2015 - JAMB Mathematics Past Questions and Answers - page 5

x² − 2x − 15

(x² − 5x) + (3x − 15)

x(x − 5)+ 3(x − 5)

(x − 5)(x + 3)

I =[ PRT ÷ 100]

[(325 × 5 × 3) ÷ (100)]

N(195 ÷ 4)

N48.75K

(0.09)² = 0.09 × 0.09

= 0.0081

= 0.008100 to 4 significant figures

Please, start counting first from the non-zero digits i.e. 8

( [1 ÷ (x^2 + 3x + 2)] + [1 ÷ (x^2 + 5x + 6)] )

= ( 1 ÷ (x^2 + 3x + 2) + [1 ÷ (x^2 +5x + 6)])

= ( [1 ÷ ((x^2 + x) + (2x + 2) )] + [1 ÷ ((x^2 + 3x) + (2x + 6) )] )

= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]

= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]

=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)

Using the L.C.M

=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)

=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)

= ( \frac{2 }{(x + 1)(x + 3)} )

One bag with 3 blue and 5 red

Pr (H) = 3/8Pr (r) 5/8

Another bag with 2 blue and 4 red (note one ball is drawn from each bag)

i.e. prob.=8 − 1 = 7

prob. (2 ball both are blue=pr (1st blue and 2nd blue)

prob. 2 balls both blue = 3/8 × 2/7

= ( \frac{3}{28} )

A ∩ B = {3, 4, ⅓}

132_x = 70₈

( 1 × x^2 + 3 × x^1 + 2 × x^0 )

( 7 × 8^1 + 0 × 8^0 )

( x^2 + 3x + 2 − 56 = 0 )

( x^2 + 3x − 54 = 0 )

x(x + 9)−6(x + 9) = 0

(x + 9)(x − 6) = 0

Either (x + 9) = 0 or (x − 6) = 0

x = − 9 or +6

The positive values for x = 6

The base number for 132_x = 132₆

Cost Price:

Selling price = 100 : (100 + 15)

= 100 : 115

Cost Price

= 100 ÷ 115 of selling price (i.e N420,000)

= 100 ÷ 15 × N420,000

= N365,217.4

∴ Cost Price = N365, 217