Paper JAMB 2003 General Maths Objective
Objective paper · 47 questions · partial
JAMB 2003 · UME Topics include Number foundations & fractions, Commercial arithmetic, Number bases, Surds, Logarithms, Sets & Venn diagrams.
Our copy of this paper is missing questions 17, 30, 38.
Sit this paper Answer every question in order, timed if you like (suggested 30 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 3 4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 39 40 41 42 43 44 45 46 47 48 49 50 Simplify 1 − ( 2 1 3 × 1 1 4 ) + 3 5 1 - \left(2\frac13 \times 1\frac14\right) + \frac35 1 − ( 2 3 1 × 1 4 1 ) + 5 3
A − 2 31 60 -2\frac{31}{60} − 2 60 31 B − 2 7 15 -2\frac7{15} − 2 15 7 C − 1 19 60 -1\frac{19}{60} − 1 60 19 D − 1 1 15 -1\frac1{15} − 1 15 1
Worked solution (try it first) Bracket first:
2 1 3 × 1 1 4 = 7 3 × 5 4 2\frac13 \times 1\frac14 = \frac73 \times \frac54 2 3 1 × 1 4 1 = 3 7 × 4 5 = 35 12 = \frac{35}{12} = 12 35 .
Now
1 − 35 12 + 3 5 1 - \frac{35}{12} + \frac35 1 − 12 35 + 5 3 .
The LCD is 60:
60 60 − 175 60 + 36 60 = − 79 60 \frac{60}{60} - \frac{175}{60} + \frac{36}{60} = -\frac{79}{60} 60 60 − 60 175 + 60 36 = − 60 79 .
− 79 60 = − 1 19 60 -\frac{79}{60} = -1\frac{19}{60} − 60 79 = − 1 60 19 , option C.
Watch out
The 3 5 \frac35 5 3 3 5 \frac35 5 3 − 151 60 = − 2 31 60 -\frac{151}{60} = -2\frac{31}{60} − 60 151 = − 2 60 31 Report a problem with this question
A cinema hall contains a certain number of people. If 22 1 2 % 22\frac12\% 22 2 1 % 47 1 2 % 47\frac12\% 47 2 1 %
Worked solution (try it first) Children and men make up
22 1 2 % + 47 1 2 % = 70 % 22\frac12\% + 47\frac12\% = 70\% 22 2 1 % + 47 2 1 % = 70% , so the 84 women are the other
30 % 30\% 30% .
So the hall holds
84 ÷ 0.3 = 280 84 \div 0.3 = 280 84 ÷ 0.3 = 280 people.
Men are
47 1 2 % 47\frac12\% 47 2 1 % of 280:
0.475 × 280 = 133 0.475 \times 280 = 133 0.475 × 280 = 133 , option A.
Watch out
22 1 2 % 22\frac12\% 22 2 1 % 47 1 2 % 47\frac12\% 47 2 1 % Report a problem with this question
Simplify 213 4 × 23 4 213_4 \times 23_4 21 3 4 × 2 3 4
A 13211 4 13211_4 1321 1 4 B 10311 4 10311_4 1031 1 4 C 10321 4 10321_4 1032 1 4 D 12231 4 12231_4 1223 1 4
Worked solution (try it first) Change to base ten:
213 4 = 32 + 4 + 3 = 39 213_4 = 32 + 4 + 3 = 39 21 3 4 = 32 + 4 + 3 = 39 and
23 4 = 8 + 3 = 11 23_4 = 8 + 3 = 11 2 3 4 = 8 + 3 = 11 .
Multiply:
39 × 11 = 429 39 \times 11 = 429 39 × 11 = 429 .
Change back:
429 = 1 × 256 + 2 × 64 + 2 × 16 + 3 × 4 + 1 429 = 1 \times 256 + 2 \times 64 + 2 \times 16 + 3 \times 4 + 1 429 = 1 × 256 + 2 × 64 + 2 × 16 + 3 × 4 + 1 , so the answer is
12231 4 12231_4 1223 1 4 , option D.
Watch out
If you multiply in columns, carry in fours, not tens. The options differ by only a digit or two, so check yours by converting to base ten. Report a problem with this question
A woman buys 270 oranges for ₦1,800.00 and sells them at 5 for ₦40.00. What is her profit?
A ₦630.00 B ₦360.00 C ₦1,620.00 D ₦2,160.00
Worked solution (try it first) 270 oranges make
270 ÷ 5 = 54 270 \div 5 = 54 270 ÷ 5 = 54 lots of 5.
At ₦40 a lot she takes
54 × 40 = 54 \times 40 = 54 × 40 = ₦2,160.
Her profit is
2160 − 1800 = 2160 - 1800 = 2160 − 1800 = ₦360.00, option B.
Watch out
₦2,160.00 (option D) is what she takes in. Profit is takings minus the ₦1,800 she paid. Report a problem with this question
Simplify 98 − 50 32 \dfrac{\sqrt{98} - \sqrt{50}}{\sqrt{32}} 32 98 − 50
A 1 2 \frac12 2 1 B 1 4 \frac14 4 1 C 1 D 3
Worked solution (try it first) Take out square factors:
98 = 7 2 \sqrt{98} = 7\sqrt2 98 = 7 2 ,
50 = 5 2 \sqrt{50} = 5\sqrt2 50 = 5 2 and
32 = 4 2 \sqrt{32} = 4\sqrt2 32 = 4 2 .
Top:
7 2 − 5 2 = 2 2 7\sqrt2 - 5\sqrt2 = 2\sqrt2 7 2 − 5 2 = 2 2 .
So the value is
2 2 4 2 = 1 2 \dfrac{2\sqrt2}{4\sqrt2} = \frac12 4 2 2 2 = 2 1 , option A.
Watch out
Simplify each surd before you subtract: 98 − 50 \sqrt{98} - \sqrt{50} 98 − 50 48 \sqrt{48} 48 2 \sqrt2 2 Report a problem with this question
The sum of four numbers is 1214 5 1214_5 121 4 5
Worked solution (try it first) Change to base ten:
1214 5 = 125 + 50 + 5 + 4 = 184 1214_5 = 125 + 50 + 5 + 4 = 184 121 4 5 = 125 + 50 + 5 + 4 = 184 .
The average of four numbers is
184 ÷ 4 = 46 184 \div 4 = 46 184 ÷ 4 = 46 .
Change back:
46 = 1 × 25 + 4 × 5 + 1 46 = 1 \times 25 + 4 \times 5 + 1 46 = 1 × 25 + 4 × 5 + 1 , so the average is 141 in base five, option C.
