JAMB Mathematics Past Questions & Answers - Page 314

By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0

Given that y = -3 - 2x + X2

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1

3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)

If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)

Let p + q = T1, then

q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1

Again, let q + r = T2, then

q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2

Using q = q

\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2

5 x 5T1 = 9 x 3T2

\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{25}\)

Giving that, T1 = 27 and T2 = 25

P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1

= (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12

q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1

= (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15

and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2

= (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10

Hence p : q : r = 12: 15 : 10

Using simple interest = \(\frac{P \times T \times R}{100}\),

where: P denoted principal = N400
T denotes time = 3 years
R denotes interest rate = ?

24 = \(\frac{400 \times 3 \times R}{100}\)

24 x 100 = 400 x 3 x R

R = \(\frac{24 \times 100}{400 \times3}\)

= 2%