Paper JAMB 1990 General Maths Objective
Objective paper · 42 questions · partial
JAMB 1990 · UME Topics include Number foundations & fractions, Quadratics & their graphs, Approximation & error, Commercial arithmetic, Indices & standard form, Surds.
Our copy of this paper is missing questions 11, 12, 13, 14, 24, 44, 47, 50.
Sit this paper Answer every question in order, timed if you like (suggested 30 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
1 2 3 4 5 6 7 8 9 10 15 16 17 18 19 20 21 22 23 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 46 48 49 Simplify 4 3 4 − 6 1 4 4 1 5 of 1 1 4 \dfrac{4\frac34 - 6\frac14}{4\frac15 \text{ of } 1\frac14} 4 5 1 of 1 4 1 4 4 3 − 6 4 1
A − 7 7 8 -7\frac78 − 7 8 7 B − 2 7 -\frac27 − 7 2 C − 10 21 -\frac{10}{21} − 21 10 D 10 21 \frac{10}{21} 21 10
Worked solution (try it first) Top:
4 3 4 − 6 1 4 = 19 4 − 25 4 4\frac34 - 6\frac14 = \frac{19}{4} - \frac{25}{4} 4 4 3 − 6 4 1 = 4 19 − 4 25 = − 6 4 = -\frac64 = − 4 6 , which is
− 3 2 -\frac32 − 2 3 .
Bottom: "of" means multiply, so
21 5 × 5 4 = 21 4 \frac{21}{5} \times \frac54 = \frac{21}{4} 5 21 × 4 5 = 4 21 .
Divide by flipping the bottom:
− 3 2 × 4 21 = − 12 42 -\frac32 \times \frac{4}{21} = -\frac{12}{42} − 2 3 × 21 4 = − 42 12 = − 2 7 = -\frac27 = − 7 2 , option B.
Watch out
The fraction line means divide the top by the bottom. Multiplying them instead gives − 3 2 × 21 4 = − 63 8 = − 7 7 8 -\frac32 \times \frac{21}{4} = -\frac{63}{8} = -7\frac78 − 2 3 × 4 21 = − 8 63 = − 7 8 7 Report a problem with this question
The H.C.F. of a 2 b x + a b x 2 a^2bx + abx^2 a 2 b x + ab x 2 a 2 b − b 3 a^2b - b^3 a 2 b − b 3
A b b b B a + b a + b a + b C a ( a + b ) a(a + b) a ( a + b ) D a b x ( a 2 − b 2 ) abx(a^2 - b^2) ab x ( a 2 − b 2 )
Worked solution (try it first) Factorise the first:
a 2 b x + a b x 2 = a b x ( a + x ) a^2bx + abx^2 = abx(a + x) a 2 b x + ab x 2 = ab x ( a + x ) .
Factorise the second:
a 2 b − b 3 = b ( a 2 − b 2 ) a^2b - b^3 = b(a^2 - b^2) a 2 b − b 3 = b ( a 2 − b 2 ) = b ( a − b ) ( a + b ) = b(a - b)(a + b) = b ( a − b ) ( a + b ) .
The only factor in both is
b b b , so the H.C.F. is
b b b , option A.
Watch out
a + b a + b a + b a + x a + x a + x Report a problem with this question
Correct 241.34 × ( 3 × 10 − 3 ) 2 241.34 \times (3 \times 10^{-3})^2 241.34 × ( 3 × 1 0 − 3 ) 2
A 0.0014 B 0.001448 C 0.0022 D 0.002172
Worked solution (try it first) Square the bracket:
( 3 × 10 − 3 ) 2 = 9 × 10 − 6 (3 \times 10^{-3})^2 = 9 \times 10^{-6} ( 3 × 1 0 − 3 ) 2 = 9 × 1 0 − 6 , squaring both the 3 and the power.
Multiply:
241.34 × 9 = 2172.06 241.34 \times 9 = 2172.06 241.34 × 9 = 2172.06 , so the value is
2172.06 × 10 − 6 = 0.00217206 2172.06 \times 10^{-6} = 0.00217206 2172.06 × 1 0 − 6 = 0.00217206 .
To 4 significant figures (starting from the 2), this is 0.002172, option D.
Watch out
Squaring 3 gives 9, not 6. Doubling instead gives 241.34 × 6 × 10 − 6 = 0.001448 241.34 \times 6 \times 10^{-6} = 0.001448 241.34 × 6 × 1 0 − 6 = 0.001448 Report a problem with this question
At what rate would a sum of ₦100.00 deposited for 5 years raise an interest of ₦7.50?
A 1 1 2 % 1\frac12\% 1 2 1 % B 2 1 2 % 2\frac12\% 2 2 1 % C 15 % 15\% 15% D 25 % 25\% 25%
Worked solution (try it first) Use
I = P R T 100 I = \dfrac{PRT}{100} I = 100 P R T and make
R R R the subject:
R = 100 I P T R = \dfrac{100I}{PT} R = P T 100 I .
Put in
I = 7.50 I = 7.50 I = 7.50 ,
P = 100 P = 100 P = 100 and
T = 5 T = 5 T = 5 :
R = 750 500 = 1.5 R = \dfrac{750}{500} = 1.5 R = 500 750 = 1.5 .
So the rate is
1 1 2 % 1\frac12\% 1 2 1 % , option A.
Watch out
Divide by the time, don't multiply. ₦7.50 over 5 years is ₦1.50 a year on ₦100, so 1.5 % 1.5\% 1.5% 15 % 15\% 15% Report a problem with this question
Three children shared a basket of mangoes such that the first child took 1 4 \frac14 4 1 3 4 \frac34 4 3
A 3 16 \frac3{16} 16 3 B 7 16 \frac7{16} 16 7 C 9 16 \frac9{16} 16 9 D 13 16 \frac{13}{16} 16 13
Worked solution (try it first) After the first child takes
1 4 \frac14 4 1 , the remainder is
3 4 \frac34 4 3 of the mangoes.
The second takes
3 4 \frac34 4 3 of the remainder:
3 4 × 3 4 = 9 16 \frac34 \times \frac34 = \frac{9}{16} 4 3 × 4 3 = 16 9 of the mangoes.
The third takes what is left:
1 − 4 16 − 9 16 = 3 16 1 - \frac{4}{16} - \frac{9}{16} = \frac{3}{16} 1 − 16 4 − 16 9 = 16 3 , option A.
