Gene Maths

7/29/2019 Gene Maths

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Prof. M A ThiruthuvadossCell: 9445204774

for Excellence in Mathematics

http://www.magicmaths.org

REFORMMuch Needed

LongOverdueand

A Genuine

TEXT BOOKS vs

GENUINE &

SMART CALCS

Put

MATH

SBACK

ONTRACK

Savethe Children&

MATHEMATICS!


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Text Book vs Genuine Maths

2

Genuine Maths Workings are with gray backgroundTamilnadu Govt. Textbook for Class 6Example 32:

An article was bought at Rs. 450 and sold for Rs. 500. Find the profit or loss.

Solution: C.P. of an article = Rs. 450

C.P. of an article = Rs. 500

S.P. is greater than C.P. so there is a proft.

Profit = S.P. C.P.

= 500 450 = Rs. 50.

Example 33:

A television set was bought for Rs. 10,500 and sold at Rs. 9,500. Find the profit or loss.

Solution: C.P. of the television set = Rs. 10500

S.P. of the television set = Rs. 9500

Here C.P. is greater than S.P. so there is a loss.Loss = C.P. S.P.

= 10,500 9,500 = Rs. 1000.

Example 34:

A bag is bought at Rs. 200 and sold at a profit of 10%. Find the selling price.

Solution: Profit = 10% of Rs. 200

=10

100x 200 = Rs. 20

S.P. = C.P. + Profit

= 200 + 20 = Rs. 220.

Discuss:

In the above example, we can simplify the procedure as follows:

Selling Price =100 10

100

x Cost Price

=110

100x 200 = Rs. 220

Article was bought for Rs.450;

Article was sold for Rs.500.

Profit = 500 450 = 50

Profit is Rs.50.

Television set was bought for Rs.10,500

Television set was sold for Rs. 9,500

Loss =10,500 9,500 = Rs. 1000

Profit = 10% of 200 = 20

S.P. = 200 + 20 = 220

Selling Price = Rs.220.

Profit 10%

cost is 100% and selling price is 110%

selling price = 110% of 200= 1.1 x 200 = Rs. 220

TraditionallythereisNOformulaforProfitandLos

s.

Nor

isthereaneedforon

e.

ThisisjustOneoftheApplicationsofPercentage

s.

Theseformulaehave

beenINVENTEDin

theLast10yearsorso.


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3

Genuine Maths Workings are with gray backgroundNCERT Testbook for Class VIIICHAPTER 4 - Prof it, Loss and Discount

Example 1: Anwar purchased 120 oranges at the rate of Rs 2 per orange. He sold 60%of the oranges at the rate of Rs 2.50 per orange and the remaining oranges at the rate of

Rs 2 per orange. Find his profit percent.

Solution: C.P. of 120 oranges = Rs 2 x 120 = Rs 240

60% of 120 oranges =60

100x 120 = 72 oranges

Now S.P. of 72 oranges = Rs 2.50 x 72 = Rs 180

and S.P. of the remaining 120 72, i.e., 48 oranges = Rs 2x48 = Rs.96

S.P. of all the 120 oranges = Rs 180 + Rs 96 = Rs 276

Therefore prof it = S.P. C.P.

= Rs (276 240) = Rs 36

Hence profit per cent =36

240x 100 = 15

Thus Anwars profit is 15%

Example 2: Maninder bought two horses at Rs 20000 each. He sold one horse at

15% gain. But he had to sell the second horse at a loss. If he had suffered a loss of Rs

1800 on the whole transaction, find the selling price of the second horse.

Solution: Total C.P. of the two horses = 2 x Rs 20000 = Rs 40000

Loss = Rs 1800

Total S.P. = Rs 40000 Rs 1800

= Rs 38200 (1)

Now S.P. of the first horse at 15% profit= C.P. x100 + Profit %

100

= Rs 20000(100 + 15)

100= Rs 23000 (2)

S.P. of the second horse = Rs 38200 Rs 23000

Thus, the selling price of the second horse is Rs 15200.

Solution:

Anwars cost price = 2 x 120 = Rs.240 --- (1)

He sold 60% of mangoes at a profit of Rs.0.50 each

60% of 120 = 0.6 x 120 = 72

Anwar made a profit of 0.50 x 72 = Rs.36 --- (2)

He sold the rest at Rs.2 only which is his cost (no profit, no loss).

Anwars total profit is Rs.36 on a C. P. of Rs.240

Profit =36

240x 100 = 15%

Anwars profit is 15%

Solution:

Cost of 2 horses isRs.40000 [2 x 20000]

Total loss Rs. 1800Total selling price Rs.38200

Profit on the f irst horse is 15% = Rs.3000 [10% is 2000, 5% is 1000]

Selling price of f irst horse: Rs.23000

So selling price of second = 38200 23000 = Rs. 15200.


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4

Genuine Maths Workings are with gray backgroundNCERT Testbook for Class VIIICHAPTER 4 - Prof it, Loss and Discount

Example 4: A farmer sold two bullocks for Rs 18000 each. On one

bullock he gained 20% and on the other he lost 20%. Find his total loss or

gain.

Solution: S.P. of the first bullock = Rs 18000

Gain = 20%

Therefore C.P. =100 x S.P.

100 + Profit %

=100 x 18000

Rs100 + 20

= Rs 15000

(1)S.P. of the second bullock = Rs 18000

Loss = 20%

Therefore C.P. =100 x S.P.

