Paper JAMB 1986 General Maths Objective
Objective paper · 44 questions · partial
JAMB 1986 · UME Topics include Number bases, Statistics: data & averages, Number foundations & fractions, Commercial arithmetic, Quadratics & their graphs, Linear & simultaneous equations.
Our copy of this paper is missing questions 7, 8, 14, 17, 28, 48.
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 9 10 11 12 13 15 16 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 50 Evaluate 212 3 − 121 3 + 222 3 212_3 - 121_3 + 222_3 21 2 3 − 12 1 3 + 22 2 3
A 313 3 313_3 31 3 3 B 1000 3 1000_3 100 0 3 C 1020 3 1020_3 102 0 3 D 1222 3 1222_3 122 2 3
Worked solution (try it first) Change to base ten:
212 3 = 18 + 3 + 2 = 23 212_3 = 18 + 3 + 2 = 23 21 2 3 = 18 + 3 + 2 = 23 ,
121 3 = 9 + 6 + 1 = 16 121_3 = 9 + 6 + 1 = 16 12 1 3 = 9 + 6 + 1 = 16 and
222 3 = 18 + 6 + 2 = 26 222_3 = 18 + 6 + 2 = 26 22 2 3 = 18 + 6 + 2 = 26 .
Work it out:
23 − 16 + 26 = 33 23 - 16 + 26 = 33 23 − 16 + 26 = 33 .
Change back:
33 = 27 + 6 33 = 27 + 6 33 = 27 + 6 = 1 × 27 + 0 × 9 + 2 × 3 + 0 = 1 \times 27 + 0 \times 9 + 2 \times 3 + 0 = 1 × 27 + 0 × 9 + 2 × 3 + 0 , so the answer is
1020 3 1020_3 102 0 3 , option C.
Watch out
Base three only has the digits 0, 1 and 2, so 313 3 313_3 31 3 3 Report a problem with this question
If Musa had scored 75 in Biology instead of 57, his average mark in four subjects would have been 60. What was his total mark?
Worked solution (try it first) With 75 in Biology the average of 4 subjects is 60, so that total would be
4 × 60 = 240 4 \times 60 = 240 4 × 60 = 240 .
His real Biology mark is
75 − 57 = 18 75 - 57 = 18 75 − 57 = 18 lower.
So his actual total is
240 − 18 = 222 240 - 18 = 222 240 − 18 = 222 , option C.
Watch out
240 (option B) is the total he would have had with 75; the question wants his real total, so take off the 18 marks he didn't get. Report a problem with this question
Divide the L.C.M. of 48, 64 and 80 by their H.C.F.
Worked solution (try it first) Write each number as a product of primes:
48 = 2 4 × 3 48 = 2^4 \times 3 48 = 2 4 × 3 ,
64 = 2 6 64 = 2^6 64 = 2 6 and
80 = 2 4 × 5 80 = 2^4 \times 5 80 = 2 4 × 5 .
The LCM takes the highest power of every prime:
2 6 × 3 × 5 = 960 2^6 \times 3 \times 5 = 960 2 6 × 3 × 5 = 960 .
The HCF takes the lowest power of the primes common to all three: only 2 is common, so the HCF is
2 4 = 16 2^4 = 16 2 4 = 16 .
So
960 ÷ 16 = 60 960 \div 16 = 60 960 ÷ 16 = 60 , option D.
Watch out
For the LCM use the highest power of 2, which is 2 6 2^6 2 6 2 4 2^4 2 4 240 ÷ 16 = 15 240 \div 16 = 15 240 ÷ 16 = 15 Report a problem with this question
Find the smallest number by which 252 can be multiplied to obtain a perfect square.
Worked solution (try it first) Write 252 as a product of primes:
252 = 4 × 63 252 = 4 \times 63 252 = 4 × 63 = 2 2 × 3 2 × 7 = 2^2 \times 3^2 \times 7 = 2 2 × 3 2 × 7 .
A perfect square has every prime to an even power.
The 2s and 3s are already paired.
Only the 7 is on its own.
So multiply by 7:
252 × 7 = 1764 = 42 2 252 \times 7 = 1764 = 42^2 252 × 7 = 1764 = 4 2 2 , option D.
Watch out
Don't pick the smallest option: 252 × 2 = 504 252 \times 2 = 504 252 × 2 = 504 2 3 2^3 2 3 Report a problem with this question
Find the reciprocal of 2 / 3 1 2 + 1 3 \dfrac{2/3}{\frac12 + \frac13} 2 1 + 3 1 2/3
A 4 5 \frac45 5 4 B 5 4 \frac54 4 5 C 2 5 \frac25 5 2 D 6 7 \frac67 7 6
Worked solution (try it first) Bottom: the LCD of 2 and 3 is 6, so
1 2 + 1 3 = 3 6 + 2 6 \frac12 + \frac13 = \frac36 + \frac26 2 1 + 3 1 = 6 3 + 6 2 Divide by flipping the second fraction:
2 3 ÷ 5 6 = 2 3 × 6 5 \frac23 \div \frac56 = \frac23 \times \frac65 3 2 ÷ 6 5 = 3 2 × 5 6 The reciprocal turns it upside down:
5 4 \frac54 4 5 , option B.
Watch out
4 5 \frac45 5 4 5 4 \frac54 4 5 Report a problem with this question
Three boys shared some oranges. The first received 1 3 \frac13 3 1 2 3 \frac23 3 2
Worked solution (try it first) After the first boy takes
1 3 \frac13 3 1 , the remainder is
1 − 1 3 = 2 3 1 - \frac13 = \frac23 1 − 3 1 = 3 2 of the oranges.
