# Frank Learning Maths Class 8 Solutions Chapter 10

## 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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Updated: June 5, 2023 — 9:13 am