MCQ Questions for Class 8 Maths Comparing Quantities Quiz 4

A shopkeeper sold a TV set for Rs.

$17940$ (price inclusive of a discount of

$8 \%$ ) and gained

$19.6 \%$ . If no discount is allowed, then what will be his gain per cent ?

0%
$25 \%$
0%
$26.4 \%$
0%
$24.8 \%$
0%
$30 \%$

Explanation

Given that discount on M.P. is

$8\%$ .

Means if M.P. is

$100$ , then S.P. is

$92$ .

Then if S.P. is Rs.

$17940$ , then M.P. is

$\dfrac {17940}{92}\times 100=19500$ .

If S.P. is

$17940$ and profit is

$19.6$

Means if C.P. is

$100$ , then S.P. is

$119.6$ .

So, when S.P. is

$17940$ , then C.P.

$=\dfrac {17940}{119.6}\times 100=15000$ .

If no discount was given, then C.P.

$=15000$ and S.P.

$=19500$ .

Profit

$=19500-15000=4500$

$\%$ profit

$=\dfrac {4500}{15000}\times 100=30\%$

If a shopkeeper sells

$\dfrac {1}{3}$ rd of his goods at a profit of

$14\%$ ,

$\dfrac {3}{5}$ th of the goods at a profit of

$17.5\%$ and remaining at a profit of

$20\%$ , then his profit on the whole is equal to

0%
$15.5\%$
0%
$16 \%$
0%
$16.5\%$
0%
$17\%$

Explanation

Let total cost of all goods be Rs. $100$ .
He sold $\dfrac{1}{3}$ of his goods that is Rs. $33.3$ at a profit of $14\%$
Then profit $= 33.3 \times \dfrac{14}{100} =$ Rs. $4.662$
Simillarly $\dfrac{3}{5} = 60$ , Profit $17.5\%$
Then profit $= 60 \times \dfrac{17.5}{100} = 0.60 \times 17.5 = 10.5$
Now $100 - (33.3+60) = 100 - 93.3 =$ Rs. $6.7$
Profit on Rs. $6.7 = \dfrac{6.7 \times 20}{100} = 1.34$
Total profit $= 4.662+10.5+1.34=$ Rs. $16.502 = 16.5\%$

I bought two tables for Rs.

$5400$ . I sold one at a loss of

$5\%$ and the other at a gain of

$7\%$ . In the whole transaction, I neither gained nor lost. Find the costs of the tables separately.

0%
Cost of the two tables are Rs. $3790$ and Rs. $2720$ .
0%
Cost of the two tables are Rs. $3150$ and Rs. $2250$ .
0%
Cost of the two tables are Rs. $2670$ and Rs. $1850$ .
0%
None of these

Explanation

Let the C.P. of one table

$=$ Rs.

$x$

Then, C.P. of the other table

$=$ Rs.

$(5400 - x)$
S.P. of the first table at a loss of

$5\%$

$=$ Rs.

$\dfrac {95}{100}\times x$

S.P. of the other table at a gain of

$7\%$

$=$ Rs.

$\dfrac {107}{100}\times (5400-x)$
Since there is no gain no loss.
Total S.P.

$=$ Total C.P.

$\Rightarrow \dfrac {95x}{100}+\dfrac {107\times (5400-x)}{100}=5400$

$\Rightarrow \dfrac {95x}{100}-\dfrac {107x}{100}+\dfrac {107\times 5400}{100}=5400$

$\Rightarrow \dfrac {-12x}{100}=\dfrac {5400\times 100-5400\times 107}{100}$

$\Rightarrow \dfrac{-12x}{100}=\dfrac{540000-577800}{100}$

$\Rightarrow -12x=-37800$

$\Rightarrow x=\dfrac{37800}{12}=$ Rs.

$3150$

$\therefore$ Cost of the two tables are Rs.

$3150$ and

$(5400-3150) =$ Rs.

$2250$ .

A machine is sold at a profit of

$10\%$ . Had it been sold for Rs.

$40$ less, there would have been a loss of

$10\%$ . What was the cost price?

0%
Rs. $175$
0%
Rs. $200$
0%
Rs. $225$
0%
Rs. $250$

Explanation

Let the cost price, C.P. be Rs.

$x$

Profit

$= 10\%$

Hence, S.P.

$= 1.1(x) =$ Rs.

$1.1x$

If a loss has been incurred, then the selling price of the machine would be Rs.

$0.9x$ .

Difference in selling prices

$= 40$

$1.1x - 0.9x = 40$

$0.2x = 40$ or

$x = 200$

The cost price of the machine is Rs.

$200$ .

If a shopkeeper sells a watch for Rs.

$552$ , he will make a profit of

$15\%$ . If he allows a discount of

$5 \%$ for cash payment his actual profit percentage is:

0%
$10\%$
0%
$5\%$
0%
$14.5\%$
0%
$9.25\%$

Explanation

S.P. of the watch is Rs.

$552$

Profit

$= 15\%$

Let the C.P. be Rs.

$x$

So, the S.P. will be Rs.

$1.15x$

Equating both selling prices, we get

$1.15x = 552$ or

$x =$ Rs.

