Paper JAMB 2004 General Maths Objective
Objective paper · 44 questions · partial
JAMB 2004 · UME Topics include Number bases, Number foundations & fractions, Commercial arithmetic, Surds, Logarithms, Sets & Venn diagrams.
Our copy of this paper is missing questions 6, 12, 31, 39, 41, 49.
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 7 8 9 10 11 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 32 33 34 35 36 37 38 40 42 43 44 45 46 47 48 50 Find x x x y y y
4 2 4 3 − 1 3 x 4 y 3 4 4 \begin{array}{rcccc} & 4 & 2 & 4 & 3 \\ - & 1 & 3 & x & 4 \\ \hline & y & 3 & 4 & 4 \end{array} − 4 1 y 2 3 3 4 x 4 3 4 4
Worked solution (try it first) Units:
3 < 4 3 < 4 3 < 4 , so borrow 5:
8 − 4 = 4 8 - 4 = 4 8 − 4 = 4 .
Fives:
4 − 1 = 3 4 - 1 = 3 4 − 1 = 3 is less than
x x x , so borrow again:
8 − x = 4 8 - x = 4 8 − x = 4 , giving
x = 4 x = 4 x = 4 .
25s:
2 − 1 = 1 < 3 2 - 1 = 1 < 3 2 − 1 = 1 < 3 , so borrow:
6 − 3 = 3 6 - 3 = 3 6 − 3 = 3 . 125s:
4 − 1 − 1 = 2 4 - 1 - 1 = 2 4 − 1 − 1 = 2 , so
y = 2 y = 2 y = 2 .
So
x = 4 x = 4 x = 4 and
y = 2 y = 2 y = 2 , option C.
Watch out
Remember the 1 borrowed from the 125s column: 4 − 1 − 1 = 2 4 - 1 - 1 = 2 4 − 1 − 1 = 2 y = 3 y = 3 y = 3 Also set as JAMB 2018 · UTME · Q1
Report a problem with this question
Find p p p 451 6 − p 7 = 305 6 451_6 - p_7 = 305_6 45 1 6 − p 7 = 30 5 6
A 611 7 611_7 61 1 7 B 142 7 142_7 14 2 7 C 116 7 116_7 11 6 7 D 62 7 62_7 6 2 7
Worked solution (try it first) Change to base ten:
451 6 = 144 + 30 + 1 = 175 451_6 = 144 + 30 + 1 = 175 45 1 6 = 144 + 30 + 1 = 175 and
305 6 = 108 + 5 = 113 305_6 = 108 + 5 = 113 30 5 6 = 108 + 5 = 113 .
So
p = 175 − 113 = 62 p = 175 - 113 = 62 p = 175 − 113 = 62 in base ten.
Write 62 in base seven:
62 = 1 × 49 + 1 × 7 + 6 62 = 1 \times 49 + 1 \times 7 + 6 62 = 1 × 49 + 1 × 7 + 6 , so
p = 116 7 p = 116_7 p = 11 6 7 , option C.
Watch out
62 is the base-ten value. p p p 62 7 62_7 6 2 7 Report a problem with this question
Simplify 1 10 × 2 3 + 1 4 1 2 ÷ 3 5 − 1 4 \dfrac{\frac1{10} \times \frac23 + \frac14}{\frac12 \div \frac35 - \frac14} 2 1 ÷ 5 3 − 4 1 10 1 × 3 2 + 4 1
A 2 25 \frac2{25} 25 2 B 19 60 \frac{19}{60} 60 19 C 7 12 \frac7{12} 12 7 D 19 35 \frac{19}{35} 35 19
Worked solution (try it first) Top: multiply first,
1 10 × 2 3 = 1 15 \frac{1}{10} \times \frac23 = \frac{1}{15} 10 1 × 3 2 = 15 1 .
Then
1 15 + 1 4 = 4 60 + 15 60 \frac{1}{15} + \frac14 = \frac{4}{60} + \frac{15}{60} 15 1 + 4 1 = 60 4 + 60 15 = 19 60 = \frac{19}{60} = 60 19 .
Bottom: divide first by flipping,
1 2 ÷ 3 5 = 1 2 × 5 3 \frac12 \div \frac35 = \frac12 \times \frac53 2 1 ÷ 5 3 = 2 1 × 3 5 Then
5 6 − 1 4 = 10 12 − 3 12 \frac56 - \frac14 = \frac{10}{12} - \frac{3}{12} 6 5 − 4 1 = 12 10 − 12 3 Top divided by bottom:
19 60 × 12 7 = 19 35 \frac{19}{60} \times \frac{12}{7} = \frac{19}{35} 60 19 × 7 12 = 35 19 , option D.
Watch out
Options B (19 60 \frac{19}{60} 60 19 7 12 \frac{7}{12} 12 7 Report a problem with this question
A farmer planted 5000 grains of maize and harvested 5000 cobs, each bearing 500 grains. What is the ratio of the number of grains sowed to the number harvested?
A 1 : 500 B 1 : 5000 C 1 : 25000 D 1 : 250000
Worked solution (try it first) Grains harvested: 5000 cobs with 500 grains each is
5000 × 500 = 2 500 000 5000 \times 500 = 2\,500\,000 5000 × 500 = 2 500 000 grains.
Grains sowed to grains harvested:
5000 : 2 500 000 5000 : 2\,500\,000 5000 : 2 500 000 .
Divide both by 5000:
1 : 500 1 : 500 1 : 500 , option A.
Watch out
When you simplify, divide both sides by the same number: 2 500 000 ÷ 5000 = 500 2\,500\,000 \div 5000 = 500 2 500 000 ÷ 5000 = 500 1 : 5000 1 : 5000 1 : 5000 1 : 50 1 : 50 1 : 50 Report a problem with this question
Three teachers shared a packet of chalk. The first teacher got 2 5 \frac25 5 2 2 15 \frac2{15} 15 2
A 11 25 \frac{11}{25} 25 11 B 12 25 \frac{12}{25} 25 12 C 13 25 \frac{13}{25} 25 13 D 8 15 \frac8{15} 15 8
Worked solution (try it first) After the first teacher takes
2 5 \frac25 5 2 , the remainder is
3 5 \frac35 5 3 of the chalk.
