BHARATHIDASANAR MATRICULATION HIGHER SECONDARY SCHOOL, ARAKKONAM VIIIth MATHEMATICS _ TERM 3 Chapter 1

LIFE MATHEMATICS

Exercise 1.1 1. Choose the correct answer. (i) There are 5 oranges in a basket of 25 fruits. The Percentage of oranges is ___ (A) 5% (B) 25% (C) 10% (D) 20 % Sol: ×100=20%

Ans=20% (ii) = _______ %. (A) 25 (B) 4 (C) 8 (D) 15

×100=8%

Ans=8%

(iii) 15% of the total number of biscuits in a bottle is 30. The total number of biscuits is _______. (A) 100 (B) 200 (C) 150 (D) 300 3 00 15% of x=30 ×x=30 × X=

Ans=20

(iv) The Price of a scooter was ` 34,000 last year. It has increased by 25% this year. Then the increase in Price is _______. (A) ` 6,500 (B) ` 8,500 (C) ` 8,000 (D) ` 7,000 3400× = 8500 Rs

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

(v) A man saves ` 3,000 Per month from his total salary of ` 20,000. The Percentage Of his savings is _______ . (A) 15% (B) 5% (C) 10% (D) 20 % ×100= =15%

2. (i) 20% of the total quantity of oil is 40 liters. Find the total quantity of oil in liters. Sol:

Let x be a total number of liters of oil 20% of x = 40 × x =40

x=

×

x = 200 liters. liters.

(ii) 25% of a journey covers 5,000 km. How long is the whole journey? Sol:

Let x be a journey 25% of x=5000 ×x=5000 × X=

X=20000 Ans=20000 (iii) 3.5% of an amount is ` 54.25. Find the amount. Sol: Let the amount be x 3.5% of an amount=54.25

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

. ×x=54.25 .× X= . . = x . =

Ans =155 Rs (iv) 60% of the total time is 30 minutes. Find the total time. Sol:

Let the total lines be x 60% of x = 30 minutes × x=30

x=

×

=5 minutes.

(v) 4% sales tax on the sale of an article is ` 2. What is the amount of sale? Sol:

Let the amount of sales be x 4% of an amount=2 × x= 2 × x=

x = 50 Rs. 3. Meenu spends ` 2000 from her salary for recreation which is 5% of her salary. What is her salary? Sol:

Given, 5% of her salary for recreation = Rs 2000 Let x be the total salary

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

5% of salary = Rs 2000 5% × x = 2000 x x = 2000

x=

×

x = 40,000 Therefore, the total salary of menu is Rs 40,000 4. 25% of the total mangoes which are rotten is 1,250. Find the total number of Mangoes in the basket. Also, find the number of good mangoes. Sol:

Given, 25% of rotten mangoes = 1,250 Let x be the total mangoes 25% of x = 1,250 × x = 1250

x=

×

x = 5000 Therefore, total mangoes in the basket is 5000Rs

5. The marks obtained by Rani in her twelfth standard exams are tabulated Below. Express these marks as Percentages. Marks (out of 100) Subjects (i) English (ii) Tamil (iii) Mathematics (iv) Physics (v) Chemistry (vi) Biology

4

Maximum Marks

Marks obtained

200 200 200 150 150 150

180 188 195 132 142 140

Percentage of

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Sol: In English Mark = 180 Total = 200 =

× 100

= 90% In Tamil Mark = 188 Total = 200 =

× 100

= 94% In math Mark = 195 Total = 200 =

× 100

= 97.5 = 98% In Physics Mark = 132 Total = 150 =

× 100

= 88%

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

In Chemistry Mark = 142 Total = 150 =

× 100

= 95% In Biology Mark = 140 Total = 150 =

× 100

= 93.33 = 93%

6. A school cricket team Played 20 matches against another school. The first school won 25% of them. How many matches did the first school win? Sol:

Total matches = 20 The first school won 25% = 25% of 20 =

× 20

= 5 matches

7. Rahim deposited ` 10,000 in a company which Pays 18% simple interest P.a. Find the interest he gets for a Period of 5 years. Sol:

Simple Interest I =

Here P = 10000, r = 18% and n = 5 years

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

I=

××

I = 100×90 I = 9000 Rs 8. The marked Price of a toy is ` 1,200. The shopkeeper gave a discount of 15%. What is the selling Price of the toy? Sol:

M.P = 1200 Rs Discount = 15% S.P =? Discount =

= 15% of 1200

× 1200

= 180 Rs S.P

= M.P – discount

= 1200 – 180 = 1020 Rs

9. In an interview for a computer firm 1,500 applicants were interviewed. If 12% of them were selected, how many applicants were selected? Also find the Number of applicants who were not selected. Sol:

Total number of applicants = 1500 Percentage of selected applicants = 12% 1500 × 12% = 1500 ×

= 180 Not selected candidates = 1500 – 180 = 1320 10. An alloy consists of 30% copper and 40% zinc and the remaining is nickel. Find the amount of nickel in 20 kilograms of the alloy.

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Sol:

Given, copper = 30%, zinc = 40% Nickel = [100 – (30 + 40)] % = 100 – 70 = 30% Amount of nickel in 20Kg of o f the alloy 20 ×

= 2 × 3

=6 Therefore, the amount of Nickel in 20Kg of the alloy = 6Kg

11. Pandian and Thamarai contested for the election to the Panchayat committee from their village. Pandian secured 11,484 votes which was 44% of the total votes. Thamarai secured 36% of the votes. Calculate (i) the number of votes cast in the village and (ii) the number of voters who did not vote for both the contestants. Sol:

Given, Pandiyan’s 44% of the total vote = 11484 Let x be the total votes 44% of x = 11484 × x = 11484 × 100 X=

X=26100 Total votes=26100 Total People=(44%+36%+20%) =100% Not voted for both =20% of total votes =

× 26100

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=5220 votes

12. A man spends 40% of his income for food, 15% for clothes and 20% for house rent and saves the rest. What is the Percentage of his saving? If his income is` 34,400, find the amount of his savings.

