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MATHS 3 Marks Question

Comparing Quantities · Gujarat Board (GSEB) STD 8 (GSEB - English Medium). 17 questions.

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Question 13 Marks
Find Cl on a sum of ₹ 8000 for 2 yr at 5% per annum compounded annually.
Answer
Given, principal $(\text{P})=$ ₹ $8000$, time $(n)=2 y r$, rate $( R )=5 \%$
We know that
$\begin{aligned}\text{Amount }(A) & =P\left(1+\frac{R}{100}\right)^n=8000\left(1+\frac{5}{100}\right)^2 \\ & =8000\left(\frac{100+5}{100}\right)^2 \\ & =8000\left(\frac{105}{100}\right)^2 \\ & =8000 \times \frac{21}{20} \times \frac{21}{20} \\ & =\frac{80 \times 21 \times 21}{2 \times 2}\end{aligned}$
$=20 \times 21 \times 21=$ ₹ $8820$
$\therefore$ Compound Interest (CI)
$=$ Amount (A) $-$ Principal (P)
$=$ ₹ $(8820-8000)$
$=$ ₹ $820$
Hence, the compound interest is ₹ 820.
Alternate Method
Principal for the Ist year,
$P_1=$ ₹ $8000, R=5 \%$ and $T=2 yr$
$\therefore \quad SI = SI$ at $5 \%$ per annum for Ist year
$=\frac{8000 \times 5 \times 1}{100}=$ ₹ $400\qquad\quad$ $\left[\because SI =\frac{P \times R \times T}{100}\right]$
Amount at the end of Ist year
$=$ Principal $+$ Simple interest
$=$ ₹ $8000+$ ₹ $400=$ ₹ $8400$
Now, principal for the IInd year
$=$ Amount at the end of Ist year
$=$ ₹ $8400$
$SI _2= SI$ at $5 \%$ per annum for IInd year
$=$ ₹ $\frac{8400 \times 5 \times 1}{100}=$ ₹ $(84 \times 5)=$ ₹ $420$
Amount at the end of IInd year
$=$ Amount at the end of Ist year $+ SI _2$
$=$ ₹ $(8400+420)=$ ₹ $8820$
$\therefore$ Compound Interest $( CI )=$ Amount $(A)-\operatorname{Principal}(P)$
$=$ ₹ $(8820-8000)=$ ₹ $820$
Hence, the compound interest is ₹ $820$.
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Question 23 Marks
In a primary school, the parents were asked about the number of hours, they spend per day in helping their children to do homework. There were 90 parents, who helped for $\frac{1}{2}\text{h}$ to $1 \frac{1}{2}\text{h}$. The distribution of parents according to the time for which, they said they helped is given in the adjoining figure, $2 0 \%$ helped for more than $1 \frac{1}{2}\text{h}$ per day, $30 \%$ helped for $\frac{1}{2}\text{h}$ to $1 \frac{1}{2} \text{h}, 50 \%$ did not help at all. Using this, answer the following :
(i) How many parents were surveyed?
(ii) How many said that they did not help?
(iii) How many said that they helped for more than $1 \frac{1}{2}\text{h}?$
Answer
(i) Here, 90 parents helped their children for $\frac{1}{2}\text{h}$ to $\frac{1}{2}\text{h}.$
Given, percentage = 30%
Let x parents be surveyed.
Then, $30 \%$ of $x$ helped for $\frac{1}{2}\text{h}$ to $1\frac{1}{2}\text{h}=90$
$\therefore \quad 30 \%$ of $x=90$
$\Rightarrow \quad \frac{30}{100} \times x=90$
$x=\frac{90 \times 100}{30}=300$
Hence, the number of parents were 300.
(ii) Here, 50% parents did not help at all.
So, number of parents did not help = 50% of 300
$\begin{array}{l}=\frac{50}{100} \times 300 \\ =50 \times 3=150\end{array}$
Hence, 150 parents did not help.
(iii) Here, $20 \%$ parents helped for more than $1 \frac{1}{2}\text{h}.$
So, number of parents who helped for more than $1 \frac{1}{2}\text{h}$
$\begin{array}{l}=20 \% \text { of } 300 \\ =\frac{20}{100} \times 300=20 \times 3=60\end{array}$
Hence, 60 parents helped for more than $1 \frac{1}{2}\text{h}.$
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Question 33 Marks
By what per cent is ₹ $2000$ less than ₹ $2400?$ Is it the same as the per cent by which ₹ 2400 is more than ₹ 2000?