Watch out
Put the powers of five in order: 46 is one 25, four 5s and one unit. Mixing up the places gives 411 or 114 (options A and D). Report a problem with this question
Evaluate log 2 4 + log 1 2 16 − log 4 32 \log_{\sqrt2} 4 + \log_{\frac12} 16 - \log_4 32 log 2 4 + log 2 1 16 − log 4 32
A − 2.5 -2.5 − 2.5 B 5.5 C − 5.5 -5.5 − 5.5 D 2.5
Worked solution (try it first) ( 2 ) 4 = 4 (\sqrt2)^4 = 4 ( 2 ) 4 = 4 , so
log 2 4 = 4 \log_{\sqrt2} 4 = 4 log 2 4 = 4 .
( 1 2 ) − 4 = 16 \left(\frac12\right)^{-4} = 16 ( 2 1 ) − 4 = 16 , so
log 1 2 16 = − 4 \log_{\frac12} 16 = -4 log 2 1 16 = − 4 .
4 2.5 = 2 5 = 32 4^{2.5} = 2^5 = 32 4 2.5 = 2 5 = 32 , so
log 4 32 = 2.5 \log_4 32 = 2.5 log 4 32 = 2.5 .
So the value is
4 + ( − 4 ) − 2.5 = − 2.5 4 + (-4) - 2.5 = -2.5 4 + ( − 4 ) − 2.5 = − 2.5 , option A.
Watch out
With base 1 2 \frac12 2 1 log 1 2 16 = − 4 \log_{\frac12} 16 = -4 log 2 1 16 = − 4 + 4 +4 + 4 4 + 4 − 2.5 = 5.5 4 + 4 - 2.5 = 5.5 4 + 4 − 2.5 = 5.5 Report a problem with this question
Given U = { even numbers between 0 and 30 } U = \{\text{even numbers between 0 and 30}\} U = { even numbers between 0 and 30 } P = { multiples of 6 between 0 and 30 } P = \{\text{multiples of 6 between 0 and 30}\} P = { multiples of 6 between 0 and 30 } Q = { multiples of 4 between 0 and 30 } Q = \{\text{multiples of 4 between 0 and 30}\} Q = { multiples of 4 between 0 and 30 } ( P ∪ Q ) c (P \cup Q)^c ( P ∪ Q ) c
A { 0 , 2 , 6 , 22 , 26 } \{0, 2, 6, 22, 26\} { 0 , 2 , 6 , 22 , 26 } B { 2 , 4 , 14 , 18 , 26 } \{2, 4, 14, 18, 26\} { 2 , 4 , 14 , 18 , 26 } C { 2 , 10 , 14 , 22 , 26 } \{2, 10, 14, 22, 26\} { 2 , 10 , 14 , 22 , 26 } D { 0 , 10 , 14 , 22 , 26 } \{0, 10, 14, 22, 26\} { 0 , 10 , 14 , 22 , 26 }
Worked solution (try it first) "Between 0 and 30" leaves out 0 and 30, so
U = { 2 , 4 , 6 , … , 28 } U = \{2, 4, 6, \dots, 28\} U = { 2 , 4 , 6 , … , 28 } .
P = { 6 , 12 , 18 , 24 } P = \{6, 12, 18, 24\} P = { 6 , 12 , 18 , 24 } and
Q = { 4 , 8 , 12 , 16 , 20 , 24 , 28 } Q = \{4, 8, 12, 16, 20, 24, 28\} Q = { 4 , 8 , 12 , 16 , 20 , 24 , 28 } , so
P ∪ Q = { 4 , 6 , 8 , 12 , 16 , 18 , 20 , 24 , 28 } P \cup Q = \{4, 6, 8, 12, 16, 18, 20, 24, 28\} P ∪ Q = { 4 , 6 , 8 , 12 , 16 , 18 , 20 , 24 , 28 } .
The complement is the even numbers left over:
( P ∪ Q ) c = { 2 , 10 , 14 , 22 , 26 } (P \cup Q)^c = \{2, 10, 14, 22, 26\} ( P ∪ Q ) c = { 2 , 10 , 14 , 22 , 26 } , option C.
Watch out
0 is not "between 0 and 30", so it is not in U U U Report a problem with this question
In a class of 40 students, 32 offer Mathematics, 24 offer Physics and 4 offer neither. How many offer both Mathematics and Physics?
Worked solution (try it first) Take away the 4 who offer neither:
40 − 4 = 36 40 - 4 = 36 40 − 4 = 36 offer at least one subject.
Both subjects:
32 + 24 − 36 = 20 32 + 24 - 36 = 20 32 + 24 − 36 = 20 , option C.
Watch out
Remove the 4 who offer neither before using the formula. 32 + 24 − 40 = 16 32 + 24 - 40 = 16 32 + 24 − 40 = 16 Report a problem with this question
Find ( 1 0.06 ÷ 1 0.042 ) − 1 \left(\frac{1}{0.06} \div \frac{1}{0.042}\right)^{-1} ( 0.06 1 ÷ 0.042 1 ) − 1
Worked solution (try it first) Dividing by
1 0.042 \frac{1}{0.042} 0.042 1 is the same as multiplying by 0.042:
1 0.06 × 0.042 = 0.042 0.06 \frac{1}{0.06} \times 0.042 = \frac{0.042}{0.06} 0.06 1 × 0.042 = 0.06 0.042 .
Multiply top and bottom by 1000:
42 60 = 0.7 \frac{42}{60} = 0.7 60 42 = 0.7 .
The power
− 1 -1 − 1 means the reciprocal:
1 0.7 = 1.428 … \frac{1}{0.7} = 1.428\ldots 0.7 1 = 1.428 … , which is 1.43 to 2 decimal places, option D.
Watch out
Don't stop at 0.7: the power − 1 -1 − 1 Report a problem with this question
If 9 2 x − 1 27 x + 1 = 1 \dfrac{9^{2x - 1}}{27^{x + 1}} = 1 2 7 x + 1 9 2 x − 1 = 1 x x x
Worked solution (try it first) Write both as powers of 3:
9 2 x − 1 = 3 4 x − 2 9^{2x - 1} = 3^{4x - 2} 9 2 x − 1 = 3 4 x − 2 and
27 x + 1 = 3 3 x + 3 27^{x + 1} = 3^{3x + 3} 2 7 x + 1 = 3 3 x + 3 .
A fraction equal to 1 has top equal to bottom, so
4 x − 2 = 3 x + 3 4x - 2 = 3x + 3 4 x − 2 = 3 x + 3 .
Solve:
x = 5 x = 5 x = 5 , option C.