100 - Loss %

=100 x 18000

Rs100 - 20

= Rs 22500

(2)

Now, total C.P. = Rs 15000 + Rs 22500

= Rs 37500

and total S.P. = 2 X Rs 18000 = Rs 36000Hence loss = Rs 37500 Rs 36000

= Rs 1500

Example 3: By selling 144 hens, Malleshwari lost the S.P. of 6 hens.

Find her loss percent.

Solution:

First Gain = 20% C. P. is 100% and S. P. is 120%

S.P. of one bullock = Rs 18000

120% is 18000, then 100% is 18000 1.2 = 15000C. P. of first bullock is Rs.15000 (1)

Second: Loss = 20% C. P. is 100% and S. P. is 80%

S.P. is Rs 18000

80% is 18000, then 100% is 18000 0.8 = 22500C. P. of second bullock is Rs.22500 (2)

Total cost of two bullocks is Rs.37500 [from (1) and (2)]

Total selling price is 18000 x 2 = Rs.36000

Total loss is Rs.1500

Example 3: By selling 144 hens, Malleshwari lost the S.P. of 6 hens.

Find her loss percent.

This is a Planted Problem, not blending with the level and intensity indicated

by the syllabus here.

Even though there is loss mentioned here, what is the practical, natural

and simple concept here which the child will learn and apply?

How does a problem of this difficulty fit in here? Check the syllabus:

Ratio & Proportion - Slightly advanced problems involving

applications on percentages, profit & loss, overhead expenses,

Discount, tax.


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5

Genuine Maths Workings are with gray backgroundEXAMPLE 1.Mohit bought a CD for Rs.750 and sold it for Rs.875. Find his gain per cent.

Solution: CP = Rs.75 and SP = Rs.875.

Since (SP) > (CP), mohit makes a gain.

Gain = Rs.(875 750) = Rs. 125

Gain%gain

= x 100 %CP

125 50 2= x 100 % = % = 16 %

750 3 3

EXAMPLE 8.

Rohit buys a geyser for Rs.3680 and sells it at a gain of 7%. For how much does he sell it?Solution: CP of the geyser = Rs. 3680.

Gain% = 7% =15

%2

.

SP of the geyser(100 + gain%)

= x CP100

15100 +

2= Rs x 3680

100

215= Rs 3680

200

= Rs 3956

x

EXAMPLE 6. If the cost price of 10 greeting cards is equal to the selling price of 8 greeting

cards, find the gain or loss per cent.

EXAMPLE 7. By selling 33 m of cloth, a draper loses an amount equal to the selling price

of 3 m of cloth. Find his gain or loss per cent.

Solution:

Buying price = Rs.750; Selling price = Rs.875or Bought for = Rs.750; Sold for = Rs.875

Profit = 875 750 = 125 (this is out of the cost: 750)

Profit percent is125 100 2

x 100 = = 16 %750 6 3

Solution

Gain = 7% of 3680= 0.075 x 3680

= 276

Gain is Rs.276

He sells it for 3680 + 276 = Rs3956

Planted Problems, not blending with the level and

intensity indicated by the syllabus here.

3 6 8 0

. 7 5

1 8 4 0 0

2 5 7 6 0 .

2 7 6 0 0 0

This calculation also can be

done in a straightline in one

step by QuickMaths


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6

Samples From Board (BORED) Textbooks

1) Convert 8.37 into an ordinary fraction.

(Tamil Nadu Std.VI Textbook 2009)

8.37 1 1= 8 x 1 + 3 x + 7 x10 100

3 7= 8 + +

10 100

8 x 100 3 x 10 7= + +

1 x 100 10 x 10 100

800 30 7= + +

100 100 100

837

= 100

Note: What is ordinaryfraction?

378

100is a mixed number.

837

100not used;

2) A Simplification. (Tamil Nadu Std.VII Textbook 2009)

9 5 9 3322 33 22 5x =

297

110

=77

2110

=7

210

378.37 = 8+

100

37= 8

100

9 5 9 33

=22 33 22 5

x

3

2

27=

107

=210

Normal

Working in

Boxes

3) Express 1000 as the product of powers of prime factors.(NCERT Std.VII Textbook 2010)

1000 = 10 x 100= 10 x 10 x 10

= (2 x 5 )x(2 x 5 )x(2 x 5 )

= 2 x 5 x 2 x 5 x 2 x 5

= 2 x 2 x 2 x 5 x 5 x 5

or 1000 = 23 x 53

4) This is a problem with Right Circular Cone.(NCERT Std.IX Textbook 2010)

2 2 22 2

2 2

Now

Therefore,

25 7

625 49

576

24

l r h

h l r

m

m

m

m

Why this PRIMITIVE Approach in every step?!?

In IX Std. why do we need the first line on top?

And why derive the relation for h?

How do they square 25?

How they find root of 576?

2 2

2 225 7

32 18

16 2 18

4 6

24

h l r

x

x x

x

m

1000 = 103

= (2 x 5)3

= 23 x 53.


