Paper JAMB 1985 General Maths Objective
Objective paper · 43 questions · partial
JAMB 1985 · UME Topics include Number foundations & fractions, Commercial arithmetic, Number bases, Approximation & error, Logarithms, Linear & simultaneous equations.
Our copy of this paper is missing questions 4, 14, 15, 22, 27, 45, 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 5 6 7 8 9 10 11 12 13 16 17 18 19 20 21 23 24 25 26 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 46 47 48 49 Arrange the numbers 6 7 \frac67 7 6 13 15 \frac{13}{15} 15 13
A 6 7 < 0.865 < 13 15 \frac67 < 0.865 < \frac{13}{15} 7 6 < 0.865 < 15 13 B 6 7 < 13 15 < 0.865 \frac67 < \frac{13}{15} < 0.865 7 6 < 15 13 < 0.865 C 13 15 < 6 7 < 0.865 \frac{13}{15} < \frac67 < 0.865 15 13 < 7 6 < 0.865 D 13 15 < 0.865 < 6 7 \frac{13}{15} < 0.865 < \frac67 15 13 < 0.865 < 7 6 E 0.865 < 6 7 < 13 15 0.865 < \frac67 < \frac{13}{15} 0.865 < 7 6 < 15 13
Worked solution (try it first) Change the fractions to decimals, to 4 places:
6 7 = 0.8571 \frac67 = 0.8571 7 6 = 0.8571 and
13 15 = 0.8667 \frac{13}{15} = 0.8667 15 13 = 0.8667 .
Write 0.865 as 0.8650 and compare digit by digit:
0.8571 < 0.8650 < 0.8667 0.8571 < 0.8650 < 0.8667 0.8571 < 0.8650 < 0.8667 .
So
6 7 < 0.865 < 13 15 \frac67 < 0.865 < \frac{13}{15} 7 6 < 0.865 < 15 13 , option A.
Watch out
Work to at least 4 decimal places. To 2 places, 0.865 and 13 15 \frac{13}{15} 15 13 0.8650 < 0.8667 0.8650 < 0.8667 0.8650 < 0.8667 Report a problem with this question
A sum of money was invested at 8 % 8\% 8%
A ₦180.00 B ₦165.00 C ₦150.00 D ₦200.00 E ₦250.00
Worked solution (try it first) Simple interest for 4 years at
8 % 8\% 8% is
4 × 8 % = 32 % 4 \times 8\% = 32\% 4 × 8% = 32% of the principal
P P P .
The amount is principal plus interest, so
1.32 P = 330 1.32P = 330 1.32 P = 330 .
Divide both sides by 1.32:
P = 250 P = 250 P = 250 .
The amount invested was ₦250.00, option E.
Watch out
Interest is worked on the principal, not on the amount. Taking 32 % 32\% 32% Report a problem with this question
In the equation 11 x = 1000 x + 101 \dfrac{11}{x} = \dfrac{1000}{x + 101} x 11 = x + 101 1000 x x x
Worked solution (try it first) Change to base ten:
11 2 = 3 11_2 = 3 1 1 2 = 3 ,
1000 2 = 8 1000_2 = 8 100 0 2 = 8 and
101 2 = 5 101_2 = 5 10 1 2 = 5 , so
3 x = 8 x + 5 \frac3x = \frac{8}{x + 5} x 3 = x + 5 8 .
Cross-multiply:
3 x + 15 = 8 x 3x + 15 = 8x 3 x + 15 = 8 x , so
5 x = 15 5x = 15 5 x = 15 and
x = 3 x = 3 x = 3 .
Change back to base two:
3 = 2 + 1 = 11 2 3 = 2 + 1 = 11_2 3 = 2 + 1 = 1 1 2 , option B.
Watch out
Solve in base ten, then convert back. Treating 101 101 101 101 2 101_2 10 1 2 Report a problem with this question
Find, correct to two decimal places, 100 + 1 100 + 3 1000 + 27 10000 100 + \frac{1}{100} + \frac{3}{1000} + \frac{27}{10000} 100 + 100 1 + 1000 3 + 10000 27
A 100.02 B 1000.02 C 100.22 D 100.01 E 100.51
Worked solution (try it first) Change each fraction to a decimal:
1 100 = 0.01 \frac{1}{100} = 0.01 100 1 = 0.01 ,
3 1000 = 0.003 \frac{3}{1000} = 0.003 1000 3 = 0.003 and
27 10000 = 0.0027 \frac{27}{10000} = 0.0027 10000 27 = 0.0027 .
Add:
100 + 0.01 + 0.003 + 0.0027 = 100.0157 100 + 0.01 + 0.003 + 0.0027 = 100.0157 100 + 0.01 + 0.003 + 0.0027 = 100.0157 .
To 2 decimal places, look at the third decimal, 5, and round up: 100.02, option A.
Watch out
A third decimal of 5 means round up. Chopping 100.0157 after two places gives 100.01 (option D). Report a problem with this question
Simplify 1 2 + 1 2 + 1 2 − 1 4 + 1 5 \dfrac12 + \cfrac{1}{2 + \cfrac{1}{2 - \cfrac{1}{4 + \frac15}}} 2 1 + 2 + 2 − 4 + 5 1 1 1 1
A 3 4 \frac34 4 3 B − 1 3 -\frac13 − 3 1 C 169 190 \frac{169}{190} 190 169 D 13 15 \frac{13}{15} 15 13 E 1 21 169 1\frac{21}{169} 1 169 21
Worked solution (try it first) Start at the bottom:
4 + 1 5 = 21 5 4 + \frac15 = \frac{21}{5} 4 + 5 1 = 5 21 , so
1 4 + 1 5 = 5 21 \dfrac{1}{4 + \frac15} = \frac{5}{21} 4 + 5 1 1 = 21 5 .
Next layer:
2 − 5 21 = 37 21 2 - \frac{5}{21} = \frac{37}{21} 2 − 21 5 = 21 37 , so one over it is
21 37 \frac{21}{37} 37 21 .
Next layer:
2 + 21 37 = 95 37 2 + \frac{21}{37} = \frac{95}{37} 2 + 37 21 = 37 95 , so one over it is
37 95 \frac{37}{95} 95 37 .
Finally,
1 2 + 37 95 = 95 + 74 190 \frac12 + \frac{37}{95} = \frac{95 + 74}{190} 2 1 + 95 37 = 190 95 + 74 = 169 190 = \frac{169}{190} = 190 169 , option C.
Watch out
Each "1 over" layer turns the fraction upside down: after 4 + 1 5 = 21 5 4 + \frac15 = \frac{21}{5} 4 + 5 1 = 5 21 5 21 \frac{5}{21} 21 5 21 5 \frac{21}{5} 5 21 Report a problem with this question
If three numbers p p p q q q r r r 6 : 4 : 5 6 : 4 : 5 6 : 4 : 5 3 p − q 4 q + r \dfrac{3p - q}{4q + r} 4 q + r 3 p − q
A 3 2 \frac32 2 3 B 2 3 \frac23 3 2 C 2 D 3 E 18
Worked solution (try it first) Only the ratio matters, so take
p = 6 p = 6 p = 6 ,
q = 4 q = 4 q = 4 and
r = 5 r = 5 r = 5 .
Top:
3 p − q = 18 − 4 = 14 3p - q = 18 - 4 = 14 3 p − q = 18 − 4 = 14 .
Bottom:
4 q + r = 16 + 5 = 21 4q + r = 16 + 5 = 21 4 q + r = 16 + 5 = 21 .
So the value is
14 21 = 2 3 \frac{14}{21} = \frac23 21 14 = 3 2 , option B.
