Question 12 Marks
Find the amount to be paid at the end of $3$ years in each case:Principal $=₹ 7,500$ at $5 \%$ p.a.
AnswerHere, Principal $(P)=₹ 7,500$, Rate $(\mathrm{R})=5 \%$ p.a., Time $(\mathrm{T})=3$ years
Simple Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{7500 \times 5 \times 3}{100}=₹ 1,125$
Now, Amount = Principal + Simple Interest $=₹ 7,500+₹ 1,125=₹ 8,625$
View full question & answer→Question 22 Marks
Find the amount to be paid at the end of $3$ years in each case:Principal $=₹ 1,200$ at $12 \%$ p.a.
AnswerHere, Principal $(P)=₹ 1,200$, Rate $(R)=12 \%$ p.a., Time $(T)=3$ years
Simple Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{1200 \times 12 \times 3}{100}=₹ 432$
Now, Amount = Principal + Simple Interest $=₹ 1200+₹ 432=₹ 1,632$
View full question & answer→Question 32 Marks
Amina buys a book for $₹ 275$ and sells it at a loss of $15 \%$. How much does she sell it for?
AnswerThe cost of a book $=₹ 275$ and loss percent $= 15\%$
Loss = Loss $\%$ of $C.P. =15 \%$ of ₹2 $275=\frac{15}{100} \times 275=₹ 41.25$
Therefore, $S.P. = C.P. - $ Loss= $₹275 -₹ 41.25=₹ 233.75$
Hence, Amina sells a book for $₹ 233.75$.
View full question & answer→Question 42 Marks
Juhi sells a washing machine for $₹ 13,500$. She loses $20 \%$ in the bargain. What was the price at which she bought it?
AnswerSelling price of washing machine $=₹ 13,500$
Loss percent $=20 \%$
Let the cost price of washing machine be $₹ x$.
Since, Loss = Loss $\%$ of $C.P. \Rightarrow $ Loss $=20 \%$ of $₹ x=\frac{20}{100} \times x=\frac{x}{5}$
Therefore, $S.P. = C.P. -$ Loss
$\Rightarrow 13500=x-\frac{x}{5}$
$\Rightarrow x=\frac{13500 \times 5}{4}=₹ 16,875$
Hence, the cost price of washing machine is $₹16,875.$
View full question & answer→Question 52 Marks
Ibuy a $T.V.$ for $₹ 10,000$ and sell it at a profit of $20 \%$. How much money do I get for it?
AnswerThe cost price of $T.V. =₹ 10,000$
Profit percent $=20 \%$
Now, Profit $=$ Profit $\%$ of $C.P. =\frac{20}{100} \times 10000=₹ 2,000$
Selling price $= C.P. +$ Profit $=₹ 10,000+₹ 2,000=₹ 12,000$
Hence, he gets $₹ 12,000$ on selling his $T.V.$
View full question & answer→Question 62 Marks
Arun bought a car for $₹ 3,50,000$. The next year, the price went upto $₹ 3,70,000$. What was the Percentage of price increase?
AnswerIncreased in price of a car from $₹ 3,50,000$ to $₹ 3,70,000$.
Amount change $=₹ 3,70,000-₹ 3,50,000=₹ 20,000$.
Therefore, increased percentage $=\frac{\text { Amount of change }}{\text { Original amount }} \times 100=\frac{20000}{350000} \times 100=5 \frac{5}{7} \%$
Hence, the percentage of price increased is $5 \frac{5}{7} \%$.
