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**Brilliant’s Composite Mathematics Class 7 Solutions Chapter 8 Percentage and Some Applications**

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Brilliant’s Composite Mathematics Class 7 Math Book, Chapter 8, Percentage and Some Applications. Here students can easily find step by step solutions of all the problems for Percentage and Some Applications, Exercise 8.1, 8.2, 8.3 and 8.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

**Percentage and Some Applications Exercise 8.1 Solution**

**Question no – (1)**

**Solution : **

Given, 5/8

= 5/8 × 100

= 62.5%

**Question no – (2)**

**Solution : **

Given, 13/625

= 13/625 × 100

= 2.08%

**Question no – (3)**

**Solution : **

Given, 0.75

= 75/100 × 100

= 75%

**Question no – (4)**

**Solution : **

Given, 0.05

= 5/100 × 100

= 5%

**Question no – (5)**

**Solution : **

Given, 11.40

= 1140/100 × 100

= 1140%

**Question no – (6)**

**Solution : **

Given, 5 : 12

= 5/12 × 100

= 41.67

**Question no – (7)**

**Solution : **

Given, 32% into simple fraction.

= 32/100

= 4/25

**Question no – (8)**

**Solution : **

Given, 56% into simple fraction.

= 56/10

= 14/25

**Question no – (9)**

**Solution : **

Given, 52% into a decimal fraction.

= 52/10

= 0.52

**Question no – (10)**

**Solution : **

Given, 0.8% into a decimal fraction.

= 8/100

= 0.08

**Question no – (11)**

**Solution : **

Given, 14% into ration.

= 14/100

= 7 : 50

**Question no – (12)**

**Solution : **

Given, 33 1/3% into ratio.

= 100/3 × 100

= 1 : 3

**Question no – (13) **

**Solution : **

**(i) 32% of Rs 850**

= 32/100 × 850

= 272

**(ii) 36% of 5 kg**

= 36/100

= 9/5

**Question no – (14) **

**Solution : **

**(i) 0.5% of x = 3**

or, 5/100 × x = 3

or, x = 3 × 100/5

= 60

**(ii) 33 1/3% of x = 15**

= or, 100/3 × x = 15

or, x = 15 × 3/100

= 9/20

**(iii) 120% of x = 36**

or, 120/100 × x = 36

or, x = 36 × 100/120

= 30

**(iv) 15% of x = Rs 24**

or, 15/100 × x = 24

or, x = 24 × 100/15

= 160

**Question no – (15) **

**Solution : **

Let, monthly income = x

According to questions,

x × 15% = 60

or, x × 15/100 = 60

or, 60 × 100/15

or, x = 400 Rs

Therefore, his monthly income will be 400 Rs.

**Question no – (16) **

**Solution : **

Let, some = 576 – 426

= 150

**∴** Same % = 150 × 100/576

= 26.04%

Therefore, he save 26.04% of his income.

**Question no – (17) **

**Solution : **

As per the question,

A girl obtained = 85%

Total of 500 marks

**∴** Girl get Mark,

= 500 × 85/10

= 425

Therefore, she will get 425 marks.

**Question no – (18) **

**Solution : **

Let, monthly income = x

According to questions,

12 1/2% = 460

or, x × 25/2 × 100 = 460

or, x = 640 × 2 × 100/25

= 3680 Rs.

Hence, her monthly income will be 3680 Rs.

**Question no – (19) **

**Solution : **

Let, total income = x

According to questions

x × 18/100 = 360

or, x = 360 × 100/18

= 2000 Rs.

**∴** Rent payment = 2000 × 11/100

= 220 Rs.

Therefore, the payment for rent will be 220 Rs.

