Paper JAMB 1984 General Maths Objective
Objective paper · 37 questions · partial
JAMB 1984 · UME Topics include Number foundations & fractions, Number bases, Commercial arithmetic, Quadratics & their graphs, Statistics: data & averages, Probability.
Our copy of this paper is missing questions 3, 15, 18, 20, 22, 25, 31, 34, 35, 38, 43, 49, 50.
Sit this paper Answer every question in order, timed if you like (suggested 25 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 4 5 6 7 8 9 10 11 12 13 14 16 17 19 21 23 24 26 27 28 29 30 32 33 36 37 39 40 41 42 44 45 46 47 48 Simplify ( 2 3 − 1 5 ) − 1 3 of 2 5 3 − 1 1 / 2 \dfrac{\left(\frac23 - \frac15\right) - \frac13 \text{ of } \frac25}{3 - \frac{1}{1/2}} 3 − 1/2 1 ( 3 2 − 5 1 ) − 3 1 of 5 2
A 1 7 \frac17 7 1 B 7 C 1 3 \frac13 3 1 D 3 E 1 5 \frac15 5 1
Worked solution (try it first) Top bracket: the LCD of 3 and 5 is 15, so
2 3 − 1 5 = 10 15 − 3 15 \frac23 - \frac15 = \frac{10}{15} - \frac{3}{15} 3 2 − 5 1 = 15 10 − 15 3 "Of" means multiply:
1 3 \frac13 3 1 of
2 5 \frac25 5 2 is
2 15 \frac{2}{15} 15 2 .
So the top is
7 15 − 2 15 = 5 15 \frac{7}{15} - \frac{2}{15} = \frac{5}{15} 15 7 − 15 2 = 15 5 , which is
1 3 \frac13 3 1 .
Bottom: dividing by
1 2 \frac12 2 1 doubles, so
1 1 / 2 = 2 \frac{1}{1/2} = 2 1/2 1 = 2 and the bottom is
3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 .
So the value is
1 3 ÷ 1 = 1 3 \frac13 \div 1 = \frac13 3 1 ÷ 1 = 3 1 , option C.
Watch out
1 1 / 2 \frac{1}{1/2} 1/2 1 1 ÷ 1 2 = 2 1 \div \frac12 = 2 1 ÷ 2 1 = 2 1 2 \frac12 2 1 1 2 \frac12 2 1 2 1 2 2\frac12 2 2 1 2 15 \frac{2}{15} 15 2 Report a problem with this question
If 263 + 441 = 714 263 + 441 = 714 263 + 441 = 714
Worked solution (try it first) Units column:
3 + 1 = 4 3 + 1 = 4 3 + 1 = 4 , with no carry.
Middle column:
6 + 4 6 + 4 6 + 4 is written as 1 carry 1, so
6 + 4 = 10 6 + 4 = 10 6 + 4 = 10 must equal one base plus 1.
The base is
10 − 1 = 9 10 - 1 = 9 10 − 1 = 9 .
Check the left column:
2 + 4 + 1 = 7 2 + 4 + 1 = 7 2 + 4 + 1 = 7 , which matches.
So the base is 9, option D.
Watch out
In base ten, 263 + 441 = 704 263 + 441 = 704 263 + 441 = 704 Report a problem with this question
P P P Q Q Q 10 % 10\% 10% Q Q Q R R R 5 % 5\% 5% P P P
A ₦200 B ₦196 C ₦180 D ₦205 E ₦150
Worked solution (try it first) Q Q Q sold at a
5 % 5\% 5% loss, so ₦209 is
95 % 95\% 95% of what
Q Q Q paid:
Q Q Q paid
209 ÷ 0.95 = 209 \div 0.95 = 209 ÷ 0.95 = ₦220.
P P P sold to
Q Q Q at a
10 % 10\% 10% profit, so ₦220 is
110 % 110\% 110% of what
P P P paid:
220 ÷ 1.1 = 200 220 \div 1.1 = 200 220 ÷ 1.1 = 200 .
So the bicycle cost
P P P ₦200, option A.
Watch out
Percentage profit and loss are always of the cost price. Adding 5 % 5\% 5% Report a problem with this question
If the price of oranges is raised by 1 2 \frac12 2 1
A 2 1 2 2\frac12 2 2 1 B 3 1 2 3\frac12 3 2 1 C 5 1 2 5\frac12 5 2 1 D 20k E 21 1 2 21\frac12 21 2 1
Worked solution (try it first) Work in kobo: ₦2.40 is 240k.
With a price of
p p p kobo you buy
240 p \frac{240}{p} p 240 oranges now and
240 p + 1 2 \frac{240}{p + \frac12} p + 2 1 240 after the rise.
The difference is 16:
240 p − 240 p + 1 2 = 16 \dfrac{240}{p} - \dfrac{240}{p + \frac12} = 16 p 240 − p + 2 1 240 = 16 .
Multiply by
p ( p + 1 2 ) p(p + \frac12) p ( p + 2 1 ) :
120 = 16 p 2 + 8 p 120 = 16p^2 + 8p 120 = 16 p 2 + 8 p .
Divide by 8 and rearrange:
2 p 2 + p − 15 = 0 2p^2 + p - 15 = 0 2 p 2 + p − 15 = 0 , which factorises as
( 2 p − 5 ) ( p + 3 ) = 0 (2p - 5)(p + 3) = 0 ( 2 p − 5 ) ( p + 3 ) = 0 .
A price is positive, so
p = 5 2 p = \frac52 p = 2 5 .
An orange costs
2 1 2 2\frac12 2 2 1 k now, option A.
Watch out
Keep the units the same. Using 2.40 with a rise of 1 2 \frac12 2 1 Report a problem with this question
A man invested a total of ₦50,000 in two companies which pay dividends of 6 % 6\% 6% 8 % 8\% 8% 8 % 8\% 8%
A ₦15,000 B ₦29,600 C ₦21,400 D ₦27,800 E ₦35,000
Worked solution (try it first) Let
x x x be the amount at
8 % 8\% 8% .
Then
50 000 − x 50\,000 - x 50 000 − x is at
6 % 6\% 6% .
The yields add up to 3700:
0.06 ( 50 000 − x ) + 0.08 x = 3700 0.06(50\,000 - x) + 0.08x = 3700 0.06 ( 50 000 − x ) + 0.08 x = 3700 , which is
3000 + 0.02 x = 3700 3000 + 0.02x = 3700 3000 + 0.02 x = 3700 .
So
0.02 x = 700 0.02x = 700 0.02 x = 700 and
x = 700 ÷ 0.02 = 35 000 x = 700 \div 0.02 = 35\,000 x = 700 ÷ 0.02 = 35 000 .
He invested ₦35,000 at
8 % 8\% 8% , option E.
