The time for which interest is calculated

The time for which loan is given

Period of conversion of simple interest into compound interest

The interest on previously accumulated interest as well as interest on principle amount

The interest on principle amount

The interest on principle for number of years

MCQ Questions for Class 8 Maths Comparing Quantities Quiz 11

Free cash flow is Rs 15000 and net investment in operating capital is Rs 9000 then net operating profit after taxes will be _____________.

0%
Rs 24, 000.00
0%
Rs 6, 000.00
0%
-Rs 6, 000.00
0%
-Rs 24, 000.00

Ramesh bought two boxes for RsHe sold one box at a profit of 20 % and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.

0%
$Rs 550, Rs 750$
0%
$Rs 550, Rs 700$
0%
$Rs 500, Rs 750$
0%
$Rs 500, Rs 750$

The compound interest on $₹\ 10000$ at $10 \%$ per annum for $3$ years, compounded annually, is

0%
1331
0%
3310
0%
3130
0%
13310

Explanation

Given:

Present value $= ₹\ 10000$

Interest rate $= 10 \%$ per annum

Time

$=3$ years

To find the amount we have the formula,

Amount $(A) = P (1+(r/100))^n$

where

$P$ is present value,

$r$ is rate of interest,

$n$ is time in years

Now substituting the values in above formula we get,

$\therefore A = 10000 (1 +10/100)^3$

$\Rightarrow A = 10000 (11/10)^3$

$\Rightarrow A = 121 (10) (11)$

$\Rightarrow A = 1331 (10)$

$\Rightarrow A = ₹\ 13310$

∴ Compound interest $= A – P$

$= 13310 – 10000= ₹\ 3310$

$40\,\%$ of ? =

$240$

0%
$60$
0%
$600$
0%
$6000$
0%
$960$

Explanation

Let us consider a number x

$40\%$ of

$x = 240$

$40/100 \times x = 240$

$x = 240 \times 100/40$

$x = 600$

What percent of

$10kg$ is

$250g$ ?

0%
$25\%$
0%
$5\%$
0%
$10\%$
0%
$2.5\%$

Explanation

We need to convert kg into grams by multiplying by

$1000$ we get,

$10kg= 10 \times 1000 = 10000g$
Now, let us consider a number x

$x\%$ of

$10000$ is

$250$

$x\% \times 10000 = 250$

$100x = 250$

$x = 250/100 = 2.5\%$

$20\%$ of

$800 =$ ?

0%
$160$
0%
$16$
0%
$1600$
0%
None of these

Explanation

Let us consider a number

$x$

$20\%$ of

$₹\ 800 = x$

$\dfrac{20}{100} \times 800 = x$

$20 \times 8 = x$

$x = 160$

$\dfrac{25\% \,of\, 50\% \,of\, 100\%}{25 \times 50}$ is equal to

0%
$1.1\%$
0%
$0.1\%$
0%
$0.01\%$
0%
$1\%$

Explanation

$\dfrac{25\% \,of\, 50\% \,of\, 100\%}{25 \times 50}$

$\Rightarrow \dfrac{\dfrac{25}{100} \times \dfrac{50}{100} \times \dfrac{100}{100}}{25 \times 50}$

$\Rightarrow \dfrac{1}{100 \times 100}$

$\Rightarrow \dfrac{0.01}{100}$

$\Rightarrow 0.01\%$

$20\%$ of

$700 \,m$ is

0%
$560\,m$
0%
$70\,m$
0%
$210\,m$
0%
$140\,m$

Explanation

$20\%$ of

$700\,m$

$=\left ( \dfrac{20}{100} \times 700 \right)$

$=140\,m$

To calculate the growth of bacteria if the rate of growth is known, the formula for calculation of amount in compound interest can be used.

0%
True
0%
False

Mark again the correct answer in each of the following:
If

$x\%$ of

$75=9$ , then the value of

$x$ is

0%
$16$
0%
$14$
0%
$12$
0%
$8$

Explanation

Because,

$=(x/100)\times 75=9$

$x=(9\times 100)/75$

$x=12$

The original price of a shampoo bottle bought for

$Rs\ 324$ . If

$8\%$ VAT is included, the resultant in the price is

$Rs\ 300.65$ ,

0%
True
0%
False

Explanation

The given statement is false.
As per given data, cost price of shampoo's bottle after

$8\%$ VAT $$ , we have

$=300+\dfrac{8}{100}\times 3000$

$=300+8\times 3$

$=Rs\ 324$

The cost of a sewing machine is

$Rs\ 7,000$ . Its value depreciates at

$8\%$\ p.a$ . Then the value of the machine after

$2$ years is

$Rs\ 5,924.80$ .