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7

Genuine Maths Workings are with gray backgroundStd. VI - Tamil NaduExample 88: (page 45)

Arrange the following in ascending order and in descending order:

2 3 5 1, , ,

3 4 6 4

Solution: Find the equivalent fractions for2 3 5 1

, , ,3 4 6 4

by taking

l.c.m. of the denominators.

l.c.m. = 12 3 3, 4, 6, 4

2 1, 4, 2, 4

2 1, 2, 1, 2 3 x 2 x 2 = 12

1, 1, 1, 1

2 2 4 8= x =

3 3 4 12

3 3 3 9= x =

4 4 3 12

5 5 2 10= x =

6 6 2 12

1 1 3 3= x =

4 4 3 12

Writing these fractions in ascending order. , , ,3 8 9 10

12 12 12 12

Therefore , , ,1 2 3 5

4 3 4 6are in ascending order

Similarly we can write these in desc. order , , ,5 3 2 1

6 4 3 4(Why?)

Std. VI - Tamil Nadu

Example 88: (page 45)

Arrange the following in ascending order and in

descending order:

2 3 5 1, , ,

3 4 6 4

SmartCalcSolution:

L. C. M. of 3, 4, 6, 4

Enough to find L. C. M. of 4 & 6 and this is 12

Taking this 12 as the COMMON DENOMINATOR,

, , ,2 3 5 1 8 9 10 3

, , ,3 4 6 4 12 12 12 12

In ascending (increasing numerators 3, 8, 9, 10) order:

1 2 3 5, , ,

4 3 4 6[By inspection]

in descending (decreasing numerators, 10, 9, 8, 3) order:

5 3 2 1, , ,

6 4 3 4[By inspection]

Note: 1. Placing of commas

2. Primitive method of finding L.C.M

3. L.C.M. to be in capitals - not l.c.m.

4. Unnecessary work for converting fractions


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Genuine Maths Workings are with gray backgroundStd. VI - Tamil Nadu (2009)Example 109: (page 56)

Convert 8.37 into an ordinary fraction.

Solution:

8.371 1

= 8 x 1 + 3 x + 7 x10 100

3 7= 8 + +

10 100

8 x 100 3 x 10 7= + +

1 x 100 10 x 10 100

800 30 7= + +100 100 100

837=

100

Std. VI - Tamil Nadu (2009)Example 45: (page 91)

A man deposits Rs.2000 in a bank. The bank pays interest at the rate of 4%

per annum. Find the interest received by him at the end of 3 years. Also find the

amount to be paid at the end of 3 years?

Solution: Principal = Rs.2000Rate of interest = 4%

Interest on Rs.100 for 1 year = Rs.4

Interest on Rs.2000 for 1 year =4

100x 2000 = Rs. 80

Interest on Rs.2000 for 3 years = 3 x 80 = Rs. 240

Amount = Principal + Interest

= Rs. 2000 + Rs. 240

= Rs. 2240

Std. VI - Tamil Nadu (2009)


Convert 8.37 into an ordinary fraction.

SmartCalcSolution:

.37 837

8 37 = 8 or 100 100


A man deposits Rs.2000 in a bank. ... Also find the amountto be paid received at the end of 3 years?

SmartCalcSolution:

Deposit, Rs.2000 (20 hundreds); Rate of Interest, 4% p.a.;

Period, 3yrs.

Using Proportion

Rate 4% for Rs.100 (1 hundred) for 1yr interest is Rs.4for Rs.2000 (20 hundreds) for 1yr Rs.80

Rs.2000 .for 3 yrs Rs.240

Interest = Rs. 240

Amount = 2000 + 240 = Rs.2240

Or In single step:Principal Time Interest

100 1 4

2000 3 ?

Interest =2000 3

4100 1

x x = Rs.240, ...


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9

Genuine Maths Workings are with gray background



Multiply: 2.1 x 1.3 x 1.2

Solution:

First multiply: 21 x 13 x 12 2 1 x 1 3

6 3

2 1

2 7 3

We have, 21 x 13 x 12 = 3276 2 7 3 x 12

Total number of decimal places in all the 5 4 6

three numbers is 3 2 7 3

3 2 7 6 2.1 x 1.3 x 1.2 = 3.276

Tamilnadu, VII (2009): Text book

9 5 9 33x

22 33 22 5 =

297

110=

772

110=

72

10



Multiply: 2.1 x 1.3 x 1.2

SmartCalcSolution:

1.3 x 1.2 = 1.56

1.56 x 2.1 = 31.2

176 (By any QMmethod)

So 2.1 x 1.3 x 1.2 = 3.276

Note:

12 x 13 = 156 (mentally) and156 x 21 is the order to be encouraged.

Taking 21 x 13 first is mischivous and with sinister intentions

156 x 21 = 156 x 7 x 3 = 1092 x 3 = 3276 (as 21 = 7 x 3) 156 x 21 = 3120 + 156 = 3276 (as 21 = 20 + 1)

Tamilnadu, VII (2009): SmartCalc:

9 5 9 33x

22 33 22 5 =

27

10=

72

10


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10




The price of 30 pens is Rs.172.50. Find the price of one pen.

Solution:

Price of 30 pens = Rs.172.50 5.75

Price of 1 pen = 172.50 30 30) 172.50= Rs.5.75 150 .

225

210 .

150

150

0

The price of 1 pen is Rs.5.75



The price of 30 pens is Rs.172.50. Find the price of

one pen.

SmartCalcSolution:

30 pens cost Rs.172.50

3 pens cost Rs.17.25

1 pen costs Rs.5.75

Do not need Long Division fordividing by 3.