Watch out
Multiply the whole index by the new power: 9 2 x − 1 = 3 2 ( 2 x − 1 ) = 3 4 x − 2 9^{2x - 1} = 3^{2(2x - 1)} = 3^{4x - 2} 9 2 x − 1 = 3 2 ( 2 x − 1 ) = 3 4 x − 2 3 4 x − 1 3^{4x - 1} 3 4 x − 1 4 x − 1 4x - 1 4 x − 1 x = 4 x = 4 x = 4 Report a problem with this question
Factorize completely 4 a b x − 2 a x y − 12 b 2 x + 6 b x y 4abx - 2axy - 12b^2x + 6bxy 4 ab x − 2 a x y − 12 b 2 x + 6 b x y
A 2 x ( 3 b − a ) ( 2 b − y ) 2x(3b - a)(2b - y) 2 x ( 3 b − a ) ( 2 b − y ) B 2 x ( a − 3 b ) ( b − 2 y ) 2x(a - 3b)(b - 2y) 2 x ( a − 3 b ) ( b − 2 y ) C 2 x ( 2 b − a ) ( 3 b − y ) 2x(2b - a)(3b - y) 2 x ( 2 b − a ) ( 3 b − y ) D 2 x ( a − 3 b ) ( 2 b − y ) 2x(a - 3b)(2b - y) 2 x ( a − 3 b ) ( 2 b − y )
Worked solution (try it first) Every term has
2 x 2x 2 x :
2 x ( 2 a b − a y − 6 b 2 + 3 b y ) 2x(2ab - ay - 6b^2 + 3by) 2 x ( 2 ab − a y − 6 b 2 + 3 b y ) .
Group the bracket in pairs:
a ( 2 b − y ) − 3 b ( 2 b − y ) a(2b - y) - 3b(2b - y) a ( 2 b − y ) − 3 b ( 2 b − y ) .
Take out the common bracket:
( a − 3 b ) ( 2 b − y ) (a - 3b)(2b - y) ( a − 3 b ) ( 2 b − y ) .
So the answer is
2 x ( a − 3 b ) ( 2 b − y ) 2x(a - 3b)(2b - y) 2 x ( a − 3 b ) ( 2 b − y ) , option D.
Watch out
− 6 b 2 + 3 b y = − 3 b ( 2 b − y ) -6b^2 + 3by = -3b(2b - y) − 6 b 2 + 3 b y = − 3 b ( 2 b − y ) a − 3 b a - 3b a − 3 b 3 b − a 3b - a 3 b − a Report a problem with this question
The sum of the first n n n − 16 -16 − 16
Worked solution (try it first) When you know the first and last terms,
S n = n 2 ( a + l ) S_n = \frac n2(a + l) S n = 2 n ( a + l ) .
So
n 2 ( − 16 + 72 ) = 252 \frac n2(-16 + 72) = 252 2 n ( − 16 + 72 ) = 252 , which is
28 n = 252 28n = 252 28 n = 252 .
Divide both sides by 28:
n = 9 n = 9 n = 9 , option B.
Watch out
Keep the sign of the first term: − 16 + 72 = 56 -16 + 72 = 56 − 16 + 72 = 56 16 + 72 = 88 16 + 72 = 88 16 + 72 = 88 44 n = 252 44n = 252 44 n = 252 Report a problem with this question
The graphs of y = x 2 + 4 y = x^2 + 4 y = x 2 + 4 P Q PQ P Q x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0 P Q PQ P Q
A y = 3 x + 2 y = 3x + 2 y = 3 x + 2 B y = 3 x − 4 y = 3x - 4 y = 3 x − 4 C y = 3 x + 4 y = 3x + 4 y = 3 x + 4 D y = 3 x − 2 y = 3x - 2 y = 3 x − 2
Worked solution (try it first) Rearrange the equation to have
x 2 x^2 x 2 alone:
x 2 = 3 x − 2 x^2 = 3x - 2 x 2 = 3 x − 2 .
Add 4 to both sides so the left side matches the curve:
x 2 + 4 = 3 x + 2 x^2 + 4 = 3x + 2 x 2 + 4 = 3 x + 2 .
So the curve
y = x 2 + 4 y = x^2 + 4 y = x 2 + 4 meets the line
y = 3 x + 2 y = 3x + 2 y = 3 x + 2 at the roots.
P Q PQ P Q is
y = 3 x + 2 y = 3x + 2 y = 3 x + 2 , option A.
Watch out
y = 3 x − 2 y = 3x - 2 y = 3 x − 2 y = x 2 y = x^2 y = x 2 x 2 + 4 x^2 + 4 x 2 + 4 Also set as JAMB 2017 · UTME · Q5
Report a problem with this question
A matrix P P P P − 1 = ( 1 − 3 0 1 ) P^{-1} = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} P − 1 = ( 1 0 − 3 1 ) P P P
A ( 1 3 0 1 ) \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} ( 1 0 3 1 ) B ( 1 − 3 0 − 1 ) \begin{pmatrix} 1 & -3 \\ 0 & -1 \end{pmatrix} ( 1 0 − 3 − 1 ) C ( 1 3 0 − 1 ) \begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix} ( 1 0 3 − 1 ) D ( − 1 3 0 − 1 ) \begin{pmatrix} -1 & 3 \\ 0 & -1 \end{pmatrix} ( − 1 0 3 − 1 )
Worked solution (try it first) P P P is the inverse of
P − 1 P^{-1} P − 1 .
The determinant of
P − 1 P^{-1} P − 1 is
1 × 1 − ( − 3 ) × 0 = 1 1 \times 1 - (-3) \times 0 = 1 1 × 1 − ( − 3 ) × 0 = 1 .
Swap the two diagonal entries (both 1, so no change) and change the signs of the other two:
− 3 -3 − 3 becomes 3, and 0 stays 0.
Divide by the determinant, 1:
P = ( 1 3 0 1 ) P = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} P = ( 1 0 3 1 ) , option A.
Watch out
Swap the diagonal entries but don't change their signs; only the off-diagonal ones change sign. Negating a diagonal 1 gives option C, and then P × P − 1 P \times P^{-1} P × P − 1 I I I Report a problem with this question
Find the values of x x x y y y 3 x − 5 y + 5 = 0 3x - 5y + 5 = 0 3 x − 5 y + 5 = 0 4 x − 7 y + 8 = 0 4x - 7y + 8 = 0 4 x − 7 y + 8 = 0
A − 4 , − 5 -4, -5 − 4 , − 5 B − 5 , − 4 -5, -4 − 5 , − 4 C 5, 4 D 4, 5
Worked solution (try it first) Move the numbers across with their signs changed:
3 x − 5 y = − 5 3x - 5y = -5 3 x − 5 y = − 5 and
4 x − 7 y = − 8 4x - 7y = -8 4 x − 7 y = − 8 .
Make the
x x x terms match: multiply the first by 4 and the second by 3, giving
12 x − 20 y = − 20 12x - 20y = -20 12 x − 20 y = − 20 and
12 x − 21 y = − 24 12x - 21y = -24 12 x − 21 y = − 24 .
Subtract the second from the first:
y = 4 y = 4 y = 4 .
Then
3 x = 5 y − 5 = 15 3x = 5y - 5 = 15 3 x = 5 y − 5 = 15 , so
x = 5 x = 5 x = 5 .
So
x , y x, y x , y are
5 , 4 5, 4 5 , 4 , option C.
