Paper JAMB 2013 General Maths Objective
Objective paper · 45 questions · partial
JAMB 2013 · UTME Topics include Number bases, Number foundations & fractions, Approximation & error, Commercial arithmetic, Logarithms, Surds.
Our copy of this paper is missing questions 1, 6, 24, 40, 41.
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.
2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 42 43 44 45 46 47 48 49 50 Convert 27 10 27_{10} 2 7 10
A 1100 3 1100_3 110 0 3 B 1000 3 1000_3 100 0 3 C 1001 3 1001_3 100 1 3 D 1010 3 1010_3 101 0 3
Worked solution (try it first) 27 = 3 × 3 × 3 = 3 3 27 = 3 \times 3 \times 3 = 3^3 27 = 3 × 3 × 3 = 3 3 .
So 27 is one lot of
3 3 3^3 3 3 and nothing else:
1 × 27 + 0 × 9 + 0 × 3 + 0 1 \times 27 + 0 \times 9 + 0 \times 3 + 0 1 × 27 + 0 × 9 + 0 × 3 + 0 .
So
27 10 = 1000 3 27_{10} = 1000_3 2 7 10 = 100 0 3 , option B.
Watch out
Check an option by converting back: 1100 3 = 27 + 9 = 36 1100_3 = 27 + 9 = 36 110 0 3 = 27 + 9 = 36 1000 3 1000_3 100 0 3 Report a problem with this question
3 girls share a number of apples in the ratio 5 : 3 : 2 5 : 3 : 2 5 : 3 : 2
Worked solution (try it first) The highest share is 5 parts, so 5 parts are 40 apples and 1 part is
40 ÷ 5 = 8 40 \div 5 = 8 40 ÷ 5 = 8 apples.
The smallest share is 2 parts:
2 × 8 = 16 2 \times 8 = 16 2 × 8 = 16 apples, option A.
Watch out
The smallest share is the 2 parts, not the 3: 3 × 8 = 24 3 \times 8 = 24 3 × 8 = 24 Report a problem with this question
Evaluate 1.25 × 0.025 0.05 \dfrac{1.25 \times 0.025}{0.05} 0.05 1.25 × 0.025
Worked solution (try it first) Top:
1.25 × 0.025 = 0.03125 1.25 \times 0.025 = 0.03125 1.25 × 0.025 = 0.03125 .
Divide by 0.05, which is the same as multiplying by 20:
0.03125 × 20 = 0.625 0.03125 \times 20 = 0.625 0.03125 × 20 = 0.625 .
To 1 decimal place, the second decimal is 2, so round down: 0.6, option C.
Watch out
Count the decimal places in 1.25 × 0.025 1.25 \times 0.025 1.25 × 0.025 Report a problem with this question
Calculate the time taken for ₦3000 to earn ₦600 if invested at 8 % 8\% 8%
A 3 1 2 3\frac12 3 2 1 B 1 1 2 1\frac12 1 2 1 C 2 1 2 2\frac12 2 2 1 D 3 years
Worked solution (try it first) One year's interest at
8 % 8\% 8% is
0.08 × 3000 = 0.08 \times 3000 = 0.08 × 3000 = ₦240.
The interest needed is ₦600, so the time is
600 ÷ 240 = 2.5 600 \div 240 = 2.5 600 ÷ 240 = 2.5 .
So it takes
2 1 2 2\frac12 2 2 1 years, option C.
Watch out
Simple interest can run for part of a year: after 2 years it has earned ₦480 and after 3 years ₦720, so ₦600 is reached at 2 1 2 2\frac12 2 2 1 Report a problem with this question
If log 10 4 = 0.6021 \log_{10} 4 = 0.6021 log 10 4 = 0.6021 log 10 4 1 3 \log_{10} 4^{\frac13} log 10 4 3 1
A 1.8063 B 0.2007 C 0.3011 D 0.9021
Worked solution (try it first) A power comes down in front:
log 10 4 1 3 = 1 3 log 10 4 \log_{10} 4^{\frac13} = \frac13 \log_{10} 4 log 10 4 3 1 = 3 1 log 10 4 .
1 3 × 0.6021 = 0.2007 \frac13 \times 0.6021 = 0.2007 3 1 × 0.6021 = 0.2007 , option B.
Watch out
The power 1 3 \frac13 3 1 Report a problem with this question
Simplify 5 ( 147 − 12 ) 15 \dfrac{\sqrt5(\sqrt{147} - \sqrt{12})}{\sqrt{15}} 15 5 ( 147 − 12 )
A 1 9 \frac19 9 1 B 9 C 5 D 1 5 \frac15 5 1
Worked solution (try it first) Take out square factors:
147 = 7 3 \sqrt{147} = 7\sqrt3 147 = 7 3 and
12 = 2 3 \sqrt{12} = 2\sqrt3 12 = 2 3 , so the bracket is
5 3 5\sqrt3 5 3 .
Multiply by
5 \sqrt5 5 :
5 3 × 5 = 5 15 5\sqrt3 \times \sqrt5 = 5\sqrt{15} 5 3 × 5 = 5 15 .
Divide by
15 \sqrt{15} 15 :
5 15 15 = 5 \dfrac{5\sqrt{15}}{\sqrt{15}} = 5 15 5 15 = 5 , option C.
Watch out
The bracket is a subtraction: 7 3 − 2 3 = 5 3 7\sqrt3 - 2\sqrt3 = 5\sqrt3 7 3 − 2 3 = 5 3 9 3 9\sqrt3 9 3 Report a problem with this question
P P P Q Q Q R R R U U U ( P ∩ Q ) ∪ R (P \cap Q) \cup R ( P ∩ Q ) ∪ R
Worked solution (try it first) ∪ R \cup R ∪ R means every region inside
R R R is shaded.
P ∩ Q P \cap Q P ∩ Q adds the overlap of
P P P and
Q Q Q , including its part outside
R R R .
Diagram C shades all of
R R R together with the overlap of
P P P and
Q Q Q , so the answer is option C.