Watch out
9 16 \frac{9}{16} 16 9 Report a problem with this question
Simplify and express in standard form 0.00275 × 0.00640 0.025 × 0.08 \dfrac{0.00275 \times 0.00640}{0.025 \times 0.08} 0.025 × 0.08 0.00275 × 0.00640
A 8.8 × 10 − 1 8.8 \times 10^{-1} 8.8 × 1 0 − 1 B 8.8 × 10 2 8.8 \times 10^2 8.8 × 1 0 2 C 8.8 × 10 − 3 8.8 \times 10^{-3} 8.8 × 1 0 − 3 D 8.8 × 10 3 8.8 \times 10^3 8.8 × 1 0 3
Worked solution (try it first) Top in standard form:
2.75 × 10 − 3 × 6.4 × 10 − 3 = 17.6 × 10 − 6 2.75 \times 10^{-3} \times 6.4 \times 10^{-3} = 17.6 \times 10^{-6} 2.75 × 1 0 − 3 × 6.4 × 1 0 − 3 = 17.6 × 1 0 − 6 .
Bottom:
2.5 × 10 − 2 × 8 × 10 − 2 = 20 × 10 − 4 2.5 \times 10^{-2} \times 8 \times 10^{-2} = 20 \times 10^{-4} 2.5 × 1 0 − 2 × 8 × 1 0 − 2 = 20 × 1 0 − 4 .
Divide:
17.6 ÷ 20 = 0.88 17.6 \div 20 = 0.88 17.6 ÷ 20 = 0.88 , and
10 − 6 ÷ 10 − 4 = 10 − 2 10^{-6} \div 10^{-4} = 10^{-2} 1 0 − 6 ÷ 1 0 − 4 = 1 0 − 2 , so the value is
0.88 × 10 − 2 0.88 \times 10^{-2} 0.88 × 1 0 − 2 .
In standard form this is
8.8 × 10 − 3 8.8 \times 10^{-3} 8.8 × 1 0 − 3 , option C.
Watch out
Keep track of the powers of 10. Ignoring them leaves 0.88 = 8.8 × 10 − 1 0.88 = 8.8 \times 10^{-1} 0.88 = 8.8 × 1 0 − 1 10 − 2 10^{-2} 1 0 − 2 Report a problem with this question
Three brothers in a business deal share the profit at the end of the contract. The first received 1 3 \frac13 3 1 2 3 \frac23 3 2
A ₦60,000.00 B ₦54,000.00 C ₦48,000.00 D ₦42,000.00
Worked solution (try it first) After the first brother takes
1 3 \frac13 3 1 , the remainder is
2 3 \frac23 3 2 of the profit.
The second takes
2 3 \frac23 3 2 of the remainder, so the third gets the other
1 3 \frac13 3 1 of it:
1 3 × 2 3 = 2 9 \frac13 \times \frac23 = \frac29 3 1 × 3 2 = 9 2 of the profit.
So
2 9 \frac29 9 2 of the profit is ₦12,000.
One ninth is ₦6,000, and the profit is
9 × 6000 = 54 000 9 \times 6000 = 54\,000 9 × 6000 = 54 000 : ₦54,000.00, option B.
Watch out
The second brother's 2 3 \frac23 3 2 Report a problem with this question
Simplify 160 r 2 + 71 r 4 + 100 r 8 \sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}} 160 r 2 + 71 r 4 + 100 r 8
A 9 r 2 9r^2 9 r 2 B 12 3 r 12\sqrt3\,r 12 3 r C 13 r 13r 13 r D 13 r \sqrt{13}\,r 13 r
Worked solution (try it first) Work from the innermost root:
100 r 8 = 10 r 4 \sqrt{100r^8} = 10r^4 100 r 8 = 10 r 4 .
Next root:
71 r 4 + 10 r 4 = 81 r 4 \sqrt{71r^4 + 10r^4} = \sqrt{81r^4} 71 r 4 + 10 r 4 = 81 r 4 , which is
9 r 2 9r^2 9 r 2 .
Outer root:
160 r 2 + 9 r 2 = 169 r 2 \sqrt{160r^2 + 9r^2} = \sqrt{169r^2} 160 r 2 + 9 r 2 = 169 r 2 , which is
13 r 13r 13 r .
So option C.
Watch out
9 r 2 9r^2 9 r 2 160 r 2 160r^2 160 r 2 Report a problem with this question
Simplify 27 + 3 3 \sqrt{27} + \dfrac{3}{\sqrt3} 27 + 3 3
A 4 3 4\sqrt3 4 3 B 4 3 \frac{4}{\sqrt3} 3 4 C 3 3 3\sqrt3 3 3 D 3 3 4 \frac{3\sqrt3}{4} 4 3 3
Worked solution (try it first) Take out the square factor:
27 = 9 × 3 \sqrt{27} = \sqrt9 \times \sqrt3 27 = 9 × 3 , which is
3 3 3\sqrt3 3 3 .
Rationalise the second term:
3 3 = 3 3 3 \dfrac{3}{\sqrt3} = \dfrac{3\sqrt3}{3} 3 3 = 3 3 3 , which is
3 \sqrt3 3 .
Add the like surds:
3 3 + 3 = 4 3 3\sqrt3 + \sqrt3 = 4\sqrt3 3 3 + 3 = 4 3 , option A.
Watch out
3 3 \frac{3}{\sqrt3} 3 3 3 \sqrt3 3 3 3 3\sqrt3 3 3 Report a problem with this question
Simplify 3 log 6 9 + log 6 12 + log 6 64 − log 6 72 3\log_6 9 + \log_6 12 + \log_6 64 - \log_6 72 3 log 6 9 + log 6 12 + log 6 64 − log 6 72
A 5 B 7776 C log 6 31 \log_6 31 log 6 31 D ( 7776 ) 6 (7776)^6 ( 7776 ) 6
Worked solution (try it first) Move the 3 up as a power:
3 log 6 9 = log 6 9 3 = log 6 729 3\log_6 9 = \log_6 9^3 = \log_6 729 3 log 6 9 = log 6 9 3 = log 6 729 .
Adding logs multiplies and subtracting divides, so the expression is
log 6 729 × 12 × 64 72 \log_6 \frac{729 \times 12 \times 64}{72} log 6 72 729 × 12 × 64 .
12 72 = 1 6 \frac{12}{72} = \frac16 72 12 = 6 1 , so the number inside is
729 × 64 6 = 7776 \frac{729 \times 64}{6} = 7776 6 729 × 64 = 7776 , which is
6 5 6^5 6 5 .
So the value is
log 6 6 5 = 5 \log_6 6^5 = 5 log 6 6 5 = 5 , option A.
Watch out
7776 (option B) is the number inside the log. The question asks for the log itself, log 6 7776 = 5 \log_6 7776 = 5 log 6 7776 = 5 Report a problem with this question
If f ( x − 4 ) = x 2 + 2 x + 3 f(x - 4) = x^2 + 2x + 3 f ( x − 4 ) = x 2 + 2 x + 3 f ( 2 ) f(2) f ( 2 )
Worked solution (try it first) f ( 2 ) f(2) f ( 2 ) needs
x − 4 = 2 x - 4 = 2 x − 4 = 2 , so
x = 6 x = 6 x = 6 .
