Paper JAMB 2016 General Maths Objective
Objective paper · 31 questions · partial
JAMB 2016 · UTME Topics include Logarithms, Quadratics & their graphs, Number foundations & fractions, Inequalities, Probability, Calculus (JAMB bridge).
Our copy of this paper is missing questions 3, 4, 11, 12, 19, 20, 25, 31, 37.
Sit this paper Answer every question in order, timed if you like (suggested 20 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
1 2 5 6 7 8 9 10 13 14 15 16 17 18 21 22 23 24 26 27 28 29 30 32 33 34 35 36 38 39 40 Without using tables, evaluate log 2 4 + log 4 2 − log 25 5 \log_2 4 + \log_4 2 - \log_{25} 5 log 2 4 + log 4 2 − log 25 5
A 1 2 \frac12 2 1 B 1 5 \frac15 5 1 C 0 D 2
Worked solution (try it first) 2 2 = 4 2^2 = 4 2 2 = 4 , so
log 2 4 = 2 \log_2 4 = 2 log 2 4 = 2 .
4 1 2 = 2 4^{\frac12} = 2 4 2 1 = 2 , so
log 4 2 = 1 2 \log_4 2 = \frac12 log 4 2 = 2 1 .
In the same way
25 1 2 = 5 25^{\frac12} = 5 2 5 2 1 = 5 , so
log 25 5 = 1 2 \log_{25} 5 = \frac12 log 25 5 = 2 1 .
So the value is
2 + 1 2 − 1 2 = 2 2 + \frac12 - \frac12 = 2 2 + 2 1 − 2 1 = 2 , option D.
Watch out
1 2 \frac12 2 1 log 4 2 \log_4 2 log 4 2 log 2 4 = 2 \log_2 4 = 2 log 2 4 = 2 Also set as JAMB 1985 · UME · Q8
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Find the values of p p p x 2 − ( p − 2 ) x + 2 p + 1 = 0 x^2 - (p - 2)x + 2p + 1 = 0 x 2 − ( p − 2 ) x + 2 p + 1 = 0
A ( 0 , 12 ) (0, 12) ( 0 , 12 ) B ( 1 , 2 ) (1, 2) ( 1 , 2 ) C ( 21 , 0 ) (21, 0) ( 21 , 0 ) D ( 4 , 5 ) (4, 5) ( 4 , 5 )
Worked solution (try it first) Equal roots means
b 2 = 4 a c b^2 = 4ac b 2 = 4 a c .
Here
a = 1 a = 1 a = 1 ,
b = − ( p − 2 ) b = -(p - 2) b = − ( p − 2 ) and
c = 2 p + 1 c = 2p + 1 c = 2 p + 1 .
So
( p − 2 ) 2 = 4 ( 2 p + 1 ) (p - 2)^2 = 4(2p + 1) ( p − 2 ) 2 = 4 ( 2 p + 1 ) , which is
p 2 − 4 p + 4 = 8 p + 4 p^2 - 4p + 4 = 8p + 4 p 2 − 4 p + 4 = 8 p + 4 .
Simplify:
p 2 − 12 p = 0 p^2 - 12p = 0 p 2 − 12 p = 0 , so
p ( p − 12 ) = 0 p(p - 12) = 0 p ( p − 12 ) = 0 .
So
p = 0 p = 0 p = 0 or
p = 12 p = 12 p = 12 , option A.
Watch out
Don't divide p 2 = 12 p p^2 = 12p p 2 = 12 p p p p p = 0 p = 0 p = 0 p ( p − 12 ) = 0 p(p - 12) = 0 p ( p − 12 ) = 0 Also set as JAMB 1985 · UME · Q13
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Simplify 3 1 3 − 1 1 4 × 2 3 + 1 2 5 3\frac13 - 1\frac14 \times \frac23 + 1\frac25 3 3 1 − 1 4 1 × 3 2 + 1 5 2
A 2 17 30 2\frac{17}{30} 2 30 17 B 3 9 10 3\frac{9}{10} 3 10 9 C 4 1 10 4\frac{1}{10} 4 10 1 D 4 11 36 4\frac{11}{36} 4 36 11
Worked solution (try it first) Multiply before you add or subtract (BODMAS):
1 1 4 = 5 4 1\frac14 = \frac54 1 4 1 = 4 5 , and
5 4 × 2 3 = 10 12 \frac54 \times \frac23 = \frac{10}{12} 4 5 × 3 2 = 12 10 Now
10 3 − 5 6 + 7 5 \frac{10}{3} - \frac56 + \frac75 3 10 − 6 5 + 5 7 .
The LCD is 30:
100 30 − 25 30 + 42 30 = 117 30 \frac{100}{30} - \frac{25}{30} + \frac{42}{30} = \frac{117}{30} 30 100 − 30 25 + 30 42 = 30 117 .
117 30 = 3 27 30 \frac{117}{30} = 3\frac{27}{30} 30 117 = 3 30 27 = 3 9 10 = 3\frac{9}{10} = 3 10 9 , option B.
