1990 - JAMB Mathematics Past Questions and Answers - page 3
21
A car painter charges N40.00 per day for himself and N10.00 per day for his assistant. if a fleet of cars were painted for N2000.00 and the painter worked 10days more than his assistant, how much did the assistant receive?
A
N32.00
B
N320.00
C
N420.00
D
N1680.00
correct option: b
Let his assistant work for x days
∴ his master worked (x + 10) day. Amount received by master = 40(x + 10),
amount got by his assistance = 10x
Total amount collected = N2000.00
∴ 40(x + 10) + 10x = 2000
= 40x + 400 + 10x
= 2000
50x + 400 = 2000
50x = 2000 - 400
50x = 1600
x = \(\frac{1600}{50}\)
x = 32 days
Users' Answers & Comments∴ his master worked (x + 10) day. Amount received by master = 40(x + 10),
amount got by his assistance = 10x
Total amount collected = N2000.00
∴ 40(x + 10) + 10x = 2000
= 40x + 400 + 10x
= 2000
50x + 400 = 2000
50x = 2000 - 400
50x = 1600
x = \(\frac{1600}{50}\)
x = 32 days
22
Simplify \(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\)
A
\(\frac{x}{x^2 - y^2}\)
B
\(\frac{y^2}{x^2 - y^2}\)
C
\(\frac{x^2}{x^2 - y^2}\)
D
\(\frac{y}{x^2 - y^2}\)
correct option: b
\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\)
\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\)
= \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\)
= \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\)
= \(\frac{y^2}{(x + y)(x - y)}\)
= \(\frac{y^2}{(x^2 - y^2)}\)
Users' Answers & Comments\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\)
= \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\)
= \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\)
= \(\frac{y^2}{(x + y)(x - y)}\)
= \(\frac{y^2}{(x^2 - y^2)}\)
23
Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3
A
\(\sqrt{10}\)
B
8
C
12.2
D
13.4
correct option: b
Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3
x = 7
∴ x2 = 49
\(\sqrt{y}\) = 3
∴ y2 = 81 = x2 + y2 + z2 = 194
49 + 81 + z2 = 194
130 + z2 = 194
z2 = 194 - 130
= 64
z = \(\sqrt{64}\)
= 8
Users' Answers & Commentsx = 7
∴ x2 = 49
\(\sqrt{y}\) = 3
∴ y2 = 81 = x2 + y2 + z2 = 194
49 + 81 + z2 = 194
130 + z2 = 194
z2 = 194 - 130
= 64
z = \(\sqrt{64}\)
= 8
24
Find the sum of the first twenty terms of the progression log a, log a2, log a3.....
A
log a20
B
log a21
C
log a200
D
log a210
correct option: d
Users' Answers & Comments25
Find the sum of the first 18 terms of the progression 3, 6, 12......
A
3(217 - 1)
B
3(218 - 1)
C
3(218 + 1)
D
3(217 - 1)
correct option: d
3 + 6 + 12 + .....18thy term
1st term = 3, common ratio \(\frac{6}{3}\) = 2
n = 18, sum og GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\)
s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\)
= 3(217 - 1)
Users' Answers & Comments1st term = 3, common ratio \(\frac{6}{3}\) = 2
n = 18, sum og GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\)
s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\)
= 3(217 - 1)
26
At what value of x is the function x2 + x + 1 minimum?
A
1
B
\(\frac{3}{4}\)
C
\(\frac{5}{3}\)
D
9
correct option: a
x + x + 1
\(\frac{dy}{dx}\) = 2x + 1
At the turning point, \(\frac{dy}{dx}\) = 0
2x + 1 = 0
x = -\(\frac{1}{2}\)
\(\frac{d^2y}{dx^2}\) = 2 > 0(min Pt)
= \(\frac{1}{4}\) + \(\frac{1}{2}\)
= 1
Users' Answers & Comments\(\frac{dy}{dx}\) = 2x + 1
At the turning point, \(\frac{dy}{dx}\) = 0
2x + 1 = 0
x = -\(\frac{1}{2}\)
\(\frac{d^2y}{dx^2}\) = 2 > 0(min Pt)
= \(\frac{1}{4}\) + \(\frac{1}{2}\)
= 1
27
The angle of a sector of s circle, radius 10.5cm, is 48 o, Calculate the perimeter of the sector
A
8.8cm
B
25.4cm
C
25.6cm
D
28.8cm
correct option: d
Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm
Perimeter = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm
Users' Answers & Comments= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm
Perimeter = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm
28
Find the length of a side of a rhombus whose diagonals are 6cm and 8cm
A
8cm
B
5cm
C
4cm
D
3cm
correct option: b
Users' Answers & Comments29
Each of the interior angles of a regular polygon is 140o. How many sides has the polygon?
A
9
B
8
C
7
D
5
correct option: a
For a regular polygon of n sides
n = \(\frac{360}{\text{Exterior angle}}\)
Exterior < = 180o - 140o
= 40o
n = \(\frac{360}{40}\)
= 9 sides
Users' Answers & Commentsn = \(\frac{360}{\text{Exterior angle}}\)
Exterior < = 180o - 140o
= 40o
n = \(\frac{360}{40}\)
= 9 sides
30
If Cos \(\theta\) = \(\frac{12}{13}\). Find \(\theta\) + cos2\(\theta\)
A
\(\frac{169}{25}\)
B
\(\frac{25}{169}\)
C
\(\frac{169}{144}\)
D
\(\frac{144}{169}\)
correct option: a
Cos \(\theta\) = \(\frac{12}{13}\)
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\)
If cos2\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\)
= 1 + \(\frac{1}{\frac{(5)^2}{12}}\)
= 1 + \(\frac{1}{\frac{25}{144}}\)
= 1 + \(\frac{144}{25}\)
= \(\frac{25 + 144}{25}\)
= \(\frac{169}{25}\)
Users' Answers & Commentsx2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\)
If cos2\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\)
= 1 + \(\frac{1}{\frac{(5)^2}{12}}\)
= 1 + \(\frac{1}{\frac{25}{144}}\)
= 1 + \(\frac{144}{25}\)
= \(\frac{25 + 144}{25}\)
= \(\frac{169}{25}\)