Watch out
Keep the top and bottom the right way round: 21 14 = 3 2 \frac{21}{14} = \frac32 14 21 = 2 3 Report a problem with this question
Without using tables, evaluate log 2 4 + log 4 2 − log 25 5 \log_2 4 + \log_4 2 - \log_{25} 5 log 2 4 + log 4 2 − log 25 5
A 1 2 \frac12 2 1 B 1 5 \frac15 5 1 C 0 D 5 E 2
Worked solution (try it first) 2 2 = 4 2^2 = 4 2 2 = 4 , so
log 2 4 = 2 \log_2 4 = 2 log 2 4 = 2 .
4 1 2 = 2 4^{\frac12} = 2 4 2 1 = 2 , so
log 4 2 = 1 2 \log_4 2 = \frac12 log 4 2 = 2 1 .
In the same way
25 1 2 = 5 25^{\frac12} = 5 2 5 2 1 = 5 , so
log 25 5 = 1 2 \log_{25} 5 = \frac12 log 25 5 = 2 1 .
So the value is
2 + 1 2 − 1 2 = 2 2 + \frac12 - \frac12 = 2 2 + 2 1 − 2 1 = 2 , option E.
Watch out
log 25 5 = 1 2 \log_{25} 5 = \frac12 log 25 5 = 2 1 2 + 1 2 − 2 = 1 2 2 + \frac12 - 2 = \frac12 2 + 2 1 − 2 = 2 1 Also set as JAMB 2016 · UTME · Q1
Report a problem with this question
John gives one third of his money to Janet, who has ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money John had at first.
A ₦45.00 B ₦48.00 C ₦52.00 D ₦58.00 E ₦60.00
Worked solution (try it first) Let John start with ₦
J J J .
After giving away
1 3 J \frac13 J 3 1 J he has
2 3 J \frac23 J 3 2 J , and Janet has
105 + 1 3 J 105 + \frac13 J 105 + 3 1 J .
John now has one fourth of Janet's money:
2 3 J = 1 4 ( 105 + 1 3 J ) \frac23 J = \frac14\left(105 + \frac13 J\right) 3 2 J = 4 1 ( 105 + 3 1 J ) .
Multiply both sides by 4:
8 3 J = 105 + 1 3 J \frac83 J = 105 + \frac13 J 3 8 J = 105 + 3 1 J , so
7 3 J = 105 \frac73 J = 105 3 7 J = 105 .
Multiply by
3 7 \frac37 7 3 :
J = 45 J = 45 J = 45 .
John had ₦45.00, option A.
Watch out
Janet's money grows by John's gift, so compare with 105 + 1 3 J 105 + \frac13 J 105 + 3 1 J 2 3 J = 26.25 \frac23 J = 26.25 3 2 J = 26.25 J = 39.375 J = 39.375 J = 39.375 Report a problem with this question
Find x x x log 9 x = 1.5 \log_9 x = 1.5 log 9 x = 1.5
Worked solution (try it first) Change to index form:
x = 9 1.5 x = 9^{1.5} x = 9 1.5 .
9 1.5 = 9 3 2 9^{1.5} = 9^{\frac32} 9 1.5 = 9 2 3 , which is the square root of 9, cubed:
3 3 = 27 3^3 = 27 3 3 = 27 .
So
x = 27.0 x = 27.0 x = 27.0 , option B.
Watch out
9 1.5 9^{1.5} 9 1.5 9 × 1.5 = 13.5 9 \times 1.5 = 13.5 9 × 1.5 = 13.5 Report a problem with this question
Write h h h a a a b b b c c c d d d a = b ( 1 − c h ) 1 − d h a = \dfrac{b(1 - ch)}{1 - dh} a = 1 − d h b ( 1 − c h )
A h = a − b a d − b c h = \dfrac{a - b}{ad - bc} h = a d − b c a − b B h = a + b a d − b c h = \dfrac{a + b}{ad - bc} h = a d − b c a + b C h = a d − b c a − b h = \dfrac{ad - bc}{a - b} h = a − b a d − b c D h = 1 − b d − b c h = \dfrac{1 - b}{d - bc} h = d − b c 1 − b E h = b − a a d − b c h = \dfrac{b - a}{ad - bc} h = a d − b c b − a
Worked solution (try it first) Multiply both sides by
1 − d h 1 - dh 1 − d h and expand:
a − a d h = b − b c h a - adh = b - bch a − a d h = b − b c h .
Collect the
h h h terms on one side:
b c h − a d h = b − a bch - adh = b - a b c h − a d h = b − a , so
h ( b c − a d ) = b − a h(bc - ad) = b - a h ( b c − a d ) = b − a .
Divide by
b c − a d bc - ad b c − a d :
h = b − a b c − a d h = \frac{b - a}{bc - ad} h = b c − a d b − a .
Multiply top and bottom by
− 1 -1 − 1 to get
h = a − b a d − b c h = \dfrac{a - b}{ad - bc} h = a d − b c a − b , option A.
Watch out
To tidy the signs, change both the top and the bottom. Changing only the bottom gives b − a a d − b c \frac{b - a}{ad - bc} a d − b c b − a Report a problem with this question
22 1 2 % 22\frac12\% 22 2 1 % 17 1 10 % 17\frac1{10}\% 17 10 1 %
A 1 M = 15 57 = \frac{15}{57} = 57 15 B 1 M = 2 11 57 = 2\frac{11}{57} = 2 57 11 C 1 M = 1 18 57 = 1\frac{18}{57} = 1 57 18 D 1 M = 38 1 4 = 38\frac14 = 38 4 1 E 1 M = 384 3 4 = 384\frac34 = 384 4 3
Worked solution (try it first) Write the percentages as decimals:
0.225 0.225 0.225 N
= 0.171 = 0.171 = 0.171 M.
Divide both sides by 0.171 to get 1 M: 1 M
= 0.225 0.171 = \dfrac{0.225}{0.171} = 0.171 0.225 N.
Clear the decimals:
225 171 = 75 57 \dfrac{225}{171} = \dfrac{75}{57} 171 225 = 57 75 = 1 18 57 = 1\frac{18}{57} = 1 57 18 .
So 1 M
= 1 18 57 = 1\frac{18}{57} = 1 57 18 N, option C.
Watch out
To find 1 M, divide by 0.171; don't multiply. Multiplying 22 1 2 22\frac12 22 2 1 1.7 1.7 1.7 38 1 4 38\frac14 38 4 1 Report a problem with this question
Find the values of p p p x 2 − ( p − 2 ) x + 2 p + 1 = 0 x^2 - (p - 2)x + 2p + 1 = 0 x 2 − ( p − 2 ) x + 2 p + 1 = 0
A ( 0 , 12 ) (0, 12) ( 0 , 12 ) B ( 1 , 2 ) (1, 2) ( 1 , 2 ) C ( 21 , 0 ) (21, 0) ( 21 , 0 ) D ( 4 , 5 ) (4, 5) ( 4 , 5 ) E ( 3 , 4 ) (3, 4) ( 3 , 4 )
Worked solution (try it first) Equal roots means
b 2 = 4 a c b^2 = 4ac b 2 = 4 a c .
Here
a = 1 a = 1 a = 1 ,
b = − ( p − 2 ) b = -(p - 2) b = − ( p − 2 ) and
c = 2 p + 1 c = 2p + 1 c = 2 p + 1 .
So
( p − 2 ) 2 = 4 ( 2 p + 1 ) (p - 2)^2 = 4(2p + 1) ( p − 2 ) 2 = 4 ( 2 p + 1 ) , which is
p 2 − 4 p + 4 = 8 p + 4 p^2 - 4p + 4 = 8p + 4 p 2 − 4 p + 4 = 8 p + 4 .
Simplify:
p 2 − 12 p = 0 p^2 - 12p = 0 p 2 − 12 p = 0 , so
p ( p − 12 ) = 0 p(p - 12) = 0 p ( p − 12 ) = 0 .
So
p = 0 p = 0 p = 0 or
p = 12 p = 12 p = 12 , option A.