$480$ .

Cost price

$=$ Rs.

$480$ .

Discount offered

$= 5\%$

So, the new S.P. will be

$0.95(552) =$ Rs.

$524.4$

Profit

$=$ Rs.

$524.4 - 480 =$ Rs.

$44.4$

Profit

$\%$

$= 44.4\times \dfrac {100}{480} = 9.25\%$

A trader lists his articles

$20\%$ above cost price and allows a discount of

$10\%$ on cash payment. His gain per cent is

0%
$10\%$
0%
$8\%$
0%
$6\%$
0%
$5\%$

The cost price of

$20$ articles is the same as selling price of

$x$ articles. If the profit is

$25$ %, then the value of

$x$ is

0%
$15$
0%
$16$
0%
$18$
0%
$25$

Explanation

$\textbf{Step 1:Calculate CP of 20 articles and SP of x Articles}$

$\text{Let the C.P. of 1 article be}$

$Rs. 1$ .

$\therefore \text{C.P. of x articles be }$

$Rs. x$ ,

$\Rightarrow \text{S.P. of x articles=C.P. of 20 articles=}$

$Rs. 20$

$\text{Profit for x Article= SP of x Article - CP of x Article}$

$\therefore \text{Profit =Rs. (20 - x)}$ .

$\Rightarrow \text{Profit} \%$

$= 25 \%$

$\Rightarrow \dfrac{SP-CP}{CP} \times100=25$

$\Rightarrow \dfrac {(20-x)}{x}\times 100=25$

$\Rightarrow 2000-100x=25x$

$\Rightarrow 125x=2000$

$\Rightarrow x=16$

$\textbf{Therefore,Value of x is 16}$

A cycle agent buys

$30$ bicycles, of which 8 are first grade and the rest are second grade, for Rs.

$3150$ . Find at what price he must sell the first grade bicycles so that if he sells the second grade bicycles at three quarters of the price, he may make a profit of

$40\%$ on his out lay?

0%
Rs. $200$
0%
Rs. $240$
0%
Rs. $180$
0%
Rs. $210$

Explanation

Let the S.P.of one

$A$ grade cycle

$=$ Rs.

$n$
Then, the S.P. of one

$B$ grade cycle

$=$ Rs.

$\dfrac {3n}{4}$

C.P. of

$30$ cycles

$=$ Rs.

$3150$ , profit

$= 40\%$ .

$\therefore$ S.P. of

$30$ cycles

$=\dfrac {140}{100}\times 3150=$ Rs.

$3150\times 1.4$

According to the question, we have

$8n+22\times \dfrac {3n}{4}=3150\times 1.4$

$\Rightarrow 32n+66n=17640$

$\Rightarrow 98n=17640$

$\Rightarrow n=180$
Therefore, he should sell the first grade bicycle at Rs.

$180$ each.

A merchant has

$1000$ kg of sugar, part of which he sells at

$8\%$ profit and the rest at

$18\%$ profit. He gains

$14\%$ on the whole. The quantity of sugar sold at

$18\%$ profit is

0%
$560$ kg
0%
$600$ kg
0%
$400$ kg
0%
$640$ kg

Explanation

Let the C.P. of sugar be Rs.

$x$ per kg.
Then, C.P.

$=$ Rs. 1

$000 x$
Let the sugar sold at

$8\%$ gain by

$y$ kg.
Then, sugar sold at

$18\%$ gain

$= (1000 - y)$ kg

$\therefore \left (\dfrac {108}{100}\times xy\right )+\left [\dfrac {118}{100}\times x(1000-y)\right ]=\dfrac {114}{100}\times 1000x$

$\Rightarrow \dfrac {108y}{100}+1180-\dfrac {118y}{100}=1140$

$\Rightarrow \dfrac {10y}{100}=40$

$\Rightarrow y=400$

$\therefore$ Quantity sold at 18% gain

$=(1000-400)$ kg

$=600$ kg.

Sanjay sold a bicycle to Salman at

$46\%$ profit. Salman spent Rs.

$40$ on repairs and sold it to Sunil for Rs.

$1500$ . In this deal, Salman made neither profit nor loss. What is the cost price for Sanjay?

0%
Rs. $900$
0%
Rs. $960$
0%
Rs. $1000$
0%
Rs. $1060$

Explanation

Let the C.P. for Sanjay be Rs. $x$
Then, S.P. for Sanjay $= x + 46\%$ of $x$
$=x+\dfrac {46}{100}x=\dfrac {146x}{100}$
$\therefore$ C.P. of Salman $=\dfrac {146x}{100}+40$
$\because$ Salman made neither profit nor loss.
$\dfrac {146x}{100}+40=1500$
$\Rightarrow \dfrac {146x}{100}=1460$

$\Rightarrow x=\dfrac {1460\times 100}{146}=$ Rs. $1000$ .

The cost of a machine is Rs.

$24500$ . After two year the value of that machine was depreciated by

$6\%$ . Find the value of machine after two year.

0%
$11560$
0%
$21560$
0%
$31560$
0%
$41560$

Explanation

Given: Principal

$= 24500$ ,

$r = 6\%, n = 2$
Reduction

$= 6\%$ of Rs.