The second takes
2 3 \frac23 3 2 of that 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 all the oranges.
So
2 9 \frac29 9 2 of the total is 12.
One ninth is
12 ÷ 2 = 6 12 \div 2 = 6 12 ÷ 2 = 6 , and the total is
6 × 9 = 54 6 \times 9 = 54 6 × 9 = 54 , option B.
Watch out
The second boy's 2 3 \frac23 3 2 Report a problem with this question
Udoh deposited ₦150.00 in the bank. At the end of 5 years the simple interest on the principal was ₦55.00. At what rate per annum was the interest paid?
A 11 % 11\% 11% B 7 1 3 % 7\frac13\% 7 3 1 % C 5 % 5\% 5% D 3 1 2 % 3\frac12\% 3 2 1 %
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 = 55 I = 55 I = 55 ,
P = 150 P = 150 P = 150 and
T = 5 T = 5 T = 5 :
R = 5500 750 = 22 3 R = \dfrac{5500}{750} = \dfrac{22}{3} R = 750 5500 = 3 22 .
So the rate is
7 1 3 % 7\frac13\% 7 3 1 % per annum, option B.
Watch out
Divide by the principal as well as the time. 55 ÷ 5 = 11 55 \div 5 = 11 55 ÷ 5 = 11 11 % 11\% 11% Report a problem with this question
A number of pencils were shared among Bisi, Sola and Tunde in the ratio 2 : 3 : 5 2 : 3 : 5 2 : 3 : 5
Worked solution (try it first) The ratio
2 : 3 : 5 2 : 3 : 5 2 : 3 : 5 has
2 + 3 + 5 = 10 2 + 3 + 5 = 10 2 + 3 + 5 = 10 parts, and Bisi's 5 pencils are 2 parts.
So one part is
5 ÷ 2 = 2.5 5 \div 2 = 2.5 5 ÷ 2 = 2.5 pencils.
All 10 parts are
10 × 2.5 = 25 10 \times 2.5 = 25 10 × 2.5 = 25 pencils, option B.
Watch out
Bisi's 5 pencils are 2 parts, not 1. Taking one part as 5 gives 10 × 5 = 50 10 \times 5 = 50 10 × 5 = 50 Report a problem with this question
The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form ( x , y ) (x, y) ( x , y ) x > y x > y x > y
A ( 12 , 9 ) (12, 9) ( 12 , 9 ) B ( 23 , 17 ) (23, 17) ( 23 , 17 ) C ( 17 , 11 ) (17, 11) ( 17 , 11 ) D ( 18 , 12 ) (18, 12) ( 18 , 12 )
Worked solution (try it first) Let the younger age be
y y y , so the older is
y + 6 y + 6 y + 6 .
Their product is
y ( y + 6 ) = 187 y(y + 6) = 187 y ( y + 6 ) = 187 .
Rearrange:
y 2 + 6 y − 187 = 0 y^2 + 6y - 187 = 0 y 2 + 6 y − 187 = 0 , which factorises as
( y + 17 ) ( y − 11 ) = 0 (y + 17)(y - 11) = 0 ( y + 17 ) ( y − 11 ) = 0 .
An age is positive, so
y = 11 y = 11 y = 11 and the older age is 17.
So the ages are
( 17 , 11 ) (17, 11) ( 17 , 11 ) , option C.
Watch out
Options B and D also differ by 6, so check the product too: 23 × 17 = 391 23 \times 17 = 391 23 × 17 = 391 18 × 12 = 216 18 \times 12 = 216 18 × 12 = 216 17 × 11 = 187 17 \times 11 = 187 17 × 11 = 187 Report a problem with this question
In 1984, Ike was 24 years old and his father was 45 years old. In what year was Ike exactly half his father's age?
Worked solution (try it first) The age gap never changes:
45 − 24 = 21 45 - 24 = 21 45 − 24 = 21 years.
When Ike is half his father's age, the father is twice Ike's age, so the gap equals Ike's age.
Ike was 21 and his father 42.
Ike was 21 three years before 1984 (when he was 24), so the year was 1981, option B.
Watch out
Don't halve today's ages: the gap stays 21, so Ike must be 21. For example 1982 (option A) makes the ages 22 and 43, and 43 is not twice 22. Report a problem with this question
Simplify ( 1 5 + 3 − 1 5 − 3 ) × 1 3 \left(\frac{1}{\sqrt5 + \sqrt3} - \frac{1}{\sqrt5 - \sqrt3}\right) \times \frac{1}{\sqrt3} ( 5 + 3 1 − 5 − 3 1 ) × 3 1
A 3 5 \frac{\sqrt3}{\sqrt5} 5 3 B − 2 3 -\frac{2}{\sqrt3} − 3 2 C − 2 -2 − 2 D − 1 -1 − 1
Worked solution (try it first) Rationalise each fraction.
1 5 + 3 = 5 − 3 5 − 3 \frac{1}{\sqrt5 + \sqrt3} = \frac{\sqrt5 - \sqrt3}{5 - 3} 5 + 3 1 = 5 − 3 5 − 3 , which is
5 − 3 2 \frac{\sqrt5 - \sqrt3}{2} 2 5 − 3 .
Likewise
1 5 − 3 = 5 + 3 2 \frac{1}{\sqrt5 - \sqrt3} = \frac{\sqrt5 + \sqrt3}{2} 5 − 3 1 = 2 5 + 3 .