**Question no – (20) **

**Solution : **

Part of copper = 13

Part of Zine = 7

Part of nickel = 5

Total part = 13 + 7 + 5 = 25

Copper % = 13/25 × 100 = 52%

Zine % = 7/25 × 100 = 28%

Nickel = 5/25 × 100 = 20%

**Question no – (21) **

**Solution : **

Total alloy = x

According to question,

39/x × 100 = 30

or, x = 39 × 100/302 = 130

Now, Let, Zinc latter consist = y

Again, According to questions

= y/130 × 100 = 70

or, y = 70 × 130/100

= 91

Therefore, there are 91 parts of zinc.

**Question no – (22) **

**Solution : **

1 kg = 100 gm

Calcium contains 1000 × 10/100 = 100 gm

Carbon contains = 1000 × 3/100 = 30 gm

Oxygen = 1000 × 12/100 = 120 gm

**Question no – (23) **

**Solution : **

We know, 9 kg = 9000 gm

Nitre contains,

= 9000 × 75/100

= 7650 gm

Sulphur contains,

= 9000 × 10/100

= 900 gm

Rest part charcoal,

= 900 – (6750 + 900)

= 9000 – 7650

= 1350 gm

**Question no – (24) **

**Solution : **

Per gold in 22 carat gold

= 22/24 × 100

= 91.67%

Therefore, percentage of pure gold will be 91.67%

**Question no – (25) **

**Solution : **

Let, total mark = x

According to question,

x × 86.875/100 = 695

or, x = 695 × 1/86.50

= 800

Hence, the total number of marks of the examination is 800

**Question no – (26) **

**Solution : **

Number of mango trees,

= 2000 × 12/100

= 240

Number of lemon trees,

= 2000 × 18/100

= 360

Number of orange trees,

= 2000 – (360 + 24)

= 2000 – 600

= 1400

Therefore, the number of orange trees will be 1400.

**Question no – (27) **

**Solution : **

Let, maximum pass mark = x

According to question,

= x × 36/100 = (123 + 39) = 162

or, x = 162 × 100/36

= 450

Thus, the maximum marks will be 450

**Question no – (28) **

**Solution : **

Let, total students = x

Number of boys = 70x/100

Number of girls = (100 – 70) = 30%

According to question,

= x × 30/100 = 504 × 100/30 = 1680

**∴** Number of boys

= 1680 × 70/100

= 1176

Therefore, the number of boys will be 1176

**Question no – (29) **

**Solution : **

Let, Total days the school opened = x days

According to question,

x × 90/100 = 216

or, x = 216 × 100/90

= 240 days

Therefore, the school will opened for 240 days.

**Question no – (30) **

**Solution : **

Let, population 1 year ago = P_{1}

P = 185220

As we know that,

= p_{1 }× (910 + 5)/100 = 185220

or, p_{1} = 185220 × 100/105

= 220500

Hence, the population one year ago was 220500

**Question no – (31) **

**Solution : **

1^{st} charge by commission

= 5000 = 5000 × 5/100

= 250

Remainder S.P.,

= 35500 – 5000

= 30500

**∴** 2 1/2% = 5/2% of remainder SP

= 3050 × 5/2 × 100

= 762.5

**∴** Total commission

= 762.5 + 250

= 1012.5

**∴** Total commission %

= 1012.5/35500 × 100

= 2.85%

Therefore, his average percentage commission will be 2.85%

**Question no – (32) **

**Solution : **

No of apples = x

According to question,

x × 60/1000 = 420

or, x = 420 × 100/60

= 700

Hence, the number of apples he had originally was 700.

**Question no – (33) **

**Solution :**

Let, Gross value = x

Commission of 5% of x = 5x/100

According to question,

x – 5x/100 = 15200

or, 100x – 5x/100 = 15200

or, 95x = 15200 × 100

or, x = 16.15200 × 100/95

= 16000 Rs

Therefore, its gross value will be 16000 Rs.

**Question no – (34) **

**Solution :**

Let, 1 year ago population = P

According to question,

p (100 + 8)/100 = 142500

or, P = 142500 = 75x/10 = 219

or, x = 219 × 100/75

= 292 days

= 100/108

= 73200

Hence, a year ago its population was 73200.

