Samacheer Class 7 Maths Percentage and Simple Interest Intext Questions Book Back Answers & Solutions

Q: 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 }

(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

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

(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

Q: What is the difference between 0.01 and 1%.

0.01 = \frac { 1 }{ 100 } = 1% 0.01 and 1% are the same.

Q: Convert these decimals to percentage. (i) 0.25 (ii) 0.07 (iii) 0.8 (iv) 0.375 (v) 3.75

(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)

Q: Write these percentage as decimals (i) 3 % (ii) 25 % (iii) 80 % (iv) 67 % (v) 17.5 % (vi) 135 % (vii) 0.5 %

(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)

Your Progress - Chapter 20% complete

1Book Back Questions78 questions

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

Answer:

Try These (Text book Page No. 29)

Q.1There 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.v

Answer:

Try These (Text book Page No. 30)

Q.1Convert 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) \(\frac { 72 }{ 90 } \)v

Answer:

(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 %

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

Answer:

(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)

Q.1What is the difference between 0.01 and 1%.v

Answer:

0.01 = \(\frac { 1 }{ 100 } \) = 1%
0.01 and 1% are the same.

Q.2In 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?v

Answer:

No. Only some of them are half of its original price.
Exercise 2.2
Try These (Text book Page No. 33)

Q.1Convert these decimals to percentage. (i) 0.25 (ii) 0.07 (iii) 0.8 (iv) 0.375 (v) 3.75v

Answer:

(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)

Q.1Write these percentage as decimals (i) 3 % (ii) 25 % (iii) 80 % (iv) 67 % (v) 17.5 % (vi) 135 % (vii) 0.5 %v

Answer:

(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 %
= \(\frac { 0.5 }{ 100 } \) = 0.005Exercise 2.3
Try These (Text book Page No. 38)

Q.1Level of water in a tank is increased from 35 litres to 50 litres in 2 minutes, what is the Percentage of increase?v

Answer:

Level of water in the tank originally = 35 litres.
Increase in the water level = amount of change = 50 – 35 = 15 litresExercise 2.4
Try These (Text book Page No. 41)

Q.1Arjun 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.v

Answer:

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

Q.2Shanti borrowed ₹ 6,000/- from a Bank for 7 years at 12 % per annum. What amount will clear off her debt?v

Answer:

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 = ₹ 11,040Think (Text book Page No. 43)

Q.1In simple interest, a sum of money doubles itself in 10 years. In how many years it will get triple itself.v

Answer:

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

Answer:

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 %

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

Answer:

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 %

Q.4Write 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 \(\frac { 3 }{ 5 } \)v

Answer:

(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 %

Q.5Anbu scored 436 marks out of 500 in his exams. What was the percentage he scored?v

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 %

Q.6Write each of the following percentage as fraction, (i) 21% (ii) 93.1 % (iii) 151 % (iv) 65 % (v) 0.64 %v

Answer:

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

Q.7Iniyan bought 5 dozen eggs. Out of that 5 dozen eggs, 10 eggs are rotten. Express the number of good eggs as percentage.v

Answer:

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 %

Q.8In an election, Candidate X secured 48% of votes. What fraction will represent his votes?v

Answer:

Percentage of votes x secured = 48% = \(\frac { 48 }{ 100 } \)
Fraction of votes x secured = \(\frac { 12 }{ 25 } \)

Q.9Ranjith total income was ₹ 7,500. He saved 25% of his total income. Find the amount saved by him.v

Answer:

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

Q.1Thendral 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 %v

Answer:

(iii) 25 %

Q.2Kavin 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 %v

Answer:

(i) 60%

Q.30.07% is (i) \(\frac { 7 }{ 10 } \) (ii) \(\frac { 7 }{ 100 } \) (iii) \(\frac { 7 }{ 1000 } \) (iv) \(\frac { 7 }{ 10,000 } \) Hint: 0.07 % = 0.07% = \(\frac { 0.07 }{ 100 } \) = \(\frac{7}{\frac{100}{100}}\) = \(\frac{7}{100 \times 100}\) = \(\frac { 7 }{ 10,000 } \)v

Answer:

(iv) \(\frac { 7 }{ 10,000 } \)
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Q.1Write each of the following percentage as decimal. (i) 21 % (ii) 93.1 % (iii) 151 % (iv) 65 % (v) 0.64 %v

Answer:

