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Frank Learning Maths Class 8 Solutions Chapter 10 Commercial Mathematics
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 10 Commercial Mathematics. Here students can easily find step by step solutions of all the problems for Commercial Mathematics. Here students will find solutions for Exercise 10.1 and 10.2. 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.
Commercial Mathematics Exercise 10.1 Solution
Question no – (1)
Solution :
According to the question,
Shopkeeper buys 80 articles for = Rs 2,400
Sells them for a profit of = 16%.
∴ Profit 16%
∴ Selling Price = 2400 × 116/100
= 2784 Rs
∴ 80 articles for
= 2784 Rs
∴ 1 articles for,
= 2784/80
= 348/10
= 34.8 Rs
Therefore, the selling price of one article is Rs. 34.8
Question no – (2)
Solution :
After repairs the article was,
= (15500 + 450)
= 15950
∴ Selling Price = 15950 × 115/100
= 183425/10
= 18342.50
Therefore, the selling price of the article is Rs. 18342.50
Question no – (3)
Solution :
Selling Price = (2 × 1980) = 3960
10% profit Cost Price,
= 1980 × 1100/110
= 1782
10% loss Cost Price,
= 1980 × 100/90
= 2200
∴ Total Cost Price,
= (1800 + 2200)
= 4000
∴ Loss,
= (4000 – 3960)
= 40
∴ Loss% = loss/C.P × 100
= 40/4000 × 100
= 1%
Question no – (4)
Solution :
100 kg rice = 924 (Selling Price)
Cost Price = 924 × 100/112
= 825
(a) 100 kg rice Cost Price = 825
1 kg rice Cost Price = 825/100
= 8.25 rupees
(b) Selling Price of wheat = 924
Cost Price 924 × 100/88
= 1050
∴ 1 kg of wheat 1050/100 = 10.5 rupees
Selling Price = 2 × 924 = 1848
Cost Price = (1050 + 825)
= 1875
(c) Loss [1875 – 1848] = 27
∴ Loss% – 27/1875 × 100
= 1.44%
Question no – (5)
Solution :
GST = 28% of list price,
= 28/100 × 13500
= 3780
Selling Price = (13500 + 3780)
= 17280
Therefore, the selling price of the DVD player will be Rs. 17280
Question no – (6)
Solution :
Let the mark price be x
∴ GST 28%
= 28/100 × x
= 28x/100
∴ S.P = MP + GST
= x + 28x/100
= 128x/100
∴ M.P = 21760 = 128x/100
= 128x = 2176000
= x = 2176000/128
= x = 17000
Therefore, the marked price will be Rs. 17000
Question no – (7)
Solution :
(a) bag for,
= (425 + 425 × 18/100)
= (425 + 76.5)
= 501.5
(b) tea set for,
= (1250 + 1250 × 12/100)
= (1250 + 150) = 1400
(c) bakery products,
= (850 + 850 × 5/100)
= 850 + 42.5
= 892.5
(d) pressure cooker for,
= (3500 + 28/100 × 3500)
= (3500 + 980) = 4480
∴ Total amount,
= (501.5 + 1400 + 892.5 + 4480)
= 7274 Rs
Question no – (8)
Solution :
Cost Price = 1470 Rs
Profit = 10%
Selling Price = 1470 × 110/100 = 1617
Let M.P = x
Discount 12.5% of x
= x × 12.5/1000 = 125x/1000
Selling Price = (x – 1.25x/1000)
= 875x/1000
∴ 875x/1000 = 1617
x = 1617 × 1000/875
= 1848 Rs
Therefore, the marked price of the article will be Rs. 1848
Question no – (9)
Solution :
Market Price = 3200
Discount = 15% of M.P
= 3200 × 15/100
= 480
∴ Selling Price
= (3200 – 480)
= 2720
∴ Profit,
= 13 1/3%
= 110/3 %
∴ Cost Price,
= (100/100 + 40/3) × 2720
= 300/340 × 2720
= 2400 Rs
Therefore, the cost price of a saree will be Rs. 2400
Question no – (11)
Solution :
M.P = 2800
Discount = 8% of M.P
= 2800 × 8/100 = 224
∴ The price,
= (2800 – 224)
= 2576
Sales tax 5%
= 2576 × 5/100
= 128.8
∴ S.P = (2576 + 128.8)
= 2704.8
Thus, the selling price of the bicycle will be Rs. 2704.8
Question no – (13)
Solution :
Price of a dinner set,
= (1888 × 18/100 + 1888)
= 339.84 + 1888
= 2227.84
Therefore, the reduction is Rs. 339.84
Question no – (14)
Solution :
8 – 15
1 – 15/8
= 1.875
= 1.9
12 – 18
1 – 18/12
= 1.5
∴ Loss (1.9 – 1.5) = .4
∴ Loss% = (4/150 × 100)%
= (132/150 × 100)%
= 21.33%
Hence, the loss per cent will be 21.33%
Question no – (15)
Solution :
S.P = 95/100 × 21000
= 19950
Umesh C.P = 19950
Spent = 2450
∴ Price, = 19950 + 2450
= 22400
Profit = 8%
S.P = 108/100 × 22400
= 24,192
Therefore, Praveen will pay Rs 24192 for the scooter.
