Paper JAMB 1993 General Maths Objective
Objective paper · 46 questions · partial
JAMB 1993 · UME Topics include Number bases, Approximation & error, Indices & standard form, Logarithms, Commercial arithmetic, Sets & Venn diagrams.
Our copy of this paper is missing questions 5, 13, 16, 25.
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 6 7 8 9 10 11 12 14 15 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Change 71 10 71_{10} 7 1 10
A 107 8 107_8 10 7 8 B 106 8 106_8 10 6 8 C 71 8 71_8 7 1 8 D 17 8 17_8 1 7 8
Worked solution (try it first) Divide by 8:
71 ÷ 8 = 8 71 \div 8 = 8 71 ÷ 8 = 8 remainder 7.
Divide again:
8 ÷ 8 = 1 8 \div 8 = 1 8 ÷ 8 = 1 remainder 0, and
1 ÷ 8 = 0 1 \div 8 = 0 1 ÷ 8 = 0 remainder 1.
Read the remainders from the bottom up:
107 8 107_8 10 7 8 , option A.
Watch out
Keep dividing until the quotient is 0. Stopping at 71 = 8 × 8 + 7 71 = 8 \times 8 + 7 71 = 8 × 8 + 7 Report a problem with this question
Evaluate 3524 0.05 \frac{3524}{0.05} 0.05 3524
Worked solution (try it first) Dividing by 0.05 is the same as multiplying by 20:
3524 × 20 = 70 480 3524 \times 20 = 70\,480 3524 × 20 = 70 480 .
To 3 significant figures, keep 7, 0, 4 and look at the next figure, 8.
Round up to 705, and keep the zeros to hold the place value:
70 500 70\,500 70 500 , option D.
Watch out
Rounding must not change the size of the number: 70 480 becomes 70 500, not 705 (option A). Fill the dropped places with zeros. Report a problem with this question
If 9 x − 1 2 = 3 x 2 9^{x - \frac12} = 3^{x^2} 9 x − 2 1 = 3 x 2 x x x
Worked solution (try it first) Write 9 as
3 2 3^2 3 2 and multiply the indices:
9 x − 1 2 = 3 2 x − 1 9^{x - \frac12} = 3^{2x - 1} 9 x − 2 1 = 3 2 x − 1 .
The bases match, so the powers are equal:
2 x − 1 = x 2 2x - 1 = x^2 2 x − 1 = x 2 , or
x 2 − 2 x + 1 = 0 x^2 - 2x + 1 = 0 x 2 − 2 x + 1 = 0 .
This is
( x − 1 ) 2 = 0 (x - 1)^2 = 0 ( x − 1 ) 2 = 0 , so
x = 1 x = 1 x = 1 , option B.
Watch out
Multiply the whole index by 2: 2 ( x − 1 2 ) = 2 x − 1 2\left(x - \frac12\right) = 2x - 1 2 ( x − 2 1 ) = 2 x − 1 2 x − 1 2 2x - \frac12 2 x − 2 1 2 x − 1 2 2x - \frac12 2 x − 2 1 Report a problem with this question
Solve for y y y 10 y × 5 2 y − 2 × 4 y − 1 = 1 10^y \times 5^{2y - 2} \times 4^{y - 1} = 1 1 0 y × 5 2 y − 2 × 4 y − 1 = 1
A 3 4 \frac34 4 3 B 2 3 \frac23 3 2 C 1 1 1 D 5 4 \frac54 4 5
Worked solution (try it first) Write each factor with bases 2 and 5:
10 y = 2 y × 5 y 10^y = 2^y \times 5^y 1 0 y = 2 y × 5 y and
4 y − 1 = 2 2 y − 2 4^{y - 1} = 2^{2y - 2} 4 y − 1 = 2 2 y − 2 .
Collect the powers:
2 y + 2 y − 2 × 5 y + 2 y − 2 = 2 3 y − 2 × 5 3 y − 2 2^{y + 2y - 2} \times 5^{y + 2y - 2} = 2^{3y - 2} \times 5^{3y - 2} 2 y + 2 y − 2 × 5 y + 2 y − 2 = 2 3 y − 2 × 5 3 y − 2 , which is
10 3 y − 2 10^{3y - 2} 1 0 3 y − 2 .
10 3 y − 2 = 1 = 10 0 10^{3y - 2} = 1 = 10^0 1 0 3 y − 2 = 1 = 1 0 0 , so
3 y − 2 = 0 3y - 2 = 0 3 y − 2 = 0 and
y = 2 3 y = \frac23 y = 3 2 , option B.
Watch out
1 = 10 0 1 = 10^0 1 = 1 0 0 3 y − 2 = 1 3y - 2 = 1 3 y − 2 = 1 y = 1 y = 1 y = 1 Report a problem with this question
If 2 log 3 y + log 3 x 2 = 4 2\log_3 y + \log_3 x^2 = 4 2 log 3 y + log 3 x 2 = 4 y y y
A 4 − log 3 x 2 2 \frac{4 - \log_3 x^2}{2} 2 4 − l o g 3 x 2 B 4 log 3 x 2 \frac{4}{\log_3 x^2} l o g 3 x 2 4 C 2 x \frac{2}{x} x 2 D ± 9 x \pm\frac{9}{x} ± x 9
Worked solution (try it first) Move the 2 up as a power,
2 log 3 y = log 3 y 2 2\log_3 y = \log_3 y^2 2 log 3 y = log 3 y 2 , then combine the logs:
log 3 ( x 2 y 2 ) = 4 \log_3 (x^2y^2) = 4 log 3 ( x 2 y 2 ) = 4 .
Change to index form:
x 2 y 2 = 3 4 = 81 x^2y^2 = 3^4 = 81 x 2 y 2 = 3 4 = 81 .
Take square roots:
x y = ± 9 xy = \pm9 x y = ± 9 , so
y = ± 9 x y = \pm\frac9x y = ± x 9 , option D.
Watch out
log 3 N = 4 \log_3 N = 4 log 3 N = 4 N = 3 4 = 81 N = 3^4 = 81 N = 3 4 = 81 x y = ± 2 xy = \pm2 x y = ± 2 Report a problem with this question
Evaluate without using tables log 5 62.5 − log 5 1 2 \log_5 62.5 - \log_5 \frac12 log 5 62.5 − log 5 2 1
Worked solution (try it first) Subtracting logs divides:
log 5 62.5 − log 5 1 2 = log 5 ( 62.5 ÷ 0.5 ) \log_5 62.5 - \log_5 \frac12 = \log_5 (62.5 \div 0.5) log 5 62.5 − log 5 2 1 = log 5 ( 62.5 ÷ 0.5 ) .
62.5 ÷ 0.5 = 125 62.5 \div 0.5 = 125 62.5 ÷ 0.5 = 125 , and
125 = 5 3 125 = 5^3 125 = 5 3 .
So the value is
log 5 125 = 3 \log_5 125 = 3 log 5 125 = 3 , option A.
