MCQ Questions for Class 8 Maths Comparing Quantities Quiz 5

What is the compound interest on an amount of Rs.

$4800$ at the rate of

$6$ percent p.a. at the end of

$2$ years?

0%
Rs. $544.96$
0%
Rs. $576$
0%
Rs. $593.28$
0%
Rs. $588$

Explanation

Given,

$P =$

$Rs.$

$4800, r = 6\%$

$p.a.$ ,

$n = 2$

$\displaystyle \therefore C.I.=A-P=P\left ( 1+\frac{r}{100} \right )^{n}-P$

$\displaystyle =4800\left ( 1+\frac{6}{100} \right )^{2}-4800$

$\displaystyle =4800\times \dfrac{106\times 106}{100\times 100}-4800$

$= 5393.28 - 4800 =$

$Rs.$

$593.28$

A shopkeeper fixes the marked price of an item 35% above its cost price. What is the percentage of discount allowed to gain 8% ?

0%
27%
0%
25%
0%
22%
0%
20%

Explanation

Let the C.P. = Rs.100 Then M.P. = Rs.135 and S.P.

$\displaystyle =\frac{108\times 100}{100}=Rs.108$

$\displaystyle \therefore Discount\: =Rs.135-Rs.108=Rs.27$ Discount %

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

The population of a village isIf the population increases by 10% in the first year, by 20% in the second year, and due to mass exodus, it decreases by 5% in the third year, what will be its population after 3 years ?

0%
10,540
0%
11,540
0%
12,540
0%
13,540

Explanation

Population after 3 years

$\displaystyle =10000\left ( 1+\frac{10}{100} \right )\left ( 1+\frac{20}{100} \right )\left ( 1-\frac{5}{100} \right )$

$\displaystyle =10000\times \frac{11}{10}\times \frac{6}{5}\times \frac{19}{20}=12540$

Two continuous discounts of 4% on any thing should be equal to

0%
8%
0%
7.92%
0%
7.84%
0%
8.08%

Explanation

Let initial cost

$=x$

After first discount cost

$=x\dfrac{96}{100}$

After second discount cost will be

$x\dfrac{96}{100}\times\dfrac{96}{100}$

Net discount will be

$\dfrac{x-x\dfrac{96}{100}\times\dfrac{96}{100}}{x} \times 100=7.84$ %

The population of a town increases annually by

$25 \%$ if the preset population is

$1$ crore, then what was the difference between the population three years ago and two years ago?

0%
$25,00,000$
0%
$12,80,000$
0%
$15,60,000$
0%
None of these

Explanation

Given that,

Population increases annually by

$25\%$

Let population

$3$ years ago be

$x$

$\Rightarrow$ Population two years ago was

$\dfrac{125}{100}x$

Population last year was

$\dfrac{125\times125}{10000}x$

Population this year is

$\dfrac{125\times125\times125}{1000000}x = 10000000$

$\Rightarrow$ Population three years ago was

$x = 5120000$

$\therefore$ Difference between population three years ago and two years ago

$= 25\%$ of

$x$

$= 1280000$

The annual increase in the population of a town is 10%. If the present population of the town is 180,000, then what will be its population after two years ?

0%
200,000
0%
210,000
0%
217,800
0%
207,800

Explanation

Population after 1 year

$= 180000 + 180000\times \dfrac{10}{100} = 198000$

Population after 2 years

$= 198000 + 198000\times \dfrac{10}{100}$

$=217800$

A machine depreciates in value each year at the rate of

$10\%$ of its previous value. However every second year there is some maintenance work, so that in that particular year depreciation is only

$5\%$ of its previous value. If at the end of the fourth year the value of the machine stands at Rs.

$146205$ , then find the value of the machine at the start at the first year?

0%
Rs. $190000$
0%
Rs. $200000$
0%
Rs. $195000$
0%
Rs. $210000$

Explanation

Let the value of the machine, at the onset, be

$x$ when

$x$ is in Rs.

Here we shall apply the rule

Final value

$=$ original value

$\times { \left( 1+rate \right) }^{ time }$ .

Here the rate will be negative since the value is depriciating.

For two yrs. the rate of depriciation is

$10\%$ .

$\therefore$ the value after

$2$ yrs.