The second gets
2 15 \frac{2}{15} 15 2 of the remainder:
2 15 × 3 5 = 6 75 \frac{2}{15} \times \frac35 = \frac{6}{75} 15 2 × 5 3 = 75 6 The third gets the rest:
1 − 10 25 − 2 25 = 13 25 1 - \frac{10}{25} - \frac{2}{25} = \frac{13}{25} 1 − 25 10 − 25 2 = 25 13 , option C.
Watch out
The second teacher's 2 15 \frac{2}{15} 15 2 3 5 \frac35 5 3 2 15 \frac{2}{15} 15 2 7 15 \frac{7}{15} 15 7 Report a problem with this question
Simplify 1 3 + 2 \dfrac{1}{\sqrt3 + 2} 3 + 2 1 a + b 3 a + b\sqrt3 a + b 3
A − 2 − 3 -2 - \sqrt3 − 2 − 3 B − 2 + 3 -2 + \sqrt3 − 2 + 3 C 2 − 3 2 - \sqrt3 2 − 3 D 2 + 3 2 + \sqrt3 2 + 3
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
2 − 3 2 - \sqrt3 2 − 3 .
Bottom:
( 2 + 3 ) ( 2 − 3 ) = 4 − 3 (2 + \sqrt3)(2 - \sqrt3) = 4 - 3 ( 2 + 3 ) ( 2 − 3 ) = 4 − 3 , which is 1.
So the value is
2 − 3 2 - \sqrt3 2 − 3 , option C.
Watch out
If you multiply by 3 − 2 \sqrt3 - 2 3 − 2 3 − 4 = − 1 3 - 4 = -1 3 − 4 = − 1 − 1 -1 − 1 − 2 + 3 -2 + \sqrt3 − 2 + 3 Similar: JAMB 1989 · UME · Q10
Report a problem with this question
If 6 log x 2 − 3 log x 3 = 3 log 5 0.2 6\log_x 2 - 3\log_x 3 = 3\log_5 0.2 6 log x 2 − 3 log x 3 = 3 log 5 0.2 x x x
A 3 8 \frac38 8 3 B 3 4 \frac34 4 3 C 4 3 \frac43 3 4 D 8 3 \frac83 3 8
Worked solution (try it first) Right side:
0.2 = 5 − 1 0.2 = 5^{-1} 0.2 = 5 − 1 , so
3 log 5 0.2 = 3 ( − 1 ) = − 3 3\log_5 0.2 = 3(-1) = -3 3 log 5 0.2 = 3 ( − 1 ) = − 3 .
Left side:
6 log x 2 = log x 64 6\log_x 2 = \log_x 64 6 log x 2 = log x 64 and
3 log x 3 = log x 27 3\log_x 3 = \log_x 27 3 log x 3 = log x 27 , so the left side is
log x 64 27 \log_x \frac{64}{27} log x 27 64 .
Change to index form:
x − 3 = 64 27 x^{-3} = \frac{64}{27} x − 3 = 27 64 , so
x 3 = 27 64 x^3 = \frac{27}{64} x 3 = 64 27 .
Take cube roots:
x = 3 4 x = \frac34 x = 4 3 , option B.
Watch out
The power is − 3 -3 − 3 x − 3 = 64 27 x^{-3} = \frac{64}{27} x − 3 = 27 64 x 3 = 27 64 x^3 = \frac{27}{64} x 3 = 64 27 4 3 \frac43 3 4 Report a problem with this question
The shaded region in the Venn diagram is
A P c ∩ ( Q ∩ R ) P^c \cap (Q \cap R) P c ∩ ( Q ∩ R ) B P ∩ Q P \cap Q P ∩ Q C P c ∪ ( Q ∩ R ) P^c \cup (Q \cap R) P c ∪ ( Q ∩ R ) D P c ∩ ( Q ∪ R ) P^c \cap (Q \cup R) P c ∩ ( Q ∪ R )
Worked solution (try it first) The shaded region lies inside both
Q Q Q and
R R R , which gives
Q ∩ R Q \cap R Q ∩ R .
It lies outside
P P P , which gives
P c P^c P c .
Together:
P c ∩ ( Q ∩ R ) P^c \cap (Q \cap R) P c ∩ ( Q ∩ R ) , option A.
Watch out
P c ∩ ( Q ∪ R ) P^c \cap (Q \cup R) P c ∩ ( Q ∪ R ) Q Q Q R R R Q Q Q R R R Report a problem with this question
In a class of 40 students, each student offers at least one of Physics and Chemistry. If the number that offer Physics is three times the number that offer both, and the number that offer Chemistry is twice the number that offer Physics, find the number of students that offer Physics only.
Worked solution (try it first) Then
3 b 3b 3 b offer Physics and
2 × 3 b = 6 b 2 \times 3b = 6b 2 × 3 b = 6 b offer Chemistry.
Everyone offers at least one, so
3 b + 6 b − b = 40 3b + 6b - b = 40 3 b + 6 b − b = 40 , which gives
8 b = 40 8b = 40 8 b = 40 and
b = 5 b = 5 b = 5 .
Physics has
3 × 5 = 15 3 \times 5 = 15 3 × 5 = 15 students, so Physics only is
15 − 5 = 10 15 - 5 = 10 15 − 5 = 10 , option C.
Watch out
15 (option B) is everyone who offers Physics. Take away the 5 who offer both. Report a problem with this question
Find the values of x x x y = x 3 + 2 x 2 − 5 x − 6 y = x^3 + 2x^2 - 5x - 6 y = x 3 + 2 x 2 − 5 x − 6 x x x
A − 2 , − 1 -2, -1 − 2 , − 1 B − 2 , 1 -2, 1 − 2 , 1 − 3 -3 − 3 C 2 , − 1 2, -1 2 , − 1 − 3 -3 − 3 D − 2 , 1 -2, 1 − 2 , 1
Worked solution (try it first) The curve crosses the
x x x -axis where
y = 0 y = 0 y = 0 .
