Comparing Quantities Test

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Comparing Quantities Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

A pair of shoe bought for Rs $$250$$ and sold for Rs $$325$$. Find profit or loss percent

A
$$10$$%

B
$$40$$%

C
$$30$$%

D
$$50$$%

Solution

Profit percentage(%)$$=\dfrac{\text{profit}}{\text{cost price}}\times 100$$
Profit$$=S.P-C.P=325-250=Rs.\ 75$$
Profit percent$$=\dfrac{75}{250}\times 100=30$$%
Question 2
1 / -0

If the simple interest on $$1700$$ rupees is $$340$$ rupees for $$2$$ years then the rate of interest must be:

A
$$12\ \%$$

B
$$15\ \%$$

C
$$4\ \%$$

D
$$10\ \%$$

Solution

Principle$$=Rs1700\quad\quad Time=2years$$

$$SI=Rs340\quad\quad Rate=?$$

$$ \cfrac{P\times R\times T}{100}=340$$

$$ \cfrac{1700\times R\times 2}{100}=340$$

$$R=\cfrac{340\times100}{1700\times2}$$

$$=\cfrac{340\times5}{170}=10\%$$
Question 3
1 / -0

The cost price of $$16$$ articles is equal to selling price of $$12$$ articles then the gain or loss per cent is

A
$$13 \frac {1}{3}$$% gain

B
$$33 \frac {1}{3}$$% gain

C
$$33 \frac {1}{3}$$% loss

D
$$13 \frac {1}{3}$$% loss

Solution

C.P. of 16 articles = S.P. of 12 articles (given)
Let SP of 1 article = Rs. x
then, SP of 12 articles = 12 x Rs.
C.P. of 16 articles = S.P. of 12 articles
CP of 16 articles = 12 x Rs.
SP of 16 articles = 16 x Rs.
SP > CP
Profit $$ = SP - CP \\ = 16 x - 12 x \\ = 4 x $$
Profit % $$ = \dfrac{\textrm{Profit}}{C} \times 100 \\ = \dfrac{4x}{12x} \times 100 \\ = 33\dfrac{1}{3} \% $$
Question 4
1 / -0

The fraction $$\dfrac{1}{5}$$ converted to percentage is

A
$$20\% $$

B
$$30\% $$

C
$$40\% $$

D
$$50\% $$

Solution

We have,
$$\dfrac{1}{5}\times 100 = 20\%$$

Hence, this is the answer.
Question 5
1 / -0

$$C.P=Rs.16000, S.P=Rs.18500, x$$% $$=\dfrac {x}{100}$$, Profit %

A
$$25$$%

B
$$15\dfrac {1}{4}$$%

C
$$25\dfrac {1}{8}$$%

D
$$15\dfrac {5}{8}$$%

Solution

Given C.P = 16000 Rs
S.P = 18.500 Rs
Profit = SP - CP = 18500-16,000
Profit = 2,500 Rs
Profit % = $$ \frac{profit}{CP}\times 100=\frac{2,500}{16000}\times 100$$
Profit % = $$ \frac{250}{16} = 15.62$$%
or
Profit % = $$ 15 \frac{5}{8}$$%
Question 6
1 / -0

Rahul and Sunjay invested $$Rs.\,\,25,000\,\,$$ and$$ \;\,Rs.\,\,35,000$$ respectively in a business. They will share the profit in the proportion.............

A
$$7:5$$

B
$$5:7$$

C
$$8:5$$

D
$$5:8$$

Solution

Given Rahul and Sunjay invested $$Rs.\,\,25,000\,\,$$ and$$ \;\,Rs.\,\,35,000$$ respectively in a business.
Now their distribution of the profit will be in the ratio of their invested amounts.
This ratio will be
$$25,000:35,000$$
$$=25:35$$
$$=5:7$$.
Question 7
1 / -0

A merchant sold an item for $$\text{Rs. } 1200$$ to earn a profit of $$50\%$$ over its cost price. what could the cost price be?