Subtract:
( 5 − 3 ) − ( 5 + 3 ) 2 = − 2 3 2 \dfrac{(\sqrt5 - \sqrt3) - (\sqrt5 + \sqrt3)}{2} = \dfrac{-2\sqrt3}{2} 2 ( 5 − 3 ) − ( 5 + 3 ) = 2 − 2 3 , which is
− 3 -\sqrt3 − 3 .
Multiply by
1 3 \frac{1}{\sqrt3} 3 1 :
− 3 × 1 3 = − 1 -\sqrt3 \times \frac{1}{\sqrt3} = -1 − 3 × 3 1 = − 1 , option D.
Watch out
Each rationalised fraction has 2 on the bottom, because 5 − 3 = 2 5 - 3 = 2 5 − 3 = 2 − 2 3 -2\sqrt3 − 2 3 − 2 -2 − 2 Report a problem with this question
Evaluate 9 1 3 × 27 − 1 2 3 − 1 6 × 3 − 2 3 \dfrac{9^{\frac13} \times 27^{-\frac12}}{3^{-\frac16} \times 3^{-\frac23}} 3 − 6 1 × 3 − 3 2 9 3 1 × 2 7 − 2 1
Worked solution (try it first) Write everything as a power of 3:
9 1 3 = 3 2 3 9^{\frac13} = 3^{\frac23} 9 3 1 = 3 3 2 and
27 − 1 2 = 3 − 3 2 27^{-\frac12} = 3^{-\frac32} 2 7 − 2 1 = 3 − 2 3 , multiplying the indices.
Top: add the powers,
2 3 − 3 2 = 4 − 9 6 \frac23 - \frac32 = \frac{4 - 9}{6} 3 2 − 2 3 = 6 4 − 9 = − 5 6 = -\frac56 = − 6 5 , so the top is
3 − 5 6 3^{-\frac56} 3 − 6 5 .
Bottom:
− 1 6 − 2 3 = − 5 6 -\frac16 - \frac23 = -\frac56 − 6 1 − 3 2 = − 6 5 , so the bottom is also
3 − 5 6 3^{-\frac56} 3 − 6 5 .
Top and bottom are equal, so the value is
3 0 = 1 3^0 = 1 3 0 = 1 , option B.
Watch out
Change the base before using the index: 27 = 3 3 27 = 3^3 27 = 3 3 27 − 1 2 = 3 − 3 2 27^{-\frac12} = 3^{-\frac32} 2 7 − 2 1 = 3 − 2 3 3 − 1 2 3^{-\frac12} 3 − 2 1 − 1 2 -\frac12 − 2 1 3 1 6 3^{\frac16} 3 6 1 3 3 3 Report a problem with this question
If x x x y 3 y^3 y 3 x = 2 x = 2 x = 2 y = 1 y = 1 y = 1 x x x y = 5 y = 5 y = 5
Worked solution (try it first) x x x varies directly as
y 3 y^3 y 3 , so
x = k y 3 x = ky^3 x = k y 3 .
Put in
x = 2 x = 2 x = 2 ,
y = 1 y = 1 y = 1 :
k = 2 k = 2 k = 2 .
When
y = 5 y = 5 y = 5 :
5 3 = 125 5^3 = 125 5 3 = 125 , so
x = 2 × 125 = 250 x = 2 \times 125 = 250 x = 2 × 125 = 250 , option D.
Watch out
Multiply by the constant: 5 3 = 125 5^3 = 125 5 3 = 125 k = 2 k = 2 k = 2 Report a problem with this question
If y = x x − 3 + x x + 4 y = \dfrac{x}{x - 3} + \dfrac{x}{x + 4} y = x − 3 x + x + 4 x y y y x = − 2 x = -2 x = − 2
A − 3 5 -\frac35 − 5 3 B 3 5 \frac35 5 3 C − 7 5 -\frac75 − 5 7 D 7 5 \frac75 5 7
Worked solution (try it first) First fraction at
x = − 2 x = -2 x = − 2 :
− 2 − 2 − 3 = − 2 − 5 \frac{-2}{-2 - 3} = \frac{-2}{-5} − 2 − 3 − 2 = − 5 − 2 , which is
2 5 \frac25 5 2 (negative over negative is positive).
Second fraction:
− 2 − 2 + 4 = − 2 2 = − 1 \frac{-2}{-2 + 4} = \frac{-2}{2} = -1 − 2 + 4 − 2 = 2 − 2 = − 1 .
So
y = 2 5 − 1 = − 3 5 y = \frac25 - 1 = -\frac35 y = 5 2 − 1 = − 5 3 , option A.
Watch out
− 2 − 5 \frac{-2}{-5} − 5 − 2 + 2 5 +\frac25 + 5 2 − 2 5 -\frac25 − 5 2 − 2 5 − 1 = − 7 5 -\frac25 - 1 = -\frac75 − 5 2 − 1 = − 5 7 Report a problem with this question
Find all the numbers x x x 1 3 ( x + 1 ) − 1 > 1 5 ( x + 4 ) \frac13(x + 1) - 1 > \frac15(x + 4) 3 1 ( x + 1 ) − 1 > 5 1 ( x + 4 )
A x < 11 x < 11 x < 11 B x < − 1 x < -1 x < − 1 C x > 6 x > 6 x > 6 D x > 11 x > 11 x > 11
Worked solution (try it first) Multiply every term by 15, the LCM of 3 and 5:
5 ( x + 1 ) − 15 > 3 ( x + 4 ) 5(x + 1) - 15 > 3(x + 4) 5 ( x + 1 ) − 15 > 3 ( x + 4 ) .