0%
True
0%
False

Explanation

The given statement is true.

Value of machine after

$1^{st}$ year

$=Rs\ 7000\left(1-\dfrac{8}{100}\right)$

$=Rs\ 7000\left(\dfrac{92}{100}\right)= Rs\ 6,440$

Value of machine after

$2^{nd}$ year

$=Rs\ 6440\left(1-\dfrac{8}{100}\right)$

$=Rs\ 6440\left(\dfrac{92}{100}\right)= Rs\ 5924.80$

Discount is reduction given on cost price of an article.

0%
True
0%
False

If the discount of

$Rs.\ y$ is available on the marked price of

$Rs.\ x$ , then the discount percent is

$\dfrac{x}{y}\times 100\%$

0%
True
0%
False

Explanation

Given, marked price

$=Rs.\ x$
Discount amount

$=Rs.\ y$
So, Discount

$\%=\dfrac{Discount}{Marked\ price}\times 100\%=\dfrac{y}{x}\times 100\%$

Hence, the given statement is false.

The buying price of

$5\ kg$ of flour with the rate

$Rs\ 20$ per kg, when

$5\%$ ST is added on the purchase is

$Rs\ 105$ .

0%
True
0%
False

Explanation

The given statement is true.

Buying price of

$5\ kg$ of flour at rate

$Rs\ 20$ per kg is

$=5\times 20=100$

Also,

$5 \%$ ST is added so, Buying price

$=100+5 \% \ of\ 100=105$

The value of a car, bought for

$Rs\ 4,40,000$ depreciates each year by

$10\%$ of its value at the beginning of that year. So its value becomes

$Rs\ 3,08,000$ after three years.

0%
True
0%
False

Explanation

The given statement is false.
As pre given data, the value of the care after depreciation in

$3$ years is:

$A=P\left(1-\dfrac{R}{100}\right)^{T}=440000\left(1-\dfrac{10}{100}\right)^{3}=440\times 729=Rs\ 320760$

Radhika bought a car for

$Rs. 2,50,000$ . Next year its price decreased by

$10\%$ and further next year it decreased by

$12\%$ . In the two years overall decrease per cent in the price of the car is

0%
$3.2\%$
0%
$22\%$
0%
$20.8\%$
0%
$8\%$

Explanation

The correct answer is option

$(c)$ .
Cost price of the car

$=Rs. 2,50,000$
It's price decreased next year by

$10\%$
Now, the price

$=Rs.(2,50,000-\dfrac{10}{100}\times 2,50,000)=Rs. 2,25,000$
Next year, further its price decreased by

$12\%$
Now, the price

$=Rs.(2,25,000-2,25,000\times \dfrac{12}{100})=Rs. 1,98,000$
Overall decreased percentage in two years,

$=(\dfrac{2,50,000-1,98,000}{2,50,000}\times 100)\%=(\dfrac{52,000}{2,50,000}\times100)\%=20.8\%$

$40\%$ of

$[100-20\%$ of

$300]$ is equal to

0%
$20$
0%
$16$
0%
$140$
0%
$64$

Explanation

The correct answer is option (b).
Given,

$40\%$ of

$[100-20\%$ of

$300]$
Now by the question,

$\dfrac{40}{100}\times \left[ 100-\dfrac{20}{100}\times 300\right]=\dfrac{40}{100}[100-60]=16$

$a\%$ is the discount per cent on a marked price

$x$ , then discount is

0%
$\dfrac xa \times 100$
0%
$\dfrac ax \times 100$
0%
$x\times \dfrac {a}{100}$
0%
$\dfrac {100}{x+a}$

Explanation

The correct option is (c).

Discount is

$=$ Dicounnt

$\%$ of marked price

If marked price of an article is

$Rs. 1,200$ and the discount is

$12\%$ then the selling price of the article is

0%
$Rs. 1,056$
0%
$Rs. 1,344$
0%
$Rs. 1,212$
0%
$Rs. 1,188$

Four alternative options are given for each of the following statements.
Select the correct option.
(g) A radio marked

$Rs. 1000$ is given away for

$Rs. 850$ . The discount is

0%
$Rs. 50$
0%
$Rs. 100$
0%
$Rs. 150$
0%
$Rs. 200$

Write the correct alternative answer for the following question.
For different types of investments what is the maximum permissible amount under section

$80\,C$ of income tax ?