Watch out
Do the multiplication first. Working out 3 1 3 − 1 1 4 3\frac13 - 1\frac14 3 3 1 − 1 4 1 2 3 + 1 2 5 \frac23 + 1\frac25 3 2 + 1 5 2 25 12 × 31 15 = 4 11 36 \frac{25}{12} \times \frac{31}{15} = 4\frac{11}{36} 12 25 × 15 31 = 4 36 11 Also set as JAMB 1991 · UME · Q1
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Factorize 1 − ( a − b ) 2 1 - (a - b)^2 1 − ( a − b ) 2
A ( 1 − a − b ) ( 1 − a + b ) (1 - a - b)(1 - a + b) ( 1 − a − b ) ( 1 − a + b ) B ( 1 − a + b ) ( 1 + a − b ) (1 - a + b)(1 + a - b) ( 1 − a + b ) ( 1 + a − b ) C ( 1 − a + b ) ( 1 − a + b ) (1 - a + b)(1 - a + b) ( 1 − a + b ) ( 1 − a + b ) D ( 1 − a − b ) ( 1 + a − b ) (1 - a - b)(1 + a - b) ( 1 − a − b ) ( 1 + a − b )
Worked solution (try it first) This is a difference of two squares, with
1 = 1 2 1 = 1^2 1 = 1 2 :
1 − ( a − b ) 2 = [ 1 − ( a − b ) ] [ 1 + ( a − b ) ] 1 - (a - b)^2 = [1 - (a - b)][1 + (a - b)] 1 − ( a − b ) 2 = [ 1 − ( a − b )] [ 1 + ( a − b )] .
Remove the inner brackets:
1 − ( a − b ) = 1 − a + b 1 - (a - b) = 1 - a + b 1 − ( a − b ) = 1 − a + b and
1 + ( a − b ) = 1 + a − b 1 + (a - b) = 1 + a - b 1 + ( a − b ) = 1 + a − b .
So the factors are
( 1 − a + b ) ( 1 + a − b ) (1 - a + b)(1 + a - b) ( 1 − a + b ) ( 1 + a − b ) , option B.
Watch out
− ( a − b ) = − a + b -(a - b) = -a + b − ( a − b ) = − a + b 1 − a − b 1 - a - b 1 − a − b Also set as JAMB 1991 · UME · Q18
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Find the range of values of x x x ( x 2 + x 3 + x 4 ) < 1 \left(\frac x2 + \frac x3 + \frac x4\right) < 1 ( 2 x + 3 x + 4 x ) < 1
A x < 12 13 x < \frac{12}{13} x < 13 12 B x < 13 x < 13 x < 13 C x < 3 x < 3 x < 3 D x < 13 12 x < \frac{13}{12} x < 12 13
Worked solution (try it first) Multiply every term by 12, the LCM of 2, 3 and 4:
6 x + 4 x + 3 x < 12 6x + 4x + 3x < 12 6 x + 4 x + 3 x < 12 .
Collect the terms:
13 x < 12 13x < 12 13 x < 12 .
Divide by 13:
x < 12 13 x < \frac{12}{13} x < 13 12 , option A.
Watch out
From 13 x < 12 13x < 12 13 x < 12 12 13 \frac{12}{13} 13 12 13 12 \frac{13}{12} 12 13 Also set as JAMB 1991 · UME · Q21
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A crate of soft drinks contains 10 bottles of Coca-Cola, 8 of Fanta and 6 of Sprite. If one bottle is selected at random, what is the probability that it is NOT a Coca-Cola bottle?
A 5 12 \frac{5}{12} 12 5 B 1 3 \frac13 3 1 C 3 4 \frac34 4 3 D 7 12 \frac{7}{12} 12 7
Worked solution (try it first) There are
10 + 8 + 6 = 24 10 + 8 + 6 = 24 10 + 8 + 6 = 24 bottles.
Not Coca-Cola means Fanta or Sprite:
8 + 6 = 14 8 + 6 = 14 8 + 6 = 14 bottles.
So the probability is
14 24 = 7 12 \frac{14}{24} = \frac{7}{12} 24 14 = 12 7 , option D.
Watch out
10 24 = 5 12 \frac{10}{24} = \frac{5}{12} 24 10 = 12 5 Also set as JAMB 1991 · UME · Q50
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The gradient of a curve is 2 x + 7 2x + 7 2 x + 7 ( 2 , 0 ) (2, 0) ( 2 , 0 )
A y = x 2 + 7 x + 9 y = x^2 + 7x + 9 y = x 2 + 7 x + 9 B y = x 2 + 7 x − 18 y = x^2 + 7x - 18 y = x 2 + 7 x − 18 C y = x 2 + 7 x + 18 y = x^2 + 7x + 18 y = x 2 + 7 x + 18 D y = x 2 + 14 x + 11 y = x^2 + 14x + 11 y = x 2 + 14 x + 11
Worked solution (try it first) Integrate the gradient to get the curve:
y = x 2 + 7 x + c y = x^2 + 7x + c y = x 2 + 7 x + c .
The curve passes through
( 2 , 0 ) (2, 0) ( 2 , 0 ) , so put in
x = 2 x = 2 x = 2 and
y = 0 y = 0 y = 0 :
0 = 4 + 14 + c 0 = 4 + 14 + c 0 = 4 + 14 + c .
So
c = − 18 c = -18 c = − 18 and the curve is
y = x 2 + 7 x − 18 y = x^2 + 7x - 18 y = x 2 + 7 x − 18 , option B.