Watch out
₦15,000 (option A) is the amount at 6 % 6\% 6% x x x Report a problem with this question
Thirty boys and x x x x x x
Worked solution (try it first) Total score = mean × number.
The boys scored
30 × 6 = 180 30 \times 6 = 180 30 × 6 = 180 and the girls
8 x 8x 8 x .
So
180 + 8 x = 468 180 + 8x = 468 180 + 8 x = 468 , which gives
8 x = 288 8x = 288 8 x = 288 .
Divide by 8:
x = 36 x = 36 x = 36 , option C.
Watch out
Don't use the average of the two means: 6 + 8 2 = 7 \frac{6 + 8}{2} = 7 2 6 + 8 = 7 30 × 6 + 8 x = 468 30 \times 6 + 8x = 468 30 × 6 + 8 x = 468 Also set as JAMB 2016 · UTME · Q34
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The cost of production of an article is made up of labour ₦70, power ₦15, materials ₦30 and miscellaneous ₦5. Find the angle of the sector representing labour in a pie chart.
A 210 ∘ 210^\circ 21 0 ∘ B 105 ∘ 105^\circ 10 5 ∘ C 175 ∘ 175^\circ 17 5 ∘ D 150 ∘ 150^\circ 15 0 ∘ E 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) Total cost: ₦70 + ₦15 + ₦30 + ₦5 = ₦120.
Labour's share of the circle is
70 120 × 360 ∘ \frac{70}{120} \times 360^\circ 120 70 × 36 0 ∘ .
That is
210 ∘ 210^\circ 21 0 ∘ , option A.
Watch out
Use the whole circle, 360 ∘ 360^\circ 36 0 ∘ 70 120 \frac{70}{120} 120 70 180 ∘ 180^\circ 18 0 ∘ 105 ∘ 105^\circ 10 5 ∘ Report a problem with this question
Bola chooses at random a number from 1 to 300. What is the probability that the number is divisible by 4?
A 1 3 \frac13 3 1 B 1 4 \frac14 4 1 C 1 5 \frac15 5 1 D 4 300 \frac4{300} 300 4 E 1 300 \frac1{300} 300 1
Worked solution (try it first) There are 300 equally likely numbers.
The multiples of 4 from 1 to 300 are
4 , 8 , … , 300 4, 8, \ldots, 300 4 , 8 , … , 300 , and
300 ÷ 4 = 75 300 \div 4 = 75 300 ÷ 4 = 75 of them.
So the probability is
75 300 = 1 4 \frac{75}{300} = \frac14 300 75 = 4 1 , option B.
Watch out
Count all the multiples of 4, not just the number 4: there are 75 of them. Taking 4 as the count gives 4 300 \frac{4}{300} 300 4 Report a problem with this question
Find without using logarithm tables the value of log 3 27 − log 1 4 64 log 3 1 81 \dfrac{\log_3 27 - \log_{\frac14} 64}{\log_3 \frac{1}{81}} log 3 81 1 log 3 27 − log 4 1 64
A 7 4 \frac74 4 7 B − 7 4 -\frac74 − 4 7 C − 3 2 -\frac32 − 2 3 D 7 3 \frac73 3 7 E − 1 4 -\frac14 − 4 1
Worked solution (try it first) 27 = 3 3 27 = 3^3 27 = 3 3 , so
log 3 27 = 3 \log_3 27 = 3 log 3 27 = 3 .
64 = 4 3 = ( 1 4 ) − 3 64 = 4^3 = \left(\frac14\right)^{-3} 64 = 4 3 = ( 4 1 ) − 3 , so
log 1 4 64 = − 3 \log_{\frac14} 64 = -3 log 4 1 64 = − 3 .
1 81 = 3 − 4 \frac{1}{81} = 3^{-4} 81 1 = 3 − 4 , so
log 3 1 81 = − 4 \log_3 \frac{1}{81} = -4 log 3 81 1 = − 4 .
The value is
3 − ( − 3 ) − 4 = 6 − 4 \frac{3 - (-3)}{-4} = \frac{6}{-4} − 4 3 − ( − 3 ) = − 4 6 = − 3 2 = -\frac32 = − 2 3 , option C.
Watch out
With base 1 4 \frac14 4 1 log 1 4 64 = − 3 \log_{\frac14} 64 = -3 log 4 1 64 = − 3 + 3 +3 + 3 Also set as JAMB 2017 · UTME · Q35
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A variable point P ( x , y ) P(x, y) P ( x , y ) ( 0 , − 3 ) (0, -3) ( 0 , − 3 ) P P P x x x y y y
A − 3 = y + 4 x + 1 -3 = \dfrac{y + 4}{x + 1} − 3 = x + 1 y + 4 B 4 y = − 3 + x 4y = -3 + x 4 y = − 3 + x C y x = − 3 4 \dfrac yx = -\dfrac34 x y = − 4 3 D y + 3 = 4 x y + 3 = 4x y + 3 = 4 x E 4 y = x + 3 4y = x + 3 4 y = x + 3
Worked solution (try it first) When
x x x goes up by 1,
y y y goes up by 4, so the gradient is
4 1 = 4 \frac41 = 4 1 4 = 4 .
The point
( 0 , − 3 ) (0, -3) ( 0 , − 3 ) is on the
y y y -axis, so the
y y y -intercept is
− 3 -3 − 3 and the line is
y = 4 x − 3 y = 4x - 3 y = 4 x − 3 .
Add 3 to both sides:
y + 3 = 4 x y + 3 = 4x y + 3 = 4 x , option D.
Watch out
The gradient is the change in y y y x x x 4 1 = 4 \frac41 = 4 1 4 = 4 1 4 \frac14 4 1 1 4 \frac14 4 1 4 y = x + … 4y = x + \ldots 4 y = x + … Report a problem with this question
A trader in a country whose currency, the Mont (M), is in base five bought 103 5 103_5 10 3 5 M 14 5 M14_5 M 1 4 5 M 24 5 M24_5 M 2 4 5
A M 103 5 M103_5 M 10 3 5 B M 1030 5 M1030_5 M 103 0 5 C M 102 5 M102_5 M 10 2 5 D M 2002 5 M2002_5 M 200 2 5 E M 3032 5 M3032_5 M 303 2 5
Worked solution (try it first) The gain on each orange is selling price minus cost price:
24 5 − 14 5 = 10 5 24_5 - 14_5 = 10_5 2 4 5 − 1 4 5 = 1 0 5 .
Multiply by the number of oranges:
103 5 × 10 5 103_5 \times 10_5 10 3 5 × 1 0 5 .
Multiplying by
10 5 10_5 1 0 5 (five) shifts every digit one place left, just as
× 10 \times 10 × 10 does in base ten:
1030 5 1030_5 103 0 5 .
So the gain is
M 1030 5 M1030_5 M 103 0 5 , option B.