$24500$ per year

$= \dfrac{24500\times 6\times 2}{100}=2940$
Value at the end of

$2$ years

$= 24500 - 2940 =$ Rs.

$21560$

A man bought two goats for Rs.

$1008$ . He sold one at a loss of

$20\%$ and the other a profit of

$44\%$ . If each goat was sold for the same price, the cost price of the goat which was sold at a loss was

0%
Rs. $648$
0%
Rs. $360$
0%
Rs. $568$
0%
Rs. $440$

Explanation

Let the C.P. of a goat be Rs. $x$

Then, C.P. of the other goat $=$ Rs. $(1008 - x)$
$\because$ S.P. of the both the goats is the same,
Therefore, $x\times \dfrac {(100-20)}{100}=(1008-x)\times \dfrac {144}{100}$
$\Rightarrow 80x=1008\times 144-144x$
$\Rightarrow 224x=1008\times 144$
$\Rightarrow x=$ Rs. $\dfrac {1008\times 144}{224}=$ Rs. $648$

A dealer buys an old cooler listed at Rs.

$950$ and gets a discount of

$10\%$ . He spends Rs.

$45$ for its repair. If he sells the cooler at a profit of

$25\%$ , then the selling price of the cooler is

0%
Rs. $1125$
0%
Rs. $1215$
0%
Rs. $1251$
0%
Rs. $1512$

Explanation

C.P. $=$ Rs. $950 - 10\%$ of Rs. $950$

$\Rightarrow 950-\dfrac{10}{100}\times 950$

$\Rightarrow 950-95=$ Rs. $855$
Repairs cost $=$ Rs. $45$
$\therefore$ Net C.P. $=$ Rs. $855+$ Rs. $45=$ Rs. $900, \text{Profit}=25\%$

S.P. $=\dfrac{100+\text{profit}\%}{100}\times \text{C.P.}$

$\therefore$ S.P. $=$ Rs. $\left (\dfrac {900\times 125}{100}\right )=$ Rs. $1125$

A person sells a table at a profit of

$20\%$ . If he had bought it at

$10\%$ less cost and sold for Rs.

$105$ more, he would have gained

$35\%$ . The cost price of the table is

0%
Rs. $7000$
0%
Rs. $7350$
0%
Rs. $9200$
0%
Rs. $8140$

Explanation

Let the C.P. of the table be Rs. x.
Then, S.P. of the table at 20% profit

$=\dfrac {120}{100}\times Rs. x=Rs \dfrac {6x}{5}$
C.P. of the table at 10% loss

$=\dfrac {90}{100}\times Rs. x$

$=Rs \dfrac {9x}{10}$
S.P. of the table at 35% profit now

$=\dfrac {135}{100}\times Rs \dfrac {9x}{10}$

Then according to the question

$\dfrac {6x}{5}+105=\dfrac {135}{100}\times \dfrac {9x}{10}$

$\Rightarrow \dfrac {6x}{5}+105=\dfrac {243}{200}x$

$\Rightarrow \dfrac {243}{200}x-\dfrac {6}{5}x=105$

$\Rightarrow \left (\dfrac {243-240}{200}\right )x=105$

$\Rightarrow x=\dfrac {105\times 200}{3}=Rs. 7000$ .

A man sells two horses for

$Rs. 1485$ . The cost price of the first is equal to the selling price of the second. If the first is sold at

$20\%$ loss and the second at

$25\%$ gain, what is his total gain or loss (in rupees)?

0%
$Rs. 80$ gain
0%
$Rs. 60$ gain
0%
$Rs. 60$ loss
0%
Neither gain nor loss

Explanation

Let the S.P. of the first horse

$=Rs. x$ . Then,
S.P. of the second horse

$=Rs. (1485 - x)$
C.P. of the first horse

$=\dfrac {x\times 100}{(100-20)}=Rs \dfrac {5x}{4}$
C.P. of the second horse

$=\dfrac {(1485-x)\times 100}{(100+25)}$

$=Rs \dfrac {4(1485-x)}{5}$
Given,

$\dfrac {5x}{4}=(1485-x)$

$\Rightarrow 5x=4\times (1485-x)\Rightarrow 9x=4\times 1485$

$\Rightarrow x=4\times \dfrac {1485}{9}=Rs. 660$

$\therefore$ Total C.P.

$=\dfrac {5\times 660}{4}+\dfrac {4(1485-660)}{5}$

$=Rs. 825+Rs. 660=Rs. 1485$
Since S.P.

$=$ C.P., there is no profit no loss.

A man bought two old scooters for Rs.

$18000$ . By selling one at a profit of

$25\%$ and the other at a loss of

$20\%$ , he neither gains nor loses. Find the cost price of each scooter.

0%
$12000, 6000$
0%
$8000, 10000$
0%
$11000, 7000$
0%
$13000, 5000$

Explanation

Let the C.P. of the

$1^{st}$ scooter

$=$ Rs.

$x$ .

Then, C.P. of the

$2^{nd}$ scooter

$=$ Rs.

$(18000 - x)$
S.P. of

$1^{st}$ scooter

$=$ Rs.