Watch out
From 0 = 18 + c 0 = 18 + c 0 = 18 + c c = − 18 c = -18 c = − 18 c = + 18 c = +18 c = + 18 ( 2 , 36 ) (2, 36) ( 2 , 36 ) ( 2 , 0 ) (2, 0) ( 2 , 0 ) Report a problem with this question
Differentiate ( cos q − sin q ) 2 (\cos q - \sin q)^2 ( cos q − sin q ) 2 q q q
A − 2 cos 2 q -2\cos 2q − 2 cos 2 q B − 2 sin 2 q -2\sin 2q − 2 sin 2 q C 1 − 2 cos 2 q 1 - 2\cos 2q 1 − 2 cos 2 q D 1 − 2 sin 2 q 1 - 2\sin 2q 1 − 2 sin 2 q
Worked solution (try it first) Expand:
( cos q − sin q ) 2 = cos 2 q + sin 2 q − 2 sin q cos q (\cos q - \sin q)^2 = \cos^2q + \sin^2q - 2\sin q\cos q ( cos q − sin q ) 2 = cos 2 q + sin 2 q − 2 sin q cos q .
Use
cos 2 q + sin 2 q = 1 \cos^2q + \sin^2q = 1 cos 2 q + sin 2 q = 1 and
2 sin q cos q = sin 2 q 2\sin q\cos q = \sin2q 2 sin q cos q = sin 2 q : the expression is
1 − sin 2 q 1 - \sin2q 1 − sin 2 q .
Differentiate: the 1 gives 0 and
sin 2 q \sin2q sin 2 q gives
2 cos 2 q 2\cos2q 2 cos 2 q .
So the derivative is
− 2 cos 2 q -2\cos2q − 2 cos 2 q , option A.
Watch out
A constant differentiates to 0, so the 1 disappears. Keeping it gives 1 − 2 cos 2 q 1 - 2\cos2q 1 − 2 cos 2 q Report a problem with this question
If y = x 2 − x − 12 y = x^2 - x - 12 y = x 2 − x − 12 x x x y ≥ 0 y \ge 0 y ≥ 0
A x < − 2 x < -2 x < − 2 x > 4 x > 4 x > 4 B x ≤ − 3 x \le -3 x ≤ − 3 x ≥ 4 x \ge 4 x ≥ 4 C − 3 < x ≤ 4 -3 < x \le 4 − 3 < x ≤ 4 D − 3 ≤ x ≤ 4 -3 \le x \le 4 − 3 ≤ x ≤ 4
Worked solution (try it first) Factorise:
x 2 − x − 12 = ( x − 4 ) ( x + 3 ) x^2 - x - 12 = (x - 4)(x + 3) x 2 − x − 12 = ( x − 4 ) ( x + 3 ) , so the roots are
x = 4 x = 4 x = 4 and
x = − 3 x = -3 x = − 3 .
The graph is a U shape, so
y ≥ 0 y \ge 0 y ≥ 0 outside the roots and at the roots themselves.
So
x ≤ − 3 x \le -3 x ≤ − 3 or
x ≥ 4 x \ge 4 x ≥ 4 , option B.
Watch out
Option D, − 3 ≤ x ≤ 4 -3 \le x \le 4 − 3 ≤ x ≤ 4 y ≤ 0 y \le 0 y ≤ 0 x = 0 x = 0 x = 0 y = − 12 y = -12 y = − 12 Report a problem with this question
A man bought a second-hand photocopying machine for ₦34,000. He serviced it at a cost of ₦2,000 and then sold it at a profit of 15 % 15\% 15%
A ₦37,550 B ₦40,400 C ₦41,400 D ₦42,400
Worked solution (try it first) The cost includes the servicing:
34 000 + 2000 = 34\,000 + 2000 = 34 000 + 2000 = ₦36,000.
A
15 % 15\% 15% profit means selling for
115 % 115\% 115% of the cost:
1.15 × 36 000 = 41 400 1.15 \times 36\,000 = 41\,400 1.15 × 36 000 = 41 400 .
So the selling price was ₦41,400, option C.
Watch out
Add the servicing to the cost before you work out the 15 % 15\% 15% Report a problem with this question
Find the radius of a sphere whose surface area is 154 cm 2 154\text{ cm}^2 154 cm 2 [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 7.00 cm 7.00\text{ cm} 7.00 cm B 3.50 cm 3.50\text{ cm} 3.50 cm C 3.00 cm 3.00\text{ cm} 3.00 cm D 1.75 cm 1.75\text{ cm} 1.75 cm
Worked solution (try it first) The surface area of a sphere is
4 π r 2 4\pi r^2 4 π r 2 :
4 × 22 7 × r 2 = 154 4 \times \frac{22}{7} \times r^2 = 154 4 × 7 22 × r 2 = 154 .
So
r 2 = 154 × 7 88 = 12.25 r^2 = \dfrac{154 \times 7}{88} = 12.25 r 2 = 88 154 × 7 = 12.25 .
Take the square root:
r = 3.50 r = 3.50 r = 3.50 cm, option B.
