Warning: Undefined array key "https://nctbsolution.com/brilliants-composite-mathematics-class-7-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
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 = P1
P = 185220
As we know that,
= p1 × (910 + 5)/100 = 185220
or, p1 = 185220 × 100/105
= 220500
Hence, the population one year ago was 220500
Question no – (31)
Solution :
1st 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%
∴ SP1 = (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 watch1
= 100/(100 + 10) × 1188
= 10/110 × 1188
= 1080 Rs
Cost Price of the watch1
= 100/100 – 10 × 1188
= 100/90 × 1188
= 1320 Rs
Selling Price of the watch1 + watch2
= 1188 × 2
= 2376 Rs
Cost Price of the watch1 + watch2
= 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 %
∴ SP1 of watch = (100 + 25/4/100) × 100 = 425 + 25/4 = 425/4
= 106.25
Now, Loss = 6 1/4% = 25/4%
∴ SP2 of watch,
= 9100 – 25/4)/100 × 100 – 25/4
= 400 – 25/4
= 375/4
= 93.75
∴ SP1 – SP2 = 106.25 – 93.75 = 12.50
∴ When (SP – SP2) = 12.50 then CP = 100
∴ (SP1 – SP2) = 1 = then CP = 100/12.50
∴ (SP1 – SP2) = 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%
∴ CP1 of manufacturer = 100x/100
wholesale = 15% gain
∴ CP2 of wholesale = 115/100 × 110x/100 = 253x/200
retail dealer = 25% gain
∴ CP3 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, SP1 = 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
∴ SP2 = 1/4 × 110.5- × X = 110.50X/4 = 27.625x
SP of last x/4 good 30% = 70/100 × 130 = 91
∴ SP3 = 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
dis1 = 5%
∴ Cost Price = 12600 × 95/100
= 119700
dis2 = 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 dis1 = 1%
∴ SP1 = 100 × 90/10 = 90
Successive dis2 = 6%
∴ SP2 =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 P1, R1 = 6.52% T = 2
∴ SI = PR1T/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, R1 = 9%, T1 = ? SI = 1050
According to question,
S.I = PRT/10
or, T1 = 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%
1st SI = P1RT1/10 = 2000 × R × 3/100
2nd SI = P2RT2/100 = 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 = PR1T1/100, SI = PR2T2/100
Now, according to the question,
PR1T/100 – PR2T/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
1st case = SI1 = PRT/100 = 100 × 12 × 2.5/100 = 30
2nd case = SI2 = PRT/100 = 100 × 10 × 3.5/100 = 35
∴ SI2 – SI1 = 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, 1st case = SI = PTR/100 = P × 1 × 15/100 = 15P/100
For, 2nd case, P1 = (1550 – P)
∴ SI = P1RT/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
P1 = 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, T2 = 3y + 4m = 3.33y
∴ SI1 = PR1T1/100 = P × 8 × 2/100 = 2P/5
∴ SI2 = PR2T2/100
= P × 18 × 3.33/100
= 360P/100
= 3P/5
∴ SI2 = SI1 = 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 :
A1 = P1 + SI1 = 2800 [After 4 years]
A2 = P2 + SI2 = 2200 [After 1 years]
∴ SI1 – SI2 = SI3 = 2800 – 2200 = 600
∴ SI3 for 3 year = 600
∴ SI3 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 1st case = P
Principle for 2nd case = P1 = (3300 – P)
S.I for 1st case = P × 6 × 9/100 × 2
S.I for 2nd 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
∴ P1 = 3300 – 1500
= 1800 Rs.
Question no – (18)
Solution :
For 1st year,
S.I = PRT/100 = 300 × 8 × 1/100 = 240
A = 3000 + 240 = 3240
For 2nd year,
Due to repaid P new = 3240 – 1200 = 2040
SI = PRT/100 = 2040 × 8 × 1/100 = 163.2
∴ A after 2nd year
= 2040 + 163.2
= 2203.2
Again, he repaid and P new1
= 2203.2 – 1300
= 903.2 Rs
For 3rd 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 👈
Very helpful site..👍