Comparing Quantities Test

Result

Comparing Quantities Test - 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

10% of 60 + 60% of 100 =?

A
90

B
46

C
66

D
70

Solution

$$\dfrac{10}{100}\times60$$ + $$\dfrac{60}{100}\times100$$ $$=6+60=66$$
Question 2
1 / -0

If the cost of a dozen soaps is $$Rs\ 285.60$$, what will be the cost of $$15$$ such soaps?

A
Rs. $$827$$

B
Rs. $$772$$

C
Rs. $$676$$

D
Rs. $$357$$

Solution

Cost of $$12$$ soaps $$=285.60$$

Cost of one soap $$=\dfrac{285.60}{12}$$

Cost of $$15$$ soaps $$=\dfrac{285.60}{12}\times{15}=Rs.357$$
Question 3
1 / -0

A housewife saved Rs. $$250$$ in buying a dress on sale, If she spent Rs. $$25$$ for the dress, she saved about:

A
$$80\%$$

B
$$90.90\%$$

C
$$89\%$$

D
$$11\%$$

E
$$12\%$$

Solution

The housewife saved $$250$$ Rs and she spent $$25$$ Rs on the dress.

$$\Rightarrow $$Actual cost $$=275$$

Saved percentage =$$\dfrac {250}{275}×100$$

$$\therefore $$saved percentage $$=90\%$$
Question 4
1 / -0

A shopkeeper earns $$15$$% profit on a shirt even after allowing $$31$$% discount on the marked price.If the market price $$1250$$, then the cost price of the shirt is

A
$$870$$

B
$$ 800$$

C
$$750$$

D
$$690$$

Solution

According to the problem the S.P and C.P is
$$\begin{array}{l} Discount=\dfrac { { 31 } }{ { 100 } } \times 1250 \\ =387.5 \\ S.P=1250-387.5 \\ S.P=862.5 \\ C.P=\dfrac { { 100 } }{ { 100+15 } } \times 862.5 \\ =\dfrac { { 100 } }{ { 115 } } \times 862.5 \\ =\dfrac { { 17250 } }{ { 23 } } \\ =750 \end{array}$$
Therefore, the cost price is Rs $$750$$.
Question 5
1 / -0

A shopkeeper sells a sweater at a loss of $$5\%$$. If he had sold it for Rs. $$260$$ more, he would have made a profit of $$15\%$$. Calculate the purchase price of the sweater.

A
Rs $$1900$$

B
Rs $$1400$$

C
Rs $$1000$$

D
Rs $$1300$$

Solution

case1:Let CP be $$x$$
Loss=5%
$$S{P_1} = x - 5\% of\,x$$
$$S{P_1} = 95\% \,of\,x = \dfrac{{95}}{{100}}x$$

Case2: CP$$=x$$,
$${P_2} = 15\% $$
$$S{P_2} = x + 15\% of\,x = \dfrac{{115}}{{100}}x$$

Difference in SP is given $$260$$
Thus,
\begin{array}{l} S{ P_{ 2 } }-S{ P_{ 1 } }=260 \\ \dfrac { { 115x-95x } }{ { 100 } } =260 \\ \Rightarrow 20x=260\times 100 \\ \Rightarrow x=\dfrac { { 26000 } }{ { 20 } } \\ \Rightarrow x=1300 \end{array}
Thus purchase price is $$1300$$ Rs.
Question 6
1 / -0

A dishonest dealer uses a scale of $$'90$$ cm instead of a metre scale and claims to sell at cost price. His profit is :

A
$$11.11 \%$$

B
$$10 \%$$

C
$$12 \%$$

D
None of these

Solution

Dealer uses a scale $$=90\ cm$$
Original scale $$=100\ cm$$

Profit in % $$=\dfrac{100}{90}\times 100$$

$$=11.11$$%
Question 7
1 / -0

Find the value:
$$6\%$$ of $$Rs.150$$

A
$$Rs.6$$

B
$$Rs.9$$

C
$$Rs.15$$

D
None of these.

Solution

To find the value of
$$6\%$$ of $$Rs.150$$

Now,
$$6\%$$ of $$Rs.150$$

$$=\dfrac{6}{100}\times Rs.150$$

$$=Rs.\dfrac{6\times150}{100}$$

$$=Rs.\dfrac{900}{100}$$

$$=Rs. 9$$

Hence, the required value is $$Rs. 9$$ .
Question 8
1 / -0

The compound interest on Rs. 50,000 at 4% per annum for two years compounded anually is :

A
$$4000$$

B
$$4080$$

C
$$4280$$

D
$$4050$$

Solution

C.I. = Amount - Principle
=> $$P((1+\dfrac{r}{100})^{T}_{} - 1)$$
CI = $$50,000(1+\dfrac{4}{100})^{2}_{} - 1)$$
C.I.= $$4080.$$
Question 9
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 10
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 % $$ = \dfrac{\textrm{Profit}}{C} \times 100 \\ = \dfrac{4x}{12x} \times 100 \\ = 33\dfrac{1}{3} \% $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 23

Result

Comparing Quantities Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

90

46

66

70

Solution

Question 2

Rs. $$827$$

Rs. $$772$$

Rs. $$676$$

Rs. $$357$$

Solution

Question 3

$$80\%$$

$$90.90\%$$

$$89\%$$

$$11\%$$

$$12\%$$

Solution

Question 4

$$870$$

$$ 800$$

$$750$$

$$690$$

Solution

Question 5

Rs $$1900$$

Rs $$1400$$

Rs $$1000$$

Rs $$1300$$

Solution

Question 6

$$11.11 \%$$

$$10 \%$$

$$12 \%$$

None of these

Solution

Question 7

$$Rs.6$$

$$Rs.9$$

$$Rs.15$$

None of these.

Solution

Question 8

$$4000$$

$$4080$$

$$4280$$

$$4050$$

Solution

Question 9

$$\text{Rs } 55$$

$$\text{Rs }607$$

$$\text{Rs }800$$

$$\text{Rs }1809$$

Solution

Question 10

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

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

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

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.