Watch out
Dividing by 1 2 \frac12 2 1 62.5 ÷ 0.5 = 125 62.5 \div 0.5 = 125 62.5 ÷ 0.5 = 125 1 2 \frac12 2 1 Report a problem with this question
If ₦225.00 yields ₦27.00 in x x x 4 % 4\% 4% x x x
Worked solution (try it first) One year's interest at
4 % 4\% 4% is
0.04 × 225 = 0.04 \times 225 = 0.04 × 225 = ₦9.
The total interest is ₦27, so the time is
27 ÷ 9 = 3 27 \div 9 = 3 27 ÷ 9 = 3 years.
So
x = 3 x = 3 x = 3 , option A.
Watch out
Use the rate. 27 ÷ 2.25 = 12 27 \div 2.25 = 12 27 ÷ 2.25 = 12 1 % 1\% 1% 4 % 4\% 4% Report a problem with this question
The shaded portion in the Venn diagram is
A X ∩ Z X \cap Z X ∩ Z B X ′ ∩ Y ∩ Z X' \cap Y \cap Z X ′ ∩ Y ∩ Z C X ∩ Y ′ ∩ Z X \cap Y' \cap Z X ∩ Y ′ ∩ Z D X ∩ Y ∩ Z ′ X \cap Y \cap Z' X ∩ Y ∩ Z ′
Worked solution (try it first) The shaded region lies inside circle
X X X and inside circle
Z Z Z , which gives
X ∩ Z X \cap Z X ∩ Z .
It lies outside circle
Y Y Y , which gives
Y ′ Y' Y ′ .
Together:
X ∩ Y ′ ∩ Z X \cap Y' \cap Z X ∩ Y ′ ∩ Z , option C.
Watch out
X ∩ Z X \cap Z X ∩ Z Y Y Y ∩ Y ′ \cap Y' ∩ Y ′ Report a problem with this question
If x 2 + 9 = x + 1 \sqrt{x^2 + 9} = x + 1 x 2 + 9 = x + 1 x x x
Worked solution (try it first) Square both sides:
x 2 + 9 = ( x + 1 ) 2 = x 2 + 2 x + 1 x^2 + 9 = (x + 1)^2 = x^2 + 2x + 1 x 2 + 9 = ( x + 1 ) 2 = x 2 + 2 x + 1 .
The
x 2 x^2 x 2 terms cancel:
9 = 2 x + 1 9 = 2x + 1 9 = 2 x + 1 , so
2 x = 8 2x = 8 2 x = 8 and
x = 4 x = 4 x = 4 .
Check:
16 + 9 = 5 \sqrt{16 + 9} = 5 16 + 9 = 5 and
4 + 1 = 5 4 + 1 = 5 4 + 1 = 5 .
So
x = 4 x = 4 x = 4 , option B.
Watch out
5 (option A) is the value of each side when x = 4 x = 4 x = 4 x x x Report a problem with this question
Make x x x 1 + a x 1 − a x = p q \dfrac{1 + ax}{1 - ax} = \dfrac pq 1 − a x 1 + a x = q p
A p + q a ( p − q ) \dfrac{p + q}{a(p - q)} a ( p − q ) p + q B p − q a ( p + q ) \dfrac{p - q}{a(p + q)} a ( p + q ) p − q C p − q a p q \dfrac{p - q}{apq} a pq p − q D p q a ( p − q ) \dfrac{pq}{a(p - q)} a ( p − q ) pq
Worked solution (try it first) Cross-multiply:
q ( 1 + a x ) = p ( 1 − a x ) q(1 + ax) = p(1 - ax) q ( 1 + a x ) = p ( 1 − a x ) , so
q + a q x = p − a p x q + aqx = p - apx q + a q x = p − a p x .
Collect the
x x x terms on the left:
a p x + a q x = p − q apx + aqx = p - q a p x + a q x = p − q , so
a x ( p + q ) = p − q ax(p + q) = p - q a x ( p + q ) = p − q .
Divide by
a ( p + q ) a(p + q) a ( p + q ) :
x = p − q a ( p + q ) x = \dfrac{p - q}{a(p + q)} x = a ( p + q ) p − q , option B.
Watch out
When you divide by a ( p + q ) a(p + q) a ( p + q ) p − q p - q p − q p + q a ( p − q ) \frac{p + q}{a(p - q)} a ( p − q ) p + q Report a problem with this question
Which of the following is a factor of 15 + 7 x − 2 x 2 15 + 7x - 2x^2 15 + 7 x − 2 x 2
A x − 3 x - 3 x − 3 B x + 3 x + 3 x + 3 C x − 5 x - 5 x − 5 D x + 5 x + 5 x + 5
Worked solution (try it first) Take out
− 1 -1 − 1 so the
x 2 x^2 x 2 term is positive:
15 + 7 x − 2 x 2 = − ( 2 x 2 − 7 x − 15 ) 15 + 7x - 2x^2 = -(2x^2 - 7x - 15) 15 + 7 x − 2 x 2 = − ( 2 x 2 − 7 x − 15 ) .
Factorise:
2 x 2 − 7 x − 15 = ( 2 x + 3 ) ( x − 5 ) 2x^2 - 7x - 15 = (2x + 3)(x - 5) 2 x 2 − 7 x − 15 = ( 2 x + 3 ) ( x − 5 ) .
So
x − 5 x - 5 x − 5 is a factor, option C.
Check:
x = 5 x = 5 x = 5 gives
15 + 35 − 50 = 0 15 + 35 - 50 = 0 15 + 35 − 50 = 0 .
Watch out
A factor x − 5 x - 5 x − 5 x = 5 x = 5 x = 5 x + 5 x + 5 x + 5 x = − 5 x = -5 x = − 5 15 − 35 − 50 = − 70 15 - 35 - 50 = -70 15 − 35 − 50 = − 70 Report a problem with this question
Solve the simultaneous equations x 2 + y − 5 = 0 x^2 + y - 5 = 0 x 2 + y − 5 = 0 y − 7 x + 3 = 0 y - 7x + 3 = 0 y − 7 x + 3 = 0 x x x
A − 2 , 4 -2, 4 − 2 , 4 B 2 , 4 2, 4 2 , 4 C − 1 , 8 -1, 8 − 1 , 8 D 1 , − 8 1, -8 1 , − 8
Worked solution (try it first) Make
y y y the subject of the linear equation:
y = 7 x − 3 y = 7x - 3 y = 7 x − 3 .
Substitute into
x 2 + y − 5 = 0 x^2 + y - 5 = 0 x 2 + y − 5 = 0 :
x 2 + 7 x − 3 − 5 = 0 x^2 + 7x - 3 - 5 = 0 x 2 + 7 x − 3 − 5 = 0 , so
x 2 + 7 x − 8 = 0 x^2 + 7x - 8 = 0 x 2 + 7 x − 8 = 0 .