$=$

$x\times { \left( 1-\dfrac { 10 }{ 100 } \right) }^{ 2 }=x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 }$ .

As given, the rate of depriciation for next

$2$ yrs

$=5\%$

$\therefore$ The value after next

$2$ yrs

$=$

$x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times { \left( 1-\dfrac { 5 }{ 100 } \right) }^{ 2 }$

$=x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 }$

But the final value after

$4$ yrs

$=$ Rs.

$146205$ .

$\therefore \quad x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 } =146205$

$\Rightarrow x=\dfrac { 146205\times 40000 }{ 81\times 19\times 19 } =5\times 40000$

$\Rightarrow x=$ Rs.

$200000$ .

So,

$4$ yrs. back, the value of the machine was Rs.

$200000$ .

How much per cent more than the cost price should a shopkeeper marks his goods so that after allowing a discount of 20% on the marked price he gains 10% ?

0%
37.5%
0%
40.0%
0%
42.5%
0%
45.0%

Explanation

Let the C.P. = Rs.100 Gain = 10%

$\displaystyle \therefore$ S.P. = Rs.110............(i)

Let the M.P. = Rs.x Discount = 20%

$\displaystyle \therefore$ S.P. = 80% of

$\displaystyle x=\frac{80x}{100}...........(ii)$
From (i) and (ii)

$\displaystyle \frac{80x}{100}=110\Rightarrow x=Rs.\frac {100\times 100}{80}=Rs.137.5$

$\displaystyle \therefore$ Marked price =

$\displaystyle37\frac{1}{2}\%$ above C.P.

The current birth rate per thousand is

$32$ , whereas the corresponding death rate is

$11$ per thousand. The net growth rate in terms of population increase in per cent is given by,

0%
$0.0021 \%$
0%
$0.021 \%$
0%
$2.1 \%$
0%
$21 \%$

Explanation

Given, birth rate per

$1000$ is

$32$ and corresponding death rate is

$11$ per thousand.

therefore, net growth on

$1000=(32-11)=21$

Hence, net growth on

$100=\dfrac {21}{1000}\times 100=2.1\%$

Therefore, net growth rate in terms of population increase in per cent is

$2.1\%$ .

400 apples were bought at Rs. 125 per hundred and were sold at a profit of Rs.Find the selling price per dozen.

0%
Rs. 20
0%
Rs. 12
0%
Rs. 18
0%
None

Explanation

Total CP =

$4\, \times\, 125\,=\, Rs. 500$

$SP = CP + profit$

$= 500 + 100 = Rs.\ 600$

SP of each apples =

$\displaystyle \frac{600}{400}\,=\, \displaystyle \frac{3}{2}$

SP of dozen apples =

$\displaystyle \frac{3}{2}\, \times\, 12$ = Rs. 18

The population of a town increases by 5% annually. If the population in 2009 is 1,38,915, what was it in 2006 ?

0%
100,000
0%
110,000
0%
120,000
0%
130,000

Explanation

Let the population in 2006 be

$x$ .
Pop. in

$\displaystyle 2009=x\times \left ( 1+\cfrac{5}{100} \right )^{3}$

$\Rightarrow 1,38,915=x\times \left ( \cfrac{21}{20} \right )^{3}$

$\Rightarrow x=\cfrac{138915\times 8000}{9261}=120000$

The population of a town was 1,60,000 three years ago. If it increased by 3%, 2.5% and 5%, respectively, in the last three years, then what is the present population?

0%
166,366
0%
177,366
0%
166,377
0%
177,377

Explanation

Present population

$\displaystyle =1,60,000\left ( 1+\frac{3}{100} \right )\left ( 1+\frac{2.5}{100} \right )\left ( 1+\frac{5}{100} \right )$

$\displaystyle =160000\times \frac{103}{100}\times \frac{102.2}{100}\times \frac{105}{100}=177366$

A seller allows a discount of

$5\%$ on a watch. If he allows a discount of

$7\%$ , he earns

$\text{Rs }15$ less in the profit. What is the marked price?

0%
$\text{ Rs }697.50$
0%
$\text{ Rs }712.50$
0%
$\text{ Rs }750$
0%
$\text{ Rs }817.50$

Explanation

Let

$x$ be the profit when the seller allows

$5\%$ discount.

Then,

$(x-15)$ is the profit when he allows

$7\%$ discount.