View full question & answer→Question 72 Marks
Convert each part of the ratio to percentage:$1: 2: 5$
Answer$1: 2: 5$Total part $=1+2+5=8$
Therefore, Fractional part $=\frac{1}{8}: \frac{2}{8}: \frac{5}{8}$
$\Rightarrow$ Percentage of parts $=\frac{1}{8} \times 100: \frac{2}{8} \times 100: \frac{5}{8} \times 100$
$\Rightarrow$ Percentage of parts $=12.5 \%: 25 \%: 62.5 \%$
View full question & answer→Question 82 Marks
Convert each part of the ratio to percentage:$1: 4$
Answer$1: 4$Total part $=1+4=5$
Therefore, Fractional part $=\frac{1}{5}: \frac{4}{5}$
$\Rightarrow$ Percentage of parts $=\frac{1}{5} \times 100: \frac{4}{5} \times 100$
$\Rightarrow$ Percentage of parts $=20 \%: 80 \%$
View full question & answer→Question 92 Marks
Convert each part of the ratio to percentage:$2: 3: 5$
Answer$2: 3: 5$Total part $=2+3+5=10$
Therefore, Fractional part $=\frac{2}{10}: \frac{3}{10}: \frac{5}{10}$
$\Rightarrow$ Percentage of parts $=\frac{2}{10} \times 100: \frac{3}{10} \times 100: \frac{5}{10} \times 100$
$\Rightarrow$ Percentage of parts $=20 \%: 30 \%: 50 \%$
View full question & answer→Question 102 Marks
Convert each part of the ratio to percentage : $3: 1$
Answer$3: 1$Total part $=3+1=4$
Therefore, Fractional part $=\frac{3}{4}: \frac{1}{4}$
$\Rightarrow \text { Percentage of parts }=\frac{3}{4} \times 100: \frac{1}{4} \times 100 $
$\Rightarrow \text { Percentage of parts }=75 \%: 25 \%$
View full question & answer→Question 112 Marks
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.A skirt bought for $₹ 250$ and sold at $₹ 150 .$
AnswerCost price of skirt = $₹ 250$ Selling price of skirt $=₹ 150$
Since, $\quad C.P. > S.P., \quad$ therefore here is loss.
$\therefore$ Loss $= C.P. - S.P. =₹ 250-₹ 150=₹ 100$
Now Loss $\%=\frac{\text { Loss }}{\text { C.P. }} \times 100=\frac{100}{250} \times 100=40 \%$
Therefore, Profit $=₹ 100$ and Profit $\%=40 \%$
View full question & answer→Question 122 Marks
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.A cupboard bought for $₹ 2,500$ and sold at $₹ 3,000$.
AnswerCost price of cupboard $=₹ 2,500$Selling price of cupboard $=₹ 3,000$
Since, $S.P. >C.P.$, therefore here is profit.
$\therefore$ Profit $= S.P. - C.P. =₹ 3,000-₹ 2,500=₹ 500$
Now $\quad$ Profit $\%=\frac{\text { Profit }}{\text { C.P. }} \times 100=\frac{500}{2500} \times 100=20 \%$
Therefore, Profit $=₹ 500$ and Profit $\%=20 \%$
View full question & answer→Question 132 Marks
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.A refrigerater bought for $₹ 12,000$ and sold at $₹ 13,500$.
AnswerCost price of refrigerator $=₹ 12,000$Selling price of refrigerator $=₹ 13,500$
Since,$S.P. >C.P$., therefore here is profit.
$\therefore$ Profit $= S.P. - C.P. =₹ 13500-₹ 12000=₹ 1,500$
Now Profit $\%=\frac{\text { Profit }}{\text { C.P. }} \times 100=\frac{1500}{12000} \times 100=12.5 \%$
Therefore, Profit $=₹ 1,500$ and Profit $\%=12.5 \%$
View full question & answer→Question 142 Marks
If Meena gives an interest of $₹ 45$ for one year at $9 \%$ rate p.a.. What is the sum she has borrowed?
AnswerSimple Interest $=₹ 45$, Rate $(R)=9 \%$ p.a., Time $(T)=1$ years
Simple Interest $=\frac{P \times R \times T}{100}$
$\Rightarrow 45=\frac{P \times 9 \times 1}{100} $
$\Rightarrow P=\frac{45 \times 100}{9 \times 1} $
$\Rightarrow P=₹ 500$
Hence, she borrowed $₹ 500 .$
View full question & answer→Question 152 Marks
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.Gardening shears bought for $₹ 250$ and sold for $₹ 325$.
AnswerCost price of gardening shears $=₹ 250$
Selling price of gardening shears $=₹ 325$
Since, $S.P. > C.P.$, therefore here is profit.
$\therefore \text { Profit }=\text { S.P. }- \text { C.P. }=₹ 325-₹ 250=₹ 75$
Now Profit $\%=\frac{\text { Profit }}{\text { C.P. }} \times 100=\frac{75}{250} \times 100=30 \%$
Therefore, Profit $= ₹75$ and Profit $\%=30 \%$
View full question & answer→Question 162 Marks
What rate gives $₹ 280$ as interest on a sum of $₹ 56,000$ in $2$ years?