0%
$1, 50, 000$ rupees
0%
$2, 50, 000$ rupees
0%
$1, 00, 000$ rupees
0%
$2, 00, 000$ rupees

Four alternative options are given for each of the following statements.
Select the correct option.
(h) A bookmarked

$Rs. 250$ was sold for

$Rs. 200$ after discount. The percentage of discount is

0%
$10$
0%
$30$
0%
$20$
0%
$25$

Rashmi buys a calculator for

$Rs. 720$ and sells it at a loss of

$6\dfrac {2}{3} \%$ . For how much does she sell it?

0%
$RS. 700$
0%
$RS. 650$
0%
$Rs. 672$
0%
None of these

Explanation

$\textbf{Given: }$

Cost price

$= Rs. 720$

Loss%

$= 6 \dfrac{2}{3}\% = \dfrac{20}{3}\%$

As Loss

$\% = \dfrac{Loss}{CP} \times 100$

$\dfrac{20}{3} = \dfrac{C.P - SP}{CP} \times 100$

$\dfrac{20}{3} = \dfrac{720 - SP}{720} \times 100$

$\Rightarrow 48 = 720-SP$

$\Rightarrow SP = 720 - 48$

$\Rightarrow SP = 672$

$\therefore$ Selling price

$= Rs. 672$

$\boxed{\text{Hence, correct answer is (C)}}$

A bookseller procures 40 books for Rs. 3200 and sells them at a profit equal to the selling price of 8 books. What is the selling price of one dozen books, if the price of each book is same?

0%
Rs. 720
0%
Rs. 960
0%
Rs. 1200
0%
Rs. 1440

The cash difference between the selling prices of an article at a profit of

$4 \%$ and

$6 \%$ is Rs.

$3$ . The ratio of the two selling prices is:

0%
$51:52$
0%
$52:53$
0%
$51:53$
0%
$52:55$

Explanation

Let the C.P be

$x$

Then S.P for

$4\%$ profit

$=\dfrac {104 \times x}{100}$

S.P for

$6\%$ profit

$=\dfrac{106x}{100}$

Then according to the question, we have

$\Rightarrow \dfrac{106x}{100}-\dfrac{104x}{100}=3$

$\Rightarrow 2x=300$

$\Rightarrow x=$ Rs.

$150$

Then S.P for

$4\%$ profit

$=\dfrac{104}{100}\times150 =$ Rs.

$156$

S.P for

$6\%$ profit

$=\dfrac{106}{100}\times 150=$ Rs.

$159$

Ratio of both S.P

$=156:159=52:53$

Mary and mike enter into a partnership by investing Rs.

$700$ and Rs.

$300$ respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received Rs.

$800$ more than Mike did, what was the profit made by their business in that year?

0%
$1000$
0%
$2000$
0%
$3000$
0%
$4000$

Explanation

Let the profit made during the year be Rs.

$3x$
Therefore, Rs.

$x$ would have been shared equally and the remaining Rs.

$2x$ would have been shared in the ratio

$7:3$ .
i.e.,

$70\%$ of

$2x$ would go to Mary and

$30\%$ of

$2x$ would go to Mike.
Hence, Mary would get

$\begin{pmatrix}70-30\end{pmatrix}\%$ of

$2x$ more than Mike.

$\Rightarrow 40\%$ of

$2x=Rs.800$
i.e.,

$\dfrac{40}{100} \times2x=800$

$\Rightarrow 2x=2000$ .
Hence, the profit made by the company during the year Rs.

$3x=$ Rs.

$3000$ .

A biology class at Central High School predicted that a local population of animals will double in size every 12 years. The population at the beginning of 2014 was estimated to be 50 animals. If P represents the population n years after 2014, then which of the following equations represents the classs model of the population over time?

0%
$\displaystyle P=12+50n$
0%
$\displaystyle P=50+12n$
0%
$\displaystyle P={ 50\left( 2 \right) }^{ 12n }$
0%
$\displaystyle P={ 50\left( 2 \right) }^{ \frac { n }{ 12 } }$

The population of a State increased from

$100$ million to

$169$ million in two decades. What is the average increase in population per decade?

0%
$20$ %
0%
$34.5$ %
0%
$69$ %
0%
$30$ %

The population of a City decreases by 3% of the initial at the beginning of every year. If the present population is 1,25,000, what will be after 2 years?