Sol: Total =(40+15+20)% =75% Overall Percentage =100% Savings =100-75 =25% Amount of savings = ×34400

=8600 Rs

13. Jyothika secured 35 marks out of 50 in English and 27 marks out of 30 in Mathematics. In which subject did she get more marks and how much? Sol:

In English Mark=35 Total=50 = ×100

=70%

In math Mark=27 Total=30

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

= ×100

=90% She get more marks in math which 20 marks greater than English. 14. A worker receives ` 11,250 as bonus, which is 15% of his annual salary. What is his monthly salary? Sol:

Given,15%of his annual salary = 11250 Let x be the annual salary

× x = 112550 X= × 11250

X=75000 Total salary is Rs 75000 His monthly salary =

=6250

15. The Price of a suit is increased from ` 2,100 to ` 2,520. Find the Percentage of Increase. Sol:

W.K.T Percentage of increase= ×100

Actual Prize=Rs 2100 Increased Prize =Rs 2520 Amount=2520-2100 =420

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=

×100

=20% Exercis e 1.2 1. Find the Cost Price / Selling Price. Cost Price selling Price Profit Loss (i) ` 7,282 ` 208 (ii) ` 572 ` 72 (iii) ` 9,684 ` 684 (iv) ` 1,973 ` 273 (v) ` 6, 76,000` 18,500 i) Sol: C.P=7282 Rs Profit=208 Rs So S.P=C.P+ Profit =7282+208 =7490 Rs ii) S.P=572 Rs Profit=72 Rs C.P=S.P-Profit =572-72 =500 Rs iii) C.P=9684 Rs Loss=684 RS S.P= C.P-loss =9684-684 =9000 Rs iv) S.P=1973 Profit=S.P-Profit =1973-273 =1700 Rs v) C.P=676000Rs Loss= 18500 S.P=S..P-Loss =676000-18500 S.P=657500 Rs

2. Fill up the appropriate boxes and leave the rest. C.P.S.P. Profit & Profit % Loss & Loss%

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

(i) ` 320 ` 384 (ii) ` 2,500 ` 2,700 (iii) ` 380 ` 361 (iv) ` 40 ` (v) ` 5,000 `

2 Loss 500 Profit.

Sol: i) C.P=320 Rs S.P=384 Rs Here S.P>C>P so there t here is a Profit Profit=S.P-C.P =384-320 Profit=64Rs Profit%= ×100 . = ×100 =20% ii C.P= 2500 Rs S.P= 2700Rs Profit=S.P-C.P =2700-2500 =200 Rs Profit%= ×100 = 8% iii) C.P=380 Rs S.P=361 Rs Here C.P>S.P so there is a loss Loss= C.P-S.P =380-361 =19 Rs Loss%= ×100 . = ×100 =5% iv) C.P=40Rs Loss= 2Rs Loss%= ×100 =5% C.P=40Rs

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Loss= 2Rs S.P= C.P-Loss = 40-2 =38 Rs v) C.P=5000 Rs Profit= 500 Rs Profit%=

×100

=10% S.P= C.P+ Profit =5000+50 =5500 Rs

3. Find the S.P. if a Profit of 5% is made on (i) a bicycle of ` 700 with ` 50 as overhead charges. Price of a bicycle=700 Rs Over head charges= 50 Rs C.P=700+50 C.P=750 Rs

Profit%=

×750

=37.50 S.P=C.P+ Profit =750 +37.50 =787.50 Rs

(ii) a computer table bought at ` 1,150 with ` 50 as transportation charges. Price of computer=1150 Rs C.P=1150+50 C.P=1200 Rs S.P= × 1200 1200 =105×12 =1260 Rs

(iii) a table-top wet grinder bought for ` 2,560 and an expense of ` 140 on repair Charges Price of wet grinder=2560 Rs Expenses made on repair =140 Rs C.P=2560+140 C.P=2700 S.P= × 2760 2760

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=105×27 =2835 Rs

4. By selling a table for ` 1,320, a trader gains 10%. Find the C.P. of the table.

S.P of a table =1320 Rs Gain=10% S.P= × 1320 1320 × 1320 1320 = C.P=1200 Rs

5. The cost Price of 16 note books is equal to the selling Price of 12 note books. Find the gain Percent.

Let S.P of each note book be x S.P of 12 note books =12x S.P of 16 note books =16x Given C.P of 16 note books=S.P of 12 note books Cost Price of 16 note books=12x S.P of 16 notebooks =16x Here S.P>C.P Profit= S.P-C.P =16x-14x =4x Profit%= ×100 = =33 % 6. A man sold two articles at ` 375 each. On the first article, he gains 25% and on the other, he loses 25%. How much does he gain or lose in the whole transaction? Also, find the gain or loss Percent in the whole transaction. S.P=first article =375 Rs Gain=25% C.P of the first article= × 375 × 375 = =300 Rs

of second article =375 RS Loss=25% C.P of the second article article = × 375 × 375 = S.P

=500 Rs C.P of two articles =300+500

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=800 Rs S.P of two articles =375 ×2=750 Here C.P > S.P so there

is a loss

Loss=C.P-S.P =800-750 =50 He lost Rs 50 in the whole transaction Loss %= × 100 =6.25 %