Put
x = 6 x = 6 x = 6 into the right-hand side:
6 2 + 2 ( 6 ) + 3 = 36 + 12 + 3 6^2 + 2(6) + 3 = 36 + 12 + 3 6 2 + 2 ( 6 ) + 3 = 36 + 12 + 3 .
So
f ( 2 ) = 51 f(2) = 51 f ( 2 ) = 51 , option D.
Watch out
Don't put x = 2 x = 2 x = 2 4 + 4 + 3 = 11 4 + 4 + 3 = 11 4 + 4 + 3 = 11 f ( − 2 ) f(-2) f ( − 2 ) x x x x − 4 x - 4 x − 4 Report a problem with this question
Factorize 9 ( x + y ) 2 − 4 ( x − y ) 2 9(x + y)^2 - 4(x - y)^2 9 ( x + y ) 2 − 4 ( x − y ) 2
A ( x + y ) ( 5 x + y ) (x + y)(5x + y) ( x + y ) ( 5 x + y ) B ( x + y ) 2 (x + y)^2 ( x + y ) 2 C ( x + 5 y ) ( 5 x + y ) (x + 5y)(5x + y) ( x + 5 y ) ( 5 x + y ) D 5 ( x + y ) 2 5(x + y)^2 5 ( x + y ) 2
Worked solution (try it first) Both terms are squares:
9 ( x + y ) 2 = [ 3 ( x + y ) ] 2 9(x + y)^2 = [3(x + y)]^2 9 ( x + y ) 2 = [ 3 ( x + y ) ] 2 and
4 ( x − y ) 2 = [ 2 ( x − y ) ] 2 4(x - y)^2 = [2(x - y)]^2 4 ( x − y ) 2 = [ 2 ( x − y ) ] 2 .
Difference of two squares: the first bracket is
3 x + 3 y − 2 x + 2 y 3x + 3y - 2x + 2y 3 x + 3 y − 2 x + 2 y , which is
x + 5 y x + 5y x + 5 y .
The second bracket is
3 x + 3 y + 2 x − 2 y 3x + 3y + 2x - 2y 3 x + 3 y + 2 x − 2 y , which is
5 x + y 5x + y 5 x + y .
So the expression is
( x + 5 y ) ( 5 x + y ) (x + 5y)(5x + y) ( x + 5 y ) ( 5 x + y ) , option C.
Watch out
Subtract the whole of 2 ( x − y ) 2(x - y) 2 ( x − y ) − 2 ( x − y ) = − 2 x + 2 y -2(x - y) = -2x + 2y − 2 ( x − y ) = − 2 x + 2 y − 2 x − 2 y -2x - 2y − 2 x − 2 y x + y x + y x + y Report a problem with this question
If a 2 + b 2 = 16 a^2 + b^2 = 16 a 2 + b 2 = 16 2 a b = 7 2ab = 7 2 ab = 7 ( a − b ) (a - b) ( a − b )
A 3 , − 3 3, -3 3 , − 3 B 2 , − 2 2, -2 2 , − 2 C 1 , − 1 1, -1 1 , − 1 D 3 , − 1 3, -1 3 , − 1
Worked solution (try it first) Expand:
( a − b ) 2 = a 2 + b 2 − 2 a b (a - b)^2 = a^2 + b^2 - 2ab ( a − b ) 2 = a 2 + b 2 − 2 ab .
Substitute the given values:
( a − b ) 2 = 16 − 7 = 9 (a - b)^2 = 16 - 7 = 9 ( a − b ) 2 = 16 − 7 = 9 .
A number has two square roots, so
a − b = 3 a - b = 3 a − b = 3 or
− 3 -3 − 3 , option A.
Watch out
2 a b 2ab 2 ab 2 × 7 = 14 2 \times 7 = 14 2 × 7 = 14 ( a − b ) 2 = 2 (a - b)^2 = 2 ( a − b ) 2 = 2 Report a problem with this question
Divide x 3 − 2 x 2 − 5 x + 6 x^3 - 2x^2 - 5x + 6 x 3 − 2 x 2 − 5 x + 6 ( x − 1 ) (x - 1) ( x − 1 )
A x 2 − x − 6 x^2 - x - 6 x 2 − x − 6 B x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6 C x 2 − 7 x + 6 x^2 - 7x + 6 x 2 − 7 x + 6 D x 2 − 5 x − 6 x^2 - 5x - 6 x 2 − 5 x − 6
Worked solution (try it first) Use synthetic division with
x = 1 x = 1 x = 1 (the root of
x − 1 x - 1 x − 1 ) on the coefficients
1 , − 2 , − 5 , 6 1, -2, -5, 6 1 , − 2 , − 5 , 6 .
Bring down 1.
Then
1 × 1 − 2 = − 1 1 \times 1 - 2 = -1 1 × 1 − 2 = − 1 , and
− 1 × 1 − 5 = − 6 -1 \times 1 - 5 = -6 − 1 × 1 − 5 = − 6 .
Last,
− 6 × 1 + 6 = 0 -6 \times 1 + 6 = 0 − 6 × 1 + 6 = 0 , so the remainder is 0 and the quotient is
x 2 − x − 6 x^2 - x - 6 x 2 − x − 6 , option A.
Watch out
Divide by x − 1 x - 1 x − 1 + 1 +1 + 1 ( x − 1 ) ( x 2 − x − 6 ) = x 3 − 2 x 2 − 5 x + 6 (x - 1)(x^2 - x - 6) = x^3 - 2x^2 - 5x + 6 ( x − 1 ) ( x 2 − x − 6 ) = x 3 − 2 x 2 − 5 x + 6 Report a problem with this question
If x + 1 x = 4 x + \dfrac1x = 4 x + x 1 = 4 x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1
Worked solution (try it first) Square both sides of
x + 1 x = 4 x + \dfrac1x = 4 x + x 1 = 4 .
The middle term is
2 × x × 1 x = 2 2 \times x \times \frac1x = 2 2 × x × x 1 = 2 , so
x 2 + 2 + 1 x 2 = 16 x^2 + 2 + \dfrac{1}{x^2} = 16 x 2 + 2 + x 2 1 = 16 .
Subtract 2:
x 2 + 1 x 2 = 14 x^2 + \dfrac{1}{x^2} = 14 x 2 + x 2 1 = 14 , option B.