Watch out
Don't divide p 2 = 12 p p^2 = 12p p 2 = 12 p p p p p = 0 p = 0 p = 0 p ( p − 12 ) = 0 p(p - 12) = 0 p ( p − 12 ) = 0 Also set as JAMB 2016 · UTME · Q2
Report a problem with this question
In a restaurant, the cost of providing a particular type of food is partly constant and partly inversely proportional to the number of people. If the cost per head for 100 people is 30k and the cost for 40 people is 60k, find the cost for 50 people.
Worked solution (try it first) Partly constant and partly inversely proportional to the number
n n n :
C = a + b n C = a + \dfrac{b}{n} C = a + n b , with
C C C in kobo.
n = 100 n = 100 n = 100 :
a + b 100 = 30 a + \frac{b}{100} = 30 a + 100 b = 30 .
n = 40 n = 40 n = 40 :
a + b 40 = 60 a + \frac{b}{40} = 60 a + 40 b = 60 .
Subtract:
b 40 − b 100 = 30 \frac{b}{40} - \frac{b}{100} = 30 40 b − 100 b = 30 , so
3 b 200 = 30 \frac{3b}{200} = 30 200 3 b = 30 and
b = 2000 b = 2000 b = 2000 .
Then
a = 30 − 20 = 10 a = 30 - 20 = 10 a = 30 − 20 = 10 .
For 50 people the varying part is
2000 50 = 40 \frac{2000}{50} = 40 50 2000 = 40 , so
C = 10 + 40 = 50 C = 10 + 40 = 50 C = 10 + 40 = 50 k, option D.
Watch out
Add the constant part back in: 2000 50 = 40 \frac{2000}{50} = 40 50 2000 = 40 Report a problem with this question
The factors of 9 − ( x 2 − 3 x − 1 ) 2 9 - (x^2 - 3x - 1)^2 9 − ( x 2 − 3 x − 1 ) 2
A − ( x − 4 ) ( x + 1 ) ( x − 1 ) ( x − 2 ) -(x - 4)(x + 1)(x - 1)(x - 2) − ( x − 4 ) ( x + 1 ) ( x − 1 ) ( x − 2 ) B ( x − 4 ) ( x − 1 ) ( x − 1 ) ( x + 2 ) (x - 4)(x - 1)(x - 1)(x + 2) ( x − 4 ) ( x − 1 ) ( x − 1 ) ( x + 2 ) C − ( x − 2 ) ( x + 1 ) ( x + 2 ) ( x + 4 ) -(x - 2)(x + 1)(x + 2)(x + 4) − ( x − 2 ) ( x + 1 ) ( x + 2 ) ( x + 4 ) D ( x − 4 ) ( x − 3 ) ( x − 2 ) ( x + 1 ) (x - 4)(x - 3)(x - 2)(x + 1) ( x − 4 ) ( x − 3 ) ( x − 2 ) ( x + 1 ) E ( x − 2 ) ( x + 2 ) ( x − 1 ) ( x + 1 ) (x - 2)(x + 2)(x - 1)(x + 1) ( x − 2 ) ( x + 2 ) ( x − 1 ) ( x + 1 )
Worked solution (try it first) This is a difference of two squares, with
9 = 3 2 9 = 3^2 9 = 3 2 :
3 2 − ( x 2 − 3 x − 1 ) 2 = ( 3 − x 2 + 3 x + 1 ) ( 3 + x 2 − 3 x − 1 ) 3^2 - (x^2 - 3x - 1)^2 = (3 - x^2 + 3x + 1)(3 + x^2 - 3x - 1) 3 2 − ( x 2 − 3 x − 1 ) 2 = ( 3 − x 2 + 3 x + 1 ) ( 3 + x 2 − 3 x − 1 ) .
Tidy each bracket:
( 4 + 3 x − x 2 ) ( x 2 − 3 x + 2 ) (4 + 3x - x^2)(x^2 - 3x + 2) ( 4 + 3 x − x 2 ) ( x 2 − 3 x + 2 ) .
Factorise:
4 + 3 x − x 2 = − ( x 2 − 3 x − 4 ) 4 + 3x - x^2 = -(x^2 - 3x - 4) 4 + 3 x − x 2 = − ( x 2 − 3 x − 4 ) = − ( x − 4 ) ( x + 1 ) = -(x - 4)(x + 1) = − ( x − 4 ) ( x + 1 ) , and
x 2 − 3 x + 2 = ( x − 1 ) ( x − 2 ) x^2 - 3x + 2 = (x - 1)(x - 2) x 2 − 3 x + 2 = ( x − 1 ) ( x − 2 ) .
So the factors are
− ( x − 4 ) ( x + 1 ) ( x − 1 ) ( x − 2 ) -(x - 4)(x + 1)(x - 1)(x - 2) − ( x − 4 ) ( x + 1 ) ( x − 1 ) ( x − 2 ) , option A.
Watch out
In 3 − ( x 2 − 3 x − 1 ) 3 - (x^2 - 3x - 1) 3 − ( x 2 − 3 x − 1 ) 4 + 3 x − x 2 4 + 3x - x^2 4 + 3 x − x 2 − 1 -1 − 1 x 2 x^2 x 2 Report a problem with this question
If 3 2 y − 6 ( 3 y ) = 27 3^{2y} - 6(3^y) = 27 3 2 y − 6 ( 3 y ) = 27 y y y
Worked solution (try it first) Then
3 2 y = ( 3 y ) 2 = u 2 3^{2y} = (3^y)^2 = u^2 3 2 y = ( 3 y ) 2 = u 2 , and the equation becomes
u 2 − 6 u − 27 = 0 u^2 - 6u - 27 = 0 u 2 − 6 u − 27 = 0 .
Factorise:
( u − 9 ) ( u + 3 ) = 0 (u - 9)(u + 3) = 0 ( u − 9 ) ( u + 3 ) = 0 , so
u = 9 u = 9 u = 9 or
u = − 3 u = -3 u = − 3 .
A power of 3 is always positive, so
3 y = − 3 3^y = -3 3 y = − 3 has no solution.
From
3 y = 9 = 3 2 3^y = 9 = 3^2 3 y = 9 = 3 2 ,
y = 2 y = 2 y = 2 , option C.
Watch out
Reject u = − 3 u = -3 u = − 3 3 y 3^y 3 y y = − 1 y = -1 y = − 1 3 − 1 = 1 3 3^{-1} = \frac13 3 − 1 = 3 1 − 3 -3 − 3 Report a problem with this question
Factorize a b x 2 + 8 y − 4 b x − 2 a x y abx^2 + 8y - 4bx - 2axy ab x 2 + 8 y − 4 b x − 2 a x y
A ( a x − 4 ) ( b x − 2 y ) (ax - 4)(bx - 2y) ( a x − 4 ) ( b x − 2 y ) B ( a x + b ) ( x − 8 y ) (ax + b)(x - 8y) ( a x + b ) ( x − 8 y ) C ( a x − 2 y ) ( b y − 4 ) (ax - 2y)(by - 4) ( a x − 2 y ) ( b y − 4 ) D ( a b x − 4 ) ( x − 2 y ) (abx - 4)(x - 2y) ( ab x − 4 ) ( x − 2 y ) E ( b x − 4 ) ( a x − 2 y ) (bx - 4)(ax - 2y) ( b x − 4 ) ( a x − 2 y )
Worked solution (try it first) Group the terms that share
b b b and those that share
y y y :
( a b x 2 − 4 b x ) + ( 8 y − 2 a x y ) (abx^2 - 4bx) + (8y - 2axy) ( ab x 2 − 4 b x ) + ( 8 y − 2 a x y ) .
Take out the common factors:
b x ( a x − 4 ) − 2 y ( a x − 4 ) bx(ax - 4) - 2y(ax - 4) b x ( a x − 4 ) − 2 y ( a x − 4 ) .
