Class 7

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.4 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 2.4

Question 1.
Find the simple interest on ₹ 35,000 at 9% per annum for 2 years?
Solution:

Principal P = ₹ 35,000
Rate of interest r = 9 % Per annum
Time (n) = 2 years
Simple Interest I = \(\frac { Pnr }{ 100 } \) = \(\frac{35000 \times 2 \times 9}{100}\) = ₹ 6300
Simple intrest I = ₹ 6300

Question 2.
Aravind borrowed a sum of ₹ 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years.
Solution:
Here Principal P = ₹ 8,000
Rate of interest r = 7% Per annum
Time (n) = 2 Years
Simple Interest (I) = \(\frac { Pnr }{ 100 } \) = \(\frac{8000 \times 2 \times 7}{100}\)
I = ₹ 1120
Amount = P + I
I = ₹ 8000 + 1120 = 9120
Interest to be paid = ₹ 1,120
Amount to be paid = ₹ 9,120

Question 3.
Sheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is ₹ 21,280. Find the sum.
Solution:
Let the Principal be ₹ P
Rate of interest r = 9.5% per annum
Time (n) = 4 years
Simple Interest I = \(\frac { Pnr }{ 100 } \)
Given I = ₹ 21,280

∴ Sum of money Sheela bought = ₹ 56,000

Question 4.
Basha borrowed ₹ 8,500 from a bank at a particular rate of simple interest. After 3 years, he paid ₹ 11,050 to settle his debt. At what rate of interest he borrowed the money?
Solution:
Let the rate of interest be r% per annum
Here Principal P = ₹ 8,500
Time n = 3 years
Total amount paid = ₹ 11,050
A = P + 1 = ₹ 11,050
i.e. 8,500 + 1 = ₹ 11,050
I = ₹ 11,050 – ₹ 8,500 = ₹ 2,550

Question 5.
In What time will ₹ 16,500 amount to ₹ 22,935 at 13% per annum?
Solution:
Rate of interest r = 13% per annum
Here Amount A = ₹ 22,935
Principal P = ₹ 16,500
A = P + I
22935 = 16,500 + I
∴ Interest I = 22935 – 16,500 = ₹ 6,435
Simple Interest I = \(\frac { pnr }{ 100 } \)

6435 = \(\frac{16500 \times n \times 13}{100}\)
n =\(\frac{6435 \times 100}{16500 \times 13}\)
n = 3 years
Required time n = 3 years

Question 6.
In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?
Solution:
Let the require time be n years
Here Principal P = ₹ 17,800
Rate of interest r = 6% per annum
Amount A = ₹ 19,936
A = P + I
19936 = 17800 + 1
19936 – 17800 = I
2136 = I
Simple Interest (I) = \(\frac { pnr }{ 100 } \)
2136 = \(\frac{17800 \times n \times 6}{100}\)
n = \(\frac{2136 \times 100}{17800 \times 6}\)

n = 2 Years
Required time = 2 years

Question 7.
A sum of ₹ 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was ₹ 55,560. Find the rate of interest per year.
Solution:
Given Principal P = ₹ 48,000
Time n = 2 years 3 months
= 2 + \(\frac { 3 }{ 12 } \) years = 2 + \(\frac { 1 }{ 4 } \) years
= \(\frac { 8 }{ 4 } \) + \(\frac { 1 }{ 4 } \) years = \(\frac { 9 }{ 4 } \) years
Amount A = ₹ 55,660
A = p + 1
55660 = 48000 + I
I = 55660 – 48000 = ₹ 7660
∴ Interest for \(\frac { 9 }{ 4 } \) years = ₹ 7660
Simple intrest = \(\frac { pnr }{ 100 } \)
7660 = 48000 × \(\frac { 9 }{ 4 } \) × \(\frac { r }{ 100 } \)
r = \(\frac{7660 \times 4 \times 100}{9 \times 48000}\) = 7.09 % = 7 %
Rate of interest = 7 % Per annum

Question 8.
A principal becomes ₹ 17,000 at the rate of 12% in 3 years. Find the principal.
Solution:
Given the Principal becomes ₹ 17,000
Let the principle initially be P
Rate of Interest r Time = 12 % Per annum
Time n = 3 years
According to the problem given I = 17000 – P = \(\frac{P \times 3 \times 12}{100}\)
17000 = \(\frac { 36 }{ 100 } \) p + p
17000 = p(\(\frac { 36 }{ 100 } \) + 1)
17000 = p(\(\frac { 136 }{ 100 } \))
p = \(\frac{17000 \times 100}{136}\) = 12,500
∴ Principal P = ₹ 12,500

Objective Type Questions

Question 9.
The interest for a principle of? 4,500 which gives an amount of? 5,000 at end of certain period is
(i) ₹ 500
(ii) ₹ 200
(iii) 20%
(iv) 15%
Hint: Interest = Amount – Principle = ₹ 5000 – ₹ 4500 = ₹ 500
Answer:
(i) ₹ 500

Question 10.
Which among the following is the simple interest for the principle of ₹ 1,000 for one year at the rate of 10% interest per annum?
(i) ₹ 200
(ii) ₹ 10
(iii) ₹ 100
(iv) ₹ 1,000
Hint: Intrest = \(\frac { pnr }{ 100 } \) = \(\frac{1000 \times 1 \times 10}{100}\) = ₹ 100
Answer:
(iii) ₹ 100

Question 11.
Which among the following rate of interest yields an interest of ₹ 200 for the principle of ₹ 2,000 for one year.
(i) 10%
(ii) 20%
(iii) 5%
(iv) 15%
Hint: r = \(\frac{I \times 100}{P \times n}\) = \(\frac{200 \times 100}{2000 \times 1}\) = 10 %
Answer:
(i) 10%

Students can Download Maths Chapter 2 Percentage and Simple Interest Additional 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 Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
72% of 25 students are good at science. How many are not good at science?
Solution:
Number of students who are good at science
= 72% of 25 = \(\frac { 72 }{ 100 } \) × 25 = 18 students
∴ Number of students who are not good at science
= 25 – 18 = 7 students

Question 2.
A flower garden has 1000 plants. 5% of the plants are roses and 1% are daisy plants. What is the total number of other plants.
Solution:
Total plants = 1000
Number of rose plants = 5 % of 1000 = \(\frac { 5 }{ 100 } \) × 1000 = 50
Number of Daisy plants = 1 % of 1000 = \(\frac { 1 }{ 100 } \) × 1000 = 10
Total of rose and daisy = 50 + 10 = 60
Number of other plants = 1000 – 60 = 940

Question 3.
Find 135 % of 80 ₹.
Solution:
135 % of 80 = \(\frac { 135 }{ 100 } \) × 80 = ₹ 108

Exercise 2.2

Question 1.
Neka bought 72.3m of cloth from a role of 100m. Express the cloth bought in terms of percentage.
Solution:
Total length of the cloth = 100 m
Length of cloth bought = 72.3 m
Percentage of cloth bought = \(\frac { 72.3 }{ 100 } \) = 72.3 %