Watch out
( x + 1 x ) 2 (x + \frac1x)^2 ( x + x 1 ) 2 x 2 + 1 x 2 x^2 + \frac1{x^2} x 2 + x 2 1 Report a problem with this question
What must be added to 4 x 2 − 4 4x^2 - 4 4 x 2 − 4
A − 1 x 2 -\frac{1}{x^2} − x 2 1 B 1 x 2 \frac{1}{x^2} x 2 1 C 1 D − 1 -1 − 1
Worked solution (try it first) Treat
− 4 -4 − 4 as the middle term of
( 2 x + k ) 2 = 4 x 2 + 4 k x + k 2 (2x + k)^2 = 4x^2 + 4kx + k^2 ( 2 x + k ) 2 = 4 x 2 + 4 k x + k 2 :
4 k x = − 4 4kx = -4 4 k x = − 4 , so
k = − 1 x k = -\frac1x k = − x 1 .
Then
( 2 x − 1 x ) 2 = 4 x 2 − 4 + 1 x 2 \left(2x - \frac1x\right)^2 = 4x^2 - 4 + \frac{1}{x^2} ( 2 x − x 1 ) 2 = 4 x 2 − 4 + x 2 1 .
So you must add
1 x 2 \frac{1}{x^2} x 2 1 , option B.
Watch out
The added term is k 2 = ( − 1 x ) 2 k^2 = \left(-\frac1x\right)^2 k 2 = ( − x 1 ) 2 − 1 x 2 -\frac{1}{x^2} − x 2 1 Report a problem with this question
Find the solution of the equation x − 8 x + 15 = 0 x - 8\sqrt x + 15 = 0 x − 8 x + 15 = 0
A 3, 5 B − 3 , − 5 -3, -5 − 3 , − 5 C 9, 25 D − 9 , 25 -9, 25 − 9 , 25
Worked solution (try it first) Let
u = x u = \sqrt x u = x , so
x = u 2 x = u^2 x = u 2 and the equation is
u 2 − 8 u + 15 = 0 u^2 - 8u + 15 = 0 u 2 − 8 u + 15 = 0 .
Factorise:
( u − 3 ) ( u − 5 ) = 0 (u - 3)(u - 5) = 0 ( u − 3 ) ( u − 5 ) = 0 , so
x = 3 \sqrt x = 3 x = 3 or
x = 5 \sqrt x = 5 x = 5 .
Square:
x = 9 x = 9 x = 9 or
x = 25 x = 25 x = 25 , option C.
Watch out
3 and 5 (option A) are the values of x \sqrt x x x x x Report a problem with this question
The lengths of the sides of a right-angled triangle are x x x ( 3 x − 1 ) (3x - 1) ( 3 x − 1 ) ( 3 x + 1 ) (3x + 1) ( 3 x + 1 ) x x x
Worked solution (try it first) The longest side,
3 x + 1 3x + 1 3 x + 1 , is the hypotenuse.
By Pythagoras,
( 3 x + 1 ) 2 = x 2 + ( 3 x − 1 ) 2 (3x + 1)^2 = x^2 + (3x - 1)^2 ( 3 x + 1 ) 2 = x 2 + ( 3 x − 1 ) 2 .
Expand:
9 x 2 + 6 x + 1 = x 2 + 9 x 2 − 6 x + 1 9x^2 + 6x + 1 = x^2 + 9x^2 - 6x + 1 9 x 2 + 6 x + 1 = x 2 + 9 x 2 − 6 x + 1 .
Cancel
9 x 2 9x^2 9 x 2 and 1, then collect the
x x x terms:
12 x = x 2 12x = x^2 12 x = x 2 , so
x ( x − 12 ) = 0 x(x - 12) = 0 x ( x − 12 ) = 0 .
A side can't be 0 cm, so
x = 12 x = 12 x = 12 (sides 12, 35 and 37), option D.
Watch out
Expand each bracket fully: ( 3 x + 1 ) 2 = 9 x 2 + 6 x + 1 (3x + 1)^2 = 9x^2 + 6x + 1 ( 3 x + 1 ) 2 = 9 x 2 + 6 x + 1 9 x 2 + 1 9x^2 + 1 9 x 2 + 1 12 x 12x 12 x x 2 = 0 x^2 = 0 x 2 = 0 Report a problem with this question
The perimeter of a rectangular lawn is 24 m. If the area of the lawn is 35 m 2 35\text{ m}^2 35 m 2
Worked solution (try it first) Half the perimeter is length + width:
24 ÷ 2 = 12 24 \div 2 = 12 24 ÷ 2 = 12 m.
You need two numbers that add up to 12 and multiply to 35: they are 7 and 5.
The width is the shorter side, 5 m, option A.
Watch out
7 m (option B) is the length. The width is the shorter side, 5 m. Report a problem with this question
Simplify x x + y + y x − y − x 2 x 2 − y 2 \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{x^2}{x^2 - y^2} x + y x + x − y y − x 2 − y 2 x 2
A x 2 x 2 − y 2 \dfrac{x^2}{x^2 - y^2} x 2 − y 2 x 2 B y 2 x 2 − y 2 \dfrac{y^2}{x^2 - y^2} x 2 − y 2 y 2 C x x 2 − y 2 \dfrac{x}{x^2 - y^2} x 2 − y 2 x D y x 2 − y 2 \dfrac{y}{x^2 - y^2} x 2 − y 2 y
Worked solution (try it first) The LCD is
x 2 − y 2 = ( x + y ) ( x − y ) x^2 - y^2 = (x + y)(x - y) x 2 − y 2 = ( x + y ) ( x − y ) .
Multiply each top by the factor its bottom is missing:
x ( x − y ) + y ( x + y ) − x 2 x(x - y) + y(x + y) - x^2 x ( x − y ) + y ( x + y ) − x 2 .
Expand:
x 2 − x y + x y + y 2 − x 2 x^2 - xy + xy + y^2 - x^2 x 2 − x y + x y + y 2 − x 2 .
The
x y xy x y terms cancel and so do the
x 2 x^2 x 2 terms, leaving
y 2 y^2 y 2 .
So the expression is
y 2 x 2 − y 2 \dfrac{y^2}{x^2 - y^2} x 2 − y 2 y 2 , option B.
Watch out
Multiply each whole top by its missing factor: y ( x + y ) = x y + y 2 y(x + y) = xy + y^2 y ( x + y ) = x y + y 2 y y y y x 2 − y 2 \frac{y}{x^2 - y^2} x 2 − y 2 y Report a problem with this question
Given that x 2 + y 2 + z 2 = 194 x^2 + y^2 + z^2 = 194 x 2 + y 2 + z 2 = 194 z z z x = 7 x = 7 x = 7 y = 3 \sqrt y = 3 y = 3
A 10 \sqrt{10} 10 B 8 C 12.2 D 13.4
Worked solution (try it first) Square both sides of
y = 3 \sqrt y = 3 y = 3 :
y = 9 y = 9 y = 9 , so
y 2 = 81 y^2 = 81 y 2 = 81 .