Factorise:
( x − 1 ) ( x + 8 ) = 0 (x - 1)(x + 8) = 0 ( x − 1 ) ( x + 8 ) = 0 , so
x = 1 x = 1 x = 1 or
x = − 8 x = -8 x = − 8 , option D.
Watch out
Each root has the opposite sign to the number in its bracket: ( x − 1 ) ( x + 8 ) = 0 (x - 1)(x + 8) = 0 ( x − 1 ) ( x + 8 ) = 0 1 1 1 − 8 -8 − 8 − 1 -1 − 1 Report a problem with this question
Solve the equation ( 3 x − 2 ) ( 5 x − 4 ) = ( 3 x − 2 ) 2 (3x - 2)(5x - 4) = (3x - 2)^2 ( 3 x − 2 ) ( 5 x − 4 ) = ( 3 x − 2 ) 2
A − 2 3 , 1 -\frac23, 1 − 3 2 , 1 B 2 3 , − 1 \frac23, -1 3 2 , − 1 C 2 3 , 1 \frac23, 1 3 2 , 1 D 2 3 , 4 5 \frac23, \frac45 3 2 , 5 4
Worked solution (try it first) Bring everything to one side:
( 3 x − 2 ) ( 5 x − 4 ) − ( 3 x − 2 ) 2 = 0 (3x - 2)(5x - 4) - (3x - 2)^2 = 0 ( 3 x − 2 ) ( 5 x − 4 ) − ( 3 x − 2 ) 2 = 0 .
Take out the common factor
3 x − 2 3x - 2 3 x − 2 :
( 3 x − 2 ) [ ( 5 x − 4 ) − ( 3 x − 2 ) ] = 0 (3x - 2)[(5x - 4) - (3x - 2)] = 0 ( 3 x − 2 ) [( 5 x − 4 ) − ( 3 x − 2 )] = 0 , which is
( 3 x − 2 ) ( 2 x − 2 ) = 0 (3x - 2)(2x - 2) = 0 ( 3 x − 2 ) ( 2 x − 2 ) = 0 .
So
x = 2 3 x = \frac23 x = 3 2 or
x = 1 x = 1 x = 1 , option C.
Watch out
The right side is not zero, so you cannot set 5 x − 4 = 0 5x - 4 = 0 5 x − 4 = 0 4 5 \frac45 5 4 Report a problem with this question
If the function f f f f ( x + 2 ) = 2 x 2 + 7 x − 5 f(x + 2) = 2x^2 + 7x - 5 f ( x + 2 ) = 2 x 2 + 7 x − 5 f ( − 1 ) f(-1) f ( − 1 )
Worked solution (try it first) f ( − 1 ) f(-1) f ( − 1 ) needs
x + 2 = − 1 x + 2 = -1 x + 2 = − 1 , so
x = − 3 x = -3 x = − 3 .
Substitute:
2 ( − 3 ) 2 + 7 ( − 3 ) − 5 = 18 − 21 − 5 2(-3)^2 + 7(-3) - 5 = 18 - 21 - 5 2 ( − 3 ) 2 + 7 ( − 3 ) − 5 = 18 − 21 − 5 .
So
f ( − 1 ) = − 8 f(-1) = -8 f ( − 1 ) = − 8 , option B.
Watch out
Don't put x = − 1 x = -1 x = − 1 2 − 7 − 5 = − 10 2 - 7 - 5 = -10 2 − 7 − 5 = − 10 f ( 1 ) f(1) f ( 1 ) x x x x + 2 = − 1 x + 2 = -1 x + 2 = − 1 Report a problem with this question
Divide the expression x 3 + 7 x 2 − x − 7 x^3 + 7x^2 - x - 7 x 3 + 7 x 2 − x − 7 x 2 − 1 x^2 - 1 x 2 − 1
A − x 3 + 7 x 2 − x − 7 -x^3 + 7x^2 - x - 7 − x 3 + 7 x 2 − x − 7 B − x 3 − 7 x + 7 -x^3 - 7x + 7 − x 3 − 7 x + 7 C x − 7 x - 7 x − 7 D x + 7 x + 7 x + 7
Worked solution (try it first) Group the terms in pairs:
x 3 + 7 x 2 − x − 7 = x 2 ( x + 7 ) − 1 ( x + 7 ) x^3 + 7x^2 - x - 7 = x^2(x + 7) - 1(x + 7) x 3 + 7 x 2 − x − 7 = x 2 ( x + 7 ) − 1 ( x + 7 ) .
Take out the common bracket:
( x 2 − 1 ) ( x + 7 ) (x^2 - 1)(x + 7) ( x 2 − 1 ) ( x + 7 ) .
So dividing by
x 2 − 1 x^2 - 1 x 2 − 1 leaves
x + 7 x + 7 x + 7 , option D.
Watch out
Taking out − 1 -1 − 1 − x − 7 -x - 7 − x − 7 + ( x + 7 ) +(x + 7) + ( x + 7 ) x + 7 x + 7 x + 7 x − 7 x - 7 x − 7 Report a problem with this question
Simplify 1 p − 1 q p q − q p \dfrac{\frac1p - \frac1q}{\frac pq - \frac qp} q p − p q p 1 − q 1
A 1 p − q \dfrac{1}{p - q} p − q 1 B − 1 p + q -\dfrac{1}{p + q} − p + q 1 C 1 p q \dfrac{1}{pq} pq 1 D 1 p q ( p − q ) \dfrac{1}{pq(p - q)} pq ( p − q ) 1
Worked solution (try it first) Top:
1 p − 1 q = q − p p q \frac1p - \frac1q = \frac{q - p}{pq} p 1 − q 1 = pq q − p .
Bottom:
p q − q p = p 2 − q 2 p q \frac pq - \frac qp = \frac{p^2 - q^2}{pq} q p − p q = pq p 2 − q 2 .
Divide.
The
p q pq pq cancels:
q − p p 2 − q 2 = − ( p − q ) ( p − q ) ( p + q ) \frac{q - p}{p^2 - q^2} = \frac{-(p - q)}{(p - q)(p + q)} p 2 − q 2 q − p = ( p − q ) ( p + q ) − ( p − q ) .
Cancel
p − q p - q p − q : the result is
− 1 p + q -\dfrac{1}{p + q} − p + q 1 , option B.
Watch out
q − p = − ( p − q ) q - p = -(p - q) q − p = − ( p − q ) + 1 p + q +\frac{1}{p + q} + p + q 1 Report a problem with this question
Solve the inequality y 2 − 3 y > 18 y^2 - 3y > 18 y 2 − 3 y > 18
A − 2 < y < 6 -2 < y < 6 − 2 < y < 6 B y < − 3 y < -3 y < − 3 y > 6 y > 6 y > 6 C y > − 3 y > -3 y > − 3 y > 6 y > 6 y > 6 D y < − 3 y < -3 y < − 3 y < 6 y < 6 y < 6
Worked solution (try it first) Bring everything to one side:
y 2 − 3 y − 18 > 0 y^2 - 3y - 18 > 0 y 2 − 3 y − 18 > 0 .