7. Anbarasan purchased a house for ` 17,75,000 and spent ` 1,25,000 on its interior decoration. He sold the house to make a Profit of 20%. Find the S.P. of the house.. house Purchase Price of the house=1775000 Amount spent for interior decoration=1775000+125000 =1900000 Profit=20% = × 19000 1900000 00 =380000 S.P of the house=1900000+380000 house=1900000+380000 =2280000

8. After spending Rupees sixty thousand for remodelling a house, Amla sold a house at a Profit of 20%. If the selling Price was Rupees forty two lakhs, how much did she spend to buy the house? S.P of house =4200000 Profit=20% S.P of the house = × 42000 4200000 00 × 420000 4200000 0 = =3500000 Amount spent for remodeling the house =60000

Purchase Price of the house =3500000-60000 =3440000

9. Jaikumar bought a Plot of land in the outskirts of the city for ` 21,00,000. He built a wall around it for which he spent ` 1,45,000. And then he wants to sell it at ` 25,00,000 by making an advertisement in the newspaper which costs him ` 5,000. Now, find his Profit Percent.

Purchase Price of the Plot=2100000 Rs

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Amount spent to build the wall=145000 Rs Amount spent for advertisement =5000 C.P of the Plot =2100000+14500000+5000 C.P=2250000 S.P=2500000 Profit=S.P-C.P =2500000-2250000 =250000 × 100 Profit%= =11 %

10. A man sold two varieties of his dog for ` 3,605 each. On one he made a gain of 15% and on the other a loss of 9%. Find his overall gain or loss. [ Hint: Hint: Find C.P. of each] S.P of first of first dog=3605% Gain=15% C.P of the first dog = × 3605 3605 × 3605 3605 = =3134.78 S.P of second dog=3605 Rs Loss=9% C.P 0f second dog = =

× 3605 3605

× 3605 3605

=3961.54 Rs C.P of the two dogs =3134.78+3961.54 =7096.32 S.P=of the two dogs =3605+3605 =7210 Rs Here S.P>C.P So there is a gain Gain=S.P-C.P =7210-7096.32 =113.68Rs

Exercise 1.3 1. Choose the correct answer: (i) The discount is always on the _______.

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

(A) Marked Price (B) Cost Price (C) Selling Price (D) Interest Ans:Marked Price (ii) If M.P. = ` 140, S.P. = ` 105, then Discount = _______. (A) ` 245 (B) ` 25 (C) ` 30 3 0 (D) ` 35 Ans:35Rs

(iii) ______ = Marked Price – Discount. (A) Cost Price (B) Selling Price (C) List Price (D) Market Price Ans:Selling Price (iv) The tax added to the value of the Product is called ______ Tax. (A) Sales Tax (B) VAT (C) Excise Tax (D) Service Tax Ans:VAT (v) If the S.P. of an article is ` 240 and the discount given on it is ` 28, then the M.P. is _______. (A) ` 212 (B) ` 228 (C) ` 268 (D) ` 258 Ans:268 Rs 2. The Price marked on a book is ` 450. The shopkeeper gives 20% discount on it a in book exhibition. What is the Selling Price?

Marked Price of the book=450Rs Discount=20% = × 450 =90 S.P of the book=450-90 =360 Rs

3. A television set was sold for ` 5,760 after giving successive discounts of 10% and 20% respectively. What was the Marked Price?

Let the marked of the television be x=100Rs First discount =10%= × 100 =10Rs S.P after

the first discount =100-10

=90

Second discount =20%= × 100

=18 Rs So S.P after second discount =90-18 =72 When S.P=72 Rs,

M.P=100 Rs

When S.P=5760 Rs, M.P= × 5760 5760

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=8000 Rs Marked Price of television is 8000Rs

MATHEMATICS 4. Sekar bought a computer for ` 38,000 and a Printer for ` 8,000. If the rate of sales tax is 7% for these items, find the Price he has to Pay to buy these two items. Price of computer =38000 =38000 RS Price of Printer =8000 Rs So the Price of computer and Printer=38000+8000 =46000Rs Sales tax=7% × 4600 46000 0 = =3220 Rs So the Price of both including sales tax=46000+3220 =49220 Rs

5. The selling Price with VAT, on a cooking range is ` 19,610. If the VAT is 6%, what is the original Price of the cooking range?

Let the original Price of the cooking range =100rs VAT=6% So S.P of the cooking range with VAT=100+6 =106 S.P=106Rs ,original Price =100 =100 RS If S.P=19610 RS then Original Price = × 1961 19610 0 =18500 RS The original Price of the cooking range =18500 Rs

6. Richard got a discount of 10% on the suit he bought. The marked Price was ` 5,000 for the suit. If he had to Pay sales tax of 10% on the Price at which he bought, how much did he Pay? Marked Price of the suit =5000 Rs Discount=10% = × 5000 5000 =500 Rs S.P of the suit =5000-500 S.P=4500Rs Sales =10%

= × 4500 4500

=450 Rs So the Purchase Price of the suit =4500+450 =4950 RS

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

7. The sales tax on a refrigerator at the rate of 9% is ` 1,170. Find the actual sale Price. Let the S.P of the refrigerator r efrigerator be x Sales tax =9% = × But sales tax=1170 Rs

× =1170 × 100 x= =>

=13000 Rs

8. A trader marks his goods 40% above the cost Price. He sells them at a discount of 5%. What is his loss or gain Percentage? Let C.P be 100Rs Marked Price =40% above C.P =100+40 =40 Rs Discount =5% × 140 = =7 S.P of the article =140-7 =133 C.P=100 Rs S.P=133 Profit =133-100 =33 RS

× 100 .

