JAMB Mathematics Past Questions & Answers - Page 149

surface area of formula = πr$^2$ 

If the radius is reduced then let its radius be x.

Its area is πx$^2$ .

x : r = 2 : 5,

so $\frac{x}{r}$ =  $\frac{2}{5}$

 → 5x = 2r

Hence, x = 0.4r.

Hence the area of the new circle is π (0.4r)$^2$  = 0.16π r$^2$ .

The ratio of the two areas is

0.16πr$^2$ : πr$^2$ 

 = 0.16 : 1 = 16 : 100

= 4 : 25.

Soln. a*b = ab + a + b,

a ♦ b = a + b + 1

a*c = ac + a + c

(a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1)

= ab + ac + 2a + b + c + 1

Let the cost of living = y.

The new cost of living = $y + \frac{15y}{100} = 1.15y$

The food bill now = $(1 - \frac{90}{100})(1.15y)$

= $1.035y$

The fractional increase in food bill = $(1.035 - 1) \times 100% = 3.5%$

= $\frac{35}{1000} = \frac{7}{200}$

y = 2x² + 9x - 35

2x² + 9x = 35

x² + $\frac{9}{2}$ = $\frac{35}{2}$

x² + $\frac{9}{2}$ + $\frac{81}{16}$ = $\frac{35}{2}$ = $\frac{81}{16}$

(x + $\frac{9}{4}$)² = $\frac{361}{16}$

x = $\frac{-9}{4}$ + $\frac{\sqrt{361}}{16}$

x = $\frac{-9}{4}$ + $\frac{19}{4}$

= 2.5 or -7

-7 < x < $\frac{5}{2}$

Let the present ages of father be x and son = y

five years ago, father = x - 5

son = y - 5

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -15 + 5 = -10 ......(i)

x + y = 110......(ii)

eqn(ii) - eqn(i)

4y = 120

y = 30

sub for y = 30 in eqn(i)

x -10 = 3(30)

x -10 = 90

x = 80