Factorise:
( y − 6 ) ( y + 3 ) > 0 (y - 6)(y + 3) > 0 ( y − 6 ) ( y + 3 ) > 0 , so the roots are
y = 6 y = 6 y = 6 and
y = − 3 y = -3 y = − 3 .
The graph of a positive
y 2 y^2 y 2 quadratic is a U shape, so it is above zero outside the roots:
y < − 3 y < -3 y < − 3 or
y > 6 y > 6 y > 6 , option B.
Watch out
"Greater than 0" means outside the roots. Between the roots, − 3 < y < 6 -3 < y < 6 − 3 < y < 6 y 2 − 3 y − 18 y^2 - 3y - 18 y 2 − 3 y − 18 Report a problem with this question
If x x x x x x x + 1 3 > 1 x + 3 \frac{x + 1}{3} > \frac{1}{x + 3} 3 x + 1 > x + 3 1
A 3 < x < 4 3 < x < 4 3 < x < 4 B − 4 < x < − 3 -4 < x < -3 − 4 < x < − 3 C − 2 < x < − 1 -2 < x < -1 − 2 < x < − 1 D − 3 < x < 0 -3 < x < 0 − 3 < x < 0
Worked solution (try it first) Bring everything to one side and use a common denominator:
x + 1 3 − 1 x + 3 = ( x + 1 ) ( x + 3 ) − 3 3 ( x + 3 ) \frac{x + 1}{3} - \frac{1}{x + 3} = \dfrac{(x + 1)(x + 3) - 3}{3(x + 3)} 3 x + 1 − x + 3 1 = 3 ( x + 3 ) ( x + 1 ) ( x + 3 ) − 3 .
Expand the top:
x 2 + 4 x + 3 − 3 = x ( x + 4 ) x^2 + 4x + 3 - 3 = x(x + 4) x 2 + 4 x + 3 − 3 = x ( x + 4 ) .
So you need
x ( x + 4 ) 3 ( x + 3 ) > 0 \dfrac{x(x + 4)}{3(x + 3)} > 0 3 ( x + 3 ) x ( x + 4 ) > 0 .
The critical values are
− 4 -4 − 4 ,
− 3 -3 − 3 and
0 0 0 .
Test a negative
x x x in each interval:
x < − 4 x < -4 x < − 4 gives a negative value,
− 4 < x < − 3 -4 < x < -3 − 4 < x < − 3 positive, and
− 3 < x < 0 -3 < x < 0 − 3 < x < 0 negative.
So
− 4 < x < − 3 -4 < x < -3 − 4 < x < − 3 , option B.
Watch out
Don't cross-multiply by x + 3 x + 3 x + 3 x < − 3 x < -3 x < − 3 Report a problem with this question
A man's initial salary is ₦540.00 a month and it increases after each period of six months by ₦36.00 a month. Find his salary in the eighth month of the third year.
A ₦828.00 B ₦756.00 C ₦720.00 D ₦684.00
Worked solution (try it first) The eighth month of the third year is month
24 + 8 = 32 24 + 8 = 32 24 + 8 = 32 .
Months 1–6 are the first six-month period, and month 32 falls in months 31–36, the sixth period.
The first period has no rise, so by the sixth period there have been 5 rises:
540 + 5 × 36 = 720 540 + 5 \times 36 = 720 540 + 5 × 36 = 720 .
So the salary is ₦720.00, option C.
Watch out
The n n n n − 1 n - 1 n − 1 n n n 540 + 6 × 36 = 756 540 + 6 \times 36 = 756 540 + 6 × 36 = 756 Report a problem with this question
If k + 1 k + 1 k + 1 2 k − 1 2k - 1 2 k − 1 3 k + 1 3k + 1 3 k + 1
A 0, 8 B − 1 , 5 3 -1, \frac53 − 1 , 3 5 C 2, 3 D 1, − 1 -1 − 1
Worked solution (try it first) For three terms of a G.P., the middle term squared equals the product of the outer two:
( 2 k − 1 ) 2 = ( k + 1 ) ( 3 k + 1 ) (2k - 1)^2 = (k + 1)(3k + 1) ( 2 k − 1 ) 2 = ( k + 1 ) ( 3 k + 1 ) .
Expand:
4 k 2 − 4 k + 1 = 3 k 2 + 4 k + 1 4k^2 - 4k + 1 = 3k^2 + 4k + 1 4 k 2 − 4 k + 1 = 3 k 2 + 4 k + 1 , so
k 2 − 8 k = 0 k^2 - 8k = 0 k 2 − 8 k = 0 and
k ( k − 8 ) = 0 k(k - 8) = 0 k ( k − 8 ) = 0 .
If
k = 0 k = 0 k = 0 the terms are
1 , − 1 , 1 1, -1, 1 1 , − 1 , 1 with ratio
− 1 -1 − 1 .
If
k = 8 k = 8 k = 8 they are
9 , 15 , 25 9, 15, 25 9 , 15 , 25 with ratio
15 9 = 5 3 \frac{15}{9} = \frac53 9 15 = 3 5 .
So the common ratio is
− 1 -1 − 1 or
5 3 \frac53 3 5 , option B.
Watch out
0 and 8 are the values of k k k k k k Report a problem with this question
A binary operation ∗ * ∗ x ∗ y = x y x * y = x^y x ∗ y = x y x x x y y y x ∗ 2 = x x * 2 = x x ∗ 2 = x x x x
Worked solution (try it first) By the definition,
x ∗ 2 = x 2 x * 2 = x^2 x ∗ 2 = x 2 , so the equation is
x 2 = x x^2 = x x 2 = x .
Bring everything to one side and factorise:
x 2 − x = x ( x − 1 ) = 0 x^2 - x = x(x - 1) = 0 x 2 − x = x ( x − 1 ) = 0 .
So
x = 0 x = 0 x = 0 or
x = 1 x = 1 x = 1 , option A.
Watch out
Don't divide both sides of x 2 = x x^2 = x x 2 = x x x x x = 1 x = 1 x = 1 x = 0 x = 0 x = 0 Report a problem with this question
A regular polygon has 150 ∘ 150^\circ 15 0 ∘
Worked solution (try it first) An interior angle and its exterior angle add up to
180 ∘ 180^\circ 18 0 ∘ , so each exterior angle is
180 ∘ − 150 ∘ = 30 ∘ 180^\circ - 150^\circ = 30^\circ 18 0 ∘ − 15 0 ∘ = 3 0 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 30 = 12 360 \div 30 = 12 360 ÷ 30 = 12 , option A.
Watch out
Work with the exterior angle 30 ∘ 30^\circ 3 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 150 ∘ 150^\circ 15 0 ∘ Report a problem with this question
Calculate the length, in cm, of the arc of a circle of diameter 8 cm which subtends an angle of 22 1 2 ∘ 22\frac12^\circ 22 2 1 ∘
A 2 π 2\pi 2 π B π \pi π C 2 3 π \frac23\pi 3 2 π D π 2 \frac\pi2 2 π
Worked solution (try it first) The diameter is 8 cm, so the radius is 4 cm and the circumference is
2 π × 4 = 8 π 2\pi \times 4 = 8\pi 2 π × 4 = 8 π cm.