Question no – (16)
Solution :
1900 × 20/100 = 380 Rs each
760 × 20/100 = 152 rupees each offer
S.P of trouser,
= 2(1900 – 380)
= 2 × 1520
= 3040
S.P of shirt,
= 2(760 – 152)
= 2 × 608
= 1216 rupees.
∴ Total amount,
= (3040 + 1216)
= 4256 rupees.
Commercial Mathematics Exercise 10.2 Solution
Question no – (1)
Solution :
Given, P = 9000, R = 8%, x = year
Compounded halfly,
∴ 1st halfly interest 9000 × 8 × 1/2 / 100 = 360
∴ amount = (9000 + 360) = 9360
∴ 2nd halfly 9360 × 8 × 1/2 /100 = 374.4
∴ C.I = 374.4
∴ Amount = (9360 + 374.4)
= 9734.4
Question no – (2)
Solution :
18 months = 18/12 years
= 1 1/2 years
P = 6000, r = 9% n = 1 1/2 year
Compounded semi annually,
∴ 1st semi annually,
= 6000 × 9 × 1/2 / 100 = 270
Amount = 6270
2nd semi annually,
= 6270 × 9 × 1/2 / 100 = 282.15
Amount = 6552.153rd semi annually
= 6552.15 × 9 × 1/2 / 10000 = 294.85
∴ Amount = (6552.15 + 294.85)
= 6847.00
∴ Compound Interest = (6847 – 6000) = 847
Question no – (3)
Solution :
Given, 9/12 year = 3/4 year, P = 15000, r = 8%
S.I = 15000 × 8 × 1/4 / 100
= 300
∴ amount = 15300
2nd quarter,
= 15300 × 8 × 1/4 / 100 = 306
∴ amount = 15606
3rd quarter,
= 15606 × 8 × 1/4 / 100
= 312.12
Therefore Amount will be = 15918.12
Question no – (4)
Solution :
Amount when compounded annually
= 60000 (1 + 8/100)1
= 60000 (1 + 2/25)1
= 60000 × 27/25
= 64800
Now the Difference,
= 64896 – 64800
= 96 rupees
Therefore, the difference in the amount will be Rs 96.
Question no – (5)
Solution :
Here, P = 50, 000, r = 12%, n = 3
∴ A = 50000 (1 + 12/100)3
= 50000 × 112 × 112 × 112/ 100 × 100 × 100
= 70246.40 Rs
∴ Amount = 70246.40 Rs
∴ Amount after 2 years
= 50000 × (1 + 12/100)2
= 50000 × 112 × 112/100 × 100
= 62720 Rs
∴ Interest 3rd year
= 70246.40 – 62720
= 7526.4 Rs
Question no – (7)
Solution :
Let, the sum of money be x rupees
∴ x (1 + 15/100)2 = 31.740
or, x (115/100)2 = 31,740
or, x (23/20)2 = 31740
or, x = 31740 × 20/23
= 24000 rupees
Therefore, the sum of money will be Rs 24000.
Question no – (8)
Solution :
Let, the time by x years
∴ 8000 (1 + 5/100)x = 9261
or, 8000 (1 + 1/20)x = 9261
or, 8000 (21/20)x = 9261
or, (21/20)x = (9261/800)
∴ x = 3 years
Therefore, the required time will be 3 years.
Question no – (10)
Solution :
C.I – S.I = 500
r = 10%, n = 2, p = x
∴ S.I = x × 10 × 2/100 = 20x/100
A = x(1 + 10/100)2 = x (110/100)2 = x(11/10)2
∴ C.I = x(11/10)2 – x
∴ x 11 × 11/ 10 × 10 – x – 20x/100
= 500
∴ 121x/100 – x – 20x/100 = 500
= 121x – 100x – 20x/100 = 500
= x/100 = 500
= x = 5000
Question no – (11)
Solution :
As per the question,
r = 8%,
P = 75000,
n = 3
∴ A = 75000 (1 – 8/100)3
= 75000 92 × 92 × 92/100 × 100 × 100
= 58401.60
Therefore, its value after 3 years will be Rs. 58401.60
Question no – (12)
Solution :
According to the question,
r = 10%,
P = 90,000,
n = 2
∴ 2018 population,
= 90000(1 + 10/100)2
= 90000 × 110/100 × 110/100
= 108900
Question no – (13)
Solution :
According to the question,
r = 4% per hour
n = 2 hour,
P = 20000
∴ Count of bacteria,
= 20000 (1 + 4/100)2
= 20000 × 104/100 × 104/100
= 21632
Question no – (14)
Solution :
As per the question,
A = 48400,
P = 40000,
r = 10%, n = ?
∴ 48400 = 40000 (1 + 10/100)n
(11/10)n = 48400/40000
(11/10)n = (11/10)2
∴ n = 2 years
Therefore, 2 years time will be taken.
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