Substitute:
49 + 81 + z 2 = 194 49 + 81 + z^2 = 194 49 + 81 + z 2 = 194 , so
z 2 = 194 − 130 = 64 z^2 = 194 - 130 = 64 z 2 = 194 − 130 = 64 .
Take the square root:
z = 8 z = 8 z = 8 , option B.
Watch out
y = 3 \sqrt y = 3 y = 3 y = 9 y = 9 y = 9 y 2 = 81 y^2 = 81 y 2 = 81 y 2 y^2 y 2 z 2 = 136 z^2 = 136 z 2 = 136 z ≈ 11.7 z \approx 11.7 z ≈ 11.7 Report a problem with this question
Find the sum of the first twenty terms of the arithmetic progression log a , log a 2 , log a 3 , … \log a, \log a^2, \log a^3, \dots log a , log a 2 , log a 3 , …
A log a 20 \log a^{20} log a 20 B log a 21 \log a^{21} log a 21 C log a 200 \log a^{200} log a 200 D log a 210 \log a^{210} log a 210
Worked solution (try it first) By the laws of logs,
log a k = k log a \log a^k = k\log a log a k = k log a , so the terms are
log a , 2 log a , 3 log a , … \log a, 2\log a, 3\log a, \dots log a , 2 log a , 3 log a , … The sum of the first twenty is
( 1 + 2 + ⋯ + 20 ) log a (1 + 2 + \dots + 20)\log a ( 1 + 2 + ⋯ + 20 ) log a , and
1 + 2 + ⋯ + 20 = 20 × 21 2 1 + 2 + \dots + 20 = \frac{20 \times 21}{2} 1 + 2 + ⋯ + 20 = 2 20 × 21 So the sum is
210 log a = log a 210 210\log a = \log a^{210} 210 log a = log a 210 , option D.
Watch out
The sum 1 + 2 + ⋯ + n 1 + 2 + \dots + n 1 + 2 + ⋯ + n n ( n + 1 ) 2 \frac{n(n + 1)}{2} 2 n ( n + 1 ) 20 × 21 2 = 210 \frac{20 \times 21}{2} = 210 2 20 × 21 = 210 20 × 20 2 \frac{20 \times 20}{2} 2 20 × 20 Report a problem with this question
Find the sum of the first 18 terms of the progression 3 , 6 , 12 , … 3, 6, 12, \dots 3 , 6 , 12 , …
A 3 ( 2 17 − 1 ) 3(2^{17} - 1) 3 ( 2 17 − 1 ) B 3 ( 2 18 ) − 1 3(2^{18}) - 1 3 ( 2 18 ) − 1 C 3 ( 2 18 + 1 ) 3(2^{18} + 1) 3 ( 2 18 + 1 ) D 3 ( 2 18 − 1 ) 3(2^{18} - 1) 3 ( 2 18 − 1 )
Worked solution (try it first) This is a G.P. with
a = 3 a = 3 a = 3 and
r = 6 ÷ 3 = 2 r = 6 \div 3 = 2 r = 6 ÷ 3 = 2 .
Since
r > 1 r > 1 r > 1 , use
S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 ) with
n = 18 n = 18 n = 18 .
So
S 18 = 3 ( 2 18 − 1 ) 2 − 1 S_{18} = \dfrac{3(2^{18} - 1)}{2 - 1} S 18 = 2 − 1 3 ( 2 18 − 1 ) = 3 ( 2 18 − 1 ) = 3(2^{18} - 1) = 3 ( 2 18 − 1 ) , option D.
Watch out
The sum formula uses r n r^n r n r n − 1 r^{n - 1} r n − 1 n n n 2 17 2^{17} 2 17 Report a problem with this question
What is the equation of the quadratic function represented by the graph shown?
A y = x 2 + x − 2 y = x^2 + x - 2 y = x 2 + x − 2 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 + x + 2 y = -x^2 + x + 2 y = − x 2 + x + 2
Worked solution (try it first) The curve opens downward, so the
x 2 x^2 x 2 term is negative: option C or D.
It crosses the
x x x -axis at
− 1 -1 − 1 and 2, so
y = − ( x + 1 ) ( x − 2 ) y = -(x + 1)(x - 2) y = − ( x + 1 ) ( x − 2 ) .
Expand:
− ( x 2 − x − 2 ) = − x 2 + x + 2 -(x^2 - x - 2) = -x^2 + x + 2 − ( x 2 − x − 2 ) = − x 2 + x + 2 , option D.
Watch out
Option B, y = x 2 − x − 2 y = x^2 - x - 2 y = x 2 − x − 2 − ( x + 2 ) ( x − 1 ) -(x + 2)(x - 1) − ( x + 2 ) ( x − 1 ) − 2 -2 − 2 Report a problem with this question
At what value of x x x x 2 + x + 1 x^2 + x + 1 x 2 + x + 1
A − 1 -1 − 1 B − 1 2 -\frac12 − 2 1 C 1 2 \frac12 2 1 D 1
Worked solution (try it first) Complete the square: half of 1 is
1 2 \frac12 2 1 , so
x 2 + x + 1 = ( x + 1 2 ) 2 + 3 4 x^2 + x + 1 = \left(x + \frac12\right)^2 + \frac34 x 2 + x + 1 = ( x + 2 1 ) 2 + 4 3 .
A square is least when it is zero, which happens when
x + 1 2 = 0 x + \frac12 = 0 x + 2 1 = 0 .
So the minimum is at
x = − 1 2 x = -\frac12 x = − 2 1 , option B.
Watch out
The turning point is at x = − b 2 a x = -\frac{b}{2a} x = − 2 a b 1 2 \frac12 2 1 Report a problem with this question
In a trapezium P Q R S PQRS P QR S Q R ∥ P S QR \parallel PS QR ∥ P S 73.5 cm 2 73.5\text{ cm}^2 73.5 cm 2 P S PS P S Q R QR QR P S PS P S
A 21 cm B 17 1 2 17\frac12 17 2 1 C 14 cm D 10 1 2 10\frac12 10 2 1
Worked solution (try it first) Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h .
Put in
Q R = 1 3 P S QR = \frac13PS QR = 3 1 P S :
1 2 ( P S + 1 3 P S ) × 10.5 = 73.5 \frac12\left(PS + \frac13PS\right) \times 10.5 = 73.5 2 1 ( P S + 3 1 P S ) × 10.5 = 73.5 .
Double both sides and divide by 10.5:
4 3 P S = 14 \frac43PS = 14 3 4 P S = 14 .