Try small numbers:
f ( 2 ) = 8 + 8 − 10 − 6 = 0 f(2) = 8 + 8 - 10 - 6 = 0 f ( 2 ) = 8 + 8 − 10 − 6 = 0 , so
x − 2 x - 2 x − 2 is a factor.
Divide out
x − 2 x - 2 x − 2 :
x 3 + 2 x 2 − 5 x − 6 = ( x − 2 ) ( x 2 + 4 x + 3 ) x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3) x 3 + 2 x 2 − 5 x − 6 = ( x − 2 ) ( x 2 + 4 x + 3 ) .
Factorise:
x 2 + 4 x + 3 = ( x + 1 ) ( x + 3 ) x^2 + 4x + 3 = (x + 1)(x + 3) x 2 + 4 x + 3 = ( x + 1 ) ( x + 3 ) , so the roots are
2 , − 1 2, -1 2 , − 1 and
− 3 -3 − 3 , option C.
Watch out
The factor x + 1 x + 1 x + 1 x = − 1 x = -1 x = − 1 x + 3 x + 3 x + 3 x = − 3 x = -3 x = − 3 − 2 , 1 -2, 1 − 2 , 1 Report a problem with this question
Factorize completely a c − 2 b c − a 2 + 4 b 2 ac - 2bc - a^2 + 4b^2 a c − 2 b c − a 2 + 4 b 2
A ( a − 2 b ) ( c + a − 2 b ) (a - 2b)(c + a - 2b) ( a − 2 b ) ( c + a − 2 b ) B ( a − 2 b ) ( c − a − 2 b ) (a - 2b)(c - a - 2b) ( a − 2 b ) ( c − a − 2 b ) C ( a − 2 b ) ( c + a + 2 b ) (a - 2b)(c + a + 2b) ( a − 2 b ) ( c + a + 2 b ) D ( a − 2 b ) ( c − a + 2 b ) (a - 2b)(c - a + 2b) ( a − 2 b ) ( c − a + 2 b )
Worked solution (try it first) Group the first two and last two terms:
c ( a − 2 b ) − ( a 2 − 4 b 2 ) c(a - 2b) - (a^2 - 4b^2) c ( a − 2 b ) − ( a 2 − 4 b 2 ) .
Factorise the difference of two squares:
a 2 − 4 b 2 = ( a − 2 b ) ( a + 2 b ) a^2 - 4b^2 = (a - 2b)(a + 2b) a 2 − 4 b 2 = ( a − 2 b ) ( a + 2 b ) .
Take out the common bracket:
( a − 2 b ) [ c − ( a + 2 b ) ] = ( a − 2 b ) ( c − a − 2 b ) (a - 2b)[c - (a + 2b)] = (a - 2b)(c - a - 2b) ( a − 2 b ) [ c − ( a + 2 b )] = ( a − 2 b ) ( c − a − 2 b ) , option B.
Watch out
− ( a + 2 b ) = − a − 2 b -(a + 2b) = -a - 2b − ( a + 2 b ) = − a − 2 b c − a + 2 b c - a + 2b c − a + 2 b Report a problem with this question
y y y x x x y = 4 y = 4 y = 4 x = 1 2 x = \frac12 x = 2 1 x x x y = 10 y = 10 y = 10
A 1 10 \frac1{10} 10 1 B 1 5 \frac15 5 1 C 2 D 10
Worked solution (try it first) Inverse proportion means
x y xy x y is constant:
k = 1 2 × 4 = 2 k = \frac12 \times 4 = 2 k = 2 1 × 4 = 2 .
When
y = 10 y = 10 y = 10 :
x = 2 10 = 1 5 x = \dfrac{2}{10} = \dfrac15 x = 10 2 = 5 1 , option B.
Watch out
Multiply for the constant: k = x y = 2 k = xy = 2 k = x y = 2 4 ÷ 1 2 = 8 4 \div \frac12 = 8 4 ÷ 2 1 = 8 x = 10 8 x = \frac{10}{8} x = 8 10 Report a problem with this question
The length L L L T T T
A 36 cm B 96 cm C 144 cm D 324 cm
Worked solution (try it first) Put in
T = 4 T = 4 T = 4 ,
L = 64 L = 64 L = 64 :
64 = 16 k 64 = 16k 64 = 16 k , so
k = 4 k = 4 k = 4 .
When
T = 9 T = 9 T = 9 :
L = 4 × 81 = 324 L = 4 \times 81 = 324 L = 4 × 81 = 324 cm, option D.
Watch out
Square the period: with L = k T L = kT L = k T k = 16 k = 16 k = 16 L = 144 L = 144 L = 144 Report a problem with this question
The shaded area in the diagram is represented by
A { ( x , y ) : y + 3 x < 6 } \{(x, y) : y + 3x < 6\} {( x , y ) : y + 3 x < 6 } B { ( x , y ) : y + 3 x < − 6 } \{(x, y) : y + 3x < -6\} {( x , y ) : y + 3 x < − 6 } C { ( x , y ) : y − 3 x < 6 } \{(x, y) : y - 3x < 6\} {( x , y ) : y − 3 x < 6 } D { ( x , y ) : y − 3 x < − 6 } \{(x, y) : y - 3x < -6\} {( x , y ) : y − 3 x < − 6 }
Worked solution (try it first) The line through
( 0 , 6 ) (0, 6) ( 0 , 6 ) and
( 2 , 0 ) (2, 0) ( 2 , 0 ) has gradient
0 − 6 2 − 0 = − 3 \frac{0 - 6}{2 - 0} = -3 2 − 0 0 − 6 = − 3 and
y y y -intercept 6, so
y = − 3 x + 6 y = -3x + 6 y = − 3 x + 6 , that is
y + 3 x = 6 y + 3x = 6 y + 3 x = 6 .
The shaded side contains the origin, where
y + 3 x = 0 < 6 y + 3x = 0 < 6 y + 3 x = 0 < 6 .
So the region is
y + 3 x < 6 y + 3x < 6 y + 3 x < 6 , option A.
