Exercise 2.5 - Chapter 2 Percentage and Simple Interest Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.5$
Miscellaneous Practice Problems
Question $1 .$
When Mathi was buying her flat she had to put down a deposit of $\frac{1}{10}$ of the value of the flat. What percentage was this?
Solution:
Percentage of $\frac{1}{10}=\frac{1}{10} \times 100 \%=10 \%$
Mathi has to put down a deposit of $10 \%$ of the value of the flat.

Question $2 .$
Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.
Solution:
Yazhini's score $=15$ out of $25=\frac{15}{25}$
Score in percentage $=\frac{15}{25} \times 100 \%=60 \%$

Question $3 .$
Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.
Solution:
Total teachers of the school $=120$
Number of male teachers $=70$
$\therefore$ Percentage of male teacher $=\frac{70}{120} \times 100 \%=\frac{700}{12} \%$
Score in percentage $=58.33 \%$
Percentage of male teachers $=58.33 \%$
 

Question $4 .$
A cricket team won 70 matches during a year and lost 28 matches and no results for two matches. Find the percentage of matches they won.

Solution:
Number of Matches won $=70$
Number of Matches lost $=28$
"No result" Matches $=2$
Total Matches $=70+28+2=100$
Percentage of Matches won $=\frac{70}{100} \times 100 \%=70 \%$
The won $70 \%$ of the matches
 

Question $5 .$
There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?
Solution:
Total number of students $=500$
Number of students who can swim $=370$
Percentage of students who can swim $=\frac{370}{500} \times 100 \%=74 \%$
Number of students who cannot swim $=500-370=130$
Percentage of students who cannot swim $=\frac{130}{500} \times 100 \%=26 \%$
i.e. $74 \%$ can swim and $26 \%$ cannot swim
 

Question $6 .$
The ratio of Saral's income to her savings is $4: 1$. What is the percentage of money saved by her?
Solution:
Total parts of money $=4+1=5$
Part of money saved = 1
$\therefore$ Percentage of money saved $=\frac{1}{5} \times 100 \%=20 \%$
$\therefore 20 \%$ of money is saved by Saral
 

Question $7 .$
A salesman is on a commission rate of $5 \%$. How much commission does he make on sales worth ₹
1,500 ?
Solution:
Total amount on sale $=₹ 1,500$
Commission rate $=5 \%$
Commission received $=5 \%$ of $₹ 1,500=\frac{5}{100} \times 1500=₹ 75$

Question 8 .
In the year 2015 ticket to the world cup cricket match was $₹ 1,500$. This year the price has been increased by $18 \%$. What is the price of a ticket this year?
Solution.
Price of a ticket in $2015=₹ 1500$
Increased price this year $=18 \%$ of price in 2015
$=18 \%$ of ₹ $1500=\frac{18}{100} \times 1500$
$=₹ 270$
Price of ticket this year $=$ last year price $+$ increased price
$=₹ 1500+₹ 270=₹ 1770$
Price of ticket this year $=₹ 1770$
 

Question $9 .$
2 is what percentage of 50 ?
Solution:
Let the required percentage be $x$
$x \%$ of $50=2$
$\frac{x}{100} \times 50=2$
$x=\frac{2 \times 100}{50}=4 \%$
$\therefore 4 \%$ of 50 is 2

Question $10 .$
What percentage of 8 is 64 ?
Solution:
Let the required percentage be $x$
So $x \%$ of $8=64$
$\frac{x}{100} \times 8=64$
$x=\frac{64 \times 100}{8}=800$
$\therefore 800 \%$ of 8 is 64
 

Question $11 .$
Stephen invested ₹ 10,000 in a savings bank account that earned $2 \%$ simple interest. Find the interest earned if the amount was kept in the hank for 4 years.
Solution:
Principal $(\mathrm{P})=₹ 10,000$
Rate of interest $(\mathrm{r})=2 \%$
Time $(n)=4$ years
$\therefore$ Simple Interest $I=\frac{p n r}{100}$
$=\frac{10000 \times 4 \times 2}{100}$
$=₹ 800$
Stephen will earn ₹ 800
 

Question 12.
Kiya bought $₹ 15,000$ from a bank to buy a car at $10 \%$ simple interest. If she paid $₹ 9,000$ as interest while clearing the loan, find the time for which the loan was given.
Solution:

Here Principal $(\mathrm{P})=₹ 15,000$
Rate of interest $(r)=10 \%$
Simple Interest (I) = ₹ 9000
$\begin{aligned}
&I=\frac{p m r}{100} \\
&9000=\frac{15000 \times n \times 10}{100} \\
&\mathrm{n}=\frac{9000 \times 100}{15000 \times 10} \\
&\mathrm{n}=6 \text { years }
\end{aligned}$
$\therefore$ The loan was given for 6 years
 

