MCQ Questions for Class 8 Maths Comparing Quantities Quiz 2

A car was sold at

$Rs 80,000$ through a broker. The broker charged

$2.5\%$ brokerage from both the seller and the buyer. What is the total amount of brokerage received by the broker?

0%
$Rs 1000$
0%
$Rs 4000$
0%
$Rs 2500$
0%
None of these

Explanation

From the buyer side, broker get charge

$=2.5\%\ of\ 80000=Rs.\ 2000$

From the seller side, broker get charge

$=2.5\%\ of\ 80000=Rs.\ 2000$

Hence total brokerage charge he get

$=2000+2000=Rs.\ 4000$

By paying

$2\%$ brokerage for buying an old scooter for

$Rs 15,000$ through a broker. At what amount is the scooter bought?

0%
$Rs 25,300$
0%
$Rs 15,300$
0%
$Rs 12,200$
0%
None of these

Deepali has bought a sofa set for Rs 14,There is an additional expense of Rs 500 on it. Since she did not like the sofa set, she sold it bearing a loss of 8.5%. For how much did she sell the sofa set?

0%
$Rs 10,725$
0%
$Rs 11,725$
0%
$Rs 13,725$
0%
None of these

Explanation

Original price =100
then
14500+500 = 15000

total cost = 15000
loss = 8.5%
then,

$\text{Total SP=[(100-Loss%)/100]*(Cost price)}$

$= (91.5/100)*15000$

$= 91.5*150$

$= 13725$

Vinodbhai sold his shop through a broker for

$Rs 7,50,000$ . The broker charged

$1\%$ brokerage for this work. What amount did Vinodbhai get on selling the shop?

0%
$Rs 8,42,500$
0%
$Rs 7,42,500$
0%
$Rs 3,42,500$
0%
None of these

Explanation

Brokerage charge

$=1\%\ of\ 7,50,000=Rs.\ 7,500$

Hence amount received by Vinodhbhai

$=7,50,000-7,500=Rs.\ 7,42,500$

Do the following sums:
C.P. = Rs 300 S.P. = Rs 350, then how much rupees profit or loss occur?

0%
Profit $Rs 50$
0%
Profit $Rs 100$
0%
Profit $Rs 40$
0%
None of these

Explanation

Given:

$\text{C.P. = Rs 300 and S.P. = Rs350}$
Since,

$S.P. > C.P.$
Therefore,

$Profit = S.P. - C.P. = 350 - 300 = Rs 50$

Jimmy's balance in a bank on 1st November was Rs.

$58709$ . He withdrew RS.

$13090$ and Rs.

$16518$ from his account and deposits Rs.

$1680$ in his account in that month. What was the balance at the end of the month?

0%
Rs. $30001$
0%
Rs. $29101$
0%
Rs. $29001$
0%
Rs. $30781$

Explanation

Amount of money in the account on 1st November

$=Rs.58709$
Total amount of money withdrawn

$=Rs.(13090+16518)=Rs.29608$
Amount of money deposited

$=Rs.1680$
Balance left at the end of the month

$=Rs.(58709-29608+1680)$

$=Rs.(60389-29608)=Rs.30781$

Vivekbhai bought a TV for Rs 16,Rs 200 was spent on transportation and labour. For earning 12 % profit, what price should it be sold?

0%
$Rs 30,144$
0%
$Rs 18,144$
0%
$Rs 20,144$
0%
None of these

The compound interest on

$Rs. 64,000$ for

$3$ years, compounded annually at

$7.5$ % per annum is

0%
$Rs. 14,400$
0%
$Rs. 15,705$
0%
$Rs. 15,507$
0%
$Rs. 15,075$

Explanation

$Principal = 64000$

$Rate = 7.5\% \,p.a$

$Time = 3 years$

$\text{Where r is the rate and t is the time.}$

$CI = 64000 (1 + 7.5/100)^3 - 64000$

$= 64000 (1 + 75/1000)^3 - 64000$

$= 64000 (1 + 3/40)^3 - 64000$

$= 64000 \times(43/40)^3 - 64000$

$= 64000 \times(43/40 \times 43/40 \times 43/40) - 64000$

$= (43 \times 43 \times 43) - 64000$

$= 79507 - 64000$

$= 15507$

$\text{Hence CI will be 15507 rupees}$

30% of 1860 + 40% of 820 = ?% of 3544

0%
30
0%
25
0%
35
0%
40

Explanation

solution:

$\dfrac{30}{100}$ x 1860 +

$\dfrac{40}{100}$ x 820 =

$x$ x 3544

$558$ +

$328$ =

$x$ x 3544

$\dfrac{886}{3544} = x$

$x = .25$

hence the correct option : B

40% of 2400+ ?% of 600=50% of 3840

0%
50
0%
40
0%
80
0%
None of these

Explanation

solution:

$\dfrac{40}{100}$ x 2400 +

$\dfrac{x}{100}$ x 600 =

$\dfrac{50}{100}$ x 3840

$960$ +

$\dfrac{x}{100}$ x 600 = 1920

$\dfrac{x}{100}$ x 600 = 960

x = 160

hence the correct opt: D

$175\times ?\ =\ 140\%$ of 1200

0%
8.4
0%
7.5
0%
13.44
0%
9.6

Explanation

solution:

175 x

$x$ =

$\dfrac{140}{100}$ of 1200

175 x

$x$ = 1680

$x$ =

$\dfrac{1680}{175}$

$x$ = 9.6

hence the correct opt: D

A shopkeeper mixes two varieties of Tea, one costing Rs.

$40$ /kg and another Rs.

$50$ /kg in the ratio

$3:2$ . if he sells the mixed variety of Tea at Rs.

$48$ /kg, his gain or loss percent is

0%
$48.4$ % gain
0%
$48.4$ % loss
0%
$10$ % gain
0%
$10$ % loss

Explanation

Average CP of Mixed Tea when Mixed in Ratio

$3:2$

$=\cfrac { 40\times 3+50\times 2 }{ 3+2 } =\cfrac { 220 }{ 5 } =Rs.44$
Wen SP=Rs.

$48/kg$

$=\cfrac { 4 }{ 44 } \times 100=9.09\approx 10$ % gain

If some articles are bought at prices ranging from Rs. 200 to Rs. 350 and are sold at prices ranging from Rs. 300 to Rs. 425, what is the maximum possible profit that might be made in selling 16 such articles?

0%
Rs. 3600
0%
Rs. 1200
0%
Rs. 800
0%
Rs. 400

Explanation

Maximum profit will be if we purchase in minimum

price & sell in maximum price.

Minimum price for purchasing = Rs 200

Maximum price for purchasing = Rs 425

Net profit on 1 article

$425 - 200 = Rs\, 225$

Net profit on 16 article

$= 225 \times 16 = Rs \,3600$

1111935_809585_ans_cd7a91071aac407e98d1d059c7256eb5.jpg

Indirect tax is:

0%
Sales Tax
0%
House Tax
0%
Water Tax
0%
Surcharge

$P$ sells a table to

$Q$ at a profit of

$10$ % and

$Q$ sells it to

$R$ at a profit of

$12$ %. If

$R$ pays Rs.

$246.40$ for it, then how much had

$P$ paid for it?

0%
$200.00$
0%
$300.00$
0%
$248.00$
0%
$346.00$

Explanation

Let

$P$ paid for table

$=$ Rs.

$P$
After all

$R$ paid =Rs.

$246.40$
According to the question,

$246.40=P\left[ \cfrac { 110 }{ 100 } \right] \left[ \cfrac { 112 }{ 100 } \right]$

$P=\cfrac { 246.40\times 100\times 100 }{ 110\times 112 }$

$=\cfrac { 246400 }{ 11\times 112 }$

$=Rs.200$

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

0%
$870$
0%
$800$
0%
$750$
0%
$690$

Explanation

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

$750$ .

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.

0%
Rs $1900$
0%
Rs $1400$
0%
Rs $1000$
0%
Rs $1300$

Explanation

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.