$x\times \dfrac {125}{100}=$ Rs.

$\dfrac {5x}{4}$
S.P. of

$2^{nd}$ scooter

$=$ Rs.

$(18000-x)\times \dfrac {80}{100}$

$=$ Rs.

$\dfrac {4}{5}(18000-x)$
Since Mohan neither gains nor loses, his total S.P.

$=$ Total C.P.

$\Rightarrow \dfrac {5x}{4}+\dfrac {4}{5}(18000-x)=18000$

$\Rightarrow \dfrac {5x}{4}-\dfrac {4x}{5}=18000-14400=3600$

$\Rightarrow \dfrac {25x-16x}{20}=3600\Rightarrow 9x=3600\times 20$

$\Rightarrow x=\dfrac {3600\times 20}{9}=8000$

$\therefore$ C.P. of

$1^{st}$ scooter

$=$ Rs.

$8000$
C.P. of

$2^{nd}$ scooter

$=$ Rs.

$10,000$ .

A tradesman sold an article at a loss of

$20\%$ . If the selling price had been increased by Rs.

$100$ , there would have been a gain of

$5\%$ . The cost price of the article was

0%
Rs. $200$
0%
Rs. $225$
0%
Rs. $400$
0%
Rs. $250$

Explanation

Let the C.P. of the article be Rs. $x$

Then, S.P. of the article at a loss of $20\%$
$=\dfrac {80}{100}\times x=$ Rs. $\dfrac {80x}{100}$
S.P. of the article at a profit of $5\%$
$=\dfrac {105}{100}\times x=$ Rs. $\dfrac {105x}{100}$
According to the question, we have
$\dfrac {80x}{100}+100=\dfrac {105x}{100}$

$\Rightarrow \dfrac {105x}{100}-\dfrac {80x}{100}=100$
$\Rightarrow x=\dfrac {100\times 100}{25}=$ Rs. $400$

A pair of articles was bought for Rs.

$37.40$ at a discount of

$15\%$ . What must be the marked price of each article?

0%
Rs. $36$
0%
Rs. $30$
0%
Rs. $26$
0%
Rs. $22$

Explanation

S.P. of each article $=$ Rs. $\left (\dfrac {37.40}{2}\right )=$ Rs. $18.70$
Discount rate $=15\%$
$\therefore$ M.P. of each article $=$ Rs. $\dfrac {18.70\times 100}{(100-15)}=$ Rs. $\dfrac {1870}{85}=$ Rs. $22$ .

At a clearance sale, all goods are on sale at

$35\%$ discount. If I buy a dress marked Rs.

$800$ , how much would I need to pay?

0%
Rs. $280$
0%
Rs. $350$
0%
Rs. $520$
0%
Rs. $600$

Explanation

M.P.

$=$ Rs.

$800$ , Discount

$= 35\%$

S.P

$=$ M.P.

$-$ discount

$\therefore$ S.P.

$=$ Rs.

$800 - 35\%$ of Rs.

$800$

$\Rightarrow 800-\dfrac{35}{100}\times 800$

$\Rightarrow 800-280=$ Rs.

$520$

Mohan bought

$20$ dining tables for Rs.

$12000$ and sold them at a profit equal to the selling price of

$4$ dining tables. The selling price of each dining table is

0%
Rs. $700$
0%
Rs. $750$
0%
Rs. $725$
0%
Rs. $775$

Explanation

Profit

$=$ (S.P. of

$20$ tables)

$-$ (C.P. of

$20$ tables)

$\Rightarrow$ S.P. of

$4$ tables

$=$ S.P. of

$20$ tables

$-$ C.P. of

$20$ tab1es

$\Rightarrow$ S.P. of

$16$ tables

$=$ C.P. of

$20$ tables

$=$ Rs.

$12000$

$\Rightarrow$ S.P. of

$1$ dining table

$=$ Rs.

$\dfrac {12000}{16}=$ Rs.

$750$

A dealer marks an item

$40\%$ above the cost price and offers a discount of

$25\%$ on the marked price. What is his profit percentage?

0%
$5\%$
0%
$7\%$
0%
$9\%$
0%
$11\%$

Explanation

Let the C.P. of the item be Rs. $100$
Then, M.P. $=$ Rs. $140$ , Discount $= 25\%$
$\therefore$ S.P. $=$ Rs. $140 - 25\%$ of Rs. $140$
$=$ Rs. $140-\dfrac {25}{100}\times$ Rs. $140$

$=140-35=$ Rs. $105$
$\therefore$ Profit $=$ Rs. $105-$ Rs. $100=$ Rs. $5$
Profit $\%$ $=\dfrac {5}{100}\times 100=5\%$ .

A shopkeeper sells his goods at

$10\%$ discount on the marked price. What price should be marked on an article that costs him Rs.

$900$ to gain

$10\%$ ?

0%
Rs. $800$
0%
Rs. $900$
0%
Rs. $1000$
0%
Rs. $1100$

After, allowing a discount of

$26\%$ on the marked price of an article, it is sold at Rs.

