2 Marks Questions - MATHS STD 8 Questions - Vidyadip

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2 Marks Questions

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Find the population of a city after 2 yr, which is at present 12 lakh, if the rate of increase is 4%.

Answer

Given, present population (P) =12 lakh = 1200000
Rate of increase (R) = 4% and time (n) = 2 yr
$\therefore$ Population after $2\text{ yr} =P\left(1+\frac{R}{100}\right)^\text{n}$
$=1200000\left(1+\frac{4}{100}\right)^2 $
$ =1200000\left(\frac{100+4}{100}\right)^2 $
$ =1200000 \times\left(\frac{104}{100}\right)^2 $
$ =1200000 \times \frac{104}{100} \times \frac{104}{100} $
$ =120 \times 104 \times 104 $
$ =1297920$
Hence, the population of a city after 2 yr will be 1297920.

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Question 22 Marks

A machinery worth ₹ 10500 depreciated by 5%. Find Its value after one year.

Answer

Here, principal (P) = ₹ 10500,
Reduction (R) = −5% and time (n) = 1 yr $\qquad\quad$ [depreciation means reduction]
$\therefore$ Reduced the value after one year
$=P\left(1-\frac{R}{100}\right)^n $
$ =10500\left(1-\frac{5}{100}\right)^1 $
$ =10500\left(\frac{100-5}{100}\right) $
$ =10500 \times \frac{95}{100} $
$ =105 \times 95 $
$=$ ₹ $9975$
Hence, the value of a machinery after one year will be ₹ 9975.

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Question 32 Marks

Find interest and amount to be paid on ₹ 15000 at 5% per annum after 2 yr.

Answer

Given, principal $(\text{P})=$₹ $15000$,
rate $(\text{R})=5 \%$ per annum and time $(\text{T})=2 \text{ yr}$
On ₹ $100$, interest charged for $1 \text{ yr} =$ ₹ $5$
On ₹ $1$, interest charged for $1 \text{ yr} =$ ₹ $\frac{5}{100}$
So, on ₹ 15000 interest charged for $1\text{ yr}$
$=\frac{5}{100} \times 15000=$ ₹ $750$
Interest for $2\text{ yr} =$ ₹ $(750 \times 2)=$ ₹ $1500$
Amount to be paid at the end of $2\text{ yr} $
$=$ Principal + Interest
$=$ ₹ $15000+$ ₹ $1500$
$=$ ₹ $16500$
Alternate Method
$\because$ Interest $=\frac{P \times R \times T}{100}=\frac{15000 \times 5 \times 2}{100}=$ ₹ $1500$
$\therefore$ Amount $=$ Principal + Interest
$=$ ₹ $(15000+1500)=$ ₹ $16500$

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Question 42 Marks

An almirah is sold at ₹ $5 2 2 5$ after allowing a discount of $5 \%$. Find its marked price.

Answer

Given, sale price of an almirah = ₹ 5225
Discount per cent $=5 \%$
Let marked price of an almirah be ₹ $x.$
$\therefore$ Discount $=5 \%$ of marked price $=5 \%$ of $x$
$=\frac{5}{100} \times x=\frac{5 x}{100}$
$\therefore$ Sale price $=$ Marked Price - Discount
$=x-\frac{5 x}{100}=\frac{100 x-5 x}{100}=\frac{95 x}{100}$
According to the question,
$\frac{95 x}{100}=5225$
$\Rightarrow x=\frac{5225 \times 100}{95}$
$\Rightarrow x=\frac{1045 \times 100}{19}=5500$
Hence, the marked price of an almirah is ₹ 5500.

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Question 52 Marks

A table marked at ₹ 15000 is available for ₹ 14400. Find the discount given and the discount percent.

Answer

$\text {Given, marked price of a table}=$ ₹ $15000$
$\text {Sale price of a table}=$ ₹ $14400$
$\therefore \quad$ Discount $=$ Marked price - Sale price
$=$ ₹ $(15000-14400)=$ ₹ $600$
and discount per cent $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right)$
$=\left(\frac{600}{15000} \times 100\right)=\frac{60}{15}=4 \%$
Hence, the discount is ₹ 600 and discount per cent is $4 \%$.

