ML Aggarwal Percentage and its Applications Exe-7.1 Class 8 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-7.1 Questions for Percentage and its Applications as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Percentage and its Applications Exe-7.1 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-7 | Percentage and its Applications |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-7.1 Questions |

Edition | 2023-2024 |

**Percentage and its Applications Exe-7.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-113

**Question 1.** **Express the following percentages as fractions:**

(i) 356%

(ii) 2 ½%

(iii) 16 2/2 %

**Answer:**

**(i) 356%**

= 356/100

= 89/25

= 3 14/25

**(ii) 2 ½%**

= 5/2%

= 5/ (2 × 100)

= 1/40

**(iii) 16 2/3 %**

= 50/3 %

= 50/3 × 1/100

= 1/6

**Question 2. Express the following fractions as percentages:**

(i) 3/2

(ii) 9/20

(iii) 1 ¼

**Answer:**

**(i) 3/2**

= 3/2 × 100%

= 150%

**(ii) 9/20**

= 9/20 × 100%

= 45%

**(iii) 1 ¼**

= 5/4 × 100%

= 125%

**Question 3.** **Express the following fractions as decimals. Then express the decimals as percentages:**

(i) ¾

(ii) 5/8

(iii) 3/16

**Answer:**

**(i) ¾ = 0.75**

¾ = ¾ × 100%

= 3 × 25%

= 75%

**(ii) 5/8 = 0.625**

5/8 = 5/8 × 100%

= 5/2 × 25%

= 125/2%

= 62.5%

**(iii) 3/16 = 0.1875**

3/16 = 3/16 × 100%

= ¾ × 25%

= 75/4%

= 18.75%

**Question 4. Express the following fractions as decimals correct to four decimal places. Then express the decimals as percentages:**

(i) 2/3

(ii) 5/6

(iii) 4/7

**Answer:**

**(i) 2/3 = 0.6667**

2/3 = 0.6667 × 100% = 66.67%

**(ii) 5/6 = 0.8333**

5/6 = 0.8333 × 100% = 83.33%

**(iii) 4/7 = 0.5714**

4/7 = 0.5714 × 100% = 57.14%

**Question 5. Express the following ratios as percentages:**

(i) 17: 20

(ii) 13: 18

(iii) 93: 80

**Answer:**

**(i) 17: 20**

17: 20 = 17/ 20

= 17/20 × 100%

= 17 × 5%

= 85%

**(ii) 13: 18**

13: 18 = 13/18

= 13/18 × 100%

= 13/9 × 50%

= 650/9 %

= 72 2/9%

**(iii) 93: 80**

93: 80 = 93/80

= 93/80 × 100%

= 93/4 × 5%

= 465/4%

= 116.25 %

**Question 6.**** Express the following percentages as decimals:**

(i) 20%

(ii) 2%

(iii) 3 ¼ %

**Answer:**

**(i) 20%**

= 20/100

= 0.20

= 0.2

**(ii) 2%**

= 2/100

= 0.02

**(iii) 3 ¼**

= 13/4

Multiply the denominator by 100

= 13/ (4 × 100)

= 13/ 400

= 3.25/100

= 0.325

**Question 7.**** ****Find the value of:**

(i) 27 % of ₹ 50

(ii) 6 ¼ % of 25 kg

**Answer:**

**(i) 27 % of ₹ 50**

= 27/100 of ₹50

= 27/100 × 50

= 27/2

= ₹ 13.50

**(ii) 6 ¼ % of 25 kg**

= 25/4% of 25 kg

= 25/ (4 × 100) of 25 kg

= (25 × 25)/ (4 × 100)

= 25/16

= 1 9/16 kg

**Question 8. ****What percent is:**

**(i) 300 g of 2 kg**

**(ii) ₹ 7.50 of ₹ 6**

**Answer:**

**(i) Required percentage = [300 gram/ 2 kg × 100] %**

= [300 gram/ (2 × 1000 gram) × 100] %

= [300/ (2 × 1000) × 100] %

= (30/2) %

= 15 %

**(ii) Required percentage = [₹ 7.50/ ₹ 6 × 100] %**

= [7.50/ 6 × 100] %

= [7.50/3 × 50] %

= [2.50 × 50] %

= 125%

**Question 9. ****What percent of:**

(i) 50 kg is 65 kg

(ii) ₹ 9 is ₹ 4

**Answer:**

**(i) Consider x% of 50 kg as 65 kg**

x% of 50 kg = 65 kg

x/ 100 × 50 = 65

x/ 2 = 65

x = 130

Therefore 130% of 50 kg is 65 kg.

**(ii) Consider x% of ₹ 9 is ₹ 4**

x% of ₹ 9 = ₹ 4

x/ 100 × 9 = 4

x = 4 × 100/9

x = 400/9

x = 44 4/9

Therefore, 44 4/9 % of ₹ 9 is ₹ 4

**Question 10.**

(i) If 16 2/3 % of a number is 25, find the number.