$444$ . Find its marked price

0%
Rs. $500$
0%
Rs. $550$
0%
Rs. $640$
0%
Rs. $600$

Explanation

S.P. $=$ Rs. $444$ , Discount $= 26\%$
Let M.P. $=$ Rs. $x$

Then, M.P. $-$ discount $=$ S.P,

$x - 26\%$ of $x = 444$
$\Rightarrow x-\dfrac{13x}{50} = 444$
$\Rightarrow \dfrac{37x}{50}=444$

$\Rightarrow x=\dfrac{444\times 50}{37}=$ Rs. $600$

Ashok buys a car at

$12\%$ discount of its value and sells it for

$20\%$ more than its value. What will be his profit percentage?

0%
$40\%$
0%
$50\%$
0%
$66$ $\dfrac {2}{3}\%$
0%
$20\%$

Explanation

Let the value of the car (M.P.)

$=$ Rs.

$100$
Then, C.P.

$= 80\%$ of Rs.

$100 =$ Rs.

$80$
S.P.

$= 120\%$ of Rs.

$100 =$ Rs.

$120$

$\therefore$ Profit

$\%$

$=\dfrac {40}{80}\times 100=50\%$

Therefore, the profit percentage is

$50\%$ .

Find the rate of discount when an article marked at Rs.

$1500$ is sold at Rs.

$1140$ .

0%
$25\%$
0%
$76\%$
0%
$24\%$
0%
$20\%$

Explanation

Given, M.P.

$=$ Rs.

$1500$ , S.P.

$=$ Rs.

$1140$

Discount

$=$ M.P.

$-$ S.P.

Discount

$=1500-1140=$ Rs.

$360$

Discount

$\%$

$=\dfrac{360}{1500}\times 100=24\%$

Thus the rate of discount is

$24\%$ .

By selling a transistor for Rs.

$572$ , a shopkeeper earns a profit equivalent to

$30\%$ of the cost price of the transistor. What is the cost price of the transistor?

0%
Rs. $340$
0%
Rs. $400$
0%
Rs. $440$
0%
Rs. $350$

Explanation

Let the C.P. of the transistor be Rs. $x$
Profit $=30\%$ of Rs. $x$ $=$ Rs. $\dfrac {3x}{10}$
S.P. $=$ Rs. $572$

C.P. $+$ Profit $=$ S.P.
Therefore, $x+\dfrac {3x}{10}=572$

$\Rightarrow \dfrac {13x}{10}=572$
$\Rightarrow x=$ Rs. $\dfrac {572\times 10}{13}=$ Rs. $440$

A discount of

$15\%$ on one article is the same as a discount of

$20\%$ on another article. The cost of the two articles can be

0%
Rs $40$ , Rs. $20$
0%
Rs. $60$ , Rs. $40$
0%
Rs. $80$ , Rs. $60$
0%
Rs. $50$ , Rs. $30$

Explanation

Let the cost of the two articles be

$x$ and

$y$ .

Then,

$15\%$ of

$x=20\%$ of

$y$

$\Rightarrow \dfrac{15}{100}\times x=\dfrac{20}{100}\times y$

$\Rightarrow \dfrac{x}{y}=\dfrac{20}{100}\times \dfrac{100}{15}$

$\Rightarrow \dfrac {x}{y}=\dfrac {20}{15}=\dfrac {4}{3}$

$\therefore$

$x$ and

$y$ should be in the ratio

$4: 3$ .

Then the cost price

$=$ Rs.

$80$ and Rs.

$60$ .

What would be the printed price of a watch purchased at Rs.

$380$ , so that after giving

$5\%$ discount, there is

$25\%$ profit?

0%
Rs. $400$
0%
Rs. $450$
0%
Rs. $500$
0%
Rs. $600$

Explanation

Given, C.P. $=$ Rs. $380$ , Profit $= 25\%$
$\therefore$ S.P. $=$ Rs. $\left (\dfrac {380\times 125}{100}\right )=$ Rs. $475$
Discount $=$ $5\%$
$\therefore$ M.P. $=$ Rs. $\left (\dfrac {475\times 100}{95}\right )=$ Rs. $500$

A shopkeeper earns a profit of

$15$ % after selling a book at

$20$ % discount on the printed price. The ratio of the cost price and the printed price of the book is

0%
$20:23$
0%
$23:20$
0%
$16:23$
0%
$23:16$

Explanation

Let the printed price of the book be Rs.

$100$
After a discount of

$20\%$ , S.P.

$=$ Rs.

$80$
Profit earned

$= 15\%$

Thus, Cost Price of Book

$=$

$\displaystyle \frac{100}{115}\times 80$ =

$\displaystyle \frac{1600}{23}$

Hence, (C.P) : (printed price)

$=$

$\displaystyle \frac{1600}{23}:100$

$=$

$16:23$

The value of an article which was purchased

$2$ years ago, depreciates at

$12$ % per annum. If its present value is Rs .

$9680$ , the price at which it was purchased is:

0%
Rs. $10,000$
0%
Rs. $12,500$
0%
Rs. $14,575$
0%
Rs. $16,250$

Explanation

Given,

$A=$ Rs.