0%
1,17,600
0%
1,17,610
0%
17,610
0%
1,17,613

A town's population increased by

$1200$ people, and then this new population decreased

$11$ %. The town now had

$32$ less people than it did before the

$1200$ increase. Find the original population.

0%
$10000$
0%
$11000$
0%
$12000$
0%
$13000$

Explanation

Let the population be $P$ .

So, the increased population will be $P + 1200$ .

The population after $11\%$ decrease in population becomes $P - 32$ .

Therefore,

${{P}} - 32 = \left( {{{P}} + 1200} \right) - \dfrac{{11}}{{100}}\left( {{{P}} + 1200} \right)$

${{P}} = 10000$

So, the total population is $10000$ .

$83.33\%$ of coins out of

$36$ coins are

$20$ paise coins and rest are

$10$ paise coins which amounts to

$6.60$ . Find the number of

$10$ paise coins.

0%
$8$
0%
$5$
0%
$6$
0%
$2$

Explanation

Given : $83.33$ % of $36$ coins are $20$ paisa coins

$83.33$ % of $36 =\left( \cfrac { 83.33 }{ 100 } \right) \times 36=30$

So, $30$ coins are $20$ paise coins and remaining $(36-30)= 6$ coins are $10$ paise coins.

Value $=30\times 20+6\times 10=660$ Paise $=Rs. 6.60$ , which is the given amount.

$\therefore$ Number of $10$ paise coins are $6$ .

Conversion period is

0%
The time for which loan is given
0%
The time for which interest is calculated
0%
Period of conversion of simple interest into compound interest
0%
None of the above

Compound interest is

0%
The interest on principle amount
0%
The interest on previously accumulated interest as well as interest on principle amount
0%
The interest on principle for number of years
0%
All of the above

A person incurs 15 % loss by selling a watch for Rs. 1530. At what price should the watch be sold to earn 15 % profit

0%
1800
0%
2070
0%
1780
0%
2200

While selling a watch, a shopkeeper gives a discount of 5%. If he gives a discount of 7%; he earns 215 less as profit. The marked price of the watch is _______.

0%
Rs. $697.50$
0%
Rs. $712.50$
0%
Rs. $787.50$
0%
Rs. $750$

Explanation

Let the marked price

$Rs.x$

$Discount\, difference = Profit \;difference$

$x\times\frac{7}{100}-x\times\frac{5}{100}=15$

$\implies x\times\frac{2}{100}=15$

$\implies x=Rs. 750$

Therefore, marked price of watch is Rs. 750

If the list price of a toy is reduced by

$Rs. 2$ a person can buy

$2$ toys more for

$Rs. 360$ . Find the original price of the toy.

0%
$Rs. 10$
0%
$Rs. 15$
0%
$Rs.20$
0%
$Rs. 25$

Explanation

Let the price of toys be ${{Rs}}\;x$ and the number of toys be $y$ . So,

$xy = 360$

$y = \dfrac{{360}}{x}$

According to the question,

$\left( {x - 2} \right)\left( {y + 2} \right) = 360$

$xy - 2y + 2x - 4 = 360$

$360 - 2\left( {\dfrac{{360}}{x}} \right) + 2x - 4 = 360$

${x^2} - 2x - 360 = 0$

${x^2} - 20x + 18x - 360 = 0$

$\left( {x - 20} \right)\left( {x + 18} \right) = 0$

$x = 20,x = - 18$

As the price cannot be negative, so the original price of the toy is ${{Rs}}\;20$ .

Hence option (C) is correct

If you plan to save Rs.

$5,000$ with a bank an interest rate of

$8\%$ what will be the worth of your amount after

$4$ years if interest is compounded annually?

0%
$Rs.5,400$
0%
$Rs.5,900$
0%
$Rs.6,600$
0%
$Rs.6,802$

Ajay has purchased a second hand car by taking a loan of Rs

$5000$ from bank at a rate of interest of

$8\%$ for

$4$ years. Calculate the total amount he will be paying to bank a loan of Rs

$5000$ at the end of

$4$ years

0%
Rs $6820$
0%
Rs $6820.44$
0%
Rs $6800.44$
0%
Rs $6802.44$

You need Rs.

$10,000$ to buy a new television. If you have Rs.

$6,000$ to invest at

$5$ percent compounded annually, how long will you have to wait to buy the telivisions?

0%
$8.42$ years
0%
$10.5$ years
0%
$15.71$ years
0%
$18.78$ years

If the price of a book is reduced by Rs.