Watch out
The line slopes down, so its gradient is − 3 -3 − 3 y + 3 x y + 3x y + 3 x y − 3 x = 6 y - 3x = 6 y − 3 x = 6 ( 0 , 6 ) (0, 6) ( 0 , 6 ) x x x ( − 2 , 0 ) (-2, 0) ( − 2 , 0 ) ( 2 , 0 ) (2, 0) ( 2 , 0 ) Report a problem with this question
What are the integral values of x x x − 1 < 3 − 2 x ≤ 5 -1 < 3 - 2x \le 5 − 1 < 3 − 2 x ≤ 5
A − 2 , 1 , 0 , − 1 -2, 1, 0, -1 − 2 , 1 , 0 , − 1 B − 1 , 0 , 1 , 2 -1, 0, 1, 2 − 1 , 0 , 1 , 2 C − 1 , 0 , 1 -1, 0, 1 − 1 , 0 , 1 D 0, 1, 2
Worked solution (try it first) Subtract 3 from all three parts:
− 4 < − 2 x ≤ 2 -4 < -2x \le 2 − 4 < − 2 x ≤ 2 .
Divide all three parts by
− 2 -2 − 2 and reverse both signs:
2 > x ≥ − 1 2 > x \ge -1 2 > x ≥ − 1 .
Read from the left:
− 1 ≤ x < 2 -1 \le x < 2 − 1 ≤ x < 2 .
So
− 1 -1 − 1 is included and 2 is not: the integers are
− 1 -1 − 1 ,
0 0 0 and
1 1 1 , option C.
Watch out
Keep track of which end is strict. The < < < − 1 , 0 , 1 , 2 -1, 0, 1, 2 − 1 , 0 , 1 , 2 Report a problem with this question
The n n n Q n = 3 ⋅ 2 n − 2 Q_n = 3 \cdot 2^{n - 2} Q n = 3 ⋅ 2 n − 2 U m = 3 ⋅ 2 2 m − 3 U_m = 3 \cdot 2^{2m - 3} U m = 3 ⋅ 2 2 m − 3 Q 2 Q_2 Q 2 U 2 U_2 U 2
Worked solution (try it first) Put
n = 2 n = 2 n = 2 into
Q n Q_n Q n :
Q 2 = 3 × 2 0 = 3 Q_2 = 3 \times 2^0 = 3 Q 2 = 3 × 2 0 = 3 .
Put
m = 2 m = 2 m = 2 into
U m U_m U m : the power is
2 × 2 − 3 = 1 2 \times 2 - 3 = 1 2 × 2 − 3 = 1 , so
U 2 = 3 × 2 = 6 U_2 = 3 \times 2 = 6 U 2 = 3 × 2 = 6 .
So
Q 2 × U 2 = 3 × 6 = 18 Q_2 \times U_2 = 3 \times 6 = 18 Q 2 × U 2 = 3 × 6 = 18 , option D.
Watch out
Any number to the power 0 is 1, so Q 2 = 3 Q_2 = 3 Q 2 = 3 2 0 2^0 2 0 Q 2 = 0 Q_2 = 0 Q 2 = 0 U 2 U_2 U 2 Report a problem with this question
Given that the first and fourth terms of a G.P. are 6 and 162 respectively, find the sum of the first three terms of the progression.
Worked solution (try it first) The 4th term of a G.P. is
a r 3 ar^3 a r 3 , so
6 r 3 = 162 6r^3 = 162 6 r 3 = 162 and
r 3 = 27 r^3 = 27 r 3 = 27 .
Take the cube root:
r = 3 r = 3 r = 3 .
The first three terms are 6, 18 and 54, and their sum is 78, option D.
Watch out
162 ÷ 6 = 27 162 \div 6 = 27 162 ÷ 6 = 27 r 3 r^3 r 3 r = 3 r = 3 r = 3 Report a problem with this question
Find the sum to infinity of the series 1 2 , 1 6 , 1 18 , … \frac12, \frac16, \frac1{18}, \dots 2 1 , 6 1 , 18 1 , …
A 1 B 3 4 \frac34 4 3 C 2 3 \frac23 3 2 D 1 3 \frac13 3 1
Worked solution (try it first) This is a G.P. with
a = 1 2 a = \frac12 a = 2 1 and
r = 1 6 ÷ 1 2 = 1 3 r = \frac16 \div \frac12 = \frac13 r = 6 1 ÷ 2 1 = 3 1 .
Since
r r r is between
− 1 -1 − 1 and 1,
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a , and
1 − 1 3 = 2 3 1 - \frac13 = \frac23 1 − 3 1 = 3 2 .
Dividing by
2 3 \frac23 3 2 means multiplying by
3 2 \frac32 2 3 :
S ∞ = 1 2 × 3 2 S_\infty = \frac12 \times \frac32 S ∞ = 2 1 × 2 3 = 3 4 = \frac34 = 4 3 , option B.
Watch out
2 3 \frac23 3 2 1 − r 1 - r 1 − r 1 2 ÷ 2 3 = 3 4 \frac12 \div \frac23 = \frac34 2 1 ÷ 3 2 = 4 3 Report a problem with this question
If the operation ∗ * ∗ p ∗ q = p q p * q = \sqrt{pq} p ∗ q = pq 4 ∗ ( 8 ∗ 32 ) 4 * (8 * 32) 4 ∗ ( 8 ∗ 32 )
Worked solution (try it first) Work out the bracket first:
8 ∗ 32 = 8 × 32 8 * 32 = \sqrt{8 \times 32} 8 ∗ 32 = 8 × 32 Then
4 ∗ 16 = 4 × 16 4 * 16 = \sqrt{4 \times 16} 4 ∗ 16 = 4 × 16 Watch out
16 (option A) is only the value of the bracket. You still have to combine it with 4: 4 ∗ 16 = 8 4 * 16 = 8 4 ∗ 16 = 8 Report a problem with this question
The inverse of the matrix ( 2 1 1 1 ) \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} ( 2 1 1 1 )
A ( 1 1 − 1 2 ) \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix} ( 1 − 1 1 2 ) B ( 1 − 1 1 2 ) \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix} ( 1 1 − 1 2 ) C ( 1 1 1 2 ) \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} ( 1 1 1 2 ) D ( 1 − 1 − 1 2 ) \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} ( 1 − 1 − 1 2 )
Worked solution (try it first) The determinant is
2 × 1 − 1 × 1 = 1 2 \times 1 - 1 \times 1 = 1 2 × 1 − 1 × 1 = 1 .
