# Frank Learning Maths Class 8 Solutions Chapter 9

## Frank Learning Maths Class 8 Solutions Chapter 9 Comparing Quantities

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 9 Comparing Quantities. Here students can easily find step by step solutions of all the problems for Comparing Quantities. Here students will find solutions for Exercise 9.1 and 9.2. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Comparing Quantities Exercise 9.1 Solution

Question no – (1)

Solution :

(a) 75 paise : 3 rupees

= 75/100 rupees : 3 rupees

= 3 : 12

= 1 : 4

(b) 2 hours 10 min : 26 min

= 2 × (60) + 10 : 26

= 130 : 262

= 5 : 1

(c) 8 months : 3 years

= 8/12 year : 3 years

= 2 : 9

(d) 45 m : 1 km 350 m

= 450 m : (1000 m + 350 m)

= 450 m : 350 m

= 1 : 3

Question no – (2)

Solution :

As per the question,

= Rs 8100 in the ratio of 7 : 2

= 8100 × 7/9 = 6300

= 8100 × 2/9 = 1800

Question no – (3)

Solution :

(a) 3 : 5

= 3/5 × 100

= 60%

(b) 7 : 8

= 7/8 × 100

= 165/2

= 82 1/2%

(c) 5 : 12

= 5/12 × 100

= 125/3

41 2/3%

(d) 11 : 15

= 11/15 × 100

= 220/3

= 66 2/3%

Question – (4)

Solution :

First, Failed candidates,

= (100 – 72)%

= 28%

Let, total candidates = x

28% = 504

= x × 28/100 = 504

= x = 504 × 200/28

= 1800

Therefore, total number of candidates are 1800

Question – (5)

Solution :

= 35 × (100 – 60)/100

= 35 × 40/100

= 14

Therefore, 14 students are not good in mathematics.

Question – (6)

Solution :

Let, total number of working days is x

x × 75/100 = 192

= x = 192 × 100/75

= x = 256 days

Therefore, the total number of working days of the school is 256 days.

Question – (7)

Solution :

Let, total of matches x,

= x × (100 – 60)/100 = 24

= x × 40/100 = 24

= x = 24 × 10/4

= 60

They played 60 matches

The required fraction,

= 24/60

= 2/5

Hence, they lost 2/5 matches.

Question – (8)

Solution :

Let, he have x mangoes originally,

x × (100 – 60)/100 = 576

= x × 40/100 = 576

= x = 576 × 100/40

= x = 1440

Therefore, the fruit seller have 1440 mangoes originally.

Question – (9)

Solution :

Let, he had x rupees originally.

x × 100 – 85/100 = 810

= x = 810 × 100/15

= x = 5400

Therefore, Mr. Sharma had Rs. 5400 originally

Question – (10)

Solution :

Let, her monthly income x

x × (100 – 30 – 60)/100 = 7560

= x × 10/100 = 7560

= x = 7560 × 100/10

= x = 75600

Hence, Ameeta’s monthly income will be Rs. 75600

Question – (11)

Solution :

As per the given question,

Mango trees = 7x

Coconut trees = 12x

7x = 259

x = 37

12 × 37

= 444

The total number of trees in garden,

= (259 + 444)

= 703

Therefore, the total numbers of trees in the garden will be 703

Question – (12)

Solution :

Let, the total distance x

x × (100 – 50 – 35 – 10)/ 100 = 144

= x × 5/100 = 144

= x = 144 × 100/5

= 2880 km

By air,

= 2880 × 50/100

= 1440 km

By train,

= 2880 × 35/100

= 1008 km

By bus,

= 2880 × 10/100

= 288 km

Question – (13)

Solution :

Total vote,

= 80000 × 90/900

= 72000 give vote.

= 72000 × 100 – 60/100

= 720 × 40

= 28800

Question – (14)

Solution :

Percentage who like other game,

= (100 – 65 – 12 – 15)%

= 8%

People who like Cricket,

= 3600000 × 65/100

= 2341000

People who like hockey,

= 3600000 × 12/100

= 432000

People who like tennis ,

= 3600000 × 15/100

= 540000

People who like other games,

= 3600000 × 8/100

= 288000

Question – (15)

Solution :

Let the number of mango trees x

50, 000 × (100 – 12 – 15 – 10)/100 = x

= x = 5000 × 63

= 31500

Therefore, the number of mango trees are Rs 31500

Comparing Quantities Exercise 9.2 Solution

Question no – (1)

Solution :

Population after 2 year,

= 1,80,000 (1 + 10/100)2

= 180000 × (11)2/100

= 1800 × 11 × 11

= 2, 17, 800

Therefore, the population of the town after 2 years will be 2, 17, 800

Question no – (2)

Solution :

People decreasing in 1 month,

= (34000 – 32,300) = 1700

= 1 month

= 1/12 year

Decrease percent,

= (1700/34000 × 100)%

= 5%

Therefore, the decrease percent in the number of visitor will be 5%

Question no – (3)

Solution :

According to the given question,

Aman’s monthly income is = Rs 32,000

Income increases every year by = 9%,

Monthly income after 2 year,

= 32000 × (1 + 9/100)2

= 32000 × (109)2/10000

= 32 × 109 × 109/10

= 380192/10

= 38019.20

Hence, Aman’s monthly income after 2 years will be 38019.20 Rs.

Question no – (4)

Solution :

Let, his salary x

44800 = x (1 + 12/100)

= 44800 = x × 112/100

= x = 44800 × 100/112

= x = 40000 Rs

Therefore, Anmol’s salary was Rs 40,000 before the increase.

Question no – (5)

Solution :

Vale of a machine depreciates (decreases) every year by = 5%

Present value of the machine is = Rs 27,500

The value of the machine after 2 years,

= 27500 × (1 – 5/100)²

= 27500 × 95 × 95/100 × 100

= 2481875/100

= 24818.75 Rs

Therefore, the value of the machine after 2 years will be 24818.75 Rs

Question no – (6)

Solution :

Let the population at 2010 is = x

1, 32300 = x (1 + 5 / 100)²

= 132300 = x 105 × 105 / 100 × 100

= x = 132300 × 100 × 100/105 × 105

= x = 120000

Therefore, population of the town in the year 2010 was 120000.

Question no – (7)

Solution :

Let the number be 100

20% increased = 20/100 × 100 = 20

New number – 120

Decreased by 20%

= 920/100 × 120

= 24

New number,

= 120 – 24

= 96

Net decrease,

= 100 – 96

= 4

Net decrease percent (%),

= 4/100 × 100

= 4%

Therefore, the net decrease percent will be 4%

Question no – (8)

Solution :

Let Alok’s salary 100

Arijit’s salary,

= (100 × 125/100)

= 125

Alok’s salary is 1, then Arijit’s salary,

= 100/125

If Alok’s salary is 100 then Arijit’s salary,

= 100 × 100/125

= 80

Hence, Alok’s salary is,

= (100 – 80)%

= 20% less then Arijit’s salary.

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Updated: June 5, 2023 — 8:48 am