Multiply by
3 4 \frac34 4 3 :
P S = 10.5 = 10 1 2 PS = 10.5 = 10\frac12 P S = 10.5 = 10 2 1 cm, option D.
Watch out
14 cm (option C) is the sum of the parallel sides, 4 3 P S \frac43PS 3 4 P S 3 4 \frac34 4 3 P S PS P S Report a problem with this question
The angle of a sector of a circle of radius 10.5 cm is 48 ∘ 48^\circ 4 8 ∘ [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 8.8 cm B 25.4 cm C 25.6 cm D 29.8 cm
Worked solution (try it first) Arc length
= θ 360 × 2 π r = \frac{\theta}{360} \times 2\pi r = 360 θ × 2 π r .
The circumference is
2 × 22 7 × 10.5 = 66 2 \times \frac{22}{7} \times 10.5 = 66 2 × 7 22 × 10.5 = 66 cm, so the arc is
48 360 × 66 = 8.8 \frac{48}{360} \times 66 = 8.8 360 48 × 66 = 8.8 cm.
The perimeter adds the two radii:
8.8 + 2 × 10.5 = 29.8 8.8 + 2 \times 10.5 = 29.8 8.8 + 2 × 10.5 = 29.8 cm, option D.
Watch out
8.8 cm (option A) is only the arc. The perimeter of a sector also includes the two radii, 21 cm. Report a problem with this question
In the figure, P S = Q S = R S PS = QS = RS P S = QS = R S ∠ Q S R = 100 ∘ \angle QSR = 100^\circ ∠ QS R = 10 0 ∘ ∠ Q P R \angle QPR ∠ QP R
A 40 ∘ 40^\circ 4 0 ∘ B 50 ∘ 50^\circ 5 0 ∘ C 80 ∘ 80^\circ 8 0 ∘ D 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) S P = S Q = S R SP = SQ = SR S P = S Q = S R , so
P P P ,
Q Q Q and
R R R lie on a circle with centre
S S S .
∠ Q S R \angle QSR ∠ QS R is at the centre and
∠ Q P R \angle QPR ∠ QP R is at the circumference, both on arc
Q R QR QR .
So
∠ Q P R = 1 2 × 100 ∘ \angle QPR = \frac12 \times 100^\circ ∠ QP R = 2 1 × 10 0 ∘ = 50 ∘ = 50^\circ = 5 0 ∘ , option B.
Watch out
The angle at the circumference is half the angle at the centre. Answering 100 ∘ 100^\circ 10 0 ∘ Report a problem with this question
In triangles X Y Z XYZ X Y Z X Q P XQP X QP X P = 4 XP = 4 X P = 4 X Q = 5 XQ = 5 X Q = 5 P Q = Q Y = 3 PQ = QY = 3 P Q = Q Y = 3 ∠ X Q P = ∠ X Z Y \angle XQP = \angle XZY ∠ X QP = ∠ X Z Y Z Y ZY Z Y
Worked solution (try it first) Triangles
X Q P XQP X QP and
X Z Y XZY X Z Y share
∠ X \angle X ∠ X and have
∠ X Q P = ∠ X Z Y \angle XQP = \angle XZY ∠ X QP = ∠ X Z Y , so they are similar, with
Q Q Q matching
Z Z Z and
P P P matching
Y Y Y .
So
P Q Y Z = X P X Y \dfrac{PQ}{YZ} = \dfrac{XP}{XY} Y Z P Q = X Y X P .
Q Q Q lies on
X Y XY X Y , so
X Y = X Q + Q Y = 5 + 3 = 8 XY = XQ + QY = 5 + 3 = 8 X Y = X Q + Q Y = 5 + 3 = 8 cm.
3 Z Y = 4 8 \dfrac{3}{ZY} = \dfrac{4}{8} Z Y 3 = 8 4 , so
Z Y = 6 ZY = 6 Z Y = 6 cm, option B.
Watch out
X P XP X P X Y XY X Y Q Q Q Z Z Z X Q XQ X Q X Q XQ X Q X Y XY X Y 3 Z Y = 5 8 \frac{3}{ZY} = \frac58 Z Y 3 = 8 5 Z Y = 4.8 ZY = 4.8 Z Y = 4.8 Report a problem with this question
Find the length of a side of a rhombus whose diagonals are 6 cm and 8 cm.
Worked solution (try it first) The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with legs
6 ÷ 2 = 3 6 \div 2 = 3 6 ÷ 2 = 3 cm and
8 ÷ 2 = 4 8 \div 2 = 4 8 ÷ 2 = 4 cm.
By Pythagoras, the side is
3 2 + 4 2 = 25 = 5 \sqrt{3^2 + 4^2} = \sqrt{25} = 5 3 2 + 4 2 = 25 = 5 cm, option B.
Watch out
Halve the diagonals first. Using 6 and 8 gives 100 = 10 \sqrt{100} = 10 100 = 10 Report a problem with this question
Each interior angle of a regular polygon is 140 ∘ 140^\circ 14 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 ∘ − 140 ∘ = 40 ∘ 180^\circ - 140^\circ = 40^\circ 18 0 ∘ − 14 0 ∘ = 4 0 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 40 = 9 360 \div 40 = 9 360 ÷ 40 = 9 , option A.
Watch out
Divide 360 ∘ 360^\circ 36 0 ∘ 40 ∘ 40^\circ 4 0 ∘ 140 ∘ 140^\circ 14 0 ∘ Report a problem with this question
In the figure, P Q R S PQRS P QR S P Q T PQT P QT S R T SRT S R T ∠ S P Q = 81 ∘ \angle SPQ = 81^\circ ∠ S P Q = 8 1 ∘ ∠ P T S = 22 ∘ \angle PTS = 22^\circ ∠ P T S = 2 2 ∘ x = ∠ P Q R x = \angle PQR x = ∠ P QR
A 59 ∘ 59^\circ 5 9 ∘ B 77 ∘ 77^\circ 7 7 ∘ C 103 ∘ 103^\circ 10 3 ∘ D 121 ∘ 121^\circ 12 1 ∘
Worked solution (try it first) The angles of triangle
P S T PST P S T add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P S R = 180 ∘ − 81 ∘ − 22 ∘ \angle PSR = 180^\circ - 81^\circ - 22^\circ ∠ P S R = 18 0 ∘ − 8 1 ∘ − 2 2 ∘ P Q R S PQRS P QR S is a cyclic quadrilateral, and
∠ P Q R \angle PQR ∠ P QR is opposite
∠ P S R \angle PSR ∠ P S R .
Opposite angles add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 77 ∘ = 103 ∘ x = 180^\circ - 77^\circ = 103^\circ x = 18 0 ∘ − 7 7 ∘ = 10 3 ∘ , option C.