The arc is
22.5 360 = 1 16 \frac{22.5}{360} = \frac1{16} 360 22.5 = 16 1 of the circle:
1 16 × 8 π = π 2 \frac1{16} \times 8\pi = \frac\pi2 16 1 × 8 π = 2 π cm, option D.
Watch out
Halve the diameter first. Using 8 as the radius doubles the answer to π \pi π Report a problem with this question
In the diagram, P Q R S PQRS P QR S O O O S Q SQ S Q P Q ∥ R T PQ \parallel RT P Q ∥ R T ∠ R T S = 32 ∘ \angle RTS = 32^\circ ∠ R T S = 3 2 ∘ ∠ P S Q \angle PSQ ∠ P S Q
A 32 ∘ 32^\circ 3 2 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 58 ∘ 58^\circ 5 8 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) P Q ∥ R T PQ \parallel RT P Q ∥ R T and
S Q T SQT S QT is a straight line, so alternate angles are equal:
∠ P Q S = ∠ R T S = 32 ∘ \angle PQS = \angle RTS = 32^\circ ∠ P QS = ∠ R T S = 3 2 ∘ .
S Q SQ S Q is a diameter, so
∠ S P Q = 90 ∘ \angle SPQ = 90^\circ ∠ S P Q = 9 0 ∘ (angle in a semicircle).
The angles of triangle
P S Q PSQ P S Q add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P S Q = 180 ∘ − 90 ∘ − 32 ∘ \angle PSQ = 180^\circ - 90^\circ - 32^\circ ∠ P S Q = 18 0 ∘ − 9 0 ∘ − 3 2 ∘ = 58 ∘ = 58^\circ = 5 8 ∘ , option C.
Watch out
32 ∘ 32^\circ 3 2 ∘ ∠ P Q S \angle PQS ∠ P QS Q Q Q ∠ P S Q \angle PSQ ∠ P S Q 90 ∘ − 32 ∘ 90^\circ - 32^\circ 9 0 ∘ − 3 2 ∘ Report a problem with this question
In the diagram, O O O P O Q POQ P O Q ∠ P O R = 96 ∘ \angle POR = 96^\circ ∠ P O R = 9 6 ∘ ∠ O R Q \angle ORQ ∠ O R Q
A 84 ∘ 84^\circ 8 4 ∘ B 48 ∘ 48^\circ 4 8 ∘ C 45 ∘ 45^\circ 4 5 ∘ D 42 ∘ 42^\circ 4 2 ∘
Worked solution (try it first) P O Q POQ P O Q is a straight line, so
∠ Q O R = 180 ∘ − 96 ∘ \angle QOR = 180^\circ - 96^\circ ∠ QO R = 18 0 ∘ − 9 6 ∘ O Q = O R OQ = OR O Q = O R (radii), so triangle
O Q R OQR O QR is isosceles and its base angles at
Q Q Q and
R R R are equal.
So
∠ O R Q = 180 ∘ − 84 ∘ 2 \angle ORQ = \dfrac{180^\circ - 84^\circ}{2} ∠ O R Q = 2 18 0 ∘ − 8 4 ∘ = 48 ∘ = 48^\circ = 4 8 ∘ , option B.
Watch out
84 ∘ 84^\circ 8 4 ∘ ∠ Q O R \angle QOR ∠ QO R ∠ O R Q \angle ORQ ∠ O R Q O Q R OQR O QR Report a problem with this question
In the diagram, Q P ∥ S T QP \parallel ST QP ∥ S T ∠ P Q R = 34 ∘ \angle PQR = 34^\circ ∠ P QR = 3 4 ∘ ∠ Q R S = 73 ∘ \angle QRS = 73^\circ ∠ QR S = 7 3 ∘ R S = R T RS = RT R S = R T ∠ S R T \angle SRT ∠ S R T
A 68 ∘ 68^\circ 6 8 ∘ B 102 ∘ 102^\circ 10 2 ∘ C 107 ∘ 107^\circ 10 7 ∘ D 141 ∘ 141^\circ 14 1 ∘
Worked solution (try it first) Draw a line through
R R R parallel to
Q P QP QP .
By alternate angles,
R Q RQ R Q makes
34 ∘ 34^\circ 3 4 ∘ with it and
R S RS R S makes
∠ R S T \angle RST ∠ R S T with it, so
73 ∘ = 34 ∘ + ∠ R S T 73^\circ = 34^\circ + \angle RST 7 3 ∘ = 3 4 ∘ + ∠ R S T .
So
∠ R S T = 39 ∘ \angle RST = 39^\circ ∠ R S T = 3 9 ∘ .
R S = R T RS = RT R S = R T , so triangle
R S T RST R S T is isosceles and
∠ R T S = ∠ R S T = 39 ∘ \angle RTS = \angle RST = 39^\circ ∠ R T S = ∠ R S T = 3 9 ∘ .
The angles of the triangle add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ S R T = 180 ∘ − 39 ∘ − 39 ∘ \angle SRT = 180^\circ - 39^\circ - 39^\circ ∠ S R T = 18 0 ∘ − 3 9 ∘ − 3 9 ∘ = 102 ∘ = 102^\circ = 10 2 ∘ , option B.
Watch out
Both base angles are 39 ∘ 39^\circ 3 9 ∘ 180 ∘ − 39 ∘ = 141 ∘ 180^\circ - 39^\circ = 141^\circ 18 0 ∘ − 3 9 ∘ = 14 1 ∘ Report a problem with this question
In the figure, P T PT P T U U U Q U ∥ R S QU \parallel RS Q U ∥ R S ∠ T U R = 35 ∘ \angle TUR = 35^\circ ∠ T U R = 3 5 ∘ ∠ S R U = 50 ∘ \angle SRU = 50^\circ ∠ S R U = 5 0 ∘ x = ∠ Q R U x = \angle QRU x = ∠ QR U
A 95 ∘ 95^\circ 9 5 ∘ B 85 ∘ 85^\circ 8 5 ∘ C 50 ∘ 50^\circ 5 0 ∘ D 35 ∘ 35^\circ 3 5 ∘
Worked solution (try it first) The angle between tangent
U T UT U T and chord
U R UR U R equals the angle in the alternate segment, so
∠ U Q R = ∠ T U R = 35 ∘ \angle UQR = \angle TUR = 35^\circ ∠ U QR = ∠ T U R = 3 5 ∘ .
Q U ∥ R S QU \parallel RS Q U ∥ R S , so alternate angles are equal:
∠ Q U R = ∠ U R S = 50 ∘ \angle QUR = \angle URS = 50^\circ ∠ Q U R = ∠ U R S = 5 0 ∘ .