Watch out
Change the sign when you move a number: 3 x − 5 y + 5 = 0 3x - 5y + 5 = 0 3 x − 5 y + 5 = 0 3 x − 5 y = − 5 3x - 5y = -5 3 x − 5 y = − 5 + 5 +5 + 5 + 8 +8 + 8 x = − 5 x = -5 x = − 5 y = − 4 y = -4 y = − 4 Report a problem with this question
Find the range of values of x x x 5 + x ≤ 8 5 + x \le 8 5 + x ≤ 8 13 + x ≥ 7 13 + x \ge 7 13 + x ≥ 7
A − 6 ≤ x ≤ 3 -6 \le x \le 3 − 6 ≤ x ≤ 3 B − 6 ≤ x ≤ − 3 -6 \le x \le -3 − 6 ≤ x ≤ − 3 C 3 ≤ x ≤ 6 3 \le x \le 6 3 ≤ x ≤ 6 D − 3 ≤ x ≤ 3 -3 \le x \le 3 − 3 ≤ x ≤ 3
Worked solution (try it first) Subtract 5 from both sides of the first:
x ≤ 3 x \le 3 x ≤ 3 .
Subtract 13 from both sides of the second:
x ≥ 7 − 13 x \ge 7 - 13 x ≥ 7 − 13 , so
x ≥ − 6 x \ge -6 x ≥ − 6 .
Both must hold:
− 6 ≤ x ≤ 3 -6 \le x \le 3 − 6 ≤ x ≤ 3 , option A.
Watch out
8 − 5 = 3 8 - 5 = 3 8 − 5 = 3 x ≤ 3 x \le 3 x ≤ 3 x ≤ − 3 x \le -3 x ≤ − 3 − 6 ≤ x ≤ − 3 -6 \le x \le -3 − 6 ≤ x ≤ − 3 Report a problem with this question
x x x U U U V V V x = 3 x = 3 x = 3 U = 3 U = 3 U = 3 V = 1 V = 1 V = 1 x x x U = 3 U = 3 U = 3 V = 3 V = 3 V = 3
Worked solution (try it first) Product on top, sum underneath:
x = k U V U + V x = \dfrac{kUV}{U + V} x = U + V k U V .
x = 3 x = 3 x = 3 ,
U = 3 U = 3 U = 3 ,
V = 1 V = 1 V = 1 :
3 = 3 k 4 3 = \dfrac{3k}{4} 3 = 4 3 k , so
k = 4 k = 4 k = 4 .
U = 3 U = 3 U = 3 ,
V = 3 V = 3 V = 3 :
x = 4 × 9 6 = 6 x = \dfrac{4 \times 9}{6} = 6 x = 6 4 × 9 = 6 , option C.
Watch out
Include the sum underneath. Leaving it out (x = k U V x = kUV x = k U V k = 1 k = 1 k = 1 x = 9 x = 9 x = 9 Report a problem with this question
Triangle O P Q OPQ O P Q
A x − 1 ≤ 0 x - 1 \le 0 x − 1 ≤ 0 y + x ≤ 0 y + x \le 0 y + x ≤ 0 y − x ≤ 0 y - x \le 0 y − x ≤ 0 B x + 1 ≥ 0 x + 1 \ge 0 x + 1 ≥ 0 y + x ≤ 0 y + x \le 0 y + x ≤ 0 y − x ≥ 0 y - x \ge 0 y − x ≥ 0 C y + x ≤ 0 y + x \le 0 y + x ≤ 0 y − x ≥ 0 y - x \ge 0 y − x ≥ 0 x − 1 ≥ 0 x - 1 \ge 0 x − 1 ≥ 0 D x − 1 ≤ 0 x - 1 \le 0 x − 1 ≤ 0 y − x ≥ 0 y - x \ge 0 y − x ≥ 0 y + x ≥ 0 y + x \ge 0 y + x ≥ 0
Worked solution (try it first) The triangle lies to the right of the vertical line
x = − 1 x = -1 x = − 1 , so
x + 1 ≥ 0 x + 1 \ge 0 x + 1 ≥ 0 .
It lies below the line
y = − x y = -x y = − x , so
y + x ≤ 0 y + x \le 0 y + x ≤ 0 , and above the line
y = x y = x y = x , so
y − x ≥ 0 y - x \ge 0 y − x ≥ 0 .
Check with the point
( − 0.5 , 0 ) (-0.5, 0) ( − 0.5 , 0 ) inside the triangle:
0.5 ≥ 0 0.5 \ge 0 0.5 ≥ 0 ,
− 0.5 ≤ 0 -0.5 \le 0 − 0.5 ≤ 0 and
0.5 ≥ 0 0.5 \ge 0 0.5 ≥ 0 .
So it is option B.
Watch out
The vertical side is the line x = − 1 x = -1 x = − 1 x + 1 x + 1 x + 1 x − 1 x - 1 x − 1 x = 1 x = 1 x = 1 Report a problem with this question
Three consecutive terms of a geometric progression are n − 2 n - 2 n − 2 n n n n + 3 n + 3 n + 3
A 2 3 \frac23 3 2 B 3 2 \frac32 2 3 C 1 2 \frac12 2 1 D 1 4 \frac14 4 1
Worked solution (try it first) For a G.P., the middle term squared equals the product of the outer two:
n 2 = ( n − 2 ) ( n + 3 ) n^2 = (n - 2)(n + 3) n 2 = ( n − 2 ) ( n + 3 ) .
Expand:
n 2 = n 2 + n − 6 n^2 = n^2 + n - 6 n 2 = n 2 + n − 6 , so
n = 6 n = 6 n = 6 .
The terms are 4, 6 and 9, so the common ratio is
6 ÷ 4 = 3 2 6 \div 4 = \frac32 6 ÷ 4 = 2 3 , option B.
Watch out
The ratio is a term divided by the one before it: 6 4 \frac64 4 6 4 6 \frac46 6 4 2 3 \frac23 3 2 Report a problem with this question
The length a person can jump is inversely proportional to his weight. If a 20 kg person can jump 1.5 m, find the constant of proportionality.
Worked solution (try it first) Inverse proportion:
L = k W L = \dfrac{k}{W} L = W k , so
k = L W k = LW k = L W .
k = 1.5 × 20 = 30 k = 1.5 \times 20 = 30 k = 1.5 × 20 = 30 , option A.
Watch out
For inverse proportion the constant is the product L W LW L W 20 1.5 \frac{20}{1.5} 1.5 20 1.5 20 \frac{1.5}{20} 20 1.5 Report a problem with this question
In the diagram, O O O P O M POM P O M ∠ M N Q = 42 ∘ \angle MNQ = 42^\circ ∠ M N Q = 4 2 ∘ ∠ Q M P \angle QMP ∠ QM P
A 138 ∘ 138^\circ 13 8 ∘ B 132 ∘ 132^\circ 13 2 ∘ C 42 ∘ 42^\circ 4 2 ∘ D 48 ∘ 48^\circ 4 8 ∘
Worked solution (try it first) Angles in the same segment are equal.
