Comparing Quantities Test

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Comparing Quantities Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Convert $$75\%$$ into vulgar fraction.

A
$$\dfrac{1}{4}$$

B
$$\dfrac{2}{4}$$

C
$$\dfrac{3}{4}$$

D
$$\dfrac{5}{4}$$

Solution

Converting a given fraction into simple fraction as:
$$75\% =$$ $$\dfrac{75}{100}$$
Reduce the fraction, we get
$$\dfrac{75}{100}=\dfrac{3}{4}$$
Question 2
1 / -0

Loss =

A
selling price - cost price

B
cost price - selling price

C
selling price + cost price

D
selling price x cost price

Solution

If the selling price is less than the cost price, the difference of the prices is the loss.
So, loss $$=$$ cost price $$-$$ selling price.
Question 3
1 / -0

The percentage profit earned by selling an article for Rs. $$1920$$ is equal to the percentage loss incurred by selling the same article for Rs. $$1280$$. At what price should the article be sold to make $$25$$% profit?

A
Rs. $$2000$$

B
Rs. $$2200$$

C
Rs. $$2400$$

D
Data inadequate

Solution

Let C.P. be $$Rs. x$$.
Then, $$\dfrac {1920 - x}{x} \times 100 = \dfrac {x - 1280}{x}\times 100$$
$$\Rightarrow 1920 - x = x - 1280$$
$$\Rightarrow 2x = 3200$$
$$\Rightarrow x = 1600$$
$$\therefore$$ Required $$S.P. = 125$$% of $$Rs. 1600 = Rs. \left (\dfrac {125}{100}\times 1600\right ) = Rs 2000$$.
Question 4
1 / -0

Sam purchased $$20$$ dozens of toys at the rate of $$Rs. 375$$ per dozen. He sold each one of them at the rate of $$Rs. 33$$. What was his percentage profit?

A
$$3.5$$

B
$$4.5$$

C
$$5.6$$

D
$$6.5$$

Solution

Cost Price of $$1$$ toy $$= Rs. \left (\dfrac {375}{12}\right ) = Rs. 31.25$$
Selling Price of $$1\ toy = Rs. 33$$
So, $$Gain = Rs. (33 - 31.25) = Rs. 1.75$$
$$\therefore Profit$$% $$= \left (\dfrac {1.75}{31.25} \times 100\right )$$% $$= \dfrac {28}{5}$$% $$= 5.6$$%
Question 5
1 / -0

A man buys oranges at $$9$$ for $$Rs. 16$$ and sells $$11$$ for $$Rs. 20$$. What does he gain or lose per cent?

A
$$16$$%

B
$$2\dfrac {1}{11}$$%

C
$$2\dfrac {3}{11}$$%

D
$$1\dfrac {3}{11}$$%

Solution

Let $$99$$ oranges be purchased (L.C.M. of $$9$$ and $$11$$)
Cost of $$99$$ oranges $$= Rs. 176$$, whereas selling price of $$99$$ oranges is $$Rs. 180$$.
$$Gain = Rs. 4$$
$$\therefore Gain$$% $$= \dfrac {4}{176}\times 100 = 2\dfrac {3}{11}$$%
Question 6
1 / -0

By selling $$8$$ dozen pencils, a shopkeeper gains the selling price of $$1$$ dozen pencils. What is the gain?

A
$$12\dfrac {1}{2}\%$$

B
$$13\dfrac {1}{7}\%$$

C
$$14\dfrac {2}{7}\%$$

D
$$87\dfrac {1}{2}\%$$

Solution

We know that S.P. $$=$$ C.P. $$+$$ Gain
$$\Rightarrow $$ S.P. of $$8$$ dozen $$=$$ C.P. of $$8$$ dozen $$+$$ S.P. of $$1$$ dozen
$$\Rightarrow$$ S.P. of $$7$$ dozen $$=$$ C.P. of $$8$$ dozen
$$\Rightarrow$$ By selling $$7$$ dozen, he gain $$1$$ dozen
Therefore, gain $$\%$$ $$= \dfrac {1}{7}\times 100\%$$ $$= 14\dfrac {2}{7}\%$$
Question 7
1 / -0

The selling price of goods which cost Rs.$$10$$ and were sold at a gain of $$10$$% is:

A
Rs. $$12$$

B
Rs. $$10$$

C
Rs. $$11$$

D
Rs. $$15$$

Solution

$$\Rightarrow$$ Cost price pf goods is Rs.$$10$$
$$\Rightarrow$$ Profit = $$10\%\,$$of Rs.$$10=$$Rs. $$1$$
$$\Rightarrow$$ Selling price of goods $$=$$ Cost price +profit
$$\Rightarrow$$ Selling price of good $$= $$Rs. $$10+$$Rs. $$1=$$Rs. $$11$$
Question 8
1 / -0

Determine the principal when time $$= 2$$ years, interest $$=$$ Rs. $$ 1000$$; rate $$= 5\%$$ p.a.