Answer
Given, ₹ 2400 is reduced to ₹ 2000 .
Case I When original value $=$ ₹ $2400$
Then, decrease value $=$ ₹ $(2400-2000)=$ ₹ $400$
$\begin{aligned} \therefore \text { Percentage decrease } & =\left(\frac{\text { Decrease value }}{\text { Original value }} \times 100\right) \% \\ & =\left(\frac{400}{2400} \times 100\right) \% \\ & =\left(\frac{4}{24} \times 100\right) \% \\ & =\left(\frac{400}{24}\right) \% \\ & =\frac{50}{3} \% \\ & =16 \frac{2}{3} \%\end{aligned}$
So, the percentage decrease is $16 \frac{2}{3} \%$.
Case II When original value $=$ ₹ $2000$
Then, ₹ 2000 is increased to ₹ 2400.
Now, increase value $=$ ₹ $(2400-2000)=$ ₹ $400$
$\begin{aligned} \therefore \text { Percentage increase } & =\left(\frac{\text { Increase value }}{\text { Original value }} \times 100\right) \\ & =\left(\frac{400}{2000} \times 100\right) \% \\ & =\left(\frac{4 \times 100}{20}\right) \% \\ & =(4 \times 5) \%=20 \%\end{aligned}$
So, percentage increase is $20 \%$.
Hence, percentage decrease and percentage increase are not same.
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Question 43 Marks
The population of a place increased to 54000 in 2003 at a rate of $5 \%$ per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
Answer
(i) Here, population in $2003=54000$,
Rate of increase $=5 \%$ and time period $(\text{n})=2\text{ yr}$
Let the population in 2001 be P.
$\therefore$ Present population $=\text{P}\left(1+\frac{5}{100}\right)^2$
$\Rightarrow 54000=\text{P} \times\left(\frac{100+5}{100}\right)^2$
$\Rightarrow 54000=P \times\left(\frac{105}{100}\right)^2$
$\Rightarrow 54000=P \times \frac{21}{20} \times \frac{21}{20}$
$\Rightarrow P \times 21 \times 21=54000 \times 20 \times 20$
$\therefore P=\frac{54000 \times 20 \times 20}{21 \times 21}$
$=\frac{21600000}{21 \times 21}$
$=48979.59$
$=48980$ (approx.)
Hence, the population in 2001 is 48980.
(ii) For population in 2005 i.e. 2 yr after 2003,
Initial population = 54000 i.e. in the year 2003
Time (n)= 2 yr
Rate of increase = 5%
Let population in 2005 be A.
$\therefore A=P\left(1+\frac{R}{100}\right)^\text{n}$
$=54000\left(1+\frac{5}{100}\right)^2 $
$ =54000\left(\frac{100+5}{100}\right)^2 $
$ =54000\left(\frac{105}{100}\right)^2 $
$ =54000 \times \frac{105}{100} \times \frac{105}{100} $
$ =27 \times 21 \times 105 $
$ =59535$
Hence, the population in 2005 is 59535.
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Question 53 Marks
Arun bought a pair of skates at a sale, where the discount given was 20%. If the amount he pays is ₹ 1600, then find the marked price.
Answer
Let the marked price of a pair of skates be ₹ $100$.
Discount per cent on a pair of skates $=20 \%$
$\therefore$ Discount amount on a pair of skates
$=20 \%$ of marked price
$=20 \%$ of ₹ $100$
$=$ ₹ $\left(\frac{20}{100} \times 100\right)=$ ₹ $20$
$\therefore$ Selling price of a pair of skates $=$ ₹ $100-$ ₹ $20$
$=$ ₹$(100-20)=$ ₹ $80$
If selling price is ₹ $80$, then marked price $=$ ₹ $100$
If selling price is ₹ $1$, then marked price $=$ ₹ $\frac{100}{80}$
If selling price is ₹ $1600$, then marked price
$=$ ₹ $\left(\frac{100}{80} \times 1600\right)$
$=$ ₹ $(100 \times 20)=$ ₹ $2000$
Hence, the marked price of a pair of skates is ₹ 2000.