Watch out
Gain is selling price minus cost price. M 2002 5 M2002_5 M 200 2 5 103 5 × 14 5 103_5 \times 14_5 10 3 5 × 1 4 5 M 3032 5 M3032_5 M 303 2 5 Report a problem with this question
Rationalize 5 7 − 7 5 7 − 5 \dfrac{5\sqrt7 - 7\sqrt5}{\sqrt7 - \sqrt5} 7 − 5 5 7 − 7 5
A − 2 35 -2\sqrt{35} − 2 35 B 4 7 − 6 5 4\sqrt7 - 6\sqrt5 4 7 − 6 5 C − 35 -\sqrt{35} − 35 D 4 7 − 8 5 4\sqrt7 - 8\sqrt5 4 7 − 8 5 E 35 \sqrt{35} 35
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
7 + 5 \sqrt7 + \sqrt5 7 + 5 .
Bottom:
( 7 − 5 ) ( 7 + 5 ) = 7 − 5 (\sqrt7 - \sqrt5)(\sqrt7 + \sqrt5) = 7 - 5 ( 7 − 5 ) ( 7 + 5 ) = 7 − 5 , which is 2.
Top:
( 5 7 − 7 5 ) ( 7 + 5 ) = 35 + 5 35 − 7 35 − 35 (5\sqrt7 - 7\sqrt5)(\sqrt7 + \sqrt5) = 35 + 5\sqrt{35} - 7\sqrt{35} - 35 ( 5 7 − 7 5 ) ( 7 + 5 ) = 35 + 5 35 − 7 35 − 35 , which is
− 2 35 -2\sqrt{35} − 2 35 .
So the value is
− 2 35 2 = − 35 \dfrac{-2\sqrt{35}}{2} = -\sqrt{35} 2 − 2 35 = − 35 , option C.
Watch out
Don't forget the bottom: after rationalising it is 7 − 5 = 2 7 - 5 = 2 7 − 5 = 2 − 2 35 -2\sqrt{35} − 2 35 Also set as JAMB 2016 · UTME · Q35
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Simplify 3 n − 3 n − 1 3 3 × 3 n − 27 × 3 n − 1 \dfrac{3^n - 3^{n-1}}{3^3 \times 3^n - 27 \times 3^{n-1}} 3 3 × 3 n − 27 × 3 n − 1 3 n − 3 n − 1
A 1 B 0 C 1 27 \frac1{27} 27 1 D 3 n − 3 n − 1 3^n - 3^{n-1} 3 n − 3 n − 1 E 2 27 \frac2{27} 27 2
Worked solution (try it first) Write
3 3 3^3 3 3 as 27, so the bottom is
27 × 3 n − 27 × 3 n − 1 27 \times 3^n - 27 \times 3^{n-1} 27 × 3 n − 27 × 3 n − 1 .
Take out the common factor 27: the bottom is
27 ( 3 n − 3 n − 1 ) 27(3^n - 3^{n-1}) 27 ( 3 n − 3 n − 1 ) .
The bracket is the same as the top, so it cancels and leaves
1 27 \frac{1}{27} 27 1 , option C.
Watch out
Factorise before you cancel. The top and the bracket cancel to 1, but the 27 stays in the bottom; forgetting it gives 1 (option A). Report a problem with this question
Factorise 6 x 2 − 14 x − 12 6x^2 - 14x - 12 6 x 2 − 14 x − 12
A 2 ( x + 3 ) ( 3 x − 2 ) 2(x + 3)(3x - 2) 2 ( x + 3 ) ( 3 x − 2 ) B 6 ( x − 2 ) ( x + 1 ) 6(x - 2)(x + 1) 6 ( x − 2 ) ( x + 1 ) C 2 ( x − 3 ) ( 3 x + 2 ) 2(x - 3)(3x + 2) 2 ( x − 3 ) ( 3 x + 2 ) D 6 ( x + 2 ) ( x − 1 ) 6(x + 2)(x - 1) 6 ( x + 2 ) ( x − 1 ) E ( 3 x + 4 ) ( 2 x + 3 ) (3x + 4)(2x + 3) ( 3 x + 4 ) ( 2 x + 3 )
Worked solution (try it first) Take out the common factor 2:
6 x 2 − 14 x − 12 = 2 ( 3 x 2 − 7 x − 6 ) 6x^2 - 14x - 12 = 2(3x^2 - 7x - 6) 6 x 2 − 14 x − 12 = 2 ( 3 x 2 − 7 x − 6 ) .
For
3 x 2 − 7 x − 6 3x^2 - 7x - 6 3 x 2 − 7 x − 6 , find two numbers with product
3 × ( − 6 ) = − 18 3 \times (-6) = -18 3 × ( − 6 ) = − 18 and sum
− 7 -7 − 7 : they are
− 9 -9 − 9 and 2.
Split the middle term and group:
3 x 2 − 9 x + 2 x − 6 = 3 x ( x − 3 ) + 2 ( x − 3 ) 3x^2 - 9x + 2x - 6 = 3x(x - 3) + 2(x - 3) 3 x 2 − 9 x + 2 x − 6 = 3 x ( x − 3 ) + 2 ( x − 3 ) = ( x − 3 ) ( 3 x + 2 ) = (x - 3)(3x + 2) = ( x − 3 ) ( 3 x + 2 ) .
So
6 x 2 − 14 x − 12 = 2 ( x − 3 ) ( 3 x + 2 ) 6x^2 - 14x - 12 = 2(x - 3)(3x + 2) 6 x 2 − 14 x − 12 = 2 ( x − 3 ) ( 3 x + 2 ) , option C.
Watch out
Check the middle term by expanding. Option A, 2 ( x + 3 ) ( 3 x − 2 ) 2(x + 3)(3x - 2) 2 ( x + 3 ) ( 3 x − 2 ) 6 x 2 + 14 x − 12 6x^2 + 14x - 12 6 x 2 + 14 x − 12 Report a problem with this question
A straight line y = m x y = mx y = m x y = x 2 − 12 x + 40 y = x^2 - 12x + 40 y = x 2 − 12 x + 40 ( 5 , 5 ) (5, 5) ( 5 , 5 )
A ( 5 , 6 ) (5, 6) ( 5 , 6 ) B ( 8 , 8 ) (8, 8) ( 8 , 8 ) C ( 8 , 5 ) (8, 5) ( 8 , 5 ) D ( 7 , 7 ) (7, 7) ( 7 , 7 ) E ( 7 , 5 ) (7, 5) ( 7 , 5 )
Worked solution (try it first) The point
( 5 , 5 ) (5, 5) ( 5 , 5 ) is on
y = m x y = mx y = m x , so
5 = 5 m 5 = 5m 5 = 5 m and
m = 1 m = 1 m = 1 .