Watch out
Keep the 4 in 4 π r 2 4\pi r^2 4 π r 2 π r 2 = 154 \pi r^2 = 154 π r 2 = 154 r = 7.00 r = 7.00 r = 7.00 Also set as JAMB 1993 · UME · Q36
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The sum of the first n n n 5 , 11 , 17 , 23 , 29 , 35 , … 5, 11, 17, 23, 29, 35, \ldots 5 , 11 , 17 , 23 , 29 , 35 , …
A n ( 3 n − 0 ) n(3n - 0) n ( 3 n − 0 ) B n ( 3 n + 2 ) n(3n + 2) n ( 3 n + 2 ) C n ( 3 n + 2.5 ) n(3n + 2.5) n ( 3 n + 2.5 ) D n ( 3 n + 5 ) n(3n + 5) n ( 3 n + 5 )
Worked solution (try it first) This is an A.P. with
a = 5 a = 5 a = 5 and
d = 6 d = 6 d = 6 .
Use
S n = n 2 ( 2 a + ( n − 1 ) d ) S_n = \frac n2\big(2a + (n - 1)d\big) S n = 2 n ( 2 a + ( n − 1 ) d ) : the bracket is
10 + 6 ( n − 1 ) = 6 n + 4 10 + 6(n - 1) = 6n + 4 10 + 6 ( n − 1 ) = 6 n + 4 .
So
S n = n 2 ( 6 n + 4 ) = n ( 3 n + 2 ) S_n = \frac n2(6n + 4) = n(3n + 2) S n = 2 n ( 6 n + 4 ) = n ( 3 n + 2 ) , option B.
Watch out
The bracket has ( n − 1 ) d (n - 1)d ( n − 1 ) d n d nd n d 6 n 6n 6 n n 2 ( 10 + 6 n ) = n ( 3 n + 5 ) \frac n2(10 + 6n) = n(3n + 5) 2 n ( 10 + 6 n ) = n ( 3 n + 5 ) n = 1 n = 1 n = 1 Report a problem with this question
What value of x x x x ( 4 − x ) x(4 - x) x ( 4 − x )
Worked solution (try it first) Expand:
x ( 4 − x ) = 4 x − x 2 x(4 - x) = 4x - x^2 x ( 4 − x ) = 4 x − x 2 .
At the maximum the gradient is zero:
4 − 2 x = 0 4 - 2x = 0 4 − 2 x = 0 , so
x = 2 x = 2 x = 2 .
The second derivative is
− 2 < 0 -2 < 0 − 2 < 0 , so it is a maximum:
x = 2 x = 2 x = 2 , option C.
Watch out
At x = 4 x = 4 x = 4 Report a problem with this question
In how many ways can a delegation of 3 be chosen from 5 men and 3 women, if at least 1 man and 1 woman must be included?
Worked solution (try it first) Two men and one woman:
5 C 2 × 3 C 1 = 10 × 3 = 30 ^5C_2 \times {^3C_1} = 10 \times 3 = 30 5 C 2 × 3 C 1 = 10 × 3 = 30 ways.
One man and two women:
5 C 1 × 3 C 2 = 5 × 3 = 15 ^5C_1 \times {^3C_2} = 5 \times 3 = 15 5 C 1 × 3 C 2 = 5 × 3 = 15 ways.
These are the only mixes with at least one of each, so add them:
30 + 15 = 45 30 + 15 = 45 30 + 15 = 45 , option D.
Watch out
"At least one man and at least one woman" allows both mixes. 30 (option C) counts only two men and one woman; add the 15 with two women. Also set as JAMB 2000 · UME · Q49
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Rationalize 2 3 + 5 5 − 3 \dfrac{2\sqrt3 + \sqrt5}{\sqrt5 - \sqrt3} 5 − 3 2 3 + 5
A 3 15 + 11 2 \dfrac{3\sqrt{15} + 11}{2} 2 3 15 + 11 B 3 15 − 11 2 \dfrac{3\sqrt{15} - 11}{2} 2 3 15 − 11 C 3 15 − 11 3\sqrt{15} - 11 3 15 − 11 D 3 15 + 11 3\sqrt{15} + 11 3 15 + 11
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
5 + 3 \sqrt5 + \sqrt3 5 + 3 .
Bottom:
( 5 − 3 ) ( 5 + 3 ) = 5 − 3 (\sqrt5 - \sqrt3)(\sqrt5 + \sqrt3) = 5 - 3 ( 5 − 3 ) ( 5 + 3 ) = 5 − 3 , which is 2.
Top:
( 2 3 + 5 ) ( 5 + 3 ) = 2 15 + 6 + 5 + 15 (2\sqrt3 + \sqrt5)(\sqrt5 + \sqrt3) = 2\sqrt{15} + 6 + 5 + \sqrt{15} ( 2 3 + 5 ) ( 5 + 3 ) = 2 15 + 6 + 5 + 15 , which is
3 15 + 11 3\sqrt{15} + 11 3 15 + 11 .
So the value is
3 15 + 11 2 \dfrac{3\sqrt{15} + 11}{2} 2 3 15 + 11 , option A.