Take out the common bracket:
( a x − 4 ) ( b x − 2 y ) (ax - 4)(bx - 2y) ( a x − 4 ) ( b x − 2 y ) , option A.
Watch out
Expand to check. Option E, ( b x − 4 ) ( a x − 2 y ) (bx - 4)(ax - 2y) ( b x − 4 ) ( a x − 2 y ) a b x 2 − 2 b x y − 4 a x + 8 y abx^2 - 2bxy - 4ax + 8y ab x 2 − 2 b x y − 4 a x + 8 y − 2 b x y -2bxy − 2 b x y − 4 a x -4ax − 4 a x Report a problem with this question
At what real value of x x x y = x 3 + x y = x^3 + x y = x 3 + x y = x 2 + 1 y = x^2 + 1 y = x 2 + 1
Worked solution (try it first) The curves meet where the
y y y -values are equal:
x 3 + x = x 2 + 1 x^3 + x = x^2 + 1 x 3 + x = x 2 + 1 , so
x 3 − x 2 + x − 1 = 0 x^3 - x^2 + x - 1 = 0 x 3 − x 2 + x − 1 = 0 .
Group the terms:
x 2 ( x − 1 ) + ( x − 1 ) = 0 x^2(x - 1) + (x - 1) = 0 x 2 ( x − 1 ) + ( x − 1 ) = 0 , so
( x − 1 ) ( x 2 + 1 ) = 0 (x - 1)(x^2 + 1) = 0 ( x − 1 ) ( x 2 + 1 ) = 0 .
x 2 + 1 x^2 + 1 x 2 + 1 is always positive, so the only real solution is
x = 1 x = 1 x = 1 , option E.
Watch out
x 2 + 1 = 0 x^2 + 1 = 0 x 2 + 1 = 0 x = − 1 x = -1 x = − 1 y = x 3 + x y = x^3 + x y = x 3 + x − 2 -2 − 2 y = x 2 + 1 y = x^2 + 1 y = x 2 + 1 Report a problem with this question
If the quadratic function 3 x 2 − 7 x + R 3x^2 - 7x + R 3 x 2 − 7 x + R R R R
A 49 24 \frac{49}{24} 24 49 B 49 3 \frac{49}{3} 3 49 C 49 6 \frac{49}{6} 6 49 D 49 12 \frac{49}{12} 12 49 E 49 36 \frac{49}{36} 36 49
Worked solution (try it first) A quadratic is a perfect square when it has equal roots, so
b 2 = 4 a c b^2 = 4ac b 2 = 4 a c .
Here
a = 3 a = 3 a = 3 ,
b = − 7 b = -7 b = − 7 and
c = R c = R c = R :
49 = 4 × 3 × R = 12 R 49 = 4 \times 3 \times R = 12R 49 = 4 × 3 × R = 12 R .
So
R = 49 12 R = \frac{49}{12} R = 12 49 , option D.
Watch out
The condition is b 2 = 4 a c b^2 = 4ac b 2 = 4 a c 49 = 3 R 49 = 3R 49 = 3 R R = 49 3 R = \frac{49}{3} R = 3 49 Report a problem with this question
Solve for ( x , y ) (x, y) ( x , y ) 2 x + y = 4 2x + y = 4 2 x + y = 4 x 2 + x y = − 12 x^2 + xy = -12 x 2 + x y = − 12
A ( 6 , − 8 ) (6, -8) ( 6 , − 8 ) ( − 2 , 8 ) (-2, 8) ( − 2 , 8 ) B ( 3 , − 4 ) (3, -4) ( 3 , − 4 ) ( − 1 , 4 ) (-1, 4) ( − 1 , 4 ) C ( 8 , − 4 ) (8, -4) ( 8 , − 4 ) ( − 1 , 4 ) (-1, 4) ( − 1 , 4 ) D ( − 8 , 6 ) (-8, 6) ( − 8 , 6 ) ( 8 , − 2 ) (8, -2) ( 8 , − 2 ) E ( − 4 , 3 ) (-4, 3) ( − 4 , 3 ) ( 4 , − 1 ) (4, -1) ( 4 , − 1 )
Worked solution (try it first) Make
y y y the subject of the linear equation:
y = 4 − 2 x y = 4 - 2x y = 4 − 2 x .
Substitute into the second:
x 2 + x ( 4 − 2 x ) = − 12 x^2 + x(4 - 2x) = -12 x 2 + x ( 4 − 2 x ) = − 12 , which is
− x 2 + 4 x = − 12 -x^2 + 4x = -12 − x 2 + 4 x = − 12 .
Rearrange and factorise:
x 2 − 4 x − 12 = 0 x^2 - 4x - 12 = 0 x 2 − 4 x − 12 = 0 , so
( x − 6 ) ( x + 2 ) = 0 (x - 6)(x + 2) = 0 ( x − 6 ) ( x + 2 ) = 0 and
x = 6 x = 6 x = 6 or
x = − 2 x = -2 x = − 2 .
Then
y = 4 − 2 x y = 4 - 2x y = 4 − 2 x gives
y = − 8 y = -8 y = − 8 and
y = 8 y = 8 y = 8 .
So the solutions are
( 6 , − 8 ) (6, -8) ( 6 , − 8 ) and
( − 2 , 8 ) (-2, 8) ( − 2 , 8 ) , option A.
Watch out
Write each pair as ( x , y ) (x, y) ( x , y ) ( − 8 , 6 ) (-8, 6) ( − 8 , 6 ) ( 8 , − 2 ) (8, -2) ( 8 , − 2 ) x = − 8 x = -8 x = − 8 2 x + y = 4 2x + y = 4 2 x + y = 4 Report a problem with this question
Solve the simultaneous equations 2 x − 3 y + 10 = 10 x − 6 y = 5 2x - 3y + 10 = 10x - 6y = 5 2 x − 3 y + 10 = 10 x − 6 y = 5
A x = 2 1 2 x = 2\frac12 x = 2 2 1 y = 3 1 3 y = 3\frac13 y = 3 3 1 B x = 3 1 2 x = 3\frac12 x = 3 2 1 y = 2 1 3 y = 2\frac13 y = 2 3 1 C x = 2 1 4 x = 2\frac14 x = 2 4 1 y = 3 y = 3 y = 3 D x = 3 1 2 x = 3\frac12 x = 3 2 1 y = 2 3 5 y = 2\frac35 y = 2 5 3 E x = 2 1 2 x = 2\frac12 x = 2 2 1 y = 2 1 3 y = 2\frac13 y = 2 3 1
Worked solution (try it first) Each part of the chain equals 5.
From
2 x − 3 y + 10 = 5 2x - 3y + 10 = 5 2 x − 3 y + 10 = 5 , take 10 from both sides:
2 x − 3 y = − 5 2x - 3y = -5 2 x − 3 y = − 5 .
The other part gives
10 x − 6 y = 5 10x - 6y = 5 10 x − 6 y = 5 .
Double the first equation:
4 x − 6 y = − 10 4x - 6y = -10 4 x − 6 y = − 10 .
Subtract it from
10 x − 6 y = 5 10x - 6y = 5 10 x − 6 y = 5 so
y y y cancels:
6 x = 15 6x = 15 6 x = 15 , so
x = 2 1 2 x = 2\frac12 x = 2 2 1 .
Then
3 y = 2 x + 5 = 10 3y = 2x + 5 = 10 3 y = 2 x + 5 = 10 , so
y = 3 1 3 y = 3\frac13 y = 3 3 1 .
The answer is option A.