Swap the two diagonal entries (2 and 1) and change the signs of the other two:
( 1 − 1 − 1 2 ) \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} ( 1 − 1 − 1 2 ) .
Divide by the determinant, 1, which changes nothing.
The inverse is option D.
Watch out
After swapping the diagonal, change the signs of the off-diagonal entries. Forgetting gives ( 1 1 1 2 ) \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} ( 1 1 1 2 ) Report a problem with this question
If P = ( 1 0 − 1 3 4 5 − 1 0 1 ) P = \begin{pmatrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ -1 & 0 & 1 \end{pmatrix} P = 1 3 − 1 0 4 0 − 1 5 1 ∣ P ∣ |P| ∣ P ∣
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
1 × ( 4 × 1 − 5 × 0 ) = 4 1 \times (4 \times 1 - 5 \times 0) = 4 1 × ( 4 × 1 − 5 × 0 ) = 4 .
The second term is 0, because the entry is 0.
The third term is
− 1 × ( 3 × 0 − 4 × ( − 1 ) ) = − 1 × 4 -1 \times (3 \times 0 - 4 \times (-1)) = -1 \times 4 − 1 × ( 3 × 0 − 4 × ( − 1 )) = − 1 × 4 So
∣ P ∣ = 4 − 4 = 0 |P| = 4 - 4 = 0 ∣ P ∣ = 4 − 4 = 0 , option B.
(Row 3 is row 1 times
− 1 -1 − 1 , and that always makes the determinant 0.)
Watch out
In the last minor, − 4 × ( − 1 ) = + 4 -4 \times (-1) = +4 − 4 × ( − 1 ) = + 4 − 1 -1 − 1 − 4 -4 − 4 + 4 +4 + 4 ∣ P ∣ = 8 |P| = 8 ∣ P ∣ = 8 Report a problem with this question
The sum of the interior angles of a pentagon is 6 x + 6 y 6x + 6y 6 x + 6 y y y y x x x
A y = 60 − x y = 60 - x y = 60 − x B y = 90 − x y = 90 - x y = 90 − x C y = 120 − x y = 120 - x y = 120 − x D y = 150 − x y = 150 - x y = 150 − x
Worked solution (try it first) The interior angles of a pentagon add up to
( 5 − 2 ) × 180 ∘ = 540 ∘ (5 - 2) \times 180^\circ = 540^\circ ( 5 − 2 ) × 18 0 ∘ = 54 0 ∘ , so
6 x + 6 y = 540 6x + 6y = 540 6 x + 6 y = 540 .
Divide by 6:
x + y = 90 x + y = 90 x + y = 90 .
Subtract
x x x :
y = 90 − x y = 90 - x y = 90 − x , option B.
Watch out
A pentagon's angles add up to 540 ∘ 540^\circ 54 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 360 360 360 y = 60 − x y = 60 - x y = 60 − x Report a problem with this question
P Q R S T V PQRSTV P QR S T V P Q R S T V PQRSTV P QR S T V [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
Worked solution (try it first) P Q R S T V PQRSTV P QR S T V has six sides, so it is a regular hexagon.
Joining the centre to each vertex makes six equilateral triangles, so the radius equals the side:
r = 7 r = 7 r = 7 cm.
Circumference
= 2 π r = 2 × 22 7 × 7 = 2\pi r = 2 \times \frac{22}{7} \times 7 = 2 π r = 2 × 7 22 × 7 .
So the circumference is 44 cm, option C.
Watch out
6 × 7 = 42 6 \times 7 = 42 6 × 7 = 42 Also set as JAMB 2017 · UTME · Q12
Report a problem with this question
P P P R R R S S S O O O Q Q Q ∠ P S O \angle PSO ∠ P S O
A 35 ∘ 35^\circ 3 5 ∘ B 40 ∘ 40^\circ 4 0 ∘ C 45 ∘ 45^\circ 4 5 ∘ D 55 ∘ 55^\circ 5 5 ∘
Worked solution (try it first) Q Q Q ,
R R R and
S S S are in a straight line, so
∠ P R S \angle PRS ∠ P R S is the exterior angle of triangle
P Q R PQR P QR :
∠ P R S = 20 ∘ + 35 ∘ \angle PRS = 20^\circ + 35^\circ ∠ P R S = 2 0 ∘ + 3 5 ∘ The angle at the centre is twice the angle at the circumference on the same arc
P S PS P S :
∠ P O S = 2 × 55 ∘ \angle POS = 2 \times 55^\circ ∠ P O S = 2 × 5 5 ∘ O P = O S OP = OS O P = O S (radii), so triangle
P O S POS P O S is isosceles and its base angles are equal.
So
∠ P S O = 180 ∘ − 110 ∘ 2 \angle PSO = \frac{180^\circ - 110^\circ}{2} ∠ P S O = 2 18 0 ∘ − 11 0 ∘ = 35 ∘ = 35^\circ = 3 5 ∘ , option A.
Watch out
55 ∘ 55^\circ 5 5 ∘ ∠ P R S \angle PRS ∠ P R S P O S POS P O S Report a problem with this question
In the diagram, P Q = 4 PQ = 4 P Q = 4 T S = 6 TS = 6 T S = 6 P Q T U PQTU P QT U P Q R S PQRS P QR S P Q T U PQTU P QT U 32 cm 2 32\text{ cm}^2 32 cm 2 P Q R U PQRU P QR U
A 24 cm 2 24\text{ cm}^2 24 cm 2 B 48 cm 2 48\text{ cm}^2 48 cm 2 C 60 cm 2 60\text{ cm}^2 60 cm 2 D 72 cm 2 72\text{ cm}^2 72 cm 2
Worked solution (try it first) Area of parallelogram
P Q T U PQTU P QT U is base × height, so the height is
32 ÷ 4 = 8 32 \div 4 = 8 32 ÷ 4 = 8 cm.
Both parallelograms give
U T = P Q = 4 UT = PQ = 4 U T = P Q = 4 cm and
S R = P Q = 4 SR = PQ = 4 S R = P Q = 4 cm.
