Comparing Quantities Test

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Comparing Quantities Test - 36

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Question 1
1 / -0

400 apples were bought at Rs. 125 per hundred and were sold at a profit of Rs. 100. Find the selling price per dozen.

A
Rs. 20

B
Rs. 12

C
Rs. 18

D
None

Solution

Total CP = $$4\, \times\, 125\,=\, Rs. 500$$

$$SP = CP + profit$$

$$= 500 + 100 = Rs.\ 600$$

SP of each apples = $$\displaystyle \frac{600}{400}\,=\, \displaystyle \frac{3}{2}$$

SP of dozen apples = $$\displaystyle \frac{3}{2}\, \times\, 12$$ = Rs. 18
Question 2
1 / -0

The C.I. on a sum of money for 2 years is Rs. 832 and the S.I. on the same sum for the same period is Rs. 800. The difference between the C.I and S.I. for 3 years will be

A
Rs. 48

B
Rs. 66.56

C
Rs. 98.56

D
None of these

Solution

Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400 and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400 and compound interest for 2nd year will be $$832 - 400 = 432$$
We can see that compound interest for 2nd year is more than simple interest for 2nd year by $$432 - 400 = Rs.32$$
Rs. 32 is the interest obtained for Rs.400 for 1 year
$$R = \frac{100\times SI}{PT} = \frac{100 \times 32}{400\times1} = 8$$%
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832 $$=\frac{PRT}{100} = \frac{832\times 8\times 1}{100} = 66.56$$
Total difference between the compound and simple interest for 3 years $$= 32 + 66.56 = Rs.98.56$$
Question 3
1 / -0

A seller allows a discount of $$5\%$$ on a watch. If he allows a discount of $$7\%$$, he earns $$\text{Rs }15$$ less in the profit. What is the marked price?

A
$$\text{ Rs }697.50$$

B
$$\text{ Rs }712.50$$

C
$$\text{ Rs }750$$

D
$$\text{ Rs }817.50$$

Solution

Let $$x$$ be the profit when the seller allows $$5\%$$ discount.
Then, $$(x-15)$$ is the profit when he allows $$7\%$$ discount.
Now,
Selling price at $$5\%$$ discount $$-$$ selling price at $$7\%$$ discount $$=x-(x-15)$$
$$\implies (100-5)\%$$ of M.P $$- (100-7)\%$$ of M.P $$= x-(x-15) = 15$$
$$\Rightarrow 2\% $$ of M.P $$= 15$$
$$\Rightarrow$$ M.P. $$=15\times \cfrac{ 100 }{ 2 } =750$$
$$\therefore$$ Marked Price is $$\text{ Rs. } 750$$
Question 4
1 / -0

Gauri want to the stationers and bought things worth Rs 25 out of which 30 paise went on sale tax on taxable purchases If the tax rate was 6% then what was the cost of the tax free item?

A
Rs 15

B
Rs 15.70

C
Rs 19.70

D
Rs 20

Solution

Given: Gauri bought $$Rs.25$$
Sales tax = $$ 30$$ paise
Sales tax rate = Rs.$$ 6$$ %
Let amount of taxable purchase = $$X$$
$$\dfrac{6}{100} \times{X} = \dfrac{30}{100}$$
$$ X =5 $$
Tax free amount = $$ 25 - 5 - 0.30 $$
= $$ 19.70$$
Question 5
1 / -0

$$A$$ began a business with Rs. $$85{,}000$$. He was joined afterwards by $$B$$ with Rs. $$42{,}500$$. For how much period does $$B$$ join, if the profits at the end of the year are divided in the ratio of $$3 : 1$$?

A
$$4$$ months

B
$$5$$ months

C
$$6$$ months

D
$$8$$ months

Solution

Suppose B joined for x months . Then, $$\dfrac{ ( 85000 \times 12 )}{(42500 \times x)} = 3$$

$$=> x = \dfrac{(85000 \times 12)}{ (42500 \times 3) }$$

$$=>x= 8$$

So, B joined for 8 months.
Question 6
1 / -0

A dealer allows a discount of $$10\%$$ and still gains $$5\%.$$ What percent above the cost price must he mark his goods?

A
$$15\%$$

B
$$20\%$$

C
$$16\dfrac{1}{3}\%$$

D
$$50\%$$

Solution

Let $$C.P. = \text{Rs }100$$
$$\because \left(\dfrac{S.P.-C.P.}{C.P.}\right)\times 100=\text{Profit} \%$$
$$\implies \left( \dfrac{S.P.-100}{100}\right)\times 100=5$$
$$\implies S.P.=\text{Rs }105$$
Discount(d) $$= 10 \%$$