(i) 21 %
= \(\frac { 21 }{ 100 } \) = 0.21
(ii) 93.1 %
= \(\frac { 93.1 }{ 100 } \) = 0.931
(iii) 151 %
= \(\frac { 151 }{ 100 } \) = 1.51
(iv) 65 %
= \(\frac { 65 }{ 100 } \) = 0.65
(v) 0.64 %
= \(\frac { 0.64 }{ 100 } \) = 0.0064

Q.2Convert each of the following decimal as percentage (i) 0.282 (ii) 1.51 (iii) 1.09 (iv) 0.71 (v) 0.858v

Answer:

(i) 0.282
= 0.282 × 100% = \(\frac { 282 }{ 1000 } \) × 100 %
= 28.2 %
(ii) 1.51
= \(\frac { 151 }{ 100 } \) × 100 %
= 151 %
(iii) 1.09
= \(\frac { 109 }{ 100 } \) × 100 %
= 109 %
(iv) 0.71
= \(\frac { 71 }{ 100 } \) × 100 %
= 71 %
(v) 0.858
= \(\frac { 858 }{ 1000 } \) × 100 %
= 85.8 %

Q.3In an examination a student scored 75% of marks. Represent the given the percentage in decimal form?v

Answer:

Student’s Score = 75% = \(\frac { 75 }{ 100 } \) = 0.75

Q.4In a village 70.5% people are literate. Express it as a decimal.v

Answer:

Percentage of literate people = 70.5%
= \(\frac { 70.5 }{ 100 } \)
= 0.705

Q.5Scoring rate of a batsman is 86%. Write his strike rate as decimal.v

Answer:

Scoring rate of the batsman = 86%
= \(\frac { 86 }{ 100 } \)
= 0.86

Q.6The height of a flag pole in school is 6.75m. Write it as percentage.v

Answer:

Height of flag pole = 6.75m
= \(\frac { 675 }{ 100 } \)
= 6.75%

Q.7The weights of two chemical substances are 20.34 g and 18.78 g. Write the difference in percentage?v

Answer:

Weight of substance 1 = 20.34g
Percentage of substance 1 = \(\frac { 2034 }{ 100 } \) = 2034 %
Weight of substance 2 = 18.78g
Percentage of substance 2 = \(\frac { 1878 }{ 100 } \) = 1878 %
Their difference = 2034 – 1878 = 156%

Q.1Decimal value of 142.5% is (i) 1.425 (ii) 0.1425 (iii) 142.5 (iv) 14.25 Hint: 142.5 % = \(\frac { 1425 }{ 10 } \) % = \(\frac { 1425 }{ 10 } \) × \(\frac { 1 }{ 100 } \) = 1.425v

Answer:

(i) 1.425

Q.2The percentage of 0.005 is (i) 0.005 % (ii) 5 % (iii) 0.5 % (iv) 0.05 % Hint: 0.005 = \(\frac { 5 }{ 1000 } \) = \(\frac { 5 }{ 1000 } \) × \(\frac { 100 }{ 100 } \) = 0.5 %v

Answer:

(iii) 0.5 %

Q.3The percentage of 4.7 is (i) 0.47 % (ii) 4.7 % (iii) 47 % (iv) 470 % Hint: 4.7 = \(\frac { 47 }{ 10 } \) = \(\frac { 47 }{ 10 } \) × \(\frac { 100 }{ 100 } \) = 470 %v

Answer:

(iv) 470 %
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Q.114 out of the 70 magazines at the bookstore are comedy magazines. What percentage of the magazines at the bookstore are comedy magazines?v

Answer:

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.

Q.2A 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?v

Answer:

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

Q.3Karun bought a pair of shoes at a sale of 25%. If the amount he paid was ₹ 1000, then find the marked price.v

Answer:

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

Q.4An 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?v

Answer:

Premium collected = ₹ 4800
Commission earned = 5% of basic premium
Commission earned for ₹ 4800 = 5% of 4800
= \(\frac { 5 }{ 100 } \) × 4800
= ₹ 240
Commission earned = ₹ 240

Q.5A 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?v

Answer:

Number of flowers examined = 40
Number of perennials = 30
Percentage = \(\frac { 30 }{ 40 } \) × 100%
= 75%
75% of the flowers were perennials.

Q.6Ismail ordered a collection of beads. He received 50 beads in all. Out of that 15 beads were brown. Find the percentage of brown beads?v

Answer:

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

Q.7Ramu 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?v

Answer:

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.