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Question 62 Marks

A shop gives $20\%$ discount. What would the sale price of each of these be
(i) a dress marked at ₹ $120?$
(ii) a pair of shoes marked at ₹ $750?$
(iii) a bag marked at ₹ $250?$

Answer

(i) Given, marked price of a dress $=$ ₹ $120$ and discount offered $=20 \%$
$\therefore$ Discount $=20 \%$ of marked price $=\left(\frac{20}{100} \times 120\right)=$ ₹ $24$
$\therefore$ Sale price of a dress $=$ Marked price - Discount
$=$ ₹ $(120-24)=$ ₹ $96$
Hence, the sale price of a dress is ₹ 96.
(ii) Given, marked price of a pair of shoes $=$ ₹ 750 and discount offered $=20 \%$
$\therefore$ Discount $=20 \%$ of marked price $=20 \%$ of ₹ $750$
$=\frac{20}{100} \times 750=\left(\frac{20 \times 750}{100}\right)=$ ₹ $150$
$\therefore$ Sale price of a pair of shoes
$=$ Marked price - Discount
$=$ ₹ $(750-150)=$ ₹ $600$
Hence, the sale price of a pair of shoes is ₹ 600.
(iii) Do same as above
The sale price of a bag is ₹ 200.

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Question 72 Marks

Two times a number is a 100% Increase in the number. If we take half the number, then what would be the decrease in percent?

Answer

Let the number be $x$ and half the number be $\frac{x}{2}$.
Now, decrease in the number $=x-\frac{x}{2}=\frac{2 x-x}{2}=\frac{x}{2}$
Decrease in per cent $=\left(\frac{\text { Decrease }}{\text { Original value }} \times 100\right) \%$
$=\left(\frac{x / 2}{x} \times 100\right) \%=\left(\frac{x}{2 \times x} \times 100\right) \%$
$=50 \%$

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Question 82 Marks

A scooter was bought at ₹ 42000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Answer

Here, principal $(\text{P})=$ ₹ $42000$,
Rate of depreciation $\left(\text R_1\right)=8 \%$ and time $(\text n)=1\text{ yr}$
$\therefore \text { Amount }(\text A)=\text P\left(1-\frac{\text R}{100}\right)^\text n$
$=42000\left(1-\frac{8}{100}\right)^1 $
$ =42000\left(\frac{100-8}{100}\right) $
$ =42000 \times \frac{92}{100} $
$ =420 \times 92 $
$=$ ₹ $38640$
Hence, the value of the scooter after one year will be ₹ 38640.

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Question 92 Marks

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 h, if the count was initially 506000.

Answer

Here, initial count of bacteria (P) = 506000, time (n)= 2h
Rate of increasing bacteria (R) = 2.5% per hour
$=\frac{25}{10} \%$ per hour
$=\frac{5}{2} \%$ per hour
Bacteria count after 2 h be A.
$\therefore A=P\left(1+\frac{R}{100}\right)^n$
$=506000\left(1+\frac{5}{2 \times 100}\right)^2 $
$ =506000\left(\frac{200+5}{200}\right)^2 $
$ =506000\left(\frac{205}{200}\right)^2 $
$ =506000\left(\frac{41}{40}\right)^2 $
$ =506000 \times \frac{41}{40} \times \frac{41}{40} $
$ =\frac{5060 \times 41 \times 41}{4 \times 4} $
$ =\frac{1265 \times 41 \times 41}{4} $
$ =\frac{2126465}{4} $
$ =531616.25 $
$ =531616\text{ (approx.)}$
Hence, the number of bacteria count at the end of 2 h are 531616.

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Question 102 Marks

An article was purchased for ₹ 1239 including GST of $18 \%$. Find the price of the article before GST was added.