Question $13 .$
In how much time will the simple interest on $₹ 3,000$ at the rate of $8 \%$ per annum be the same as simple interest on 24,000 at $12 \%$ per annum for 4 years?
Solution:
Let the required number of years be $\mathrm{x}$
Simple Interest I = $\frac{p n r}{100}$
Principal $\mathrm{P}_{1}=₹ 3000$
Rate of interest (r) = $8 \%$
Time $\left(\mathrm{n}_{1}\right)=\mathrm{n}_{1}$ years
Simple Interest $I_{1}=\frac{3000 \times 8 \times n_{1}}{100}=240 \mathrm{n}_{1}$
Principal $\left(\mathrm{P}_{2}\right)=₹ 4000$
Rate of interest (r) $=12 \%$
Time $\mathrm{n}_{2}=4$ years
Simple Interest $\mathrm{I}_{2}=\frac{4000 \times 12 \times 4}{100}$
$\mathrm{I}_{2}=1920$
If $\mathrm{I}_{1}=\mathrm{I}_{2}$
$240 \mathrm{n}_{1}=1920$
$\mathrm{n}_{1}=\frac{1920}{240}=8$
$\therefore$ The required time $=8$ years
 

Challenge Problems
Question 14.
A man travelled $80 \mathrm{~km}$ by car and $320 \mathrm{~km}$ by train to reach his destination. Find what percent of total journey did he travel by car and what per cent by train?
Solution:
Distance travelled by car $=80 \mathrm{~km}$.
Distance travelled by train $=320 \mathrm{~km}$
Total distance $=80+320 \mathrm{~km}=400 \mathrm{~km}$
Percentage of distance travelled by car $=\frac{80}{400} \times 100 \%=20 \%$
Percentage of distance travelled by train $=\frac{320}{800} \times 100 \%=40 \%$
 

Question $15 .$
Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?
Solution:
Number of correct answers $=35$
Number of incorrect answers $=10$
Total number of answers $=35+10=45$
Percentage of correct answers $=\frac{35}{45} \times 100 \%$
$=77.777 \%=77.78 \%$
 

Question $17 .$
The population of a village is 8000 . Out of these, $80 \%$ are literate and of these literate people, $40 \%$ are women. Find the percentage of literate women to the total population?
Solution:
Population of the village $=8000$ people
literate people $=80 \%$ of population
$=80 \%$ of $8000=\frac{80}{100} \times 8000$
literate people $=6400$
Percentage of women $=40 \%$
Number of women $=40 \%$ of literate people
$=\frac{40}{100} \times 6400=2560$
$\therefore$ literate women : Total population
$=8000: 2560$
$=25: 8$

Question $18 .$
A student earned a grade of $80 \%$ un a math test that had 20 problems. Huw many problens on this test did the student answer correctly?

Solution:

Total number of problems in the test $=20$
Students score $=80 \%$
Number of problem answered $=\frac{80}{100} \times 20=16$
 

Question $19 .$
A metal bar weighs $8.5 \mathrm{~kg} .85 \%$ of the bar is silver. How many kilograms of silver are in the bar? Solution:
Total weight of the metal $=8.5 \mathrm{~kg}$
Percentage of silver in the metal $=85 \%$
Weight of silver in the metal $=85 \%$ of total weight
$\begin{aligned}
&=\frac{85}{150} \times 8.5 \mathrm{~kg} \\
&=7.225 \mathrm{~kg}
\end{aligned}$
$7.225 \mathrm{~kg}$ of silver are in the bar.
 

Question $20 .$
Concession card holders pay $₹ 120$ for a train ticket. Full fare is $₹ 230$. What is the percentage of discount for concession card holders?
Solution:
Train ticket fare $=₹ 230$
Ticket fare on concession $=₹ 120$
Discount $=$ Ticket fare $-$ concession fare $=230-120=₹ 110$
Percentage of discount $=\frac{\text { Discount }}{\text { Original rate }} \times 100 \%=\frac{110}{230} \times 100=47.826 \%=47.83 \%$

Question 21.
A tank can hold 200 litres of water. At present, it is only $40 \%$ full. How many litres of water to fill in the tank, so that it is $75 \%$ full?
Solution:
Capacity of the water $\operatorname{tank}=200$ litres
Percentage of water in the tank $=40 \%$
Percentage of water to fill $=$ Upto $75 \%$
Difference in percentage $=75 \%-40 \%=35 \%$
$\therefore$ Volume of water to be filled $=$ Percentage of difference $\times$ total capacity
$=\frac{35}{100} \times 200=701$
701 of water to be filled
 

Question $22 .$
Which is greater $16 \frac{2}{3}$ or $\frac{2}{5}$ or $0.17$ ?
Solution:
$\begin{aligned}
&16 \frac{2}{3}=\frac{50}{30} \\
&=\frac{50}{30} \times 100 \%=1666.67 \% \\
&\Rightarrow \frac{2}{5} \\
&=\frac{2}{5} \times 100=40 \% \\
&0.17=\frac{17}{100}=17 \% \\
&\therefore 1666.67 \text { is greater } \\
&\therefore 16 \frac{2}{3} \text { is greater }
\end{aligned}$