**Question no – (35) **

**Solution :**

Let, school opened for x days

According to question,

= 75x/10 = 219

or, x = 219 × 100/75

= 292 days

Therefore, the school was opened for 292 days.

**Question no – (36) **

**Solution :**

Value after 2 years

= 225000 (100-20)/100 × (100-20)/100

= 225000 × 80/100 × 80/100

= 144000 Rs.

Therefore, the value of the car after two years will be 144000 Rs.

**Question no – (37) **

**Solution :**

Let, Thomas’s income = 100

John’s income,

= 100 + (20/1000 × 100)

= 100 + 20 + 120

Difference of both income,

= 120 – 100

= 20%

Required Percent (%),

= 20/120 × 100

= 50/3

= 16 2/3%

Hence, the income of Mr. Thomas was 16 2/3% less.

**Percentage and Some Applications Exercise 8.2 Solution**

**Question no – (1) **

**Solution : **

As per the question,

Ram buys a fan for = Rs 155,

and sells it at a gain of = 20%.

**∴** SP of fan

= 155 × 120/100

= 186 Rs.

Therefore, the selling pries of the fan will be 186 Rs.

**Question no – (2) **

**Solution : **

Given, Profit = 8%,

Selling price = Rs 2160.

**∴** Cost price of article

= 2160 × 100/108

= 2000 Rs.

Therefore, the cost price of the article 2000 Rs.

**Question no – (3) **

**Solution : **

**∴** Selling Price of paper,

= 50 × 120/100

= 60

**∴** Price of paper,

= 60/20

= 3

Hence, he should he sell Rs 3 per quire.

**Question no – (4) **

**Solution :**

Given, Selling Price = 3852

Let, cost = CP

According to question,

Gain % = (SP – CP)/CP × 100

or, 7 = (3852 – CP) × 100/CP

or, 7 CP = 385200 – 100 CP

or, 100 CP + 7 CP = 385200

or, 107 CP = 385200

or, CP = 385200/107 = 3600

**∴** Gain %

= 4050 – 3600/3600 × 100

= 450/3600 × 100

= 12.5%

Therefore, his gain per cent will be 12.5%

**Question no – (6) **

**Solution :**

Let, CP = 100 × 100 – 5 × 1140

= 140 × 1140/95

= 1200

Now, Profit = 5%

**∴** SP_{1} = (100 + 5)/100 × 1200

= 105/100 × 120

= 1260 Rs.

Therefore, the original price of the cycle was 1260 Rs.

**Question no – (7) **

**Solution :**

As per the given question,

Cost Price of 1 lemon = 1/6

Selling Price of 1 lemon = 1/5

**∴** Profit = SP = CP

= 1/5 – 1/6

= 6- 5/30

= 1/30

**∴** Now the Profit %

= 1/30/1/6 × 100

= 6/30 × 100

= 20%

Therefore, the gain percent will be 20%

**Question no – (8) **

**Solution :**

According to the question,

Cost Price of 1 orange = 1/52

Selling Price of 1 orange = 3/5

**∴** Profit,

= 3/5- 1/2

= 6 – 5/10

= 1/10

**∴** Now the Profit %

= 1/10/1/2 × 100

= 2/10 × 100

= 20%

Therefore, his gain per cent will be 20%

**Question no – (9) **

**Solution :**

Let, CP of 1 article = 1

Cost Price of 12 article = 12

Selling Price of 15 article = 15

**∴** Gain = 15 – 12 = 3

**∴** Gain % = 3/12 × 100

= 25%

Hence, his gain per cent will be 25%

**Question no – (10) **

**Solution :**

Cost Price of 1 article = 1

Cost Price of 15 article = 15

= SP of 12 article

Loss = 15 – 12 = 3

Loss = 15 – 12 = 3

**∴** Loss % = 3/15 × 100

= 20%

Therefore, the loss per cent will be 20%

**Question no – (12) **

**Solution :**

504 oranges = 504/12 = 42 dozen

250 oranges = 240/12 = 20 dozen

Cost Price of 42 dozen oranges = 42 × 4 = 168

Selling Price of 20 dozen oranges = 20 × 5 = 100

**∴** Selling Price of (42 – 20) = 22 dozen oranges = 22 × 6 = 132

Total SP of 42 oranges = (100 + 132)