Answer

Let the price of article before including GST be ₹ x.
Price of article including GST = ₹ 1239
GST (Good and Services Tax) = $18\%$ of x
$=\frac{18}{100}\times\text{x}=0.18\text{x}$
$\therefore$ Price of article including GST
$=$ Price before GST + GST
$\Rightarrow$ $1239=x+0.18 x$
$\Rightarrow 1.18 x=1239$
$\Rightarrow x=\frac{1239}{118}=$ ₹ $1050$
So, the price of article before GST was added is ₹ 1050.

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Question 112 Marks

I purchased a hair-dryer for ₹ 5400 Including 8% VAT. Find the price before VAT was added.

Answer

Given, cost price of hair-dryer including VAT $=$ ₹ 5400
Rate of VAT $=8 \%$
Let the price of a hair-dryer before VAT was added be ₹ 100.
Price after adding VAT $=$ ₹ $(100+8)=$ ₹ $108$
If price after adding VAT is ₹ 108, then the price before
$\text{VAT}=$ ₹ $100$
If price after adding VAT is ₹ 1, then the price before
$\text{VAT}=$ ₹ $\frac{100}{108}$
If price after adding VAT is ₹ 5400, then the price before
$\text{VAT}$ $=$ ₹ $\left(\frac{100}{108} \times 5400\right)=$ ₹ $(100 \times 50)=$ ₹ $5000$
Hence, the price of a hair-dryer before VAT is ₹ 5000.

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Question 122 Marks

The price of a TV Is? ₹ 13000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay, If he buys it.

Answer

Given, price of a TV $=$ ₹ $13000$
and rate of sale tax $=12 \%$
$\therefore$ Sale tax $=12 \%$ of price of a TV $=12 \%$ of ₹ $13000$
$=$ ₹$\left(\frac{12}{100} \times 13000\right)=$ ₹ $(12 \times 130)=$ ₹ $1560$
So, the total amount, which Vinod will have to pay for purchasing a TV = Price of a TV + Sale tax
$=$ ₹ $(13000+1560)$
$=$ ₹ $14560$
Hence, Vinod will have to pay for buying a TV is ₹ 14560.

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Question 132 Marks

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 ₹ $1450$ and two shirts marked at ₹ $850$ each?

Answer

Given, marked price of a pair of jeans $=$ ₹ $1450$
and marked price of one shirt $=$ ₹ $850$
Now, marked price of two shirts $=$ ₹ $(2 \times 850)=$ ₹ $1700$
$\therefore$ Total marked price for jeans and shirts
$=$ ₹ $1450+$ ₹ $1700=$ ₹ $3150$
Also, discount offered by a shop = 10%
$\therefore$ Discount = 10% of total marked price
$=10 \%$ of ₹ $3150=$ ₹$\left(\frac{10}{100} \times 3150\right)=$ ₹ $315$
Thus, customer would have to pay for a pair of jeans and two shirts = Total marked price - Discount
$=$ ₹ $3150~-$ ₹ $315=$ ₹ $(3150-315)=$ ₹ $2835$

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Question 142 Marks

If $60 \%$ people in a city like cricket, $30 \%$ like football and the remaining like other games, then what percent of the people like other games? If the total number of people are 50 lakh, then find the exact number who like each type of games.

Answer

Given people who like cricket $=60\%$
and people who like football $=30\%$
$\therefore$ People who like other games $=[100-(60+30)]\%$
$=(100-90)\%=10\%$
Total number of people $=50$ lakh $=5000000$
Now, people who like cricket $=60 \%$ of $5000000$
$\begin{array}{l}=\frac{60}{100} \times 5000000=60 \times 50000 \\ =3000000=30\text{ lakh} \end{array}$
$\begin{aligned} \text { People who like football } & =30 \% \text { of } 5000000 \\ & =\frac{30}{100} \times 5000000 \\ & =1500000=15 \text { lakh }\end{aligned}$
and people who like other type of games
$\begin{array}{l}=10 \% \text { of } 5000000 \\ =\frac{10}{100} \times 5000000 \\ =500000=5 \text { lakh }\end{array}$

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Question 152 Marks

A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Answer

Number of matches won by the football team = 10
Let x matches be played by the team.
$\therefore 40 \%$ of $x=10$
$\Rightarrow \frac{40}{100} \times x=10$
$\Rightarrow \quad x=\frac{10 \times 100}{40}=25$
Hence, a football team played 25 matches in all.