Where the line meets the curve,
x = x 2 − 12 x + 40 x = x^2 - 12x + 40 x = x 2 − 12 x + 40 , so
x 2 − 13 x + 40 = 0 x^2 - 13x + 40 = 0 x 2 − 13 x + 40 = 0 .
Factorise:
( x − 5 ) ( x − 8 ) = 0 (x - 5)(x - 8) = 0 ( x − 5 ) ( x − 8 ) = 0 , so
x = 5 x = 5 x = 5 (the point you know) or
x = 8 x = 8 x = 8 .
On
y = x y = x y = x ,
x = 8 x = 8 x = 8 gives
y = 8 y = 8 y = 8 .
The other point is
( 8 , 8 ) (8, 8) ( 8 , 8 ) , option B.
Watch out
Find the new y y y y = x y = x y = x ( 8 , 5 ) (8, 5) ( 8 , 5 ) Report a problem with this question
In a racing competition, Musa covered 5 x 5x 5 x ( x + 10 ) (x + 10) ( x + 10 )
A 0 < − x < 15 0 < -x < 15 0 < − x < 15 B − 3 < x < 3 -3 < x < 3 − 3 < x < 3 C 15 < x < 18 15 < x < 18 15 < x < 18 D 0 < x < 15 0 < x < 15 0 < x < 15 E 0 < x < 18 0 < x < 18 0 < x < 18
Worked solution (try it first) Add the two hours: Musa covered
5 x + ( x + 10 ) = 6 x + 10 5x + (x + 10) = 6x + 10 5 x + ( x + 10 ) = 6 x + 10 km.
He came second to Ngozi, so he covered less than her 118 km:
6 x + 10 < 118 6x + 10 < 118 6 x + 10 < 118 .
Subtract 10 from both sides:
6 x < 108 6x < 108 6 x < 108 .
Divide by 6:
x < 18 x < 18 x < 18 .
A distance must be positive, so
x > 0 x > 0 x > 0 as well.
So
0 < x < 18 0 < x < 18 0 < x < 18 , option E.
Watch out
Take the 10 km away before you divide: 6 x < 108 6x < 108 6 x < 108 x < 18 x < 18 x < 18 x x x x > 0 x > 0 x > 0 Report a problem with this question
Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x x x x x x
A x 2 − 5 x − 18 = 0 x^2 - 5x - 18 = 0 x 2 − 5 x − 18 = 0 B x 2 − 20 x + 360 = 0 x^2 - 20x + 360 = 0 x 2 − 20 x + 360 = 0 C x 2 − 21 x − 270 = 0 x^2 - 21x - 270 = 0 x 2 − 21 x − 270 = 0 D 2 x 2 + 42 x − 190 = 0 2x^2 + 42x - 190 = 0 2 x 2 + 42 x − 190 = 0 E 3 x 2 − 31 x + 150 = 0 3x^2 - 31x + 150 = 0 3 x 2 − 31 x + 150 = 0
Worked solution (try it first) In one day Tunde does
1 x \frac1x x 1 of the work and Shola does
1 x + 15 \frac{1}{x + 15} x + 15 1 .
Together they do
1 18 \frac{1}{18} 18 1 .
So
1 x + 1 x + 15 = 1 18 \dfrac1x + \dfrac{1}{x + 15} = \dfrac{1}{18} x 1 + x + 15 1 = 18 1 .
Multiply through by
18 x ( x + 15 ) 18x(x + 15) 18 x ( x + 15 ) :
18 ( x + 15 ) + 18 x = x ( x + 15 ) 18(x + 15) + 18x = x(x + 15) 18 ( x + 15 ) + 18 x = x ( x + 15 ) .
Expand:
36 x + 270 = x 2 + 15 x 36x + 270 = x^2 + 15x 36 x + 270 = x 2 + 15 x .
Bring everything to one side:
x 2 − 21 x − 270 = 0 x^2 - 21x - 270 = 0 x 2 − 21 x − 270 = 0 , option C.
Watch out
Add the rates (the fraction of the job done per day), not the times. Writing x + ( x + 15 ) = 18 x + (x + 15) = 18 x + ( x + 15 ) = 18 Report a problem with this question
The quadratic equation whose roots are 1 − 13 1 - \sqrt{13} 1 − 13 1 + 13 1 + \sqrt{13} 1 + 13
A x 2 + ( 1 − 13 ) x + 1 + 13 = 0 x^2 + (1 - \sqrt{13})x + 1 + \sqrt{13} = 0 x 2 + ( 1 − 13 ) x + 1 + 13 = 0 B x 2 + ( 1 − 13 ) x + 1 − 13 = 0 x^2 + (1 - \sqrt{13})x + 1 - \sqrt{13} = 0 x 2 + ( 1 − 13 ) x + 1 − 13 = 0 C x 2 + 2 x + 12 = 0 x^2 + 2x + 12 = 0 x 2 + 2 x + 12 = 0 D x 2 − 2 x + 12 = 0 x^2 - 2x + 12 = 0 x 2 − 2 x + 12 = 0 E x 2 − 2 x − 12 = 0 x^2 - 2x - 12 = 0 x 2 − 2 x − 12 = 0
Worked solution (try it first) The sum of the roots is
( 1 − 13 ) + ( 1 + 13 ) = 2 (1 - \sqrt{13}) + (1 + \sqrt{13}) = 2 ( 1 − 13 ) + ( 1 + 13 ) = 2 .
The product is a difference of two squares:
( 1 − 13 ) ( 1 + 13 ) = 1 − 13 (1 - \sqrt{13})(1 + \sqrt{13}) = 1 - 13 ( 1 − 13 ) ( 1 + 13 ) = 1 − 13 The equation is
x 2 − ( sum ) x + ( product ) = 0 x^2 - (\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0 , which is
x 2 − 2 x − 12 = 0 x^2 - 2x - 12 = 0 x 2 − 2 x − 12 = 0 , option E.
Watch out
The product is 1 − 13 = − 12 1 - 13 = -12 1 − 13 = − 12 + 12 +12 + 12 13 × 13 = 13 \sqrt{13} \times \sqrt{13} = 13 13 × 13 = 13 + 12 +12 + 12 Report a problem with this question
Find a factor which is common to all three expressions 4 a 2 − 9 b 2 4a^2 - 9b^2 4 a 2 − 9 b 2 a 3 + 27 b 3 a^3 + 27b^3 a 3 + 27 b 3 ( 4 a + 6 b ) 2 (4a + 6b)^2 ( 4 a + 6 b ) 2
A 4 a + 6 b 4a + 6b 4 a + 6 b B 4 a − 6 b 4a - 6b 4 a − 6 b C 2 a + 3 b 2a + 3b 2 a + 3 b D 2 a − 3 b 2a - 3b 2 a − 3 b E none
Worked solution (try it first) Factorise the first and third:
4 a 2 − 9 b 2 = ( 2 a − 3 b ) ( 2 a + 3 b ) 4a^2 - 9b^2 = (2a - 3b)(2a + 3b) 4 a 2 − 9 b 2 = ( 2 a − 3 b ) ( 2 a + 3 b ) and
( 4 a + 6 b ) 2 = 4 ( 2 a + 3 b ) 2 (4a + 6b)^2 = 4(2a + 3b)^2 ( 4 a + 6 b ) 2 = 4 ( 2 a + 3 b ) 2 .
