LOG J1

 

If log 2= 0.3010; find the value of log 1600000 without using tables.

 

solution

 

=log1600000

 

=log(16 x 100,000)

 

=log16 + log100,000

 

=log2⁴ + log10⁵

 

=4log2 + 5log10

 

=4(0.3010) + (5 x 1)  [since log10=1 and Log3=0.3010]

 

=(0.6020) +  5

 

=5.6020

 

Hence log1600,000=5.6020 answer.

 

 

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If log 2= 0.3010; find the value of log 40,000,000 without using tables.

 

FACTORIZE J1

 

Factorize completely tc + tr - cr - c².  

 

Solution

 

= tc + tr - cr - c².

 

= (tc + tr) – (cr - c²).      [Grouping the factors]

 

= t(c + r) - c(r + c).   [after factoring out]

 

 = (c + r)(t - c) answer

 

  Hence tc + tr - cr - c². =  (c + r)(t - c).   

 

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Factorize completely yw + yr - rw - w². 

SETS J1

 If n(A)= 85 , n(B)= 96 and n(AuB)= 129, find n(AnB).

 

Solution

 

n(AuB) = n(A) + n(B) - n(AnB)

 

129 = 85 + 96 - n(AnB)

 

129 = 181 - n(AnB)

 

129 - 181 = - n(AnB)

 

-58 = -n(AnB)

 

n(AnB) = 58 [after dividing by -1 both sides]

 

Hence n(AnB) = 58 answer

 

 

TRY THIS………….

 

 

If n(A)= 91 , n(B)= 94 and n(AuB)= 127, find n(AnB).

G.PROGRESSION J1

 The 1st term of a geometric progression is 7 and the 5th term is 112. Find i)the common ratio. ii) 12^th term

 

Solution

 

G₁=7, G₅=112, n=5

 

i) G_n = G₁r^n-1;

 

G₅ = G₁r^5-1;

 

G₅ = G₁r⁴;

 

112 = 7 x r⁴;

 

112 = 7r⁴;

 7       7

 

16 = r⁴;  Finding the fourth root of 16,

 

r = 2

 

 

 

Hence the common ratio is 2.

 

ii)  12^th term

 

G_n = G₁r^n-1;

 

G₁₂ = G₁r^12-1;

 

G₁₂ = G₁r¹¹;

 

G₁₂ = 7 x 2¹¹;

 

G₁₂ = 7 x 2048;

 

G₁₂ = 14336

 

Hence the 12th term is 14336

 

TRY THIS………………………….. 

 

The 1st term of a geometric progression is 2 and the 6th term is 486. Find i) the common ratio. ii) 10^th term

QUADRATICS - I1

 

ALGEBRA - I1

 If │4x – 7 │= 19; Find  x

 

Solution

 

±(4x – 11 )= 19

 

4x -11= 19   OR   –(4x-11) = 19

 

4x – 11 = 19    OR   -4x+11=19

 

4x = 19 + 11    OR   -4x= 19-11

 

4x = 30     OR   -4x  =   8

4      4               -4        -4

 

x = ⁷/2  OR  x = -2

 

 

Hence x = ⁷/2  OR  x = -2

ARITHMETIC - I1

 Evaluate 18 x 369 + 331 x 18.

 

Solution

 

= 18 x 369 + 331 x 18.

 

= 18 x (369 + 331)     factoring out the common number

 

= 18 x 700   

 

= 12600      [after multiplying 18and 7 and adding two zeros on the answer]

 

Hence 18 x 369 + 331 x 18= 12600

 

TRY THIS………….

 

 

Evaluate 137 x 516 + 484 x 137.

ALGEBRA - I1

 

The sum of four consecutive numbers is 474. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+1	n+2	n+3	474

 

Then, n + (n+1) + (n+2) + (n+3) = 474

 

4n + 1+2+3= 474

 

4n + 6 = 474

 

4n = 474 – 6

 

4n = 468 

 

4n = 468

 4        4

 

n = 117

 

3rd number = n +2

 

                   = 117 + 2

 

                   = 119

 

Hence the 3rd number is 119

 

TRY THIS………………….. 

 

The sum of four consecutive numbers is 1006. Find the largest number.

LOGARITHMS - I1

 Evaluate Log₂(512 ÷ 8).

 

Solution

 

= Log₂(512 ÷ 8)

 

= Log₂512 -  Log₂8          (applying the product rule)

 

= Log₂2⁹ -  Log₂2³             ( 512= 2⁹ and 8=2³ )

 

= 9Log₂2 -  3Log₂2       ( remember  Log_aa^c = cLog_aa )

 

= (9 x 1) -  (3 x 1)          ( remember  Log_aa = 1 )

 

= 9 - 3

 

= 6 answer

 

 

TRY THIS...............

 

Evaluate Log₂(2048 - 32).

 

FUNCTIONS -I1

 If f(x) = |2x - 10| evaluate f(-70)

 

Solution

 

f(x) = |2x - 10|

 

f(-70) = |(2 x -70) - 10|

 

f(-70) = |-140 - 10|

 

f(-70) = |-150| = 150 (since any number out of absolute signs is +ve).

 

Hence f(-70) = 150

 

TRY THIS..................

 

If f(x) = |5x - 300| evaluate f(-60)