## Wednesday, 31 July 2013

## Tuesday, 30 July 2013

### QUADRATIC-4

Length of a
rectangle is

*a+5*while the width is*2a-2*. If the area is 14cm, find its perimeter.__Solution__

Area = length
x width

14 = (a+5) x
(2a-2)

(a+5) (2a-2) =
14

2a

^{2}– 2a + 10a – 10 = 14
2a

^{2}+ 8a – 10 = 14
2a

^{2}+ 8a – 10 - 14 = 0
2a

^{2}+ 8a - 24 = 0 - - - - - - - (i)
We are going
to solve equation (i) by factorization.

2a

^{2}+ 8a - 24 = 0
2a

^{2}+ 12a – 4a - 24 = 0 [splitting 8a into 12a-4a]
(2a

^{2}+ 12a) – (4a – 24) = 0
2a(a + 6) – 4(a
+ 6) = 0

(2a - 4)(a + 6) = 0

2a – 4 = 0 OR
a + 6 = 0

a = 2 OR a =
-6

Length = a+5

= 2+5

= 7

Width = 2a-2

= 2(2)- 2

= 4 – 2

= 2

Perimeter =
2(length + width)

Perimeter =
2(7 + 2)

= 2 x 14

= 28

__Hence perimeter is 28cm__

## Monday, 29 July 2013

### SETS-5

In a certain school, 180 students take
either Chemistry or physics. 130 take chemistry and 100 take both subjects.
Find those who take physics only.

__Solution__
In most cases, OR stands for union whereas
AND/BOTH, stands for intersection.

Let Chemistry = n(C), physics = n(P).

n(C)= 130 ,

n(P)= ?

n(CuP) = 180,

n(CnP)= 100

__Solution__
n(CuP) = n(C) + n(P) - n(CnP)

180
= 130 + n(P) – 100

180
= n(P) + 30

180 - 30
= n(P)

n(P)= 150

physics only = n(P) – n(CnP)

= 150 – 100

= 50

__Hence those taking only physics are 50.__

## Saturday, 27 July 2013

## Friday, 26 July 2013

### AREA OF TRIANGLE

Find the base of the following triangle
if its area is 10.5 cm

__Solution__

Area
=

__1__x base x height
2

10.5
=

__1__x (a-4) x a
2

2
x 10.5 = ~~2~~

__1__x (a-4) x a x^{1}
21
= 1
x (a-4)a

a (a-4)
= 21

a

^{2}– 4a = 21
a

^{2}– 4a - 21= 0 [solving it by factorization]
a

^{2}+ 3a - 7a - 21= 0
(a

^{2}+ 3a) – (7a – 21)= 0
a(a
+ 3) –7 (a + 3)= 0

(a – 7)(a + 3) = 0

a – 7=0
OR a + 3 = 0

a =7
OR a = -3

Since there is no negative length, we take
a=7.

Base = a – 4

= 7 – 4

= 3

__Hence the base is 3cm.__

## Thursday, 25 July 2013

### LOGARITHMS-5

Simplify

__Log0.000001__
Log(1/100)

__Solution__

=

__Log0.000001__
Log(1/100)

=

__Log10__^{-6}
Log100

^{-1}
=

__Log10__^{-6}
Log(10

^{2})^{-1}
=

__Log10__^{-6}
Log10

^{-2}
=

__-6Log10__
-2Log10

=

__-6__
-2

= 3

Hence

__Log0.000001__= 3
Log(1/100)

## Wednesday, 24 July 2013

### PERIMETER

Perimeter of
a square below is 44mm. Find the length of one side.

__Solution__

Perimeter = 4
x side

44 = 4(a - 5)

44 = 4a – 20

44 + 20 = 4a

64 = 4a

__64__=

__4a__

4 4

16 = a

Now the side
= a – 5

= 16 – 5

= 11

**Hence length of one side is 11mm**## Tuesday, 23 July 2013

### FUNCTIONS-5

Find the
inverse of { (2,6), (7,1), (-9,8), (-3,-6), (11,-4) }

__Solution__

Here, the inverse
is the opposite of a given expression. Therefore, the values will be swapped.

__Hence, inverse is { (6,2), (1,7), (8,-9), (-6,-3), (-4,11) }__

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