Watch out
Union means everything in either part, so all of R R R P P P Q Q Q R R R Report a problem with this question
If P = { x : x is odd , − 1 < x ≤ 20 } P = \{x : x \text{ is odd}, -1 < x \le 20\} P = { x : x is odd , − 1 < x ≤ 20 } Q = { y : y is prime , − 2 < y ≤ 25 } Q = \{y : y \text{ is prime}, -2 < y \le 25\} Q = { y : y is prime , − 2 < y ≤ 25 } P ∩ Q P \cap Q P ∩ Q
A { 3 , 5 , 7 , 11 , 13 , 17 , 19 } \{3, 5, 7, 11, 13, 17, 19\} { 3 , 5 , 7 , 11 , 13 , 17 , 19 } B { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } \{2, 3, 5, 7, 11, 13, 17, 19\} { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } C { 3 , 5 , 7 , 11 , 17 , 19 } \{3, 5, 7, 11, 17, 19\} { 3 , 5 , 7 , 11 , 17 , 19 } D { 3 , 5 , 11 , 13 , 17 , 19 } \{3, 5, 11, 13, 17, 19\} { 3 , 5 , 11 , 13 , 17 , 19 }
Worked solution (try it first) P P P is the odd numbers from 1 to 19:
{ 1 , 3 , 5 , … , 19 } \{1, 3, 5, \dots, 19\} { 1 , 3 , 5 , … , 19 } .
Q Q Q is the primes up to 25:
{ 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } \{2, 3, 5, 7, 11, 13, 17, 19, 23\} { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } .
The common elements are the odd primes up to 19:
{ 3 , 5 , 7 , 11 , 13 , 17 , 19 } \{3, 5, 7, 11, 13, 17, 19\} { 3 , 5 , 7 , 11 , 13 , 17 , 19 } , option A.
Watch out
2 is prime but it is even, so it is not in P P P Report a problem with this question
If S = t 2 − 4 t + 4 S = \sqrt{t^2 - 4t + 4} S = t 2 − 4 t + 4 t t t S S S
A S − 2 S - 2 S − 2 B S 2 + 2 S^2 + 2 S 2 + 2 C S 2 − 2 S^2 - 2 S 2 − 2 D S + 2 S + 2 S + 2
Worked solution (try it first) The expression under the root is a perfect square:
t 2 − 4 t + 4 = ( t − 2 ) 2 t^2 - 4t + 4 = (t - 2)^2 t 2 − 4 t + 4 = ( t − 2 ) 2 .
So
S = ( t − 2 ) 2 = t − 2 S = \sqrt{(t - 2)^2} = t - 2 S = ( t − 2 ) 2 = t − 2 .
Add 2 to both sides:
t = S + 2 t = S + 2 t = S + 2 , option D.
Watch out
Squaring both sides gives S 2 = ( t − 2 ) 2 S^2 = (t - 2)^2 S 2 = ( t − 2 ) 2 S 2 = t − 2 S^2 = t - 2 S 2 = t − 2 t = S 2 + 2 t = S^2 + 2 t = S 2 + 2 Report a problem with this question
If x − 4 x - 4 x − 4 x 2 − x − k x^2 - x - k x 2 − x − k k k k
Worked solution (try it first) By the factor theorem,
x − 4 x - 4 x − 4 is a factor, so the expression is 0 at
x = 4 x = 4 x = 4 .
16 − 4 − k = 0 16 - 4 - k = 0 16 − 4 − k = 0 , so
k = 12 k = 12 k = 12 , option D.
Watch out
x − 4 x - 4 x − 4 x = + 4 x = +4 x = + 4 x = − 4 x = -4 x = − 4 16 + 4 − k = 0 16 + 4 - k = 0 16 + 4 − k = 0 k = 20 k = 20 k = 20 Report a problem with this question
The remainder when 6 p 3 − p 2 − 47 p + 30 6p^3 - p^2 - 47p + 30 6 p 3 − p 2 − 47 p + 30 p − 3 p - 3 p − 3
Worked solution (try it first) By the remainder theorem, the remainder on dividing by
p − 3 p - 3 p − 3 is the value at
p = 3 p = 3 p = 3 .
6 ( 27 ) − 9 − 47 ( 3 ) + 30 6(27) - 9 - 47(3) + 30 6 ( 27 ) − 9 − 47 ( 3 ) + 30 , which is
162 − 9 − 141 + 30 162 - 9 - 141 + 30 162 − 9 − 141 + 30 .
So the remainder is 42, option D.
Watch out
p − 3 p - 3 p − 3 p = + 3 p = +3 p = + 3 p = − 3 p = -3 p = − 3 − 162 − 9 + 141 + 30 = 0 -162 - 9 + 141 + 30 = 0 − 162 − 9 + 141 + 30 = 0 Report a problem with this question
P P P m m m u u u q q q P = 4 P = 4 P = 4 m = 3 m = 3 m = 3 u = 2 u = 2 u = 2 q = 1 q = 1 q = 1 P P P m = 6 m = 6 m = 6 u = 4 u = 4 u = 4 q = 8 5 q = \frac85 q = 5 8
A 10 B 288 5 \frac{288}{5} 5 288 C 128 5 \frac{128}{5} 5 128 D 15
Worked solution (try it first) Jointly as
m m m and
u u u (on top), inversely as
q q q (underneath):
P = k m u q P = \dfrac{kmu}{q} P = q k m u .
P = 4 P = 4 P = 4 ,
m = 3 m = 3 m = 3 ,
u = 2 u = 2 u = 2 ,
q = 1 q = 1 q = 1 :
4 = 6 k 4 = 6k 4 = 6 k , so
k = 2 3 k = \frac23 k = 3 2 .
Now
m u = 24 mu = 24 m u = 24 , so
P = 2 3 × 24 ÷ 8 5 P = \frac23 \times 24 \div \frac85 P = 3 2 × 24 ÷ 5 8 , which is
16 × 5 8 = 10 16 \times \frac58 = 10 16 × 8 5 = 10 , option A.
Watch out
Divide by q q q 8 5 \frac85 5 8 5 8 \frac58 8 5 8 5 \frac85 5 8 128 5 \frac{128}{5} 5 128 Report a problem with this question
If r r r s s s t t t s s s r r r t t t
A s s s r 2 r^2 r 2 t 2 t^2 t 2 B s s s r r r t t t C s s s r r r t 2 t^2 t 2 D s s s r 2 r^2 r 2 t t t
Worked solution (try it first) r = k s t r = \dfrac{k}{\sqrt{st}} r = s t k .
Square both sides:
r 2 = k 2 s t r^2 = \dfrac{k^2}{st} r 2 = s t k 2 .