∠ M P Q \angle MPQ ∠ M P Q and
∠ M N Q \angle MNQ ∠ M N Q both stand on arc
M Q MQ M Q , so
∠ M P Q = 42 ∘ \angle MPQ = 42^\circ ∠ M P Q = 4 2 ∘ .
P M PM P M is a diameter, so
∠ M Q P = 90 ∘ \angle MQP = 90^\circ ∠ M QP = 9 0 ∘ (angle in a semicircle).
Triangle
Q M P QMP QM P :
∠ Q M P = 180 ∘ − 90 ∘ − 42 ∘ \angle QMP = 180^\circ - 90^\circ - 42^\circ ∠ QM P = 18 0 ∘ − 9 0 ∘ − 4 2 ∘ = 48 ∘ = 48^\circ = 4 8 ∘ , option D.
Watch out
42 ∘ 42^\circ 4 2 ∘ ∠ M P Q \angle MPQ ∠ M P Q P P P ∠ Q M P \angle QMP ∠ QM P 90 ∘ − 42 ∘ 90^\circ - 42^\circ 9 0 ∘ − 4 2 ∘ Report a problem with this question
The locus of a point P P P X Y XY X Y ∠ X P Y = 90 ∘ \angle XPY = 90^\circ ∠ X P Y = 9 0 ∘
A the perpendicular bisector of X Y XY X Y B a circle C a semicircle D an arc of a circle through X X X Y Y Y
Worked solution (try it first) The angle in a semicircle is a right angle, so if
∠ X P Y = 90 ∘ \angle XPY = 90^\circ ∠ X P Y = 9 0 ∘ then
P P P lies on the circle with diameter
X Y XY X Y .
P P P stays on one side of
X Y XY X Y , so it traces only half of that circle: a semicircle, option C.
Watch out
"An arc through X X X Y Y Y X Y XY X Y Report a problem with this question
In the diagram, P Q PQ P Q R S RS R S α + β + γ \alpha + \beta + \gamma α + β + γ
A 180 ∘ 180^\circ 18 0 ∘ B 90 ∘ 90^\circ 9 0 ∘ C 200 ∘ 200^\circ 20 0 ∘ D 360 ∘ 360^\circ 36 0 ∘
Worked solution (try it first) Draw a line through the apex parallel to
P Q PQ P Q and
R S RS R S .
It splits
β \beta β into two parts.
α \alpha α and the left part of
β \beta β are co-interior angles between
P Q PQ P Q and the new line, so they add up to
180 ∘ 180^\circ 18 0 ∘ .
In the same way
γ \gamma γ and the right part of
β \beta β add up to
180 ∘ 180^\circ 18 0 ∘ .
So
α + β + γ = 180 ∘ + 180 ∘ \alpha + \beta + \gamma = 180^\circ + 180^\circ α + β + γ = 18 0 ∘ + 18 0 ∘ = 360 ∘ = 360^\circ = 36 0 ∘ , option D.
Watch out
The three angles are not the angles of one triangle, so they don't add up to 180 ∘ 180^\circ 18 0 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
Which of the following is the graph of sin θ \sin\theta sin θ − π 2 ≤ θ ≤ 3 π 2 -\frac{\pi}{2} \le \theta \le \frac{3\pi}{2} − 2 π ≤ θ ≤ 2 3 π
Worked solution (try it first) Work out the key values:
sin ( − π 2 ) = − 1 \sin(-\frac{\pi}{2}) = -1 sin ( − 2 π ) = − 1 ,
sin 0 = 0 \sin 0 = 0 sin 0 = 0 ,
sin π 2 = 1 \sin\frac{\pi}{2} = 1 sin 2 π = 1 ,
sin π = 0 \sin\pi = 0 sin π = 0 and
sin 3 π 2 = − 1 \sin\frac{3\pi}{2} = -1 sin 2 3 π = − 1 .
So the graph starts at
− 1 -1 − 1 , passes through the origin, peaks at 1 when
θ = π 2 \theta = \frac{\pi}{2} θ = 2 π and falls back to
− 1 -1 − 1 at
3 π 2 \frac{3\pi}{2} 2 3 π .
Only graph C does this, so the answer is option C.
Watch out
Graph D has its peak of 1 at θ = 0 \theta = 0 θ = 0 cos θ \cos\theta cos θ θ = 0 \theta = 0 θ = 0 Report a problem with this question
In the diagram, P Q R PQR P QR P S PS P S Q R S QRS QR S P S = S R PS = SR P S = S R ∠ S P R = 40 ∘ \angle SPR = 40^\circ ∠ S P R = 4 0 ∘ ∠ P S Q \angle PSQ ∠ P S Q
A 20 ∘ 20^\circ 2 0 ∘ B 10 ∘ 10^\circ 1 0 ∘ C 40 ∘ 40^\circ 4 0 ∘ D 30 ∘ 30^\circ 3 0 ∘
Worked solution (try it first) P S = S R PS = SR P S = S R , so triangle
P S R PSR P S R is isosceles and the angles opposite the equal sides are equal:
∠ S R P = ∠ S P R = 40 ∘ \angle SRP = \angle SPR = 40^\circ ∠ S R P = ∠ S P R = 4 0 ∘ .
P Q R PQR P QR is a straight line, so
∠ S R Q = 40 ∘ \angle SRQ = 40^\circ ∠ S R Q = 4 0 ∘ .
The angle between tangent
P S PS P S and chord
S Q SQ S Q equals the angle in the alternate segment:
∠ P S Q = ∠ S R Q = 40 ∘ \angle PSQ = \angle SRQ = 40^\circ ∠ P S Q = ∠ S R Q = 4 0 ∘ , option C.
Watch out
The equal sides P S PS P S S R SR S R R R R P P P 40 ∘ 40^\circ 4 0 ∘ R R R 20 ∘ 20^\circ 2 0 ∘ Report a problem with this question
If π 2 ≤ θ ≤ 2 π \frac\pi2 \le \theta \le 2\pi 2 π ≤ θ ≤ 2 π f ( θ ) = 4 6 + 2 cos θ f(\theta) = \dfrac{4}{6 + 2\cos\theta} f ( θ ) = 6 + 2 cos θ 4
A 1 B 1 2 \frac12 2 1 C 4 D 2 3 \frac23 3 2
Worked solution (try it first) The top is fixed at 4, so the fraction is largest when the bottom,
6 + 2 cos θ 6 + 2\cos\theta 6 + 2 cos θ , is smallest.
cos θ \cos\theta cos θ is smallest (
− 1 -1 − 1 ) at
θ = π \theta = \pi θ = π , which is inside the range.
The bottom is then
6 − 2 = 4 6 - 2 = 4 6 − 2 = 4 .