They share
2 a + 3 b 2a + 3b 2 a + 3 b .
The second is a sum of cubes:
a 3 + 27 b 3 = ( a + 3 b ) ( a 2 − 3 a b + 9 b 2 ) a^3 + 27b^3 = (a + 3b)(a^2 - 3ab + 9b^2) a 3 + 27 b 3 = ( a + 3 b ) ( a 2 − 3 ab + 9 b 2 ) .
It has no factor
2 a + 3 b 2a + 3b 2 a + 3 b : putting
a = − 3 2 b a = -\frac32 b a = − 2 3 b gives
− 27 8 b 3 + 27 b 3 -\frac{27}{8}b^3 + 27b^3 − 8 27 b 3 + 27 b 3 , which is not zero.
So no factor is common to all three, option E.
Watch out
2 a + 3 b 2a + 3b 2 a + 3 b Report a problem with this question
If ( x − 2 ) (x - 2) ( x − 2 ) ( x + 1 ) (x + 1) ( x + 1 ) x 3 + p x 2 + q x + 1 x^3 + px^2 + qx + 1 x 3 + p x 2 + q x + 1 p p p q q q
A 0 B − 3 -3 − 3 C 3 D − 17 3 -\frac{17}{3} − 3 17 E − 2 3 -\frac23 − 3 2
Worked solution (try it first) By the factor theorem, the expression is 0 at
x = − 1 x = -1 x = − 1 :
− 1 + p − q + 1 = 0 -1 + p - q + 1 = 0 − 1 + p − q + 1 = 0 , so
p = q p = q p = q .
It is also 0 at
x = 2 x = 2 x = 2 :
8 + 4 p + 2 q + 1 = 0 8 + 4p + 2q + 1 = 0 8 + 4 p + 2 q + 1 = 0 , so
4 p + 2 q = − 9 4p + 2q = -9 4 p + 2 q = − 9 .
Since
q = p q = p q = p , this is
6 p = − 9 6p = -9 6 p = − 9 , so
p = q = − 3 2 p = q = -\frac32 p = q = − 2 3 .
So
p + q = − 3 p + q = -3 p + q = − 3 , option B.
Watch out
Include the constant term 1 in both equations. Leaving it out of the x = 2 x = 2 x = 2 6 p = − 8 6p = -8 6 p = − 8 p + q = − 8 3 p + q = -\frac83 p + q = − 3 8 Report a problem with this question
A cone is formed by bending a sector of a circle with an angle of 210 ∘ 210^\circ 21 0 ∘
A 7.00 cm B 1.75 cm C 21 \sqrt{21} 21 D 3.50 cm E 2 21 2\sqrt{21} 2 21
Worked solution (try it first) The circle has diameter 12 cm, so its radius is 6 cm.
This radius becomes the slant height of the cone.
The arc of the sector becomes the circumference of the base:
210 360 × 2 π × 6 = 2 π r \frac{210}{360} \times 2\pi \times 6 = 2\pi r 360 210 × 2 π × 6 = 2 π r .
Divide both sides by
2 π 2\pi 2 π :
r = 6 × 210 360 = 3.5 r = 6 \times \frac{210}{360} = 3.5 r = 6 × 360 210 = 3.5 cm, option D.
Watch out
Halve the diameter first: the circle's radius is 6 cm. Using 12 cm gives r = 7 r = 7 r = 7 Report a problem with this question
In triangle X Y Z XYZ X Y Z ∠ Y X Z = 120 ∘ \angle YXZ = 120^\circ ∠ Y X Z = 12 0 ∘ X Z = 3 XZ = 3 X Z = 3 Y Z = 5 YZ = 5 Y Z = 5 ∠ X Y Z \angle XYZ ∠ X Y Z
A 29 ∘ 29^\circ 2 9 ∘ B 31 ∘ 20 ′ 31^\circ20' 3 1 ∘ 2 0 ′ C 31 ∘ 31^\circ 3 1 ∘ D 31 ∘ 18 ′ 31^\circ18' 3 1 ∘ 1 8 ′ E 59 ∘ 59^\circ 5 9 ∘
Worked solution (try it first) Use the sine rule.
Y Z = 5 YZ = 5 Y Z = 5 faces the
120 ∘ 120^\circ 12 0 ∘ angle at
X X X , and
X Z = 3 XZ = 3 X Z = 3 faces
∠ Y \angle Y ∠ Y , so
sin Y 3 = sin 120 ∘ 5 \dfrac{\sin Y}{3} = \dfrac{\sin120^\circ}{5} 3 sin Y = 5 sin 12 0 ∘ .
Multiply both sides by 3:
sin Y = 3 × 0.8660 5 \sin Y = \dfrac{3 \times 0.8660}{5} sin Y = 5 3 × 0.8660 So
Y = 31.31 ∘ Y = 31.31^\circ Y = 31.3 1 ∘ .
The decimal part is
0.31 × 60 ≈ 18 ′ 0.31 \times 60 \approx 18' 0.31 × 60 ≈ 1 8 ′ , so
∠ X Y Z = 31 ∘ 18 ′ \angle XYZ = 31^\circ18' ∠ X Y Z = 3 1 ∘ 1 8 ′ , option D.
Watch out
The sine rule gives sin Y \sin Y sin Y sin − 1 \sin^{-1} sin − 1 cos − 1 ( 0.5196 ) \cos^{-1}(0.5196) cos − 1 ( 0.5196 ) 59 ∘ 59^\circ 5 9 ∘ Report a problem with this question
The sides of a triangle are ( x + 4 ) (x + 4) ( x + 4 ) x x x ( x − 4 ) (x - 4) ( x − 4 ) 1 5 \frac15 5 1 x x x
A 24 cm B 20 cm C 28 cm D 88 7 \frac{88}{7} 7 88 E 0 cm
Worked solution (try it first) The largest angle faces the longest side,
x + 4 x + 4 x + 4 .
The cosine rule gives
( x + 4 ) 2 = x 2 + ( x − 4 ) 2 − 2 x ( x − 4 ) × 1 5 (x + 4)^2 = x^2 + (x - 4)^2 - 2x(x - 4) \times \frac15 ( x + 4 ) 2 = x 2 + ( x − 4 ) 2 − 2 x ( x − 4 ) × 5 1 .
