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8 COMPARING QUANTITIES

Exercise 8.1

Q.1. Find the ratio of the following. (a) Speed of a cycle 15 km per hour to the speed of

scooter 30 km per hour. (b) 5 m to 10 km (c) 50 paise to Rs 5

Ans. (a) Required ratio = Speed of cycleSpeed of scooter

= 15 1 30 2

=

= 1 : 2

(b) Required ratio = 5 m10 km

(1 km = 1000 m)

= 5 m10000 m

= 1 m2000 m

= 1 : 2000

(c) Required ratio = 50 paiseRs 5

(1 Rs = 100 paise)

= 50 paise500 paise

= 110

= 1 : 10



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Q.2. Convert the following ratios to percentages. (a) 3 : 4 (b) 2 : 3

Ans. (a) Ratio = 3 : 4

Fraction = 34

∴ Required percentage = 3 10 × 100 100

0 = 75%

(b) Ratio = 2 : 3

Fraction = 23

(Divide and Multiply by 100)

Required percentage = 2 100 200 1 × = × 3 100 3 100

= 2663

%

Q.3. 72% of 25 students are good in mathematics. How many are not good in mathematics?

Ans. Percentage of students who are good in mathematics = 72% Thus, percentage of students not good in mathematics = (100 – 72)% = 28% Hence, number of students not good in mathematics = 28% of 25

= 28100

× 25 = 7

Q.4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Ans. Let total number of matches played by the football team = x Win percentage = 40 ⇒ 40% of x = 10



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⇒ 40100

× x = 10

∴ x = 10 × 10040

= 25

Hence, total number of matches they played = 25.

Q.5. If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning?

Ans. Let Chameli had Rs x in the beginning Chameli spent 75% of her money. So, percentage of left money = (100 – 75)% of her money = 25% of her money ⇒ 25% of x = Rs 600

⇒ 25100

× x = Rs 600

∴ x = 600 × 10025

= 2400

Hence, Chameli had Rs 2400 in the beginning.

Q.6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.

Ans. Let the total number be 100 Percentage of people who like cricket = 60% Percentage of people who like football = 30% Percentage of people who like other game

= 100 – (60 + 30) = 10%



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If total no. of people = 50 lakh Number of people who like cricket = 60% of 50 lakh

= 60100

× 50 lakh = 30 lakh

Number of people who like football = 30% of 50 lakh

= 30100

× 50 lakh = 15 lakh

∴ Number of people who like other game = 10% of 50 lakh

= 10100

× 50 lakh = 5 lakh

Hence, the number of people who like cricket is 30 lakh, people who football is 15 lakh and the people who like other games is 5 lakh.

Exercise 8.2

Q.1. A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his original salary.

Ans. 10% increase in salary = 100 + 10100

× 100 = 100 + 10 = 110

If previous salary is Rs 100, then increased salary = Rs 110 If new salary is Rs 110, then original salary is Rs 100 If new salary is Rs 1,54,000, then original salary

= 100110

× 154000 = 1,40,000

Hence, original salary = Rs 1,40,000 OR Let, x be the original salary increased in salary = 10% of



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x = 10100

x

New salary = Original salary + Increased in salary

⇒ 154000 = x + 10100

x

⇒ 154000 = 1110

x

∴ x = 154000 × 1011

= 148000

Hence, Original salary is Rs 1,40,000

Q.2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

Ans. Number of people visiting zoo on Sunday = 845 persons Number of people visiting zoo on Monday = 169 persons Decrease in the number of people visiting zoo on Monday = 845 – 169 = 676 Required percentage decrease in number of visitor

= Decrease in the number of people visiting zoo on MondayNumber of visitors on Sunday

= 676845

× 100 = 80%

Q.3. A shopkeeper buys 80 articles for Rs 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Ans. C.P. of all 80 articles = Rs 2400 16% profit means if C.P. is Rs 100, then S.P



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= 100 + 16 = Rs 116

If C.P. is Rs 2400, then S.P. = 116100

× 2400 = Rs 2784

S.P. of 80 articles = Rs 2784

S.P. of one article = Rs 278480

= Rs 34.80

Hence, S.P. of one article = Rs 34.80 OR

CP of 80 articles = 2400 Profit = 16%

Then SP of 80 articles = 2400 + 2400 × 16100

= 2400 + 384 = 2784 S.P of 80 articles = Rs 2784

S.P of article = Rs 278480

= Rs 34.80

Q.4. The cost of an article was Rs 15,500. Rs 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Ans. C.P. of an article = Rs 15500 Charge on its repairing = Rs 450 So, the amount he had to pay = Rs 15500 + Rs 450 = Rs 15950 At profit of 15% means if C.P. is Rs 100, then S.P. = 100 + 15 = Rs 115



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If C.P. is Rs 15950, then S.P.