Make
s s s the subject:
s = k 2 r 2 t s = \dfrac{k^2}{r^2t} s = r 2 t k 2 .
So
s s s varies inversely as
r 2 r^2 r 2 and
t t t , option D.
Watch out
Squaring removes the root from s s s t t t r r r t t t Report a problem with this question
Solve 3 ( x + 2 ) > 6 ( x + 3 ) 3(x + 2) > 6(x + 3) 3 ( x + 2 ) > 6 ( x + 3 )
A x < − 4 x < -4 x < − 4 B x > 4 x > 4 x > 4 C x < 4 x < 4 x < 4 D x > − 4 x > -4 x > − 4
Worked solution (try it first) Expand both brackets:
3 x + 6 > 6 x + 18 3x + 6 > 6x + 18 3 x + 6 > 6 x + 18 .
Subtract
6 x 6x 6 x and 6 from both sides:
− 3 x > 12 -3x > 12 − 3 x > 12 .
Divide by
− 3 -3 − 3 and reverse the sign:
x < − 4 x < -4 x < − 4 , option A.
Watch out
Dividing by − 3 -3 − 3 > > > x > − 4 x > -4 x > − 4 x = 0 x = 0 x = 0 6 > 18 6 > 18 6 > 18 Report a problem with this question
A curve crosses the x x x − 1 -1 − 1 2 2 2 y y y − 2 -2 − 2
A y = x 2 − x − 1 y = x^2 - x - 1 y = x 2 − x − 1 B y = x 2 + x − 2 y = x^2 + x - 2 y = x 2 + x − 2 C y = x 2 − x − 2 y = x^2 - x - 2 y = x 2 − x − 2 D y = x 2 − 3 x + 2 y = x^2 - 3x + 2 y = x 2 − 3 x + 2
Try it on a graph The graph crosses the x-axis at −1 and 2 and the y-axis at −2.
Open the interactive graph Worked solution (try it first) The curve crosses the
x x x -axis at
− 1 -1 − 1 and 2, so
y = k ( x + 1 ) ( x − 2 ) y = k(x + 1)(x - 2) y = k ( x + 1 ) ( x − 2 ) .
It crosses the
y y y -axis at
− 2 -2 − 2 : at
x = 0 x = 0 x = 0 ,
k ( 1 ) ( − 2 ) = − 2 k(1)(-2) = -2 k ( 1 ) ( − 2 ) = − 2 , so
k = 1 k = 1 k = 1 .
So
y = ( x + 1 ) ( x − 2 ) = x 2 − x − 2 y = (x + 1)(x - 2) = x^2 - x - 2 y = ( x + 1 ) ( x − 2 ) = x 2 − x − 2 , option C.
Watch out
Option B, x 2 + x − 2 = ( x + 2 ) ( x − 1 ) x^2 + x - 2 = (x + 2)(x - 1) x 2 + x − 2 = ( x + 2 ) ( x − 1 ) − 2 -2 − 2 Report a problem with this question
Solve for x x x ∣ x − 2 ∣ < 3 |x - 2| < 3 ∣ x − 2∣ < 3
A − 1 < x < 5 -1 < x < 5 − 1 < x < 5 B x < 1 x < 1 x < 1 C x < 5 x < 5 x < 5 D − 1 < x < 3 -1 < x < 3 − 1 < x < 3
Worked solution (try it first) ∣ x − 2 ∣ < 3 |x - 2| < 3 ∣ x − 2∣ < 3 means
x − 2 x - 2 x − 2 is less than 3 away from 0:
− 3 < x − 2 < 3 -3 < x - 2 < 3 − 3 < x − 2 < 3 .
Add 2 to all three parts:
− 1 < x < 5 -1 < x < 5 − 1 < x < 5 , option A.
Watch out
The modulus gives two limits, not one. Option C, x < 5 x < 5 x < 5 x = − 4 x = -4 x = − 4 ∣ x − 2 ∣ = 6 |x - 2| = 6 ∣ x − 2∣ = 6 Report a problem with this question
If the sum of the first two terms of a G.P. is 3, and the sum of the second and third terms is − 6 -6 − 6
Worked solution (try it first) Write the sums in terms of
a a a and
r r r :
a + a r = a ( 1 + r ) = 3 a + ar = a(1 + r) = 3 a + a r = a ( 1 + r ) = 3 and
a r + a r 2 = a r ( 1 + r ) = − 6 ar + ar^2 = ar(1 + r) = -6 a r + a r 2 = a r ( 1 + r ) = − 6 .
Divide the second by the first:
r = − 6 ÷ 3 = − 2 r = -6 \div 3 = -2 r = − 6 ÷ 3 = − 2 .
Then
a ( 1 − 2 ) = 3 a(1 - 2) = 3 a ( 1 − 2 ) = 3 , so
a = − 3 a = -3 a = − 3 , and
a + r = − 3 + ( − 2 ) = − 5 a + r = -3 + (-2) = -5 a + r = − 3 + ( − 2 ) = − 5 , option A.
Watch out
The question asks for a + r a + r a + r − 3 -3 − 3 a a a − 2 -2 − 2 r r r Report a problem with this question
The n n n 4 2 , 7 3 , 10 4 , 13 5 , … \frac42, \frac73, \frac{10}{4}, \frac{13}{5}, \ldots 2 4 , 3 7 , 4 10 , 5 13 , …
A 3 n + 1 n − 1 \dfrac{3n + 1}{n - 1} n − 1 3 n + 1 B 3 n − 1 n + 1 \dfrac{3n - 1}{n + 1} n + 1 3 n − 1 C 1 − 3 n n + 1 \dfrac{1 - 3n}{n + 1} n + 1 1 − 3 n D 3 n + 1 n + 1 \dfrac{3n + 1}{n + 1} n + 1 3 n + 1
Worked solution (try it first) Look at the tops and bottoms separately.
The tops
4 , 7 , 10 , 13 4, 7, 10, 13 4 , 7 , 10 , 13 go up by 3, starting at 4, so they are
3 n + 1 3n + 1 3 n + 1 .
The bottoms
2 , 3 , 4 , 5 2, 3, 4, 5 2 , 3 , 4 , 5 are each one more than the term number, so they are
n + 1 n + 1 n + 1 .
So the
n n n th term is
3 n + 1 n + 1 \dfrac{3n + 1}{n + 1} n + 1 3 n + 1 , option D.