Profit%= =33%

9. A T.V. with marked Price ` 11,500 is sold at 10% discount. Due to festival season, the shop keeper allows a further discount of 5%. Find the net selling Price of the T.V. M.P of TV =11500Rs First discount=10%= × 1150 11500 0 =1150 Rs S.P after the first discount =11500-1150 =10350 Second discount =5% =

× 1035 10350 0

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=517.50 So S.P of the T.V after second discount =10350-517.50 =9832.50 The net selling Price of the T.V=9832.50

10. A Person Pays ` 2,800 for a cooler listed at ` 3,500. Find the discount Percent offered.

Marked Price of cooler=3500 RS S.P of cooler=2800 RS Discount =M.P-S.P =3500-2800 700 Rs Discount%= × 100 =20%

11. DeePa Purchased 15 shirts at the rate of ` 1,200 each and sold them at a Profit of 5%. If the customer has to Pay sales tax at the rate of 4%, how much will one shirt cost to the customer?

C.P of 1 shirt =1200Rs Profit=5% = × 1200 1200 =60 Rs So S.P of 1 shirt =1200+60 =1260 Rs Sales tax=4% =

× 1260 1260

=50.40 Rs

So the amount Paid to buy one shirt =1260+50.40 =1310.40 RS 12. Find the discount, discount Percent, selling Price and the marked Price. Sl. No Items M. P Rate of Discount Amount of Discount S. P (i) Saree` 2,300 20% (ii) Pen set ` 140 ` 105 (iii) Dining table 20% ` 16,000 (iv) Washing Machine ` 14,500 ` 13,775 (v) Crockery set ` 3,224 12½%

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

i)M.P of saree =2300 Rs Discount=20% × 2300 2300 = =460Rs Amount of discount =460 RS S.P of saree of =2300-460 =1840 Rs ii) M.P= Pen set =140Rs S.P=105 Rs Discount=140-105 =35 Rs Rate of discount =

× 100

=25% iii) Let the marked Price =100 Rs Discount=20% S.P of dining table =100-20=80

S.P=16000 RS M.P= × 1600 16000 0 If

M.P=2000Rs Now M.P=20000 S.P=16000 Discount=20000-16000 =4000Rs iv) M.P of washing machine =14500 Rs S.P=13775 Rs Discount =M.P- S.P =14500-13775 =725 Rs

× 100 100

Discount%= =5% v)

M.P of crockery crockery set =3224 Rs Rate of discount =12(1\2)%

. × 3224 3224

Discount= =403 Rs

S.P=M.P-discount =3224-403 =2821 Rs

Exercis 1.4 1. Find the Amount and Compound Interest in the following cases:

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Sl. No. Principal in Rupees Rate% Per annum Time in years (i) 1000 5% 3

Find the C.I and amount if compounded annually A=P(1+

ņ )

=1000(1+ )3 =1000(1+ )3 =1000( )3 =1000X( × × )

=9261/8 A=1157.63 C.I=A-P =1157.63

(ii)

A= P(1+

4000

10% 2

ņ )

= 4000(1+

2 )

=4000( )2

=4000X(

× × ) ×

=40X121 A=4840 C.I=A-P Rs.4840-4000 C.I=Rs.840

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

(iii)

18,000 10%

2(1\2)

P=18000 R=10% N=2 ½ years.. A= P(1+

ņ )

=18000(1+

2 ) ( 1 + ½ ( ))

=18000( )2 ( )) = 18000X X( )

= 9X121X21 = 1089X21 Amount=Rs.22869 Compound Interest=A-P =Rs.22869-Rs.18000 =Rs.4869

2. Sangeetha borrowed ` 8,000 from Alex for 2 years at 12½% Per annum. What interest did Sangeetha Pay to Alex if the interest is compounded annually?

Amount borrowed by Sangeetha P=Rs.8000, n=2 years r=12 ½ %. A= P(1+

ņ )

=8000(1+(

))2 ×

=8000(1+ )2

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=8000( )2 =8000X ×

=8000×

=81000/8 A=1012 Amount=Rs.10125 C.I=A-P =Rs.10125-Rs.8000 =Rs.2125 3. Maria invested ` 80,000 in a business. She would be Paid interest at 5% Per annum compounded annually. Find (i) the amount standing to her credit at the end of second year and (ii) the interest for the third year.

P=Rs80,000 N= 2 years R=5% A=80000(1 + =80000(1 +

)

)

=80000X × = 80000X ×

=200X441 A=88200 Amount at the end of second year=Rs.88200 ii. So the interest for third year

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

P=Rs.88200 N=1 year R=5% So interest =88200×

=4410 Amount at the end of second year =88200 Rs Interest for the third years=4410 Rs.

4. Find the compound interest on ` 24,000 compounded half - yearly for 1½ years at the rate of 10% Per annum.

P=24000 Rs N=1 Years r=10% A=P(1 + ) =24000((1 + ) ×× =24000× ×× × =

A=27783 Rs C.I=A-P =27783-24000 =3783 Rs 5. Find the amount that Dravid would receive if he invests ` 8,192 for 18 months at 12½% Per annum, the interest being compounded half - yearly. P=8192 Rs N=1 Years r=12 %

=8192((1 +

25

)

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

×× =8192× ×× ××× =

A=9826Rs

6. Find the compound interest on ` 15,625 for 9 months, at 16% Per annum compounded quarterly.

P=15625Rs N=9months r=16% A=P(1 + ) =15625((1 +

)

×× =15625× ××

=2 6 × 2 6 × 2 6 A=17576Rs C.I=A-P =17576-15625 =1951Rs

7. Find the Principal that will yield a compound interest of ` 1,632 in 2 years at 4% rate of interest Per annum.

A=P+I C.I=A-P A-P=C.I [P(1 + ) -P]=1632 [P(1 +

) -1]=1632

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

[P(1 + ) -P]=1632 [P( ) -1]=1632 [P( -1]=1632 [P( ]=1632 P( )=1632 × P= P=2000Rs

8. Vicky borrowed ` 26,400 from a bank to buy a scooter at the rate of 15% P.a. compounded yearly. What amount will he Pay at the end of 2 years and 4 months to clear the loan?