Expand the brackets:
5 x + 5 − 15 > 3 x + 12 5x + 5 - 15 > 3x + 12 5 x + 5 − 15 > 3 x + 12 , so
5 x − 10 > 3 x + 12 5x - 10 > 3x + 12 5 x − 10 > 3 x + 12 .
Subtract
3 x 3x 3 x and add 10 to both sides:
2 x > 22 2x > 22 2 x > 22 .
Divide by 2:
x > 11 x > 11 x > 11 , option D.
Watch out
You only reverse the sign when you multiply or divide by a negative. Here you divide by + 2 +2 + 2 x > 11 x > 11 x > 11 x < 11 x < 11 x < 11 Report a problem with this question
Factorize x 2 + 2 a + a x + 2 x x^2 + 2a + ax + 2x x 2 + 2 a + a x + 2 x
A ( x + 2 a ) ( x + 1 ) (x + 2a)(x + 1) ( x + 2 a ) ( x + 1 ) B ( x + 2 a ) ( x − 1 ) (x + 2a)(x - 1) ( x + 2 a ) ( x − 1 ) C ( x 2 − 1 ) ( x + a ) (x^2 - 1)(x + a) ( x 2 − 1 ) ( x + a ) D ( x + 2 ) ( x + a ) (x + 2)(x + a) ( x + 2 ) ( x + a )
Worked solution (try it first) Reorder so the pairs share a factor:
x 2 + 2 x + a x + 2 a x^2 + 2x + ax + 2a x 2 + 2 x + a x + 2 a .
Take out the common factors:
x ( x + 2 ) + a ( x + 2 ) x(x + 2) + a(x + 2) x ( x + 2 ) + a ( x + 2 ) .
Take out the common bracket:
( x + 2 ) ( x + a ) (x + 2)(x + a) ( x + 2 ) ( x + a ) , option D.
Watch out
Expand to check. Option A, ( x + 2 a ) ( x + 1 ) (x + 2a)(x + 1) ( x + 2 a ) ( x + 1 ) x 2 + x + 2 a x + 2 a x^2 + x + 2ax + 2a x 2 + x + 2 a x + 2 a 2 a x 2ax 2 a x x x x a x ax a x 2 x 2x 2 x Report a problem with this question
Solve the equation 3 x 2 + 6 x − 2 = 0 3x^2 + 6x - 2 = 0 3 x 2 + 6 x − 2 = 0
A x = − 1 ± 3 3 x = -1 \pm \frac{\sqrt3}{3} x = − 1 ± 3 3 B x = − 1 ± 15 3 x = -1 \pm \frac{\sqrt{15}}{3} x = − 1 ± 3 15 C x = − 2 ± 2 3 3 x = -2 \pm \frac{2\sqrt3}{3} x = − 2 ± 3 2 3 D x = − 2 ± 2 15 3 x = -2 \pm \frac{2\sqrt{15}}{3} x = − 2 ± 3 2 15
Worked solution (try it first) Use the formula with
a = 3 a = 3 a = 3 ,
b = 6 b = 6 b = 6 and
c = − 2 c = -2 c = − 2 .
The discriminant is
b 2 − 4 a c = 36 + 24 = 60 b^2 - 4ac = 36 + 24 = 60 b 2 − 4 a c = 36 + 24 = 60 .
So
x = − 6 ± 60 6 x = \dfrac{-6 \pm \sqrt{60}}{6} x = 6 − 6 ± 60 .
Simplify the surd:
60 = 4 × 15 \sqrt{60} = \sqrt{4 \times 15} 60 = 4 × 15 Divide each term by 6:
x = − 1 ± 15 3 x = -1 \pm \frac{\sqrt{15}}{3} x = − 1 ± 3 15 , option B.
Watch out
Watch the sign of c c c − 4 a c = − 4 ( 3 ) ( − 2 ) = + 24 -4ac = -4(3)(-2) = +24 − 4 a c = − 4 ( 3 ) ( − 2 ) = + 24 36 + 24 = 60 36 + 24 = 60 36 + 24 = 60 36 − 24 = 12 36 - 24 = 12 36 − 24 = 12 − 1 ± 3 3 -1 \pm \frac{\sqrt3}{3} − 1 ± 3 3 Report a problem with this question
Simplify 1 5 x + 5 + 1 7 x + 7 \dfrac{1}{5x + 5} + \dfrac{1}{7x + 7} 5 x + 5 1 + 7 x + 7 1
A 12 35 + 7 \dfrac{12}{35 + 7} 35 + 7 12 B 1 35 ( x + 1 ) \dfrac{1}{35(x + 1)} 35 ( x + 1 ) 1 C 12 x 35 ( x + 1 ) \dfrac{12x}{35(x + 1)} 35 ( x + 1 ) 12 x D 12 35 x + 35 \dfrac{12}{35x + 35} 35 x + 35 12
Worked solution (try it first) Factorise each bottom:
5 x + 5 = 5 ( x + 1 ) 5x + 5 = 5(x + 1) 5 x + 5 = 5 ( x + 1 ) and
7 x + 7 = 7 ( x + 1 ) 7x + 7 = 7(x + 1) 7 x + 7 = 7 ( x + 1 ) .
The LCD is
35 ( x + 1 ) 35(x + 1) 35 ( x + 1 ) .