Watch out
Check n = 1 n = 1 n = 1 n − 1 = 0 n - 1 = 0 n − 1 = 0 4 2 \frac42 2 4 n + 1 n + 1 n + 1 Report a problem with this question
If a binary operation ∗ * ∗ x ∗ y = x + 2 y x * y = x + 2y x ∗ y = x + 2 y 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) 2 ∗ ( 3 ∗ 4 )
Worked solution (try it first) Work out the bracket first, doubling the second number:
3 ∗ 4 = 3 + 2 × 4 = 11 3 * 4 = 3 + 2 \times 4 = 11 3 ∗ 4 = 3 + 2 × 4 = 11 .
Then
2 ∗ 11 = 2 + 2 × 11 = 24 2 * 11 = 2 + 2 \times 11 = 24 2 ∗ 11 = 2 + 2 × 11 = 24 , option C.
Watch out
It is the second number that is doubled. Doubling the first gives 3 ∗ 4 = 10 3 * 4 = 10 3 ∗ 4 = 10 2 ∗ 10 = 14 2 * 10 = 14 2 ∗ 10 = 14 Report a problem with this question
If P = ( 5 3 2 1 ) P = \begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix} P = ( 5 2 3 1 ) Q = ( 4 2 3 5 ) Q = \begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix} Q = ( 4 3 2 5 ) 2 P + Q 2P + Q 2 P + Q
A ( 7 7 8 14 ) \begin{pmatrix} 7 & 7 \\ 8 & 14 \end{pmatrix} ( 7 8 7 14 ) B ( 8 14 7 7 ) \begin{pmatrix} 8 & 14 \\ 7 & 7 \end{pmatrix} ( 8 7 14 7 ) C ( 7 7 14 8 ) \begin{pmatrix} 7 & 7 \\ 14 & 8 \end{pmatrix} ( 7 14 7 8 ) D ( 14 8 7 7 ) \begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix} ( 14 7 8 7 )
Worked solution (try it first) Double every entry of
P P P :
2 P = ( 10 6 4 2 ) 2P = \begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix} 2 P = ( 10 4 6 2 ) .
Add
Q Q Q entry by entry, in matching positions:
( 10 + 4 6 + 2 4 + 3 2 + 5 ) \begin{pmatrix} 10 + 4 & 6 + 2 \\ 4 + 3 & 2 + 5 \end{pmatrix} ( 10 + 4 4 + 3 6 + 2 2 + 5 ) .
So
2 P + Q = ( 14 8 7 7 ) 2P + Q = \begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix} 2 P + Q = ( 14 7 8 7 ) , option D.
Watch out
Each entry of the answer comes from the same position in 2 P 2P 2 P Q Q Q 10 + 4 = 14 10 + 4 = 14 10 + 4 = 14 Report a problem with this question
Find the inverse of ( 5 3 6 4 ) \begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix} ( 5 6 3 4 )
A ( 2 3 2 − 3 − 5 2 ) \begin{pmatrix} 2 & \frac32 \\ -3 & -\frac52 \end{pmatrix} ( 2 − 3 2 3 − 2 5 ) B ( 2 3 2 − 3 5 2 ) \begin{pmatrix} 2 & \frac32 \\ -3 & \frac52 \end{pmatrix} ( 2 − 3 2 3 2 5 ) C ( 2 − 3 2 − 3 − 5 2 ) \begin{pmatrix} 2 & -\frac32 \\ -3 & -\frac52 \end{pmatrix} ( 2 − 3 − 2 3 − 2 5 ) D ( 2 − 3 2 − 3 5 2 ) \begin{pmatrix} 2 & -\frac32 \\ -3 & \frac52 \end{pmatrix} ( 2 − 3 − 2 3 2 5 )
Worked solution (try it first) The determinant is
5 × 4 − 3 × 6 = 20 − 18 = 2 5 \times 4 - 3 \times 6 = 20 - 18 = 2 5 × 4 − 3 × 6 = 20 − 18 = 2 .
Swap the two diagonal entries (5 and 4) and change the signs of the other two:
( 4 − 3 − 6 5 ) \begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix} ( 4 − 6 − 3 5 ) .
Divide every entry by 2:
( 2 − 3 2 − 3 5 2 ) \begin{pmatrix} 2 & -\frac32 \\ -3 & \frac52 \end{pmatrix} ( 2 − 3 − 2 3 2 5 ) , option D.
Watch out
Change the signs of both off-diagonal entries: 3 becomes − 3 -3 − 3 − 6 -6 − 6 Report a problem with this question
In the diagram, P Q ∥ T S PQ \parallel TS P Q ∥ T S ∠ P Q R = 110 ∘ \angle PQR = 110^\circ ∠ P QR = 11 0 ∘ ∠ T S R = 120 ∘ \angle TSR = 120^\circ ∠ T S R = 12 0 ∘ x x x
A 70 ∘ 70^\circ 7 0 ∘ B 130 ∘ 130^\circ 13 0 ∘ C 110 ∘ 110^\circ 11 0 ∘ D 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) Draw a line through
R R R parallel to
P Q PQ P Q and
T S TS T S .
∠ P Q R \angle PQR ∠ P QR and the angle between
R Q RQ R Q and that line are co-interior, so the upper part of
x x x is
180 ∘ − 110 ∘ = 70 ∘ 180^\circ - 110^\circ = 70^\circ 18 0 ∘ − 11 0 ∘ = 7 0 ∘ .
∠ T S R \angle TSR ∠ T S R and the angle between
R S RS R S and that line are co-interior, so the lower part of
x x x is
180 ∘ − 120 ∘ = 60 ∘ 180^\circ - 120^\circ = 60^\circ 18 0 ∘ − 12 0 ∘ = 6 0 ∘ .
So
x = 70 ∘ + 60 ∘ = 130 ∘ x = 70^\circ + 60^\circ = 130^\circ x = 7 0 ∘ + 6 0 ∘ = 13 0 ∘ , option B.