$9680,T=2$ years,

$R=12\%$

$\because A = P\left (1 - \displaystyle \frac{R}{100} \right)$

$9680 = P \left(1 - \displaystyle \frac{12}{100} \right)^{2}$

$9680=P\left(\dfrac{88}{100}\right)^2$

$P = 9680 \times \displaystyle \frac{25}{22} \times \frac{25}{22}$

$=$ Rs.

$12,500$

Successive discounts of

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

$7\dfrac {1}{2}$ % are given on the marked price of a cupboard. If the customer pays Rs

$2590$ , then what is the marked price?

0%
Rs. $3108$
0%
Rs. $3148$
0%
Rs. $3200$
0%
Rs. $3600$

Explanation

Let the M.P. of the cupboard be Rs $x$ .

Then, $(100-12.5)\%$ of $(100-7.5)\%$ of Rs. $x=2590$

$\Rightarrow \dfrac {87.5}{100}\times \dfrac {92.5}{100}\times x=2590$
$\Rightarrow x=\dfrac {2590\times 100\times 100}{87.5\times 92.5}=$ Rs. $3200$

An article when sold at a gain of

$5$ % yield Rs.

$15$ more than when sold at a loss of

$5$

$\%$ . Its cost price would be:

0%
Rs. $150$
0%
Rs. $200$
0%
Rs. $250$
0%
Rs. $300$

Explanation

Suppose the CP of the article

$=$ Rs.

$100$
I case :
Profit

$= 5\%$
SP

$=$

$100 + 5$

$=$ Rs.

$105$
II case :
Loss

$= 5\%$
SP

$=$ CP

$-$ loss

$=$

$100 - 5$

$=$ Rs.

$95$
Difference between two SP's

$=$

$105 - 95$

$=$ Rs.

$10$
If the difference

$=$ Rs.

$10$ , then CP

$=$ Rs.

$100$
If the difference

$=$ Rs.

$15$ , then CP

$= \frac{100}{10}\times 15=$ Rs.

$150$

Which type of tax is paid by a landowner to the local government or the municipal corporation of his area?

0%
income tax
0%
sales tax
0%
property tax
0%
VAT

A shopkeeper buys an article for RsAfter allowing a discount of 10% on his marked price, he wants to earn a profit of 20%. His marked price must be

0%
Rs 210
0%
Rs 240
0%
Rs 270
0%
Rs 300

Explanation

Given,

Cost price of an article, $CP=$ Rs $180$
Profit or gain = $20$ %
Discount on the marked price of an article $=10$ %

Marked price of an article, $MP=?$

Let the marked priice of an article be $x$ .

Discount on the marked price on an article $=x-0.10 x$

$=0.9 x$

$\therefore$ Selling Price of an article is, $SP=0.9x$

We know that,

$\dfrac{SP-CP}{CP}=Gain$ %

Put $SP=0.9x$ , $CP=180$ , and $Gain=20$ %

$\dfrac{0.9x-180}{180}=20$ %

$\dfrac{0.9x-180}{180}=\dfrac{20}{100}$

$\dfrac{0.9x-180}{180}=\dfrac{1}{5}$

$\dfrac{0.9x-180}{36}=1$

$0.9x=36+180$

$0.9x=216$

$x=\dfrac{216}{0.9}$

$x=240$

$\therefore$ Marked Price of an article is Rs 240.

A trader marks his goods at

$40$ % above the cost price but allows a discount of

$20$ % on the marked price. His profit percentage is

0%
$20$ %
0%
$10$ %
0%
$8$ %
0%
$12$ %

Explanation

Let the C.P. of the goods $=Rs 100$ . Then,
Marked price of the goods $= Rs 140$
Discount $=$ 20%
$\therefore$ S.P. of the goods $=$ 80% of $Rs 140=Rs 112$
$\therefore$ Profit % $=\dfrac {(112-100)}{100}\times 100=12$ %

Evaluate approximate value of

$15.2$ % of

$726\times 12.8$ % of

$643.$

0%
$9562$
0%
$9324$
0%
$9082$
0%
$9710$

Explanation

the approximate value of

$15.2\%$ of

$726 \times 12.8\%$ of

$643=$

$\dfrac{15.2}{100}\times 726\times \dfrac{12.8}{100}\times 643$

$=9082\cdot 411$

$=9082$ (approx)

By selling an article at

$\left (\dfrac {2}{5}\right)^{th}$ of the marked price, there is a loss of

$25\%$ . The ratio of the marked price and the cost price of the article is

0%
$2:5$
0%
$5:2$
0%
$8:15$
0%
$15:8$

Explanation

Let M.P. $=$ Rs. $100$

Then S.P. $=\dfrac{2}{5}$ of M.P. $=\dfrac{2}{5}\times 100=$ Rs. $40$

Loss $=25\%$

Then, C.P. $=\dfrac{100}{100- \text{loss} \%}\times$ S.P.

$\Rightarrow \dfrac{100}{100-25}\times 40$

$\Rightarrow \dfrac{100}{75}\times 40=$ Rs. $\dfrac{160}{3}$

M.P. : C.P. $=100:\dfrac{160}{3}=300 : 160$

$=15:8$

The cost of a vehicle is Rs.