So
U R = 4 + 6 + 4 = 14 UR = 4 + 6 + 4 = 14 U R = 4 + 6 + 4 = 14 cm.
Trapezium
P Q R U PQRU P QR U :
1 2 ( 4 + 14 ) × 8 = 72 cm 2 \frac12(4 + 14) \times 8 = 72\text{ cm}^2 2 1 ( 4 + 14 ) × 8 = 72 cm 2 , option D.
Watch out
The base U R UR U R U T UT U T T S TS T S S R SR S R 1 2 ( 4 + 10 ) × 8 = 56 cm 2 \frac12(4 + 10) \times 8 = 56\text{ cm}^2 2 1 ( 4 + 10 ) × 8 = 56 cm 2 Report a problem with this question
An arc of a circle of length 22 cm subtends an angle of 3 x ∘ 3x^\circ 3 x ∘ x x x [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 30 ∘ 30^\circ 3 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 180 ∘ 180^\circ 18 0 ∘
Worked solution (try it first) The circumference is
22 7 × 14 = 44 \frac{22}{7} \times 14 = 44 7 22 × 14 = 44 cm.
An arc of 22 cm is half of it.
So the arc subtends half of
360 ∘ 360^\circ 36 0 ∘ :
3 x = 180 3x = 180 3 x = 180 .
So
x = 60 ∘ x = 60^\circ x = 6 0 ∘ , option B.
Watch out
180 ∘ 180^\circ 18 0 ∘ 3 x 3x 3 x x x x Report a problem with this question
Determine the locus of a point inside a square P Q R S PQRS P QR S P Q PQ P Q Q R QR QR
A The diagonal P R PR P R B The diagonal Q S QS QS C Side S R SR S R D The perpendicular bisector of P Q PQ P Q
Worked solution (try it first) P Q PQ P Q and
Q R QR QR meet at the corner
Q Q Q .
Points equidistant from two lines that meet lie on the bisector of the angle between them, here
∠ P Q R \angle PQR ∠ P QR .
In a square, the diagonal from
Q Q Q bisects the right angle at
Q Q Q .
That diagonal is
Q S QS QS , option B.
Watch out
Bisect the angle at Q Q Q P Q PQ P Q Q R QR QR P R PR P R Q Q Q Report a problem with this question
The locus of a point which is 5 cm from the line L M LM L M
A pair of lines on opposite sides of L M LM L M L M LM L M B line parallel to L M LM L M L M LM L M C pair of parallel lines on one side of L M LM L M L M LM L M D line 10 cm from L M LM L M L M LM L M
Worked solution (try it first) Points 5 cm from a line can be on either side of it, so there are two lines, each parallel to
L M LM L M and 5 cm from it.
So the locus is a pair of lines on opposite sides of
L M LM L M , each 5 cm from it, option A.
Watch out
Don't forget the other side of the line. One parallel line (option B) is only half the locus. Report a problem with this question
Find the midpoint of the line joining P ( − 3 , 5 ) P(-3, 5) P ( − 3 , 5 ) Q ( 5 , − 3 ) Q(5, -3) Q ( 5 , − 3 )
A ( 4 , − 4 ) (4, -4) ( 4 , − 4 ) B ( 4 , 4 ) (4, 4) ( 4 , 4 ) C ( 2 , 2 ) (2, 2) ( 2 , 2 ) D ( 1 , 1 ) (1, 1) ( 1 , 1 )
Worked solution (try it first) The midpoint is the average of the ends: add the coordinates and halve.
x x x :
− 3 + 5 2 = 1 \frac{-3 + 5}{2} = 1 2 − 3 + 5 = 1 .
y y y :
5 + ( − 3 ) 2 = 1 \frac{5 + (-3)}{2} = 1 2 5 + ( − 3 ) = 1 .
So the midpoint is
( 1 , 1 ) (1, 1) ( 1 , 1 ) , option D.
Watch out
Add the coordinates, don't subtract them: 5 − ( − 3 ) 2 = 4 \frac{5 - (-3)}{2} = 4 2 5 − ( − 3 ) = 4 − 3 − 5 2 = − 4 \frac{-3 - 5}{2} = -4 2 − 3 − 5 = − 4 ( 4 , − 4 ) (4, -4) ( 4 , − 4 ) ( 2 , 2 ) (2, 2) ( 2 , 2 ) Report a problem with this question
In the triangle shown, the base angles are 45 ∘ 45^\circ 4 5 ∘ 60 ∘ 60^\circ 6 0 ∘ 60 ∘ 60^\circ 6 0 ∘ 45 ∘ 45^\circ 4 5 ∘ x x x x x x
A 20 6 20\sqrt6 20 6 B 15 6 15\sqrt6 15 6 C 5 6 5\sqrt6 5 6 D 3 6 3\sqrt6 3 6
Worked solution (try it first) Sine rule:
x x x faces the
45 ∘ 45^\circ 4 5 ∘ angle and 15 cm faces the
60 ∘ 60^\circ 6 0 ∘ angle, so
x sin 45 ∘ = 15 sin 60 ∘ \dfrac{x}{\sin45^\circ} = \dfrac{15}{\sin60^\circ} sin 4 5 ∘ x = sin 6 0 ∘ 15 .
So
x = 15 × 2 2 3 2 x = \dfrac{15 \times \frac{\sqrt2}{2}}{\frac{\sqrt3}{2}} x = 2 3 15 × 2 2 = 15 2 3 = \dfrac{15\sqrt2}{\sqrt3} = 3 15 2 .
Rationalise: multiply top and bottom by
3 \sqrt3 3 to get
15 6 3 = 5 6 \dfrac{15\sqrt6}{3} = 5\sqrt6 3 15 6 = 5 6 cm, option C.
Watch out
After rationalising, divide by the 3 underneath: 15 6 3 = 5 6 \dfrac{15\sqrt6}{3} = 5\sqrt6 3 15 6 = 5 6 15 6 15\sqrt6 15 6 Report a problem with this question
The shadow of a pole 5 3 5\sqrt3 5 3
A 30 ∘ 30^\circ 3 0 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 75 ∘ 75^\circ 7 5 ∘
Worked solution (try it first) The pole is opposite the angle of elevation and the shadow is adjacent, so
tan θ = 5 3 5 = 3 \tan\theta = \frac{5\sqrt3}{5} = \sqrt3 tan θ = 5 5 3 = 3 .
tan 60 ∘ = 3 \tan60^\circ = \sqrt3 tan 6 0 ∘ = 3 , so
θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘ , option C.