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Question 63 Marks
If Chameli had ₹ 600 left after spending 75% of her money, then how much did she have in the beginning?
Answer
Let Chameli had total money be ₹ $\text x.$
Percentage of money spent by Chameli = 75%
Chameli had left money after spending
$=(100-75) \%=25 \%$
But money left $=$ ₹ $600$ $\qquad$ [given]
$\therefore \quad 25 \%$ of $\text x=600$
$\Rightarrow \quad \frac{25}{100} \times \text{x}=600$
$\Rightarrow \quad \text{x}=\frac{600 \times 100}{25}=2400$
Hence, Chameli had ₹ 2400 in the beginning.
Alternate Method
Let total money with Chameli at the beginning be ₹ 100.
Expenditure money = 75% of total money
$=$ ₹$\left(\frac{75}{100} \times 100\right)=$ ₹ $75$
Money left $=$ ₹ $100-$ ₹ $75=$ ₹ $25$
Now, if saving is ₹ $25$, then money at the beginning $=$ ₹ $100$
If saving is ₹ $1$, then money at the beginning $=$ ₹ $\frac{100}{25}$
If saving is ₹ $600$, then money at the beginning
$=$ ₹ $(\frac{100}{25} \times 600)$
$=$ ₹ $(4 \times 600) $
$ =$ ₹ $2400$
Hence, total money at the beginning with Chameli was ₹ 2400.
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Question 73 Marks
Find the ratio of the following.
(i) Speed of a cycle $15 \text{ km/h}$ to the speed of scooter $30\text{ km/h}$
(ii) 5 m to 10 km
(iii) 50 paise to ₹ 5
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Question 83 Marks
The table shows the cost of sunscreen of two brands with and without sales tax :
BrandsCost (in ₹)Cost + Tax (in ₹)
1.X (100 g)7075
2.Y (100 g)6265
Which brand has a greater sales tax rate ? Give the sales tax rate of each brand.
Answer
For Brand X,
Cost price $=$ ₹ $70$ and selling price $=$ ₹ $75$
$\therefore$ Sales tax $=$ Selling Price - Cost Price
$=$ ₹ $75~-$ ₹ $70=$ ₹ $5$
We know that
Sales $\text{tax}= \text{Tax} \%$ on cost price
$\therefore 5=\frac{x}{100} \times 70$ [let $x$ be the sales tax percentage]
$\Rightarrow 500=70 \times x$
$\Rightarrow x=\frac{500}{70} \%=7 \frac{1}{7} \%=7.14 \%$
For Brand Y,
Cost price $=$ ₹ $62$ and selling price $=$ ₹ $65$
$\therefore$ Sales $\operatorname{tax}=65-62=$ ₹ $3$
Let $\text{y%}$ be the sales tax of Brand Y.
$\therefore \text y \%$ of $62=3 \Rightarrow \frac{\text y}{100} \times 62=3$
$\Rightarrow \text y=\frac{300}{62}=4.838 \approx 4.84 \%$
Clearly, for Brand X sales tax is greater.
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Question 93 Marks
In the year $2001,$ the number of malaria patients admitted in the hospitals of a state was $4375.$ Every year this number decreases by $8 \%$. Find the number of patients in $2003.$
Answer
In 2001, the number of malaria patients admitted in the hospitals of a state was 4375.
Every year this number decreases by $8\%.$
Let $\text{P} = 4375, \text{R} = 8\%,\text n=2$
$\therefore$ Number of patients in 2003
$=\text P(1-\frac{\text R}{100})^\text n$
$=4375\left(1-\frac{8}{100}\right)^2=4375\left(\frac{92}{100}\right)^2 $
$ =\frac{4375 \times 23 \times 23}{25 \times 25} $
$ =\frac{2314375}{625}=3703$
Hence, there are 3703 patients in 2003.
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Question 103 Marks
Ranjana bought a product for ₹ $3155$ including $4.5 \%$ sales tax. Find the price before tax was added.