= 115100

× 15950

= Rs 18342.50 Hence, S.P. = Rs 18342.50

Q.5. A VCR and TV were bought for Rs 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Ans. CP of T.V. = Rs 8000 ------------- (i) CP of VCR = Rs 8000 ------------- (ii) 4% loss on VCR means If CP is Rs 100, then S.P = 100 – 4 = Rs 96

So, if CP is Rs 8000, S.P. will be = 96100

× 8000 = Rs 7680

∴ SP of VCR = Rs 7680 ------------ (iii) 8% profit on TV means If CP is Rs 100, then S.P. = 100 + 8 = Rs 108

So, if CP is Rs 8000, SP will be = 108100

× 8000 = Rs 8640

SP of TV = Rs 8640 Overall CP on TV and VCR = Rs 8000 + Rs 8000

= Rs 16,000 Overall SP on TV and VCR = Rs 7680 + Rs 8640

= Rs 16320 ⇒ Overall S.P. > Overall C.P So, there is gain of 16320 – 16000 = Rs 320



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∴ Gain % = gain 100total c.p

⎛ ⎞×⎜ ⎟

⎝ ⎠%

= 320 × 10016000

⎛ ⎞⎜ ⎟⎝ ⎠

= 2%

Q.6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

Ans. 10% discount means if an article of M.P. Rs 100, then S.P. = Rs 100 – 10 = Rs 90 So, if M.P. for a pair of jeans is Rs 1450, then S.P.

= Rs 90100

× 1450 = Rs 1305

M.P for two shirts = Rs 850 × 2 = Rs 1700 If M.P for two shirts is Rs 1700, then S.P.

= Rs 90100

× 1700 = Rs. 1530

Hence, the amount the customer will have to pay for jeans and two shirts

= Rs 1305 + Rs 1530 = Rs 2835.

Q.7. A milkman sold two of his buffaloes for Rs 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint : Find CP of each)

Ans. S.P. of each buffalo = Rs 20,000 Case I 5% gain means = profit of Rs 5 on Rs 100



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If SP is of Rs 105 then CP = Rs 100

If SP is Rs 20,000 then CP = Rs 100105

× 20,000

= Rs 19047.6190 ∴ CP of one of buffalo = Rs 19047.62 Case II 10% loss on other buffalo means If SP is Rs 90 then CP is Rs 100

If SP is Rs 20, 000 then CP = Rs 10090

× 20,000

∴ CP = Rs 22222.22 Overall SP = Rs 20,000 + Rs 20,000 = Rs 40,000 Overall CP = Rs19047.62+ 22222.22 = Rs 41269.84 ⇒ CP > S.P. So, loss = C.P. – S.P. = Rs 41269.84 – Rs 40000 Hence, the required loss = Rs 1269.84

Q.8. The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Ans. Price of TV = Rs 13,000 Sale tax charged on it = 12% of T.V. price = 12% of Rs 13,000

= Rs 12100

× 13000 = Rs 1560

Amount paid by Vinod for TV



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= Rs 13000 + Rs 1560 = Rs 14560

Q.9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs 1,600, find the marked price.

Ans. At 20% discount if M.P. is Rs 100, then S.P. = Rs (100 – 20)

= Rs 80 If S.P. is Rs 80, then M.P. = Rs 100

If S.P. is Rs 1600, then M.P. will be = Rs 10080

× 1600

= Rs 2000 Hence, marked price of the skates is Rs 2,000

Q.10. I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

Ans. 8% VAT included means Rs 8 is added to original price of Rs 100

If price after including VAT is Rs 108 then original price = Rs 100 If price after including VAT is Rs 5400

then original price = Rs 100108

× 5400 = Rs 5000

Hence, price of the hair-dryer before VAT is Rs 5,000.

Exercise 8.3



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Q.1. Calculate the amount and compound interest on

(a) Rs 10,800 for 3 years at 1122

% per annum

compounded annually.

(b) Rs 18,000 for 122

years at 10% per annum

compound annually.

(c) Rs 62,500 for 112

years at 8% per annum

compounded half yearly. (d) Rs 8,000 for 1 year at 9% per annum compounded

half yearly. (You could use the year by year calculation using SI formula to verify)

(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.

