JAMB Mathematics Past Questions & Answers - Page 395

1,971.

Three consecutive terms of a geometric progression are given as n-2, n and n+3. Find the common ratio

A.

1/4

B.

1/2

C.

2/3

D.

3/2

Correct answer is D

\(r=\frac{n}{n-2}\hspace{1mm}and\hspace{1mm}r=\frac{n+3}{n}\\∴\frac{n}{n-2}=\frac{n+3}{n}\\n^2 = n^2 +3n - 2n-6\\0=n-6\\∴n=6\\But\hspace{1mm}r = \frac{n}{n-2}\\r=\frac{6}{6-2}\\\frac{6}{4}=\frac{3}{2}\)

1,972.

Find the value of x and y respectively if 3x - 5y + 5 = 0 and 4x - 7y + 8 = 0

A.

-5, -4

B.

-4,. -5

C.

4, 5

D.

5, 4

Correct answer is D

3x - 5y + 5 = 0 → eqn1
4x - 7y + 8 = 0 → eqn2
eqn1 * 4; 12x - 20y + 20 = 0 → eqn3
eqn2 * 3; 12x - 21y + 24 = 0 → eqn4
eqn3 - eqn4 = y - 4 = 0
∴ y = 4
From eqn1,
3x - 5y + 5 = 0
3x - 5(4) + 5 = 0
3x - 20 + 5 = 0
3x - 15 = 0
3x = 15
x = 5
x and y = 5, 4 respectively

1,974.

The graph of the function y = x\(^2\) + 4 and a straight line PQ are drawn to solve the equation x\(^2\) - 3x + 2 = 0. What is the equation of PQ?

A.

y = 3x - 2

B.

y = 3x + 2

C.

y = 3x - 4

D.

y = 3x + 4

Correct answer is B

x\(^2\) + 4 = x\(^2\) - 3x + 2
3x + 2 = y

1,975.

Find the range of the value of x satisfying the inequalities 5 + x \(\leq\) 8 and 13 + x \(\geq\) 7

A.

-3 \(\leq\) x \(\leq\) 3

B.

3 \(\leq\) x \(\leq\) 6

C.

-6 \(\leq\) x \(\leq\) 3

D.

-6 \(\leq\) x \(\leq\) -3

Correct answer is C

5 + x \(\leq\) 8 and 13 + x \(\geq\) 7
5 + x \(\leq\) 8
X \(\leq\) 8 – 5
X \(\leq\) 3
And 13 + x \(\geq\) 7
X \(\geq\) 7 – 15
X \(\geq\) -7
Combining the two together
-6 \(\leq\) x \(\leq\) 3