Watch out
70 ∘ 70^\circ 7 0 ∘ x x x 60 ∘ 60^\circ 6 0 ∘ Report a problem with this question
If the angles of a quadrilateral are ( 3 y + 10 ) ∘ (3y + 10)^\circ ( 3 y + 10 ) ∘ ( 2 y + 30 ) ∘ (2y + 30)^\circ ( 2 y + 30 ) ∘ ( y + 20 ) ∘ (y + 20)^\circ ( y + 20 ) ∘ 4 y ∘ 4y^\circ 4 y ∘ y y y
A 66 ∘ 66^\circ 6 6 ∘ B 12 ∘ 12^\circ 1 2 ∘ C 30 ∘ 30^\circ 3 0 ∘ D 42 ∘ 42^\circ 4 2 ∘
Worked solution (try it first) The angles of a quadrilateral add up to
360 ∘ 360^\circ 36 0 ∘ :
( 3 y + 10 ) + ( 2 y + 30 ) + ( y + 20 ) + 4 y = 360 (3y + 10) + (2y + 30) + (y + 20) + 4y = 360 ( 3 y + 10 ) + ( 2 y + 30 ) + ( y + 20 ) + 4 y = 360 .
Collect terms:
10 y + 60 = 360 10y + 60 = 360 10 y + 60 = 360 , so
10 y = 300 10y = 300 10 y = 300 .
Divide by 10:
y = 30 y = 30 y = 30 , option C.
Watch out
A quadrilateral's angles add up to 360 ∘ 360^\circ 36 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 10 y = 120 10y = 120 10 y = 120 y = 12 y = 12 y = 12 Report a problem with this question
A square tile has side 30 cm 30\text{ cm} 30 cm 7.2 m 7.2\text{ m} 7.2 m 4.2 m 4.2\text{ m} 4.2 m
Worked solution (try it first) Work in centimetres: the floor is 720 cm by 420 cm.
Along the length,
720 ÷ 30 = 24 720 \div 30 = 24 720 ÷ 30 = 24 tiles fit.
Along the width,
420 ÷ 30 = 14 420 \div 30 = 14 420 ÷ 30 = 14 tiles.
So
24 × 14 = 336 24 \times 14 = 336 24 × 14 = 336 tiles, option B.
Watch out
Put everything in the same units first. The floor is 30.24 m 2 30.24\text{ m}^2 30.24 m 2 0.3 2 = 0.09 m 2 0.3^2 = 0.09\text{ m}^2 0. 3 2 = 0.09 m 2 0.3 m 2 0.3\text{ m}^2 0.3 m 2 Also set as JAMB 1989 · UME · Q43
Report a problem with this question
Find the length of a chord which subtends an angle of 90 ∘ 90^\circ 9 0 ∘ 8 cm 8\text{ cm} 8 cm
A 8 3 cm 8\sqrt3\text{ cm} 8 3 cm B 4 cm 4\text{ cm} 4 cm C 8 cm 8\text{ cm} 8 cm D 8 2 cm 8\sqrt2\text{ cm} 8 2 cm
Worked solution (try it first) The chord and the two radii make a triangle with a right angle at the centre and two sides of 8 cm.
Pythagoras: the chord is
8 2 + 8 2 = 128 \sqrt{8^2 + 8^2} = \sqrt{128} 8 2 + 8 2 = 128 .
128 = 64 × 2 = 8 2 \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt2 128 = 64 × 2 = 8 2 cm, option D.
Watch out
The chord equals the radius, 8 cm (option C), only when the angle at the centre is 60 ∘ 60^\circ 6 0 ∘ 90 ∘ 90^\circ 9 0 ∘ 8 2 8\sqrt2 8 2 Report a problem with this question
A chord of a circle subtends an angle of 120 ∘ 120^\circ 12 0 ∘ 4 3 cm 4\sqrt3\text{ cm} 4 3 cm
A 32 π cm 2 32\pi\text{ cm}^2 32 π cm 2 B 4 π cm 2 4\pi\text{ cm}^2 4 π cm 2 C 8 π cm 2 8\pi\text{ cm}^2 8 π cm 2 D 16 π cm 2 16\pi\text{ cm}^2 16 π cm 2
Worked solution (try it first) The radius is half of
4 3 4\sqrt3 4 3 , so
r = 2 3 r = 2\sqrt3 r = 2 3 cm and
r 2 = 12 r^2 = 12 r 2 = 12 .
The major sector has angle
360 ∘ − 120 ∘ = 240 ∘ 360^\circ - 120^\circ = 240^\circ 36 0 ∘ − 12 0 ∘ = 24 0 ∘ , which is
2 3 \frac23 3 2 of the circle.
Area
= 2 3 × 12 π = 8 π cm 2 = \frac23 \times 12\pi = 8\pi\text{ cm}^2 = 3 2 × 12 π = 8 π cm 2 , option C.
Watch out
The major sector is the larger part, 240 ∘ 240^\circ 24 0 ∘ 120 ∘ 120^\circ 12 0 ∘ 4 π cm 2 4\pi\text{ cm}^2 4 π cm 2 Also set as JAMB 2002 · UME · Q12
Report a problem with this question
The locus of the points which are equidistant from the line P Q PQ P Q
A perpendicular line to P Q PQ P Q B circle centre P P P C circle centre Q Q Q D pair of parallel lines to P Q PQ P Q
Worked solution (try it first) Points at a fixed distance from a straight line lie on two lines, one on each side of it.
Both are parallel to
P Q PQ P Q , so the locus is a pair of lines parallel to
P Q PQ P Q , option D.
Watch out
A fixed distance from a line gives parallel lines. A circle (options B and C) is what you get for a fixed distance from a point. Report a problem with this question
If the midpoint of the line P Q PQ P Q ( 2 , 3 ) (2, 3) ( 2 , 3 ) P P P ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) Q Q Q
A ( 8 , 6 ) (8, 6) ( 8 , 6 ) B ( 5 , 6 ) (5, 6) ( 5 , 6 ) C ( 0 , 4 ) (0, 4) ( 0 , 4 ) D ( 6 , 5 ) (6, 5) ( 6 , 5 )
Worked solution (try it first) The midpoint is the average of the ends, so each coordinate of
Q Q Q is twice the midpoint's minus
P P P 's.
x x x :
2 ( 2 ) − ( − 2 ) = 6 2(2) - (-2) = 6 2 ( 2 ) − ( − 2 ) = 6 .
y y y :
2 ( 3 ) − 1 = 5 2(3) - 1 = 5 2 ( 3 ) − 1 = 5 .
So
Q = ( 6 , 5 ) Q = (6, 5) Q = ( 6 , 5 ) , option D.