= 232

**∴** CP > SP

**∴** Gain = 232 – 168 = 64

**∴ **Gain % = 64/168 × 100

= 38.09%

Therefore, his gain per cent will be 38.09%

**Question no – (13) **

**Solution :**

Let, Selling Price of 1 table = x

Selling Price of 20 table = 20 x

Selling Price of 4 table = 4x

Cost Price of 20 table = 12000

According to question,

= 20x – 1200 = 4x

or, 20x – 4x = 12000

or, 16x = 12000

or, x = 12000/16

= 750 Rs.

Therefore, the S.P. of 1 table will be 750 Rs.

**Question no – (15) **

**Solution :**

Cost Price of 1 tables = 40/100 = 2/5

Selling Price of 20 tables

= 2/5 × 20 = 2/5 × 20 × 5/100

= 8 + 2 × 20 × 5/5 × 100 = 8 + 2/5

= 40 + 2/5

= 42/5

**∴** Whole Profit = 20%

**∴** Profit = 20/100 × 40 = 8

**∴** Cost Price of (100 – 20) = 80

= Tables = 80 × 2/5 = 38

**∴** Profit = (8 – 2/5)/38 × 100

= 38/5/38 × 100

= 38 × 100/38 × 5

= 20%

Therefore, he must sells the remainder at 20% gain.

**Question no – (16) **

**Solution :**

Cost Price of 80 kg rice @ 6.75 = 6.75 × 80 = 540

Cost Price of 120 kg rice @ 8

= 8 × 120 = 960

**∴** Total Cost Price of (120 + 80)

= 200 rice mix = 540 + 960 = 1500

**∴** Total Cost Price of (120 + 80)

= 1 rice mix = 1500/200

= 7.5 kg

**∴** Total Selling Price of (120 + 80)

= 1 rice = 7.5 × 120/100

= 9 kg

**Question no – (17) **

**Solution :**

Cost Price of the watch_{1}

= 100/(100 + 10) × 1188

= 10/110 × 1188

= 1080 Rs

Cost Price of the watch_{1}

= 100/100 – 10 × 1188

= 100/90 × 1188

= 1320 Rs

Selling Price of the watch_{1 }+ watch_{2}

= 1188 × 2

= 2376 Rs

Cost Price of the watch_{1} + watch_{2}

= 1080 + 1320

= 2400 Rs

**∴** CP > SP,

**∴** Loss = 2400 – 2376 = 24

**∴** Loss = 24/2400 × 100

= 1%

**Question no – (18) **

**Solution :**

Let, Cost Price of watch = 100

Profit = 6 1/4% = 25/4 %

**∴** SP_{1} of watch = (100 + 25/4/100) × 100 = 425 + 25/4 = 425/4

= 106.25

Now, Loss = 6 1/4% = 25/4%

**∴** SP_{2} of watch,

= 9100 – 25/4)/100 × 100 – 25/4

= 400 – 25/4

= 375/4

= 93.75

**∴** SP_{1} – SP_{2} = 106.25 – 93.75 = 12.50

**∴** When (SP – SP_{2}) = 12.50 then CP = 100

**∴ **(SP_{1} – SP_{2}) = 1 = then CP = 100/12.50

**∴** (SP_{1} – SP_{2}) = 95 then CP

= 100/12.5 × 7

= 560 Rs

Therefore, the cost price of the watch will be 560 Rs.