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Question 162 Marks

72% of 25 students are good in Mathematics. How many students are not good in Mathematics?

Answer

Total number of students $=25$
Students good in Mathematics $=72 \%$
Students, who are not good in Mathematics
$=(100-72) \%=28 \%$
Now, number of students who are not good in Mathematics $=28 \%$ of $25=\frac{28}{100} \times 25=7$
Hence, 7 students are not good in Mathematics.

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Question 172 Marks

Convert the following ratlos to percentages.
2 : 4

Answer

Given ratio $=2: 3$
$\therefore \text { Percentage }=\left(\frac{2}{3} \times 100\right)=\left(\frac{200}{3}\right)=66 \frac{2}{3} \%$

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Question 182 Marks

Convert the following ratlos to percentages.
3 : 4

Answer

Given ratio $=3: 4$
$\therefore \text { Percentage }=\left(\frac{3}{4} \times 100\right)=(3 \times 25)=75 \%$

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Question 192 Marks

Using the formula $\text A=\text P\left(1+\frac{\text R}{100}\right)^\text n,$ calculate the amount and the compound interest for the following
(i) $\text{P}=$ ₹ $500, \text R=12 \%$ and $\text n=2 \text{ yr}.$
(ii) $\text P=$ ₹ $1000, \text R=10 \%$ and $\text{n}=2\text{ yr}.$

Answer

(i) $\because A=P\left(1+\frac{R}{100}\right)^n$
$\therefore A=500\left(1+\frac{12}{100}\right)^2$
$=500\left(\frac{112}{100}\right)^2$
$=500 \times \frac{12544}{10000}=$ ₹ $627.2$
$\therefore C I=A-P=$ ₹ $(627.2-500)=$ ₹ $127.2$
(ii) $\because A=P\left(1+\frac{R}{100}\right)^n$
$\therefore A=1000\left(1+\frac{10}{100}\right)^2$
$=1000 \times\left(\frac{110}{100}\right)^2$
$=1000 \times \frac{121}{100}=1210$
$\therefore CI =A-P=$ ₹ $(1210-1000)=$ ₹ $210$

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Question 202 Marks

After increasing $15\%$ of the price of an article, its price is ₹ 1725. Find the increased amount.

Answer

Let the initial price of the article be ₹ $x.$
Then, $x+15 \% \text { of } x=1725$
$\Rightarrow x+\frac{15}{100} \times x=1725$
$\Rightarrow \frac{100 x+15 x}{100}=1725 \Rightarrow 115 x=1725 \times 100$
$\Rightarrow x=\frac{1725 \times 100}{115}=1500$
$\therefore $ Increased amount $=$ ₹ $ (1725-1500)=$ ₹ $225$

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Question 212 Marks

By how much per cent $2800$ is greater than $2400?$

Answer

Let 2800 be $x \%$ greater than $2400.$
Then, $2400+x \%$ of $2400=2800$
$\Rightarrow 2400+2400 \times \frac{x}{100}=2800$
$\Rightarrow 2400 \times \frac{x}{100}=2800-2400=400$
$\Rightarrow x=\frac{400 \times 100}{2400}=\frac{50}{3} \%$
So, 2800 is $\frac{50}{3} \%$ greater than 2400.

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Question 222 Marks

Navin has a cell phone of fixed MP, ₹ $\text{12000}$ and sold it for ₹ $\text{8000}.$ Find discount per cent.

Answer

Given, $\text{MP} =$ ₹ $12000, \text{SP} =$ ₹ $8000$
Discount $= \text{MP}-\text{SP} =12000-8000=$ ₹ $4000$
$\therefore$ Discount $\%=\frac{\text { Discount }}{\text{MP}} \times 100$
$=\frac{4000}{12000} \times 100=\frac{100}{3} \%=33 \frac{1}{3} \%$

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