Or30 pens cost Rs.172.50

1 pen cost =172.50

30

=17.25

3

= 5.75

The price of 1 pen is Rs.5.75


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11




Area of the four walls of a hall is 360 sq. m. Its length and breadth are 20m and

10m respectively. Find its height.

Solution:

ph = 360 m2 ; l= 20m ; b = 10m ; h = ?

p = 2l+ 2b

= 2 x 20 + 2 x 10

= 40 + 20 = 60p x h = 360

i.e. 60 x h = 360

Dividing both sides by 60,

60 360=

60 60

xh

h = 6 m

height = 6 m.



Area of the four walls of a hall is 360 sq. m. The halls

length and breadth are 20m and 10m respectively. Find the

height of the hall.

SmartCalcSolution:

A = 360 m2; l= 20m; b = 10m; h = ?

P = 2 (l + b) = 2 (20 + 10) = 60 m

We know: Ph = A60 h = 360

h = 6 m

Height of the hall is 6 m.

Following comments refer to page on the left (terror-book)

What are the two its in the problem supposed to referto?

Perimeter is Pand notp On the first lineA = 360m2and notph = 360m2

In algebra P= 2(l+ b); not p = 2l+ 2b and not p = 2(l+ b) Dividing both sides by 60 etc. is very wicked thinking.


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12




A square room of side 16m is 4 m high.1

4parts of the area of the walls is

occupied by the doors and windows. Find the cost of white washing the walls of

the room at Rs.3.50 persq.m.

Solution:

Perimeter of the square roomp = 4 a

= 4 x 16m = 64 m

height = 4 mArea of the four walls = ph

= 64 m x 4 m = 256 m2

Area of doors and windows =1

4x 256 m2

= 64 m2

Area to be white washed = 256 m2 64 m2

= 192 m2

Cost of white washing 1 m2 = Rs. 3.50

Cost of white washing 192 m2 = Rs.192 x 3.50

= Rs.672Hence the cost of white washing the room = Rs.672



A square room of side 16m is 4 m high. ... Find the

cost of ...

SuperCalcSolution:

side (a) = 16m; height (h) = 4m; cost of ... = Rs.3.50 perm2.

A of walls = Ph

= (4 x 16) x 4

= 16 x 16

Area of doors & windows =1

4of walls

= 4 x 16

Area to be white washed = 12 x 16 m2

Cost of white washing = 12 x 16 x 3.50= 12 x 8 x 7 = Rs.672

OrFor bright thinking:

Area to paint =3

4of walls as Doors & windows =

1

4of walls

=3

4Ph

=3

4x (4 x 16) x 4 = 16 x 3 x 4 m2

Cost of white washing @ Rs.3.50 is 16 x 3 x 4 x 3.5

= 8 x 3 x 4 x 7 = Rs.672

How muchwork & timeto get 672?

?


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13


Std. VII - N.C.E.R.T


Convert the given decimals to per cents: (a) 0.75, (b) 0.09, (c) 0.2 page 161

Solution:

( ) 0.75 0.75 100%

75100% 75%

100

9( ) 0.09 9%

100

2( ) 0.2 100% 20%

10

a x

x

b

c x

[eqn 59]



Convert the given decimals to per cents: (a) 0.75, (b) 0.09,

(c) 0.2 page 161

SmartCalcSolution:

Move decimal point 2 places to the right

(a) 0.75 = 75%

(b) 0.09 = 9%

(c) 0.2 = 20%


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14




The area of a square and a rectangle are equal. If the side of the square is 40 cm

and the breadth of the rectangle is 25 cm, find the length of the rectangle.page 207

Solution:

Area of square = (side)2

= 40 cm x 40 cm = 1600 cm2

It is given that,

The area of the rectangle = The area of the squareArea of the rectangle = 1600 cm2, breadth of the rectangle = 25 cm.

Area of the rectangle = l x b

or 1600 = l x 25

or1600

25= l or l= 64 cm

So the length of rectangle is 64 cm.



The area of a square and a rectangle are equal. If the side of

the square is 40 cm and the breadth of the rectangle is 25 cm,

find the length of the rectangle.page 207

SmartCalcSolution:

Area of a sq. and a rectangle are equal.

Square: s = 40cm; Rectangle: b = 25cm,

l= ?

Area of rectangle = Area of sq.lb = s2

25 l = 40 x 40

l =40 x 40

25

= 8 x 8 = 64 cm

40 cm

40 cm

25 cm

l cm

How muchwork & timeto get 64?

?


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15

Genuine Maths Workings are with gray backgroundStd. VII - N.C.E.R.T


Express the following numbers as a product of powers of prime factors: page 251(i) 72 (ii) 432 (iii) 1000 (iv) 16000

Solution:

(i)

3 2

3 2

72 2 36 2 2 18

2 2 2 9

2 2 2 3 3 2 3

,72 2 3

x x x

x x x

x x x x x

Thus x

[eqn 63]

(ii)

4 3

432 2 216 2 2 108 2 2 2 542 2 2 2 27 2 2 2 2 3 9

2 2 2 2 3 3 3

or,432 2 3 (required form)

x x x x x xx x x x x x x x x

x x x x x x

x

[eqn 64]

(iii)

3 3

1000 2 500 2 2 250 2 2 2 125

2 2 2 5 25 2 2 2 5 5 5


x x x x x x

x x x x x x x x x

x

[eqn 65]

or NCERTs supermanAtul:

3 3

1000 10 100 10 10 10

(2 5) (2 5) (2 5) (since 10=2x5)

2 5 2 5 2 5 2 2 2 5 5 5


x x x

x x x x x

x x x x x x x x x x

x

[eqn 66]

Is Atuls method correct?