Watch out
The bottom becomes 5 − 3 = 2 5 - 3 = 2 5 − 3 = 2 3 15 + 11 3\sqrt{15} + 11 3 15 + 11 Also set as JAMB 2010 · UTME · Q9
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Solve the inequalities − 6 ≤ 4 − 2 x < 5 − x -6 \le 4 - 2x < 5 - x − 6 ≤ 4 − 2 x < 5 − x
A − 1 ≤ x < 6 -1 \le x < 6 − 1 ≤ x < 6 B − 1 < x ≤ 5 -1 < x \le 5 − 1 < x ≤ 5 C − 1 < x < 5 -1 < x < 5 − 1 < x < 5 D − 1 ≤ x ≤ 6 -1 \le x \le 6 − 1 ≤ x ≤ 6
Worked solution (try it first) Split it into two inequalities.
First,
− 6 ≤ 4 − 2 x -6 \le 4 - 2x − 6 ≤ 4 − 2 x : add
2 x 2x 2 x and 6 to both sides to get
2 x ≤ 10 2x \le 10 2 x ≤ 10 , so
x ≤ 5 x \le 5 x ≤ 5 .
Second,
4 − 2 x < 5 − x 4 - 2x < 5 - x 4 − 2 x < 5 − x : add
2 x 2x 2 x and subtract 5 from both sides to get
− 1 < x -1 < x − 1 < x .
Both must hold:
− 1 < x ≤ 5 -1 < x \le 5 − 1 < x ≤ 5 , option B.
Watch out
Each end keeps its own sign. The first part has ≤ \le ≤ − 1 < x < 5 -1 < x < 5 − 1 < x < 5 Also set as JAMB 2010 · UTME · Q20
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A cylindrical pipe 50 m 50\text{ m} 50 m 7 m 7\text{ m} 7 m
A 100 π m 2 100\pi\text{ m}^2 100 π m 2 B 98 π m 2 98\pi\text{ m}^2 98 π m 2 C 350 π m 2 350\pi\text{ m}^2 350 π m 2 D 749 π m 2 749\pi\text{ m}^2 749 π m 2
Worked solution (try it first) One end open means one end is closed: the curved surface plus one circle.
Curved surface:
2 π r h = 2 π × 7 × 50 2\pi rh = 2\pi \times 7 \times 50 2 π r h = 2 π × 7 × 50 One end:
π r 2 = 49 π \pi r^2 = 49\pi π r 2 = 49 π .
Total:
700 π + 49 π = 749 π m 2 700\pi + 49\pi = 749\pi\text{ m}^2 700 π + 49 π = 749 π m 2 , option D.
Watch out
Most of the area is the curved side. 98 π m 2 98\pi\text{ m}^2 98 π m 2 Also set as JAMB 2010 · UTME · Q33
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Find the standard deviation of 2, 3, 5 and 6.
A 5 2 \sqrt{\frac52} 2 5 B 10 \sqrt{10} 10 C 6 \sqrt6 6 D 2 5 \sqrt{\frac25} 5 2
Worked solution (try it first) The mean is
2 + 3 + 5 + 6 4 = 4 \frac{2 + 3 + 5 + 6}{4} = 4 4 2 + 3 + 5 + 6 = 4 .
The squared deviations are 4, 1, 1, 4, which add up to 10.
The variance is
10 4 = 5 2 \frac{10}{4} = \frac52 4 10 = 2 5 , so the standard deviation is
5 2 \sqrt{\frac52} 2 5 , option A.
Watch out
Divide by the number of values before taking the root. Leaving out the division gives 10 \sqrt{10} 10 2 5 \sqrt{\frac25} 5 2 Also set as JAMB 2010 · UTME · Q48
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Given that log 4 ( y + 1 ) + log 4 ( 1 2 x ) = 1 \log_4(y + 1) + \log_4\left(\frac12x\right) = 1 log 4 ( y + 1 ) + log 4 ( 2 1 x ) = 1 log 2 ( y − 1 ) + log 2 x = 2 \log_2(y - 1) + \log_2 x = 2 log 2 ( y − 1 ) + log 2 x = 2 x x x y y y
A 2 , 3 2, 3 2 , 3 B 3 , 2 3, 2 3 , 2 C − 2 , − 3 -2, -3 − 2 , − 3 D − 3 , − 2 -3, -2 − 3 , − 2
Worked solution (try it first) First equation: combine the logs,
log 4 ( x ( y + 1 ) 2 ) = 1 \log_4\left(\frac{x(y + 1)}{2}\right) = 1 log 4 ( 2 x ( y + 1 ) ) = 1 , so
x ( y + 1 ) 2 = 4 \frac{x(y + 1)}{2} = 4 2 x ( y + 1 ) = 4 and
x ( y + 1 ) = 8 x(y + 1) = 8 x ( y + 1 ) = 8 .
Second equation:
log 2 [ x ( y − 1 ) ] = 2 \log_2 [x(y - 1)] = 2 log 2 [ x ( y − 1 )] = 2 , so
x ( y − 1 ) = 2 2 = 4 x(y - 1) = 2^2 = 4 x ( y − 1 ) = 2 2 = 4 .
Subtract the second from the first:
x y + x − ( x y − x ) = 8 − 4 xy + x - (xy - x) = 8 - 4 x y + x − ( x y − x ) = 8 − 4 , so
2 x = 4 2x = 4 2 x = 4 and
x = 2 x = 2 x = 2 .