The angles of triangle
Q U R QUR Q U R add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 35 ∘ − 50 ∘ x = 180^\circ - 35^\circ - 50^\circ x = 18 0 ∘ − 3 5 ∘ − 5 0 ∘ = 95 ∘ = 95^\circ = 9 5 ∘ , option A.
Watch out
The 35 ∘ 35^\circ 3 5 ∘ ∠ U Q R \angle UQR ∠ U QR Q Q Q U R UR U R x x x R R R Report a problem with this question
In the diagram, ∠ Q P S = ∠ S P R \angle QPS = \angle SPR ∠ QP S = ∠ S P R P R = 9 PR = 9 P R = 9 P Q = 4 PQ = 4 P Q = 4 Q S = 3 QS = 3 QS = 3 S R SR S R
A 6 3 4 6\frac34 6 4 3 B 3 3 8 3\frac38 3 8 3 C 4 3 8 4\frac38 4 8 3 D 2 2 3 2\frac23 2 3 2
Worked solution (try it first) P S PS P S bisects
∠ Q P R \angle QPR ∠ QP R .
The angle bisector divides the opposite side in the ratio of the other two sides:
Q S S R = P Q P R \dfrac{QS}{SR} = \dfrac{PQ}{PR} S R QS = P R P Q .
So
3 S R = 4 9 \dfrac{3}{SR} = \dfrac{4}{9} S R 3 = 9 4 , which gives
4 × S R = 27 4 \times SR = 27 4 × S R = 27 .
S R = 27 4 = 6 3 4 SR = \frac{27}{4} = 6\frac34 S R = 4 27 = 6 4 3 cm, option A.
Watch out
Q S QS QS P Q PQ P Q Q S : S R = 4 : 9 QS : SR = 4 : 9 QS : S R = 4 : 9 S R = 4 3 SR = \frac43 S R = 3 4 Report a problem with this question
The three sides of an isosceles triangle are of lengths x + 3 x + 3 x + 3 2 x + 3 2x + 3 2 x + 3 2 x − 3 2x - 3 2 x − 3 x x x
Worked solution (try it first) Two of the sides are equal.
2 x + 3 2x + 3 2 x + 3 and
2 x − 3 2x - 3 2 x − 3 can never be equal, so one of them equals
x + 3 x + 3 x + 3 .
x + 3 = 2 x + 3 x + 3 = 2x + 3 x + 3 = 2 x + 3 gives
x = 0 x = 0 x = 0 , which makes a side
− 3 -3 − 3 : impossible.
x + 3 = 2 x − 3 x + 3 = 2x - 3 x + 3 = 2 x − 3 gives
x = 6 x = 6 x = 6 , with sides 9, 15 and 9.
So
x = 6 x = 6 x = 6 , option D.
Watch out
Check the sides are positive. x = 0 x = 0 x = 0 − 3 -3 − 3 Report a problem with this question
In the figure, the line segment S T ST S T S S S T T T O O O Q Q Q O S = 5 OS = 5 O S = 5 Q T = 2 QT = 2 QT = 2 O Q = 14 OQ = 14 O Q = 14 S T ST S T
A 7 3 7\sqrt3 7 3 B 12 cm C 8 7 8\sqrt7 8 7 D 7 cm
Worked solution (try it first) Radii meet a tangent at
90 ∘ 90^\circ 9 0 ∘ , so
O S ⊥ S T OS \perp ST O S ⊥ S T and
Q T ⊥ S T QT \perp ST QT ⊥ S T .
The tangent crosses between the circles, so the radii point opposite ways.
Slide
Q T QT QT along to the end of
O S OS O S : this makes a right-angled triangle with hypotenuse
O Q = 14 OQ = 14 O Q = 14 , one side
5 + 2 = 7 5 + 2 = 7 5 + 2 = 7 and the other side equal to
S T ST S T .
Pythagoras:
S T 2 = 14 2 − 7 2 = 147 ST^2 = 14^2 - 7^2 = 147 S T 2 = 1 4 2 − 7 2 = 147 , so
S T = 147 = 7 3 ST = \sqrt{147} = 7\sqrt3 S T = 147 = 7 3 cm, option A.
Watch out
The tangent crosses between the circles, so add the radii (5 + 2 = 7 5 + 2 = 7 5 + 2 = 7 196 − 9 = 187 \sqrt{196 - 9} = \sqrt{187} 196 − 9 = 187 Report a problem with this question
In the figure, the area of the square P Q R S PQRS P QR S 100 cm 2 100\text{ cm}^2 100 cm 2 T U Y S TUYS T U Y S X Q V U XQVU X Q V U 1 : 16 1 : 16 1 : 16 Y R YR Y R
Worked solution (try it first) P Q R S PQRS P QR S has area 100 cm², so its side is 10 cm.
The areas of the small squares are in the ratio
1 : 16 1 : 16 1 : 16 , so their sides are in the ratio
1 : 4 1 : 4 1 : 4 .
Call them
a a a and
4 a 4a 4 a .
Along
S R SR S R :
S Y + Y R = a + 4 a = 10 SY + YR = a + 4a = 10 S Y + Y R = a + 4 a = 10 , so
a = 2 a = 2 a = 2 .
Y R YR Y R is a side of the big square
X Q V U XQVU X Q V U :
Y R = 4 a = 8 YR = 4a = 8 Y R = 4 a = 8 cm, option C.
Watch out
The ratio of the sides is the square root of the ratio of the areas: 1 : 4 1 : 4 1 : 4 1 : 16 1 : 16 1 : 16 17 a = 10 17a = 10 17 a = 10 Report a problem with this question
Find the radius of a sphere whose surface area is 154 cm 2 154\text{ cm}^2 154 cm 2 [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 7.00 cm B 3.50 cm C 3.00 cm D 1.75 cm
Worked solution (try it first) The surface area of a sphere is
4 π r 2 4\pi r^2 4 π r 2 :
4 × 22 7 × r 2 = 154 4 \times \frac{22}{7} \times r^2 = 154 4 × 7 22 × r 2 = 154 .
So
r 2 = 154 × 7 88 = 12.25 r^2 = \dfrac{154 \times 7}{88} = 12.25 r 2 = 88 154 × 7 = 12.25 .
Take the square root:
r = 3.5 r = 3.5 r = 3.5 cm, option B.
Watch out
Keep the 4 in 4 π r 2 4\pi r^2 4 π r 2 π r 2 = 154 \pi r^2 = 154 π r 2 = 154 r = 7 r = 7 r = 7 Also set as JAMB 2016 · UTME · Q15
Report a problem with this question
Find the area of the sector of a circle with radius 3 m if the angle of the sector is 60 ∘ 60^\circ 6 0 ∘
A 4.0 m 2 4.0\text{ m}^2 4.0 m 2 B 4.1 m 2 4.1\text{ m}^2 4.1 m 2 C 4.7 m 2 4.7\text{ m}^2 4.7 m 2 D 5.0 m 2 5.0\text{ m}^2 5.0 m 2
Worked solution (try it first) Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 .