Now,

Selling price at

$5\%$ discount

$-$ selling price at

$7\%$ discount

$=x-(x-15)$

$\implies (100-5)\%$ of M.P

$- (100-7)\%$ of M.P

$= x-(x-15) = 15$

$\Rightarrow 2\%$ of M.P

$= 15$

$\Rightarrow$ M.P.

$=15\times \cfrac{ 100 }{ 2 } =750$

$\therefore$ Marked Price is

$\text{ Rs. } 750$

The present population of a town is

$35,000$ . If the population becomes

$35,700$ next year, the rate of growth is

0%
$2\%$
0%
$20\%$
0%
$70\%$
0%
$7\%$

Explanation

Population next year,

$P = 35700$
Current population,

$P_o = 35000$

$P = P_o \left(1 + \cfrac{R}{100}\right)^T$

$35700 = 35000 \left(1+ \cfrac{R}{100}\right)^1$

$35700 - 35000 = \cfrac{35000 R}{100}$

$R = \cfrac{700 \times 100}{35000}$

$R = 2\%$

What will be the difference between simple and compound interest at 10% per annum on a sum of Rs. 1000 after 4 years?

0%
Rs. 31
0%
Rs. 32.10
0%
Rs. 40.40
0%
Rs. 64.10

Explanation

Principal=1000 Rs.
Time=4 years
Rate of interes=10%
Then Simple interest

$=\frac{1000\times 10\times 4}{100}=400 Rs.$
Compound interest

$=$ Amount

$-$ Principal

Amount

$=1000(1+\frac{10}{100})^4$

$\Rightarrow 1000\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\times$

$\Rightarrow 1464.10 Rs.$

$C.I.=1464.10-4000=464.10 Rs.$
Difference between C.I and S.I.=

$464.10-400=64.10 Rs.$

A dealer allows a discount of

$10\%$ and still gains

$5\%.$ What percent above the cost price must he mark his goods?

0%
$15\%$
0%
$20\%$
0%
$16\dfrac{1}{3}\%$
0%
$50\%$

Explanation

Let

$C.P. = \text{Rs }100$

$\because \left(\dfrac{S.P.-C.P.}{C.P.}\right)\times 100=\text{Profit} \%$

$\implies \left( \dfrac{S.P.-100}{100}\right)\times 100=5$

$\implies S.P.=\text{Rs }105$
Discount(d)

$= 10 \%$

We have,

$S.P.\,=\, \displaystyle\dfrac{M.P.\, (100\,-\,d)}{100}$ (where

$d,$ is discount)

$M.P.\,=\, \displaystyle\dfrac{105\, \times\, 100}{100\, -\, 10}\,=\, \dfrac{105\, \times\, 100}{90}\,=\, \dfrac{350}{3}$

$M.P.\,-\, C.P.\,=\, \displaystyle\dfrac{350}{3}\,-\, 100\,=\, \dfrac{50}{3}$

$\therefore \text{Required }\%=\, \displaystyle\dfrac{\dfrac{50}{3}}{100}\, \times\, 100\,=\, 16\frac{2}{3}\%$

$\therefore$ He should mark his goods

$16\displaystyle\frac{2}{3} \%$ above the C.P.

A dealer allows

$25\%$ discount on the marked price of articles and earns a profit of

$20\%$ on them. What is the marked price of the article on which he gains

$\text{Rs.}\ 800$ ?

0%
Rs. $6000$
0%
Rs. $6400$
0%
Rs. $7200$
0%
Rs. $7000$

Explanation

Let the M.P.=

$\text{Rs.}\ 100$

Discount

$= 25\%$

$\therefore \text{S.P.} =\text{Rs.} 75$

$\text{Profit} = 20 \%$

$\displaystyle \text{C.P.}=\text{Rs.}\cfrac{75\times 100}{120}=\text{Rs.}\ 62.50$

$\displaystyle \therefore \text{Profit} = \text{Rs.}75 - \text{Rs.}62.50 = \text{Rs.}12.50$

If the gain is Rs. 12.50, M.P.

$=\text{Rs.}\ 100$

If the gain is Rs.800, M.P.

$=\text{Rs.}\displaystyle \cfrac{100}{12.50}\times 800=\text{Rs.}\ 6400$

Ramesh bought a calculator with 20% discount on the tag-price. He obtained 10% profit by selling it for Rs.What was the tag-price?