P=26400 Rs r=15% n=2 Years 4 months =2(4\12) years =2 (1\3)years A= 26400((1 + ) (1 + ) =26400× × × ××× = ××× = =36659.70rs Vicky has to Pay Rs 36659.70 to clear the loan. 9. Arif took a loan of ` 80,000 from a bank. If the rate of interest is 10% P. a., find the difference in amounts he would be Paying after 1½ years if the interest is (i) compounded annually and (ii) compounded half - yearly.

When interested compounded annually P=80000 Rs r=10%

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

n=1 years A=80000[(1+ )(1+ ( )] =80000( )( )

=800× 1 1 × 2 1 A=92400 Rs ii) when interest is is compounded half yearly P=80000 r=10% n=1 years A=80000(1+ )3 =80000(1+ )3 =80000( )3 =

80000( × × )

9261 =10× 9261 =92610 Rs The amount between the amount is Rs 92610-92400 =210Rs 10. Find the difference between simple interest and compound interest on ` 2,400 at 2 years at 5% Per annum compounded annually. The difference between S.I and C.I For 2 years is P( )2 =2400( )2 =2400( ) × =6 The difference between C.I and S.I is Rs 6

11. Find the difference between simple interest and compound interest on ` 6,400

28

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

for 2 years at 6 ¼% P. a. compounded annually. The difference between between S.I and C.I for for 2 Years is is P( )2 =6400( )2 =6400( 2 )

=6400(

× ) ×

=25 Rs

12. The difference between C. I. and S. I. for 2 years on a sum of money lent at 5% P.a. is ` 5. Find the sum of money lent.

Let the Principle be P N=2years R=5% The difference between between S.I and C.I for for 2 Years is is P( )2 2 =P( ) =5 Rs P(

)=5 ×

P=

5×(

× ) ×

P=2000 Rs 13. Sujatha borrows ` 12,500 at 12% Per annum for 3 years at simple interest and Radhika borrows the same amount for the same Period at 10% Per annum compounded annually. Who Pays more interest and by how much?

For sujatha P= 12500 RS n=3years r=12% S.I= ×× = S.I=4500 Rs For radhika P=12500 Rs N=3 years R=10% A=12500(1+ )3 3 = 12500( )

29

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=12500×××

××× =×

A=16637.50 C.I=A-P =16637.50-12500 =4137.50 Rs Sujatha is Paying more i nterest=4500-4137.50 =362.50 Rs

14. What sum is invested for 1½ years at the rate of 4% P.a. compounded half -yearly which amounts to ` 1,32,651? Let the Principal be P N=1 years R=4% A=P(1+ )3 132651=

P(1+ )3

P()3=132651

P= ×××

×× P=2500

× 50

P=125000 So the amount invested =125000

15. Gayathri invested a sum of ` 12,000 at 5% P.a. at compound interest. She received an amount of ` 13,230 after ‘n’ years. Find the value of ‘ n’.

P= 12000 RS n=3years A=13230 Rs A= P(1+ )n P(1+ )n =A (1+ )n=13230 12000 n=13230 ( ) n= ( ) ( )n= 12000

30

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

( )n=( )2

n=2 years

16. At what rate Percent compound interest Per annum will ` 640 amount to ` 774.40 in 2 years?

P=640 Rs A=774.40 N=2 years A= P(1+ )2 774.40= = 640(1+ )2 (1+ )2 = 2= ) (1+ )2= (1+ )2=( )2 == ( ) ( -1) r= ×100

(1+

r=10% 17. Find the rate Percent Per annum, if ` 2,000 amount to ` 2,315.25 in an year and a half, interest being compounded half-yearly. A= P(1+ )2n 2n =A P(1+ ) 2000(1+ )3 =2315.25 3 = (1+ ) × 3 == 3 (1+ ) ( ) 1+ = => = r= ×200 r=10%

Exercis e 1.5 1. The number of students enrolled in a school is 2000. If the enrollment increases by 5% every year, how many students will be there after two years?

The Present strength of the school P= 2000 Rs Its gets increased by 5% every year r=5%

31

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

To find the strength after 2 years n=2 Now A= P(1+ )n 2 = 2000(1+ ) = 2000(1+ )2 2 =2000( ) =2000 × × × =5 × 441 =2205 So the number 0f students students after 2 year is 2205

2. A car which costs ` 3,50,000 depreciates by 10% every year. What will be the worth of the car after three years?

Present strength of the car P=3,50,000 Rate of deprecation r=10% Number of year =3 year So the value of the car after 3 years Now A= P(1- )n 3 = 350000(1- ) 3 = 350000( ) 2 =2000( ) ×× =350000 × ×× =350 ×81×9 =255150

3. A motorcycle was bought at ` 50,000. The value depreciated at the rate of 8% Per annum. Find the value after one year.

Value of motor cycle p=50000 Rate of depreciation r=8% To find value of the motorcycle after 1 year Now A= P(1- )n = 50000(1- ) = 50000( ) = 500 × 92 =46000

4. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% Per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000. The initial amount of bacteria P=506000

32

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Percentage of increase r=2.5% Per hour The amount of bacteria at the end of 2 hours Now A= P(1+ )n 2 = 506000(1+ ) 2 = 506000( ) =506000 × × × =5060 × =531616.25

5. From a village People started migrating to nearby cities due to unemployment Problem. The Population of the village two years ago was 6,000. The migration is taking Place at the rate of 5% Per annum. Find the Present Population.

