\%$$

B
$$56\%$$

C
$$66\%$$

D
$$38\%$$

Solution

Let the total students be $$x$$.
Given that, $$70\%$$ students passed in half yearly examination.
Then, passed students in half yearly $$=\dfrac{70}{100}\times x=0.70x$$

Out of these passed in half yearly, $$60\%$$ students passed in annual exam.
Then, passed students in annual examination $$=\dfrac{60}{100}\times 0.70x=0.42x$$

The number of students who did not pass in half yearly examination $$=x-0.70x=0.30x$$

Given out of these who did not pass the half yearly examination, $$80\%$$ students passed in annual exam.
Then, out of these student who did not pass in half yearly examination, number of students who passed in annual exam $$=\dfrac{80}{100}\times 30x=0.24x$$

Then, total number of students who passed annual examination $$=0.42x+0.24x=0.66x$$

Therefore, $$\%$$ of students who passed the annual exam$$=\dfrac{0.66x}{x}\times 100=66\%$$

Hence, $$66\%$$ students passed the annual exam.

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 50

Result

Comparing Quantities Test - 50

Score

-

Rank

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SHARING IS CARING

Question 1

$$\$106.80$$

$$\$118.80$$

$$\$120.00$$

$$\$121.20$$

Solution

Question 2

$$2$$% gain

$$1.5$$% gain

$$2.5$$% loss

$$2.5$$% gain

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

Rs. $$60$$

Rs. $$70$$

Rs. $$72$$

Rs. $$75$$

Solution

Question 5

8748

8750

8752

8744

Solution

Question 6

$$30$$%

$$70$$%

$$100$$%

$$250$$%

Solution

Question 7

Only I and II

Only II and III

Only I and III

All I, II and III

None of these

Solution

Question 8

$$5\%$$ gain

$$2\%$$ gain

$$1\%$$ loss

$$3\%$$ loss

Solution

Question 9

$$\$72.80$$

$$\$77$$

$$\$80$$

$$\$82.60$$

Solution

Question 10

$$42\%$$

$$56\%$$

$$66\%$$

$$38\%$$

Solution

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