$5$ , a person can buy

$5$ more books for Rs

$300$ . Find the original list price of the book.

0%
$15$
0%
$20$
0%
$25$
0%
$30$

Explanation

$\textbf{Step 1: Stating given conditions}$

$\text{Given that the person can buy 5 more books for Rs.300 is the price is reduced by Rs.5}$

$\text{Let the original price of the book be }Rs.x$

$\Rightarrow \text{Number of books he can buy before the price is reduced}=\dfrac{300}{x}$

$\Rightarrow \text{Number of books he can buys after the price is reduced}=\dfrac{300}{x-5}$

$\Rightarrow \dfrac{300}{x}=\dfrac{300}{x-5}+5$

$\textbf{Step 2: Solving}$

$\Rightarrow \dfrac{300}{x}+5=\dfrac{300}{x-5}$

$\Rightarrow 300[\dfrac1{x-5}-\dfrac1{x}]=5$

$\Rightarrow 300[\dfrac{x-x+5}{x(x-5)}]=5$

$\Rightarrow x(x-5)=300$

$\Rightarrow x^2-5x-300=0$

$\Rightarrow x=\dfrac{5\pm\sqrt{1225}}{2}$

$\Rightarrow x=\dfrac{5\pm35}{2}$

$\Rightarrow x=Rs.20$

$\textbf{Hence , the cost of one book is Rs.20. Option B is correct.}$

A shirt was sold at a profit of

$15\%$ . If its cost had been

$5\%$ less and it had been sold for

$Rs.\; 21$ less, then profit would have been

$10\%$ . Find the cost of the shirt.

0%
$Rs.\;210$
0%
$Rs.\;220$
0%
$Rs.\;100$
0%
$Rs.\;200$

Sales tax on the cloth is reduced by

$10\%$ but the sales increase by

$10\%$ . What is the effect on the revenue earned as sales tax?

0%
Remains the same
0%
Increases by $5.5\%$
0%
Decreases by $5.5\%$
0%
Decreases by $1\%$

Explanation

Let the sales be of

$x$ and the sales tax be

$y%$

Then,

Revenue=

$x+\dfrac{y}{100}x~~~~~~~~~~~-(1)$

When Sales tax recued by

$10%$ but sales increase by

$10%$ then

Revenue

$=(x+\dfrac{10}{100}x+\dfrac{(y-10)}{100}x$

$=x+\dfrac{10}{100}x+\dfrac{y}{100}x-\dfrac{10}{100}x$

$=x+\dfrac{y}{100}x~~~~-(2)$

from eqs.(1) and (2) we

Revenue remains the same.

Shyam purchases a scooter costing

$Rs.36,450$ and the rate of sales tax is

$9%$ , then the total amount by her is:

0%
$Rs.36,490.50$
0%
$Rs.39,730.50$
0%
$Rs.36,454.50$
0%
$Rs.33,169.50$

A species has an initial population

${ 4 }^{ 10 }.$ At the end of second day, it decreases by the same percentage. If the process continues in the same pattern, the number of days for the population to reach

${ 3 }^{ 10 }.$ is

0%
10
0%
20
0%
50
0%
100

A shopkeeper prefer to dell his goods at the cost price but uses a weight of 800 gm instead of 1kg. He earns a profit of:

0%
2%
0%
80%
0%
20%
0%
25%

A reduction of

$20%$ in the price of wheat enables Bhuvnesh to buy

$5\ kg$ more wheat for

$Rs.320$ . The original rate (in rupees per

$kg$ ) of wheat was:

0%
$16$
0%
$18$
0%
$20$
0%
$21$

A dishonest dealer professes to sell his goods at cost price but uses a weight of

$960\ gms$ instead of a

$Kg$ weight. Find his gain

$%$ ?

0%
$\dfrac { 27 }{ 4 }\%$
0%
` $\dfrac { 8 }{ 3 }\%$
0%
$\dfrac { 25 }{ 6 }\%$
0%
$\dfrac { 21 }{ 4 }\%$

A bookseller sells a book at a gain of 10%.If he had brought it at 4% less sold it for Rs.6 more,he would have gained

$18\frac {3}{4}$ %.The cost price of the book is

0%
Rs.130
0%
Rs.140
0%
Rs.150
0%
Rs.160

On selling an article for Rs. 42, there is a percent profit of 5%. If the same article is sold for Rs.What is the percent profit?

0%
10
0%
15
0%
20
0%
25

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 11 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 11 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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