If the cost of a dozen soaps is

$Rs\ 285.60$ , what will be the cost of

$15$ such soaps?

0%
Rs. $827$
0%
Rs. $772$
0%
Rs. $676$
0%
Rs. $357$

Explanation

Cost of

$12$ soaps

$=285.60$

Cost of one soap

$=\dfrac{285.60}{12}$

Cost of

$15$ soaps

$=\dfrac{285.60}{12}\times{15}=Rs.357$

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

0%
$4000$
0%
$4080$
0%
$4280$
0%
$4050$

Explanation

C.I. = Amount - Principle

$P((1+\dfrac{r}{100})^{T}_{} - 1)$

CI =

$50,000(1+\dfrac{4}{100})^{2}_{} - 1)$

C.I.=

$4080.$

If a banana's cost is

$Rs. 1.25$ and apple's cost is

$Rs. 1.75$ what will be the cost of

$2$ Dozen of Banana and

$3$ Dozen of apple?

0%
$Rs. 93$
0%
$Rs. 83$
0%
$Rs. 85$
0%
$Rs. 70$

Explanation

$\because$ Cost of one banana

$=Rs.1.25$

$\therefore 2$ dozens banana's cost

$=24\times 1.25=Rs.30$

$\because$ One apple's cost

$=Rs.1.75$

$\therefore 3$ dozens banana's cost

$=36\times 1.75=Rs.63$

Total cost

$=63+30=Rs.93$

Hence, option (A) is the correct option

The cost price of

$16$ articles is equal to selling price of

$12$ articles then the gain or loss per cent is

0%
$13 \frac {1}{3}$ % gain
0%
$33 \frac {1}{3}$ % gain
0%
$33 \frac {1}{3}$ % loss
0%
$13 \frac {1}{3}$ % loss

Explanation

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 %

$= \dfrac{\textrm{Profit}}{C} \times 100 \\ = \dfrac{4x}{12x} \times 100 \\ = 33\dfrac{1}{3} \%$

Rahul and Sunjay invested

$Rs.\,\,25,000\,\,$ and

$\;\,Rs.\,\,35,000$ respectively in a business. They will share the profit in the proportion.............

0%
$7:5$
0%
$5:7$
0%
$8:5$
0%
$5:8$

Explanation

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

$25,000:35,000$

$=25:35$

$=5:7$ .

If 16 shirts were sold for 20 notes of Rs 100, double its notes of Rs 50 and 200 notes of Rs 20, then what is the selling price of a shirt ?

0%
Rs 250
0%
Rs 400
0%
Rs 350
0%
Rs 500

Explanation

Total amount obtained for 16 shirts

$= 20 \times 100 + 40 \times 50 + 200 \times 20$

$= 8000 \ Rs$

So, Selling Price of a shirt

$= \dfrac{8000}{16} = 500 \ Rs$

Thus, D is the correct answer.

State whether the statements are True or False.

$12\%$ of

$120$ is

$100.$

0%
True
0%
False

Explanation

$12\%$ of

$120 = \dfrac{12}{100} \times 120 = 14.40$
Hence the stattement is false.

Compound interest is the interest calculated on the previous year's amount.

0%
True
0%
False

If the marked price of an article is Rs. x, and the selling price is Rs. y, then what is the discount percentage?

0%
$\frac{(x-y)100}{x}$
0%
$\frac{(y-x)100}{x}$
0%
$(\frac{y-x}{y})100$
0%
$\frac{x-y}{100}$

The manufacturer A of a certain item sells it to a wholesaler at a profit of 20% on his manufacturing cost. The wholesaler sells it to a retailer at a profit of 25% and the retailer sells it to a consumer at a profit of 20%. The price paid by the consumer over and above manufacturing cost will be

0%
65%
0%
80%
0%
85%
0%
90%

Explanation

Let the manufacturing cost be Rs.

$x$ .