Watch out
Put the height on top: tan θ = height shadow \tan\theta = \frac{\text{height}}{\text{shadow}} tan θ = shadow height 1 3 \frac{1}{\sqrt3} 3 1 30 ∘ 30^\circ 3 0 ∘ Also set as JAMB 2017 · UTME · Q13
Report a problem with this question
Find the derivative of ( 2 + 3 x ) ( 1 − x ) (2 + 3x)(1 - x) ( 2 + 3 x ) ( 1 − x ) x x x
A 6 x − 1 6x - 1 6 x − 1 B 1 − 6 x 1 - 6x 1 − 6 x C 6 D 3
Worked solution (try it first) Expand first:
( 2 + 3 x ) ( 1 − x ) = 2 − 2 x + 3 x − 3 x 2 (2 + 3x)(1 - x) = 2 - 2x + 3x - 3x^2 ( 2 + 3 x ) ( 1 − x ) = 2 − 2 x + 3 x − 3 x 2 = 2 + x − 3 x 2 = 2 + x - 3x^2 = 2 + x − 3 x 2 .
Differentiate term by term:
1 − 6 x 1 - 6x 1 − 6 x , option B.
Watch out
Watch the signs when expanding: − 2 x + 3 x = + x -2x + 3x = +x − 2 x + 3 x = + x 3 x × ( − x ) = − 3 x 2 3x \times (-x) = -3x^2 3 x × ( − x ) = − 3 x 2 6 x − 1 6x - 1 6 x − 1 Report a problem with this question
Find the derivative of the function y = 2 x 2 ( 2 x − 1 ) y = 2x^2(2x - 1) y = 2 x 2 ( 2 x − 1 ) x = − 1 x = -1 x = − 1
Worked solution (try it first) Expand:
y = 4 x 3 − 2 x 2 y = 4x^3 - 2x^2 y = 4 x 3 − 2 x 2 .
Differentiate:
d y d x = 12 x 2 − 4 x \frac{dy}{dx} = 12x^2 - 4x d x d y = 12 x 2 − 4 x .
At
x = − 1 x = -1 x = − 1 :
12 ( 1 ) − 4 ( − 1 ) = 12 + 4 = 16 12(1) - 4(-1) = 12 + 4 = 16 12 ( 1 ) − 4 ( − 1 ) = 12 + 4 = 16 , option C.
Watch out
− 4 x -4x − 4 x x = − 1 x = -1 x = − 1 + 4 +4 + 4 − 4 -4 − 4 12 − 4 = 8 12 - 4 = 8 12 − 4 = 8 Report a problem with this question
If y = 3 cos x 3 y = 3\cos\frac x3 y = 3 cos 3 x d y d x \frac{dy}{dx} d x d y x = 3 π 2 x = \frac{3\pi}{2} x = 2 3 π
Worked solution (try it first) Chain rule:
cos x 3 \cos\frac x3 cos 3 x differentiates to
− 1 3 sin x 3 -\frac13\sin\frac x3 − 3 1 sin 3 x , so
d y d x = 3 × ( − 1 3 ) sin x 3 \frac{dy}{dx} = 3 \times \left(-\frac13\right)\sin\frac x3 d x d y = 3 × ( − 3 1 ) sin 3 x = − sin x 3 = -\sin\frac x3 = − sin 3 x .
At
x = 3 π 2 x = \frac{3\pi}{2} x = 2 3 π ,
x 3 = π 2 \frac x3 = \frac\pi2 3 x = 2 π and
sin π 2 = 1 \sin\frac\pi2 = 1 sin 2 π = 1 .
So
d y d x = − 1 \frac{dy}{dx} = -1 d x d y = − 1 , option C.
Watch out
cos differentiates to − sin -\sin − sin Report a problem with this question
What is the rate of change of the volume V V V r r r r = 2 r = 2 r = 2
A 2 π 2\pi 2 π B 4 π 4\pi 4 π C 8 π 8\pi 8 π D 16 π 16\pi 16 π
Worked solution (try it first) A hemisphere is half a sphere, so
V = 1 2 × 4 3 π r 3 V = \frac12 \times \frac43\pi r^3 V = 2 1 × 3 4 π r 3 = 2 3 π r 3 = \frac23\pi r^3 = 3 2 π r 3 .
Differentiate:
d V d r = 2 π r 2 \frac{dV}{dr} = 2\pi r^2 d r d V = 2 π r 2 .
At
r = 2 r = 2 r = 2 :
2 π × 4 = 8 π 2\pi \times 4 = 8\pi 2 π × 4 = 8 π , option C.
Watch out
Halve the sphere's volume first. Using the whole sphere, d V d r = 4 π r 2 = 16 π \frac{dV}{dr} = 4\pi r^2 = 16\pi d r d V = 4 π r 2 = 16 π Report a problem with this question
The pie chart shows the distribution of the crops harvested from a farmland in a year. If 3000 tonnes of millet is harvested, what amount of beans is harvested?
A 9000 tonnes B 6000 tonnes C 1500 tonnes D 1200 tonnes
Worked solution (try it first) Maize is marked with a right angle, so it is
90 ∘ 90^\circ 9 0 ∘ .
Beans is the rest:
360 ∘ − 60 ∘ − 90 ∘ − 150 ∘ = 60 ∘ 360^\circ - 60^\circ - 90^\circ - 150^\circ = 60^\circ 36 0 ∘ − 6 0 ∘ − 9 0 ∘ − 15 0 ∘ = 6 0 ∘ .
Millet's
150 ∘ 150^\circ 15 0 ∘ stands for 3000 tonnes, so
1 ∘ 1^\circ 1 ∘ stands for
3000 ÷ 150 = 20 3000 \div 150 = 20 3000 ÷ 150 = 20 tonnes.
Beans:
60 × 20 = 1200 60 \times 20 = 1200 60 × 20 = 1200 tonnes, option D.