A
$$\text{Rs } 55$$

B
$$\text{Rs }607$$

C
$$\text{Rs }800$$

D
$$\text{Rs }1809$$

Solution

$$\Rightarrow$$ The selling price of an item $$=Rs.1200$$
$$\Rightarrow$$ Profit $$=50\%$$
Let the cost price be $$Rs.x$$
$$\Rightarrow$$ Profit $$=x\times\dfrac{50}{100}=\dfrac{x}{2}$$
$$\Rightarrow$$ Selling price of an item $$=x+\dfrac{x}{2}=\dfrac{3x}{2}$$ $$[\because $$ selling price $$=$$ cost price $$+$$ profit $$]$$
Now,
$$\Rightarrow$$ $$\dfrac{3x}{2}=1200$$
$$\Rightarrow$$ $$3x=2400$$
$$\Rightarrow$$ $$x=800$$
$$\Rightarrow$$ The required cost price is $$Rs.800$$
Question 8
1 / -0

What rate percent of S.I. will a sum of money double itself

A
1/2%

B
8.33%

C
both 1 &2

D
none of these

Solution

In case of simple Interest, total amount is $$ A = P \left( 1 + \dfrac{rt}{100} \right) $$
Where,
P = Principal
R = Rate
T = time
After 12 year sum of money double itself ?
that is A = 2 P
Now apply the formula
$$ A = P \left( 1 + \dfrac{r \times 12}{100} \right) $$
$$ 2P = P \left( 1 + \dfrac{r \times 12}{100} \right) $$
$$ 2 = 1 + \dfrac{ 12 r}{100} $$
$$ 1 = \dfrac{ 12 r}{100} \\ \therefore r = \dfrac{100}{12} = 8 \dfrac{1}{3} \%$$
Question 9
1 / -0

The fractional equivalent of 57.12% (approx.) is

A
$$\frac{359}{625}$$

B
$$\frac{357}{625}$$

C
$$\frac{347}{625}$$

D
$$\frac{349}{625}$$

Solution

$$57.12\% = \dfrac{57.12}{100} = \dfrac{5712}{10000} = \dfrac{357}{625} $$
Question 10
1 / -0

$$\dfrac x4=\dfrac 43$$ then $$x=?$$

A
$$\dfrac {16}3$$

B
$$\dfrac 34$$

C
$$\dfrac {16}9$$

D
None.

Solution

Given relation is $$\dfrac x4=\dfrac 43$$

$$\implies x=4\times \dfrac 43=\dfrac {16}3$$

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 21

Result

Comparing Quantities Test - 21

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SHARING IS CARING

Question 1

$$10$$%

$$40$$%

$$30$$%

$$50$$%

Solution

Question 2

$$12\ \%$$

$$15\ \%$$

$$4\ \%$$

$$10\ \%$$

Solution

Question 3

$$13 \frac {1}{3}$$% gain

$$33 \frac {1}{3}$$% gain

$$33 \frac {1}{3}$$% loss

$$13 \frac {1}{3}$$% loss

Solution

Question 4

$$20\% $$

$$30\% $$

$$40\% $$

$$50\% $$

Solution

Question 5

$$25$$%

$$15\dfrac {1}{4}$$%

$$25\dfrac {1}{8}$$%

$$15\dfrac {5}{8}$$%

Solution

Question 6

$$7:5$$

$$5:7$$

$$8:5$$

$$5:8$$

Solution

Question 7

$$\text{Rs } 55$$

$$\text{Rs }607$$

$$\text{Rs }800$$

$$\text{Rs }1809$$

Solution

Question 8

1/2%

8.33%

both 1 &2

none of these

Solution

Question 9

$$\frac{359}{625}$$

$$\frac{357}{625}$$

$$\frac{347}{625}$$

$$\frac{349}{625}$$

Solution

Question 10

$$\dfrac {16}3$$

$$\dfrac 34$$

$$\dfrac {16}9$$

None.

Solution

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