Watch out
Move the 10 across with its sign changed: 2 x − 3 y = 5 − 10 = − 5 2x - 3y = 5 - 10 = -5 2 x − 3 y = 5 − 10 = − 5 2 x − 3 y = 5 2x - 3y = 5 2 x − 3 y = 5 x = − 5 6 x = -\frac56 x = − 6 5 Report a problem with this question
If f ( x − 2 ) = 4 x 2 + x + 7 f(x - 2) = 4x^2 + x + 7 f ( x − 2 ) = 4 x 2 + x + 7 f ( 1 ) f(1) f ( 1 )
Worked solution (try it first) f ( 1 ) f(1) f ( 1 ) means the bracket
x − 2 x - 2 x − 2 must equal 1, so
x = 3 x = 3 x = 3 .
Put
x = 3 x = 3 x = 3 into the right-hand side:
4 ( 3 ) 2 + 3 + 7 = 36 + 10 4(3)^2 + 3 + 7 = 36 + 10 4 ( 3 ) 2 + 3 + 7 = 36 + 10 .
So
f ( 1 ) = 46 f(1) = 46 f ( 1 ) = 46 , option D.
Watch out
Don't put x = 1 x = 1 x = 1 4 x 2 + x + 7 4x^2 + x + 7 4 x 2 + x + 7 f ( − 1 ) f(-1) f ( − 1 ) x x x x − 2 = 1 x - 2 = 1 x − 2 = 1 Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z X Y = 13 XY = 13 X Y = 13 Y Z = 9 YZ = 9 Y Z = 9 X Z = 11 XZ = 11 X Z = 11 ∠ X Y Z = θ ∘ \angle XYZ = \theta^\circ ∠ X Y Z = θ ∘ cos θ \cos\theta cos θ
A 4 39 \frac4{39} 39 4 B 43 39 \frac{43}{39} 39 43 C 209 286 \frac{209}{286} 286 209 D 1 6 \frac16 6 1 E 43 78 \frac{43}{78} 78 43
Worked solution (try it first) θ \theta θ is at
Y Y Y , between
X Y = 13 XY = 13 X Y = 13 and
Y Z = 9 YZ = 9 Y Z = 9 .
The side facing it is
X Z = 11 XZ = 11 X Z = 11 .
Cosine rule:
cos θ = 13 2 + 9 2 − 11 2 2 × 13 × 9 \cos\theta = \dfrac{13^2 + 9^2 - 11^2}{2 \times 13 \times 9} cos θ = 2 × 13 × 9 1 3 2 + 9 2 − 1 1 2 , which is
169 + 81 − 121 234 = 129 234 \dfrac{169 + 81 - 121}{234} = \dfrac{129}{234} 234 169 + 81 − 121 = 234 129 .
Divide top and bottom by 3:
cos θ = 43 78 \cos\theta = \frac{43}{78} cos θ = 78 43 , option E.
Watch out
Keep the 2 in 2 a c 2ac 2 a c 2 × 13 × 9 = 234 2 \times 13 \times 9 = 234 2 × 13 × 9 = 234 13 × 9 = 117 13 \times 9 = 117 13 × 9 = 117 129 117 = 43 39 \frac{129}{117} = \frac{43}{39} 117 129 = 39 43 Report a problem with this question
The number of goals scored by a football team in 20 matches is shown below. What are the mean and the mode respectively?
No. of goals
0
1
2
3
4
5
No. of matches
3
5
7
4
1
0
A ( 1.75 , 5 ) (1.75, 5) ( 1.75 , 5 ) B ( 1.75 , 2 ) (1.75, 2) ( 1.75 , 2 ) C ( 1.75 , 1 ) (1.75, 1) ( 1.75 , 1 ) D ( 2 , 2 ) (2, 2) ( 2 , 2 ) E ( 2 , 1 ) (2, 1) ( 2 , 1 )
Worked solution (try it first) Total goals:
0 ( 3 ) + 1 ( 5 ) + 2 ( 7 ) + 3 ( 4 ) + 4 ( 1 ) + 5 ( 0 ) = 0 + 5 + 14 + 12 + 4 0(3) + 1(5) + 2(7) + 3(4) + 4(1) + 5(0) = 0 + 5 + 14 + 12 + 4 0 ( 3 ) + 1 ( 5 ) + 2 ( 7 ) + 3 ( 4 ) + 4 ( 1 ) + 5 ( 0 ) = 0 + 5 + 14 + 12 + 4 , which is 35.
Mean
= 35 20 = 1.75 = \frac{35}{20} = 1.75 = 20 35 = 1.75 .
The mode is the number of goals with the highest frequency: 2 goals, in 7 matches.
So the mean and mode are
( 1.75 , 2 ) (1.75, 2) ( 1.75 , 2 ) , option B.
Watch out
The mode is the number of goals that happened most often (2), not the highest frequency (7) or the largest number of goals (5, option A). Report a problem with this question
If the hypotenuse of a right-angled isosceles triangle is 2, what is the length of each of the other sides?
A 2 \sqrt2 2 B 1 2 \frac1{\sqrt2} 2 1 C 2 2 2\sqrt2 2 2 D 1 E 2 − 1 \sqrt2 - 1 2 − 1
Worked solution (try it first) Isosceles means the two shorter sides are equal.
Pythagoras:
a 2 + a 2 = 2 2 a^2 + a^2 = 2^2 a 2 + a 2 = 2 2 , so
2 a 2 = 4 2a^2 = 4 2 a 2 = 4 and
a 2 = 2 a^2 = 2 a 2 = 2 .
So
a = 2 a = \sqrt2 a = 2 , option A.
Watch out
Don't halve the hypotenuse: the two sides are not half of it each. Halving gives 1 (option D), but 1 2 + 1 2 = 2 1^2 + 1^2 = 2 1 2 + 1 2 = 2 Report a problem with this question
If two fair coins are tossed, what is the probability of getting at least one head?
A 1 4 \frac14 4 1 B 1 2 \frac12 2 1 C 1 D 2 3 \frac23 3 2 E 3 4 \frac34 4 3
Worked solution (try it first) Two coins give 4 equally likely outcomes: HH, HT, TH, TT.
"At least one head" is every outcome except TT, so it is the complement of no heads:
1 − 1 4 1 - \frac14 1 − 4 1 .
So the probability is
3 4 \frac34 4 3 , option E.
Watch out
"At least one" includes two heads. Counting only HT and TH (exactly one head) gives 2 4 = 1 2 \frac24 = \frac12 4 2 = 2 1 Report a problem with this question
The ratio of the lengths of two similar rectangular blocks is 2 : 3 2 : 3 2 : 3 351 cm 3 351\text{ cm}^3 351 cm 3
A 234.00 cm 3 234.00\text{ cm}^3 234.00 cm 3 B 526.50 cm 3 526.50\text{ cm}^3 526.50 cm 3 C 166.00 cm 3 166.00\text{ cm}^3 166.00 cm 3 D 729.75 cm 3 729.75\text{ cm}^3 729.75 cm 3 E 104.00 cm 3 104.00\text{ cm}^3 104.00 cm 3
Worked solution (try it first) For similar solids, volumes are in the ratio of the cubes of the lengths:
2 3 : 3 3 = 8 : 27 2^3 : 3^3 = 8 : 27 2 3 : 3 3 = 8 : 27 .
So the smaller volume is
351 × 8 27 = 13 × 8 351 \times \frac{8}{27} = 13 \times 8 351 × 27 8 = 13 × 8 = 104 cm 3 = 104\text{ cm}^3 = 104 cm 3 , option E.
Watch out
Cube the length ratio for volumes. Using 2 3 \frac23 3 2 351 × 2 3 = 234 cm 3 351 \times \frac23 = 234\text{ cm}^3 351 × 3 2 = 234 cm 3 Report a problem with this question
The bearing of a bird on a tree from a hunter on the ground is N72 ∘ 72^\circ 7 2 ∘
A S18 ∘ 18^\circ 1 8 ∘ B S72 ∘ 72^\circ 7 2 ∘ C S72 ∘ 72^\circ 7 2 ∘ D S27 ∘ 27^\circ 2 7 ∘ E S27 ∘ 27^\circ 2 7 ∘
Worked solution (try it first) The hunter is in exactly the opposite direction from the bird, so turn the bearing round by
180 ∘ 180^\circ 18 0 ∘ .
