MATRIX-1

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

2a² – 2a + 10a – 10 = 14

2a² + 8a – 10 = 14

2a² + 8a – 10 - 14 = 0

2a² + 8a - 24 = 0 - - - - - - - (i)

We are going to solve equation (i) by factorization.

2a² + 8a - 24 = 0

2a² + 12a – 4a - 24 = 0 [splitting 8a into 12a-4a]

(2a² + 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

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.       

QUADRATIC-3

SOLUTION

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 =  1 x (a-4) x  a x 2¹

                    2 

21 =  1  x  (a-4)a

a (a-4)   = 21

a² – 4a = 21       

           

a² – 4a - 21= 0 [solving it by factorization]

a² + 3a - 7a - 21= 0

(a² + 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.

LOGARITHMS-5

Simplify        Log0.000001

                        Log(1/100)

Solution

=    Log0.000001

       Log(1/100)

=    Log10^-6

       Log100^-1

=    Log10^-6

       Log(10²)^-1

=    Log10^-6

       Log10^-2

=    -6Log10

       -2Log10

=    -6

       -2

=    3

Hence           Log0.000001         =  3

                        Log(1/100)

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

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)  }