Continuedon Next Page



Express the following numbers as a product of powers of prime

factors: page 251

(i) 72 (ii) 432 (iii) 1000 (iv) 16000

SmartCalcSolution:

(i) 72 = 4 x 18 = 22 x 2 x 32 = 23 x 32

(ii) 432 = 2 x 216 = 2 x 63 = 2 x 23 x 33 = 24 x 33

(iii) 1000 = 103 = 23 x 53

(iv) 16000 = 42 x 103 = 24 x 23 x 53 = 27 x 53

Prof. Doss

NOTE:

Maths is taught to make the children FOOLS!


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16

Genuine Maths Workings are with gray backgroundContinued from previous page.

(iv)4 3

4 3 3

7 3

16,000 =16 1000 =(2 2 2 2) 1000=2 10 (as 16=2x2x2x2)

=(2 2 2 2) (2 2 2 5 5 5) 2 2 5

(since 1000 is 2x2x2x5x5x5)

=(2 2 2 2 2 2 2) (5 5 5)

or,16,000=2 5

x x x x xx

x x x x x x x x x x x

x x x x x x x x x

x


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17


Std. VIII - N.C.E.R.T


Find the compound interest on Rs.25600 for 2 years at 6.25% per annum.

Solution:

P= Rs.25600, n = 2, r= 6.25

Compound interest

2

2

= 1 1100

6.25= Rs 25600 1 1

100

1= Rs 25600 1 1

16

17 17= Rs 25600 1

16 16

289 256= Rs 25600

25625600x33

= Rs25

nr

P

33006

Rs

How muchwork & time to

do thesecalculations?

?


Example 5: (page 77)Find the compound interest on Rs.25600 for 2 years at

6.25% per annum.

SmartCalcSolution:

P= Rs.25600, n = 2, r% = 6.25% = 1/16

rA = P 1+

100

n

21

A = 25600 x 1+ 16

17 x 17= 25600 x

16 x 16

=100 x 17 x 17

= 28900

[Note that 256 = 162]

Compound interest is 28900 25600 = Rs.3300

Or

n

2

22

2

rC. I. = P 1+ -1

100

1= 25600 x 1+ -1

16

17=16 x 100 x -1

16

= 100 x (172 162)

= 100 x 33 x 1

= 3300

Look at the HUGE

numbers on left!

Compare with

working on right.


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18




A certain sum amounts to Rs.5832 in 2 years at 8% compound interest. Find the

sum.

Solution:

Here A = Rs.5832, r= 8, n = 2.

Let P be the required sum in rupees. Then

2

2

85832 1100

1085832=

100

27 275832=

25 25

5832 25 25=

27 27

= 5000

P

P

P

x xP

x

Hence, the required sum is Rs.5000.



A certain sum amounts to Rs.5832 in 2 years at 8% compound

interest. Find the sum.

SmartCalcSolution:

A = Rs.5832, r= 8, n = 2; P = ?

2

2

2

81 = 5832

100

1.08 = 5832

5832=

1.08

5832=

1.1664

5832= 10000

11664

1

= 10000 50002

P

P

P

P

P x

P x

OrWhat amounts to Rs.5832 in 2 years @ 8% C. I.?

Rs.1000 amounts to 1000x1.082 = 1166.40

Ans. = Rs.5000(By estimation 5 x 1166.40 = 5832)

By intuition double Nr.,

you get denominator

How muchwork & time to

do thesecalculations?

?


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19

Genuine Maths Workings are with gray backgroundStd. VIII - N.C.E.R.TExample 2: (page 263)

A metallic cylindrical pipe has thickness 0.5cm and outside diameter 4.5cm. If 1

cm3

of the metal has mass of 8g find the mass of 77cm long pipe. page 262

Solution:

We first find the vloume of the metallic part of the cylinderical pipe. This is the

difference of the volume of two solid cylinders one of diameter 4.5 cm and the

other of diameter (4.5 1.0) cm or 3.5 cm.

For outer cylinder, radius =1

4.52x and height = 77 cm

Volume =

2

3 322 4.5 4.5 4.577cm 242 cm

7 2 2 2

x x x x

(1) [eqn 11]

For inner cylinder, radius =4.5 3.5

0.5 cm= cm2 2

, and height = 77 cm [eqn 12]

Volume =2

3 322 3.5 3.5 3.577 cm 242 cm7 2 2 2

x x x x

(2) [eqn 13]

The required bolume of the pipe =Volume of the outer cylinder Volume of the inner cylinder

2 2

3

3

3 3

4.5 3.5242 cm2 2

4.5 3.5 4.5 3.5242 cm

2 2 2 2

242 4 0.5 cm 484cm

x

x x

x x

[eqn 14]

Mass of the pipe = 484 x 8 g [Since 1 cm3 of the metal has mass 8g]= 3872 g

=3.872 kg



A metallic cylindrical pipe has thickness 0.5cm and outside

diameter 4.5cm. If 1 cm3 of the metal has mass of 8g f ind the

mass of 77cm long pipe. page 262

SmartCalcSolution:

Outer diameter = 4.5 cm,

thickness of pipe = 0.5 cm

Mass of 1 cm3 of metal is 8g.