Sine and cosine are both over the hypotenuse, 5, so find it by Pythagoras first.
1 1 3 1\frac13 1 3 1 (option A) is
4 3 = 1 tan θ \frac43 = \frac{1}{\tan\theta} 3 4 = t a n θ 1 , not
sin θ + cos θ \sin\theta + \cos\theta sin θ + cos θ .
Watch out
A smaller bottom makes a bigger fraction. Taking cos θ = 1 \cos\theta = 1 cos θ = 1 1 2 \frac12 2 1 Report a problem with this question
An aeroplane flies due north from airport P P P Q Q Q R R R Q Q Q P P P R R R P P P R R R
A 270 ∘ 270^\circ 27 0 ∘ B 090 ∘ 090^\circ 09 0 ∘ C 135 ∘ 135^\circ 13 5 ∘ D 225 ∘ 225^\circ 22 5 ∘
Worked solution (try it first) P Q PQ P Q is due north and
Q R QR QR is due east, and
P Q = Q R PQ = QR P Q = QR , so triangle
P Q R PQR P QR is right-angled and isosceles.
So
R R R is
45 ∘ 45^\circ 4 5 ∘ east of north from
P P P : the bearing of
R R R from
P P P is
045 ∘ 045^\circ 04 5 ∘ .
The back bearing adds
180 ∘ 180^\circ 18 0 ∘ : the bearing of
P P P from
R R R is
225 ∘ 225^\circ 22 5 ∘ , option D.
Watch out
A back bearing adds 180 ∘ 180^\circ 18 0 ∘ 180 ∘ − 45 ∘ = 135 ∘ 180^\circ - 45^\circ = 135^\circ 18 0 ∘ − 4 5 ∘ = 13 5 ∘ P P P R R R Report a problem with this question
Find the equation of the locus of a point P ( x , y ) P(x, y) P ( x , y ) Q ( 0 , 0 ) Q(0, 0) Q ( 0 , 0 ) R ( 2 , 1 ) R(2, 1) R ( 2 , 1 )
A 2 x + y = 5 2x + y = 5 2 x + y = 5 B 2 x + 2 y = 5 2x + 2y = 5 2 x + 2 y = 5 C 4 x + 2 y = 5 4x + 2y = 5 4 x + 2 y = 5 D 4 x − 2 y = 5 4x - 2y = 5 4 x − 2 y = 5
Worked solution (try it first) P Q = P R PQ = PR P Q = P R , so square both distances:
x 2 + y 2 = ( x − 2 ) 2 + ( y − 1 ) 2 x^2 + y^2 = (x - 2)^2 + (y - 1)^2 x 2 + y 2 = ( x − 2 ) 2 + ( y − 1 ) 2 .
Expand the right side:
x 2 − 4 x + 4 + y 2 − 2 y + 1 x^2 - 4x + 4 + y^2 - 2y + 1 x 2 − 4 x + 4 + y 2 − 2 y + 1 .
Cancel
x 2 x^2 x 2 and
y 2 y^2 y 2 :
0 = − 4 x − 2 y + 5 0 = -4x - 2y + 5 0 = − 4 x − 2 y + 5 .
So
4 x + 2 y = 5 4x + 2y = 5 4 x + 2 y = 5 , option C.
Watch out
The middle term of ( x − 2 ) 2 (x - 2)^2 ( x − 2 ) 2 − 4 x -4x − 4 x − 2 x -2x − 2 x ( y − 1 ) 2 (y - 1)^2 ( y − 1 ) 2 y 2 − y + 1 y^2 - y + 1 y 2 − y + 1 2 x + y = 5 2x + y = 5 2 x + y = 5 Report a problem with this question
An arc of a circle subtends an angle of 30 ∘ 30^\circ 3 0 ∘ [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
Worked solution (try it first) The angle at the centre is twice the angle at the circumference, so the arc subtends
60 ∘ 60^\circ 6 0 ∘ at the centre.
The circumference is
2 × 22 7 × 21 = 132 2 \times \frac{22}{7} \times 21 = 132 2 × 7 22 × 21 = 132 cm.
The arc is
60 360 = 1 6 \frac{60}{360} = \frac16 360 60 = 6 1 of it:
132 ÷ 6 = 22 132 \div 6 = 22 132 ÷ 6 = 22 cm, option C.
Watch out
The 30 ∘ 30^\circ 3 0 ∘ Report a problem with this question
A trapezium has two parallel sides of length 5 cm and 9 cm. If the area is 21 cm 2 21\text{ cm}^2 21 cm 2
Worked solution (try it first) Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h :
1 2 ( 5 + 9 ) h = 21 \frac12(5 + 9)h = 21 2 1 ( 5 + 9 ) h = 21 .
So
7 h = 21 7h = 21 7 h = 21 and
h = 3 h = 3 h = 3 cm, option B.
Watch out
7 (option A) is half the sum of the parallel sides. Divide the area by it to get the height, 3 cm. Report a problem with this question
X Y Z XYZ X Y Z O O O ∠ X Z Y = 45 ∘ \angle XZY = 45^\circ ∠ X Z Y = 4 5 ∘ X Y XY X Y [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 14 cm 2 14\text{ cm}^2 14 cm 2 B 38 cm 2 38\text{ cm}^2 38 cm 2 C 77 cm 2 77\text{ cm}^2 77 cm 2 D 84 cm 2 84\text{ cm}^2 84 cm 2
Worked solution (try it first) The angle at the centre is twice the angle at the circumference, so
∠ X O Y = 2 × 45 ∘ = 90 ∘ \angle XOY = 2 \times 45^\circ = 90^\circ ∠ X O Y = 2 × 4 5 ∘ = 9 0 ∘ .
The quarter-circle sector is
1 4 × 22 7 × 7 2 = 38.5 cm 2 \frac14 \times \frac{22}{7} \times 7^2 = 38.5\text{ cm}^2 4 1 × 7 22 × 7 2 = 38.5 cm 2 .
The right-angled triangle
X O Y XOY X O Y is
1 2 × 7 × 7 = 24.5 cm 2 \frac12 \times 7 \times 7 = 24.5\text{ cm}^2 2 1 × 7 × 7 = 24.5 cm 2 .
Segment = sector − triangle:
38.5 − 24.5 = 14 cm 2 38.5 - 24.5 = 14\text{ cm}^2 38.5 − 24.5 = 14 cm 2 , option A.
Watch out
The sector alone is about 38 cm 2 38\text{ cm}^2 38 cm 2 X O Y XOY X O Y Report a problem with this question
A triangle has vertices P ( − 1 , 6 ) P(-1, 6) P ( − 1 , 6 ) Q ( − 3 , − 4 ) Q(-3, -4) Q ( − 3 , − 4 ) R ( 1 , − 4 ) R(1, -4) R ( 1 , − 4 ) P Q PQ P Q Q R QR QR
A ( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) ( − 1 , − 1 ) (-1, -1) ( − 1 , − 1 ) B ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) ( − 1 , − 4 ) (-1, -4) ( − 1 , − 4 ) C ( 0 , − 1 ) (0, -1) ( 0 , − 1 ) ( − 1 , − 4 ) (-1, -4) ( − 1 , − 4 ) D ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) ( 0 , 1 ) (0, 1) ( 0 , 1 )
Worked solution (try it first) The midpoint averages the coordinates.