**Question no – (19) **

**Solution :**

Cost Price of 6 lemon = 1

**∴** Cost Price of 1 lemon = 1/6

**∴** Cost Price of 8 lemon 1/9

Cost Price of 8 lemon = 1/8

Selling Price of 8 lemon = 1

**∴** SP of 1 lemon = 1/8

**∴** Total = 1/6 + 1/9 = 9 + 6/54

= 15/54 = 5/18

**∴** Total SP = 1/8 × 2 = 1/4

**∴** Loos = 5/18 – 1/4 = 20 – 5/72 = 5/24

**∴** Loss = 5/24 × 100/5/18

= 5 × 100 × 18/5 × 24

= 75%

Therefore, his loss per cent will be 75%

**Question no – (20) **

**Solution :**

Let, Cost Price of watch Ram = x

**∴** Selling Price of watch Ram

= x/10 + x = x + 20/10

= 11x/10

**∴** Cost Price of watch Mohan = 11x/10

Again of watch Mohan,

= 11/10 x + 4/100 × 11/10x

= 11/10x (1 + 4/100)

= 11/10x × (100 + 4)/100

= 11x × 104/100

Now, Cost Price of watch Sunil = 11x × 104/100

According to question,

11x × 104/100 = 1430

or, x = 1430 × 1000/11 × 1054

= 1250 Rs

Therefore, Ram purchase it in 1250 Rs.

**Question no – (21) **

**Solution :**

Selling Price of TV = 18000 × 95/100 = 17100

Selling Price of W.M 12000 × 120/100 = 14400

Total Cost Price = 18000 + 12000 = 30000

Total Selling Price = 17100 + 14400 + 14400 = 31500

Gain = 31500 – 30000 = 1500 %

**∴** Now the gain%,

= 1500/30000 × 100

= 5%

Therefore, his total gain percent will be 5%.

**Question no – (22) **

**Solution :**

Let, CP of pump-set = x

manufacturer = 10%

**∴** CP_{1} of manufacturer = 100x/100

wholesale = 15% gain

**∴** CP_{2} of wholesale = 115/100 × 110x/100 = 253x/200

retail dealer = 25% gain

**∴** CP_{3} of dealer = 125/100 × 253x/200

**∴** According to question,

= 125/100 × 253x/200 = 1265

or, x = 1265 × 200 × 100/253 × 125

= 800

Therefore, the cost of production of a pump-set will be Rs 800.

**Percentage and Some Applications Exercise 8.3 Solution**

**Question no – (1) **

**Solution : **

Given, Market Price = 14200,

Discount = 10%

**∴** Now, the Cost Price,

= 90/100 × 14200

= 127800 Rs.

Therefore, retailer pay for the car 127800 Rs.

**Question no – (2) **

**Solution : **

Given, Cost Price = 153, gain = 20%

**∴** Selling Price = 120/100 × 153 = 183.6

**∴** Let, MP = x, dis = 15%

According to question,

= x – 15x/100 = 183.6

or, 100x – 15x/100 = 183.6

or, 85x = 183.6 × 100

or, x = 183.6 × 100/85

= 216 Rs.

Hence, The tradesman should mark the article at 216 Rs.

**Question no – (3) **

**Solution : **

As per the question,

Market Price = 240,

Discount = 30%

**∴** Cost Price of machine,

= 70/100 × 240

= 168 Rs.

Thus, the retailer pay for the machine 168 Rs.

**Question no – (4) **

**Solution : **

Let, good quality = x

Let, CP = 100

**∴** SP = 100 + 30 = 130

CP of x good = 100x

Now, SP_{1} = 130x × 1/2 = 65x [∴ 1/2 of good 130 sells]

again Now, SP of remaining = x/4 good

15% = (100 – 15)/100 × 130 = 85/100 × 130

= 110.50

**∴** SP_{2 }= 1/4 × 110.5- × X = 110.50X/4 = 27.625x

SP of last x/4 good 30% = 70/100 × 130 = 91

**∴** SP_{3} = 1/4 × 91 × x = 22.75x

**∴** Total SP= 65x + 27.625x + 22.75 = 115.375x

**∴** Profit = SP – CP = 115.37x – 100x = 15.375x

**∴** Profit% = 15.375/100x × 100

= 15.375%

Therefore, his gain per cent altogether will be 15.375%.