Then
2 ( y + 1 ) = 8 2(y + 1) = 8 2 ( y + 1 ) = 8 gives
y = 3 y = 3 y = 3 .
So
x = 2 x = 2 x = 2 and
y = 3 y = 3 y = 3 , option A.
Watch out
"x x x y y y x x x y y y Report a problem with this question
When the expression p m 2 + q m + 1 pm^2 + qm + 1 p m 2 + q m + 1 ( m − 1 ) (m - 1) ( m − 1 ) ( m + 1 ) (m + 1) ( m + 1 ) p p p q q q
A 2 , − 1 2, -1 2 , − 1 B − 1 , 2 -1, 2 − 1 , 2 C 3 , − 2 3, -2 3 , − 2 D − 2 , 3 -2, 3 − 2 , 3
Worked solution (try it first) By the remainder theorem, dividing by
m − 1 m - 1 m − 1 leaves the value at
m = 1 m = 1 m = 1 :
p + q + 1 = 2 p + q + 1 = 2 p + q + 1 = 2 , so
p + q = 1 p + q = 1 p + q = 1 .
Dividing by
m + 1 m + 1 m + 1 leaves the value at
m = − 1 m = -1 m = − 1 :
p − q + 1 = 4 p - q + 1 = 4 p − q + 1 = 4 , so
p − q = 3 p - q = 3 p − q = 3 .
Add the equations:
2 p = 4 2p = 4 2 p = 4 , so
p = 2 p = 2 p = 2 and
q = − 1 q = -1 q = − 1 , option A.
Watch out
"Respectively" means give p p p q q q − 1 , 2 -1, 2 − 1 , 2 Also set as JAMB 1998 · UME · Q10
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Divide 2 x 3 + 11 x 2 + 17 x + 6 2x^3 + 11x^2 + 17x + 6 2 x 3 + 11 x 2 + 17 x + 6 2 x + 1 2x + 1 2 x + 1
A x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6 B 2 x 2 + 5 x + 6 2x^2 + 5x + 6 2 x 2 + 5 x + 6 C 2 x 2 − 5 x + 6 2x^2 - 5x + 6 2 x 2 − 5 x + 6 D x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6
Worked solution (try it first) Long division:
2 x 3 ÷ 2 x = x 2 2x^3 \div 2x = x^2 2 x 3 ÷ 2 x = x 2 .
Subtract
x 2 ( 2 x + 1 ) = 2 x 3 + x 2 x^2(2x + 1) = 2x^3 + x^2 x 2 ( 2 x + 1 ) = 2 x 3 + x 2 to leave
10 x 2 + 17 x + 6 10x^2 + 17x + 6 10 x 2 + 17 x + 6 .
10 x 2 ÷ 2 x = 5 x 10x^2 \div 2x = 5x 10 x 2 ÷ 2 x = 5 x .
Subtract
5 x ( 2 x + 1 ) = 10 x 2 + 5 x 5x(2x + 1) = 10x^2 + 5x 5 x ( 2 x + 1 ) = 10 x 2 + 5 x to leave
12 x + 6 12x + 6 12 x + 6 .
12 x ÷ 2 x = 6 12x \div 2x = 6 12 x ÷ 2 x = 6 , and
6 ( 2 x + 1 ) = 12 x + 6 6(2x + 1) = 12x + 6 6 ( 2 x + 1 ) = 12 x + 6 leaves 0.
So the quotient is
x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6 , option A.
Watch out
Divide the leading terms: 2 x 3 ÷ 2 x = x 2 2x^3 \div 2x = x^2 2 x 3 ÷ 2 x = x 2 2 x 2 2x^2 2 x 2 Also set as JAMB 1998 · UME · Q15
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⊗ \otimes ⊗ p p p q q q r r r s s s
p p p r r r p p p r r r p p p
q q q p p p q q q r r r s s s
r r r r r r r r r r r r r r r
s s s q q q s s s r r r q q q
The identity element with respect to the operation shown in the table is
Worked solution (try it first) The identity leaves every element unchanged, so its row and its column must repeat the headings
p , q , r , s p, q, r, s p , q , r , s .
The row for
q q q reads
p , q , r , s p, q, r, s p , q , r , s , and so does the column under
q q q .
So the identity is
q q q , option B.
Watch out
r r r r r r r r r Also set as JAMB 1998 · UME · Q21
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In the figure, P Q S T PQST P QS T T S R TSR T S R △ Q R S \triangle QRS △ QR S 20 cm 2 20\text{ cm}^2 20 cm 2 P Q R T PQRT P QR T
A 35 cm 2 35\text{ cm}^2 35 cm 2 B 65 cm 2 65\text{ cm}^2 65 cm 2 C 70 cm 2 70\text{ cm}^2 70 cm 2 D 140 cm 2 140\text{ cm}^2 140 cm 2
Worked solution (try it first) △ Q R S \triangle QRS △ QR S has base
S R = 8 SR = 8 S R = 8 cm:
1 2 × 8 × h = 20 \frac12 \times 8 \times h = 20 2 1 × 8 × h = 20 , so
h = 5 h = 5 h = 5 cm.
This is also the height of the trapezium.
P Q S T PQST P QS T is a parallelogram, so
P Q = T S = 10 PQ = TS = 10 P Q = T S = 10 cm.