Question 2.
Convert (i) 88 % (ii) 1.86 % into decimals.
Solution:
(i) 88 % = \(\frac { 88 }{ 100 } \) = 0.88
(ii) 1.86 % = \(\frac { 1.86 }{ 100 } \) = 0.0186

Question 3.
Convert (i) 3.35 (ii) 0.5 into percentage.
Solution:
(i) 3.35 = \(\frac { 335 }{ 100 } \) × 100 % = 335 %
(ii) 0.5 = \(\frac { 5 }{ 10 } \) × 100 % = 50 %

Exercise 2.3

Question 1.
If Gayathri had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Suppose Gayathri had ₹ X in the beginning.
Then money Spend = 75 % of X = \(\frac { 75 }{ 100 } \) X = \(\frac { 3X }{ 4 } \)
Money left with her = X – \(\frac { 3X }{ 4 } \) = \(\frac { 4X-3X }{ 4 } \) = \(\frac { X }{ 4 } \)
But it is given that money left = ₹ 600
i.e. \(\frac { X }{ 4 } \) = 600
X = 600 × 4 = 2400
∴ Gayathri had ₹ 2,400

Question 2.
Mohan gets 98 marks in her exams. This amounts to 56% of the total marks, What are the maximum marks?
Solution:
Let the maximum marks be X. 56 % of X = 98
\(\frac { 56 }{ 100 } \) × (X) = 98
⇒ X = 98 × \(\frac { 100 }{ 56 } \)
X = 175
∴ Maximum marks = 175

Exercise 2.4

Question 1.
On what sum of money lent out at 9% per annum for 6 years does the simple interest amount to ₹ 810?
Solution:
Given Simple Interest I = ₹ 810
Let the sum of money (Principal) be P
Rate of interest r = 9 % Per annum.
Time n = 6 years
I = \(\frac { pnr }{ 100 } \)
810 = \(\frac{P \times 6 \times 9}{100}\)
P = \(\frac{810 \times 100}{6 \times 9}\)
P = ₹ 1500
Sum of money required = ₹ 1500

Question 2.
Find the simple interest on ₹ 1120 for 2 \(\frac { 2 }{ 5 } \) years at the rate of 5% per annum.
Solution:
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P = ₹ 1120
Time n = 2 \(\frac { 2 }{ 5 } \) years
= \(\frac { 12 }{ 5 } \) years
Rate of Interest r = 5 %

∴ I = 1120 × \(\frac { 12 }{ 5 } \) × \(\frac { 5 }{ 100 } \)
= \(\frac { 672 }{ 5 } \)
= ₹ 134.4
Simple interest = = ₹ 134.4

Students can Download Maths Chapter 2 Percentage and Simple Interest Intext 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 Intext Questions

Exercise 2.1
Try These (Text book Page No. 28)

Question 1.
Find the percentage of children whose scores fall in different categories given in table below.

Solution:

Try These (Text book Page No. 29)

Question 1.
There are 50 students in class VII of a school. The number of students involved in these activities are :
Scout: 7
Red Ribbon Club : 6
Junior Red Cross : 9
Green Force : 3
Sports : 14
Cultural activity : 11
Find the percentage of students who involved in various activities.
Solution:

Try These (Text book Page No. 30)

Question 1.
Convert the fractions as percentage.
(i) \(\frac { 1 }{ 20 } \)
(ii) \(\frac { 13 }{ 25 } \)
(iii) (i) \(\frac { 45 }{ 50 } \)
(iv) \(\frac { 18 }{ 5 } \)
(v) \(\frac { 27 }{ 10 } \)
(vi) \(\frac { 72 }{ 90 } \)
Solution:
(i) \(\frac { 1 }{ 20 } \)
= \(\frac { 1 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 20 } \) × 100 %
= 5 %

(ii) \(\frac { 13 }{ 25 } \)
= \(\frac { 13 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 13 }{ 25 } \) × 100 %
= 52 %

(iii) \(\frac { 45 }{ 50 } \)
= \(\frac { 45 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 45 }{ 50 } \) × 100 %
= 90 %

(iv) \(\frac { 18 }{ 5 } \)
= \(\frac { 18 }{ 5 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 18 }{ 50 } \) × 100 %
= 360 %

(iv) \(\frac { 27 }{ 10 } \)
= \(\frac { 27 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 27 }{ 10 } \) × 100 %
= 270 %

(iv) \(\frac { 27 }{ 10 } \)
= \(\frac { 27 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 27 }{ 10 } \) × 100 %
= 270 %

(vi) \(\frac { 72 }{ 90 } \)
= \(\frac { 72 }{ 90 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 72 }{ 90 } \) × 100 %
= 80 %

Question 2.
Convert the following percentage as fractions.
(i) 50%
(ii) 75%
(iii) 250%
(iv) 30 \(\frac { 1 }{ 5 } \) %
(v) \(\frac { 7 }{ 20 } \) %
(vi) 90 %
Solution:
(i) 50 %
= \(\frac { 50 }{ 100 } \)
= \(\frac { 5 }{ 10 } \)
= \(\frac { 1 }{ 2 } \)

(ii) 75 %
= \(\frac { 75 }{ 100 } \)
= \(\frac { 3 }{ 4 } \)

(iii) 250 %
= \(\frac { 250 }{ 100 } \)
= \(\frac { 25 }{ 10 } \)
= \(\frac { 5 }{ 2 } \)

(iv) 30 \(\frac { 1 }{ 5 } \) %
= \(\frac{30 \frac{1}{5}}{100}=\frac{\left(\frac{151}{5}\right)}{100}\)
= \(\frac { 151 }{ 500 } \)

(v) \(\frac { 7 }{ 20 } \) %
= \(\frac{\frac{7}{20}}{100}=\frac{7}{20 \times 100}\)
= \(\frac { 7 }{ 2000 } \)

(vi) 90 % = \(\frac { 90 }{ 100 } \) = \(\frac { 9 }{ 10 } \)

Think (Text book Page No. 32)

Question 1.
What is the difference between 0.01 and 1%.
Solution:
0.01 = \(\frac { 1 }{ 100 } \) = 1%
0.01 and 1% are the same.

Question 2.
In a readymade shop there will be a board showing upto 50% off. Most of the people will realize that everything is half of its original price, Is that true?
Solution:
No. Only some of them are half of its original price.