0%
Rs.500
0%
Rs.400
0%
Rs.480
0%
Rs.360

Explanation

Let the tag price of the calculator be Rs.

$x$

Then C.P. of Ramesh after 20% discount

$= 80\%$ of

$\displaystyle Rs.x=Rs.\cfrac{80x}{100}=Rs.\frac{4x}{5}$ ........(i)

Also, given S.P. = Rs.440 and Profit = 10%

$\displaystyle \therefore C.P.=Rs.\left ( \frac{440\times 100}{110} \right )Rs.400$ ............(ii)
From (i) and (ii)

$\displaystyle \therefore \frac{4x}{5}=400\Rightarrow x=Rs.500$

The list price of a slipper is Rs.The discount offered is 10%.Find the selling Price of a slipper.

0%
144
0%
124
0%
100
0%
134

Explanation

List price of slipper

$=160$
discount

$=10\%$

$10\%$ of

$160=\cfrac{10}{100}\times160=16$
Selling Price of slipper aftter discount=

$160-16=144$

$A$ started a business with

$Rs. 21, 000$ and is joined afterwards by

$B$ with

$Rs. 36, 000$ . After how many months did B join if the profit at the end of the year are divided equally ?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Suppose B joined after x months

then,

$21000\times12=36000\times (12-x)$

$=> 36x = 180$

$=> x = 5$

The C.I. on a sum of money for 2 years is Rs. 832 and the S.I. on the same sum for the same period is Rs. The difference between the C.I and S.I. for 3 years will be

0%
Rs. 48
0%
Rs. 66.56
0%
Rs. 98.56
0%
None of these

Explanation

Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400 and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400 and compound interest for 2nd year will be

$832 - 400 = 432$
We can see that compound interest for 2nd year is more than simple interest for 2nd year by

$432 - 400 = Rs.32$
Rs. 32 is the interest obtained for Rs.400 for 1 year

$R = \frac{100\times SI}{PT} = \frac{100 \times 32}{400\times1} = 8$ %
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

$=\frac{PRT}{100} = \frac{832\times 8\times 1}{100} = 66.56$
Total difference between the compound and simple interest for 3 years

$= 32 + 66.56 = Rs.98.56$

Gauri want to the stationers and bought things worth Rs 25 out of which 30 paise went on sale tax on taxable purchases If the tax rate was 6% then what was the cost of the tax free item?

0%
Rs 15
0%
Rs 15.70
0%
Rs 19.70
0%
Rs 20

Explanation

Given: Gauri bought

$Rs.25$
Sales tax =

$30$ paise
Sales tax rate = Rs.

$6$ %
Let amount of taxable purchase =

$X$

$\dfrac{6}{100} \times{X} = \dfrac{30}{100}$

$X =5$
Tax free amount =

$25 - 5 - 0.30$
=

$19.70$

$A$ began a business with Rs.

$85{,}000$ . He was joined afterwards by

$B$ with Rs.

$42{,}500$ . For how much period does

$B$ join, if the profits at the end of the year are divided in the ratio of

$3 : 1$ ?

0%
$4$ months
0%
$5$ months
0%
$6$ months
0%
$8$ months

Explanation

Suppose B joined for x months . Then, $\dfrac{ ( 85000 \times 12 )}{(42500 \times x)} = 3$

$=> x = \dfrac{(85000 \times 12)}{ (42500 \times 3) }$

$=>x= 8$

So, B joined for 8 months.

A scooter was bought at

$\text{Rs. } 42000$ . Its value depreciated at the rate of

$8\%$ per annum, find its value after one year.