For a compass bearing this means swapping N with S and E with W, and keeping the angle.
So N
72 ∘ 72^\circ 7 2 ∘ E becomes S
72 ∘ 72^\circ 7 2 ∘ W, option B.
Watch out
Swap both letters. Changing only N to S gives S72 ∘ 72^\circ 7 2 ∘ Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z ∠ X K Z = 90 ∘ \angle XKZ = 90^\circ ∠ X K Z = 9 0 ∘ X K = 15 XK = 15 X K = 15 X Z = 25 XZ = 25 X Z = 25 Y K = 8 YK = 8 Y K = 8 △ X Y Z \triangle XYZ △ X Y Z
A 180 sq. cm B 210 sq. cm C 160 sq. cm D 320 sq. cm E 390 sq. cm
Worked solution (try it first) In the right-angled triangle
X K Z XKZ X K Z , by Pythagoras,
K Z = 25 2 − 15 2 KZ = \sqrt{25^2 - 15^2} K Z = 2 5 2 − 1 5 2 , which is
400 = 20 \sqrt{400} = 20 400 = 20 cm.
The base of
△ X Y Z \triangle XYZ △ X Y Z is
Y Z = 8 + 20 = 28 YZ = 8 + 20 = 28 Y Z = 8 + 20 = 28 cm, and its height is
X K = 15 XK = 15 X K = 15 cm.
Area
= 1 2 × 28 × 15 = 210 = \frac12 \times 28 \times 15 = 210 = 2 1 × 28 × 15 = 210 sq. cm, option B.
Watch out
The base is the whole of Y Z YZ Y Z K Z KZ K Z △ X K Z \triangle XKZ △ X K Z Report a problem with this question
Without using tables, calculate the value of 1 + sec 2 30 ∘ 1 + \sec^2 30^\circ 1 + sec 2 3 0 ∘
A 2 1 3 2\frac13 2 3 1 B 2 C 1 1 3 1\frac13 1 3 1 D 3 4 \frac34 4 3 E 3 7 \frac37 7 3
Worked solution (try it first) sec 30 ∘ = 1 cos 30 ∘ \sec30^\circ = \dfrac{1}{\cos30^\circ} sec 3 0 ∘ = cos 3 0 ∘ 1 = 1 3 / 2 = \dfrac{1}{\sqrt3/2} = 3 /2 1 = 2 3 = \dfrac{2}{\sqrt3} = 3 2 .
Square it:
sec 2 30 ∘ = 4 3 \sec^2 30^\circ = \dfrac{4}{3} sec 2 3 0 ∘ = 3 4 .
Add 1:
1 + 4 3 = 7 3 = 2 1 3 1 + \frac43 = \frac73 = 2\frac13 1 + 3 4 = 3 7 = 2 3 1 , option A.
Watch out
Remember to add the 1 at the end: sec 2 30 ∘ \sec^2 30^\circ sec 2 3 0 ∘ 1 1 3 1\frac13 1 3 1 Report a problem with this question
What is the probability that a number chosen at random from the integers 1 to 10 inclusive is either a prime or a multiple of 3?
A 7 10 \frac7{10} 10 7 B 3 5 \frac35 5 3 C 4 5 \frac45 5 4 D 1 2 \frac12 2 1 E 3 10 \frac3{10} 10 3
Worked solution (try it first) The primes from 1 to 10 are 2, 3, 5, 7, and the multiples of 3 are 3, 6, 9.
Combine them, writing 3 only once:
{ 2 , 3 , 5 , 6 , 7 , 9 } \{2, 3, 5, 6, 7, 9\} { 2 , 3 , 5 , 6 , 7 , 9 } , which has 6 members.
So the probability is
6 10 = 3 5 \frac{6}{10} = \frac35 10 6 = 5 3 , option B.
Watch out
3 is both a prime and a multiple of 3, so count it once. Adding 4 + 3 4 + 3 4 + 3 7 10 \frac{7}{10} 10 7 Report a problem with this question
Find the area of a regular hexagon inscribed in a circle of radius 8 cm.
A 16 3 cm 2 16\sqrt3\text{ cm}^2 16 3 cm 2 B 96 3 cm 2 96\sqrt3\text{ cm}^2 96 3 cm 2 C 192 3 cm 2 192\sqrt3\text{ cm}^2 192 3 cm 2 D 16 cm 2 16\text{ cm}^2 16 cm 2 E 32 cm 2 32\text{ cm}^2 32 cm 2
Worked solution (try it first) Joining the centre to the six corners splits the hexagon into six equilateral triangles, each of side 8 cm (the radius).
One triangle has area
3 4 × 8 2 = 16 3 cm 2 \frac{\sqrt3}{4} \times 8^2 = 16\sqrt3\text{ cm}^2 4 3 × 8 2 = 16 3 cm 2 .
The hexagon is six of them:
6 × 16 3 = 96 3 cm 2 6 \times 16\sqrt3 = 96\sqrt3\text{ cm}^2 6 × 16 3 = 96 3 cm 2 , option B.
Watch out
16 3 cm 2 16\sqrt3\text{ cm}^2 16 3 cm 2 Report a problem with this question
In the figure, M N Q P MNQP M N QP M N MN M N P Q PQ P Q X X X N Q NQ N Q M P MP M P Y Y Y ∠ M N Q = 86 ∘ \angle MNQ = 86^\circ ∠ M N Q = 8 6 ∘ ∠ N Q P = 122 ∘ \angle NQP = 122^\circ ∠ N QP = 12 2 ∘ ( x ∘ , y ∘ ) (x^\circ, y^\circ) ( x ∘ , y ∘ ) X X X Y Y Y
A ( 28 ∘ , 36 ∘ ) (28^\circ, 36^\circ) ( 2 8 ∘ , 3 6 ∘ ) B ( 36 ∘ , 28 ∘ ) (36^\circ, 28^\circ) ( 3 6 ∘ , 2 8 ∘ ) C ( 43 ∘ , 61 ∘ ) (43^\circ, 61^\circ) ( 4 3 ∘ , 6 1 ∘ ) D ( 61 ∘ , 43 ∘ ) (61^\circ, 43^\circ) ( 6 1 ∘ , 4 3 ∘ ) E ( 36 ∘ , 43 ∘ ) (36^\circ, 43^\circ) ( 3 6 ∘ , 4 3 ∘ )
Worked solution (try it first) Opposite angles of a cyclic quadrilateral add up to
180 ∘ 180^\circ 18 0 ∘ .
So
∠ N M P = 180 ∘ − 122 ∘ \angle NMP = 180^\circ - 122^\circ ∠ N M P = 18 0 ∘ − 12 2 ∘ = 58 ∘ = 58^\circ = 5 8 ∘ and
∠ M P Q = 180 ∘ − 86 ∘ \angle MPQ = 180^\circ - 86^\circ ∠ M P Q = 18 0 ∘ − 8 6 ∘ X X X is where
M N MN M N and
P Q PQ P Q meet, so use triangle
M X P MXP M X P :
x = 180 ∘ − 58 ∘ − 94 ∘ x = 180^\circ - 58^\circ - 94^\circ x = 18 0 ∘ − 5 8 ∘ − 9 4 ∘ Y Y Y is where
N Q NQ N Q and
M P MP M P meet, so use triangle
M N Y MNY M N Y :
y = 180 ∘ − 58 ∘ − 86 ∘ y = 180^\circ - 58^\circ - 86^\circ y = 18 0 ∘ − 5 8 ∘ − 8 6 ∘ So
( x ∘ , y ∘ ) = ( 28 ∘ , 36 ∘ ) (x^\circ, y^\circ) = (28^\circ, 36^\circ) ( x ∘ , y ∘ ) = ( 2 8 ∘ , 3 6 ∘ ) , option A.