Exercise 2.2
Try These (Text book Page No. 33)

Question 1.
Convert these decimals to percentage.
(i) 0.25
(ii) 0.07
(iii) 0.8
(iv) 0.375
(v) 3.75
Solution:
(i) 0.25
= \(\frac { 25 }{ 100 } \) = 25 %

(ii) 0.07
= \(\frac { 7 }{ 100 } \) = 7 %

(iii) 0.8
= \(\frac { 80 }{ 100 } \) = 80 %

(iv) 0.375
= \(\frac { 375 }{ 1000 } \)
= \(\frac { 375 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= 37.5 %

(v) 3.75
= \(\frac { 375 }{ 100 } \) = 375 %

Try These (Text book Page No. 34)

Question 1.
Write these percentage as decimals
(i) 3 %
(ii) 25 %
(iii) 80 %
(iv) 67 %
(v) 17.5 %
(vi) 135 %
(vii) 0.5 %
Solution:
(i) 3 %
= \(\frac { 3 }{ 100 } \) = 0.03

(ii) 25 %
= \(\frac { 25 }{ 100 } \) = 0.25

(iii) 80 %
= \(\frac { 80 }{ 100 } \) = 0.8

(iv) 67 %
= \(\frac { 67 }{ 100 } \) = 0.67

(v) 17.5 %
= \(\frac { 17.5 }{ 100 } \) = 0.175

(vi) 135 %
= \(\frac { 135 }{ 100 } \) = 1.35

(vii) 0.5 %
= \(\frac { 0.5 }{ 100 } \) = 0.005

Exercise 2.3
Try These (Text book Page No. 38)

Question 1.
Level of water in a tank is increased from 35 litres to 50 litres in 2 minutes, what is the Percentage of increase?
Solution:
Level of water in the tank originally = 35 litres.
Increase in the water level = amount of change = 50 – 35 = 15 litres

Exercise 2.4
Try These (Text book Page No. 41)

Question 1.
Arjun borrowed a sum of ₹ 5,000 from a bank at 5% per annum. Find the interest and amount to be paid at the end of three year.
Solution:
Here Principal (P) = ₹ 5,000
Rate of interest (r) = 5 % Per annum
Time (n) = 3 years
Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{5000 \times 3 \times 5}{100}\)
= ₹ 750
Amount to be paid A = P + I = ₹ 5,000 + ₹ 750 = ₹ 5,750
I = ₹ 750 ; A = ₹ 5,750

Question 2.
Shanti borrowed ₹ 6,000/- from a Bank for 7 years at 12 % per annum. What amount will clear off her debt?
Solution:
Here principal (P) = ₹ 6,000
Rate of Interest (r) = 12 % Per annum
Time (n) = 4 Years
Simple Interest (I) = \(\frac { pnr }{ 100 } \) =
= \(\frac{6000 \times 7 \times 12}{100}\)
I = ₹ 5,040
Amount to be paid A = P + I = 6,000 + 5,040 = ₹ 11,040

Think (Text book Page No. 43)

Question 1.
In simple interest, a sum of money doubles itself in 10 years. In how many years it will get triple itself.
Solution:
Let the Principal be P and Rate of interest be r % per annum.
Here the number of years n = 10 years
Given in 10 years P becomes 2 P.
A = P + I
After 2 years A = 2P
i.e. 2P = P + I
2P – P = I

Now if the amount becomes triple then A = P + I = 3P
3P = P + I
3P – P = I
2P = I

∴ After 20 years the amount get tripled.

Students can Download Maths Chapter 3 Algebra Ex 3.2 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 3 Algebra Ex 3.2

Question 1.
Given that x > y. Fill in the blanks with suitable inequality signs.
(i) y [ ] x
(ii) x+ 6 [ ] y + 6
(iii) x² [ ] xy
(iv) -xy [ ] – y²
(v) x – y [ ] 0
Answer:
(i) y [<] x
(ii) x+ 6 [>] y + 6
(iii) x² [>] xy
(iv) -xy [<] – y²
(v) x – y [>] 0

Question 2.
Say True or False.

(i) Linear inequation has almost one solution.
Answer:
False

(ii) When x is an integer, the solution set for x < 0 are -1, -2,..
Answer:
False

(iii) An inequation, -3 < x < -1, where x is an integer, cannot be represented in the number line.
Answer:
True

(iv) x < -y can be rewritten as – y < x
Ans :
False

Question 3.
Solve the following inequations.
(i) x < 7, where x is a natural number.
(ii) x – 6 < 1, where x is a natural number.
(iii) 2a + 3 < 13, where a is a whole number.
(iv) 6x – 7 > 35, where x is an integer.
(v) 4x – 9 > -33, where x is a negative integer.
Solution:
(i) x < 7, where x is a natural number.
Since the solution belongs to the set of natural numbers, that are less than or equal to 7, we take the values of x as 1,2, 3, 4, 5, 6 and 7.

(ii) x – 6 < 1, where x is a natural number.
x – 6 < 1 Adding 6 on the both the sides x – 6 + 6 < 1 + 6
x < 7
Since the solutions belongs to the set of natural numbers that are less than 7, we take the values of x as 1,2, 3, 4, 5 and 6

(iii) 2a + 3 < 13, where a is a whole number.
2a + 3 < 13
Subtracting 3 from both the sides 2a + 3 – 3 < 13 – 3
2a < 10
Dividing both the side by 2. \(\frac { 2a }{ 2 } \) < \(\frac { 10 }{ 2 } \)
a < 5
Since the solutions belongs to the set of whole numbers that are less than or equal to 5 we take the values of a as 0, 1, 2, 3, 4 and 5

(iv) 6x – 7 > 35, where x is an integer.
6x – 7 > 35 Adding 7 on both the sides
6x – 7 + 7 > 35 + 7
6x > 42
Dividing both the sides by 6 we get \(\frac { 6x }{ 6 } \) > \(\frac { 42 }{ 6 } \)
x > 7
Since the solution belongs to the set of integers that are greater than or equal to 7, we take the values of x as 7, 8, 9, 10…

(v) 4x – 9 > -33, where x is a negative integer.
4x – 9 > – 33 + 9 Adding 9 both the sides
4x – 9 + 9 > -33 + 9
4x > – 24
Dividing both the sides by 4
\(\frac { 4x }{ 4 } \) > \(\frac { -24 }{ 4 } \)
x > -6
Since the solution belongs to a negative integer that are greater than -6, we take values of u as -5, -4, -3, -2 and -1

Question 4.
Solve the following inequations and represent the solution on the number line:
(i) k > -5, k is an integer.
(ii) -7 < y, y is a negative integer.
(iii) -4 < x < 8, x is a natural number.
(iv) 3m – 5 < 2m + 1, m is an integer.
Solution:
(i) k > -5, k is an integer.
Since the solution belongs to the set of integers, the solution is -4, -3, -2, -1, 0,… It’s graph on number line is shown below.