Radius: Outer (R) = 2.25cm, inner (r) = 1.75cm, h = 77cm

Volume of metal in pipe = h (R2 r2)

=22

7x 77 x (2.252 1.752)

= 22 x 11 x 4 x 0.5

= 22 x 11 x 2

= 484

Volume of metal = 484 cm3.

Mass = 484 x 8 = 3872 g = 3.872 kg

Where isDiagram?

?4.5 cm

0.5 cm

77 cm


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Genuine Maths Workings are with gray backgroundStd. VIII - N.C.E.R.T

Example 3: (page 245)It is required to make a closed cylindrical tank of height 1m and

base diameter 140cm from a metal sheet. How many square

metres of the metal sheet are required for the same?page 245

SmartCalcSolution:

We shall convert all units, and work with, metresd= 140cm = 1.4m, r= 0.7m; h = 1m

T.S.A.= 2r (r + h)

= 2 x 227

x 0.7 (0.7 + 1)

= 2 x22

7x 7 x 0.17

= 4 x 11 x 0.17

= 0.68 x 11

= 7.48

Area of metal sheet required is 7.48m2

Hence 7.48 sq. metres of metal sheet is required.



It is required to make a closed cylindrical tank of height 1m and base diameter140cm from a metal sheet. How many square metres of the metal sheet are required

for the same? page 245

Solution: Here, diameter = 140 cm

Radius r=140 70 7

cm 70cm m m2 100 10

Height h = 1m

Total surface area of the tank

2

2 2

2 ( )

22 7 72 17 10 10

2 22 177.48

100

r h r

x x m

x xm m

Thus, the area of the metal sheet required is 7.48 m2

Hence, 7.48 square metres of metal sheet are required.

Where isDiagram?

?

How muchwork & timeto get 7.48?

?d = 140 cm

h = 1 m

r = 0.7 m


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21




Diameter of the base of a cone is 10.5 cm and its slant height is

10cm. Find the curved surface area of the cone.page 251

SmartCalcSolution:

d= 10.5 cm, r= 5.25 cm; l= 10 cm

C.S.A. = rl

=

22

7 x 5.25 x 10

= 22 x 0.75 x 10

= 11 x 1.5 x 10

= 165

Curved Surface Area = 165 cm2.

d = 10.5 cm

l = 10 cm



Diameter of the base of a cone is 10.5 cm and its slant height is 10cm. Find the

curved surface area of the cone. page 251

Solution: Diameter = 10.5 cm

Therefore, base radius r10.5

cm2

Slant height l= 10 cm

Hence, curved surface area

2 2

2

22 10.5 22 10510cm 10cm

7 2 7 20

165cm

rl

x x x x

Quick Maths:22 10.5 22 3

10 10 11 3 5 11 15 1657 2 2 2

x x x x x x x ]


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22

Genuine Maths Workings are with gray backgroundStd. IX - N.C.E.R.TExample 16: (page 232)

Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical

tent made, with a base radius of 7 m. Assuming that all the stitching margins andthe wastage incurred while cutting, amounts to approximately 1 m3, find the volume

of the tent.page 232

Solution:

Since the area of the canvas = 551 m2 and area of the canvas lost in wastage is 1

m2, therefore the area of canvas available for making the tent is (551 1) m2 = 550

m2.

Now the surface area of the tent = 550 m 2 and the required base radius of the

conical tent = 7 m.

Note that a tent has only a curved surface (the floor of a tent is not covered bycanvas!!)

Therefore, curved surface area of tent = 550 m2.

That is, rl = 550

or,22

7 = 5507

x x l

or,550

3 2522

l m m

Now, l2 = r2 + h2

Therefore, 2 2 2 2= = 25 7 m = 625 49 m = 576 mh l r

= 24 m.

So, the volume of the conical tent2 2 21 1 22 7 7 24 1232

3 3 7r h x x x x m m

NOTE:See how the sq rt is handled on the Right-sides

Quick Maths:1 22

7 7 24 = 2 11 7 8 = 11 112 = 12323 7x x x x x x x x

Std. IX - N.C.E.R.T


Monica has a piece of canvas whose area is 551 m2. ... find the

volume of the tent.

SmartCalcSolution: Base radius r= 7 m

Canvas used only for the curved surface and not for the floor.

So C.S.A. = 551 1 = 550 m2.

550

550 7 1 550 50550 2522 7 22 2

rl

l x x mr

Now,2 2

2 2

h = l - r

= 25 - 7

= 32 x 18

= 16 x 36

= 4 x 6 = 24m

Volume of tent 2

2

3

1

3

1 227 24

3 7

22 7 8 11 112 1232

r h

x x x

x x x m

7 m

lh


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Genuine Maths Workings are with gray backgroundStd. X - N.C.E.R.T


In figure, two circular flower beds have been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circular flower bed is the point of intersection

O of the diagonals of the square lawn, find the sum of the areas of the lawn and

the flower beds.