Watch out
Keep the order: x x x X X X M N MN M N P Q PQ P Q y y y Y Y Y N Q NQ N Q M P MP M P ( 36 ∘ , 28 ∘ ) (36^\circ, 28^\circ) ( 3 6 ∘ , 2 8 ∘ ) Report a problem with this question
If cos θ = 3 2 \cos\theta = \frac{\sqrt3}{2} cos θ = 2 3 θ \theta θ 90 ∘ 90^\circ 9 0 ∘ cot ( 90 ∘ − θ ) sin 2 θ \dfrac{\cot(90^\circ - \theta)}{\sin^2\theta} sin 2 θ cot ( 9 0 ∘ − θ )
A 4 3 3 \frac{4\sqrt3}{3} 3 4 3 B 4 3 4\sqrt3 4 3 C 3 2 \frac{\sqrt3}{2} 2 3 D 1 3 \frac1{\sqrt3} 3 1 E 2 3 \frac2{\sqrt3} 3 2
Worked solution (try it first) cos θ = 3 2 \cos\theta = \frac{\sqrt3}{2} cos θ = 2 3 with
θ \theta θ acute, so
θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ .
Then
cot ( 90 ∘ − θ ) = cot 60 ∘ \cot(90^\circ - \theta) = \cot60^\circ cot ( 9 0 ∘ − θ ) = cot 6 0 ∘ = 1 3 = \frac{1}{\sqrt3} = 3 1 , and
sin 2 30 ∘ = ( 1 2 ) 2 \sin^2 30^\circ = \left(\frac12\right)^2 sin 2 3 0 ∘ = ( 2 1 ) 2 Divide:
1 / 3 1 / 4 = 4 3 \dfrac{1/\sqrt3}{1/4} = \dfrac{4}{\sqrt3} 1/4 1/ 3 = 3 4 .
Rationalise by multiplying top and bottom by
3 \sqrt3 3 :
4 3 3 \dfrac{4\sqrt3}{3} 3 4 3 , option A.
Watch out
Square the sine: sin 2 30 ∘ = 1 4 \sin^2 30^\circ = \frac14 sin 2 3 0 ∘ = 4 1 sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 2 3 \frac{2}{\sqrt3} 3 2 Report a problem with this question
A solid sphere of radius 4 cm has a mass of 64 kg. What will be the mass of a shell of the same metal whose internal and external radii are 2 cm and 3 cm respectively?
A 5 kg B 16 kg C 19 kg D 25 kg E 48 kg
Worked solution (try it first) Same metal, so mass is proportional to volume, and the volume of a sphere is proportional to
r 3 r^3 r 3 .
The shell's volume is the outer sphere minus the inner one, which goes with
3 3 − 2 3 = 27 − 8 = 19 3^3 - 2^3 = 27 - 8 = 19 3 3 − 2 3 = 27 − 8 = 19 .
The solid sphere goes with
4 3 = 64 4^3 = 64 4 3 = 64 .
So the shell's mass is
64 × 19 64 = 19 64 \times \frac{19}{64} = 19 64 × 64 19 = 19 kg, option C.
Watch out
Volume goes with the cube of the radius, not the radius. Using the thickness 3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 64 × 1 4 = 16 64 \times \frac14 = 16 64 × 4 1 = 16 Report a problem with this question
In the figure, P O Q POQ P O Q P Q R S PQRS P QR S ∠ P S R = 145 ∘ \angle PSR = 145^\circ ∠ P S R = 14 5 ∘ x ∘ x^\circ x ∘ Q P R QPR QP R
A 25 ∘ 25^\circ 2 5 ∘ B 35 ∘ 35^\circ 3 5 ∘ C 45 ∘ 45^\circ 4 5 ∘ D 55 ∘ 55^\circ 5 5 ∘ E 25 ∘ 25^\circ 2 5 ∘
Worked solution (try it first) P Q R S PQRS P QR S is a cyclic quadrilateral, so opposite angles add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P Q R = 180 ∘ − 145 ∘ \angle PQR = 180^\circ - 145^\circ ∠ P QR = 18 0 ∘ − 14 5 ∘ P Q PQ P Q is a diameter, and the angle in a semicircle is
90 ∘ 90^\circ 9 0 ∘ , so
∠ P R Q = 90 ∘ \angle PRQ = 90^\circ ∠ P R Q = 9 0 ∘ .
The angles of triangle
P Q R PQR P QR add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 90 ∘ − 35 ∘ x = 180^\circ - 90^\circ - 35^\circ x = 18 0 ∘ − 9 0 ∘ − 3 5 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option D.
Watch out
35 ∘ 35^\circ 3 5 ∘ ∠ P Q R \angle PQR ∠ P QR x x x P Q R PQR P QR 90 ∘ − 35 ∘ 90^\circ - 35^\circ 9 0 ∘ − 3 5 ∘ Report a problem with this question
G H I J K L M N GHIJKLMN G H I J K L M N a a a H N HN H N H N HN H N
A 3 a 3\sqrt a 3 a B 3 a 3a 3 a C 3 a 2 3a^2 3 a 2 D a 2 a\sqrt2 a 2 E a 3 a\sqrt3 a 3
Worked solution (try it first) First cross one face: the face diagonal has length
a 2 + a 2 = a 2 \sqrt{a^2 + a^2} = a\sqrt2 a 2 + a 2 = a 2 .
H N HN H N is the hypotenuse of a right-angled triangle with that face diagonal and a vertical edge
a a a :
H N 2 = 2 a 2 + a 2 = 3 a 2 HN^2 = 2a^2 + a^2 = 3a^2 H N 2 = 2 a 2 + a 2 = 3 a 2 .
So
H N = a 3 HN = a\sqrt3 H N = a 3 , option E.
Watch out
a 2 a\sqrt2 a 2 H N HN H N Report a problem with this question
P Q R S PQRS P QR S 14 cm 2 14\text{ cm}^2 14 cm 2 P Q ∥ R S PQ \parallel RS P Q ∥ R S P Q = 4 PQ = 4 P Q = 4 S R = 3 SR = 3 S R = 3 △ S Q R \triangle SQR △ S QR cm 2 \text{cm}^2 cm 2
Worked solution (try it first) Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h :
1 2 ( 4 + 3 ) h = 14 \frac12(4 + 3)h = 14 2 1 ( 4 + 3 ) h = 14 , so
3.5 h = 14 3.5h = 14 3.5 h = 14 and
h = 4 h = 4 h = 4 cm.
△ S Q R \triangle SQR △ S QR has base
S R = 3 SR = 3 S R = 3 cm, and its height is the distance between the parallel sides, 4 cm.
Area
= 1 2 × 3 × 4 = \frac12 \times 3 \times 4 = 2 1 × 3 × 4 = 6.0 cm 2 = 6.0\text{ cm}^2 = 6.0 cm 2 , option B.
Watch out
The diagonal S Q SQ S Q Report a problem with this question
In the figure, P Q PQ P Q P P P Q R S QRS QR S S R SR S R S R P SRP S R P ∠ P Q R = θ \angle PQR = \theta ∠ P QR = θ ∠ Q P R = ϕ \angle QPR = \phi ∠ QP R = ϕ
A θ + ϕ = 90 ∘ \theta + \phi = 90^\circ θ + ϕ = 9 0 ∘ B ϕ = 90 ∘ − 2 θ \phi = 90^\circ - 2\theta ϕ = 9 0 ∘ − 2 θ C θ = ϕ \theta = \phi θ = ϕ D ϕ = 2 θ \phi = 2\theta ϕ = 2 θ E θ + 2 ϕ = 120 ∘ \theta + 2\phi = 120^\circ θ + 2 ϕ = 12 0 ∘
Worked solution (try it first) The angle between the tangent
P Q PQ P Q and the chord
Q R QR QR equals the angle in the alternate segment, so
∠ Q S R = θ \angle QSR = \theta ∠ QS R = θ .