Population of village P=6000 Rs Percentage of migration r=5% So the Present Population A= P(1- )n = 6000(1- )2 =6000 × × × A=5415

6. The Present value of an oil engine is ` 14,580. What was the worth of the engine 3 years before if the value depreciates at the rate of 10% every year?

Let the worth of the engine before 3 years be P Percentage of depreciation r=10% Per annum So Present value =A n ) 14580= P(1- )3 3 14580= P( ) ×× P=14580 × ×× =×

A= P(1-

=20000 Rs

7. The Population of a village increases by 9% every year which is due to the job opportunities available in that vill village. age. If the Present Population of the village is 11,881, what was the Population two years ago?

33

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Let P be the Population of the village 2 years years ago Percentage of increase r=9% So the Present Population A= P(1+ )n 11881= P(1+ )2 11881= 2 P( ) ×× P=14580 × ×

=10000 Rs

Exercise 1.6 1. Ponmani makes a fixed deposit of ` 25,000 in a bank for 2 years. If the rate of interest is 4% Per annum, find the maturity value.

P=25000 Rs n=2 years r=4% Interest= ×× =

=250× 8 Interest= 2000Rs Maturity value=P+I =25000+2000 Maturity value =27000 Rs 2. Deva makes a fixed deposit of ` 75,000 in a bank for 3 years. If the rate of interest is 5% Per annum, find the maturity value.

P=75000 Rs n=3years r=5% Interest= ×× =

=750× 15 Interest=11250Rs Maturity value=P+I

34

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=75000+11250 Maturity value =86250 Rs

3. Imran deposits ` 400 Per month in a Post office off ice as R.D. for 2 years. If the rate of interest is 12%, find the amount he will receive at the end of 2 years.

Amount deposited every month P=400 Rs Number of month= 2 years N=2×12=24 month Rate of interest r=12% ( ) Total Deposit made=N= [ ] ×() = [ ] 25 years Interest= ×× =

=1200

Total amount due=P n + =9600+1200 =10800 Rs

4. The cost of a microwave oven is ` 6,000. Poorani wants to buy it in 5 instalments. If the company offers it at the rate of 10% P. a. Simple Interest, find the E.M.I. and the total amount Paid by her.

Cost of microwave P=6000 Rs Interest=10% N=5 months = years ×× Interest= × =250 Rs Total amount Paid for the Oven =6000+250 =6250 Rs Number of installment=5 So EMI= =1250 Rs Total amount Paid by Poorani=6250 Rs

35

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

EMI=1250 RS

5. The cost Price of a refrigerator is ` 16,800. Ranjith wants to buy the refrigerator at 0% finance scheme Paying 3 E.M.I. in advance. A Processing fee of 3% is also collected from Ranjith. Find the E.M.I. E.M. I. and the total amount Paid by him h im for a Period of 24 months.

Cost of refrigerator =16800 Rs N=24 Interest= 0% EMI= =700 Rs 3 EMI is Paid in advance so amount amount Paid Paid in advance=700 ×3=2100 Processing fees=3% = ×16800 =504 Rs Total amount Paid =16800+2100+504 =19404 Rs 6. The cost of a dining table is ` 8,400. Venkat wants to buy it in 10 installments’. If the company offers it for a S.I. of 5% P. a., find the E.M.I. and the total amount Paid by him.

Cost of dining table =8400 Rs Number of installments’=10 So n= years Rate of interest r=5% × Interest =8400× × = 350 Rs =Total amount to be Paid =8400+350 =8750 Rs Number of installments’ =10 E.M.I= =875 Rs

Exercise 1.7 1. Twelve carpenters working 10 hours a day complete a furniture work in 18 days.

36

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

How long would it take for 15 carpenters working for 6 hours Per day to complete the same Piece of work?

No of carpenters 12 15

No of hours in a day 10 6

No of days 18 x

Step 1: Consider carpenter and days The multiplying facto factorr is Step 2: Consider no of hour Per day and no of days. The Multiplying factor is ×× X= × =24 days 2. Eighty machines can Produce 4,800 identical mobiles in 6 hours. How many mobiles can one machine Produce in one hour? How many mobiles would 25 machines Produce in 5 hours? No of machines No of Mobiles No of hours 4800 80 6 1 x 1 25 y 5

Step 1: Consider the number of machines and no of mobiles Produced The Multiplying Factor is SteP2: Consider number of hours the machine works and the no of mobiles Produced ×× The Multiplying factor =X= × =10 ii) The Multiplying Multiplying factor is The Multiplying factors is ×× X= × =1250 3. If 14 compositors can compose 70 Pages of a book in 5 hours, how many compositors will compose 100 Pages of this book in 10 hours?

37

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

No of compositors 14 x Step1 :

no of Pages in a book 70 100

No of hours 5 10

No of compositors and no of Papers of a book The multiplying factors = Step2: The multiplying factor is 5\10 So the number of compositors needed ×× X= × =10 compositors 4. If 2,400 sq.m. of land can be tilled by 12 workers in 10 days, how many workers are needed to till 5,400 sq.m. of land in 18 days? Area of land 2400sqm 5400

No of workers 12 x

No of days 10 18

Step1 : Area of land to be tiled and number of days The multiplying factors = Step 2: The multiplying factor is 10\18 So the number of workers needed ×× X= × =15 workers

5. Working 4 hours daily, Swati can embroid 5 sarees in 18 days. How many days will it take for her to embroid 10 sarees working 6 hours daily?

Step1 : No of hours Per day and number of days The multiplying factors = Step2: The multiplying factor is 10\5

38

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

So the number of days needed ×× X= × =24 days

6. A sum of ` 2,500 deposited in a bank gives an interest of ` 100 in 6 months. What will be the interest on ` 3,200 for 9 months at the same rate of interest?