Multiply through by 5 and expand:
5 x 2 + 40 x + 80 = 10 x 2 − 40 x + 80 − 2 x 2 + 8 x 5x^2 + 40x + 80 = 10x^2 - 40x + 80 - 2x^2 + 8x 5 x 2 + 40 x + 80 = 10 x 2 − 40 x + 80 − 2 x 2 + 8 x .
Collect terms:
3 x 2 − 72 x = 0 3x^2 - 72x = 0 3 x 2 − 72 x = 0 , so
3 x ( x − 24 ) = 0 3x(x - 24) = 0 3 x ( x − 24 ) = 0 and
x = 0 x = 0 x = 0 or
x = 24 x = 24 x = 24 .
A side of
x − 4 x - 4 x − 4 must be positive, so
x = 24 x = 24 x = 24 cm, option A.
Watch out
Reject x = 0 x = 0 x = 0 − 4 -4 − 4 Report a problem with this question
If a = 2 x 1 − x a = \dfrac{2x}{1 - x} a = 1 − x 2 x b = 1 + x 1 − x b = \dfrac{1 + x}{1 - x} b = 1 − x 1 + x a 2 − b 2 a^2 - b^2 a 2 − b 2
A 3 x + 1 x − 1 \dfrac{3x + 1}{x - 1} x − 1 3 x + 1 B 3 x 2 − 1 ( x − 1 ) 2 \dfrac{3x^2 - 1}{(x - 1)^2} ( x − 1 ) 2 3 x 2 − 1 C 3 x 2 + 1 ( 1 − x ) 2 \dfrac{3x^2 + 1}{(1 - x)^2} ( 1 − x ) 2 3 x 2 + 1 D 5 x 2 − 1 ( 1 − x ) 2 \dfrac{5x^2 - 1}{(1 - x)^2} ( 1 − x ) 2 5 x 2 − 1 E 5 x 2 − 2 x − 1 ( 1 − x ) 2 \dfrac{5x^2 - 2x - 1}{(1 - x)^2} ( 1 − x ) 2 5 x 2 − 2 x − 1
Worked solution (try it first) Both fractions have bottom
1 − x 1 - x 1 − x , so
a 2 − b 2 = 4 x 2 − ( 1 + x ) 2 ( 1 − x ) 2 a^2 - b^2 = \dfrac{4x^2 - (1 + x)^2}{(1 - x)^2} a 2 − b 2 = ( 1 − x ) 2 4 x 2 − ( 1 + x ) 2 .
Expand the top:
4 x 2 − ( 1 + 2 x + x 2 ) = 3 x 2 − 2 x − 1 4x^2 - (1 + 2x + x^2) = 3x^2 - 2x - 1 4 x 2 − ( 1 + 2 x + x 2 ) = 3 x 2 − 2 x − 1 , which factorises as
( 3 x + 1 ) ( x − 1 ) (3x + 1)(x - 1) ( 3 x + 1 ) ( x − 1 ) .
The bottom
( 1 − x ) 2 (1 - x)^2 ( 1 − x ) 2 equals
( x − 1 ) 2 (x - 1)^2 ( x − 1 ) 2 , so one factor
x − 1 x - 1 x − 1 cancels.
So
a 2 − b 2 = 3 x + 1 x − 1 a^2 - b^2 = \dfrac{3x + 1}{x - 1} a 2 − b 2 = x − 1 3 x + 1 , option A.
Watch out
( 1 + x ) 2 = 1 + 2 x + x 2 (1 + x)^2 = 1 + 2x + x^2 ( 1 + x ) 2 = 1 + 2 x + x 2 2 x 2x 2 x 3 x 2 − 1 3x^2 - 1 3 x 2 − 1 Report a problem with this question
In the figure, P Q R S T W PQRSTW P QR S T W Q S QS QS R T RT R T V V V ∠ T V S \angle TVS ∠ T V S
A 60 ∘ 60^\circ 6 0 ∘ B 90 ∘ 90^\circ 9 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 30 ∘ 30^\circ 3 0 ∘ E 80 ∘ 80^\circ 8 0 ∘
Worked solution (try it first) Each interior angle of a regular hexagon is
120 ∘ 120^\circ 12 0 ∘ .
Triangle
Q R S QRS QR S is isosceles (
Q R = R S QR = RS QR = R S ), so
∠ R S Q = 180 ∘ − 120 ∘ 2 \angle RSQ = \frac{180^\circ - 120^\circ}{2} ∠ R S Q = 2 18 0 ∘ − 12 0 ∘ In the same way triangle
R S T RST R S T is isosceles (
R S = S T RS = ST R S = S T ), so
∠ S R T = 30 ∘ \angle SRT = 30^\circ ∠ S R T = 3 0 ∘ .
∠ T V S \angle TVS ∠ T V S is an exterior angle of triangle
V R S VRS V R S , so it equals the sum of the two opposite interior angles:
30 ∘ + 30 ∘ = 60 ∘ 30^\circ + 30^\circ = 60^\circ 3 0 ∘ + 3 0 ∘ = 6 0 ∘ , option A.
Watch out
120 ∘ 120^\circ 12 0 ∘ ∠ R V S \angle RVS ∠ R V S V R S VRS V R S ∠ T V S \angle TVS ∠ T V S R V T RVT R V T 180 ∘ − 120 ∘ = 60 ∘ 180^\circ - 120^\circ = 60^\circ 18 0 ∘ − 12 0 ∘ = 6 0 ∘ Report a problem with this question
Find the integral values of x x x − 3 < 2 − 5 x < 12 -3 < 2 - 5x < 12 − 3 < 2 − 5 x < 12
A − 2 , − 1 -2, -1 − 2 , − 1 B − 2 , 2 -2, 2 − 2 , 2 C − 1 , 0 -1, 0 − 1 , 0 D 0 , 1 0, 1 0 , 1 E 1 , 2 1, 2 1 , 2
Worked solution (try it first) Subtract 2 from all three parts:
− 5 < − 5 x < 10 -5 < -5x < 10 − 5 < − 5 x < 10 .
Divide all three parts by
− 5 -5 − 5 .
Dividing by a negative reverses the signs:
1 > x > − 2 1 > x > -2 1 > x > − 2 .
Read from the left:
− 2 < x < 1 -2 < x < 1 − 2 < x < 1 .
Both ends are strict, so the integers are
− 1 -1 − 1 and
0 0 0 , option C.