$1,75,000$ . If its value depreciates at the rate of

$20$ % per annum, then the total depreciation after

$3$ years will be:

0%
Rs. $86,400$
0%
Rs. $82,500$
0%
Rs. $84,500$
0%
Rs. $85,400$

Explanation

Value of the vehicle after 3 years

$= 1,75,000 \times \left(1 - \displaystyle \frac{20}{100} \right)^{3}$

$= 1,75,000 \times \displaystyle \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}$
= Rs. 89, 600

$\therefore$ Total depreciation = 1, 75, 000 - 89, 600

$= Rs. 85,400$

If SP of an article is

$\displaystyle\frac{4}{3}$ of its CP then the profit in the transaction is

0%
$\displaystyle\frac{1}{3}\,\%$
0%
$\displaystyle20\frac{1}{2}\,\%$
0%
$\displaystyle33\frac{1}{3}\,\%$
0%
$\displaystyle25\frac{1}{2}\,\%$

Explanation

Suppose CP of article = Rs. 100
SP =

$100\,\times\, \displaystyle\frac{4}{3}\,=\, Rs. \frac{400}{3}$
Profit = SP - CP =

$\displaystyle\frac{400}{3}\, -\, 100\, =\, Rs. \frac{100}{3}$

Profit % =

$\displaystyle\frac{\frac{100}{3}}{100}\, \times\, 100\, =\, 33\frac{1}{3}$ %

Jatin bought a refrigerator with

$20\%$ discount on the labelled price. If he woud have bought it with

$25\%$ discount, he could have saved Rs.

$500$ . At what price did he buy the refrigerator?

0%
Rs. $5,000$
0%
Rs. $10,000$
0%
Rs. $12,500$
0%
Rs. $15,000$

Explanation

$\Rightarrow$ Let actual price of refrigerator is

$Rs.x$

$\Rightarrow$ The cost of refrigerator to jatin after 20% discount

$=\dfrac{80}{100}\times x=\dfrac{4x}{5}$

$\Rightarrow$ Cost of refrigerator to jatin after 25% discount

$=\dfrac{75}{100}\times x=\dfrac{3x}{4}$

$\Rightarrow$ Then according to the question,

$\Rightarrow$

$\dfrac{4x}{5}-\dfrac{3x}{4}=500$

$\Rightarrow$

$\dfrac{16x-15x}{20}=500$

$\therefore$

$x=20\times 500=10000$

If a company sells a car with a marked price of Rs.

$2,72,000$ and gives a discount of

$4\%$ on Rs.

$2,00,000$ and

$2.5\%$ on the remaining amount of Rs.

$72,000$ , then the actual price charged by the company for the car is

0%
Rs. $2,50,000$
0%
Rs. $2,55,000$
0%
Rs. $2,60,100$
0%
Rs. $2,62,200$

Explanation

M.P. $=$ Rs. $2,72,000$ .
Discount $=$ Rs. $[(4\%$ of $2,00,000)+ (2.5\%$ of $72,000)]$

$=\left [\left (\dfrac{4}{100}\times 200000)+(\dfrac{2.5}{100}\times 72000\right)\right]$
$=$ Rs. $(8,000 + 1,800)$
$=$ Rs. $9,800.$
Therefor, actual price $=$ Rs. $(2,72,000 - 9,800)$
$=$ Rs. $2,62,200.$

A shopkeeper fixes the MP of an item

$35$ % above its CP. The percentage of discount allowed to gain

$8$ % is:

0%
$43$ %
0%
$30$ %
0%
$20$ %
0%
$31$ %

Explanation

Let

$C.P.$ of article = Rs.

$100$

$M.P.\,=\, 100\, \times\, \displaystyle\frac{135}{100}\,=\, Rs. 135$

$SP$ required =

$\displaystyle100\times\frac{108}{100}\,=\, Rs. 108$
On Rs.

$135$ discount allowed
= Rs.

$135 -$ Rs.

$108 =$ Rs.

$27$
Hence on Rs.

$100$ discount allowed is

$\displaystyle\frac{27}{135}\, \times\, 100\,=\, 20\%$

The shopkeeper gives a discount of

$12$ % on the pair of shoes marked for Rs.

$1425$ , then S.P is:

0%
Rs. $1245$
0%
Rs. $1445$
0%
Rs. $1254$
0%
Rs. $1524$

Explanation

Discount

$=12$ %

$=1425$

SP =MP -Discount

$=1425-\dfrac{12}{100} \times 1425$

$=1254$

$5$ % more is gained by selling an article for Rs.

$350$ than by selling it for Rs.

$340$ , the cost of the article is:

0%
Rs. $50$
0%
Rs. $160$
0%
Rs. $200$
0%
Rs. $225$

Explanation

Let the C.P is =Rs. x

When S.P is Rs.340 then gain % $=\dfrac{340-x}{x}\times 100$

When S.P is rs.350 then gain % $=\dfrac{350-x}{x}\times 100$

According to the question

$\Rightarrow \left[\dfrac{350-x}{x}\times 100 \right]-\left[\dfrac{340-x}{x}\times 100 \right]=5$

$\Rightarrow \dfrac{100}{x}[350-x-340+x]=5$

$\Rightarrow \dfrac{100}{x}[10]=5$

$\Rightarrow x=\dfrac{1000}{5}=200$

Hence C.P of article is Rs.200.

