Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

To convert CGPA to percentage, all you need to do is multiply your CGPA by 9.5.

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 \(\frac { 1 }{ 10 } \) = \(\frac { 1 }{ 10 } \) × 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 = \(\frac { 15 }{ 25 } \) × 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
∴ Percentage of male teacher = \(\frac { 70 }{ 120 } \) × 100 % = \(\frac { 700 }{ 12 } \) %
Score in percentage = 58.33%
Percentage of male teachers = 58.33%

The percentage difference calculator is here to help you compare two numbers.

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 = \(\frac { 70 }{ 100 } \) × 100 % = 70 %
The won 70% of the matches

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

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 = \(\frac { 370 }{ 500 } \) × 100 % = 74 %
Number of students who cannot swim = 500 – 370 = 130
Percentage of students who cannot swim = \(\frac { 130 }{ 500 } \) × 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
∴ Percentage of money saved = \(\frac { 1 }{ 5 } \) × 100% = 20%
∴ 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 = \(\frac { 5 }{ 100 } \) × 1500 = ₹ 75
∴ Commission received = ₹ 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 = \(\frac { 18 }{ 100 } \) × 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
\(\frac { x }{ 100 } \) × 50 = 2
x = \(\frac{2 \times 100}{50}\) = 4 %
∴ 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
\(\frac { x }{ 100 } \) × 8 = 64
x = \(\frac{64 \times 100}{8}\) = 800
∴ 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 bank for 4 years.
Solution:
Principal (P) = ₹ 10,000
Rate of interest (r) = 2%
Time (n) = 4 years
∴ Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{10000 \times 4 \times 2}{100}\)
= ₹ 800
Stephen will earn ₹ 800

Question 12.
Riya 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 (P) = ₹ 15,000
Rate of interest (r) = 10 %
Simple Interest (I) = ₹ 9000
I = \(\frac { pnr }{ 100 } \)
9000 = \(\frac{15000 \times n \times 10}{100}\)
n = \(\frac{9000 \times 100}{15000 \times 10}\)
n = 6 years
∴ 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 ?4,000 at 12% per annum for 4 years?
Solution:
Let the required number of years be x
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P₁ = ₹ 3000
Rate of interest (r) = 8 %
Time (n₁) = n₁ years
Simple Interest I₁ = \(\frac{3000 \times 8 \times n_{1}}{100}\) = 240 n₁
Principal (P₂) = ₹ 4000
Rate of interest (r) = 12 %
Time n₂ = 4 years
Simple Interest I₂ = \(\frac{4000 \times 12 \times 4}{100}\)
I₂ = 1920
If I₁ = I₂
240 n₁ = 1920
n₁ = \(\frac { 1920 }{ 240 } \) = 8
∴ The required time = 8 years

Challenge Problems

Question 14.
A man travelled 80 km by car and 320 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 km.
Distance travelled by train = 320 km
Total distance = 80 + 320 km = 400 km
Percentage of distance travelled by car = \(\frac { 80 }{ 400 } \) × 100 % = 20 %
Percentage of distance travelled by train = \(\frac { 320 }{ 800 } \) × 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 = \(\frac { 35 }{ 45 } \) × 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 = \(\frac { 80 }{ 100 } \) × 8000
literate people = 6400
Percentage of women = 40 %
Number of women = 40 % of literate people
= \(\frac { 40 }{ 100 } \) × 6400 = 2560
∴ literate women : Total population
= 8000 : 2560
= 25 : 8

Question 18.
A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly?
Solution:
Total number of problems in the test = 20
Students score = 80 %
Number of problem answered = \(\frac { 80 }{ 100 } \) × 20 = 16

Question 19.
A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?
Solution:
Total weight of the metal = 8.5 kg
Percentage of silver in the metal = 85%
Weight of silver in the metal = 85% of total weight
= \(\frac { 85 }{ 100 } \) × 8.5 kg
= 7.225 kg
7.225 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 = 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 tank = 200 litres
Percentage of water in the tank = 40%
Percentage of water to fill = Upto 75%
Difference in percentage = 75 % – 40 % = 35 %
∴ Volume of water to be filled = Percentage of difference × total capacity
= \(\frac { 35 }{ 100 } \) × 200 = 70 l
70 l of water to be filled

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 22.
Which is greater 16 \(\frac { 2 }{ 3 } \) or \(\frac { 2 }{ 5 } \) or 0.17 ?
Solution:
16 \(\frac { 2 }{ 3 } \) = \(\frac { 50 }{ 30 } \)
= \(\frac { 50 }{ 30 } \) × 100 % = 1666.67 %
⇒ \(\frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \) × 100 = 40 %
0.17 = \(\frac { 17 }{ 100 } \) = 17 %
∴ 1666.67 is greater
∴ 16 \(\frac { 2 }{ 3 } \) is greater

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 one year = Worth of Machine – Depreciation
= 1,67,000 – 16,200 = 1,45,800
Depreciation of the machine for 2nd year = 145800 × 1 × \(\frac { 10 }{ 100 } \) = 14580
Worth of the machine after 2 years = 1,45,800 – 14,580 = 1,31,220
∴ 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 P = ₹ 100
If A = 6200
⇒ Principal + Interest for 2 years = 6200
A = ₹ 7400
⇒ Principal + Interest for 3 years = 7400
∴ Difference gives the Interest for 1 year
∴ Interest for 1 year = 7400 – 6200
I = 1200
\(\frac { pnr }{ 100 } \) = 1200 ⇒ \(\frac{P \times 1 \times r}{100}\) = 1200
If the Principal = 10,000 then
\(\frac{10,000 \times 1 \times r}{100}\) = 1200 ⇒ 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 P = ₹ 46900
Time n = 2 years
Amount A = ₹ 53466
Let r n be the rate of interest per year p
Intrest I = \(\frac { pnr }{ 100 } \)
A = P + I
53466 = 46900 + \(\frac{46900 \times 2 \times r}{100}\)
53466 – 46900 = \(\frac{46900 \times 2 \times r}{100}\)
6566 = 469 × 2 × r
r = \(\frac{6566}{2 \times 469}\) % = 7 %
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 P₁ = ₹ 5000
Time n₁ = 2 years
Let r be the rate of interest per year
Simple interest got from Balaji = \(\frac { pnr }{ 100 } \) ⇒ I₁ = \(\frac{5000 \times 25 \times r}{100}\)
Again principal let to Charles P₂ = ₹ 3000
Time (n₂) = 4 years
Simple interest got from Charles (I₂) = \(\frac{3000 \times 4 \times r}{100}\)
Altogether Arun got ₹ 2200 as interest.
∴ I₁ + I₂ = 2200
\(\frac{5000 \times 2 \times r}{100}+\frac{3000 \times 4 \times r}{100}\) = 2200
100r + 120r = 2200
220r = 2200 = \(\frac { 2200 }{ 220 } \)
r = 10 %
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 P = ₹ 100)
Solution:
Let the principal P = ₹ 100
Given it is doubled after 4 years
i.e. Time n = 4 years
After 4 years A = ₹ 200
∴ A = P + I
A – P = I
200 – 100 = I
After 4 years interest I = 100
I = \(\frac { pnr }{ 100 } \) ⇒ 100 = \(\frac{100 \times 4 \times r}{100}\)
4r = 100 ⇒ r = 25 %
Rate of interest r = 25 %