Answer
Given, billed amount of the product $=$ ₹ $3155$
and sales $\text{tax}=4.5 \%$
Let the price before tax be ₹ $\text x.$
Then, $\text x+4.5 \%$ of $\text x=3155$
$\Rightarrow x+\frac{4.5}{100} \times x=3155 $
$ \Rightarrow \frac{100 x+4.5 x}{100}=3155 $
$ \Rightarrow 104.5 x=315500$
$\therefore$ $x=\frac{315500}{104.5}$
$=3019.138=$ ₹ $3019.14$
Therefore, the price before added tax was ₹ $3019.14.$
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Question 113 Marks
Ayesha announced a festival discount of $25 \%$ on all the items in her mobile phone shop. Ramandeep bought a mobile phone for himself. He got a discount of ₹ $1960.$ What was the marked price of the mobile phone?
Answer
Given that, discount $=$ ₹ $1960$ and discount $=25 \%$
When discount is 25, then marked price $=$ ₹ $100$
$\therefore$ When discount is ₹ $1$, then marked price $=$ ₹ $\frac{100}{25}$
$\therefore$ When discount of ₹ 1960, then marked price
$=\frac{100}{25} \times 1960=$ ₹ $7840$
So, the marked price of the mobile phone is ₹ $7840.$
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Question 123 Marks
Ramender offers a discount of $20 \%$ on all the items at his shop. What is the selling price of an article marked at ₹ $2800?$
Answer
Given, marked price $=$ ₹ $2800$
Discount $=20 \%$ of $2800$
$=\frac{20}{100} \times 2800=$ ₹ $560$
$\therefore$ Selling price $=$ ₹ $(2800-560)=$ ₹ $2240$
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Question 133 Marks
In a public school, $70 \%$ of the student population are boys. If there are $120$ girls, then find the total enrollment of the school.
Answer
Percentage of girls $=100 \%-70 \%=30 \%$
$30 \%$ of the student population $=120$
$\therefore 1 \%$ of the student population $=\frac{120}{30}$
$\therefore 100 \%$ of the student population $=\frac{120}{30} \times 100=400$
So, the total enrollment of the school is $400.$
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Question 143 Marks
Ranjit works in a multinational company as a senior executive with annual package of 12 lakh.
He pays income tax regularly and donates 5% of his annual earning to an orphanage. What amount of money he donate?
Answer
Total annual income $=$ ₹ $1200000$
$\therefore 5 \%$ of the total income $=1200000 \times \frac{ 5 }{100}=$ ₹ $60000$
Thus, he donates ₹ $60000$ to an orphanage.
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Question 153 Marks
If marked price of an article is ₹ 800 and it is sold at ₹ $760,$ then find the discount rate.
Answer
$\text {We have, marked price}=$ ₹ $800$
$ \text {and selling price}=$ ₹ $760 $
$\therefore \text {Discount} =\text {Marked Price}- \text { Selling price } $
$ =$ ₹ $800~-$ ₹ $760=$ ₹ $40$
Let the discount rate be $\text x\%$
Then, $\text x\%$ of marked price $= 40$
$\Rightarrow \frac{x}{100} \times 800=40 \Rightarrow x=\frac{40 \times 100}{800}=5 \%$
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Question 163 Marks
Madhu's room measures $6\text{ m} \times 3\text{ m}.$ Her carpet cover is $8\text{ m}^2.$ What per cent of floor is covered by the carpet?
Answer
Area of floor $=(6 \times 3)\text{ m} ^2=18\text{ m}^2$
Area covered by carpet $=8\text{ m}^2$
When $18\text{ m}^2$ area will covered by carpet, then floor will be covered $=100 \%$
or when $1\text{ m}^2$ area covered by carpet, then floor will be covered $=\frac{100}{18} \%$
So, when $8\text{ m}^2$ area covered by carpet, then floor will be covered $=\left(\frac{100}{18} \times 8\right) \%=44.44 \%$
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Question 173 Marks
A student used the proportion $\frac{\text n}{100}=\frac{5}{32}$ to find $5 \%$ of 32. What did the student do wrong?
Answer
Given that student used proportion.
$\frac{\text n}{100}=\frac{5}{32} \Rightarrow\text n=\frac{5 \times 100}{32}$
While $5 \%$ of $32=\frac{5}{100} \times 32$
So, the student did wrong in selection of formula.
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