Watch out
77 ∘ 77^\circ 7 7 ∘ ∠ P S R \angle PSR ∠ P S R Report a problem with this question
In a regular pentagon P Q R S T PQRST P QR S T P R PR P R Q S QS QS O O O ∠ R Q S \angle RQS ∠ R QS
A 36 ∘ 36^\circ 3 6 ∘ B 72 ∘ 72^\circ 7 2 ∘ C 108 ∘ 108^\circ 10 8 ∘ D 144 ∘ 144^\circ 14 4 ∘
Worked solution (try it first) Each interior angle of a regular pentagon is
( 5 − 2 ) × 180 ∘ 5 = 108 ∘ \frac{(5 - 2) \times 180^\circ}{5} = 108^\circ 5 ( 5 − 2 ) × 18 0 ∘ = 10 8 ∘ , so
∠ Q R S = 108 ∘ \angle QRS = 108^\circ ∠ QR S = 10 8 ∘ .
Q R = R S QR = RS QR = R S , so triangle
Q R S QRS QR S is isosceles with equal base angles at
Q Q Q and
S S S .
∠ R Q S = 180 ∘ − 108 ∘ 2 \angle RQS = \frac{180^\circ - 108^\circ}{2} ∠ R QS = 2 18 0 ∘ − 10 8 ∘ = 36 ∘ = 36^\circ = 3 6 ∘ , option A.
Watch out
∠ R Q S \angle RQS ∠ R QS Q R S QRS QR S 108 ∘ 108^\circ 10 8 ∘ 72 ∘ 72^\circ 7 2 ∘ Report a problem with this question
If cos θ = 12 13 \cos\theta = \frac{12}{13} cos θ = 13 12 1 + cot 2 θ 1 + \cot^2\theta 1 + cot 2 θ
A 169 25 \frac{169}{25} 25 169 B 25 169 \frac{25}{169} 169 25 C 169 144 \frac{169}{144} 144 169 D 144 169 \frac{144}{169} 169 144
Worked solution (try it first) Draw a right-angled triangle with adjacent side 12 and hypotenuse 13.
Pythagoras gives the opposite side:
169 − 144 = 5 \sqrt{169 - 144} = 5 169 − 144 = 5 .
So
sin θ = 5 13 \sin\theta = \frac{5}{13} sin θ = 13 5 .
Use the identity
1 + cot 2 θ = cosec 2 θ 1 + \cot^2\theta = \text{cosec}^2\theta 1 + cot 2 θ = cosec 2 θ = 1 sin 2 θ = \dfrac{1}{\sin^2\theta} = sin 2 θ 1 , which is
169 25 \dfrac{169}{25} 25 169 .
That is option A.
Watch out
1 + cot 2 θ 1 + \cot^2\theta 1 + cot 2 θ cosec 2 θ \text{cosec}^2\theta cosec 2 θ sec 2 θ \sec^2\theta sec 2 θ sec 2 θ = 1 cos 2 θ \sec^2\theta = \frac{1}{\cos^2\theta} sec 2 θ = c o s 2 θ 1 169 144 \frac{169}{144} 144 169 Report a problem with this question
In the figure, ∠ Y X Z = 30 ∘ \angle YXZ = 30^\circ ∠ Y X Z = 3 0 ∘ ∠ X Y Z = 105 ∘ \angle XYZ = 105^\circ ∠ X Y Z = 10 5 ∘ X Y = 8 XY = 8 X Y = 8 Y Z YZ Y Z
A 16 2 16\sqrt2 16 2 B 8 2 8\sqrt2 8 2 C 4 2 4\sqrt2 4 2 D 2 2 2\sqrt2 2 2
Worked solution (try it first) The angles add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ Z = 180 ∘ − 30 ∘ − 105 ∘ \angle Z = 180^\circ - 30^\circ - 105^\circ ∠ Z = 18 0 ∘ − 3 0 ∘ − 10 5 ∘ Sine rule:
Y Z YZ Y Z faces the
30 ∘ 30^\circ 3 0 ∘ at
X X X , and
X Y = 8 XY = 8 X Y = 8 faces the
45 ∘ 45^\circ 4 5 ∘ at
Z Z Z .
So
Y Z sin 30 ∘ = 8 sin 45 ∘ \dfrac{YZ}{\sin30^\circ} = \dfrac{8}{\sin45^\circ} sin 3 0 ∘ Y Z = sin 4 5 ∘ 8 .
So
Y Z = 8 × 1 2 2 2 YZ = \dfrac{8 \times \frac12}{\frac{\sqrt2}{2}} Y Z = 2 2 8 × 2 1 = 8 2 = \dfrac{8}{\sqrt2} = 2 8 , which is
4 2 4\sqrt2 4 2 cm, option C.
Watch out
Pair each side with the angle facing it. Putting sin 45 ∘ \sin45^\circ sin 4 5 ∘ sin 30 ∘ \sin30^\circ sin 3 0 ∘ 8 2 8\sqrt2 8 2 Report a problem with this question
In the figure, P Q R PQR P QR P R PR P R O P R S OPRS O P R S O P = 11 OP = 11 O P = 11 O T S OTS O T S O T = 6 OT = 6 O T = 6 T S = 8 TS = 8 T S = 8 T T T [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 125 2 7 cm 2 125\frac27\text{ cm}^2 125 7 2 cm 2 B 149 2 7 cm 2 149\frac27\text{ cm}^2 149 7 2 cm 2 C 243 1 7 cm 2 243\frac17\text{ cm}^2 243 7 1 cm 2 D 267 1 7 cm 2 267\frac17\text{ cm}^2 267 7 1 cm 2
Worked solution (try it first) In the right-angled triangle
O T S OTS O T S , by Pythagoras,
O S = 6 2 + 8 2 = 10 OS = \sqrt{6^2 + 8^2} = 10 O S = 6 2 + 8 2 = 10 cm.
So
P R = 10 PR = 10 P R = 10 cm and the semicircle has radius 5 cm.
Rectangle:
10 × 11 = 110 cm 2 10 \times 11 = 110\text{ cm}^2 10 × 11 = 110 cm 2 .
Triangle
O T S OTS O T S :
1 2 × 6 × 8 = 24 cm 2 \frac12 \times 6 \times 8 = 24\text{ cm}^2 2 1 × 6 × 8 = 24 cm 2 .
Semicircle:
1 2 × 22 7 × 5 2 = 275 7 \frac12 \times \frac{22}{7} \times 5^2 = \frac{275}{7} 2 1 × 7 22 × 5 2 = 7 275 = 39 2 7 cm 2 = 39\frac27\text{ cm}^2 = 39 7 2 cm 2 .