(ii) -7 < y, y is a negative integer.
-7 < y
Since the solution set belongs to the set of negative integers, the solution is
-7, -6, -5, -4, -3, -2, -1.
Its graph on the number line is shown below

(iii) -4 < x < 8, x is a natural number.
-4 < x < 8
Since the solution belongs to the set of natural numbers, the solution is
1, 2, 3, 4, 5, 6, 7 and 8.
Its graph on number line is shown below

(iv) 3m – 5 < 2m + 1, m is an integer.
3m – 5 < 2m + 1
Subtracting 1 on both the sides
3m – 5 – 1 < 2m + 1 + 1
3m – 6 < 2m
Subtracting 2m on both the sides 3m- 6 – 2m < 2m -2m
m – 6 < 0
Adding 6 on both the sides m – 6 + 6 < 0 + 6
m < 6
Since the solution belongs to the set of integers, the solution is
6, 5, 4, 3, 2, 1, 0,-1,…
Its graph on number line is shown below

Question 5.
An artist can spend any amount between ₹ 80 to ₹ 200 on brushes. If cost of each brush is ₹ 5 and there are 6 brushes in each packet, then how many packets of brush can the artist buy?
Solution:
Given the artist can spend any amount between ₹ 80 to ₹ 200
Let the number of packets of brush he can buy be x
Given cost of 1 brush = ₹ 5
Cost of 1 packet brush (6 brushes) = ₹ 5 × 6 = ₹ 30
∴ Cost of x packets of brushes = 30 x
∴ The inequation becomes 80 < 30x < 200
Dividing throughout by 30 we get \(\frac { 80 }{ 30 } \) < \(\frac { 30x }{ 30 } \) < \(\frac { 200 }{ 30 } \)
\(\frac { 8 }{ 3 } \) < x < \(\frac { 20 }{ 3 } \) ;
2 \(\frac { 2 }{ 3 } \) < x < 6 \(\frac { 2 }{ 3 } \)
brush packets cannot get in fractions.
∴ The artist can buy 3 < x < 6 packets of brushes,
or x = 3, 4, 5 and 6 packets of brushes.

Objective Type Questions

Question 1.
The solutions set of the inequation 3 < p < 6 are (where p is a natural number)
(i) 4,5 and 6
(ii) 3,4 and 5
(iii) 4 and 5
(iv) 3,4,5 and 6
Answer:
(iv) 3,4,5 and 6

Question 2.
The solution of the inequation 5x + 5 < 15 are (where x is a natural number)
(i) 1 and 2
(ii) 0,1 and 2
(iii) 2, 1,0, -1,-2
(iv) 1, 2, 3..
Answer:
(i) 1 and 2
Hint: 5x + 5 < 15
5x < 15 – 5 = 10
x < \(\frac { 10 }{ 5 } \) = 2

Question 3.
The cost of one pen is ₹ 8 and it is available in a sealed pack of 10 pens. If Swetha has only ₹ 500, how many packs of pens can she buy at the maximum?
(i) 10
(ii) 5
(iii) 6
(iv) 8
Answer:
(iii) 6
Hint:
Price of 1 pen = ₹ 8
Price of 1 pack = 10 × 8 = 80
Number of packs Swetha can buy = x
80x < 500
8x < 50
x < \(\frac { 50 }{ 8 } \) = 6.25
x is a natural number x = 1, 2, 3, 4, 5, 6

Question 4.
The inequation that is represented on the number line as shown below is _______.

(i) -4 < x < 0
(ii) -4 < x < 0
(iii) -4 < x < 0
(iv) -4 < x < 0
(v) -4 < x < 2
Answer:
(v) -4 < x < 2

Students can Download Maths Chapter 1 Number System Ex 1.3 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 1 Number System Ex 1.3

Question 1.
Find the product of the following
(i) 0.5 × 3
(ii) 3.75 × 6
(iii) 50.2 × 4
(iv) 0.03 × 9
(v) 453.03 × 7
(vi) 4 × 0.7
Solution:
(i) 0.5 × 3
5 × 3 = 15
∴ 0.5 × 3 = 1.5

(ii) 3.75 × 6

375 × 6 = 2250
3.75 × 6 = 22.50

(iii) 50.2 × 4

502 × 4 = 2008
50.2 × 4 = 200.8

(iv) 0.03 × 9
3 × 9 = 27
0.03 × 9 = 0.27

(v) 453.03 × 7

45303 × 7 = 317121
453.03 × 7 = 3171.21

(vi) 4 × 0.7
4 × 7 = 28
4 × 0.7 = 2.8

Question 2.
Find the area of the parallelogram whose base is 6.8 cm and height is 3.5 cm.
Solution:

Base of the parallelogram b = 6.8 cm
Height of the parallelogram h = 3.5 cm
Area of the parallelogram A = b × h sq.units = 6.8 × 3.5 cm²
Area of the parallelogram = 23.80 cm²

Question 3.
Find the area of the rectangle whose length is 23.7 cm and breadth is 15.2 cm.
Solution:

Length of the rectangle l = 23.7 cm
Breadth of the rectangle b= 15.2 cm
Area of the rectangle A = l × b sq.units
= 23.7 × 15.2 cm²
Area of the rectangle = 360.24 cm²

Question 4.
Multiply the following

1. 2.57 × 10
2. 0.51 × 10
3. 125.367 × 100
4. 34.51 × 100
5. 62.735 × 100
6. 0.7 × 10
7. 0.03 × 100
8. 0.4 × 1000

Solution:

1. 2.57 × 10 = 25.7
2. 0.51 × 10 = 5.1
3. 125.367 × 100 = 12536.7
4. 34.51 × 100 = 3451
5. 62.735 × 100 = 6273.5
6. 0.7 × 10 = 7.0
7. 0.03 × 100 = 3
8. 0.4 × 1000 = 400

Question 5.
A wheel of a baby cycle covers 49.7 cm in one rotation. Find the distance covered in 10 rotations.
Solution:
Length covered in 1 rotation = 49.7 cm
Length covered in 10 rotations = 49.7 × 10 cm = 497 cm

Question 6.
A picture chart costs ₹ 1.50. Radha wants to buy 20 charts to make an album. How much does she have to pay?
Solution:

Cost of 1 chart = ₹ 1.50
Cost of 20 charts = ₹ 1.50 × 20 = ₹ 30.00
Cost of 20 charts = ₹ 30

Question 7.
Find the product of the following.
(i) 3.6 × 0.3
(ii) 52.3 × 0.1
(iii) 537.4 × 0.2
(iv) 0.6 × 0.06
(v) 62.2 × 0.23
(vi) 1.02 × 0.05
(vii) 10.05 × 1.05
(viii) 101.01 × 0.01
(ix) 100.01 × 1.1
Solution:
(i) 3.6 × 0.3

36 × 3 = 108
3.6 × 0.3 = 1.08

(ii) 52.3 × 0.1
523 × 1 = 523
52.3 × 0.1 = 5.23

(iii) 537.4 × 0.2

5374 × 2 = 10748
537.4 × 0.2 = 107.48

(iv) 0.6 × 0.06
6 × 6 = 36
0.6 × 0.06 = 0.036

(v) 62.2 × 0.23

622 × 23 = 14306
62.2 × 0.23 = 14.306

(vi) 1.02 × 0.05
102 × 5 = 510
1.02 × 0.05 = 0.0510

(vii) 10.05 × 1.05

1005 × 105 = 105525
10.05 × 1.05 = 10.5525

(viii) 101.01 × 0.01
10101 × 1 = 10101
101.01 × 0.01 = 1.0101

(ix) 100.01 × 1.1
1001 × 11 = 110011
100.01 × 1.1 = 110.011

Objective Type Questions

Question 1.
1.07 × 0.1 _______
(i) 1.070
(ii) 0.107
(iii) 10.70
(iv) 11.07
Answer:
(ii) 0.107
Hint:
107 × 1 = 107
1.07 × 0.1 = 0.107