Ans. (a) We know, A = P R1 100

n⎛ ⎞+⎜ ⎟⎝ ⎠

Where, principle P = Rs 10,800,

rate, R = 1122

% = 252

%

Number of years (n) = 3

A = Rs 10,800 3251 +

2 × 100⎛⎜⎝ ⎠

⎞⎟ = Rs 10800

398

⎛ ⎞⎜ ⎟⎝ ⎠

= Rs 10,800 × 9 9 9 8 8 8

× ×

= Rs 10,800 × 729512

= 7873200512

A = Rs 15,377.34 Hence, amount is Rs 15,377.34



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C.P. = A – P = Rs 15377.34 – Rs 10,800 = Rs 4,577.34

(b) A = Rs 18,000 R = 10% n = 122

year

21 × 1010 21 1 +

100 100

⎛ ⎞⎜ ⎟

Amount = Rs 18,000 ⎛ ⎞+ × ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟

⎝ ⎠

= Rs 18,000 211 5× 1 +

10 100⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= Rs 18,000 211 21×

10 20⎛ ⎞⎜ ⎟⎝ ⎠

= Rs 18,000 × 11 11 21× × 10 10 20

= Rs 22,869

Compound interest = Amount – Principle = Rs 4869 (c) P = Rs 62,500, R = 8% per annum = 4% half yearly

n = 1 12

years.

= 3 half yearly Therefore, compounding has to be done 3 times. A = ?, C.I = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 62500 341 +

100⎛ ⎞⎜ ⎟⎝ ⎠



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= 62500 × 26 26 26× × 25 25 25

= Rs 70,304 C.I. = A – P = Rs 70304 – Rs 62500 = Rs 7,804 (d) P = Rs 8000, n = 1 year = 2 half years,

R = 9% per annum = 92

% per half yearly, A = ?, C.I. = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 291 +

2 × 100⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 × 209 209 × 200 200

= Rs 8736.20 C.I. = A – P = Rs 8736.20 – Rs 8000 = Rs 736.20 (e) P = Rs 10000, n = 1 year = 2 half years, R = 8% per annum = 4% per half yearly, A = ?, C.I. = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 10000 241 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 10000 × 26 26× 25 25



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= Rs 10816 C.I. = A – P = Rs 10816 – Rs 10000 = Rs 816

Q.2. Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

(Hint: Find A for 2 years with interest is compounded

yearly and then find SI on the 2 year amount for 412

years).

Ans. P = Rs 26,400, R = 15% per annum,

n = 2 years and 4 months, A = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 26400 2151 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 26400 231 +

20⎛ ⎞⎜ ⎟⎝ ⎠

= 26400 × 23 23× 20 20

= 66 × 23 × 23 = Rs 34914 Rs 34914 would act as principal for next 4 months i.e.

13

year



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So, S.I = P × R × T100

(where T = 13

year)

= 34914 × 15 1× 100 3

= Rs 1745.70 ∴ Amount paid by Kamla after 2 years and 4 months = Rs 34914 + Rs 1745.70 = Rs 36659.70

Q.3. Fabina borrows Rs 12, 500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Ans. Case I For Fabina P = Rs 12,500, R = 12% per annum, T = 3 years

S.I. = P × R × T100

= 12500 × 12 × 3100

= Rs 4500

Case II For Radha P = Rs 12500, R = 10% per annum,

n = 3 years, C.I. = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 12500 3101 +

100⎛ ⎞⎜ ⎟⎝ ⎠



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= 12500 × 11 11 11× × 10 10 10

= Rs 16637.50 ∴ C.I. = A – P = Rs 16637.50 – Rs 12500 = Rs 4137.50 ⇒ Interest of Fabina > Interest of Radha Difference = (Rs 4500 – Rs 4137.50) = Rs 362.50 Hence, Fabina paid of Rs 362.50 more than Radha.

Q.4. I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Ans. P = Rs 12000, R = 6% per annum, T = 2 years, S.I. = ?, C.I. = ?

Case I : In case of simple interest :

S.I. = P × R × T100

= 12000 × 6 × 2100

= Rs 1440

Case II : For compound interest :

A = P nR1 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 12000 261 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 12000 × 106 106 × 100 100



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= Rs 13483.20 ∴ Amount = Rs 13483.20 ⇒ C.I. = A – P = Rs 13483.20 – Rs 12000 = Rs 1483.20 So, C.I. > S.I. So, extra amount to be paid in case of C.I. = Rs 1483.20 – Rs 1440 = Rs 43.20 Hence, I will have to pay Rs 43.20 extra.

Q.5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months (ii) after 1 year?

Ans. P = Rs 60,000 R = 12% per annum compounded half yearly R = 6% half yearly ⇒ A6 = ? (Where A6 = Amount after 6 month) A12 = ? (A12 = Amount after 1 year) T6 = 6 months = 1 half year T12 = 1 year = 2 half years (i) Amount after 6 months

⇒ A6 = P 6R1 +

100

T⎛ ⎞⎜ ⎟⎝ ⎠

= 60000 161 +

100⎛ ⎞⎜ ⎟⎝ ⎠



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= 60000 × 106100

= Rs 63,600

Hence, amount after 6 months is Rs 63,600.

(ii) Amount after 1 year, A12 = P 12R1 +

100

T⎛ ⎞⎜ ⎟⎝ ⎠

= 60000 261 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 60000 × 106 106 × 100 100

= Rs 67416 A12 = Rs 67,416 Hence, amount after 1 year is Rs 67416.