(ii) If 13.25 % of a number is 159, find the number.

**Answer:**

**(i) Consider the number as x**

16 2/3 % of x = 25

50/3 % of x = 25

50/3 × 1/100 of x = 25

x = (25 × 3 × 100)/ 50

x = 150

Therefore, the number is 150.

**(ii) Consider the number as x**

13.25% of x = 159

13.25/ 100 of x = 159

x = (159 × 100)/ 13.25

Multiply and divide by 100

x = (159 × 100 × 100)/ 1325

x = (159 × 4 × 100)/ 53

x = 3 × 4 × 100

x = 1200

Therefore, the number is 1200.

**Question 11.**

(i) Increase the number 60 by 30 %

(ii) Decrease the number 750 by 10%

**Answer:**

**(i) New number = (1 + 30/100) of 60**

= (1 + 3/10) × 60

= 13/10 × 60

= 78

**(ii) New number = (1 – 10/100) of 750**

= (1 – 1/10) × 750

= 9/10 × 750

= 9 × 75

= 675

**Question 12.**

(i) What number when increased by 15% becomes 299?

(ii) On decreasing the number by 18%, it becomes 697. Find the number.

**Answer:**

**(i) Consider the original number as x**

New number = (1 + 15/100) of original number

299 = (1 + 3/20) × x

Taking LCM

299 = [(20 + 3)/ 20] × x

299 = 23/20 × x

x = (299 × 20)/ 23

x = 13 × 20

x = 260

Therefore, the original number is 260.

**(ii) Consider the original number as x**

New number = (1 – 18/100) of original number

697 = (1 – 18/100) of x

Taking LCM

697 = [(100 – 18)/ 100] × x

697 = 82/100 × x

x = (697 × 100)/ 82

x = (697 × 50)/ 41

x = 17 × 50

x = 850

Therefore, the original number is 850

**Question 13.**** Mr. Khanna spent 83% of his salary and saved ₹ 1870. Calculate his monthly salary.**

**Answer:**

Mr. Khanna spent 83% of his salary

Savings = 100 – 83 = 17%

So 17% of his salary = ₹ 1870

His salary = ₹ (1870 × 100)/ 17

= ₹ 11000

**Question 14.**** In school, 38% of the students are girls. If the number of boys is 1023, find the total strength of the school.**

**Answer:**

No. of girls in school = 38%

No. of boys in school = (100 – 38) % = 62%

Consider x as the total strength of school

62% of x = 1023

62/100 × x = 1023

x = 1023 × 100/62

x = 1023 × 50/31

x = 33 × 50

x = 1650

Therefore, the total strength of the school is 1650.

**Percentage and its Applications Exe-7.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-114

**Question 15.**** The price of an article increases from ₹ 960 to ₹ 1080. Find the percentage increase in the price.**

**Answer:**

Increase in the price of an article = 1080 – 960 = ₹ 120

Percentage increase in the price = 120/960 × 100%

= 1/8 × 100%

= 100/8 %

= 25/2 %

= 12.5 %

**Question 16.**** In a straight contest, the loser polled 42% votes and lost by 14400 votes. Find the total number of votes polled. If the total number of eligible voters was 1 lakh, find what percentage of voters did not vote.**

**Answer:**

Losing candidate got 42% of the votes polled

Votes secured by winning candidate = (100 – 42) % of the votes polled

= 58 % of the votes polled

So the difference of votes = 58% – 42%

= 16% of the votes polled

16% of the votes polled = 14400

16%/100 of the votes polled = 14400

So the votes polled = 14400 × 100/16

= 900 × 100

= 90000

Total number of eligible voters = 100000

No. of voters who did not vote = 100000 – 90000

= 10000

Percentage of voters did not vote = [10000/ 100000 × 100] %

= 10000/1000 %

= 10 %

**Question 17.**** Out of 8000 candidates, 60% were boys. If 80% of the boys and 90% of the girls passed the exam, find the number of candidates who failed.**

**Answer:**

Total number of candidates = 8000

No. of boys = 60% of 8000

= 60/100 × 8000

= 60 × 80

= 4800

No. of girls = 8000 – 4800 = 3200

No. of passed boys = 80% of No. of boys

= 80/100 × 4800

= 80 × 48

= 3840

No. of passed girls = 90% of No. of girls

= 90/100 × 3200

= 90 × 32

= 2880

No. of passed candidates = 3840 + 2880 = 6720

No. of failed candidates = 8000 – 6720 = 1280

Hence, the number of candidates who failed is 1280.

**Question 18.**** On increasing the price of an article by 16%, it becomes ₹ 1479. What was its original price?**

**Answer:**

Consider the original price of an article = ₹ x

1479 = (1 + 16/100) of original price

1479 = [(100 + 16)/ 100] × ₹ x

1479 = 116/100 × x

116x/100 = 1479

By separating the terms

x = (1479 × 100)/ 116

x = (1479 × 25)/ 29

x = 51 × 25

x = 1275

Hence, the original price of an article is ₹ 1275.