The other parallel side is
T R = 10 + 8 = 18 TR = 10 + 8 = 18 T R = 10 + 8 = 18 cm.
Area of the trapezium
= 1 2 ( 10 + 18 ) × 5 = \frac12(10 + 18) \times 5 = 2 1 ( 10 + 18 ) × 5 = 70 cm 2 = 70\text{ cm}^2 = 70 cm 2 , option C.
Watch out
Keep the 1 2 \frac12 2 1 ( 10 + 18 ) × 5 = 140 cm 2 (10 + 18) \times 5 = 140\text{ cm}^2 ( 10 + 18 ) × 5 = 140 cm 2 Also set as JAMB 1998 · UME · Q27
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Find the equation of the curve which passes through the point ( 2 , 5 ) (2, 5) ( 2 , 5 ) 6 x − 5 6x - 5 6 x − 5
A 6 x 2 − 5 x 6x^2 - 5x 6 x 2 − 5 x B 6 x 2 + 5 x + 5 6x^2 + 5x + 5 6 x 2 + 5 x + 5 C 3 x 2 − 5 x − 5 3x^2 - 5x - 5 3 x 2 − 5 x − 5 D 3 x 2 − 5 x + 3 3x^2 - 5x + 3 3 x 2 − 5 x + 3
Worked solution (try it first) Integrate the gradient:
y = 3 x 2 − 5 x + c y = 3x^2 - 5x + c y = 3 x 2 − 5 x + c .
The curve passes through
( 2 , 5 ) (2, 5) ( 2 , 5 ) :
5 = 12 − 10 + c 5 = 12 - 10 + c 5 = 12 − 10 + c , so
c = 3 c = 3 c = 3 .
So
y = 3 x 2 − 5 x + 3 y = 3x^2 - 5x + 3 y = 3 x 2 − 5 x + 3 , option D.
Watch out
6 x 6x 6 x 3 x 2 3x^2 3 x 2 6 x 2 6x^2 6 x 2 Also set as JAMB 1998 · UME · Q41
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0.0001432 1940000 = k × 10 n \dfrac{0.0001432}{1940000} = k \times 10^n 1940000 0.0001432 = k × 1 0 n 1 ≤ k < 10 1 \le k < 10 1 ≤ k < 10 n n n k k k n n n
A 7.381 and − 11 -11 − 11 B 2.34 and 10 C 3.871 and 2 D 7.831 and − 11 -11 − 11
Worked solution (try it first) Write both in standard form:
0.0001432 = 1.432 × 10 − 4 0.0001432 = 1.432 \times 10^{-4} 0.0001432 = 1.432 × 1 0 − 4 and
1 940 000 = 1.94 × 10 6 1\,940\,000 = 1.94 \times 10^6 1 940 000 = 1.94 × 1 0 6 .
Divide the numbers,
1.432 ÷ 1.94 = 0.7381 1.432 \div 1.94 = 0.7381 1.432 ÷ 1.94 = 0.7381 , and subtract the powers,
10 − 4 − 6 = 10 − 10 10^{-4 - 6} = 10^{-10} 1 0 − 4 − 6 = 1 0 − 10 .
0.7381 × 10 − 10 = 7.381 × 10 − 11 0.7381 \times 10^{-10} = 7.381 \times 10^{-11} 0.7381 × 1 0 − 10 = 7.381 × 1 0 − 11 , so
k = 7.381 k = 7.381 k = 7.381 and
n = − 11 n = -11 n = − 11 , option A.
Watch out
Check the digits carefully: 1.432 ÷ 1.94 = 0.7381 1.432 \div 1.94 = 0.7381 1.432 ÷ 1.94 = 0.7381 Report a problem with this question
Thirty boys and x x x x x x
Worked solution (try it first) Total score = mean × number.
The boys scored
30 × 6 = 180 30 \times 6 = 180 30 × 6 = 180 and the girls
8 x 8x 8 x .
So
180 + 8 x = 468 180 + 8x = 468 180 + 8 x = 468 , which gives
8 x = 288 8x = 288 8 x = 288 .
Divide by 8:
x = 36 x = 36 x = 36 , option C.
Watch out
Take the boys' total off first: 8 x = 468 − 180 8x = 468 - 180 8 x = 468 − 180 Also set as JAMB 1984 · UME · Q7
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Rationalize 5 7 − 7 5 7 − 5 \dfrac{5\sqrt7 - 7\sqrt5}{\sqrt7 - \sqrt5} 7 − 5 5 7 − 7 5
A − 2 35 -2\sqrt{35} − 2 35 B 7 − 6 5 \sqrt7 - 6\sqrt5 7 − 6 5 C − 35 -\sqrt{35} − 35 D 4 7 4\sqrt7 4 7
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
7 + 5 \sqrt7 + \sqrt5 7 + 5 .
Bottom:
( 7 − 5 ) ( 7 + 5 ) = 7 − 5 (\sqrt7 - \sqrt5)(\sqrt7 + \sqrt5) = 7 - 5 ( 7 − 5 ) ( 7 + 5 ) = 7 − 5 , which is 2.