Question 2.
2.08 × 10 = ______
(i) 20.8
(ii) 208.0
(iii) 0.208
(iv) 280.0
Answer:
(i) 20.8
Hint:
208 × 10 = 2080
2.08 × 10 = 20.80 = 20.8

Question 3.
A frog jumps 5.3 cm in one jump. The distance travelled by the frog in 10 jumps is _______
(i) 0.53 cm
(ii) 530 cm
(iii) 53.0 cm
(iv) 53.5 cm
Answer:
(iii) 53.0 cm
Hint:
53 × 10 = 530
5.3 × 10 = 53.0

Students can Download Maths Chapter 1 Number System Additional 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 1 Number System Additional Questions

Additional Questions and Answers

Exercise 1.1

Question 1.
Match the following:

Solution:
1 – (v)
2 – (iv)
3 – (i)
4 – (iii)
5 – (ii)

Question 2.
Round 89.357 to the nearest whole number.
Solution:
Underlining the digit to be rounded 89.357. Since the digit next to the underlined digit 3 which is less than 5, the underlined digit remains the same.
∴ The nearest whole number 89.357 rounds to 89.

Question 3.
Round 110.929 to the nearest tenths place.
Solution:
Underlining the digit to be rounded 110.929. Since the digit next to the underlined digit is 2 which is less than 5.
∴ The underlined digit 9 remains the same. Hence the rounded number is 110.9

Question 4.
Round 87.777 upto 2 places of decimal.
Solution:
Rounding 87.777 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.777 gives 87.777. Since the digit after the hundredth place value is 7 which is more than 5, we add 1 to the underlined digit. So the rounded value of 87.777 upto 2 places of decimal is 87.78

Exercise 1.2

Question 1.
If Sheela bought 2.083 kg of grapes and 3.752 kg of orange. What is the total weight of fruits
Solution:

Weight of grapes = 2.083 Kg
Weight of orange = 2.752 Kg
Total weight = (2.083 + 2.752) Kg = 4.835 Kg

Question 2.
Kathir bought 8.72 kg of sugar, 7.302 kg of grains. His carry bag can contain only 15 kg of weight. What is the remaining weight of items bought?
Solution:

Question 3.
Use place value grid to add 7.357 and 13.92.
Solution:
Let as use place value grid.

Exercise 1.3

Question 1.
Cost of 1m cloth is ₹ 6.75. Find the cost of 14.75m correct to two places of decimal.
Solution:

Cost of 1 m cloth = ₹ 6.75
Cost of 14.75 m cloth = 14.75 × 6.75
= ₹ 99.5625
= ₹ 99.56

Question 2.
Length of a side of a square is 18.35 cm. Find its Area.
Solution:

Side of a square = 18.35 cm
Area of a square = (Side × Side) sq.units
= 18.35 × 18.35 cm²
= 336.7225 cm²

Exercise 1.4

Question 1.
A wire of length 363.987m is cut into 30 pieces. What is the length of each piece?
Solution:
Length of the wire = 363.987m
i.e Total length of 30 pieces = \(\frac { 363987 }{ 1000 } \) m

∴ Length of 1 piece
= 12132.9 × \(\frac { 1 }{ 1000 } \)
Length of 1 piece of wire = 12.1329 m

Question 2.
A cake of 50kg needs 23.4 kg sugar. Find the weight of cake made by 1 kg of sugar.
Solution:
Weight of cake made using 23.4 kg sugar = 50 kg
Weight of cake made using 1 kg sugar = \(\frac { 50 }{ 23.4 } \)
= \(\frac { 50 }{ 23.4 } \) x \(\frac { 10 }{ 10 } \) = \(\frac { 500 }{ 234 } \) = 2.1367 Kg
= 2.14 Kg
Weight of cake made using 1 kg sugar = 2.14 Kg

Question 3.
A pack of 20 pencils cost ₹ 94.4. What is the cost of each pencil?
Solution:
Cost of 20 pencils = ₹ 94.4

∴ Cost of 1 pencil = ₹ 4.72

Students can Download Maths Chapter 1 Number System Ex 1.2 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 1 Number System Ex 1.2

Question 1.
Add by using grid 0.51 + 0.25.
Solution:
Here 0.51 = \(\frac { 51 }{ 100 } \) and 0.25 = \(\frac { 25 }{ 100 } \)

First we shade the region 0.51 and then 0.25.
The sum is the total shaded area. 0.51 + 0.25 = 0.76

Question 2.
Add the following by using place value grid.
(i) 25.8 + 18.53
(ii) 17.4 + 23.435
Solution:
(i) 25.8 + 18.53.
Using place value grid.

Therefore 25.8 + 18.53 = 44.33

(ii) 17.4 + 23.435
Lets use the place value grid.

Therefore 17.4 + 23.435 = 40.835

Question 3.
Find the value of 0.46 – 0.13 by grid model.
Solution:
Here 0.46 = \(\frac { 46 }{ 100 } \) and 0.13 = \(\frac { 13 }{ 100 } \)

Shading the region 0.46 and then crossing out 0.13 from the shaded area. The left out shaded region without cross marks is the difference. So 0.46 – 0.13 = 0.33

Question 4.
Subtract the following by using place value grid, (i) 6.567 from 9.231 (ii) 3.235 from 7
Solution:
(i) Let as use place value grid

Therefore 9.231 – 6.567 = 2.664

(ii) Let as use place value grid.

Therefore 7 – 3.235 = 3.765

Question 5.
Simplify: 23.5 – 27.89 + 35.4 – 17.
Solution:
23.5 – 27.89 + 35.4 – 17 = 14.01

Question 6.
Sulaiman bought 3.350 kg of Potato, 2.250 kg of Tomato and some onions. If the weight of the total items are 10.250 kg, then find the weight of onions?
Solution:
Weight of Potato = 3.350 kg
Weight of Tomato = 2.250 kg
Total weight of Potato and Tomato = (3.350 + 2.250 kg)
= 5.600 kg
Weight of potato, tomato and onions = 10.250
Weight of potato and tomato = 5.600
∴ Weight of onions = (10.250 – 5.600) Kg = 4.650 Kg
Weight of onions = 4.650 Kg

Question 7.
What should be subtracted from 7.1 to get 0.713?
Solution:
To get the number to be subtracted

We have 7.1 – 0.713 = 6.387
∴ The number to be subtracted = 6.387

Question 8.
How much is 35.6 km less than 53.7 km?
Solution:
To get the answer we must subtract 53.7km – 35.6 km = 18.1 km

So 35.6 km is 18.1 km less than 53.7 km.