Watch out
3000 tonnes is the millet sector, not the whole harvest, so scale by 60 150 \frac{60}{150} 150 60 60 360 × 3000 = 500 \frac{60}{360} \times 3000 = 500 360 60 × 3000 = 500 Report a problem with this question
The graph shows the cumulative frequency curve of the distribution of marks in a class test. What percentage of the students scored more than 20 marks?
A 68 % 68\% 68% B 28 % 28\% 28% C 17 % 17\% 17% D 8 % 8\% 8%
Worked solution (try it first) The curve ends at 25, so there are 25 students.
Read the curve at 20 marks: the cumulative frequency is 18, so 18 students scored 20 or less.
So
25 − 18 = 7 25 - 18 = 7 25 − 18 = 7 students scored more than 20.
As a percentage:
7 25 × 100 = 28 % \frac{7}{25} \times 100 = 28\% 25 7 × 100 = 28% , option B.
Watch out
A cumulative frequency counts the students at or below a mark. Using 18 directly gives 72 % 72\% 72% Report a problem with this question
The mean age of a group of students is 15 years. When the age of a teacher, 45 years old, is added, the mean becomes 18 years. Find the number of students in the group.
Worked solution (try it first) Let there be
n n n students.
Their total age is
15 n 15n 15 n .
With the teacher it is
15 n + 45 15n + 45 15 n + 45 for
n + 1 n + 1 n + 1 people.
The new mean is 18, so
15 n + 45 = 18 ( n + 1 ) = 18 n + 18 15n + 45 = 18(n + 1) = 18n + 18 15 n + 45 = 18 ( n + 1 ) = 18 n + 18 .
Collect terms:
27 = 3 n 27 = 3n 27 = 3 n , so
n = 9 n = 9 n = 9 , option B.
Watch out
With the teacher there are n + 1 n + 1 n + 1 n n n 15 n + 45 = 18 n 15n + 45 = 18n 15 n + 45 = 18 n n = 15 n = 15 n = 15 Report a problem with this question
The weights of 10 pupils in a class are 15 kg, 16 kg, 17 kg, 18 kg, 16 kg, 17 kg, 17 kg, 17 kg, 18 kg and 16 kg. What is the range of this distribution?
Worked solution (try it first) The heaviest pupil is 18 kg and the lightest is 15 kg.
The range is
18 − 15 = 3 18 - 15 = 3 18 − 15 = 3 kg, option C.
Watch out
15 kg appears only once, at the start. Missing it and using 16 kg as the lightest gives 2 (option B). Report a problem with this question
Find the mean deviation of 1, 2, 3 and 4.
Worked solution (try it first) The mean of 1, 2, 3, 4 is 2.5.
The distances from 2.5 are 1.5, 0.5, 0.5 and 1.5, which add up to 4.
The mean deviation is
4 4 = 1.0 \frac44 = 1.0 4 4 = 1.0 , option A.
Watch out
The mean deviation is the average distance from the mean, not the mean itself. 2.5 (option D) is the mean. Report a problem with this question
In how many ways can 2 students be selected from a group of 5 students in a debating competition?
A 10 ways B 15 ways C 20 ways D 25 ways
Worked solution (try it first) The two students form a pair, so their order doesn't matter: use
5 C 2 ^5C_2 5 C 2 .
5 C 2 = 5 × 4 2 × 1 ^5C_2 = \dfrac{5 \times 4}{2 \times 1} 5 C 2 = 2 × 1 5 × 4 = 10 = 10 = 10 ways, option A.
Watch out
Selecting is not arranging. 5 P 2 = 5 × 4 = 20 ^5P_2 = 5 \times 4 = 20 5 P 2 = 5 × 4 = 20 Report a problem with this question
A committee of six is to be formed by a state governor from nine state commissioners and three members of the state house of assembly. In how many ways can the committee be chosen so as to include one member of the house of assembly?
A 924 ways B 840 ways C 462 ways D 378 ways
Worked solution (try it first) Choose the one assembly member from three:
3 C 1 = 3 ^3C_1 = 3 3 C 1 = 3 ways.
The other 5 places go to commissioners, 5 from 9:
9 C 5 = 126 ^9C_5 = 126 9 C 5 = 126 ways.
Multiply:
3 × 126 = 378 3 \times 126 = 378 3 × 126 = 378 ways, option D.
Watch out
12 C 6 = 924 ^{12}C_6 = 924 12 C 6 = 924 Report a problem with this question
Some white balls were put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white ball from the basket is 3 7 \frac37 7 3
Worked solution (try it first) Let
w w w white balls be added.
The total becomes
12 + 16 + w = 28 + w 12 + 16 + w = 28 + w 12 + 16 + w = 28 + w .
So
w 28 + w = 3 7 \frac{w}{28 + w} = \frac37 28 + w w = 7 3 .
Cross-multiply:
7 w = 84 + 3 w 7w = 84 + 3w 7 w = 84 + 3 w .
Take
3 w 3w 3 w from both sides:
4 w = 84 4w = 84 4 w = 84 , so
w = 21 w = 21 w = 21 , option C.
Watch out
The white balls are part of the total too. Writing w 28 = 3 7 \frac{w}{28} = \frac37 28 w = 7 3 w = 12 w = 12 w = 12 Report a problem with this question
A container has 30 gold medals, 22 silver medals and 18 bronze medals. If one medal is selected at random, what is the probability that it is not a gold medal?
A 4 7 \frac47 7 4 B 3 7 \frac37 7 3 C 11 35 \frac{11}{35} 35 11 D 9 35 \frac9{35} 35 9
Worked solution (try it first) There are
30 + 22 + 18 = 70 30 + 22 + 18 = 70 30 + 22 + 18 = 70 medals.
Not gold means silver or bronze:
22 + 18 = 40 22 + 18 = 40 22 + 18 = 40 medals.
So the probability is
40 70 = 4 7 \frac{40}{70} = \frac47 70 40 = 7 4 , option A.
Watch out
30 70 = 3 7 \frac{30}{70} = \frac37 70 30 = 7 3 Also set as JAMB 2017 · UTME · Q15
Report a problem with this question