Watch out
Double the midpoint, then subtract P P P P P P ( 0 , 4 ) (0, 4) ( 0 , 4 ) Report a problem with this question
Find the equation of the perpendicular bisector of the line joining P ( 2 , 3 ) P(2, 3) P ( 2 , 3 ) Q ( − 5 , 1 ) Q(-5, 1) Q ( − 5 , 1 )
A 4 y + 14 x + 13 = 0 4y + 14x + 13 = 0 4 y + 14 x + 13 = 0 B 4 y − 14 x + 13 = 0 4y - 14x + 13 = 0 4 y − 14 x + 13 = 0 C 4 y − 14 x − 13 = 0 4y - 14x - 13 = 0 4 y − 14 x − 13 = 0 D 4 y + 14 x − 13 = 0 4y + 14x - 13 = 0 4 y + 14 x − 13 = 0
Worked solution (try it first) Midpoint of
P Q PQ P Q :
( 2 + ( − 5 ) 2 , 3 + 1 2 ) = ( − 3 2 , 2 ) \left(\frac{2 + (-5)}{2}, \frac{3 + 1}{2}\right) = \left(-\frac32, 2\right) ( 2 2 + ( − 5 ) , 2 3 + 1 ) = ( − 2 3 , 2 ) .
Gradient of
P Q PQ P Q :
1 − 3 − 5 − 2 = − 2 − 7 \frac{1 - 3}{-5 - 2} = \frac{-2}{-7} − 5 − 2 1 − 3 = − 7 − 2 The perpendicular gradient is its negative reciprocal,
− 7 2 -\frac72 − 2 7 .
Line through the midpoint:
y − 2 = − 7 2 ( x + 3 2 ) y - 2 = -\frac72\left(x + \frac32\right) y − 2 = − 2 7 ( x + 2 3 ) .
Multiply by 4:
4 y − 8 = − 14 x − 21 4y - 8 = -14x - 21 4 y − 8 = − 14 x − 21 .
Bring everything to the left:
4 y + 14 x + 13 = 0 4y + 14x + 13 = 0 4 y + 14 x + 13 = 0 , option A.
Watch out
When you move terms across, change their signs: 4 y + 14 x = − 13 4y + 14x = -13 4 y + 14 x = − 13 4 y + 14 x + 13 = 0 4y + 14x + 13 = 0 4 y + 14 x + 13 = 0 − 13 -13 − 13 Report a problem with this question
In triangle P Q R PQR P QR q = 8 cm q = 8\text{ cm} q = 8 cm r = 6 cm r = 6\text{ cm} r = 6 cm cos P = 1 12 \cos P = \frac{1}{12} cos P = 12 1 p p p
A 108 cm \sqrt{108}\text{ cm} 108 cm B 9 cm \sqrt9\text{ cm} 9 cm C 92 cm \sqrt{92}\text{ cm} 92 cm D 10 cm 10\text{ cm} 10 cm
Worked solution (try it first) p p p faces angle
P P P , which lies between sides
q q q and
r r r .
Cosine rule:
p 2 = q 2 + r 2 − 2 q r cos P p^2 = q^2 + r^2 - 2qr\cos P p 2 = q 2 + r 2 − 2 q r cos P .
Put in the numbers:
p 2 = 64 + 36 − 2 ( 8 ) ( 6 ) × 1 12 p^2 = 64 + 36 - 2(8)(6) \times \frac{1}{12} p 2 = 64 + 36 − 2 ( 8 ) ( 6 ) × 12 1 .
The last term is
96 ÷ 12 = 8 96 \div 12 = 8 96 ÷ 12 = 8 .
So
p 2 = 100 − 8 = 92 p^2 = 100 - 8 = 92 p 2 = 100 − 8 = 92 and
p = 92 p = \sqrt{92} p = 92 cm, option C.
Watch out
The cosine rule has a minus: p 2 = 100 − 8 p^2 = 100 - 8 p 2 = 100 − 8 108 \sqrt{108} 108 Report a problem with this question
If tan θ = 3 4 \tan\theta = \frac34 tan θ = 4 3 sin θ + cos θ \sin\theta + \cos\theta sin θ + cos θ
A 1 1 3 1\frac13 1 3 1 B 1 2 3 1\frac23 1 3 2 C 1 3 5 1\frac35 1 5 3 D 1 2 5 1\frac25 1 5 2
Worked solution (try it first) tan θ = 3 4 \tan\theta = \frac34 tan θ = 4 3 , so draw a right-angled triangle with opposite side 3 and adjacent side 4.
Pythagoras gives the hypotenuse
9 + 16 = 5 \sqrt{9 + 16} = 5 9 + 16 = 5 .
So
sin θ = 3 5 \sin\theta = \frac35 sin θ = 5 3 and
cos θ = 4 5 \cos\theta = \frac45 cos θ = 5 4 .
Add:
3 5 + 4 5 = 7 5 = 1 2 5 \frac35 + \frac45 = \frac75 = 1\frac25 5 3 + 5 4 = 5 7 = 1 5 2 , option D.
Watch out
Sine and cosine are both over the hypotenuse, 5. Using the sides 3 and 4 as bottoms, for example 3 4 + 4 4 \frac34 + \frac44 4 3 + 4 4 1 + 1 3 1 + \frac13 1 + 3 1 1 1 3 1\frac13 1 3 1 Report a problem with this question
If y = ( 2 x + 2 ) 3 y = (2x + 2)^3 y = ( 2 x + 2 ) 3 d y d x \dfrac{dy}{dx} d x d y
A 3 ( 2 x + 2 ) 3(2x + 2) 3 ( 2 x + 2 ) B 6 ( 2 x + 2 ) 2 6(2x + 2)^2 6 ( 2 x + 2 ) 2 C 3 ( 2 x + 2 ) 2 3(2x + 2)^2 3 ( 2 x + 2 ) 2 D 6 ( 2 x + 2 ) 6(2x + 2) 6 ( 2 x + 2 )
Worked solution (try it first) Chain rule: bring down the power 3 and reduce it by one, giving
3 ( 2 x + 2 ) 2 3(2x + 2)^2 3 ( 2 x + 2 ) 2 .
Multiply by the derivative of
2 x + 2 2x + 2 2 x + 2 , which is 2:
d y d x = 6 ( 2 x + 2 ) 2 \frac{dy}{dx} = 6(2x + 2)^2 d x d y = 6 ( 2 x + 2 ) 2 , option B.