Solution:

Area of the square lawn ABCD = 56 x 56 m2 (1)

Let OA = OB = xmetres

So, x2+ x2 = 562

or, 2x2 = 56 x 56

or, x2 = 28 x 56 (2)

Now, area of sector OAB 2 2

2

90 1

360 4

1 2228 56m [From (2)] (3)

4 7

x x x x

x x x

Also, Area of OAB21 56 56m ( 90 ) (4)

4

Ox x AOB

So, area of flower bed AB 2

2

2

1 22 128 56 56 56 m4 7 4

[From (3) and (4)]

1 2228 56 2 m

4 7

1 828 56 m (5)

4 7

x x x x x

x x

x x x

[Continued on next left page!!!]

Std. X - N.C.E.R.TExample 4: (page 231)

In figure, two circular flower beds have been shown ...find the

sum of the areas of the lawn and the flower beds.

SmartCalcSolution:

The diagonals cut at 900.

The total area can be split as

1) two equalsectors OAB, OCD [making

a semi-circle, with radius OA] and

2) two equaltriangles ODA, OBC[making half the square]

Diagonal AC = 56 2

r of sector, OA = 28 2

Total area = halfcircle + halfsquare

2 2

2

1 1=

2 2

1 22 1= 2 28 28 56

2 7 2

r side

x x x x x

= (44 x 56) + (28 x 56)

= 72 x 56

= 4032m2 [by QM Rainbow method]


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Genuine Maths Workings are with gray background[Continued from previous left page!!!]

Similarly, area of the other f lower bed

21 828 56 m (6)4 7

x x x

Therefore, total area 2

2

2 2

1 8 1 856 56 28 56 28 56 m

4 7 4 7

[From (1), (5) and (6)]

2 228 56 2 m

7 7

1828 56 m 4032 m7

x x x x x x x

x

x x


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Find the area of the shaded region in figure, where ABCD is a square of side 14cm.Solution:

Area of square ABCD = 14 x 14 cm2 = 196 cm2.

Diameter of each circle =14

2cm = 7 cm.

So radius of each circle =7

2cm

So area of one circle2

2

2 2

22 7 7cm

7 2 2

154 77cm cm

4 2

r

x x

Therefore, area of the four circles is2 2774 cm 154 cm

2x

Hence, area of the shaded region = (196 154) cm2 = 42 cm2.

Std. X - N.C.E.R.T


Find the area of the shaded

region in figure, where ABCD is

a square of side 14 cm.

SmartCalcSolution:

Side of square = 14

Diameter of a circle = 7Radius of a circle = 3.5cm

Shaded area = square 4 circles

= 142 4 x22

7x 3.5 x 3.5

= 142 11 x 14

= 14(14 11)

= 42

Shaded area = 42cm2


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Std. X - N.C.E.R.T


A toy is in the shape of a

... Determine the volume

of the toy. (= 3.14)

SmartCalcSolution:

Volume of the toy h3 22 1

= r + r3 3

2 1= x 8 + x 8

3 3

2 1= 8 +

3 3

= 8 = 8 x 3.14 = 25.12 cm3.

Std. X - N.C.E.R.T

Example 7: (page 247)A toy is in the shape of a right circular cone

on top of a hemisphere as in the diagram.

The radius of the sphere, as well as the base

of the cone is 2 cm and height of the toy is

4 cm. Determine the volume of the toy.(= 3.14)

Solution:

Volume of the toy

3 2

3 2 3 3

2 1

3 3

2 13.14 2 3.14 2 2 cm 25.12 cm

3 3

r r h

x x x x x

How much work &time to get 25.12?

?


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27



A hemispherical tank full of water is emptied by a pipe at the rate of 437

litres per

second. How much time will it take to empty half the tank, if it is 3m in diameter?

22Take =

7

page 250 [eqn 69, 70]

Solution:

Radius of the hemispherical tank

3

2

m

Volume of the tank

3

3 32 22 3 99

3 7 2 14x x m m

So, the volume of the water to be emptied31 99 99 1000 litres

2 14 28

99000litres

28

x m x

Since,25

7 litres of water is emptied in 1 second,99000

28 litres of water will be

emptied in99000 7

28 25x seconds, i.e. in 16.5 minutes. ?

Std. X - N.C.E.R.T


A hemispherical tank full of water is emptied by a pipe at the ...

... empty half the tank, if it is 3m in diameter?

SmartCalcSolution:

d = 3 m, r =2

3m;

EmtyingRate =4

37

=25

7litres

=25 1

7x1000 7x40 m3 per sec.

Vol. of hemi-sph. tank =2

3r3

Vol. to be emptied = Half of2

3r3 =

1

3r3

=1

3x

22

7x

3

2x

3

2x

3

2

=11 x 9

7 x 4m3

Time to empty =volume

rate

=11 x 9

7 x 4x

7 x 40

1= 990 secs.

= 16.5 min. (as dividing by 60 MUST be mental)

Seethesimplicityofthecalculati

on!


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Genuine Maths Workings are with gray backgroundStd. X - Tamilnadu Matric

Example 2.12: (page 53)

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipeof diameter 3.5 cm through which water flows at the rate of 2 m/sec. Calculate in

minutes the time it takes to fill the tank. page 53

Solution: Cylindrical tank

Diameter = 1.4 m

Radius (r1) = 0.7 m

Height (h1) = 2.1 m

=21

10m

Cylindrical pipe

Diameter = 3.5 cm

35cm

10

7cm

2

Radius (r2) 7 cm47

m400

Speed (h2) = 2 m/sec

Continued on next page ...