0%
$\text{Rs. } 38640$
0%
$\text{Rs. } 37740$
0%
$\text{Rs. } 38145$
0%
$\text{Rs. } 39640$

A tradesman sold an article at a loss of

$20\%$ . Had he sold it for

$Rs. 100$ more, he should have gained

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

0%
Rs. $360$
0%
Rs. $400$
0%
Rs. $425$
0%
Rs. $450$

Explanation

Let the cost price

$=Rs. x$
Loss

$= 20 \%$
Loss

$=\cfrac{20}{100} \times x$ =

$\dfrac{x}{5}$
Now,

$SP = CP - Loss = x - \cfrac{x}{5} = \cfrac{4x}{5}$
If SP was Rs 100 more,
New

$SP = \cfrac{4x}{5} + 100$
Profit

$= 5\%$ of CP

$=\cfrac{5}{100} \times x = New \, SP - CP$
Profit

$=\cfrac{x}{20} = \cfrac{4x}{5} + 100 - x$

$\cfrac{20x - 16x + x}{20} = 100$

$5x = 2000$

$x = Rs. 400$

The present population of a town is

$5000$ . If the population decreases at the rate of

$10\%$ per year, the population after

$2$ years will be

0%
$4820$
0%
$4050$
0%
$4980$
0%
$4500$

Explanation

Present population,

$P_o = 5000$
Rate

$= 10\%$
Time

$= 2$ years
Let after

$2$ years population

$= P$
Thus,

$P = P_o(1 - \frac{R}{100})^T$

$P = 5000 (1 - \frac{10}{100})^2$

$P = 4050$
Thus after two years the population will be

$4050$

The population of a town increases from

$10000$ to

$10400$ . The growth rate is

0%
$4\%$
0%
$40\%$
0%
$2\%$
0%
$20\%$

Explanation

Population an year ago,

$P_o = 100000$

Population after one year

$P = 10400$

Let the growth rate be R,

then

$P = P_o \left(1 + \cfrac{R}{100}\right)^T$

$10400 = 10000\left(1 + \cfrac{R}{100}\right)$

$10400 = 10000 + 100 R$

$R = \cfrac{400}{100} = 4\%$

Thus, growth rate

$= 4\%$

By selling a pen for

$\text{Rs. } 15$ , a man loses one-sixteenth of what it costs him. Find the cost price of one pen.

0%
$\text{Rs. } 16$
0%
$\text{Rs. } 18$
0%
$\text{Rs. } 20$
0%
$\text{Rs. } 21$

Explanation

Let's assume the cost price of the pen is

$x$
According to the question, Loss

$=\cfrac { x }{ 16 }$
Cost Price

$=$ Selling Price

$+$ Loss
Hence,

$15+\dfrac { x }{ 16 } =x$

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

$15=\dfrac { 15x }{ 16 }$

$x=16$
Hence, the cost of

$1$ pen is

$\text{Rs. } 16$

A shopkeeper sold sarees at

$\text{Rs. }266$ each after giving a

$5 \%$ discount on labeled price. Had he not given the discount, he would have earned a profit of

$12 \%$ on the cost price. what was the cost price of each saree?

0%
$\text{Rs. } 240$
0%
$\text{Rs. } 260$
0%
$\text{Rs. } 280$
0%
Date indepequate
0%
None of these

Explanation

Let us assume that the labeled price is equal to

$x$ .

$\dfrac{95}{100}\times x=266$

$\Rightarrow x = \text{Rs. } 280$

Now consider that the cost price of the saree is equal to

$y$ . Then,

$\text{Profit}\% = 12=\dfrac{ (280-y)\times 100 }{ y }$

$\begin{array}{l}\Rightarrow12y = 28000 - 100y\\\Rightarrow112y = 28000\\\Rightarrow y =\text{Rs. } 250\end{array}$

Hence, option

$D$ is the correct answer.

By selling a book for Rs. 10, the publisher loses

$\displaystyle \frac{1}{11}$ of what it costs him. His cost price is ....

0%
Rs. 9
0%
Rs. 10
0%
Rs. 11
0%
Rs. 12

Explanation

Let's assume cost price

$=x$
Selling price

$=\text{Rs.}\ 10$

Loss

$=\dfrac{1}{11}.x$

$\text{Cost Price=Selling Price + Loss}$

$x=10+\dfrac { x }{ 11 }$

$\dfrac { 10x }{ 11 } =10$

$x=\text{Rs.}\ 11$
Answer (C) Rs. 11

Samant brought microwave oven and paid 10% less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did Samant earn on the original price?

0%
17%
0%
20%
0%
27%
0%
32%

Explanation

Lets assume original price of the microwave oven = Rs. 100
So, buying price for samant = Rs. 90
Profit

$\% = 30\% = \dfrac{p\times100}{Cost price}$

$30\% = \dfrac{p\times100}{90}$

${30\times90}{100}=p$

$profit=27$
Selling price =90+27=117 Rs
Profit =117-100=17 Rs
profit

$\%=\dfrac{17\times100}{100} = 17%$

A person has enough money to buy

$25$ cycles worth

$Rs.\,500$ each. How many cycles will the person be able to buy, if each cycle now costs

$Rs.\,125$ more ?