S R SR S R is a diameter, so
∠ S Q R = 90 ∘ \angle SQR = 90^\circ ∠ S QR = 9 0 ∘ (angle in a semicircle).
The whole angle
S Q P SQP S QP is
90 ∘ + θ 90^\circ + \theta 9 0 ∘ + θ .
The angles of triangle
S Q P SQP S QP add up to
180 ∘ 180^\circ 18 0 ∘ :
θ + ( 90 ∘ + θ ) + ϕ = 180 ∘ \theta + (90^\circ + \theta) + \phi = 180^\circ θ + ( 9 0 ∘ + θ ) + ϕ = 18 0 ∘ .
So
ϕ = 90 ∘ − 2 θ \phi = 90^\circ - 2\theta ϕ = 9 0 ∘ − 2 θ , option B.
Watch out
At Q Q Q S Q P SQP S QP 90 ∘ + θ 90^\circ + \theta 9 0 ∘ + θ 90 ∘ 90^\circ 9 0 ∘ 90 ∘ 90^\circ 9 0 ∘ θ + ϕ = 90 ∘ \theta + \phi = 90^\circ θ + ϕ = 9 0 ∘ Report a problem with this question
A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?
A 1 3 \frac13 3 1 B 2 9 \frac29 9 2 C 2 15 \frac2{15} 15 2 D 1 5 \frac15 5 1 E 3 5 \frac35 5 3
Worked solution (try it first) The first ball is red with probability
6 10 \frac{6}{10} 10 6 .
Without replacement, 5 red balls are left out of 9, so the second is red with probability
5 9 \frac59 9 5 .
Multiply:
6 10 × 5 9 = 30 90 \frac{6}{10} \times \frac59 = \frac{30}{90} 10 6 × 9 5 = 90 30 = 1 3 = \frac13 = 3 1 , option A.
Watch out
The first ball is not put back, so the second draw is from 9 balls with 5 red. Using 6 10 \frac{6}{10} 10 6 9 25 \frac{9}{25} 25 9 Report a problem with this question
Two points X X X Y Y Y 60 ∘ 60^\circ 6 0 ∘ 147 ∘ 147^\circ 14 7 ∘ 153 ∘ 153^\circ 15 3 ∘ X X X Y Y Y 2 π R = 4 × 10 4 2\pi R = 4 \times 10^4 2 π R = 4 × 1 0 4 R R R
A 28,850 km B 16,667 km C 8,333 km D 6,667 km E 3,333 km
Worked solution (try it first) X X X is east and
Y Y Y is west, so the longitude difference one way is
147 ∘ + 153 ∘ = 300 ∘ 147^\circ + 153^\circ = 300^\circ 14 7 ∘ + 15 3 ∘ = 30 0 ∘ .
The short way round is
360 ∘ − 300 ∘ = 60 ∘ 360^\circ - 300^\circ = 60^\circ 36 0 ∘ − 30 0 ∘ = 6 0 ∘ .
The parallel of latitude
60 ∘ 60^\circ 6 0 ∘ has radius
R cos 60 ∘ R\cos60^\circ R cos 6 0 ∘ , so its length is
2 π R cos 60 ∘ = 40 000 × 0.5 2\pi R\cos60^\circ = 40\,000 \times 0.5 2 π R cos 6 0 ∘ = 40 000 × 0.5 = 20 000 = 20\,000 = 20 000 km.
The arc for
60 ∘ 60^\circ 6 0 ∘ is
60 360 \frac{60}{360} 360 60 of that:
1 6 × 20 000 ≈ 3333 \frac16 \times 20\,000 \approx 3333 6 1 × 20 000 ≈ 3333 km, option E.
Watch out
Multiply by cos 60 ∘ \cos60^\circ cos 6 0 ∘ 1 6 × 40 000 ≈ 6667 \frac16 \times 40\,000 \approx 6667 6 1 × 40 000 ≈ 6667 Report a problem with this question
In the figure, a sector of a circle of radius 3 has an angle of 120 ∘ 120^\circ 12 0 ∘
A 3 π 3\pi 3 π B 9 3 4 \frac{9\sqrt3}{4} 4 9 3 C 3 ( π − 3 3 4 ) 3\left(\pi - \frac{3\sqrt3}{4}\right) 3 ( π − 4 3 3 ) D 3 ( 3 − π ) 4 \frac{3(\sqrt3 - \pi)}{4} 4 3 ( 3 − π ) E π + 9 3 4 \pi + \frac{9\sqrt3}{4} π + 4 9 3
Worked solution (try it first) The sector is
120 360 = 1 3 \frac{120}{360} = \frac13 360 120 = 3 1 of the circle:
1 3 × π × 3 2 = 3 π \frac13 \times \pi \times 3^2 = 3\pi 3 1 × π × 3 2 = 3 π .
The triangle
O A B OAB O A B has two sides of 3 with
120 ∘ 120^\circ 12 0 ∘ between them: area
= 1 2 × 3 × 3 × sin 120 ∘ = \frac12 \times 3 \times 3 \times \sin120^\circ = 2 1 × 3 × 3 × sin 12 0 ∘ = 9 3 4 = \frac{9\sqrt3}{4} = 4 9 3 .
Segment = sector − triangle:
3 π − 9 3 4 = 3 ( π − 3 3 4 ) 3\pi - \frac{9\sqrt3}{4} = 3\left(\pi - \frac{3\sqrt3}{4}\right) 3 π − 4 9 3 = 3 ( π − 4 3 3 ) , option C.
Watch out
3 π 3\pi 3 π O A B OAB O A B Report a problem with this question
In a class of 120 students, 18 of them scored an A grade in Mathematics. If the sector representing the A-grade students on a pie chart has angle Z ∘ Z^\circ Z ∘ Z Z Z
Worked solution (try it first) The A-grade students are
18 120 \frac{18}{120} 120 18 of the class.
Their sector is that fraction of
360 ∘ 360^\circ 36 0 ∘ :
18 120 × 360 = 54 \frac{18}{120} \times 360 = 54 120 18 × 360 = 54 .
So
Z = 54 Z = 54 Z = 54 , option E.
Watch out
Multiply the fraction by 360, not 100: 18 120 × 100 = 15 \frac{18}{120} \times 100 = 15 120 18 × 100 = 15 Report a problem with this question
In the figure, find the angle x x x
A 100 ∘ 100^\circ 10 0 ∘ B 120 ∘ 120^\circ 12 0 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 110 ∘ 110^\circ 11 0 ∘ E 140 ∘ 140^\circ 14 0 ∘
Worked solution (try it first) The interior angles of a quadrilateral add up to
360 ∘ 360^\circ 36 0 ∘ .
The three marked angles add up to
20 ∘ + 80 ∘ + 40 ∘ = 140 ∘ 20^\circ + 80^\circ + 40^\circ = 140^\circ 2 0 ∘ + 8 0 ∘ + 4 0 ∘ = 14 0 ∘ .
So the interior angle at the inward-pointing vertex is
360 ∘ − 140 ∘ = 220 ∘ 360^\circ - 140^\circ = 220^\circ 36 0 ∘ − 14 0 ∘ = 22 0 ∘ , a reflex angle.
x x x is the angle on the other side of that vertex, and angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ :
x = 360 ∘ − 220 ∘ = 140 ∘ x = 360^\circ - 220^\circ = 140^\circ x = 36 0 ∘ − 22 0 ∘ = 14 0 ∘ , option E.
Watch out
At a dent the interior angle is reflex (220 ∘ 220^\circ 22 0 ∘ x x x x x x 140 ∘ 140^\circ 14 0 ∘ Report a problem with this question