If the discount is

$10$ %, an item bought for Rs.

$9$ is priced at:

0%
Rs. $11$
0%
Rs. $10$
0%
Rs. $8.10$
0%
Rs. $12$

Explanation

let the M.P $=$ Rs. $x$

Then $S.P=x-10\%$ of $x$

$\Rightarrow x-\dfrac{10x}{100}$

$\Rightarrow x-\dfrac{x}{10}$

$\Rightarrow \dfrac{9x}{10}$
Then $\dfrac{9x}{10}=9$

$\Rightarrow 9x=10\times 9$

$\Rightarrow x=10$

The shopkeeper gives a discount of

$12\%$ on the pair of shoes marked for Rs.

$1425$ , then S.P is:

0%
Rs. $1245$
0%
Rs. $1425$
0%
Rs. $1524$
0%
Rs. $1254$

Explanation

Given M.P.

$=$ Rs.

$1425$

Discount

$=12\%$ of Rs.

$1425 =$

$\displaystyle\frac{12}{100}\,\times\, 1425$

$=$ Rs.

$171$

Selling price of shoes

$=$ Rs.

$(1425 - 171) =$ Rs.

$1254$ .

A cycle is sold for Rs. 880 at a loss of 20%. For how much should it be sold to gain 10%?

0%
Rs. $1400$
0%
Rs. $1210$
0%
Rs. $1100$
0%
Rs. $1000$

Explanation

SP of cycle = Rs. 880
Loss = 20%
CP of cycle =

$\displaystyle880\, \times\, \frac{100}{100\, -\, 20}\, =\, Rs. 1100$
If gain required is 10 % then

$SP\,=\, 1100\, \times\, \displaystyle\frac{100\, +\, 10}{100}\, =\, Rs. 1210$

A tradesman gives 4% discount on his marked price and gives 1 article free for buying every 15 articles and thus gains 35%. By what per cent is the marked price increased above the cost price ?

0%
50%
0%
40%
0%
60%
0%
80%

Explanation

Let the C.P. of each article be Rs.x

Then C.P. of 16 articles = Rs.16x

S.P. of 15 articles = 135% of Rs.16x = Rs.

$\left ( \cfrac{135\times 16x}{100} \right )=Rs.\cfrac{108x}{5}$

S.P. of 1 article

$=Rs.\left ( \cfrac{108x}{5}\times \cfrac{1}{15} \right )=Rs.\cfrac{36x}{25}$

Let M.P. = Rs.100 Then S.P. =Rs.96 after a discount of 4%

$\therefore$ If S.P. =Rs.96 then M.P. = Rs.100

If S.P. =Rs.

$\cfrac{36x}{25}$ then M.P.

$=Rs.\left ( \cfrac{100}{96}\times \cfrac{36x}{25} \right )=Rs.\cfrac{3x}{2}$

$\therefore$ % increase in M.P. over C.P.

$=\cfrac{\cfrac{3x}{2}-x}{x}\times 100=\left ( \cfrac{x}{2}\times \cfrac{1}{x}\times 100 \right )\%=50\%$

A shopkeeper allows a discount of 10% on the marked price. How much above the cost price must he mark his goods to gain 8%?

0%
20%
0%
100%
0%
80%
0%
None of these

Explanation

let Marked price of an article be

$x$

Discount on article

$=x-10$ %of

$x$

$=x-\frac{10}{100}x$

$=0.9x$

Therefore, selling price of the article

$=0.9x$

Now, Gain %

$=\frac{sp-cp}{cp}$

Put the value of gain % and

$sp$ in the above equation and calculate

$cp$ .

$\frac{8}{100}=\frac{0.9x-cp}{cp}$

$8cp=90x-100cp$

$108cp=90x$

$cp=\frac{90x}{108}$

$cp=\frac{5x}{6}$

Therefore, cost price

$cp$ marked above marked price(

$mp$ )

$=\frac{mp-cp}{cp}\times100$

Put the valus of

$cp$ and

$mp$ in the above equation

$=\frac{x-\frac{5x}{6}}{\frac{5x}{6}}\times100$

$=\frac{\frac{x}{6}}{\frac{5x}{6}}$

$20$ %

Profit earned by selling an article for Rs.

$1,060$ is

$20$ % more than the loss incurred by selling the article for Rs.

$950$ . At what price should the article be sold to earn

$20$ % profit?

0%
Rs. $1000$
0%
Rs. $1150$
0%
Rs. $1,250$
0%
Rs. $1,200$

Explanation

Let C.P. be Rs $x$

Then $(1060 - x ) = \dfrac{120}{100} \times (x-950)$
$\Rightarrow 10600 - 10x = 120x- 120 \times 950$
$\Rightarrow 220x = 220000$
$\Rightarrow x = 1000$
$\therefore$ Desired S.P. $=$ Rs. $\left (\dfrac{120}{100} \times 1000\right) =$ Rs. $1200$

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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