P Q PQ P Q :
( − 1 − 3 2 , 6 − 4 2 ) = ( − 2 , 1 ) \left(\frac{-1 - 3}{2}, \frac{6 - 4}{2}\right) = (-2, 1) ( 2 − 1 − 3 , 2 6 − 4 ) = ( − 2 , 1 ) .
Q R QR QR :
( − 3 + 1 2 , − 4 − 4 2 ) = ( − 1 , − 4 ) \left(\frac{-3 + 1}{2}, \frac{-4 - 4}{2}\right) = (-1, -4) ( 2 − 3 + 1 , 2 − 4 − 4 ) = ( − 1 , − 4 ) .
So the midpoints are
( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) and
( − 1 , − 4 ) (-1, -4) ( − 1 , − 4 ) , option B.
Watch out
Use the right pair of vertices for the second midpoint. ( 0 , 1 ) (0, 1) ( 0 , 1 ) P R PR P R Q R QR QR Report a problem with this question
Evaluate ∫ 2 3 ( x 2 − 2 x ) d x \displaystyle\int_2^3 (x^2 - 2x)\,dx ∫ 2 3 ( x 2 − 2 x ) d x
A 4 3 \frac43 3 4 B 1 3 \frac13 3 1 C 2 D 4
Worked solution (try it first) Integrate:
[ x 3 3 − x 2 ] 2 3 \left[\frac{x^3}{3} - x^2\right]_2^3 [ 3 x 3 − x 2 ] 2 3 .
At
x = 3 x = 3 x = 3 :
9 − 9 = 0 9 - 9 = 0 9 − 9 = 0 .
At
x = 2 x = 2 x = 2 :
8 3 − 4 = − 4 3 \frac83 - 4 = -\frac43 3 8 − 4 = − 3 4 .
Subtract:
0 − ( − 4 3 ) = 4 3 0 - \left(-\frac43\right) = \frac43 0 − ( − 3 4 ) = 3 4 , option A.
Watch out
Taking away a negative adds it, so the answer is + 4 3 +\frac43 + 3 4 2 x 2x 2 x x 2 x^2 x 2 2 x 2 2x^2 2 x 2 Report a problem with this question
If y = 3 sin ( − 4 x ) y = 3\sin(-4x) y = 3 sin ( − 4 x ) d y d x \frac{dy}{dx} d x d y
A − 12 cos ( − 4 x ) -12\cos(-4x) − 12 cos ( − 4 x ) B 12 sin ( − 4 x ) 12\sin(-4x) 12 sin ( − 4 x ) C 12 x cos ( 4 x ) 12x\cos(4x) 12 x cos ( 4 x ) D − 12 x cos ( − 4 x ) -12x\cos(-4x) − 12 x cos ( − 4 x )
Worked solution (try it first) Chain rule:
sin ( − 4 x ) \sin(-4x) sin ( − 4 x ) differentiates to
cos ( − 4 x ) \cos(-4x) cos ( − 4 x ) times the derivative of
− 4 x -4x − 4 x , which is
− 4 -4 − 4 .
So
d y d x = 3 × ( − 4 ) cos ( − 4 x ) \frac{dy}{dx} = 3 \times (-4)\cos(-4x) d x d y = 3 × ( − 4 ) cos ( − 4 x ) = − 12 cos ( − 4 x ) = -12\cos(-4x) = − 12 cos ( − 4 x ) , option A.
Watch out
The chain rule multiplies by the derivative of − 4 x -4x − 4 x − 4 -4 − 4 − 4 x -4x − 4 x x x x Report a problem with this question
Find the slope of the curve y = 2 x 2 + 5 x − 3 y = 2x^2 + 5x - 3 y = 2 x 2 + 5 x − 3 ( 1 , 4 ) (1, 4) ( 1 , 4 )
Worked solution (try it first) The slope is
d y d x = 4 x + 5 \frac{dy}{dx} = 4x + 5 d x d y = 4 x + 5 .
At
x = 1 x = 1 x = 1 :
4 + 5 = 9 4 + 5 = 9 4 + 5 = 9 , option B.
Watch out
2 x 2 2x^2 2 x 2 4 x 4x 4 x 2 x 2x 2 x 2 + 5 = 7 2 + 5 = 7 2 + 5 = 7 Report a problem with this question
The histogram shows the ages of the victims of a pollution. How many people were involved in the pollution?
Worked solution (try it first) Each bar's height is the number of people in that age group.
Read the bars: 3, 4, 5, 6 and 2.
Add them:
3 + 4 + 5 + 6 + 2 = 20 3 + 4 + 5 + 6 + 2 = 20 3 + 4 + 5 + 6 + 2 = 20 people, option D.
Watch out
Read each bar's height on the frequency axis and add them all; missing the short last bar of 2 gives 18 (option A). Report a problem with this question
Find the mean of the distribution below.
Value
0
1
2
3
4
Frequency
1
2
2
1
9
Worked solution (try it first) Multiply each value by its frequency and add:
0 + 2 + 4 + 3 + 36 = 45 0 + 2 + 4 + 3 + 36 = 45 0 + 2 + 4 + 3 + 36 = 45 .
Total frequency:
1 + 2 + 2 + 1 + 9 = 15 1 + 2 + 2 + 1 + 9 = 15 1 + 2 + 2 + 1 + 9 = 15 .
Mean
= 45 15 = 3 = \frac{45}{15} = 3 = 15 45 = 3 , option B.
Watch out
4 (option A) is the mode, the value with frequency 9. The mean needs ∑ f x ÷ ∑ f \sum fx \div \sum f ∑ f x ÷ ∑ f Report a problem with this question
The mean of the numbers 3, 6, 4, x x x
Worked solution (try it first) The mean is 5, so the five numbers add up to 25:
20 + x = 25 20 + x = 25 20 + x = 25 and
x = 5 x = 5 x = 5 .
The squared deviations from 5 are 4, 1, 1, 0, 4, which add up to 10.
The variance is
10 5 = 2 \frac{10}{5} = 2 5 10 = 2 , so the standard deviation is
2 \sqrt2 2 , option D.
Watch out
2 is the variance (option A). The standard deviation is its square root, 2 \sqrt2 2 Report a problem with this question
A bag contains 5 black balls and 3 red balls. Two balls are picked at random without replacement. What is the probability that a black and a red ball are picked?