Multiply the first top by 7 and the second by 5:
7 + 5 35 ( x + 1 ) \frac{7 + 5}{35(x + 1)} 35 ( x + 1 ) 7 + 5 .
So the sum is
12 35 ( x + 1 ) = 12 35 x + 35 \dfrac{12}{35(x + 1)} = \dfrac{12}{35x + 35} 35 ( x + 1 ) 12 = 35 x + 35 12 , option D.
Watch out
The common factor x + 1 x + 1 x + 1 12 x 35 ( x + 1 ) \frac{12x}{35(x + 1)} 35 ( x + 1 ) 12 x x x x Report a problem with this question
The curve y = − x 2 + 3 x + 4 y = -x^2 + 3x + 4 y = − x 2 + 3 x + 4
A ( 4 , 0 ) , ( 0 , 0 ) , ( − 1 , 0 ) (4, 0), (0, 0), (-1, 0) ( 4 , 0 ) , ( 0 , 0 ) , ( − 1 , 0 ) B ( − 4 , 0 ) , ( 0 , 4 ) , ( 1 , 1 ) (-4, 0), (0, 4), (1, 1) ( − 4 , 0 ) , ( 0 , 4 ) , ( 1 , 1 ) C ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) (0, 0), (0, 1), (1, 0) ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) D ( 0 , 4 ) , ( 4 , 0 ) , ( − 1 , 0 ) (0, 4), (4, 0), (-1, 0) ( 0 , 4 ) , ( 4 , 0 ) , ( − 1 , 0 )
Worked solution (try it first) On the
y y y -axis
x = 0 x = 0 x = 0 , so
y = 4 y = 4 y = 4 .
The point is
( 0 , 4 ) (0, 4) ( 0 , 4 ) .
On the
x x x -axis
y = 0 y = 0 y = 0 :
− x 2 + 3 x + 4 = 0 -x^2 + 3x + 4 = 0 − x 2 + 3 x + 4 = 0 .
Multiply by
− 1 -1 − 1 :
x 2 − 3 x − 4 = 0 x^2 - 3x - 4 = 0 x 2 − 3 x − 4 = 0 .
Factorise:
( x − 4 ) ( x + 1 ) = 0 (x - 4)(x + 1) = 0 ( x − 4 ) ( x + 1 ) = 0 , so
x = 4 x = 4 x = 4 or
x = − 1 x = -1 x = − 1 .
So the points are
( 0 , 4 ) (0, 4) ( 0 , 4 ) ,
( 4 , 0 ) (4, 0) ( 4 , 0 ) and
( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) , option D.
Watch out
( x − 4 ) ( x + 1 ) = 0 (x - 4)(x + 1) = 0 ( x − 4 ) ( x + 1 ) = 0 x = 4 x = 4 x = 4 x = − 1 x = -1 x = − 1 − 4 -4 − 4 Report a problem with this question
Factorize ( 4 a + 3 ) 2 − ( 3 a − 2 ) 2 (4a + 3)^2 - (3a - 2)^2 ( 4 a + 3 ) 2 − ( 3 a − 2 ) 2
A ( a + 1 ) ( a + 5 ) (a + 1)(a + 5) ( a + 1 ) ( a + 5 ) B ( a − 5 ) ( 7 a − 1 ) (a - 5)(7a - 1) ( a − 5 ) ( 7 a − 1 ) C ( a + 5 ) ( 7 a + 1 ) (a + 5)(7a + 1) ( a + 5 ) ( 7 a + 1 ) D a ( 7 a + 1 ) a(7a + 1) a ( 7 a + 1 )
Worked solution (try it first) Use the difference of two squares,
A 2 − B 2 = ( A − B ) ( A + B ) A^2 - B^2 = (A - B)(A + B) A 2 − B 2 = ( A − B ) ( A + B ) , with
A = 4 a + 3 A = 4a + 3 A = 4 a + 3 and
B = 3 a − 2 B = 3a - 2 B = 3 a − 2 .
A − B = 4 a + 3 − 3 a + 2 A - B = 4a + 3 - 3a + 2 A − B = 4 a + 3 − 3 a + 2 , which is
a + 5 a + 5 a + 5 .
A + B = 4 a + 3 + 3 a − 2 A + B = 4a + 3 + 3a - 2 A + B = 4 a + 3 + 3 a − 2 , which is
7 a + 1 7a + 1 7 a + 1 .
So the expression is
( a + 5 ) ( 7 a + 1 ) (a + 5)(7a + 1) ( a + 5 ) ( 7 a + 1 ) , option C.
Watch out
Subtract the whole of 3 a − 2 3a - 2 3 a − 2 − ( 3 a − 2 ) = − 3 a + 2 -(3a - 2) = -3a + 2 − ( 3 a − 2 ) = − 3 a + 2 A − B = a + 5 A - B = a + 5 A − B = a + 5 − 3 a − 2 -3a - 2 − 3 a − 2 a + 1 a + 1 a + 1 Report a problem with this question
If 5 x + 2 y = 5 5^{x + 2y} = 5 5 x + 2 y = 5 4 x + 3 y = 16 4^{x + 3y} = 16 4 x + 3 y = 16 3 x + y 3^{x + y} 3 x + y
Worked solution (try it first) 5 x + 2 y = 5 1 5^{x + 2y} = 5^1 5 x + 2 y = 5 1 , so
x + 2 y = 1 x + 2y = 1 x + 2 y = 1 .
