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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.
∴ Votes Received by B
= 72000 × 100 – 60/100
= 720 × 40
= 28800
Therefore, the number of votes received by B will be 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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