Here
60 360 = 1 6 \frac{60}{360} = \frac16 360 60 = 6 1 and
r 2 = 9 r^2 = 9 r 2 = 9 .
So the area is
1 6 × 9 π = 1.5 π \frac16 \times 9\pi = 1.5\pi 6 1 × 9 π = 1.5 π ≈ 4.71 m 2 \approx 4.71\text{ m}^2 ≈ 4.71 m 2 .
To one decimal place that is
4.7 m 2 4.7\text{ m}^2 4.7 m 2 , option C.
Watch out
Square the radius: r 2 = 9 r^2 = 9 r 2 = 9 r = 3 r = 3 r = 3 r 2 r^2 r 2 0.5 π ≈ 1.6 m 2 0.5\pi \approx 1.6\text{ m}^2 0.5 π ≈ 1.6 m 2 Report a problem with this question
The angle between latitudes 30 ∘ 30^\circ 3 0 ∘ 13 ∘ 13^\circ 1 3 ∘
A 17 ∘ 17^\circ 1 7 ∘ B 33 ∘ 33^\circ 3 3 ∘ C 43 ∘ 43^\circ 4 3 ∘ D 53 ∘ 53^\circ 5 3 ∘
Worked solution (try it first) 30 ∘ 30^\circ 3 0 ∘ S and
13 ∘ 13^\circ 1 3 ∘ N are on opposite sides of the equator, so add the latitudes.
The angle is
30 ∘ + 13 ∘ = 43 ∘ 30^\circ + 13^\circ = 43^\circ 3 0 ∘ + 1 3 ∘ = 4 3 ∘ , option C.
Watch out
Subtract only when both latitudes are on the same side of the equator. Here one is south and one north, so 30 ∘ − 13 ∘ = 17 ∘ 30^\circ - 13^\circ = 17^\circ 3 0 ∘ − 1 3 ∘ = 1 7 ∘ Report a problem with this question
If sin θ = cos θ \sin\theta = \cos\theta sin θ = cos θ θ \theta θ 0 ∘ 0^\circ 0 ∘ 360 ∘ 360^\circ 36 0 ∘
A 45 ∘ , 225 ∘ 45^\circ, 225^\circ 4 5 ∘ , 22 5 ∘ B 135 ∘ , 315 ∘ 135^\circ, 315^\circ 13 5 ∘ , 31 5 ∘ C 45 ∘ , 315 ∘ 45^\circ, 315^\circ 4 5 ∘ , 31 5 ∘ D 135 ∘ , 225 ∘ 135^\circ, 225^\circ 13 5 ∘ , 22 5 ∘
Worked solution (try it first) Divide both sides by
cos θ \cos\theta cos θ :
tan θ = 1 \tan\theta = 1 tan θ = 1 .
The acute angle with
tan θ = 1 \tan\theta = 1 tan θ = 1 is
45 ∘ 45^\circ 4 5 ∘ .
Tangent is positive in the first and third quadrants, so
θ = 45 ∘ \theta = 45^\circ θ = 4 5 ∘ or
180 ∘ + 45 ∘ = 225 ∘ 180^\circ + 45^\circ = 225^\circ 18 0 ∘ + 4 5 ∘ = 22 5 ∘ , option A.
Watch out
Tangent is positive in the first and third quadrants, not the fourth. At 315 ∘ 315^\circ 31 5 ∘ Report a problem with this question
In the figure, P P P Q Q Q H H H P Q = 5 PQ = 5 P Q = 5 T H TH T H T T T P P P Q Q Q 30 ∘ 30^\circ 3 0 ∘ 45 ∘ 45^\circ 4 5 ∘ T H TH T H
A 5 3 + 1 \frac{5}{\sqrt3 + 1} 3 + 1 5 B 5 3 − 1 \frac{5}{\sqrt3 - 1} 3 − 1 5 C 5 3 \frac{5}{\sqrt3} 3 5 D 3 5 \frac{\sqrt3}{5} 5 3
Worked solution (try it first) At
Q Q Q the angle is
45 ∘ 45^\circ 4 5 ∘ , so
Q H = h QH = h Q H = h .
At
P P P :
tan 30 ∘ = h P H \tan30^\circ = \frac{h}{PH} tan 3 0 ∘ = P H h , so
P H = h tan 30 ∘ = h 3 PH = \frac{h}{\tan30^\circ} = h\sqrt3 P H = t a n 3 0 ∘ h = h 3 .
P P P is further from
H H H than
Q Q Q by
P Q PQ P Q :
h 3 − h = 5 h\sqrt3 - h = 5 h 3 − h = 5 , so
h ( 3 − 1 ) = 5 h(\sqrt3 - 1) = 5 h ( 3 − 1 ) = 5 .
So
T H = 5 3 − 1 TH = \dfrac{5}{\sqrt3 - 1} T H = 3 − 1 5 m, option B.
Watch out
In the figure P P P Q Q Q H H H P Q = P H − Q H PQ = PH - QH P Q = P H − Q H h ( 3 + 1 ) = 5 h(\sqrt3 + 1) = 5 h ( 3 + 1 ) = 5 Report a problem with this question
If two angles of a triangle are 30 ∘ 30^\circ 3 0 ∘
A 5 cm B 4 cm C 3 3 3\sqrt3 3 3 D 10 3 3 \frac{10\sqrt3}{3} 3 10 3
Worked solution (try it first) The third angle is
180 ∘ − 30 ∘ − 30 ∘ = 120 ∘ 180^\circ - 30^\circ - 30^\circ = 120^\circ 18 0 ∘ − 3 0 ∘ − 3 0 ∘ = 12 0 ∘ .
The longest side, 10 cm, faces it.
Each of the other sides faces a
30 ∘ 30^\circ 3 0 ∘ angle.
Sine rule:
a sin 30 ∘ = 10 sin 120 ∘ \dfrac{a}{\sin30^\circ} = \dfrac{10}{\sin120^\circ} sin 3 0 ∘ a = sin 12 0 ∘ 10 .
So
a = 10 × 1 2 3 2 a = \dfrac{10 \times \frac12}{\frac{\sqrt3}{2}} a = 2 3 10 × 2 1 = 10 3 = \dfrac{10}{\sqrt3} = 3 10 .
Rationalise:
10 3 3 \dfrac{10\sqrt3}{3} 3 10 3 cm, option D.
Watch out
The largest angle is 120 ∘ 120^\circ 12 0 ∘ 90 ∘ 90^\circ 9 0 ∘ 10 sin 30 ∘ = 5 10\sin30^\circ = 5 10 sin 3 0 ∘ = 5 Report a problem with this question
Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7.
A 20 ∘ 20^\circ 2 0 ∘ B 80 ∘ 80^\circ 8 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 140 ∘ 140^\circ 14 0 ∘
Worked solution (try it first) Add the proportions:
1 + 4 + 6 + 7 = 18 1 + 4 + 6 + 7 = 18 1 + 4 + 6 + 7 = 18 parts.