4 x + 3 y = 16 = 4 2 4^{x + 3y} = 16 = 4^2 4 x + 3 y = 16 = 4 2 , so
x + 3 y = 2 x + 3y = 2 x + 3 y = 2 .
Subtract the first equation from the second:
y = 1 y = 1 y = 1 .
Then
x = 1 − 2 = − 1 x = 1 - 2 = -1 x = 1 − 2 = − 1 .
So
x + y = 0 x + y = 0 x + y = 0 and
3 x + y = 3 0 = 1 3^{x + y} = 3^0 = 1 3 x + y = 3 0 = 1 , option B.
Watch out
Any non-zero number to the power 0 is 1, not 0. Stopping at x + y = 0 x + y = 0 x + y = 0 Report a problem with this question
Simplify 1 x − 2 + 1 x + 2 + 2 x x 2 − 4 \dfrac{1}{x - 2} + \dfrac{1}{x + 2} + \dfrac{2x}{x^2 - 4} x − 2 1 + x + 2 1 + x 2 − 4 2 x
A 2 x ( x − 2 ) ( x + 2 ) ( x 2 − 4 ) \dfrac{2x}{(x - 2)(x + 2)(x^2 - 4)} ( x − 2 ) ( x + 2 ) ( x 2 − 4 ) 2 x B 2 x x 2 − 4 \dfrac{2x}{x^2 - 4} x 2 − 4 2 x C x x 2 − 4 \dfrac{x}{x^2 - 4} x 2 − 4 x D 4 x x 2 − 4 \dfrac{4x}{x^2 - 4} x 2 − 4 4 x
Worked solution (try it first) Since
x 2 − 4 = ( x − 2 ) ( x + 2 ) x^2 - 4 = (x - 2)(x + 2) x 2 − 4 = ( x − 2 ) ( x + 2 ) , the LCD is just
x 2 − 4 x^2 - 4 x 2 − 4 .
Change each top to match:
1 x − 2 = x + 2 x 2 − 4 \frac{1}{x - 2} = \frac{x + 2}{x^2 - 4} x − 2 1 = x 2 − 4 x + 2 and
1 x + 2 = x − 2 x 2 − 4 \frac{1}{x + 2} = \frac{x - 2}{x^2 - 4} x + 2 1 = x 2 − 4 x − 2 .
Add the tops:
( x + 2 ) + ( x − 2 ) + 2 x = 4 x (x + 2) + (x - 2) + 2x = 4x ( x + 2 ) + ( x − 2 ) + 2 x = 4 x .
So the sum is
4 x x 2 − 4 \dfrac{4x}{x^2 - 4} x 2 − 4 4 x , option D.
Watch out
Don't multiply all the bottoms together: x 2 − 4 x^2 - 4 x 2 − 4 x − 2 x - 2 x − 2 x + 2 x + 2 x + 2 ( x − 2 ) ( x + 2 ) ( x 2 − 4 ) (x - 2)(x + 2)(x^2 - 4) ( x − 2 ) ( x + 2 ) ( x 2 − 4 ) Report a problem with this question
Make v v v S 2 = 6 v − w 2 S^2 = \frac{6}{v} - \frac{w}{2} S 2 = v 6 − 2 w
A v = 6 S 2 − 12 w v = \frac{6}{S^2} - \frac{12}{w} v = S 2 6 − w 12 B v = 12 2 S 2 − w v = \frac{12}{2S^2 - w} v = 2 S 2 − w 12 C v = 12 w − 2 S 2 v = \frac{12}{w} - 2S^2 v = w 12 − 2 S 2 D v = 12 2 S 2 + w v = \frac{12}{2S^2 + w} v = 2 S 2 + w 12
Worked solution (try it first) Multiply every term by 2 to clear the halves:
2 S 2 = 12 v − w 2S^2 = \frac{12}{v} - w 2 S 2 = v 12 − w .
Add
w w w to both sides:
12 v = 2 S 2 + w \frac{12}{v} = 2S^2 + w v 12 = 2 S 2 + w .
Turn both sides upside down and multiply by 12:
v = 12 2 S 2 + w v = \dfrac{12}{2S^2 + w} v = 2 S 2 + w 12 , option D.
Watch out
w w w + w +w + w 12 2 S 2 − w \frac{12}{2S^2 - w} 2 S 2 − w 12 Report a problem with this question
If a b = c d = k \dfrac ab = \dfrac cd = k b a = d c = k 3 a 2 − a c + c 2 3 b 2 − b d + d 2 \dfrac{3a^2 - ac + c^2}{3b^2 - bd + d^2} 3 b 2 − b d + d 2 3 a 2 − a c + c 2 k k k
A 3 k 2 3k^2 3 k 2 B 3 k − k 2 3k - k^2 3 k − k 2 C 17 k 2 4 \frac{17k^2}{4} 4 17 k 2 D k 2 k^2 k 2
Worked solution (try it first) From
a b = c d = k \frac ab = \frac cd = k b a = d c = k , write
a = k b a = kb a = k b and
c = k d c = kd c = k d .
Substitute in the top:
3 k 2 b 2 − k 2 b d + k 2 d 2 = k 2 ( 3 b 2 − b d + d 2 ) 3k^2b^2 - k^2bd + k^2d^2 = k^2(3b^2 - bd + d^2) 3 k 2 b 2 − k 2 b d + k 2 d 2 = k 2 ( 3 b 2 − b d + d 2 ) .
The bracket is exactly the bottom, so it cancels and the value is
k 2 k^2 k 2 , option D.