Q.6. Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in

amounts he would be paying after 112

years if the

interest is (i) compounded annually. (ii) compounded half yearly.

Ans. P = Rs 80000, R = 10% per annum,

T = 112

years = 3 half years

(i) When the rate of interest is compounded annually

A = P R1 + 100

T⎛⎜⎝ ⎠

⎞⎟ (Where T = 1)

= 80000 1101 +

100⎛ ⎞⎜ ⎟⎝ ⎠



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= 80000 × 1110

= Rs 88000 Rs 88000 would be principal for half year.

Calculate S.I. for 12

year

S.I. = P × R × T

= 88000 × 10 1× 100 2

= Rs 4400

∴ C.I. for 112

on given principal

= (Rs 88000 – Rs 80000) + Rs 4400 = Rs 12400 Amount = Principal + Compound interest ⇒ A = P + C.I. = Rs 80000 + Rs 12400 = Rs 92400 When interest is compounded half yearly.

(ii) A = P R1 + 100

T⎛ ⎞⎜ ⎟⎝ ⎠

= 80000 351 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 80000 × 21 21 21× × 20 20 20

= Rs 92610

Q.7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find



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(i) The amount credited against her name at the end of the second year.

(ii) The interest for the 3rd year.

Ans. P = Rs 8000, R = 5% per annum, n1 = 2 year, n2 = 3 year, A2 = ?, A3 = ?

(i) A2 = Amount after 2 years = P 1R1 +

100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 251 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 × 21 21× 20 20

= Rs 8,820 Hence, amount credited after 2 years = Rs 8820

(ii) A3 = Amount after 3 years = P 2R1 +

100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 351 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 8000 × 21 21 21× × 20 20 20

= Rs 9261 ∴ Interest for 3rd year = A3 – A2

= Rs 9261 – Rs 8820 = Rs 441.

Q.8. Find the amount and the compound interest on

Rs 10,000 for 112

years at 10% per annum, compounded



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half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans. P = Rs 10000, n = 112

years = 3 half years

R = 10% per annum = 5% half yearly A = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 10000 351 +

100⎛ ⎞⎜ ⎟⎝ ⎠

Amount = 10000 × 21 21 21× × 20 20 20

= Rs 11576.25

⇒ Amount = Rs 11576.25 ∴ C.I. = A – P Hence, interest = Rs 11576.25 – Rs 10000 = Rs 1,576.25

Q.9. Find the amount which Ram will get on Rs 4096, if he

gave it for 18 months at 1122

% per annum, interest

being compounded half yearly.

Ans. P = Rs 4096, n = 18 months = 3 half years

R = 1122

% per annum = 254

% per half year

A = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠



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= 4096 3251 +

4 × 100⎛ ⎞⎜ ⎟⎝ ⎠

= 4096 × 17 17 17× × 16 16 16

= Rs 4913.

Q.10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001. (ii) what would be its population in 2005?

Ans. (i) P = 54000, R = 5% per annum, A = ?, n = 2 years Population before 2 years

A = P R1 – 100

T⎛⎜⎝ ⎠

⎞⎟ = 54000

2S1 – 100

⎛ ⎞⎜ ⎟⎝ ⎠

= 54000 × 19 19× 20 20

= 48735

Hence, the population in 2001 is 48735 (ii) For population in 2005 i.e., 2 years after 2003 P = 54000, n = 2 years, R = 5% per annum, A = ?

A = P R1 + 100

T⎛ ⎞⎜ ⎟⎝ ⎠

= 54000 2R1 +

100⎛ ⎞⎜ ⎟⎝ ⎠

= 54000 × 21 21× 20 20

= 59535

Hence, population in 2005 will be 59535.

Q.11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour.



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Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Ans. (i) Initial count of bacteria, P = 5,06,000, n = 2 hour Rate of increasing bacteria, R = 2.5% per hour

= 52

% per hour, A = ?

A = P R1 + 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= 506000 251 +

2 × 100⎛ ⎞⎜ ⎟⎝ ⎠

= 506000 × 41 41× 40 40

= 531616.25 = 531616 (approx)

Hence, count of bacteria after 2 hour = 531616. (approx)

Q.12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans. P = Rs 42,000, R = 8% per annum, n = 1 year, A = ?

A = P R1 – 100

n⎛ ⎞⎜ ⎟⎝ ⎠

= Rs 42,000 181 –

100⎛ ⎞⎜ ⎟⎝ ⎠

= Rs 42,000 × 92100

= Rs 38640.00 Hence, value of the scooter after one year = Rs 38640.00

8 COMPARING QUANTITIES - Testlabztestlabz.com/ModelPapers/1_6_28_180.pdf · Speed of scooter = 15 1...

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