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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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Text Book vs Genuine Maths
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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Text Book vs Genuine Maths
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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Text Book vs Genuine Maths
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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Text Book vs Genuine Maths
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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Text Book vs Genuine Maths
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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Text Book vs Genuine Maths
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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)
Example 109: (page 56)
Convert 8.37 into an ordinary fraction.
SmartCalcSolution:
.37 837
8 37 = 8 or 100 100
Example 45: (page 91)
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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Text Book vs Genuine Maths
9
Genuine Maths Workings are with gray background
Std. VI - Tamil Nadu
Example 117: (page 61)
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
Std. VI - Tamil Nadu
Example 117: (page 61)
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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Text Book vs Genuine Maths
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Genuine Maths Workings are with gray background
Std. VI - Tamil Nadu
Example 122: (page 62)
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
Std. VI - Tamil Nadu
Example 122: (page 62)
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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Text Book vs Genuine Maths
11
Genuine Maths Workings are with gray background
Std. VI - Tamil Nadu
Example 54: (page 132)
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.
Std. VI - Tamil Nadu
Example 54: (page 132)
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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Text Book vs Genuine Maths
12
Genuine Maths Workings are with gray background
Std. VI - Tamil Nadu
Example 56: (page 133)
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
Std. VI - Tamil Nadu
Example 56: (page 133)
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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Text Book vs Genuine Maths
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Genuine Maths Workings are with gray background
Std. VII - N.C.E.R.T
Example 5: (page 161)
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]
Std. VII - N.C.E.R.T
Example 5: (page 161)
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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Text Book vs Genuine Maths
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Genuine Maths Workings are with gray background
Std. VII - N.C.E.R.T
Example 5: (page 207)
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.
Std. VII - N.C.E.R.T
Example 5: (page 207)
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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Text Book vs Genuine Maths
15
Genuine Maths Workings are with gray backgroundStd. VII - N.C.E.R.T
Example 5: (page 251)
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
or,1000 2 5 (required form)
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
or,1000 2 5 (required form)
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
Std. VII - N.C.E.R.T
Example 5: (page 251)
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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Text Book vs Genuine Maths
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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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Genuine Maths Workings are with gray background
Std. VIII - N.C.E.R.T
Example 5: (page 77)
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?
?
Std. VIII - N.C.E.R.T
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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Text Book vs Genuine Maths
18
Genuine Maths Workings are with gray background
Std. VIII - N.C.E.R.T
Example 9: (page 79)
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.
Std. VIII - N.C.E.R.T
Example 9: (page 79)
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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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
Std. VIII - N.C.E.R.T
Example 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 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.
Std. VIII - N.C.E.R.T
Example 3: (page 245)
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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Std. VIII - N.C.E.R.T
Example 5: (page 251)
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
Std. VIII - N.C.E.R.T
Example 5: (page 251)
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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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
Example 16: (page 232)
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
Example 4: (page 231)
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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24
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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Genuine Maths Workings are with gray backgroundStd. X - N.C.E.R.T
Example 5: (page 233)
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
Example 5: (page 233)
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
Example 7: (page 247)
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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Genuine Maths Workings are with gray backgroundStd. X - N.C.E.R.T
Example 11: (page 250)
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
Example 11: (page 250)
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
Example 2.12: (page 53)
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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29
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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Genuine Maths Workings are with gray backgroundStd. X - Tamilnadu Matric
Example 2.6: (page 50)
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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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
Example 6.23: (page 175)
3 5 9 5 14564 12 12 12 ???
3 5 48 15 33 331
4 12 48 48x
Example 6.28: (page 182)
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
Example 6.22: (page 175)
(ii)
13
6 12 11 2 3
1 32
SmartCalc
x
Example 6.23: (page 175)
3 5
36 20 564 123 5 48 15 33
14 12
SmartCalc
x
Example 6.28: (page 182)
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
Example 10.17: (page 166)
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
Example 10.17: (page 166)
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