0%
$40$ cycles
0%
$70$ cycles
0%
$20$ cycles
0%
$65$ cycles

Explanation

Total money=25*500=12500
New cost of cycle=500+125=625
No of cycle person is able to buy

$N=\frac { Total\quad money }{ cost\quad of\quad 1\quad cycle }$

$N=\frac { 12500 }{ 625 } =20$
Answer (C) 20

A scientific calculator is available at Universal shop in Hazratganz at

$20\%$ discount and the same is available at only

$15\%$ discount at Bhootnath market. Pooja has just sufficient amount of

$Rs.\,800$ to purchase it at Universal shop, Hazratganz. What is the amount that Pooja has less than the required amount to purchase it at Bhootnath?

0%
$Rs.\,70$
0%
$Rs.\,50$
0%
$Rs.\,100$
0%
Data insufficient

Explanation

Let cost of calculator is

$Rs. 100$
calculator is available at Universal shop in Hazratganj at

$20\%$ discount then sell cost

$=100-20=80\ RS$
Pooja paid

$Rs. 800$ at Universal shop
So actual cost of calculator

$=\dfrac{800}{80}\times 100= 1000$ Rs
same as 15% discount at Bhootnath market
Then selling price of calculator at bhootnath market

$=\dfrac{85}{100}\times 1000= 850$ Rs
amount that Pooja has less paid

$=850-800=Rs.50$

A student purchased a pen for RsAt the end of 8 years it was valued at Rs 20 Assuming the yearly depreciation is a constant amount find the annual depreciation

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

Explanation

Cost Price of the pen

$=$

$Rs.100$

After

$8$ years, Value of the pen

$=Rs. 20$

Since, depreciation is yearly and constant

So, Annual depreciation

$=\dfrac{Rs.100-Rs.20}{8}=\dfrac{Rs.80}{8}=Rs.10$

Hence, C is the correct option.

Harry goes to a shop to buy a watch costing Rs. 1404, including sales tax of 8

$\%$ . He asks the shopkeeper to reduce the price of the watch so that he can save an amount equal to the sales tax. The reduction in the price of the watch is _____ .

0%
$Rs.\ 108$
0%
$Rs.\ 104$
0%
$Rs.\ 112.32$
0%
$\text{None of these}$

Explanation

Let the original cost of the watch be

$Rs.\ x$ .
Given that, after a sales tax of

$8\%$ , the price of the watch is

$Rs.\ 1404$

$\therefore \ x+\dfrac{8}{100}\times x=1404$

$\Rightarrow \dfrac{108}{100}\times x=1404$

$\Rightarrow x=\dfrac{1404}{1.08}=1300$

Hence, the sales tax on the watch is

$=1404-1300=Rs.\ 104$ .

Harry saves an amount equal to the sales tax.

Thus, Harry saves

$Rs.\ 104$ .

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

0%
Rs 1275
0%
Rs 1280
0%
Rs 1210
0%
Rs 1260

Explanation

let the new S.P be rs x.then
(100-loss%) : (1 st S.P)=(100+gain%) (2nd S.P)

$\frac{100-5}{1140}=\frac{100+5}{x}$

$x=\frac{105\times 1140}{95}$

$x=1260 Rs.$

Latika bought a radio set and spent Rs. 110 on its repairs. She then sold it to Garima at 20% profit, Garima sold it to Megha at a loss of 10% and Megha sold it for Rs. 1188 at a profit of 10%. What is the amount for which Latika bought the radio set?

0%
Rs. 850
0%
Rs. 890
0%
Rs. 930
0%
Rs. 950

Explanation

Let's assume the amount for which Latika bought the radio set=x
So, As per question,
Latika's selling price=(x+110)*1.20
Megha's buying price=(x+110)*1.20*0.90
Megha's selling price=

$(x+110)\times 1.20\times 0.90\times 1.10=1188$

$1.19x+130.68=1188$

$x=888.50$
Answer (B) Rs. 890

The taxi charges in a city comprise of a fixed charge, together with the charge of the distance covered. For a journey of 16 km, the charges paid are Rs. 156 and for a journey of 24 km, the charge paid are Rs,What will a person have to pay for travelling a distance of 30 km?

