BODMAS 1

Evaluate 190 + 50 x 30 – 160 ÷ 4

Solution

We apply BODMAS.

= 190 + 50 x 30 – 160 ÷ 4

= 190 + 50 x 30 – 40    [After dividing]

= 190 + 1500 – 40    [After multiplying]

= 1690 – 40    [After adding]

= 1650       [After subtracting]

Hence 190 + 50 x 30 – 160 ÷ 4= 1650      

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

Evaluate 900 + 40 x 30 – 200 ÷ 5

PROBABILITY 1

A number is chosen at random from 1 – 15 inclusive. Find the probability that it is a multiple of five or an even number.

Solution

THIS IS A NON-MUTUALLY EXCLUSIVE EVENT.

Let n(S) represent sample space

P(E) = probability of even number

P(M) = probability of a multiple of 5

n(S) = {1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12, 13, 14, 15} = 15

n(E) = {2, 4, 6, 8, 10, 12, 14} = 7

n(M) = {5, 10, 15 } = 3

But 14 appears on both categories.

P(E) = ⁷/15

P(M) = ³/15

P(EnM) = ¹/15

………………………….

P(EuM) = P(E) + P(M) - P(EnM) 

P(EuM) = ⁷/15 + ³/15 -¹/15

            = ⁹/15

            = ³/5

∴P(EuM) = ³/5

TRY THIS………………

A number is chosen at random from 4 – 30 inclusive. Find the probability that it is a multiple of 5 or an even number. 

FUNCTIONS 1

Given that F(x) = 3x  +  17. Find F(-4)

Solution

F(x) = 3x  +  17

F(-4) = 3(-4)  +  17

F(-4) = -12  +  17

F(-4) = 5

Hence  F(-4) = 5

TRY THIS………………..

Given that F(x) = 44x  -  3. Find F(-4)

SETS 1

If n(AnB)=46, n(B)= 96 and n(AuB)= 140, find n(A)

Solution

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

140 = n(A)  + 96 - 46

140 = n(A)  +  50

140 - 50 = n(A) 

90 = n(A) 

Hence n(A)  = 90 answer

TRY THIS………….

If n(AnB)=40, n(B)= 98 and n(AuB)= 175, find n(A)