Std. X - Tamilnadu Matric


A cylindrical water tank of diameter 1.4 m and height 2.1 m is

being fed by a pipe of diameter 3.5 cm through which water

flows at the rate of 2 m/sec. Calculate in minutes the time it

takes to fill the tank. page 53

SmartCalcSolution:

Tank: d1

= 1.4 m r1

= 0.7 m; h1

= 2.1 m

Pipe: d2

= 3.5 cm r2

= 1.75 cm; h2

= 2 m

= 0.0175 m

Time taken

2

1 1

2

2 2

Vol. of cylindrical tank

Vol. of water in 1 sec.

0.7 0.7 2.1

0.0175 0.0175 2

7 7 21 100000

175 175 221 100 100 10

25 25 2

21 2 4 10secs

21 2 4 10

60

7 4

28 minutes

r h

r h

x x

x x

x x x

x xx x x

x x

x x x

x x x

x

1.4 m

2.1 m

3.5 cm

2 m


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Genuine Maths Workings are with gray backgroundContinued from previous page:

Time taken

2

1 1

2

2 2

Vol. of cylindrical tank

Vol. of cylindrical pipe

7 7 21

10 10 107 7

2400 400

7 7 21 400 400 1

10 10 10 7 7 2

1680 secs

1680min 28 minutes.

60

r h

r h

x x x

x x x

x x x x x


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30

Genuine Maths Workings are with gray backgroundStd. X - Tamilnadu Matric


A solid cylinder has a total surface area of 231 cm

2

. Its curved surface area istwo-thirds of the total surface area. Find the volume of the cylinder. page 50

SmartCalcSolution:

Total surface area = 231 cm2

Curved surface area =2

3of TSA

2

2

2

2

2

2

2

22 231 154

3T ota l s urfa ce are a 2 31

2 ( ) 231

2 2 231

154 2 231

2 231 154

222 77

7

1 22 49772 7 4

7

2

3.5 cm .

rh x cm

cm

r h r

h r

r

r

x xr

r x x

r

r

Continued on next page ...

Std. X - Tamilnadu Matric

Example 2.6: (page 50)A solid cylinder has a total surface area of 231 cm2. Its curved

surface area is two-thirds ... Find the volume of the cylinder.

SmartCalcSolution:

TSA = 231 cm2, CSA is two-thirds of 231

So, 2 x base area = one-third of 231 = 77 cm2

So, base area: 2

2

22 77

7 2

2 7

7 2

7

2

x r

x r

r cm

CSA: 2prh = 2 x 77

22 7x x h 77

7 2

h = 7 cm

Volume of cylinder =Ah77 539

7 269.52 2

x cm3

ORVolume of cylinder =

2 322 7 7 49 117 24.5 11 269.57 2 2 2

xr h x x x x cm


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31


Continued from previous page:

2

2

Curved surface area 154cm

2 154

22 72 154

7 2

1547 cm.

22

Volume of the cylinder r h cu. units.

22 7 7 5397 269.5 cu. cm.

7 2 2 2

rh

x x xh

h

x x x


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Genuine Maths Workings are with gray backgroundStd. XI - TamilnaduExample 6.22: (page 175)

ii)

1 5

3 2 2 11 5

1 32 2

x


3 5 9 5 14564 12 12 12 ???

3 5 48 15 33 331

4 12 48 48x


3

2

3

2

2

33

2

2 2

2 2

3 2

2

3

cot 3cotShow that cot3

3cot 1

cot 3cot. . .

3cot 1

1 3 tan1 3

tantantan

3 3 tan1tan tan

1 3 tan tan

tan 3 tan

1 3 tan

3 tan tan

1cot . .

tan3

A AA

A

A AR H S

A

A

AAA

AA A

A Ax

A A

A

A A

A L HA

.S

Std. XI - Tamilnadu


(ii)

13

6 12 11 2 3

1 32

SmartCalc

x


3 5

36 20 564 123 5 48 15 33

14 12

SmartCalc

x


03) Show that3

2

cot 3cotcot3

3cot 1

A AA

A

SmartCalcSolution:

2

3

33 3

2

1cot3

tan3

1 3tan

3 tan tan

cot 3cot( tan or cot )

3cot 1

AA

A

A A

A Adivide by A multiply by A

A

Multiplyeverything

by the LCM of

all

denominators


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Genuine Maths Workings are with gray backgroundStd. XI - Tamilnadu


2

70 4

0.0280 28100 100/ = = =

30 3 70 4 0.0370 37

100 100 100 100

x

P A B

x x

Note:0s are not written at the end of decimals like 0.0280 and 0.0370

except to imply the accuracy of the number..

70 4

100 100x

= 0.70 x 0.04 = 0.0280 is improper

70 4

100 100x

= 0.7 x 0.04 = 0.028 is proper

70 4 7 4 280.028

100 100 10 100 1000x x

is proper

Std. XI - Tamilnadu


SmartCalcSolution:

2

70 4

100 100/ =

30 3 70 4

100 100 100 100

70 4 7 4

= 30 3 70 4 (3 3) (7 4)

280 28 28= =

370 37 37

x

P A B

x x

x x

orx x x x

Or

2

70 4

100 100/ =

30 3 70 4

100 100 100 100

0.7x0.04=

0.3 x 0.03 + 0.7 x 0.04

0.028=

0.009 + 0.028

0.028 28= =

0.037 37

x

P A B

x x

Gene Maths

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