0%
Rs. 236
0%
Rs. 240
0%
Rs. 248
0%
Rs. 252

Explanation

Let's take fixed charge=F
charge of the distance covered=x
So, As per question,

$F+16x=156$ .................eq1

$F+24x=204$ .......................eq2
Eq2-eq1

$8x=48$

$x=6$
from eq 1

$F=156-96=60$
Charge a distance of 30 km=

$F+30x=60+30*6=240Rs$
Answer (B) Rs. 240

The marked price of a book is Rs.

$100$ . The shopkeeper gave

$25\%$ discount on it. Then the sale price of the book is

0%
Rs. $100$
0%
Rs. $75$
0%
Rs. $25$
0%
Rs. $125$

If the manufacturer gains 10%, the wholesale dealer 15% and the retailer 25% then find the cost of production of a table the retail price of which is Rs 1265?

0%
600
0%
800
0%
700
0%
900

Explanation

Let the cost of production of the table be Rs
Then ,125% of 115% of 110% of x=1265

$\Rightarrow \frac {125}{100}\times \frac{115}{100}\times \frac{110}{100}\times x=1265$

$x=\frac{1265\times 100\times 100\times 100}{125\times 115\times 110}$

$x=800 Rs.$

A tempo is insured to the extent of

$\displaystyle \frac{4}{5}$ of its original value. If the premium on it at the rate of 1.3 percent amounts to Rs 910 the original value of the tempo is

0%
Rs 78500
0%
Rs 80000
0%
Rs 82500
0%
Rs 87500

Explanation

Let the original value of the tempo be Rs. x. Then,

1.3 Percent of 4/5 of x = 910

$\Rightarrow \frac{13}{10}\times\frac{1}{100}\times\frac{4}{5}\times x=910$

$\Rightarrow x = \left(\frac{910\times10\times100\times5}{13\times4}\right)=87500$

A trader purchases 70 kg of tea at Rs. 15 per kg and 30 kg of tea at Rs. 18.50 per kg. If the packing charges are 2 percent, then at what price he must sell the mixture of two to gain 15%?

0%
Rs. 18.82 per kg
0%
Rs. 18 per kg
0%
Rs. 18.50 per kg
0%
Rs. 17.80 per kg

Explanation

Cost of 70 kg of tea at Rs. 15 per kg

$= 70\times 15=1050 Rs.$
Cost of 30 kg of tea at Rs. 18.50 per kg

$=30\times 18.5 = 555Rs.$ .
total cost

$=1050 Rs.+555 Rs.=Rs. 1605$

Selling price includes packing charges are 2 percent and gain 15%:
Therefore,

$S= Cost price\times ({1+\dfrac{2}{100}})\times ({1+\dfrac{15}{100}})$

$S=1605\times 1.02\times1.15$

$S=Rs. 1882.67$
total weight

$=70+30=100kg$

Selling price per kg

$=\dfrac{1882.67}{100}= 18.82Rs.$ per kg

Answer (A) is correct.

Arun purchased 30 kg of wheat at the rate of Rs 11.50 per kg and 20 kg of wheat at the rate of Rs 14.25 per kg . He mixed the two and sold the mixture. Approximately what price per kg should he sell the mixture to make 30% profit?

0%
Rs 14.80
0%
Rs 15.40
0%
Rs 15.60
0%
Rs 16.40

Explanation

C.P. of $50 Kg$ wheat = $(30 \times 11.50 + 20 \times 14.25)$

$= Rs. (345 + 285) = Rs. 630.$

S.P of $50 Kg$ wheat=130% of 630= $\frac{130}{100}\times 630=819 Rs$

S.P. per kg= $\frac{819}{30+20}$

= $\frac{819}{50}$

= $16.40 Rs.$

After getting two successive discounts a shirt with a list price of Rs. 150 is available at Rs.If the second discount is 12.5%, find the first discount.