Watch out
Every term on top gains k 2 k^2 k 2 a c = ( k b ) ( k d ) = k 2 b d ac = (kb)(kd) = k^2bd a c = ( k b ) ( k d ) = k 2 b d 3 k 2 3k^2 3 k 2 Report a problem with this question
At what points does the straight line y = 2 x + 1 y = 2x + 1 y = 2 x + 1 y = 2 x 2 + 5 x − 1 y = 2x^2 + 5x - 1 y = 2 x 2 + 5 x − 1
A ( − 2 , − 3 ) (-2, -3) ( − 2 , − 3 ) ( 1 2 , 2 ) (\frac12, 2) ( 2 1 , 2 ) B ( − 1 2 , 0 ) (-\frac12, 0) ( − 2 1 , 0 ) ( 2 , 5 ) (2, 5) ( 2 , 5 ) C ( 1 2 , 2 ) (\frac12, 2) ( 2 1 , 2 ) ( 1 , 3 ) (1, 3) ( 1 , 3 ) D ( 1 , 3 ) (1, 3) ( 1 , 3 ) ( 2 , 5 ) (2, 5) ( 2 , 5 )
Worked solution (try it first) Where they meet,
2 x 2 + 5 x − 1 = 2 x + 1 2x^2 + 5x - 1 = 2x + 1 2 x 2 + 5 x − 1 = 2 x + 1 .
Bring everything to one side:
2 x 2 + 3 x − 2 = 0 2x^2 + 3x - 2 = 0 2 x 2 + 3 x − 2 = 0 .
Factorise:
( 2 x − 1 ) ( x + 2 ) = 0 (2x - 1)(x + 2) = 0 ( 2 x − 1 ) ( x + 2 ) = 0 , so
x = 1 2 x = \frac12 x = 2 1 or
x = − 2 x = -2 x = − 2 .
Find
y y y from the line
y = 2 x + 1 y = 2x + 1 y = 2 x + 1 :
x = 1 2 x = \frac12 x = 2 1 gives
y = 2 y = 2 y = 2 , and
x = − 2 x = -2 x = − 2 gives
y = − 3 y = -3 y = − 3 .
So the points are
( − 2 , − 3 ) (-2, -3) ( − 2 , − 3 ) and
( 1 2 , 2 ) (\frac12, 2) ( 2 1 , 2 ) , option A.
Watch out
Every point in every option lies on the line y = 2 x + 1 y = 2x + 1 y = 2 x + 1 ( 1 , 3 ) (1, 3) ( 1 , 3 ) y = 2 + 5 − 1 = 6 y = 2 + 5 - 1 = 6 y = 2 + 5 − 1 = 6 Report a problem with this question
A regular polygon of n n n 160 ∘ 160^\circ 16 0 ∘ n n n
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 ∘ − 160 ∘ = 20 ∘ 180^\circ - 160^\circ = 20^\circ 18 0 ∘ − 16 0 ∘ = 2 0 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so
n = 360 ÷ 20 = 18 n = 360 \div 20 = 18 n = 360 ÷ 20 = 18 , option A.
Watch out
Divide 360 ∘ 360^\circ 36 0 ∘ 20 ∘ 20^\circ 2 0 ∘ 360 ÷ 160 360 \div 160 360 ÷ 160 Report a problem with this question
If cos θ = a b \cos\theta = \dfrac ab cos θ = b a 1 + tan 2 θ 1 + \tan^2\theta 1 + tan 2 θ
A b 2 a 2 \dfrac{b^2}{a^2} a 2 b 2 B a 2 b 2 \dfrac{a^2}{b^2} b 2 a 2 C a 2 + b 2 b 2 − a 2 \dfrac{a^2 + b^2}{b^2 - a^2} b 2 − a 2 a 2 + b 2 D 2 a 2 + b 2 a 2 + b 2 \dfrac{2a^2 + b^2}{a^2 + b^2} a 2 + b 2 2 a 2 + b 2
Worked solution (try it first) Use the identity
1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ .
sec θ = 1 cos θ \sec\theta = \dfrac{1}{\cos\theta} sec θ = cos θ 1 = b a = \dfrac ba = a b , so
sec 2 θ = b 2 a 2 \sec^2\theta = \dfrac{b^2}{a^2} sec 2 θ = a 2 b 2 , option A.
Watch out
sec θ \sec\theta sec θ cos θ \cos\theta cos θ a b \frac ab b a cos θ \cos\theta cos θ a 2 b 2 \frac{a^2}{b^2} b 2 a 2 Report a problem with this question
In the diagram, P Q PQ P Q R S RS R S O O O T T T T P = 24 TP = 24 T P = 24 T Q = 8 TQ = 8 T Q = 8 T S = 12 TS = 12 T S = 12 T R TR T R
O P Q R S T
Worked solution (try it first) For two secants from an outside point
T T T : (outside part) × (whole line) is the same on both.
So
T Q × T P = T S × T R TQ \times TP = TS \times TR T Q × T P = T S × T R .
Put in the lengths:
8 × 24 = 12 × T R 8 \times 24 = 12 \times TR 8 × 24 = 12 × T R , so
192 = 12 × T R 192 = 12 \times TR 192 = 12 × T R .
Divide both sides by 12:
T R = 16 TR = 16 T R = 16 cm, option A.