**Question no – (5) **

**Solution : **

According to the given question,

MP = 12600

dis_{1} = 5%

**∴** Cost Price = 12600 × 95/100

= 119700

dis_{2} = 2%

**∴** Cost Price = 11970 = 98/100

= 11730 Rs.

Hence, cash payment 11730 Rs is to be made for buying it.

**Question no – (6) **

**Solution : **

As per the given question,

Market Price = 800,

discount = 40%

**∴** Cost Price = 800 × 60/100

= 480

Selling Price = 600

**∴** gain = 600 – 480 = 120

**∴** gain% = 120/480 × 100

= 25%

Therefore, his gain per cent will be 25%.

**Question no – (7) **

**Solution : **

Let, add price = x

Profit % = 20%

According to question,

= 20/100 × x = 60

or, x 60 × 100/20

= 300 Rs.

Therefore, his advertised price of a cycle will be 300 Rs.

**Question no – (8) **

**Solution : **

As per the question,

Cost Price = 36,

Profit = 10%

Selling Price,

= 36 × 110/100

= 39.60

Market Price,

= 39.60 × 100/90 [∴ Dis = 10%]

= 44 Rs.

Therefore, its marked price will be 44 Rs.

**Question no – (10) **

**Solution : **

Let, MP = 100

Successive dis_{1} = 1%

**∴** SP_{1} = 100 × 90/10 = 90

Successive dis_{2} = 6%

**∴** SP_{2} =m 90 × 94/100 = 84.6

Another, Discount = 15%

**∴** SP 15% = 100 × 85/100

= 85 Rs

Therefore, Single discount 15% is batter.

**Percentage and Some Applications Exercise 8.4 Solution**

**Question no – (1) **

**Solution : **

**(a) Principal = Rs = 500;**

Rate = 8% per annum;

Time = 3 years

= I = PRT/100 = 500 × 8 × 3/100 = 120

**∴** A = 500 + 120

= 620 Rs.

**(b) Principal = Rs 1000;**

Rate = 13% per annum;

Time = 2 years

= I = 100 × 13 × 2/100 = 260

**∴** A = 1000 + 260

= 1260 Rs.

**(c) Principal = Rs 400;**

Rate = 20 paise per rupee per annum ;

Time = 6 months

= I = PRT/100 = 400 × 20 × 1/2/100

= 400 × 20/100 × 2 = 40

**∴** A = 400 + 40

= 440 Rs.

**(d) Principal = Rs 300;**

Rate = 12% per annum;

Time = 8 months

= I = PRT/100

= 300 × 12 × 8/100 × 12 = 24

**∴** A = 300 + 24

= 324 Rs.

**Question no – (2) **

**Solution : **

As per the given question,

Simple Interest = 150,

Rate = 10%

Time = 5, P = ?

**∴** Simple Interest = PRT/100

or, P = SI × 100/R × T

= 150 × 100/10 × 5

= 300 Rs.

Therefore, the required sum is 300 Rs.

**Question no – (3) **

**Solution : **

Given, Principal = 1500,

Rate = 5%,

Time = 1 year 4 month = 12 + 4

= 16 month = 16/12 = 4/3 year

**∴** Simple Interest = PRT/100

= 150 × 5 × 4/100 × 3

= 100 Rs.

Let, Sum P_{1}, R_{1} = 6.52% T = 2

**∴** SI = PR_{1}T/100 = P × 6.25 × 2/100

or, P = SI × 100/6.25 × 2 = 100 × 100/6.225 × 2

= 800 Rs.

Therefore, the required sum will be 800 Rs.