Question 9.
Akilan purchased a geometry box for ₹ 25.75, a pencil for ₹ 3.75 and a pen for ₹ 17.90. He gave ₹ 50 to the shopkeeper. What amount did he get back?
Solution:
Cost of geometry box = ₹ 25.75 (+)
Cost of Pencil box = ₹ 3.75

Question 10.
Find the perimeter of an equilateral triangle with a side measuring 3.8 cm.
Solution:
Perimeter of an equilateral triangle = (Side + Side + Side) Sq. units.

Given side = 3.8
∴ Perimeter = 3.8 + 3.8 + 3.8
Perimeter of the triangle = 11.4 cm

Objective Type Questions

Question 1.
1.0 + 0.83 = ?
(i) 0.17
(ii) 0.71
(iii) 1.83
(iv) 1.38
Answer:

(iii) 1.83

Question 2.
7.0 – 2.83 = ?
(i) 3.47
(ii) 4.17
(iii) 7.34
(iv) 4.73
Answer:

(ii) 4.17

Question 3.
Subtract 1.35 from 3.51
(i) 6.21
(ii) 4.86
(iii) 8.64
(iv) 2.16
Answer:

(iv) 2.16

Question 4.
Sum of two decimals is 4.78 and one decimal is 3.21 then the other one is
(i) 1.57
(ii) 1.75
(iii) 1.59
(iv) 1.58
Answer:

(i) 1.57

Question 15.
The difference of two decimals is 86.58 and one of the decimal is 42.31 Find the other one
(i) 128.89
(ii) 128.69
(iii) 128.36
(iv) 128.39
Answer:

(i) 128.89

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.3 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 2.3

Question 1.
14 out of the 70 magazines at the bookstore are comedy magazines. What percentage of the magazines at the bookstore are comedy magazines?
Solution:
Total number of magazines in the bookstore = 100 m
Number of comedy magazines = 14
Percentage of comedy magzines = \(\frac { 14 }{ 70 } \) × 100% = 20%
20% of the magazines are comedy magazines.

Question 2.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water will fill the tank, so that it is 50% full?
Solution:
Capacity of the tank = 50 litres
Amount of water filled = 30% of 50 litres = \(\frac { 30 }{ 100 } \) × 50 = 15 litres
Amount of water to be filled = 50 – 15 = 35 litres

Question 3.
Karun bought a pair of shoes at a sale of 25%. If the amount he paid was ₹ 1000, then find the marked price.
Solution:
Let the marked price of the raincoat be ₹ P
Amount he paid at a discount of 25% = ₹ 1000
(Marked Price) – (25% of P) = 1000
P – (\(\frac { 25 }{ 100 } \) × P) = 1000
P – \(\frac { 1 }{ 4 } \) × P = 1000
P (1 – \(\frac { 1 }{ 4 } \)) = 1000
\(\frac { 3 }{ 4 } \) P = 1000
P = 1000 × \(\frac { 4 }{ 3 } \)
= \(\frac { 4000 }{ 3 } \)
P = 1333.33
∴ Marked price of the shoes = ₹ 1333

Question 4.
An agent of an insurance company gets a commission of 5% on the basic premium he collects. What will be the commission earned by him if he collects ₹ 4800?
Solution:
Premium collected = ₹ 4800
Commission earned = 5% of basic premium
Commission earned for ₹ 4800 = 5% of 4800
= \(\frac { 5 }{ 100 } \) × 4800
= ₹ 240
Commission earned = ₹ 240

Question 5.
A biology class examined some flowers in a local Grass land. Out of the 40 flowers they saw, 30 were perennials. What percentage of the flowers were perennials?
Solution:
Number of flowers examined = 40
Number of perennials = 30
Percentage = \(\frac { 30 }{ 40 } \) × 100%
= 75%
75% of the flowers were perennials.

Question 6.
Ismail ordered a collection of beads. He received 50 beads in all. Out of that 15 beads were brown. Find the percentage of brown beads?
Solution:
Number of beads received = 50
Number of brown beads = 5
Percentage of brown beads = \(\frac { 15 }{ 50 } \) × 100 %
= 10 %
10% of the beads was brown

Question 7.
Ramu scored 20 out of 25 marks in English, 30 out of 40 marks in Science and 68 out of 80 marks in mathematics. In which subject his percentage of marks is best?
Solution:
Ramu’s score in English = 20 out of 25
Percentage scored in English = \(\frac { 20 }{ 25 } \) × 100 % = 80 %
Ramu’s Score in Science = 30 out of 40
Percentage scored in Science = \(\frac { 30 }{ 40 } \) × 100 % = 75%
Ramu’s score in Mathematics = 68 out of 80
Percentage scored in Maths = \(\frac { 68 }{ 80 } \) × 100 % = 85 %
85% > 80% > 75%.
∴ In Mathematics his percentage of marks is the best.

Question 8.
Peter requires 50% to pass. If he gets 280 marks and falls short by 20 marks, what would have been the maximum marks of the exam?
Solution:
Peters score = 280 marks
Marks needed for a pass = 20
∴ Total marks required to get a pass = 280 + 20 = 300
i.e. 50% of total marks = 300
\(\frac { 50 }{ 100 } \) × Total marks = 300
\(\frac { 1 }{ 2 } \) × Total Marks = 300
Total Marks = 300 × 2 = 600
Total marks of the exam = 600

Question 9.
Kayal scored 225 marks out of 500 in revision test 1 and 265 out of 500 marks in revision test 2. Find the percentage of increase in her score.
Solution:
Marks scored in revision I = 225
Marks scored in revision II = 265
Change in marks = 265 – 225 = 40

Percentage of increase in marks = 8%

Question 10.
Roja earned ₹ 18,000 per month. She utilized her salary in the ratio 2 : 1 : 3 for education, savings and other expenses respectively. Express her usage of income in percentage.
Solution:
Amount of Salary = ₹ 18,000
(i) Total number of parts of salary = 2 + 1 + 3 = 6
Salary is divided into 3 portions as \(\frac { 2 }{ 6 } \),\(\frac { 1 }{ 6 } \) and \(\frac { 3 }{ 6 } \)
Portion of salary used for education = \(\frac { 2 }{ 6 } \)
Salary used for education = \(\frac { 2 }{ 6 } \) × 18,000 = ₹ 6,000
Percentage for Education = \(\frac { 6000 }{ 18000 } \) × 100 = 33.33%

(ii) Usage of salary for savings = \(\frac { 1 }{ 6 } \) × 18,000 = ₹ 3,000
Percentage for savings = \(\frac { 3000 }{ 18000 } \) × 100 = 16.67 %