**Question 19.**** Pratibha reduced her weight by 15%. If now she weighs 59.5 kg, what was her earlier weight?**

**Answer:**

Weight reduced by Pratibha = 15%

Present weight of Pratibha = 59.5 kg

Consider her original weight = 100

Reduced weight = 100 – 15 = 85%

85% of her original weight = 59.5 kg

So her original weight = (59.5 × 100)/ 85

= 0.7 × 100

= 70 kg

**Question 20. ****In a sale, a shop reduces all its prices by 15%. Calculate:**

(i) the cost of an article which was originally priced at ₹ 40.

(ii) the original price of an article which was sold for ₹ 20.40.

**Answer:**

Rate of reduction = 15%

**(i) Original price of an article = ₹ 40**

Rate of reduction = 15%

Reduction = (40 × 15)/ 100 = ₹ 6

So the sale price = 40 – 6 = ₹ 34

**(ii) Sale price = ₹ 20.40**

Rate of reduction = 15%

Cost price = (SP × 100)/ (100 – reduction %)

= (20.40 × 100)/ (100 – 15)

= (2040 × 100)/ (100 × 85)

= ₹ 24

**Question 21.**** Increase the price of ₹ 200 by 10% and then decrease the new price by 10%. Is the final price same as the original one?**

**Answer:**

Rate of increase = 10%

Rate of decrease = 10%

Price of article = ₹ 200

Increased price = ₹ 200 × (100 + 10)/ 100

= ₹ 200 × 110/100

= ₹ 220

Decreased price = ₹ 200 × (100 – 10)/ 100

= ₹ 220 × 90/100

= ₹ 198

**No, the final price is not as same as the original one.**

**Question 22.**** Chandani purchased some parrots. 20% flew away and 5% died. Of the remaining, 45% were sold. Now 33 parrots remain. How many parrots had Chandani purchased?**

**Answer:**

Consider Chandani purchased x parrots

No. of parrots flew away = 20% of x

= 20/100 × x

= 1/5 × x

= x/5

No. of parrots died = 5% of x

= 5/100 × x

= x/20

No. of parrots remaining = x – (x/5 + x/20)

Taking LCM

= x – [(4x + x)/ 20]

= x – 5x/20

= x – x/4

Taking LCM

= (4x – x)/ 4

= 3x/4

No. of sold parrots = 45% of 3x/4

= 45/100 × 3x/4

= 9/20 × 3x/4

= 27x/80

No. of parrots which are not sold = 3x/4 – 27/80

Taking LCM

= (60x – 27x)/ 80

= 33x/80

33x/80 = 33

By cross multiplication

33x = 33 × 80

x = (33 × 80)/ 33

x = 80

hence, Chandani purchased 80 parrots.

**Question 23.**** A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks, gets 18 more marks than the minimum pass marks. Find the maximum marks and the percentage of pass marks.**

**Answer:**

Consider x as the maximum marks

Marks secured by the first candidate = 36% of x

= 36/100 × x

= 36x/ 100

Marks secured by another candidate = 43% of x

= 43/100 × x

= 43x/ 100

The qualifying marks are same for both the candidates

36x/100 + 24 = 43x/100 – 18

24 + 18 = 43x/100 – 36x/ 100

Taking LCM

42 = (43x – 36x)/ 100

42 = 7x/100

x = 42 × 100/7

x = 6 × 100

x = 600

Here the maximum marks = 600

Marks secured by first candidate = 36/100 × 600 = 36 × 6 = 216

Qualifying marks = 216 + 24 = 240

So the percentage of qualifying marks = (240/600 × 100) %

= 240/6 %

= 40 %

therefore, the maximum mark is 600 and the percentage of pass marks is 40%.

**Question 24.**** The side of rectangle are 20 cm and 15 cm. If each side is increased by 20 %, find the percentage increase in the area.**

**Answer:**

Length of rectangle is 20cm

Breadth is 15cm

Area of rectangle is length *breadth

Area is 20*15 = 300cm².

A1= 300 cm²

If the length and breadth are increased by 20percent

Then length = 20+ 20*0.20= 24cm

Breath = 15+15*0.20= 18Cm.

Area is 24*18 =432cm²

A2= 432cm²

Difference between two areas is = 432–300

132cm²

Percentage of area increased is (132/300)*100

**Is 44% Ans **

**Question 25.**** Imran gives 1 % of his monthly income to his two sons as pocket money. The elder son gets 80 % of the total amount given and he spends 80 % of his share. If he saves 60 Rs. per month determine imrans’s monthly income.**

**Answer:**

Let, Imran’s monthly income be *x*

Pocket money given to the two sons = *x*/100

Money given to the elder son = (80/100 x *x*/100)

Savings on his share = (100-80)% = 20%

ATP

**Imran’s monthly income is ₹37500.**

— : End of ML Aggarwal Percentage and its Applications Exe-7.1 Class 8 ICSE Maths Solutions :–

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