**Question no – (4) **

**Solution : **

Let, Time = T

R = 12.5%, P = 8000, SI = 2500

According to question,

Simple Interest = PRT/100

or, T = SI × 100/P × R

= 2500 × 100/8000 × 12.5

= 2.5 Years

Therefore, the required time will be 2.5 Years.

**Question no – (5) **

**Solution : **

Let, Principle = P, Time = T

R = 18.75% A = 2 × P = 2P

**∴** S.I = 2P – P = P

**∴** According to question,

S.I = PRT/100

or, P = PRT/100

or, T = 100/R = 100/18.75 = 16/3

= 5 years 4 months

Hence, in 5 years 4 months the sum of money will double.

**Question no – (6) **

**Solution : **

As per the given question,

P = 2500, R = 10.5%, T = 4 years

**∴ **S.I = PRT/100 = 2500 × 10.5 × 4/100

= 1050

Now, R_{1} = 9%, T_{1} = ? SI = 1050

According to question,

S.I = PRT/10

or, T_{1} = SI × 100/PR

= 1050 × 100/2500 × 9 = 203

= 6 years 8 months

Hence, the required time will be 6 years 8 months

**Question no – (7) **

**Solution : **

Let, P = 100

Sum double in 8 years,

**∴** A = 2 ∴ 100 = 200

**∴** SI = A – P = 200 – 100 = 100

Let, rate = 2%

According to questions

Simple Interest = PRT/100

or, R = SI × 100/P × T = 100 × 100/100 × 8

= 12.5%

Therefore, the rate per cent per annum will be 12.5%.

**Question no – (8) **

**Solution : **

Let, P = 100

**∴** Sum treble in 16 years

**∴** A = 100 × 3 = 300

**∴** SI = 300 – 100 = 200

Let, rate = R%

According to the question,

S.I = PRT/100

or, R = SI × 100/P × T = 2090 × 100/100 × 16

= 12.5%

Therefore, the rate percent will be 12.5%

**Question no – (9)**

**Solution : **

T= 2, P = 800, A = 1000, R = ?

**∴** S.I = 1000 – 800 = 200

**∴** S.I = PRT/100

or, R = SI × 100/P × T

= 200 × 100/800 × 2

= 12.5%

Hence, the required rate percent will be 12.5%.

**Question no – (10) **

**Solution : **

Let, Rate = R%

**1 ^{st}** SI = P

_{1}RT

_{1}/10 = 2000 × R × 3/100

**2**^{nd }SI = P_{2}RT_{2}/100 = 1600 (R + 2) × 3/100

According to question,

= (20 × R × 3) + (16 × (R + 2) × 3] = 996

or, 60R + 48R + 96 = 996

or, 108R = 996 – 96 = 900

or, 108R = 996 – 96 = 900

or, R = 900/108 = 25/3

= 8 1/3%

Therefore, the two rates of interest will be 8 1/3%

**Question no – (11) **

**Solution : **

Let, principle = P

SI = PR_{1}T_{1}/100, SI = PR_{2}T_{2}/100

Now, according to the question,

PR_{1}T/100 – PR_{2}T/100 = 420

or, P × 12 × 3/100 – P × 8 × 3/100 = 4200

or, 36P – 24P = 420 × 100

or, 12P = 420 × 100

or, P = 240 × 100/12

= 3500 Rs.

Therefore, the Sum will be 3500 Rs.

**Question no – (12) **

**Solution : **

Let, Sum = 100

**1 ^{st}** case = SI

_{1}= PRT/100 = 100 × 12 × 2.5/100 = 30

**2 ^{nd}** case = SI

_{2}= PRT/100 = 100 × 10 × 3.5/100 = 35

**∴** SI_{2} – SI_{1} = 35 – 30 = 5

When, Difference = 5, Then principle = 100

**∴** When, Difference = 1, then principle = 100/5

**∴** When, Difference = 40, then principle,

= 100 × 40/5

= 800 Rs.

Therefore, the required Sum will be 800 Rs.