(iii) Usage of salary for other expenses = \(\frac { 3 }{ 6 } \) × 18,000 = ₹ 9,000
Percentage for other expenses = \(\frac { 9000 }{ 18000 } \) × 100 = 50 %

Students can Download Maths Chapter 1 Number System Ex 1.5 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 1 Number System Ex 1.5

Miscellaneous Practice problems

Question 1.
Malini bought three ribbon of lengths 13.92 m, 11.5 m and 10.64 m. Find the total length of the ribbons?
Solution:
Length of ribbon 1 = 13.92 m
Length of ribbon 2 = 11.50 m
Length of ribbon 3 = 10.64 m

Total Length of the ribbons = 13.92 m + 11.5 m + 10.64 m = 36.06 m
Totla length of the ribbons = 36.06 m

Question 2.
Chitra has bought 10 kg 35 g of ghee for preparing sweets. She used 8 kg 59 g of ghee. How much ghee will be left?
Solution:
Total weight of ghee bought = 10 kg 35 g
Weight of ghee used = 8 kg 59 g
Weight of ghee left = 10.35 kg – 8.59 kg = 1.76 kg

∴ Weight of ghee left= 1 kg 76 g = 1.76 kg

Question 3.
If the capacity of a milk can is 2.53 l, then how much milk is required to fill 8 such cans?
Solution:
Capacity of 1 milk can= 2.53 l
∴ Capacity of 8 milk cans= 2.53 l × 8 = 20.24 l

To fill 8 cans 20.24 l of milk is required.

Question 4.
A basket of orange weighs 22.5 kg. If each family requires 2.5 kg of orange, families can share?
Solution:
Total weight of orange = 22.5 kg
Weight of orange required for 1 family = 2.5 kg
∴ Number of families sharing orange = 22.5 kg ÷ 2.5 kg
= \(\frac { 22.5 }{ 2.5 } \) = \(\frac { 22.5 }{ 2.5 } \) × \(\frac { 10 }{ 10 } \) = \(\frac { 225 }{ 25 } \) = 9
∴ 9 families can share the oranges.

Question 5.
A baker uses 3.924 kg of sugar to bake 10 cakes of equal size. How much sugar is used in each cake?
Solution:
For 10 cakes sugar required = 3.924 kg
For 1 cake sugar required = 3.924 ÷ 10 = \(\frac { 3.924 }{ 10 } \) = 0.3924 kg
For 1 cake sugar required = 0.3924 kg.

Question 6.
Evaluate:
(i) 26.13 × 4.6
(ii) 3.628 + 31.73 – 2.1
Solution:
(i) 26.13 × 4.6
26.13 × 4.6 = 120.198

(ii) 3.628 + 31.73 – 2.1 = 33.258

Question 7.
Murugan bought some bags of vegetables. Each bag weighs 20.55 kg. If the total weight of all the bags is 308.25 kg, how many bags did he buy?
Solution:
Total weight of all bags = 308.25 kg
Weight of 1 bag = 20.55 kg

Question 8.
A man walks around a circular park of distance 23.761 m. How much distance will he cover in 100 rounds?
Solution:
In 1 round distance covered = 23.761 m
∴ In 100 rounds distance = 23.761 × 100
= 2376.1 m
∴ In 100 round he covers 2376.1 m.

Question 9.
How much 0.0543 is greater than 0.002?
Solution:

∴ Required answer is 0.0523

Question 10.
A printer can print 15 pages per minute. How many pages can it print in 4.6 minutes?
Solution.
In 1 minute the pages printed = 15
In 4.6 minutes the pages printed = 15 × 4.6
= 69

The printer prints 69 pages.

Challenge Problems

Question 1.
The distance travelled by Prabhu from home to Yoga centre is 102 m and from Yoga centre to school is 165 m. What is the total distance travelled by him in kilometres (in decimal form)?
Solution:

∴ 267 metres = \(\frac { 267 }{ 1000 } \) km = 0.267 km
∴ Total distance travelled = 0.267 km

Question 2.
Anbu and Mala travelled from A to C in two different routes. Anbu travelled from place A to place B and from there to place C. A is 8.3 km from B and B is 15.6 km from C. Mala travelled from place A to place D and from there to place C. D is 7.5 km from A and C is 16.9 km from D. Who travelled more and by how much distance?
Solution:
Distance travelled by Anbu:

From place A to place B = 8.3 km
Distance from place B to place C = 15.6 km
∴ Total distance travelled by Anbu = 8.3 + 15.6
= 23.9 km

Distance travlled by Mala:

Distance travelled place A to D = 7.5 km
Distance from place D to place C = 16.9 km
Total distance travelled by mala = (7.5 + 16.9) km = 24.4 km
24.4 > 23.9
∴ Mala travelled more distance. She travelled (24.4 – 23.9) km more i.e she travelled 0.5 km more

Question 3.
Ramesh paid ₹ 97.75 per hour for a taxi and he used 35 hours in a week. How much he has to pay totally as taxi fare for a week?
Solution:
Payment for the taxi for an hour = ₹ 97.75
Total hours the taxi was used = 35 hrs
∴ Total payment for the taxi for the week

= 97.75 × 35
= 3421.25
Total payment for a week = ₹ 3421.25

Question 4.
An Aeroplane travelled 2781.20 kms in 6 hours. Find the average speed of the aeroplane in Km/hr.
Solution:
In 6 hours the distance travelled = 2781.20 km
In 1 hour the distance travelled = \(\frac { 2781.20 }{ 6 } \) km

Average speed of the aroplane = 463.53 km/hr.

Question 5.
Kumar’s car gives 12.6 km mileage per litre. If his fuel tank holds 25.8 litres then how far can he travel?
Solution.
Distance travelled with 1 litre fuel = 12.6 km
∴ with 25.8 litres distance travelled = 12.6 × 25.8
= 325.08 km

The car can travel 325.08 km

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.1 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 2.1

Question 1.
In each of the following grid, find the numbers of coloured squares and express it as a fraction, decimal and percentage.

Solution:
Number of coloured square = 58
Total number of squares = 100
∴ Fraction : \(\frac { 58 }{ 100 } \)
Decimal : 0.58
Percentage : 58%

(ii) Number of coloured square = 53
Total number of squares = 100
∴ Fraction : \(\frac { 53 }{ 100 } \)
Decimal : 0.53
Percentage : 53%

(iii) Number of coloured square = 25
Total number of squares = 50
∴ Fraction : \(\frac { 25 }{ 50 } \)

Decimal : \(\frac { 25 }{ 50 } \) × \(\frac { 2 }{ 2 } \)
= \(\frac { 50 }{ 100 } \)
= 0.50

Percentage : \(\frac { 25 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 25 }{ 50 } \) × 100% = 50%

(iv) Number of coloured square = 17
Total number of squares = 25
∴ Fraction : \(\frac { 17 }{ 25 } \)
Decimal : \(\frac { 17 }{ 25 } \) × \(\frac { 4 }{ 4 } \)
= \(\frac { 68 }{ 100 } \) = 0.68
Percentage : \(\frac { 17 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 17 }{ 25 } \) × 100%
= 68%

(v) Number of coloured square = 15
Total number of squares = 30
∴ Fraction : \(\frac { 15 }{ 30 } \)
Decimal : \(\frac { 15 }{ 30 } \)
= \(\frac { 1 }{ 2 } \) × \(\frac { 50 }{ 50 } \)
= \(\frac { 50 }{ 100 } \) = 0.50
Percentage : \(\frac { 15 }{ 30 } \)
= \(\frac { 15 }{ 30 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 15 }{ 30 } \) × 100%
= 50 %

Question 2.
A picture of chess board is given.
(i) Find the percentage of the white coloured squares
(ii) Find the percentage of gray coloured squares
(iii) Find the percentage of the squares that have the pieces and
(iv) The squares that do not have the pieces.