Then the wholesaler's cost price

$= x+\cfrac{20}{100}x = Rs. \cfrac{6x}{5}$

Retailer's cost price

$= \cfrac{6x}{5}+\cfrac{25}{100}.\cfrac{6x}{5} = Rs. \cfrac{3x}{2}$

Finally, consumer's purchasing price

$= \cfrac{3x}{2}+\cfrac{20}{100}\times \cfrac{3x}{2} = Rs. \cfrac{9x}{5}$

Thus, on the manufacturing cost of Rs.

$x$ , the consumer pays

$(\cfrac{9x}{5}-x) = Rs. \cfrac{4x}{5}$ over and above.

$\therefore$ % price paid by consumer over and above manufacturing cost

$= \cfrac{4x}{5}\times\cfrac{1}{x} \times 100 = 80$ .

A person purchases

$20$ litres of juice at

$Rs. 2.20$ per litre and diluted it with water to make the contents

$22$ litres. In order to earn

$10\%$ profit he should sell the juice at

0%
$Rs. 2.20$
0%
$Rs. 2.00$
0%
$Rs.2.35$
0%
$Rs. 2.40$

Explanation

Volume of juice purchased

$= 20$ litres

Rate at which juice purchased

$= Rs.2.20$ per litre

$\Rightarrow$ Cost price of juice

$= 20 \times 2.2 = Rs..44$

In order to gain

$10\%$ profit,

Selling price

$= \dfrac{110}{100} \times 44 = 48.4$

Final volume of juice after dilution

$= 22$ litres

$\therefore$ Rate of selling

$= \dfrac{48.4}{22} = Rs. 2.20$

A and B invest Rs. 200 and Rs. 300, respectively, in a business for a period of three years and two years, respectively. Then the profit will be divided into the ratio of

0%
$4 : 3$
0%
$2 : 3$
0%
$1 : 1$
0%
$6 : 5$

Explanation

Interest earned

$\propto$ Amount invested

$\times$ Time

$\therefore$ Profit of A

$:$ Profit of B

$= 200\times 3 : 300 \times 2$

$= 1:1$

How many

$\text{kg}$ of sugar costing

$\text{Rs.}\ 5.75$ per

$\text{kg}$ should be mixed with

$75\ \text{kg}$ of cheaper sugar costing

$\text{Rs.}\ 4.50$ per

$\text{kg}$ so that the mixture is worth

$\text{Rs.}\ 5.50$ per

$\text{kg}$ ?

0%
$350\ \text{kg}$
0%
$300\ \text{kg}$
0%
$250\ \text{kg}$
0%
$325\ \text{kg}$

Explanation

Let say

$x\ \text{kg}$ sugar costing

$\text{Rs.}\ 5.75$ per

$\text{kg}$ is added.

Total cost

$= \text{Rs.}\ (5.75x + 75) \times 4.50$

$=\text{Rs.}\ (5.75x+337.5)$

Total sugar

$= (x + 75)\ \text{kg}$

Average cost of sugar mixture

$=\dfrac{\text{Total cost of sugar}}{\text{Total sugar}}$

$5.5 = \dfrac{\left(5.75x + 337.5\right)}{\left(x + 75\right)}$

$5.5x + 412.5 = 5.75x + 337.5$

$75 = 0.25x$

$x =\dfrac{75}{0.25}$

$\therefore\,x = 300\ \text{kg}$

The list price of a parker pen is Rs. 160 and a customer buys it for Rs. 122.40 after two successive discounts. If first is 10%, then the second is

0%
18%
0%
17%
0%
16%
0%
15%

Explanation

$\Rightarrow$ The list price of parker pen is

$Rs.160$

$\Rightarrow$ Selling price of parker pen is

$Rs.122.40$

$\Rightarrow$ Total discount =

$Rs.160-Rs.122.40=Rs.37.60$

$\Rightarrow$ Fist discount is

$10\%$ on list price

$\therefore$ Fist discount =

$\dfrac{10}{100}\times 160=Rs.16$

$\Rightarrow$ Second discount =

$Rs.37.60-Rs.16=Rs.21.60$

$\Rightarrow$ Pen price after 1st discount =

$Rs.160-Rs.16=Rs.144$

$\Rightarrow$ Second discount

$\%$ =

$\dfrac{21.60}{144}\times 100=15\%$

A shopkeeper professes to sell the goods at cost price, but he uses a weight of 900 grams for a Kilogram. Then his gain percent is

0%
$\dfrac{100}{9}$ %
0%
$\dfrac{100}{8}$ %
0%
$\dfrac{100}{7}$ %
0%
$\dfrac{100}{6}$ %

Explanation

Shopkeeper uses weight of 900 grams for a kilogram or 1000 grams.