Watch out
Multiply by the derivative of the bracket. Stopping at 3 ( 2 x + 2 ) 2 3(2x + 2)^2 3 ( 2 x + 2 ) 2 Report a problem with this question
If y = x sin x y = x\sin x y = x sin x d y d x \dfrac{dy}{dx} d x d y
A cos x + x sin x \cos x + x\sin x cos x + x sin x B sin x + x cos x \sin x + x\cos x sin x + x cos x C sin x − cos x \sin x - \cos x sin x − cos x D cos x − x sin x \cos x - x\sin x cos x − x sin x
Worked solution (try it first) Product rule with
u = x u = x u = x and
v = sin x v = \sin x v = sin x :
u ′ = 1 u' = 1 u ′ = 1 and
v ′ = cos x v' = \cos x v ′ = cos x .
d y d x = u ′ v + u v ′ \frac{dy}{dx} = u'v + uv' d x d y = u ′ v + u v ′ = sin x + x cos x = \sin x + x\cos x = sin x + x cos x , option B.
Watch out
Each derivative goes with the other factor: 1 × sin x 1 \times \sin x 1 × sin x x × cos x x \times \cos x x × cos x cos x + x sin x \cos x + x\sin x cos x + x sin x Similar: JAMB 1992 · UME · Q37 · Also set as JAMB 2010 · UTME · Q41
Report a problem with this question
The radius of a circle is increasing at the rate of 0.02 cm s − 1 0.02\text{ cm s}^{-1} 0.02 cm s − 1 7 cm 7\text{ cm} 7 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 0.35 cm 2 s − 1 0.35\text{ cm}^2\text{s}^{-1} 0.35 cm 2 s − 1 B 0.88 cm 2 s − 1 0.88\text{ cm}^2\text{s}^{-1} 0.88 cm 2 s − 1 C 0.75 cm 2 s − 1 0.75\text{ cm}^2\text{s}^{-1} 0.75 cm 2 s − 1 D 0.55 cm 2 s − 1 0.55\text{ cm}^2\text{s}^{-1} 0.55 cm 2 s − 1
Worked solution (try it first) The area is
A = π r 2 A = \pi r^2 A = π r 2 , so
d A d r = 2 π r \frac{dA}{dr} = 2\pi r d r d A = 2 π r .
Chain rule:
d A d t = 2 π r × d r d t \frac{dA}{dt} = 2\pi r \times \frac{dr}{dt} d t d A = 2 π r × d t d r = 2 × 22 7 × 7 × 0.02 = 2 \times \frac{22}{7} \times 7 \times 0.02 = 2 × 7 22 × 7 × 0.02 .
2 × 22 × 0.02 = 0.88 2 \times 22 \times 0.02 = 0.88 2 × 22 × 0.02 = 0.88 , so the area increases at
0.88 cm 2 s − 1 0.88\text{ cm}^2\text{s}^{-1} 0.88 cm 2 s − 1 , option B.
Watch out
d A d r \frac{dA}{dr} d r d A 2 π r 2\pi r 2 π r π r \pi r π r 0.44 0.44 0.44 Report a problem with this question
Integrate 1 + x x 3 d x \dfrac{1 + x}{x^3}\,dx x 3 1 + x d x
A 2 x 2 − 1 x + k 2x^2 - \frac1x + k 2 x 2 − x 1 + k B − 1 2 x 2 − 1 x + k -\frac{1}{2x^2} - \frac1x + k − 2 x 2 1 − x 1 + k C − x 2 2 − 1 x + k -\frac{x^2}{2} - \frac1x + k − 2 x 2 − x 1 + k D x 2 − 1 x + k x^2 - \frac1x + k x 2 − x 1 + k
Worked solution (try it first) Split the fraction into powers of
x x x :
1 + x x 3 = x − 3 + x − 2 \dfrac{1 + x}{x^3} = x^{-3} + x^{-2} x 3 1 + x = x − 3 + x − 2 .
Add one to each power and divide by the new power:
x − 3 x^{-3} x − 3 gives
x − 2 − 2 = − 1 2 x 2 \frac{x^{-2}}{-2} = -\frac{1}{2x^2} − 2 x − 2 = − 2 x 2 1 , and
x − 2 x^{-2} x − 2 gives
x − 1 − 1 = − 1 x \frac{x^{-1}}{-1} = -\frac1x − 1 x − 1 = − x 1 .
So the integral is
− 1 2 x 2 − 1 x + k -\frac{1}{2x^2} - \frac1x + k − 2 x 2 1 − x 1 + k , option B.
Watch out
Adding one to − 3 -3 − 3 − 2 -2 − 2 x − 2 x^{-2} x − 2 x 2 x^2 x 2 − x 2 2 -\frac{x^2}{2} − 2 x 2 Report a problem with this question
Evaluate ∫ 0 π / 2 sin x d x \displaystyle\int_0^{\pi/2} \sin x\,dx ∫ 0 π /2 sin x d x
Worked solution (try it first) sin x \sin x sin x integrates to
− cos x -\cos x − cos x .
[ − cos x ] 0 π / 2 = − cos π 2 − ( − cos 0 ) \left[-\cos x\right]_0^{\pi/2} = -\cos\frac\pi2 - (-\cos0) [ − cos x ] 0 π /2 = − cos 2 π − ( − cos 0 ) Watch out
The integral of sin x \sin x sin x − cos x -\cos x − cos x cos x \cos x cos x − 1 -1 − 1 sin x ≥ 0 \sin x \ge 0 sin x ≥ 0 Report a problem with this question
Find the mean of t + 2 t + 2 t + 2 2 t − 4 2t - 4 2 t − 4 3 t + 2 3t + 2 3 t + 2 2 t 2t 2 t
A 2 t + 1 2t + 1 2 t + 1 B t t t C t + 1 t + 1 t + 1 D 2 t 2t 2 t
Worked solution (try it first) Add the four expressions: the
t t t terms give
t + 2 t + 3 t + 2 t = 8 t t + 2t + 3t + 2t = 8t t + 2 t + 3 t + 2 t = 8 t and the numbers give
2 − 4 + 2 = 0 2 - 4 + 2 = 0 2 − 4 + 2 = 0 .