Top:
( 5 7 − 7 5 ) ( 7 + 5 ) = 35 + 5 35 − 7 35 − 35 (5\sqrt7 - 7\sqrt5)(\sqrt7 + \sqrt5) = 35 + 5\sqrt{35} - 7\sqrt{35} - 35 ( 5 7 − 7 5 ) ( 7 + 5 ) = 35 + 5 35 − 7 35 − 35 , which is
− 2 35 -2\sqrt{35} − 2 35 .
So the value is
− 2 35 2 = − 35 \dfrac{-2\sqrt{35}}{2} = -\sqrt{35} 2 − 2 35 = − 35 , option C.
Watch out
Don't forget the bottom: after rationalising it is 7 − 5 = 2 7 - 5 = 2 7 − 5 = 2 − 2 35 -2\sqrt{35} − 2 35 Also set as JAMB 1984 · UME · Q13
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If 2 x + 3 y = 1 2x + 3y = 1 2 x + 3 y = 1 x − 2 y = 11 x - 2y = 11 x − 2 y = 11 ( x + y ) (x + y) ( x + y )
Worked solution (try it first) Make
x x x the subject of the second equation:
x = 11 + 2 y x = 11 + 2y x = 11 + 2 y .
Substitute into the first:
2 ( 11 + 2 y ) + 3 y = 1 2(11 + 2y) + 3y = 1 2 ( 11 + 2 y ) + 3 y = 1 , so
22 + 7 y = 1 22 + 7y = 1 22 + 7 y = 1 and
y = − 3 y = -3 y = − 3 .
Then
x = 11 − 6 = 5 x = 11 - 6 = 5 x = 11 − 6 = 5 .
So
x + y = 5 + ( − 3 ) = 2 x + y = 5 + (-3) = 2 x + y = 5 + ( − 3 ) = 2 , option D.
Watch out
Finish the question: it asks for x + y x + y x + y x = 5 x = 5 x = 5 5 − ( − 3 ) = 8 5 - (-3) = 8 5 − ( − 3 ) = 8 x − y x - y x − y Report a problem with this question
Simplify ( 64000 3 ) − 1 \left(\sqrt[3]{64000}\right)^{-1} ( 3 64000 ) − 1
A 80 B 40 C 1 40 \frac{1}{40} 40 1 D 1 80 \frac{1}{80} 80 1
Worked solution (try it first) Split the number:
64 000 = 64 × 1000 64\,000 = 64 \times 1000 64 000 = 64 × 1000 , and
64 3 = 4 \sqrt[3]{64} = 4 3 64 = 4 ,
1000 3 = 10 \sqrt[3]{1000} = 10 3 1000 = 10 .
So
64 000 3 = 40 \sqrt[3]{64\,000} = 40 3 64 000 = 40 .
The power
− 1 -1 − 1 means the reciprocal:
1 40 \frac{1}{40} 40 1 , option C.
Watch out
Don't forget the index − 1 -1 − 1 1 40 \frac{1}{40} 40 1 Report a problem with this question
Find the value of p p p ( p , 4 ) (p, 4) ( p , 4 ) ( 6 , − 2 ) (6, -2) ( 6 , − 2 ) ( 2 , p ) (2, p) ( 2 , p ) ( − 1 , 3 ) (-1, 3) ( − 1 , 3 )
Worked solution (try it first) Gradient of the first line:
− 2 − 4 6 − p = − 6 6 − p \dfrac{-2 - 4}{6 - p} = \dfrac{-6}{6 - p} 6 − p − 2 − 4 = 6 − p − 6 .
Gradient of the second line:
3 − p − 1 − 2 = p − 3 3 \dfrac{3 - p}{-1 - 2} = \dfrac{p - 3}{3} − 1 − 2 3 − p = 3 p − 3 .
Perpendicular gradients multiply to
− 1 -1 − 1 :
− 6 ( p − 3 ) 3 ( 6 − p ) = − 1 \dfrac{-6(p - 3)}{3(6 - p)} = -1 3 ( 6 − p ) − 6 ( p − 3 ) = − 1 , so
2 ( p − 3 ) = 6 − p 2(p - 3) = 6 - p 2 ( p − 3 ) = 6 − p .
Then
3 p = 12 3p = 12 3 p = 12 , so
p = 4 p = 4 p = 4 , option C.
Check: the gradients are
− 3 -3 − 3 and
1 3 \frac13 3 1 , and
− 3 × 1 3 = − 1 -3 \times \frac13 = -1 − 3 × 3 1 = − 1 .
Watch out
For perpendicular lines the product of the gradients is − 1 -1 − 1 p 2 − 9 p = 0 p^2 - 9p = 0 p 2 − 9 p = 0 p = 0 p = 0 p = 0 Also set as JAMB 2001 · UME · Q30
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Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
Worked solution (try it first) Let the exterior angle be
e e e .
The interior angle is
2 e 2e 2 e , and the two add up to
180 ∘ 180^\circ 18 0 ∘ :
3 e = 180 ∘ 3e = 180^\circ 3 e = 18 0 ∘ , so
e = 60 ∘ e = 60^\circ e = 6 0 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 60 = 6 360 \div 60 = 6 360 ÷ 60 = 6 , option C.
Watch out
It is the interior angle that is twice the exterior. The other way round gives e = 120 ∘ e = 120^\circ e = 12 0 ∘ Also set as JAMB 2001 · UME · Q21
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