$\Rightarrow$

$\text{Profit} = 1000-900=100$

$\text{Profit}\ \%=\dfrac{100}{900}\times 100=\dfrac{100}{9}\%$

A began business with Rs. 4,500 and was joined afterward by B with Rs. 5,If the profit at the end of the year was divided in the ratio of 2:1, then the time of joining B was after:

0%
5 months
0%
7 months
0%
8 months
0%
9 months

Explanation

$\Rightarrow$ Let B remained in the business for

$x$ months. Then,

$\Rightarrow$

$\dfrac{A}{B}=\dfrac{4500\times 12}{5400\times x}=\dfrac{54000}{5400x}=\dfrac{10}{x}$

$\therefore$

$\dfrac{10}{x}=\dfrac{2}{1}$

$\Rightarrow$

$2x=10$

$\therefore$

$x=5$

$\Rightarrow$ Thus, B remained in the business for 5 months.

$\Rightarrow$ So, B joined the business after

$7\, months.$

The population of a small town is increased by

$\cfrac{15}{2}$ % per annum for two years. If the present population is

$73, 960$ , then two years before the population was

0%
$68,000$
0%
$70,000$
0%
$64,000$
0%
$60,000$

Explanation

$A=P(1+\cfrac{R}{100})^n$

$73960=P(1+0.075)^2$

$\therefore P=64000$

During a sale, a shop offered a discount of

$10\%$ on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs

$1450$ and two shirts marked at Rs

$850$ each

0%
Rs $2835$
0%
Rs $3000$
0%
Rs $2830$
0%
Rs $2800$

Explanation

Given Marked Price of Jeans = Rs 1,450
Market Price of two Shirts= Rs 1,700

Total Price of a pair of jeans and two shirts

$=1450+2\times 850=3150$

Now,
Discount = Marked Price x % Discount

$=3150\times \dfrac { 10 }{ 100 }$

$=315$

$\therefore$ Effective Price After Discount = Rs.

$\left( 3150-315 \right) =$ Rs.

$2,835$

If the population of a town is 64,000 and its annual increase is 10%, then its correct population at the end of 3 years will be

0%
80,000
0%
85,000
0%
85,100
0%
85,184

Explanation

Population (P)

$= 64,000$
Annual increase rate

$(r) = 10\%$
Hence, population after

$n=3$ years

$= P(1+\cfrac{r}{100})^n$

$= 64,000(1+\cfrac{10}{100})^3$

$= 64,000\times (\cfrac{11}{10})^3$

$= 64,000\times \cfrac{11\times 11\times 11}{1000}$

$= 85,184$

A shopkeeper sells a badminton racket marked at Rs. 30 at 15% discount and gives a shuttle cock costing Rs. 1.50 free with each badminton racket. He then makes a profit of 20%. His cost price per badminton racket is

0%
Rs. 35
0%
Rs. 30
0%
Rs. 25
0%
Rs. 20

Explanation

$\Rightarrow$ Marked price =

$Rs.30$

$\Rightarrow$ Selling price =

$Rs.[(\dfrac{85}{100}\times 30)-1.50]$

$\Rightarrow$ Selling price =

$Rs.(25.50-1.50)=Rs.24$ .

$\Rightarrow$ Let Cost price be

$Rs.x$

$\Rightarrow$ Then,

$120\%$ of

$x=24$

$\Rightarrow$

$x=\dfrac{24\times 100}{120}=Rs.20$ .

