Comparing Quantities Test

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Comparing Quantities Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

By selling a jeans for Rs. $$432$$, John loses $$4\%$$. For how much should John sell it to gain $$6\%$$?

A
Rs. $$458$$

B
Rs. $$428$$

C
Rs. $$477$$

D
Rs. $$436$$

Solution

SP of the shirt $$=432$$
Loss $$=4\%$$
Therefore, $$CP$$ of the shirt $$=\dfrac {100}{100-\text{loss} \%}\times SP$$
$$=\dfrac {100}{100-4}\times 432$$
$$=450$$
Now,$$CP=450$$, desired gain $$=6\%$$
Desired$$,SP=\dfrac {100+\text{gain} \%}{100}\times CP$$
$$=\dfrac {100+6}{100}\times 450$$
$$=477$$
Hence, the desired selling price is Rs. $$477$$.
Question 2
1 / -0

A merchant marks his goods in such a way that the profit on sale of $$50$$ articles is equal to the selling price of $$25$$ articles. What is his profit margin ?

A
$$25\%$$

B
$$50\%$$

C
$$100\%$$

D
$$66.67\%$$

Solution

Let the selling price per article be $$=$$ Rs. $$1$$
Therefore, selling price of $$50$$ articles $$=$$ Rs. $$50$$
Profit on sale of $$50$$ articles $$=$$ selling price of $$25$$ articles $$=$$ Rs. $$25$$.
S.P $$=$$Rs. $$ 50$$, Profit $$=$$ Rs. $$25$$
Therefore, CP $$=$$ Rs.$$(50-25)=$$ Rs. $$25$$
Profit $$\%=\dfrac{25}{25}\times100\%=100\%$$
Question 3
1 / -0

The cost of an article is decreased by $$15$$ $$\%$$. If the original cost is Rs. $$80$$, find the new cost.

A
$$68$$

B
$$65$$

C
$$57$$

D
$$55$$

Solution

Original cost of an article $$=$$ Rs. $$80$$
Decrease in cost is $$=15$$ $$\%$$ of Rs. $$80=\dfrac{15}{100}\times80=\dfrac{1200}{100}=$$ Rs. $$12$$
Therefore, new cost of an article $$=$$ Rs. $$80-$$ Rs. $$12=$$ Rs. $$68$$.
Question 4
1 / -0

The price of a TV is Rs. $$13,000$$. The sales tax on it is $$12\%$$. Find the amount that Vinod will have to pay if he buys it.

A
Rs. $$14560$$

B
Rs. $$2620$$

C
Rs. $$1456$$

D
Rs. $$1560$$

Solution

Sales tax $$=$$ Price $$\times$$ Tax Rate
$$=\dfrac{12}{100} \times 13000=1560$$
Final price $$=13000+1560=14560$$
Therefore, the amount that Vinod will have to pay will be Rs. $$14560$$.
Question 5
1 / -0

The present population of town is $$12500$$ and it is increasing at the rate of $$8\%$$ per annum. What will be the population of town after two years?

A
$$14000$$

B
$$12500$$

C
$$14580$$

D
$$25000$$

Solution

Given, present population of town $$=12500, R=8\%$$
Therefore, population after $$2$$ years $$=12500\times\begin{Bmatrix}1+\dfrac{8}{100}\end{Bmatrix}^2$$
$$=12500\times\begin{Bmatrix}\dfrac{108}{100}\end{Bmatrix}^2$$
$$=14580$$
Question 6
1 / -0

In a culture, the present bacteria count was found as $$67,60,000$$. It was found that the number was increased by $$4\%$$ per hour. Find the number of hours before which the bacteria count was $$62,50,000$$.

A
$$1$$ hr

B
$$2$$ hr

C
$$3$$ hr

D
$$4$$ hr

Solution

Let the number of years be $$n$$
$$\Rightarrow 67,60,000=62,50,000 \left [1+\dfrac{4}{100}\right]^n$$
$$\Rightarrow \dfrac{676}{625}=[1+\dfrac{1}{25}]^n$$
$$\Rightarrow \left [\dfrac{26}{25}\right]^2=\left [\dfrac{26}{25}\right]^n$$
$$n=2$$ hrs
Thus, the number of hours are $$2$$.
Question 7
1 / -0

The present population of a city is $$18522000$$. If it has been increased at the rate of $$5\%$$ per annum, find the population(approx) after $$3$$ years.