We have,
$$S.P.\,=\, \displaystyle\dfrac{M.P.\, (100\,-\,d)}{100}$$ (where $$d,$$ is discount)

$$M.P.\,=\, \displaystyle\dfrac{105\, \times\, 100}{100\, -\, 10}\,=\, \dfrac{105\, \times\, 100}{90}\,=\, \dfrac{350}{3}$$

$$M.P.\,-\, C.P.\,=\, \displaystyle\dfrac{350}{3}\,-\, 100\,=\, \dfrac{50}{3}$$

$$\therefore \text{Required }\%=\, \displaystyle\dfrac{\dfrac{50}{3}}{100}\, \times\, 100\,=\, 16\frac{2}{3}\%$$

$$\therefore$$ He should mark his goods $$16\displaystyle\frac{2}{3} \%$$ above the C.P.
Question 7
1 / -0

A dealer allows $$25\%$$ discount on the marked price of articles and earns a profit of $$20\%$$ on them. What is the marked price of the article on which he gains $$\text{Rs.}\ 800$$ ?

A
Rs.$$6000$$

B
Rs.$$6400$$

C
Rs.$$7200$$

D
Rs.$$7000$$

Solution

Let the M.P.=$$\text{Rs.}\ 100$$
Discount $$= 25\%$$
$$\therefore \text{S.P.} =\text{Rs.} 75 $$

$$ \text{Profit} = 20 \%$$

$$\displaystyle \text{C.P.}=\text{Rs.}\cfrac{75\times 100}{120}=\text{Rs.}\ 62.50$$

$$\displaystyle \therefore \text{Profit} = \text{Rs.}75 - \text{Rs.}62.50 = \text{Rs.}12.50$$

If the gain is Rs. 12.50, M.P.$$=\text{Rs.}\ 100$$

If the gain is Rs.800, M.P.$$=\text{Rs.}\displaystyle \cfrac{100}{12.50}\times 800=\text{Rs.}\ 6400$$
Question 8
1 / -0

The list price of a slipper is Rs. 160. The discount offered is 10%.Find the selling Price of a slipper.

A
144

B
124

C
100

D
134

Solution

List price of slipper $$=160$$
discount $$=10\%$$
$$10\%$$ of $$160=\cfrac{10}{100}\times160=16$$
Selling Price of slipper aftter discount=$$160-16=144$$
Question 9
1 / -0

A tradesman sold an article at a loss of $$20\%$$. Had he sold it for $$Rs. 100$$ more, he should have gained $$5\%$$. The cost price of the article was

A
Rs. $$360$$

B
Rs. $$400$$

C
Rs. $$425$$

D
Rs. $$450$$

Solution

Let the cost price $$=Rs. x$$
Loss $$= 20 \%$$
Loss $$=\cfrac{20}{100} \times x$$ = $$\dfrac{x}{5}$$
Now, $$SP = CP - Loss = x - \cfrac{x}{5} = \cfrac{4x}{5}$$
If SP was Rs 100 more,
New $$SP = \cfrac{4x}{5} + 100$$
Profit $$= 5\%$$ of CP $$=\cfrac{5}{100} \times x = New \, SP - CP$$
Profit $$=\cfrac{x}{20} = \cfrac{4x}{5} + 100 - x$$
$$\cfrac{20x - 16x + x}{20} = 100$$
$$5x = 2000$$
$$x = Rs. 400$$
Question 10
1 / -0

The present population of a town is $$5000$$. If the population decreases at the rate of $$10\%$$ per year, the population after $$2$$ years will be

A
$$4820$$

B
$$4050$$

C
$$4980$$

D
$$4500$$

Solution

Present population, $$P_o = 5000$$
Rate $$= 10\%$$
Time $$= 2$$ years
Let after $$2$$ years population $$= P$$
Thus, $$P = P_o(1 - \frac{R}{100})^T$$
$$P = 5000 (1 - \frac{10}{100})^2$$
$$P = 4050$$
Thus after two years the population will be $$4050$$

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 36

Result

Comparing Quantities Test - 36

Score

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Rank

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SHARING IS CARING

Question 1

Rs. 20

Rs. 12

Rs. 18

None

Solution

Question 2

Rs. 48

Rs. 66.56

Rs. 98.56

None of these

Solution

Question 3

$$\text{ Rs }697.50$$

$$\text{ Rs }712.50$$

$$\text{ Rs }750$$

$$\text{ Rs }817.50$$

Solution

Question 4

Rs 15

Rs 15.70

Rs 19.70

Rs 20

Solution

Question 5

$$4$$ months

$$5$$ months

$$6$$ months

$$8$$ months

Solution

Question 6

$$15\%$$

$$20\%$$

$$16\dfrac{1}{3}\%$$

$$50\%$$

Solution

Question 7

Rs.$$6000$$

Rs.$$6400$$

Rs.$$7200$$

Rs.$$7000$$

Solution

Question 8

144

124

100

134

Solution

Question 9

Rs. $$360$$

Rs. $$400$$

Rs. $$425$$

Rs. $$450$$

Solution

Question 10

$$4820$$

$$4050$$

$$4980$$

$$4500$$

Solution

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