An agent makes a profit of 20% even after a discount of 10% on the advertised price. If he makes a profit of Rs. 900 on the sale of a scooter, then the advertised price is

0%
Rs. 4,500
0%
Rs. 4,000
0%
Rs. 6,000
0%
Rs. 5,050

Explanation

The advertising price is basically the marked price.

Agent makes a profit of

$20\%$ after giving a discount of

$10\%$

Marked Price

$\times$

$(1$

$-$ Discount

$\%) =$ Cost Price

$\times (1 +$ Profit

$\%)$

$\therefore MP\times \dfrac{90}{100} = CP \times \dfrac{120}{100} = SP$

Profit

$= SP - CP$

$\therefore 900 = \dfrac{20}{100} \times CP$

$CP = 4500$

$\Rightarrow$ Advertised Price

$= \dfrac{120}{90} \times 4500 =$ Rs.

$6000$

Answer

$=$ Rs

$6000$

A sum of money becomes Rs. 13380 after 3 years and Rs. 20070 after 6 years on compound interest. The sum is

0%
Rs. 8800
0%
Rs. 8890
0%
Rs. 8920
0%
Rs. 9040

Explanation

Let the sum be Rs.

$x$ . Then ...(i)

$x (1+\cfrac{R}{100})^3 = 13380$
and

$x (1+\cfrac{R}{100})^6 = 20070$ ...(ii)
Dividing equation (ii) by (i), we get

$(1+\cfrac{R}{100})^3 = (\cfrac{200070}{13380}) = (\cfrac{3}{2})$

$\therefore x\cfrac{3}{2} = 13380$

$\Rightarrow x = (13380\times \cfrac{2}{3}) = 8920$
Hence, the sum is Rs. 8920

A shopkeeper gives a discount of 25% on the marked price of an article and still earns a profit of 20% on his outlay, if he had not given any discount, then the profit on his outlay would have been

0%
45%
0%
50%
0%
55%
0%
60%

Explanation

Let the Market Price

$=$ Rs. 100, Discount

$=$ 25%
Discounted Price

$= Rs. 100\times \cfrac{100-25}{100}$

$Rs. 100\times \cfrac{75}{100} = Rs. 75$

$Profit = 20$ %

$\therefore$ Cost Price

$= Rs. 75\times \cfrac{100}{100+20}$

$= Rs. 75\times \cfrac{100}{120} = Rs. \cfrac{125}{2}$
Had he not given discount, his profit would have been

$= Rs. 100-Rs. \cfrac{125}{2} = Rs. \cfrac{75}{2}$
Profit on Rs.

$\cfrac{125}{2} = Rs. \cfrac{75}{2}$

$\therefore$ Profit on Rs. 100

$= \cfrac{75}{2}\times \cfrac{2}{125}\times 100 = 60\%$

An article marked at Rs 135 is sold for Rs 118.Then the rate of discount offered is

0%
10%
0%
12%
0%
14%
0%
16%

Explanation

Given: Marked price of article is

$Rs.135$ .

and Selling price of article is

$Rs.118.80$

Price of article after discount

$=$ Marked price of article - Selling price of article

$=Rs.135-Rs.118.8$

$=Rs16.2$

$\Rightarrow$ Rate of discount

$=\dfrac{16.2}{135}\times 100$

$=12\%$

A tree increases annually by one-fourth of its height. What will be it's height after

$2$ years, if it stands

$64$ cm high today?

0%
$76$ cm
0%
$80$ cm
0%
$100$ cm
0%
$105$ cm

Explanation

Initial height

$=64\ cm$

Every year it increases one-fourth of it's height.