AnswerHere, Principal $(P)=₹ 56,000$,
Simple Interest $(S.I.) = ₹ 280$, Time $(T)=2$ years
Simple Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} $
$\Rightarrow 280=\frac{56000 \times \mathrm{R} \times 2}{100}$
$\Rightarrow \mathrm{R}=\frac{280 \times 100}{56000 \times 2} $
$\Rightarrow \mathrm{R}=0.25 \%$
Hence, the rate of interest on sum is $0.25 \%$.
View full question & answer→Question 172 Marks
Meeta saves $₹ 4000$ from her salary. If this is $10 \%$ of her salary. What is her salary?
AnswerLet Meera's salary be $₹ x$.
Now, $\quad 10 \%$ of salary $=₹ 400$
$\Rightarrow 10 \% \text { of } x=₹ 400 $
$\Rightarrow \frac{10}{100} \times x=400 $
$\Rightarrow x=\frac{400 \times 100}{10} $
$\Rightarrow x=4,000$
Hence, Meera's salary is $₹ 4,000..$
View full question & answer→Question 182 Marks
Out of $15,000$ voters in a constituency, $60 \%$ voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
AnswerTotal voters $=15,000$
Percentage of voted candidates $=60 \%$
Percentage of not voted candidates $=100-60=40 \%$
Actual candidates, who did not vote $=40 \%$ of $15000=\frac{40}{100} \times 15000=6,000$
Hence, 6,000 candidates did not vote.
View full question & answer→Question 192 Marks
In a city, $30 \%$ are females, $40 \%$ are males and remaining are children. What per cent are children?
AnswerGiven: Percentage of females $=30 \%$Percentage of males $=40 \%$
Total percentage of females and males $=30+40=70 \%$
Percentage of children
$=$ Total percentage - Percentage of males and females
$=100 \%-70 \%=30 \%$
Hence, $30 \%$ are children.
View full question & answer→Question 202 Marks
If $\triangle ABC ≅ \triangle FED$ under the correspondence $ABC \leftrightarrow FED$, write all the corresponding congruent parts of the triangles.
AnswerWe know that If two triangles are congruent then corresponding parts of both the triangles are equal:
Therefore, corresponding parts of triangle $ABC$ and $FED$ are as;
$\angle A \leftrightarrow \angle F$
$ \angle B \leftrightarrow \angle E$
$ \angle C \leftrightarrow \angle D$
$AB \leftrightarrow FE$
$CA \leftrightarrow DF$
$CB \leftrightarrow DE$ View full question & answer→Question 212 Marks
A local cricket team played $20$ matches in one season. It won $25 \%$ of them. How many matches did they win?
AnswerNumber of matches played by cricket team $=20$
Percentage of won matches $=25 \%$
Total matches won by them $=25 \%$ of $20=\frac{25}{100} \times 20=5$
Hence, they won $5$ matches.
View full question & answer→Question 222 Marks
A survey of $40$ children showed that $25\%$ liked playing football. How many children liked playing football?
AnswerGiven that the total number of children is $40.$
Out of these, $25\%$ like playing football.
So out of $40$, the number of children who like playing football $= \frac{25}{100} \times 40 = 10$
View full question & answer→Question 232 Marks
What percent of the adjoining figure is shaded?
AnswerThe given figure has three parts out of which one and a half are shaded. i,e, half of the total figure is shaded.
Here, $\frac{1}{2}=\frac{1}{2} \times 100\% = 50\%$
Thus, $50 \%$ of the figure is shaded.
View full question & answer→Question 242 Marks
If Manohar pays an interest of $₹750$ for $2$ years on a sum of $₹4,500$, find the rate of interest
AnswerWe know that, $I=\frac{P \times T \times R}{100}$
Therefore, $750=\frac{4500 \times 2 \times R}{100}$
or $\frac{750}{45 \times 2}=R$
Therefore, Rate $=8 \frac{1}{3} \%$
View full question & answer→Question 252 Marks
A school team won $6$ games this year against $4$ games won last year. What is the percent increase?
AnswerHere, the increase in the number of wins (or amount of change) $= 6 – 4 = 2.$
Thus, the percentage increase = $\frac{\text { amount of change }}{\text { original amount or base }} \times 100$
$=\frac{\text { increase in the number of wins }}{\text { original number of wins }} \times 100=\frac{2}{4} \times 100=50%$
View full question & answer→