A 5 14 \frac5{14} 14 5 B 13 28 \frac{13}{28} 28 13 C 3 14 \frac3{14} 14 3 D 15 28 \frac{15}{28} 28 15
Worked solution (try it first) Black then red:
5 8 × 3 7 = 15 56 \frac58 \times \frac37 = \frac{15}{56} 8 5 × 7 3 = 56 15 , since after one black, 7 balls are left.
Red then black:
3 8 × 5 7 = 15 56 \frac38 \times \frac57 = \frac{15}{56} 8 3 × 7 5 = 56 15 .
Add the two orders:
30 56 = 15 28 \frac{30}{56} = \frac{15}{28} 56 30 = 28 15 , option D.
Watch out
The red ball can come first or second, so add both orders. Using only black then red gives 15 56 \frac{15}{56} 56 15 Report a problem with this question
On a pie chart there are four sectors, three of which have angles 45 ∘ 45^\circ 4 5 ∘ 90 ∘ 90^\circ 9 0 ∘ 135 ∘ 135^\circ 13 5 ∘
A ₦48.00 B ₦96.00 C ₦42.00 D ₦84.00
Worked solution (try it first) The four angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the fourth is
360 ∘ − ( 45 ∘ + 90 ∘ + 135 ∘ ) = 90 ∘ 360^\circ - (45^\circ + 90^\circ + 135^\circ) = 90^\circ 36 0 ∘ − ( 4 5 ∘ + 9 0 ∘ + 13 5 ∘ ) = 9 0 ∘ .
So the largest sector is
135 ∘ 135^\circ 13 5 ∘ , which is 3 times the smallest,
45 ∘ 45^\circ 4 5 ∘ .
It represents
3 × 28 = 84 3 \times 28 = 84 3 × 28 = 84 naira: ₦84.00, option D.
Watch out
Work out the fourth angle before deciding which sector is largest: it is 90 ∘ 90^\circ 9 0 ∘ 135 ∘ 135^\circ 13 5 ∘ 135 ∘ 135^\circ 13 5 ∘ Report a problem with this question
The range of 4, 3, 11, 9, 6, 15, 19, 23, 27, 24, 21 and 16 is
Worked solution (try it first) The largest number in the list is 27 and the smallest is 3.
The range is
27 − 3 = 24 27 - 3 = 24 27 − 3 = 24 , option B.
Watch out
The smallest number is 3, not the first number 4. Using 4 gives 27 − 4 = 23 27 - 4 = 23 27 − 4 = 23 Report a problem with this question
The result of tossing a fair die 120 times is summarized below. Find the value of x x x
Number
1
2
3
4
5
6
Frequency
12
20
x x x 21
x − 1 x - 1 x − 1 28
Worked solution (try it first) The frequencies add up to the 120 tosses:
12 + 20 + x + 21 + ( x − 1 ) + 28 = 120 12 + 20 + x + 21 + (x - 1) + 28 = 120 12 + 20 + x + 21 + ( x − 1 ) + 28 = 120 .
Collect terms:
80 + 2 x = 120 80 + 2x = 120 80 + 2 x = 120 , so
2 x = 40 2x = 40 2 x = 40 .
So
x = 20 x = 20 x = 20 , option D.
Watch out
x − 1 = 19 x - 1 = 19 x − 1 = 19 x x x Report a problem with this question
If n P 3 − 6 ( n C 4 ) = 0 ^nP_3 - 6\,(^nC_4) = 0 n P 3 − 6 ( n C 4 ) = 0 n n n
Worked solution (try it first) Write both in terms of
n n n :
n P 3 = n ( n − 1 ) ( n − 2 ) ^nP_3 = n(n - 1)(n - 2) n P 3 = n ( n − 1 ) ( n − 2 ) and
6 ( n C 4 ) = 6 n ( n − 1 ) ( n − 2 ) ( n − 3 ) 24 6\,(^nC_4) = \dfrac{6\,n(n - 1)(n - 2)(n - 3)}{24} 6 ( n C 4 ) = 24 6 n ( n − 1 ) ( n − 2 ) ( n − 3 ) = n ( n − 1 ) ( n − 2 ) ( n − 3 ) 4 = \dfrac{n(n - 1)(n - 2)(n - 3)}{4} = 4 n ( n − 1 ) ( n − 2 ) ( n − 3 ) .
Set them equal and divide both sides by
n ( n − 1 ) ( n − 2 ) n(n - 1)(n - 2) n ( n − 1 ) ( n − 2 ) , which is not zero:
1 = n − 3 4 1 = \dfrac{n - 3}{4} 1 = 4 n − 3 .
So
n − 3 = 4 n - 3 = 4 n − 3 = 4 and
n = 7 n = 7 n = 7 , option D.
Watch out
n C 4 ^nC_4 n C 4 4 ! = 24 4! = 24 4 ! = 24 n − 3 = 2 3 n - 3 = \frac23 n − 3 = 3 2 Report a problem with this question
Two dice are thrown. What is the probability that the sum of the numbers is divisible by 3?
A 1 2 \frac12 2 1 B 1 3 \frac13 3 1 C 1 4 \frac14 4 1 D 2 3 \frac23 3 2
Worked solution (try it first) Two dice give 36 equally likely outcomes, with sums from 2 to 12.
The sums divisible by 3 are 3, 6, 9 and 12.
They occur 2, 5, 4 and 1 times, which is 12 outcomes.
So the probability is
12 36 = 1 3 \frac{12}{36} = \frac13 36 12 = 3 1 , option B.
Watch out
Count outcomes, not sums: the sums 2 to 12 are not equally likely. A sum of 6 comes up 5 ways, but 12 only one way. Report a problem with this question
Find the number of committees of three that can be formed consisting of two men and one woman from four men and three women.
Worked solution (try it first) Two men from four:
4 C 2 = 6 ^4C_2 = 6 4 C 2 = 6 ways.
One woman from three:
3 C 1 = 3 ^3C_1 = 3 3 C 1 = 3 ways.
Each pair of men can go with each woman, so multiply:
6 × 3 = 18 6 \times 3 = 18 6 × 3 = 18 , option B.
Watch out
4 C 2 = 6 ^4C_2 = 6 4 C 2 = 6 Report a problem with this question
By how much is the mean of 30, 56, 31, 55, 43 and 44 less than the median?
Worked solution (try it first) The numbers add up to 259, so the mean is
259 6 ≈ 43.17 \frac{259}{6} \approx 43.17 6 259 ≈ 43.17 .
In order: 30, 31, 43, 44, 55, 56.
The median is halfway between the 3rd and 4th:
43 + 44 2 = 43.5 \frac{43 + 44}{2} = 43.5 2 43 + 44 = 43.5 .
The difference is
43.5 − 43.17 = 0.33 43.5 - 43.17 = 0.33 43.5 − 43.17 = 0.33 , option D.
Watch out
Order the numbers before finding the median. The 3rd and 4th as written are 31 and 55, which gives 43 and a difference of 0.17 (option C). Report a problem with this question