Watch out
Dividing by − 5 -5 − 5 − 5 ÷ ( − 5 ) = 1 -5 \div (-5) = 1 − 5 ÷ ( − 5 ) = 1 10 ÷ ( − 5 ) = − 2 10 \div (-5) = -2 10 ÷ ( − 5 ) = − 2 − 1 < x < 2 -1 < x < 2 − 1 < x < 2 Report a problem with this question
If p q + 1 = q 2 pq + 1 = q^2 pq + 1 = q 2 t = 1 p − 1 p q t = \dfrac1p - \dfrac1{pq} t = p 1 − pq 1 t t t q q q
A 1 p − q \dfrac1p - q p 1 − q B 1 q − 1 \dfrac1q - 1 q 1 − 1 C 1 q + 1 \dfrac{1}{q + 1} q + 1 1 D 1 + q 1 + q 1 + q E 1 1 − q \dfrac{1}{1 - q} 1 − q 1
Worked solution (try it first) Put
t t t over one denominator:
t = 1 p − 1 p q t = \frac1p - \frac{1}{pq} t = p 1 − pq 1 = q − 1 p q = \frac{q - 1}{pq} = pq q − 1 .
From
p q + 1 = q 2 pq + 1 = q^2 pq + 1 = q 2 , subtract 1:
p q = q 2 − 1 pq = q^2 - 1 pq = q 2 − 1 .
So
t = q − 1 q 2 − 1 t = \frac{q - 1}{q^2 - 1} t = q 2 − 1 q − 1 = q − 1 ( q − 1 ) ( q + 1 ) = \frac{q - 1}{(q - 1)(q + 1)} = ( q − 1 ) ( q + 1 ) q − 1 , and cancelling
q − 1 q - 1 q − 1 leaves
1 q + 1 \dfrac{1}{q + 1} q + 1 1 , option C.
Watch out
q 2 − 1 = ( q − 1 ) ( q + 1 ) q^2 - 1 = (q - 1)(q + 1) q 2 − 1 = ( q − 1 ) ( q + 1 ) q − 1 q - 1 q − 1 q + 1 q + 1 q + 1 1 q − 1 \frac{1}{q - 1} q − 1 1 1 1 − q \frac{1}{1 - q} 1 − q 1 Report a problem with this question
The cumulative frequency function of the data below is y = c f ( x ) y = cf(x) y = c f ( x ) c f ( 5 ) cf(5) c f ( 5 )
Score
3
4
5
6
7
Frequency
30
32
30
35
20
Worked solution (try it first) c f ( 5 ) cf(5) c f ( 5 ) is the number of scores that are 5 or less, so add the frequencies for scores 3, 4 and 5.
30 + 32 + 30 = 92 30 + 32 + 30 = 92 30 + 32 + 30 = 92 , so
c f ( 5 ) = 92 cf(5) = 92 c f ( 5 ) = 92 , option E.
Watch out
Cumulative frequency is a running total. The frequency of the score 5 alone is 30 (option A); c f ( 5 ) cf(5) c f ( 5 ) Report a problem with this question
A right circular cone has base radius r r r 2 y ∘ 2y^\circ 2 y ∘
A r tan y ∘ r\tan y^\circ r tan y ∘ B r sin y ∘ r\sin y^\circ r sin y ∘ C r cot y ∘ r\cot y^\circ r cot y ∘ D r cos y ∘ r\cos y^\circ r cos y ∘ E r cosec y ∘ r\,\text{cosec}\,y^\circ r cosec y ∘
Worked solution (try it first) The height drops from the vertex to the centre of the base and cuts the vertical angle in half, so the half-angle at the vertex is
y ∘ y^\circ y ∘ .
In that right-angled triangle,
r r r is opposite the angle
y ∘ y^\circ y ∘ and the height
h h h is adjacent:
tan y ∘ = r h \tan y^\circ = \dfrac rh tan y ∘ = h r .
Rearrange:
h = r tan y ∘ h = \dfrac{r}{\tan y^\circ} h = tan y ∘ r = r cot y ∘ = r\cot y^\circ = r cot y ∘ cm, option C.
Watch out
The angle y ∘ y^\circ y ∘ tan y = r h \tan y = \frac rh tan y = h r h r \frac hr r h r tan y ∘ r\tan y^\circ r tan y ∘ Report a problem with this question
Two fair dice are rolled. What is the probability that both show the same number?
A 1 36 \frac1{36} 36 1 B 7 36 \frac7{36} 36 7 C 1 2 \frac12 2 1 D 1 3 \frac13 3 1 E 1 6 \frac16 6 1
Worked solution (try it first) Two dice give
6 × 6 = 36 6 \times 6 = 36 6 × 6 = 36 equally likely outcomes.
The same number on both is a double:
( 1 , 1 ) , ( 2 , 2 ) , … , ( 6 , 6 ) (1, 1), (2, 2), \ldots, (6, 6) ( 1 , 1 ) , ( 2 , 2 ) , … , ( 6 , 6 ) , which is 6 outcomes.
So the probability is
6 36 = 1 6 \frac{6}{36} = \frac16 36 6 = 6 1 , option E.
Watch out
1 36 \frac{1}{36} 36 1 ( 6 , 6 ) (6, 6) ( 6 , 6 ) Report a problem with this question
The larger value of y y y ( y − 1 ) 2 = 4 y − 7 (y - 1)^2 = 4y - 7 ( y − 1 ) 2 = 4 y − 7
Worked solution (try it first) Expand the left side:
y 2 − 2 y + 1 = 4 y − 7 y^2 - 2y + 1 = 4y - 7 y 2 − 2 y + 1 = 4 y − 7 .
Bring everything to one side:
y 2 − 6 y + 8 = 0 y^2 - 6y + 8 = 0 y 2 − 6 y + 8 = 0 .
Factorise:
( y − 2 ) ( y − 4 ) = 0 (y - 2)(y - 4) = 0 ( y − 2 ) ( y − 4 ) = 0 , so
y = 2 y = 2 y = 2 or
y = 4 y = 4 y = 4 .
The larger value is 4, option B.
Watch out
Both 2 and 4 solve the equation; the question asks for the larger one. Choosing 2 (option A) gives the smaller root. Report a problem with this question
Find the x x x y = 2 x + 1 y = 2x + 1 y = 2 x + 1 y = x 2 − 2 x + 1 y = x^2 - 2x + 1 y = x 2 − 2 x + 1
A 1, 1 B 0, − 4 -4 − 4 C 4, 9 D 0, 0 E 0, 4
Worked solution (try it first) At the points where the graphs meet, the
y y y -values are equal:
x 2 − 2 x + 1 = 2 x + 1 x^2 - 2x + 1 = 2x + 1 x 2 − 2 x + 1 = 2 x + 1 .
Bring everything to one side:
x 2 − 4 x = 0 x^2 - 4x = 0 x 2 − 4 x = 0 .
Factorise:
x ( x − 4 ) = 0 x(x - 4) = 0 x ( x − 4 ) = 0 , so
x = 0 x = 0 x = 0 or
x = 4 x = 4 x = 4 , option E.
