ALGEBRA 4

The sum of four consecutive numbers is 362. Find the 4th 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	362

Then, n + (n+1) + (n+2) + (n+3) = 362

4n + 1+2+3= 362

4n + 6 = 362

4n = 362 – 6

4n = 356 

4n = 356

 4       4

n = 89

4th number = n + 3

                 = 89 + 3

                 = 92

Hence the 4th number is 92

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

The sum of four consecutive numbers is 358. Find the 4th number.  

INEQUALITIES-1

If │4x – 3 │= 17; Find  x

Solution

±(4x – 3 )= 17

4x -3= 17    OR   –(4x-3) = 17

4x – 3 = 17    OR   -4x+3=17

4x = 17 + 3    OR   -4x= 17-3

4x = 20     OR   -4x  =  14

4      4                4        4

x = 5  OR  x = ⁷/2

Hence x = 5 OR x = 7/2     

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

If │4x – 7 │= 19; Find  x

LOGARITHMS 18

Evaluate Log1000000000 +  log 0.001

Solution

= Log1000000000 +  log 0.001

= Log10⁹ +  log 10^-3

= 9Log10 +  (-3log 10)

= (9x1) + (-3x1)

= 9 + (-3)

= 6

Hence Log1000000000 +  log 0.001 = 6

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

Evaluate Log100000000 +  log 0.00001

SETS 4

In a certain school, 186 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) = 186,

n(CnP)= 100

Solution

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

186  =  130 + n(P)  – 100

186  =  n(P) + 30

186 - 30  =  n(P)

n(P)= 156

physics only = n(P) – n(CnP)

                         = 156 – 100

                         = 56 

Hence those taking only physics are 56.       

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

In a certain school, 186 students take either Chemistry or physics. 134 take chemistry and 110 take both subjects. Find those who take Chemistry only.