Step1 : Amount deposited and interest received The multiplying factors = Step2: Interest received and number of months The multiplying factor is 9\6 So interested received ×× X= ×

=192 Rs

Exercise 1.8 1. A man can complete a work in 4 days, whereas a woman can complete it in only 12 days. If they work together, in how many days, can the work be completed?

Time taken by a man to finish finish the work = 4 days. days. Work done by a man in 1 day = 1/4 Time taken by a woman to finish the work = 12days. Work done by a woman in 1 day = 1/12 Work done by both in 1 day = 1/4 + 1/12 = 4/12= 1/3

39

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Time taken by both to finish the work = 1/3days, days. Hence both can finish the work in 1/3days

2. Two boys can finish a work in 10 days when they work together. The first boy can do it alone in 15 days. Find in how many days will the second boy do it all by himself?

Time taken by (A + B) to finish the work = 1/10days. Time taken by first boy to finish the work 15 days. So the Portion of work done by the second=1/10-1/15 =

1

/30

so time taken by

the second boy to finish the work=1(1/30)days

=30 days

3. Three men A, B and C can complete a job in 8, 12 and 16 days respectively. A and B work together for 3 days; then B leaves and C joins. In how many days, can A and C finish the work?

Time taken by A to finish the work = 1/8days. Time taken by B to finish the work= 1/12 days Time taken by C to finish finish the work= 1/16days So the Portion Portion of work done by A and B=1/8+1/12 =

5

/24

So the Portion Portion of work done by A and Bin Bin 3 days= =3(

5

/24)= 5/8

The Portion of work left out unfinished=1-

5

/8=3/8

The Portion of work done by A and B=1/8+1/1 =

40

3

/16

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Number of days days needed to complete =

3

3

/8th Portion of the work =3/8: 3/16

/8x 16/3

=2 days

4. A tap A can fill a drum in 10 minutes. A second tap B can fill in 20 minutes. A third tap C can empty in 15 minutes. If initially the drum is empty, find when it will be full if all taps are opened op ened together?

Time taken by tap A to fill the drum = 10hours. Work done by tap A in 1 hour = 1/10 Time taken by tap B to fill the drum= 20 hours. Work done by tap B in 1 hour = 1/20 Time taken by tap A to fill the drum = 15hours. Work done by tap A in 1 hour = 1/15 When all the taps opened then the Portion of drum filled in one minute=

1

/10+1/20-1/15

5 1 = /60= /12

So the time taken for the taps to fill the drum =1(1/12)=12 minutes 5. A can finish a job in 20 days and B can complete it in 30 days. They work together and finish the job. If ` 600 is Paid as wages, find the share of each.

Time taken by A to finish the work = 1/20days. Time taken by B to finish the work= 1/30 days The ratio of work done by b y A, B in 1 day= da y=1/20: 1/30 The ratio of their wages=3:2 Type

equation here.

Total wages Paid for the work=600Rs

41

So the wage received by A==

3

So the wage received by B= =

2

/5×600 =360 Rs

/5×600 =240 Rs

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

6. A, B and C can do a work in 12, 24 and 8 days respectively. They all work for one day. Then C leaves the group. In how many days will A and B complete the rest of the work?

Time taken by A,B,C to finish the work = 1/12, 1/24, 1/18

So the Portion Portion of work done by A and B and C Together in One Day=1/12+1/24+1/8

=

=

6

=

/24

1

/4

1 Remaining Portion=1- /4 =

3

/4

The Portion of work done done by A,B A,B together is

1

/12+1/24 =3/24=1/8

Time taken for A and B together to complete

of the work= 3

3

3

/4

/4÷ 1/8

/4×8 = 6 days

7. A tap can fill a tank in 15 minutes. Another tap can empty it in 20 minutes. Initially the tank is empty. If both b oth the taps start functioning, when will the tank become full?

Time taken by tap A to fill the tank = 15 hours. Work done by tap A in 1 min= 1/15 Time taken by tap B to fill the tank = 20 hours. Work done by tap B in 1 min = 1/20 Work done by (A + B) in 1 min = (1/15- 1/20) = 1/60 1 So the time taken to fill the ta nk=1( /60) =60min

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BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Chapter 2: Geometry

Geometry Exercise 2.1 1. Choose the correct answer (i) The Point of concurrency of the medians of a triangle is known as (A) In centre (B) circle centre (C) orthocenter (D) centroid Ans: centroid centroid (ii) The Point of concurrency of the altitudes of a triangle is known as (A) in centre (B) circle centre (C) orthocenter (D) centroid Ans: orthocentre orthocentre (iii) The Point of concurrency of the angle bisectors of a triangle is known as (A) in centre (B) circle centre (C) orthocenter (D) centroid Ans: incentre incentre (iv) The Point of concurrency of the Perpendicular bisectors of a triangle is known as (A) in centre (B) circumcentre (C) orthocenter (D) centroid Ans: circumcent circumcentre re 2. In an isosceles triangle AB = AC and +B = 65c. Which is the shortest side.

A A

C Sol:

B

C

Given ∠B=65 °, ∠C=65° Sum of the angles of a triangle is 180 ˚∠ A+∠B+∠C=180˚

∠ A+65˚+65˚=180˚ ∠ A+130=180˚ ∠ A=180˚-130˚ ∠ A=50˚ In ABC ∠ A =50˚,,∠B=∠C=65˚ 43

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

∠ A is smallest smallest angle. The side opposite to smallest angle is smallest. BC is smallest side.

3. PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

In PQR P=90 The hypotenuse is QRR Given PR= 24cm PQ=10cm QR= + 24cm =√ 24 24 + 10 =√ 576 576 + 100 P

x Q

= 676 676 10cm QR=26cm 4. Check whether the following can be the sides of a right angledtriangle AB = 25 cm, BC = 24 cm, AC = 7cm. AB=25 cm AB2=252=625

BC2=242=576 AC=7, AC2=7 AC2=72 =49 Now 576+49=625 BC2+AC2=AB2 So the sides AB, BC, AC from the sides of a right angled Triangle. 5. Q and R of a triangle PQR are 25° and 65°. Is DPQR a right angledtriangle?

Moreover PQ is 4cm and PR is 3 cm. Find QR. In PQR ∟Q= 25˚ 25˚ R=65˚ ∟R=65˚ ∟P+∟Q+∟R=180˚ ∟P=180-( ∟P=180-(∟Q+∟R) ∟Q+∟R) 180-(25+65) =180-90 =90 =>PQR is a right triangle Given PQ=4cm PR=3cm ∟P=90˚, QR is the hypotenuse ∟QR= +

44

R

3cm

P

4cm

Q

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

=√ 4 + 3 =√ 1 6 + 9 =√ 25 25 =5cm 6. A 15 m long ladder reached a window 12m high from the ground. On Placing it Against a wall at a distance x m. Find x.

A

Length of ladder AC=15cm Height of window AB= 12m ABC is a right angled triangle t riangle with ∟B=90 2

2

15cm

12cm

2

So AC =AB +BC

Here BC =x, 152=122+x2 X2=152-122 =225-144 X2=81 X=9 m

C

B

7. Find the altitude of an equilateral triangle of side 10 cm . A ABC is an equilateral triangle with side 10 cm AD is drawn Perpendicular to BC 10cm 10cm BD=DC BD=DC=10\2 =5cm From the right angled triangle ABD right angledat D. B 5cm 5cm C AB2=AD2+BD2 2 2 2 A 10 =AD +5 2 2 2 AD =10 -5 AD2=100-25 10cm 10cm AD2=75 AD=√ 75 75 =5√ 3 cm B C The length of the altitude is 5√ 3 cm. 8. Are the numbers 12, 5 and 13 form a Pythagorean Triplet? 122=144 52=25 132=169 144+25=169 122+52=132 The numbers 12,5,13 form a Pythagoras Triplet. MATHEMATICS

45

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

9. A Painter sets a ladder up to reach the bottom of a Second store window 16 feet above the ground. The base of the ladder is 12 feet from the house. While the Painter mixes the Paint a neighbor’s Dog bumps the ladder which moves the base 2 Feet farther away from the house. How far up side Of the house does the ladder reach?

In ABC ∟B=90° AC is the length of ladder Given AB=16m, BC=12m AC=√ + ==√ 16 16 + 12 =√ 256 + 144 =√ 400 400 =20m Length of ladder is 20m Now AC=20m BC=14m AB2+BC2=AC2 AB2+142=202 AB2+196=400 AB2=400-196 =204 AB=√ 5 1 × 4 =2√ 51 51 ft The ladder reaches upside of the house at 2 √ 51 51 ft

Exercise 2.2 1. Choose the correct answer: (i) The _______ of a circle is the distance from the centre to the circumference. (A) Sector (B) segment (C) diameters (D) radius Ans: Radius (ii) The relation between radius and diameter of a circle is ______ (A) Radius = 2 × diameters (B) radius = diameter + 2 (C) Diameter = radius + 2 (D) diameter = 2 (radius) Ans: diameter diameter = 2 (radius) (radius)

46

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

(iii) The longest chord of a circle is (A) radius (B) secant (C) diameter (D) tangent

Ans: ) diameter diameter 2. If the sum of the two diameters is 200 mm, find the radius of the circle in cm.

Sum of 2 diameters=200mm So length of 1 diameters= =100mm

=2xradius Radius= =50mm=5cm

3. Define the circle segment and sector of a circle. Segment of a Circle A chord of a circle divides divides the circular circular region region Into two Parts. Each Part is called as segment of The circle.

Sector of a Circle The circular region enclosed by an arc of a circle and the two radii at its end Points is known as Sector of a circle 4. Define the arc of a circle.

AB is a chord. chord. The chord chord AB divides the Circle into two Parts. The curved Parts ALB and AMB are known as Arcs 5. Define the tangent of a circle and secant of a circle. Secant of a Circle A line Passing Passing through a circle and intersecting intersecting the the circle at two Points Points is called called The secant of the circle Tangent Tangent is a line that touches a circle at exactly one Point, and the Point is Known as Point of contact

47

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Chapter 3: Practical Geometry Exercise 4.1 Draw a concentric circles for the following measurements of radii, find out the width of each circular ring. 1. 4cm and 6cm.

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 5.5cm. With O as centre draw a circle OB = 3.5cm. Now the concentric circle is drawn. The width of the circular ring

= OA – OB

= 5.5 – 3.5 = 2cm

48

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 5.5cm. With O as centre draw a circle OB = 3.5cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 5.5 – 3.5 = 2cm

49

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 6.8cm. With O as centre draw a circle OB = 4.2cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 6.8 – 4.2 = 2.6cm

50

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 6.5cm. With O as centre draw a circle OB = 5cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 6.5 – 5 = 1.5cm

51

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 8.1cm. With O as centre draw a circle OB = 6.2cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 8.1 – 6.2 = 1.9cm

52

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 7cm. With O as centre draw a circle OB = 5.3cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 7 – 5.3 = 1.7cm

53

BHARATHIDASANAR MATRIC HIGHER SECONDARY SCHOOL, ARAKKONAM_VIIIth MATHEMATICS TERM-3

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