A
$$2522200$$

B
$$16000000$$

C
$$21441530$$

D
$$252200$$

Solution

Given, Principal amount $$18522000$$, rate of interest $$=5\%$$ and period $$=3$$ years
Total Population $$=18522000\left [1+\dfrac{5}{100}\right]^3$$
$$=18522000\left [1+\dfrac{1}{20}\right]^3$$
$$=21441530$$(approx)
Question 8
1 / -0

A plot is sold for Rs.$$18,700$$ with a loss of $$15\%$$. At what price it should be sold to get profit of $$15\%$$?

A
$$25300Rs$$

B
$$54000Rs$$

C
$$25600Rs$$

D
$$25900Rs$$

Solution

Let at Rs. $$ x$$

It can earn $$15\%$$ profit

$$\Rightarrow 85:18700=115:x$$ [as, loss $$=100-15$$, profit $$=100+15$$]

$$\Rightarrow x=\dfrac{18700\times1}{85}$$

$$\Rightarrow x=$$ Rs. $$25300$$

So, the plot is sold of Rs. $$25300$$
Question 9
1 / -0

What is the percentage approximate for the fraction $$\dfrac{12}{35}$$?

A
$$34.28\%$$

B
$$23.78\%$$

C
$$30.12\%$$

D
$$29.10\%$$

Solution

To get the percent you will divide the numerator by the denominator.
Then multiply $$100$$ to the answer and add the percent sign.
therefore, the percentage approximate value if $$\dfrac{12}{35}\times 100 = 34.28$$ $$\%$$.
Question 10
1 / -0

A person incurs a loss of $$5\%$$ be selling a watch for $$Rs.1140$$. At what price should the watch be sold to earn $$5\%$$ profit ?

A
Rs.$$1200$$

B
Rs.$$1230$$

C
Rs.$$1260$$

D
Rs.$$1290$$

Solution

Formulas required for this sum:
$$Cost\ price(C.P)=\dfrac{100}{100-Loss\%}\times Selling\ price$$

$$Selling\ price(S.P)=\dfrac{100+Profit\%}{100\%}\times Cost\ price$$
$$Loss\%=5\%$$
$$S.P=Rs\ 1140$$

$$C.P=\dfrac{100}{100-5}\times1140$$

$$C.P=\dfrac{100}{95}\times1140$$
$$C.P=1200$$
We found that the $$Cost\ price\ is\ Rs\ 1200$$
$$Profit\%=5\%$$

$$S.P=\dfrac{100+5}{100}\times 1200$$

$$S.P=\dfrac{105}{100}\times 1200$$
$$S.P=Rs\ 1260$$

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 47

Result

Comparing Quantities Test - 47

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SHARING IS CARING

Question 1

Rs. $$458$$

Rs. $$428$$

Rs. $$477$$

Rs. $$436$$

Solution

Question 2

$$25\%$$

$$50\%$$

$$100\%$$

$$66.67\%$$

Solution

Question 3

$$68$$

$$65$$

$$57$$

$$55$$

Solution

Question 4

Rs. $$14560$$

Rs. $$2620$$

Rs. $$1456$$

Rs. $$1560$$

Solution

Question 5

$$14000$$

$$12500$$

$$14580$$

$$25000$$

Solution

Question 6

$$1$$ hr

$$2$$ hr

$$3$$ hr

$$4$$ hr

Solution

Question 7

$$2522200$$

$$16000000$$

$$21441530$$

$$252200$$

Solution

Question 8

$$25300Rs$$

$$54000Rs$$

$$25600Rs$$

$$25900Rs$$

Solution

Question 9

$$34.28\%$$

$$23.78\%$$

$$30.12\%$$

$$29.10\%$$

Solution

Question 10

Rs.$$1200$$

Rs.$$1230$$

Rs.$$1260$$

Rs.$$1290$$

Solution

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