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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.

**Previous Chapter Solution : **