0%
25%
0%
12%
0%
15%
0%
20%

Explanation

Let the first discount be

$x \%$

Then,

$87.5\; \%$ of

$(100-x)\;\%$ of

$150=105$

$\Rightarrow \dfrac{87.5}{100}\times \dfrac{100-x}{100}\times 150=105$

$\Rightarrow 100-x=\dfrac{105\times 100\times 100}{87.5\times 150}$

$\Rightarrow 100-x=80$

$\Rightarrow x=20\;\%$

The difference between simple interest and compound interest is that

0%
Simple interest is easier to compute.
0%
Simple interest yields a lot of money.
0%
Simple interest is the interest on original principal only.
0%
None of these

Explanation

$\Rightarrow Simple Interest = \dfrac {P \times N\times R} {100}$

$where, P = Principal \:or\: the\: sum$

$N = Number \:of\: period \:or \:Time$

$R = Rate\:of\:interest$

$\Rightarrow Easier\: to\: compute.$

$\Rightarrow Simple\: interest\: is \:calculated \:on \: original \: principal.$

A and B entered into partnership with capitals in the ratio 4 : 5 After 3 months A withdraw

$\displaystyle \frac{1}{4}$ of his capital and B withdraw

$\displaystyle \frac{1}{5}$ of his capital. The gain at the end of 10 months was RsA's share in this profit is

0%
Rs 330
0%
Rs 360
0%
Rs 380
0%
Rs 430

Explanation

Ratio of the initial capital of A and B = 4 : 5
Hence the initial capitals of A and B is 4x and 5x
Ratio in which profit will be divided

$(4x\times 3)+(\frac{3}{4}\times 4x\times 7):(5x\times 3)+(\frac{4}{5}\times 5x\times 7)$

$33:43$
A's share=

$\frac{33}{76}\times 760=Rs 330$

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

0%
Rs $19.75$
0%
Rs $20$
0%
Rs $21$
0%
Rs $21.25$

Explanation

Marked price

$=$ Rs.

$30$ .
Discount

$=15 \%$ of

$30$

$\Rightarrow$ Discount

$= \dfrac{15}{100}\times 30=$ Rs

$4.5$
Then S.P

$=30-4.50=$ Rs

$25.50 .$
Rs.

$1.50$ shuttle cock is given free so actual S.P

$=$ Rs

$25.50-Rs1.50=Rs24$
Profit percentage

$=20%$
CP

$= \dfrac{100}{100 + Percentage \; Profit} \times SP$
Then

$C.P.=\dfrac{100}{120}\times 24=Rs20 .$

$A, B$ and

$C$ enter into partnership

$A$ invests some money at the beginning ,

$B$ invests double the amount after

$6$ months and

$C$ invests thrice the amount after

$8$ months If the annual profit be

$Rs. 27000\ C's$ share is

0%
$Rs. 8625$
0%
$Rs. 9000$
0%
$Rs. 10800$
0%
$Rs. 11250$

Explanation

Let

$A$ invested

$Rs. x$ for

$12$ months
Then according to the question

$B$ invested

$Rs. 2x$ for

$6$ months
And

$C$ invested

$Rs.3x$ for

$4$ months
Then Ratio of their capital

$=12\times x:2x\times 6:3x\times 4$

$=12x:12x:12x$

$=1:1:1$

$C's$ share of profit

$=\dfrac{1}{3}\times 27000=Rs.9000.$

The prices of a scooter and a TV are in the ratio 7 : 5 If the scooter costs Rs 8000 more than a TV set then the price of a TV set is

0%
Rs 20000
0%
Rs 24000
0%
Rs 28000
0%
Rs 32000

Explanation

Let the prices of scooter be 7x and tv set be 5x.
the difference between the two prices is 7x-5x=2x,
According to the question.

$2x=8000$

$x=4000.$
so price of scooter is

$7x=7\times 4000=28000$
and price of tv set is

$5x=5\times 4000=20000$

A invested Rs 76000 in a business After few months B joined him with Rs 57000 At the end of the year the total profit was divided between them in the ratio of 2 :1 After how many months did B join?

0%
4
0%
8
0%
9
0%
None of these

Explanation

Let A invested Rs.76000 for 12 months and B invested Rs.57000 for x months
Then

$\frac{7600\times 12}{5700\times x}=\frac{2}{1}$

$\Rightarrow x=\frac{76\times 12}{57\times 2}$

$\Rightarrow x=8$
hence B invested for 8 months

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Comparing Quantities Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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