Question 23 .
The value of a machine depreciates at $10 \%$ per year. If the present value is $₹ 1,62,000$, what is the worth of the machine after two years.
Solution:
Present value of the machine $=₹ 1,67,000$
Rate of depreciation $=10 \%$ Per annum
Time $(n)=2$ years
For 1 year depreciation amount $=\frac{1,62,000 \times 1 \times 10}{100}=₹ 16,200$
Worth of the machine after ane year = Warth of Machine - Depreciation $=1,67,000-16,200=1,45,800$
Depreciation of the machine for 2 nd year $=145800 \times 1 \times \frac{10}{100}=14580$
Worth of the machine after 2 years $=1,45,800-14,580=1,31,220$
$\therefore$ Worth of the machine after 2 years $=₹ 1,31,220$

Question $24 .$
In simple interest, a sum of money amounts to ₹ 6,200 in 2 years and ₹ 6,800 in 3 years. Find the principal and rate of interest.
Solution:
Let the principal $\mathrm{P}=₹ 100$
If $\mathrm{A}=6200$
$\Rightarrow$ Principal + Interest for 2 years $=6200$
$\mathrm{A}=₹ 7400$
$\Rightarrow$ Principal + Interest for 3 years $=7400$
$\therefore$ Difference gives the Interest for 1 year
$\therefore$ Interest for l year $=7400-6200$
$\begin{aligned}
&I=1200 \\
&\frac{p n r}{100}=1200 \Rightarrow \frac{P \times 1 \times r}{100}=1200
\end{aligned}$
If the Principal $=10,000$ then

$\frac{10,000 \times 1 \times r}{100}=1200 \Rightarrow \mathrm{r}=12 \%$
Rate of interest $=12 \%$ Per month
 

Question $25 .$
A sum of ₹ 46,900 was lent out at simple interest and at the end of 2 years, the total amount was ₹ 53,466 . Find the rate of interest per year.
Solution:
Here principal $\mathrm{P}=₹ 46900$
Time $n=2$ years
Amount A = ₹ 53466
Let $\mathrm{r} \mathrm{n}$ be the rate of interest per year $\mathrm{p}$
Intrest $\mathrm{I}=\frac{p n r}{100}$
$\begin{aligned}
&\mathrm{A}=\mathrm{P}+\mathrm{I} \\
&53466=46900+\frac{46900 \times 2 \times r}{100} \\
&53466-46900=\frac{46900 \times 2 \times r}{100} \\
&6566=469 \times 2 \times \mathrm{r} \\
&\mathrm{r}=\frac{6566}{2 \times 469} \%=7 \%
\end{aligned}$
Rate of interest $=7 \%$ Per Year
 

Question 26 .
Arun lent $₹ 5,000$ to Balaji for 2 years and $₹ 3,000$ to Charles for 4 years on simple interest at the same rate of interest and received $₹ 2,200$ in all from both of them as interest. Find the rate of interest per year.
Solution:
Principal lent to Balaji $\mathrm{P}_{1}=₹ 5000$
Time $\mathrm{n}_{1}=2$ years
Let r be the rate of interest per year
Simple interest got from Balaji $=\frac{p n r}{100} \Rightarrow I_{1}=\frac{5000 \times 25 \times r}{100}$
Again principal let to Charles $\mathrm{P}_{2}=₹ 3000$
Time $\left(n_{2}\right)=4$ years
Simple interest got from Charles $\left(\mathrm{I}_{2}\right)=\frac{3900 \times 4 \times \mathrm{r}}{100}$
Altogether Arun got $₹ 2200$ as interest.
$\therefore \mathrm{I}_{1}+\mathrm{I}_{2}=2200$
$\frac{\frac{5000 \times 2 \times r}{100}+\frac{3000 \times 4 \times r}{100}}{100 \mathrm{r}+120 \mathrm{r}=2200}=2200$

$\begin{aligned}
&220 \mathrm{r}=2200=\frac{2200}{220} \\
&\mathrm{r}=10 \%
\end{aligned}$
Rate of interest per year $=10 \%$
 

Question $27 .$

If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let $\mathrm{P}=₹$ 100)
Solution:
Let the principal $\mathrm{P}=₹ 100$
Given it is doubled after 4 years
i.e. Time $\mathrm{n}=4$ years
After 4 years $A=₹ 200$
$200-100=I$ After 4 years interest $I=100$
$\begin{aligned}
&\therefore \mathrm{A}=\mathrm{P}+\mathrm{I} \\
&\mathrm{A}-\mathrm{P}=\mathrm{I}
\end{aligned}$
$\begin{aligned}
&\mathrm{I}=\frac{p n r}{100} \Rightarrow 100=\frac{100 \times 4 \times r}{100} \\
&4 \mathrm{r}=100 \Rightarrow \mathrm{r}=25 \%
\end{aligned}$
Rate of interest $\mathrm{r}=25 \%$