Increase

$\%= (\cfrac{1}{4}\times 100\%)=25\%$

Height after

$2$ years

$= 64\times \left(1+\cfrac{25}{100}\right)^2\ cm$

$= 64\times \left(1+\cfrac{1}{4}\right)^2\ cm$

$= 64\times \left(\cfrac{5}{4}\right)^2\ cm$

$= (64\times \cfrac{5}{4}\times \cfrac{5}{4})\ cm$

$= 100\ cm$

Hence, the height of the tree after

$2$ years will be

$100\ cm$

A started business with Rs. 4,500 and B joined afterward with Rs. 3,If the profits at the end of one year were divided in the ratio 2 : 1, respectively, then B would have joined A for business after

0%
1 month
0%
2 months
0%
3 months
0%
4 months

Explanation

A's 1 month's investment

$= Rs. (4500\times 12)$

$= Rs. 54,000$
Let B joins after

$x$ months.

$\therefore$ B's 1 month's investment

$= Rs.3,000 (12 -x)$
By hypothesis

$54,000 : 3,000 (12 -x) = 2 : 1$
or

$\cfrac{18}{12-x} = 2$
or

$18 = 24-2x$
or

$2x = 6$

$\therefore x = 3$ months

The current population of a town is 10,If the population increases by

$10\%$ every year, then the population of the town after three years will be

0%
13,310
0%
13,500
0%
14,000
0%
14,500

Explanation

$A=P(1+\cfrac{R}{100})^n$

$A=10000(1+0.1)^3=13310$

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 12000, the compound interest on the same sum for the same period at the same rate, is

0%
Rs. 12600
0%
Rs. 12610
0%
Rs. 12640
0%
Rs. 12650

Explanation

Clearly, Rate =

$5\; \%$ , Time =

$3$ years and S.I. =

$\text{Rs.}\;12000$

Sum

$= \text{Rs.} \left (\cfrac{100\times 12000}{3\times 5}\right ) = \text{Rs.}\; 80000$
Amount

$= \text{Rs. } \left [80000\times \left (1+\cfrac{5}{100}\right )^3\right ]$

$= \text{Rs. } 92610$

We know that C.I. = Amount - Sum
So,

C.I.

$= 92610-80000$

$= \text{Rs. } 12610$

What was the cost price of the suitcase purchased by Samir?
(1) Samir got 20 per cent concession on the labelled price.
(2) Samir sold the suitcase for Rs. 2000 with 25 per cent profit on the labelled price.

0%
1280
0%
2301
0%
1602
0%
1222

Explanation

(2)

$\Rightarrow$ Labelled price

$=\frac {2000}{125}\times 100=Rs. 1600$
(1)

$\Rightarrow$ C.P. of suitcase

$=1600(1-\frac {20}{100})=Rs. 1280$
Hence, both the statements together are necessary to answer the question.

A shopkeeper purchases 11 pens for Rs. 10 and sells them at the rate of Rs. 11, then the profit is

0%
8%
0%
9%
0%
10%
0%
11%

A shopkeeper mixes 80 kg sugar worth of Rs. 6.75 per kg with 120 kg of sugar worth of Rs. 8 per kg. He earns a profit of 20% by selling the mixture. He sells it at the rate.

0%
Rs. 7.50 per kg
0%
Rs. 9 per kg
0%
Rs. 8.20 per kg
0%
Rs. 8.85 per kg

Explanation

Total cost

$=80\times 6.75+120\times 8=Rs. 1500$
Selling price

$=1500\times \dfrac {120}{100}=Rs. 1800$
S.P. per kg

$=\dfrac {1800}{200}=Rs.9$

If the cost of a pen is Rs. 12.60 and the gain was

$10\%$ of the marked price, then the marked price of the pen was

0%
Rs. 15
0%
Rs. 14
0%
Rs. 13
0%
Rs. 12

Explanation

Given cost price 12.60

Profit

$= 10\%$

Selling price = Cost price + Profit

Profit

$=12.60 \times \cfrac{10}{100}=1.26$

Selling price

$= 12.60+1.26=13.86$ , which is approximately 14.

So, the correct answer is option B.

Marked price of a Saree is Rs. 600 and is available on Rs.Rate of discount is :

0%
25%
0%
30%
0%
15%
0%
40%

Explanation

$MP=600, SP=450$

$Disc=\dfrac{600-450}{600}=\dfrac {150}{600}\times 100=25$ %

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 2 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 2 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.