Shaded = rectangle − triangle + semicircle:
110 − 24 + 39 2 7 = 125 2 7 cm 2 110 - 24 + 39\frac27 = 125\frac27\text{ cm}^2 110 − 24 + 39 7 2 = 125 7 2 cm 2 , option A.
Watch out
Take the unshaded triangle away. Forgetting it gives 110 + 39 2 7 = 149 2 7 cm 2 110 + 39\frac27 = 149\frac27\text{ cm}^2 110 + 39 7 2 = 149 7 2 cm 2 Report a problem with this question
A cylindrical metal pipe is 3 cm thick. If the internal radius of the pipe is 10 cm, find the volume of metal used in making 3 m of the pipe.
A 153 π cm 3 153\pi\text{ cm}^3 153 π cm 3 B 207 π cm 3 207\pi\text{ cm}^3 207 π cm 3 C 15,300 π cm 3 15{,}300\pi\text{ cm}^3 15 , 300 π cm 3 D 20,700 π cm 3 20{,}700\pi\text{ cm}^3 20 , 700 π cm 3
Worked solution (try it first) Internal radius 10 cm, wall 3 cm, so the external radius is 13 cm.
The length is 3 m, which is 300 cm.
Area of the ring:
π ( 13 2 − 10 2 ) = π ( 169 − 100 ) \pi(13^2 - 10^2) = \pi(169 - 100) π ( 1 3 2 − 1 0 2 ) = π ( 169 − 100 ) = 69 π cm 2 = 69\pi\text{ cm}^2 = 69 π cm 2 .
Volume:
69 π × 300 = 20 700 π cm 3 69\pi \times 300 = 20\,700\pi\text{ cm}^3 69 π × 300 = 20 700 π cm 3 , option D.
Watch out
Change 3 m into 300 cm before multiplying. Using 3 gives 207 π cm 3 207\pi\text{ cm}^3 207 π cm 3 Report a problem with this question
The heights of two circular cylinders are in the ratio 2 : 3 2 : 3 2 : 3 9 : 8 9 : 8 9 : 8
A 27 : 32 27 : 32 27 : 32 B 27 : 23 27 : 23 27 : 23 C 23 : 32 23 : 32 23 : 32 D 21 : 27 21 : 27 21 : 27
Worked solution (try it first) Volume of a cylinder is
π r 2 h \pi r^2h π r 2 h , so the ratio of volumes is (ratio of radii)
2 ^2 2 × (ratio of heights).
( 9 8 ) 2 × 2 3 = 81 64 × 2 3 \left(\frac98\right)^2 \times \frac23 = \frac{81}{64} \times \frac23 ( 8 9 ) 2 × 3 2 = 64 81 × 3 2 = 162 192 = \frac{162}{192} = 192 162 .
Divide top and bottom by 6:
27 32 \frac{27}{32} 32 27 , so the ratio is
27 : 32 27 : 32 27 : 32 , option A.
Watch out
Square the ratio of the radii. Using 9 8 \frac98 8 9 9 8 × 2 3 = 3 4 \frac98 \times \frac23 = \frac34 8 9 × 3 2 = 4 3 Report a problem with this question
The locus of a point which moves so that it is equidistant from two intersecting straight lines is the
A perpendicular bisector of the two lines B angle bisector of the two lines C bisector of the two lines D line parallel to the two lines
Worked solution (try it first) A point the same distance from two intersecting lines sits on a line that splits the angle between them into two equal parts.
So the locus is the bisector of the angles between the two lines, option B.
Watch out
The perpendicular bisector (option A) is the locus for two points, not two lines. For two intersecting lines, bisect the angle between them. Report a problem with this question
The numbers 4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the sectors representing all numbers equal to or greater than 16.
A 48 ∘ 48^\circ 4 8 ∘ B 84 ∘ 84^\circ 8 4 ∘ C 92 ∘ 92^\circ 9 2 ∘ D 276 ∘ 276^\circ 27 6 ∘
Worked solution (try it first) Total:
4 + 16 + 30 + 20 + 10 + 14 + 26 = 120 4 + 16 + 30 + 20 + 10 + 14 + 26 = 120 4 + 16 + 30 + 20 + 10 + 14 + 26 = 120 .
The numbers equal to or greater than 16 are 16, 30, 20 and 26, which add up to 92.
Their angles together:
92 120 × 360 ∘ = 276 ∘ \frac{92}{120} \times 360^\circ = 276^\circ 120 92 × 36 0 ∘ = 27 6 ∘ , option D.
Watch out
92 (option C) is the sum of the numbers, not of the angles; each unit is worth 3 ∘ 3^\circ 3 ∘ Report a problem with this question
The scores of a group of students in a test are shown. If the average score is 3.5, find the value of x x x
Score
1
2
3
4
5
6
No. of students
1
4
5
6
x x x 2
Worked solution (try it first) Number of students:
1 + 4 + 5 + 6 + x + 2 = 18 + x 1 + 4 + 5 + 6 + x + 2 = 18 + x 1 + 4 + 5 + 6 + x + 2 = 18 + x .
Total score:
1 + 8 + 15 + 24 + 5 x + 12 = 60 + 5 x 1 + 8 + 15 + 24 + 5x + 12 = 60 + 5x 1 + 8 + 15 + 24 + 5 x + 12 = 60 + 5 x .
Mean = total ÷ number:
60 + 5 x 18 + x = 3.5 \frac{60 + 5x}{18 + x} = 3.5 18 + x 60 + 5 x = 3.5 , so
60 + 5 x = 63 + 3.5 x 60 + 5x = 63 + 3.5x 60 + 5 x = 63 + 3.5 x .
Collect terms:
1.5 x = 3 1.5x = 3 1.5 x = 3 , so
x = 2 x = 2 x = 2 , option B.
Watch out
Multiply each score by its number of students: score 5 contributes 5 x 5x 5 x x x x Report a problem with this question
Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?
A 2 3 \frac23 3 2 B 1 2 \frac12 2 1 C 1 3 \frac13 3 1 D 1 4 \frac14 4 1
Worked solution (try it first) List the pairs:
{ 1 , 2 } , { 1 , 3 } , { 1 , 4 } , { 2 , 3 } , { 2 , 4 } , { 3 , 4 } \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\} { 1 , 2 } , { 1 , 3 } , { 1 , 4 } , { 2 , 3 } , { 2 , 4 } , { 3 , 4 } , which is 6 pairs.
An even sum needs two odds or two evens: only
{ 1 , 3 } \{1, 3\} { 1 , 3 } and
{ 2 , 4 } \{2, 4\} { 2 , 4 } .
So the probability is
2 6 = 1 3 \frac26 = \frac13 6 2 = 3 1 , option C.
Watch out
Don't assume even and odd sums are equally likely. List the pairs: only 2 of the 6 have an even sum, not half (option B). Report a problem with this question