Watch out
Give the x x x ( 4 , 9 ) (4, 9) ( 4 , 9 ) Report a problem with this question
In the figure, ∠ T S P = ∠ P R Q \angle TSP = \angle PRQ ∠ T S P = ∠ P R Q Q R = 8 QR = 8 QR = 8 P R = 6 PR = 6 P R = 6 S T = 12 ST = 12 S T = 12 S P SP S P
A 4 cm B 16 cm C 9 cm D 14 cm E Impossible: insufficient data
Worked solution (try it first) Triangles
P R Q PRQ P R Q and
P S T PST P S T share
∠ P \angle P ∠ P and have
∠ P R Q = ∠ P S T \angle PRQ = \angle PST ∠ P R Q = ∠ P S T , so they are similar, with
R R R matching
S S S and
Q Q Q matching
T T T .
Corresponding sides are in the same ratio:
P R P S = Q R T S \dfrac{PR}{PS} = \dfrac{QR}{TS} P S P R = T S QR , so
6 S P = 8 12 \dfrac{6}{SP} = \dfrac{8}{12} S P 6 = 12 8 .
Cross-multiply:
8 × S P = 72 8 \times SP = 72 8 × S P = 72 , so
S P = 9 SP = 9 S P = 9 cm, option C.
Watch out
Pair sides that lie opposite equal angles: Q R QR QR T S TS T S P R PR P R P S PS P S P R PR P R T S TS T S 6 12 = 8 S P \frac{6}{12} = \frac{8}{SP} 12 6 = S P 8 S P = 16 SP = 16 S P = 16 Report a problem with this question
The bar chart shows the mark distribution in a class test. Find the number of students in the class.
Worked solution (try it first) The height of each bar is the number of students with marks in that class.
Read the bars: 4, 6, 8, 9, 5 and 2.
Add them:
4 + 6 + 8 + 9 + 5 + 2 = 34 4 + 6 + 8 + 9 + 5 + 2 = 34 4 + 6 + 8 + 9 + 5 + 2 = 34 students, option E.
Watch out
Add the heights of the bars; 9 (option A) is only the tallest bar, and 60 (option C) is the top of the mark scale, not a number of students. Report a problem with this question
In the figure, O O O P Q R S PQRS P QR S P R PR P R S Q SQ S Q P S ∥ R T PS \parallel RT P S ∥ R T ∠ P R T = 135 ∘ \angle PRT = 135^\circ ∠ P R T = 13 5 ∘ ∠ P S Q \angle PSQ ∠ P S Q
A 67 1 2 ∘ 67\frac12^\circ 67 2 1 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 90 ∘ 90^\circ 9 0 ∘ D 33 3 4 ∘ 33\frac34^\circ 33 4 3 ∘ E 22 1 2 ∘ 22\frac12^\circ 22 2 1 ∘
Worked solution (try it first) P S ∥ R T PS \parallel RT P S ∥ R T , so co-interior angles add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ S P R = 180 ∘ − 135 ∘ \angle SPR = 180^\circ - 135^\circ ∠ S P R = 18 0 ∘ − 13 5 ∘ O P = O S OP = OS O P = O S (radii), so triangle
O P S OPS O P S is isosceles and
∠ O S P = ∠ O P S = 45 ∘ \angle OSP = \angle OPS = 45^\circ ∠ O S P = ∠ O P S = 4 5 ∘ .
S Q SQ S Q is a diameter through
O O O , so
∠ P S Q = ∠ P S O = 45 ∘ \angle PSQ = \angle PSO = 45^\circ ∠ P S Q = ∠ P S O = 4 5 ∘ , option B.
Watch out
The 45 ∘ 45^\circ 4 5 ∘ P P P O P S OPS O P S S S S 45 ∘ 45^\circ 4 5 ∘ O O O 180 ∘ − 45 ∘ 2 = 67 1 2 ∘ \frac{180^\circ - 45^\circ}{2} = 67\frac12^\circ 2 18 0 ∘ − 4 5 ∘ = 67 2 1 ∘ Report a problem with this question
X Y Z XYZ X Y Z X W XW X W Y Z YZ Y Z W W W X Z = 5 XZ = 5 X Z = 5 W Z = 4 WZ = 4 W Z = 4 Y Z = 10 YZ = 10 Y Z = 10 X Y XY X Y
A 5 3 5\sqrt3 5 3 B 3 5 3\sqrt5 3 5 C 3 3 3\sqrt3 3 3 D 5 cm E 6 cm
Worked solution (try it first) In the right-angled triangle
X W Z XWZ X W Z , Pythagoras gives
X W = 5 2 − 4 2 = 9 = 3 XW = \sqrt{5^2 - 4^2} = \sqrt9 = 3 X W = 5 2 − 4 2 = 9 = 3 cm.
W W W lies on
Y Z YZ Y Z , so
Y W = Y Z − W Z = 10 − 4 = 6 YW = YZ - WZ = 10 - 4 = 6 Y W = Y Z − W Z = 10 − 4 = 6 cm.
In the right-angled triangle
X W Y XWY X W Y :
X Y = 6 2 + 3 2 = 45 XY = \sqrt{6^2 + 3^2} = \sqrt{45} X Y = 6 2 + 3 2 = 45 .
Simplify:
45 = 9 × 5 = 3 5 \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt5 45 = 9 × 5 = 3 5 cm, option B.
Watch out
Simplify the surd by taking out the square factor: 45 = 9 × 5 45 = 9 \times 5 45 = 9 × 5 3 5 3\sqrt5 3 5 5 × 3 5 \times 3 5 × 3 5 3 5\sqrt3 5 3 75 \sqrt{75} 75 Report a problem with this question
The diameters in centimetres of 20 copper spheres are distributed as shown. What is the mean diameter?
Class boundary (cm)
3.35–3.45
3.45–3.55
3.55–3.65
3.65–3.75
Frequency
3
6
7
4
A 3.40 cm B 3.58 cm C 3.56 cm D 3.62 cm E 3.63 cm
Worked solution (try it first) Use the mid-point of each class: 3.40, 3.50, 3.60 and 3.70.
Multiply by the frequencies and add:
3 ( 3.40 ) + 6 ( 3.50 ) + 7 ( 3.60 ) + 4 ( 3.70 ) = 10.2 + 21 + 25.2 + 14.8 3(3.40) + 6(3.50) + 7(3.60) + 4(3.70) = 10.2 + 21 + 25.2 + 14.8 3 ( 3.40 ) + 6 ( 3.50 ) + 7 ( 3.60 ) + 4 ( 3.70 ) = 10.2 + 21 + 25.2 + 14.8 , which is 71.2.
Divide by the total frequency:
71.2 20 = 3.56 \frac{71.2}{20} = 3.56 20 71.2 = 3.56 cm, option C.
Watch out
Divide ∑ f x \sum fx ∑ f x Report a problem with this question