Solution:
(i) Total number of squares in the chess board = 64
Number of white coloured squares = 32
Percentage = \(\frac { 32 }{ 64 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 32 }{ 64 } \) × 100 %
= 50%

(ii) Grey coloured squares = 64
Percentage = \(\frac { 32 }{ 64 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 32 }{ 64 } \) } [/latex] × 100 %
= 50 %

(iii) Number of squares having pieces = 20
Total number of squares = 64

(iv) Number of squares do not have pieces = 44

Question 3.
A picture of dart board is given. Find the percentage of white coloured portion and black coloured portion.

Solution:
Total sector = 20
White coloured sector = 10
Black coloured sector = 10
Percentage of white : \(\frac { 10 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
Decimal : \(\frac { 10 }{ 20 } \) × 100 %
= 50 %
Percentage of black colour : \(\frac { 10 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
Decimal : \(\frac { 10 }{ 20 } \) × 100 %
= 50 %

Question 4.
Write each of the following fraction as percentage.
(i) \(\frac { 36 }{ 50 } \)
(ii) \(\frac { 81 }{ 30 } \)
(iii) \(\frac { 42 }{ 56 } \)
(iv) 2 \(\frac { 1 }{ 4 } \)
(v) 1 \(\frac { 3 }{ 5 } \)
Solution:
(i) \(\frac { 36 }{ 50 } \)
= \(\frac { 36 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 36 }{ 50 } \) × 100 %
= 72 %

(ii) \(\frac { 81 }{ 30 } \)
= \(\frac { 81 }{ 30 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 81 }{ 30 } \) × 100 %
= 270 %

(iii) \(\frac { 42 }{ 56 } \)
= \(\frac { 42 }{ 56 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 42 }{ 56 } \) × 100 %
= \(\frac { 21 }{ 28 } \) × 100 %
= 75 %

(iv) 2 \(\frac { 1 }{ 4 } \)
= \(\frac { 9 }{ 4 } \)
= \(\frac { 9 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 9 }{ 4 } \) × 100 %
= 225 %

(v) 1 \(\frac { 3 }{ 5 } \)
= \(\frac { 8 }{ 5 } \)
= \(\frac { 8 }{ 5 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 8 }{ 5 } \) × 100 %
= 160 %

Question 5.
Anbu scored 436 marks out of 500 in his exams. What was the percentage he scored?
Answer:
Total marks = 500
Anbu’s Score = 436
Percentage = \(\frac { 436 }{ 500 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 436 }{ 500 } \) × 100 %
= 87.2 %
Anbu’s Score = 87.2 %

Question 6.
Write each of the following percentage as fraction,
(i) 21%
(ii) 93.1 %
(iii) 151 %
(iv) 65 %
(v) 0.64 %
Solution:
(i) 21%
= \(\frac { 21 }{ 100 } \)

(ii) 93.1 %
= \(\frac { 93.1 }{ 100 } \)
= \(\frac{93.1 \times 10}{100 \times 10}\)
= \(\frac { 931 }{ 1000 } \)

(iii) 151 %
= \(\frac { 151 }{ 100 } \)

(iv) 65 %
= \(\frac { 65 }{ 100 } \)
= \(\frac { 13 }{ 20 } \)

(v) 0.64 %
= \(\frac { 0.64 }{ 100 } \)
= \(\frac{0.64 \times 100}{100 \times 100}\)
= \(\frac { 64 }{ 10000 } \)
= \(\frac { 4 }{ 625 } \)

Question 7.
Iniyan bought 5 dozen eggs. Out of that 5 dozen eggs, 10 eggs are rotten. Express the number of good eggs as percentage.
Solution:
1 dozen eggs = 12
5 dozen = 5 × 12
Total eggs = 60 eggs
Rotten eggs = 10 Good
eggs = 60 – 10 = 50
Fraction of good eggs = \(\frac { 50 }{ 60 } \)
Percentage of good eggs = \(\frac { 50 }{ 60 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 50 }{ 60 } \) × 100 %
= \(\frac { 5 }{ 6 } \) × 100 %
= 83.33 %
Percentage of good eggs = 83.33 %

Question 8.
In an election, Candidate X secured 48% of votes. What fraction will represent his votes?
Solution:
Percentage of votes x secured = 48% = \(\frac { 48 }{ 100 } \)
Fraction of votes x secured = \(\frac { 12 }{ 25 } \)

Question 9.
Ranjith total income was ₹ 7,500. He saved 25% of his total income. Find the amount saved by him.
Solution:
Total income of Ranjith = ₹ 7500
His savings = 25 % of 7500
= \(\frac { 25 }{ 100 } \) of 7500
= \(\frac { 25 }{ 100 } \) × 7500
= ₹ 1,875
∴ Amount saved by Ranjith = ₹ 1,875

Objective Type Questions

Question 1.
Thendral saved one fourth of her salary. Her savings percentage is
(i) \(\frac { 3 }{ 4 } \)
(ii) \(\frac { 1 }{ 4 } \) %
(iii) 25 %
(iv) 1 %
Hint: \(\frac { 1 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 4 } \) × 100 %
= 25 %
Answer:
(iii) 25 %

Question 2.
Kavin scored 15 out of 25 in a test. The percentage of his marks is
(i) 60%
(ii) 15%
(iii) 25%
(iv) 15/25
Hint: \(\frac { 15 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 15 }{ 25 } \) × 100 %
= 60 %
Answer:
(i) 60%

Question 3.
0.07% is
(i) \(\frac { 7 }{ 10 } \)
(ii) \(\frac { 7 }{ 100 } \)
(iii) \(\frac { 7 }{ 1000 } \)
(iv) \(\frac { 7 }{ 10,000 } \)
Hint: 0.07 %
= 0.07%
= \(\frac { 0.07 }{ 100 } \)
= \(\frac{7}{\frac{100}{100}}\)
= \(\frac{7}{100 \times 100}\)
= \(\frac { 7 }{ 10,000 } \)
Answer:
(iv) \(\frac { 7 }{ 10,000 } \)