Q.8Peter 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?v

Answer:

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

Q.9Kayal 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.v

Answer:

Marks scored in revision I = 225
Marks scored in revision II = 265
Change in marks = 265 – 225 = 40Percentage of increase in marks = 8%

Q.10Roja 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.v

Answer:

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

Answer:

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

Q.2Aravind 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.v

Answer:

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

Q.3Sheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is ₹ 21,280. Find the sum.v

Answer:

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

Q.4Basha 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?v

Answer:

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

Q.5In What time will ₹ 16,500 amount to ₹ 22,935 at 13% per annum?v

Answer:

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

Q.6In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?v

Answer:

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

Q.7A 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.v

Answer:

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

Q.8A principal becomes ₹ 17,000 at the rate of 12% in 3 years. Find the principal.v

Answer:

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 = ₹ 12,500Objective Type Questions

Q.9The 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 = ₹ 500v

Answer:

(i) ₹ 500

Q.10Which 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 = \(\frac { pnr }{ 100 } \) = \(\frac{1000 \times 1 \times 10}{100}\) = ₹ 100v

Answer:

(iii) ₹ 100

Q.11Which 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 %v

Answer:

(i) 10%
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Q.1When 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?v

Answer:

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.

Q.2Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.v

Answer:

Yazhini’s score = 15 out of 25 = \(\frac { 15 }{ 25 } \)
Score in percentage = \(\frac { 15 }{ 25 } \) × 100% = 60%

Q.3Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.v

Answer:

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%

Q.4A 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.v

Answer:

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

Q.5There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?v

Answer:

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

Q.6The ratio of Saral’s income to her savings is 4 : 1. What is the percentage of money saved by her?v

Answer:

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

Q.7A salesman is on a commission rate of 5%. How much commission does he make on sales worth ₹ 1,500?v

Answer:

Total amount on sale = ₹ 1,500
Commission rate = 5 %
Commission received = 5 % of ₹ 1,500 = \(\frac { 5 }{ 100 } \) × 1500 = ₹ 75
∴ Commission received = ₹ 75

Q.10What percentage of 8 is 64?v

Answer:

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

Q.11Stephen 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.v

Answer:

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

Q.12Riya 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.v

Answer:

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

Q.13In 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?v

Answer:

Let the required number of years be x
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P 1 = ₹ 3000
Rate of interest (r) = 8 %
Time (n 1 ) = n 1 years
Simple Interest I 1 = \(\frac{3000 \times 8 \times n_{1}}{100}\) = 240 n 1
Principal (P 2 ) = ₹ 4000
Rate of interest (r) = 12 %
Time n 2 = 4 years
Simple Interest I 2 = \(\frac{4000 \times 12 \times 4}{100}\)
I 2 = 1920
If I 1 = I 2
240 n 1 = 1920
n 1 = \(\frac { 1920 }{ 240 } \) = 8
∴ The required time = 8 years
Challenge Problems

Q.14A 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?v

Answer:

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 %

Q.15Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?v

Answer:

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 %

Q.17The 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?v

Answer:

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

Q.18A 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?v

Answer:

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

Q.19A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?v

Answer:

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.

Q.20Concession card holders pay ₹ 120 for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession card holders?v

Answer:

Train ticket fare = ₹ 230
Ticket fare on concession = ₹ 120
Discount = Ticket fare – concession fare = 230 – 120 = ₹ 110Percentage of discount = 47.83%

Q.21A 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?v

Answer:

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

Q.22Which is greater 16 \(\frac { 2 }{ 3 } \) or \(\frac { 2 }{ 5 } \) or 0.17 ?v

Answer:

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

Q.23The 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.v

Answer:

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

Q.24In 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.v

Answer:

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

Q.25A 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.v

Answer:

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

Q.26Arun 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.v

Answer:

Principal lent to Balaji P 1 = ₹ 5000
Time n 1 = 2 years
Let r be the rate of interest per year
Simple interest got from Balaji = \(\frac { pnr }{ 100 } \) ⇒ I 1 = \(\frac{5000 \times 25 \times r}{100}\)
Again principal let to Charles P 2 = ₹ 3000
Time (n 2 ) = 4 years
Simple interest got from Charles (I 2 ) = \(\frac{3000 \times 4 \times r}{100}\)
Altogether Arun got ₹ 2200 as interest.
∴ I 1 + I 2 = 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 %

Q.27If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let P = ₹ 100)v

Answer:

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 %
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