Each part is worth
360 ∘ 18 = 20 ∘ \frac{360^\circ}{18} = 20^\circ 18 36 0 ∘ = 2 0 ∘ .
So proportion 7 has
7 × 20 ∘ = 140 ∘ 7 \times 20^\circ = 140^\circ 7 × 2 0 ∘ = 14 0 ∘ , option D.
Watch out
20 ∘ 20^\circ 2 0 ∘ Report a problem with this question
The bar chart shows the distribution of marks in a class test. How many students took the test?
Worked solution (try it first) Each bar's height is the number of students with that mark.
Read the bars: 2, 5, 3, 4, 3, 1 and 2.
Add them:
2 + 5 + 3 + 4 + 3 + 1 + 2 = 20 2 + 5 + 3 + 4 + 3 + 1 + 2 = 20 2 + 5 + 3 + 4 + 3 + 1 + 2 = 20 students, option B.
Watch out
Add the heights of the bars, not the marks on the horizontal axis. Marks with no bar (2, 5, 6 and 9) have no students. Report a problem with this question
The following marks were obtained by twenty students in an examination: 53, 30, 70, 84, 59, 43, 90, 20, 78, 48, 44, 60, 81, 73, 50, 37, 67, 68, 64, 52. Find the number of students who scored at least 50 marks.
Worked solution (try it first) It is quicker to count the marks below 50: 30, 43, 20, 48, 44 and 37, which is 6 students.
So
20 − 6 = 14 20 - 6 = 14 20 − 6 = 14 students scored at least 50, option D.
Watch out
"At least 50" includes a score of exactly 50. Leaving out the student with 50 gives 13 (option C). Report a problem with this question
Estimate the mode of the frequency distribution below.
Weight (g)
0–10
10–20
20–30
30–40
40–50
No. of coconuts
10
27
19
6
2
A 13.2 g B 15.0 g C 16.8 g D 17.5 g
Worked solution (try it first) The modal class has the highest frequency: 10–20 g, with 27 coconuts.
Differences from the neighbouring classes:
Δ 1 = 27 − 10 = 17 \Delta_1 = 27 - 10 = 17 Δ 1 = 27 − 10 = 17 and
Δ 2 = 27 − 19 = 8 \Delta_2 = 27 - 19 = 8 Δ 2 = 27 − 19 = 8 .
Mode
= L + Δ 1 Δ 1 + Δ 2 × c = L + \frac{\Delta_1}{\Delta_1 + \Delta_2} \times c = L + Δ 1 + Δ 2 Δ 1 × c = 10 + 17 25 × 10 = 10 + \frac{17}{25} \times 10 = 10 + 25 17 × 10 , which is
10 + 6.8 10 + 6.8 10 + 6.8 .
So the mode is about 16.8 g, option C.
Watch out
The estimate of the mode is not the mid-point of the modal class (15.0 g, option B); it leans towards the bigger neighbouring class, here 20–30. Report a problem with this question
The mean of the ages of ten secondary school pupils is 16, but when the age of their teacher is added, the mean becomes 19. Find the age of the teacher.
Worked solution (try it first) Total age of the 10 pupils:
10 × 16 = 160 10 \times 16 = 160 10 × 16 = 160 years.
With the teacher there are 11 people:
11 × 19 = 209 11 \times 19 = 209 11 × 19 = 209 years.
The teacher's age is the difference:
209 − 160 = 49 209 - 160 = 49 209 − 160 = 49 , option D.
Watch out
With the teacher there are 11 people, not 10. Using 10 × 19 = 190 10 \times 19 = 190 10 × 19 = 190 Report a problem with this question
Find the median of the observations in the table below.
Class
1–5
6–10
11–15
16–20
21–25
26–30
31–35
36–40
Frequency
2
4
5
2
3
2
1
1
Worked solution (try it first) The cumulative frequencies are 2, 6, 11, 13, 16, 18, 19, 20, so
N = 20 N = 20 N = 20 and the median is the
N 2 = 10 \frac{N}{2} = 10 2 N = 10 th value.
The 10th value lies in the class 11–15, where the running total goes from 6 to 11.
Its lower class boundary is 10.5, its frequency is 5 and its width is 5.
Median
= 10.5 + 10 − 6 5 × 5 = 10.5 + \frac{10 - 6}{5} \times 5 = 10.5 + 5 10 − 6 × 5 , which is
10.5 + 4 = 14.5 10.5 + 4 = 14.5 10.5 + 4 = 14.5 , option D.
Watch out
Start from the lower class boundary 10.5, not the class limit 11. Starting from 11 gives 15, which is not an option. Report a problem with this question
A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime.
A 2 11 \frac2{11} 11 2 B 5 11 \frac5{11} 11 5 C 6 11 \frac6{11} 11 6 D 8 11 \frac8{11} 11 8
Worked solution (try it first) From 20 to 30 inclusive there are 11 numbers.
The primes among them are 23 and 29.
21, 25 and 27 are odd but divisible by 3, 5 and 3.
So the probability is
2 11 \frac{2}{11} 11 2 , option A.
Watch out
Not every odd number is prime: 21, 25 and 27 all have factors. Taking the 5 odd numbers as primes gives 5 11 \frac{5}{11} 11 5 Report a problem with this question
Calculate the standard deviation of the data 7, 8, 9, 10, 11, 12, 13.
Worked solution (try it first) There are 7 values and they add up to 70, so the mean is 10.
The deviations are
− 3 -3 − 3 to 3.
Their squares are 9, 4, 1, 0, 1, 4, 9, which add up to 28.
The variance is
28 7 = 4 \frac{28}{7} = 4 7 28 = 4 , so the standard deviation is
4 = 2 \sqrt4 = 2 4 = 2 , option A.
Watch out
Take the square root at the end: 4 is the variance (option B), and the standard deviation is 4 = 2 \sqrt4 = 2 4 = 2 Report a problem with this question
The chances of three independent events X X X Y Y Y Z Z Z 1 2 \frac12 2 1 2 3 \frac23 3 2 1 4 \frac14 4 1 Y Y Y Z Z Z
A 1 8 \frac18 8 1 B 1 24 \frac1{24} 24 1 C 1 12 \frac1{12} 12 1 D 1 4 \frac14 4 1
Worked solution (try it first) "
Y Y Y and
Z Z Z only" means
Y Y Y happens,
Z Z Z happens and
X X X does not.
X X X fails with probability
1 − 1 2 = 1 2 1 - \frac12 = \frac12 1 − 2 1 = 2 1 .
The events are independent, so multiply:
1 2 × 2 3 × 1 4 = 2 24 \frac12 \times \frac23 \times \frac14 = \frac{2}{24} 2 1 × 3 2 × 4 1 = 24 2 .
So the probability is
1 12 \frac{1}{12} 12 1 , option C.
Watch out
"Only" means X X X 1 2 \frac12 2 1 X X X 2 3 × 1 4 = 1 6 \frac23 \times \frac14 = \frac16 3 2 × 4 1 = 6 1 Report a problem with this question