So the sum is
8 t 8t 8 t , and the mean is
8 t 4 = 2 t \frac{8t}{4} = 2t 4 8 t = 2 t , option D.
Watch out
Keep the sign of − 4 -4 − 4 2 − 4 + 2 = 0 2 - 4 + 2 = 0 2 − 4 + 2 = 0 + 4 +4 + 4 8 t + 8 8t + 8 8 t + 8 2 t + 2 2t + 2 2 t + 2 Report a problem with this question
The mean of seven numbers is 10. If six of the numbers are 2, 4, 8, 14, 16 and 18, find the mode.
Worked solution (try it first) Total of all seven numbers:
7 × 10 = 70 7 \times 10 = 70 7 × 10 = 70 .
The six given add up to
2 + 4 + 8 + 14 + 16 + 18 = 62 2 + 4 + 8 + 14 + 16 + 18 = 62 2 + 4 + 8 + 14 + 16 + 18 = 62 , so the seventh is
70 − 62 = 8 70 - 62 = 8 70 − 62 = 8 .
Now 8 appears twice and every other number once, so the mode is 8, option D.
Watch out
Find the missing number first: with only the six given numbers, every value appears once and there is no mode. Report a problem with this question
Age
20
25
30
35
40
45
No. of people
3
5
1
1
2
5
Calculate the median age of the frequency distribution.
Worked solution (try it first) There are
3 + 5 + 1 + 1 + 2 + 5 = 17 3 + 5 + 1 + 1 + 2 + 5 = 17 3 + 5 + 1 + 1 + 2 + 5 = 17 people, so the median is the
17 + 1 2 = 9 \frac{17 + 1}{2} = 9 2 17 + 1 = 9 th age.
Running totals: 3 (age 20), 8 (age 25), 9 (age 30).
The 9th person is 30, so the median age is 30, option D.
Watch out
Only 8 people are aged 25 or under, so the 9th is in the next group. Stopping at 25 gives option C. Report a problem with this question
If the variance of 3 + x 3 + x 3 + x 6 6 6 4 4 4 x x x 7 − x 7 - x 7 − x
Worked solution (try it first) The standard deviation is the square root of the variance, so it is
4 = 2 \sqrt4 = 2 4 = 2 , option D.
As a check: the mean is
20 + x 5 = 5 \frac{20 + x}{5} = 5 5 20 + x = 5 , so
x = 5 x = 5 x = 5 .
The numbers are 8, 6, 4, 5, 2, whose squared deviations 9, 1, 1, 0, 9 add up to 20, and
20 5 = 4 \frac{20}{5} = 4 5 20 = 4 .
Watch out
You don't need x x x 4 = 2 \sqrt4 = 2 4 = 2 2 \sqrt2 2 Report a problem with this question
Score
3
4
5
6
7
8
9
10
Freq.
1
0
7
5
2
3
1
1
The table shows the scores of 20 students in a further mathematics test. What is the range of the distribution?
Worked solution (try it first) The range uses the scores, not the frequencies.
The highest score anyone got is 10 and the lowest is 3 (one student).
The range is
10 − 3 = 7 10 - 3 = 7 10 − 3 = 7 , option C.
Watch out
The score 3 counts, even though nobody scored 4. Starting from 4 gives 10 − 4 = 6 10 - 4 = 6 10 − 4 = 6 Report a problem with this question
In how many ways can a student select 2 subjects from 5 subjects?
A 5 ! 2 ! 3 ! \dfrac{5!}{2!3!} 2 ! 3 ! 5 ! B 5 ! 2 ! \dfrac{5!}{2!} 2 ! 5 ! C 5 ! 3 ! \dfrac{5!}{3!} 3 ! 5 ! D 5 ! 2 ! 2 ! \dfrac{5!}{2!2!} 2 ! 2 ! 5 !
Worked solution (try it first) The order of the two subjects doesn't matter, so this is
5 C 2 ^5C_2 5 C 2 .
n C r = n ! r ! ( n − r ) ! ^nC_r = \dfrac{n!}{r!\,(n - r)!} n C r = r ! ( n − r )! n ! , so
5 C 2 = 5 ! 2 ! 3 ! ^5C_2 = \dfrac{5!}{2!\,3!} 5 C 2 = 2 ! 3 ! 5 ! , option A.
Watch out
5 ! 3 ! \frac{5!}{3!} 3 ! 5 ! 5 P 2 ^5P_2 5 P 2 2 ! 2! 2 ! Report a problem with this question
In how many ways can 3 seats be occupied if 5 people are willing to sit?
Worked solution (try it first) The seats are different, so order matters: this is
5 P 3 ^5P_3 5 P 3 .
The first seat has 5 choices of person, the second 4 and the third 3.
So there are
5 × 4 × 3 = 60 5 \times 4 \times 3 = 60 5 × 4 × 3 = 60 ways, option C.
Watch out
Only 3 seats are filled, so stop after three factors. 5 ! = 120 5! = 120 5 ! = 120 Report a problem with this question
What is the probability that an integer x x x 1 ≤ x ≤ 25 1 \le x \le 25 1 ≤ x ≤ 25
A 4 25 \frac{4}{25} 25 4 B 3 4 \frac34 4 3 C 1 25 \frac{1}{25} 25 1 D 1 5 \frac15 5 1
Worked solution (try it first) Divisible by both 2 and 3 means divisible by 6.
The multiples of 6 up to 25 are 6, 12, 18, 24, which is 4 numbers.
So the probability is
4 25 \frac{4}{25} 25 4 , option A.
Watch out
"Both" means multiples of 6, not multiples of 2 or 3. Counting all multiples of 2 and of 3 answers a different question. Report a problem with this question
A basket contains 9 apples, 8 bananas and 7 oranges. A fruit is picked from the basket. Find the probability that it is neither an apple nor an orange.
A 7 24 \frac{7}{24} 24 7 B 2 3 \frac23 3 2 C 3 8 \frac38 8 3 D 1 3 \frac13 3 1
Worked solution (try it first) There are
9 + 8 + 7 = 24 9 + 8 + 7 = 24 9 + 8 + 7 = 24 fruits.
Neither an apple nor an orange means a banana: 8 fruits.
So the probability is
8 24 = 1 3 \frac{8}{24} = \frac13 24 8 = 3 1 , option D.
Watch out
16 24 = 2 3 \frac{16}{24} = \frac23 24 16 = 3 2 Report a problem with this question