**Question no – (13) **

**Solution : **

Let, Principle = P

For, 1^{st} case = SI = PTR/100 = P × 1 × 15/100 = 15P/100

For, 2^{nd} case, P_{1} = (1550 – P)

**∴** SI = P_{1}RT/100 = (1550 – P) × 1 × 24/100

= 37200 – 24P

According to question,

15P/100 + (37200 – 24P/100) = 300

or, 15P + 37200 – 24P = 300 × 100

or, 37200 – 9P = 30000

or, 9P = 37200 – 30000 = 7200

or, P = 7200/9 = 800

P_{1} = 1500 – 800

= 750 Rs.

**Question no – (14) **

**Solution : **

Let, first part be x Rs

Second part = (3000-x) Rs

S.I = x × 8 × 7/100 = 8x/25 Rs

S.I on second part

= (3000-x) × 9 × 2/100

= 27000-9x/50 Rs

Therefore, 8x/25 = 27000-9x/50

= 16x = 27000-9x

= 25x = 27000

= x = 27000/25

= 1080 Rs

**∴** 1st part = 1080 Rs

**∴** 2nd part (3000 – 1080) Rs = 1920 Rs

**Question no – (15) **

**Solution : **

Let, Principle = P, T_{2} = 3y + 4m = 3.33y

**∴** SI_{1} = PR_{1}T_{1}/100 = P × 8 × 2/100 = 2P/5

**∴** SI_{2} = PR_{2}T_{2}/100

= P × 18 × 3.33/100

= 360P/100

= 3P/5

**∴** SI_{2} = SI_{1} = 3P/5 – 2P/5 = P/5

According to question,

P/5 = 200

or, = 200 × 5

= 1000 Rs.

Therefore, the required sum will be 1000 Rs.

**Question no – (16) **

**Solution : **

A_{1} = P_{1} + SI_{1} = 2800 [After 4 years]

A_{2} = P_{2} + SI_{2} = 2200 [After 1 years]

**∴** SI_{1} – SI_{2} = SI_{3} = 2800 – 2200 = 600

**∴** SI_{3} for 3 year = 600

**∴** SI_{3} for 1 year = 600/3 = 200

**∴** P = 2200 – 200 [for 1 year]

= 2000 Rs

Now, SI = PRT/100

or, R = SI × 100/PT

= 100 × 200/2000 × 1

= 10%

Therefore, the sum of money is 2000 Rs and the rate of interest is 10%.

**Question no – (17) **

**Solution : **

Let, Principle for **1 ^{st}** case = P

Principle for **2 ^{nd}** case = P

_{1}= (3300 – P)

S.I for **1 ^{st}** case = P × 6 × 9/100 × 2

S.I for **2 ^{nd}** year case = (3300 – P) × 5 × 9/100 × 2

According to question,

= P × 6 × 9/100 × 2 = (3300 – P) × 5 × 9/100 × 2

or, 6P = 16500 – 5P

or, 6P + 5P = 16500

or, 11p = 1600

or, P = 16500/11

or, P = 1500

**∴** P_{1} = 3300 – 1500

= 1800 Rs.

**Question no – (18) **

**Solution : **

For 1^{st} year,

S.I = PRT/100 = 300 × 8 × 1/100 = 240

A = 3000 + 240 = 3240

**For 2 ^{nd} year,**

Due to repaid P new = 3240 – 1200 = 2040

SI = PRT/100 = 2040 × 8 × 1/100 = 163.2

**∴ **A after 2^{nd} year

= 2040 + 163.2

= 2203.2

Again, he repaid and P new_{1}

= 2203.2 – 1300

= 903.2 Rs

**For 3 ^{rd} year,**

S.I = PRT/100

= 903.2 × 8 ×1/100

= 72.256

**∴** A to clear off the debt

= 903.2 + 72.256

= 975.46 Rs

Therefore, he should pay Rs. 975.46

**Next Chapter Solution : **

👉 Chapter 9 👈

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