A
Rs. $$10000$$

B
Rs. $$20000$$

C
Rs. $$30000$$

D
Rs. $$40000$$

Solution

Here, time $$=2$$ years, interest $$=$$ Rs. $$1000$$, rate $$=5\%$$ p.a.

We know $$S.I=\dfrac{P\times R\times T}{100}$$

$$\Rightarrow$$ $$1000=\dfrac{P\times 5\times 2}{100}$$

$$\Rightarrow$$ $$P=\dfrac{1000\times 100}{5\times 2}=$$ Rs. $$10,000$$

Therefore, principal is Rs. $$10,000$$.
Question 9
1 / -0

A shopkeeper expects a gain of $$22.5\%$$ on his cost price. If in a week, his sale was of Rs. 392, what was his profit?

A
$$Rs. 18.20$$

B
$$Rs. 70$$

C
$$Rs. 72$$

D
$$Rs. 88.25$$

Solution

Given that,
$$S.P.=Rs.\ 392$$ and $$\text{gain}\%=22.5\%$$
To find out: profit
We know that, $$C.P.=\dfrac{S.P.\times 100}{100+\text{gain}\%}$$

$$\therefore \ C.P. = Rs. \left (\dfrac {100}{122.5}\times 392\right ) \\$$

$$= Rs. \left (\dfrac {1000}{1225}\times 392\right ) \\$$

$$= Rs. 320\\$$

We also know that, $$\text{profit}=S.P.-\ C.P.$$

$$\therefore \ \text{Profit} = Rs. (392 - 320) \\$$
$$= Rs. 72$$.

Hence, the profit of the shopkeeper on the sale of $$Rs.\ 392$$ is $$Rs.\ 72$$.
Question 10
1 / -0

A shopkeeper sells some toys at Rs. 250 each. What percent profit does he make? To find the answer, which of the following information given in Statements I and II is/are necessary?
I. Number of toys sold.
II. Cost price of each toy.

A
Only I is necessary

B
Only II is necessary

C
Both I and II are necessary

D
Either I or II is necessary

E
None of these

Solution

Given that, $$S.P. = Rs. \ 250 $$ each.
We know that,
$$\text{Profit}\%=\dfrac{(S.P.- \ C.P.)}{C.P.}\times 100$$
We already have $$S.P.$$ and hence need only $$C.P.$$ to calculate percentage profit.
Hence, option B is correct.

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 15

Result

Comparing Quantities Test - 15

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SHARING IS CARING

Question 1

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

$$\dfrac{2}{4}$$

$$\dfrac{3}{4}$$

$$\dfrac{5}{4}$$

Solution

Question 2

selling price - cost price

cost price - selling price

selling price + cost price

selling price x cost price

Solution

Question 3

Rs. $$2000$$

Rs. $$2200$$

Rs. $$2400$$

Data inadequate

Solution

Question 4

$$3.5$$

$$4.5$$

$$5.6$$

$$6.5$$

Solution

Question 5

$$16$$%

$$2\dfrac {1}{11}$$%

$$2\dfrac {3}{11}$$%

$$1\dfrac {3}{11}$$%

Solution

Question 6

$$12\dfrac {1}{2}\%$$

$$13\dfrac {1}{7}\%$$

$$14\dfrac {2}{7}\%$$

$$87\dfrac {1}{2}\%$$

Solution

Question 7

Rs. $$12$$

Rs. $$10$$

Rs. $$11$$

Rs. $$15$$

Solution

Question 8

Rs. $$10000$$

Rs. $$20000$$

Rs. $$30000$$

Rs. $$40000$$

Solution

Question 9

$$Rs. 18.20$$

$$Rs. 70$$

$$Rs. 72$$

$$Rs. 88.25$$

Solution

Question 10

Only I is necessary

Only II is necessary

Both I and II are necessary

Either I or II is necessary

None of these

Solution

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