Watch out
Multiply the outside part by the whole secant, T Q × T P TQ \times TP T Q × T P Q P QP QP Q P = 16 QP = 16 QP = 16 T R = 128 12 ≈ 10.7 TR = \frac{128}{12} \approx 10.7 T R = 12 128 ≈ 10.7 Report a problem with this question
The angle of elevation of the top of a vertical tower 50 m high from a point X X X 30 ∘ 30^\circ 3 0 ∘ Y Y Y 60 ∘ 60^\circ 6 0 ∘ X X X Y Y Y
A 14.43 m B 57.73 m C 101.03 m D 115.47 m
Worked solution (try it first) X X X and
Y Y Y are on opposite sides, so
X Y XY X Y is the sum of their distances from the foot of the tower.
From
X X X :
tan 30 ∘ = 50 d \tan30^\circ = \frac{50}{d} tan 3 0 ∘ = d 50 , so
d = 50 tan 30 ∘ d = \frac{50}{\tan30^\circ} d = t a n 3 0 ∘ 50 ≈ 86.60 \approx 86.60 ≈ 86.60 m.
From
Y Y Y :
d = 50 tan 60 ∘ d = \frac{50}{\tan60^\circ} d = t a n 6 0 ∘ 50 = 50 3 = \frac{50}{\sqrt3} = 3 50 ≈ 28.87 \approx 28.87 ≈ 28.87 m.
So
X Y = 86.60 + 28.87 = 115.47 XY = 86.60 + 28.87 = 115.47 X Y = 86.60 + 28.87 = 115.47 m, option D.
Watch out
The points are on opposite sides of the tower, so add the distances. Subtracting them gives 57.73 m (option B), which is the answer when both points are on the same side. Report a problem with this question
A girl walks 45 m in the direction 050 ∘ 050^\circ 05 0 ∘ Q Q Q X X X 140 ∘ 140^\circ 14 0 ∘ X X X Y Y Y Q Q Q
Worked solution (try it first) At
X X X the direction back to
Q Q Q is
050 ∘ + 180 ∘ = 230 ∘ 050^\circ + 180^\circ = 230^\circ 05 0 ∘ + 18 0 ∘ = 23 0 ∘ , and she sets off on
140 ∘ 140^\circ 14 0 ∘ .
These differ by
90 ∘ 90^\circ 9 0 ∘ , so the angle at
X X X is a right angle.
By Pythagoras,
Q Y 2 = 45 2 + 24 2 QY^2 = 45^2 + 24^2 Q Y 2 = 4 5 2 + 2 4 2 , which is
2025 + 576 = 2601 2025 + 576 = 2601 2025 + 576 = 2601 .
So
Q Y = 2601 = 51 QY = \sqrt{2601} = 51 Q Y = 2601 = 51 m, option C.
Watch out
Distances in different directions don't simply add. 45 + 24 = 69 45 + 24 = 69 45 + 24 = 69 Q Q Q Report a problem with this question
A solid prism has the trapezium P Q R S PQRS P QR S P Q = 6 PQ = 6 P Q = 6 S R = 11 SR = 11 S R = 11 Q R = 12 QR = 12 QR = 12
A 102 m 3 102\text{ m}^3 102 m 3 B 576 m 3 576\text{ m}^3 576 m 3 C 816 m 3 816\text{ m}^3 816 m 3 D 1056 m 3 1056\text{ m}^3 1056 m 3
Worked solution (try it first) The cross-section is a trapezium with parallel sides 6 m and 11 m, 12 m apart.
Its area is half the sum of the parallel sides times the distance between them:
1 2 ( 6 + 11 ) × 12 = 102 m 2 \frac12(6 + 11) \times 12 = 102\text{ m}^2 2 1 ( 6 + 11 ) × 12 = 102 m 2 .
Volume of a prism = cross-section area × length:
102 × 8 = 816 m 3 102 \times 8 = 816\text{ m}^3 102 × 8 = 816 m 3 , option C.
Watch out
102 102 102 Report a problem with this question
P Q PQ P Q P R PR P R P P P O O O ∠ Q R P = 34 ∘ \angle QRP = 34^\circ ∠ QR P = 3 4 ∘ x x x ∠ Q O R \angle QOR ∠ QO R
A 34 ∘ 34^\circ 3 4 ∘ B 56 ∘ 56^\circ 5 6 ∘ C 68 ∘ 68^\circ 6 8 ∘ D 112 ∘ 112^\circ 11 2 ∘
Worked solution (try it first) ∠ Q R P \angle QRP ∠ QR P is the angle between the tangent
R P RP R P and the chord
R Q RQ R Q .
It equals the angle that
Q R QR QR makes at any point on the major arc.
The angle at the centre is twice the angle at the circumference, so
x = 2 × 34 ∘ = 68 ∘ x = 2 \times 34^\circ = 68^\circ x = 2 × 3 4 ∘ = 6 8 ∘ , option C.
Check: tangents from
P P P are equal, so
∠ P Q R = 34 ∘ \angle PQR = 34^\circ ∠ P QR = 3 4 ∘ and
∠ Q P R = 112 ∘ \angle QPR = 112^\circ ∠ QP R = 11 2 ∘ .
In
O Q P R OQPR O QP R the radii meet the tangents at
90 ∘ 90^\circ 9 0 ∘ , so
x = 360 ∘ − 90 ∘ − 90 ∘ − 112 ∘ x = 360^\circ - 90^\circ - 90^\circ - 112^\circ x = 36 0 ∘ − 9 0 ∘ − 9 0 ∘ − 11 2 ∘ Watch out
112 ∘ 112^\circ 11 2 ∘ ∠ Q P R \angle QPR ∠ QP R 180 ∘ − 112 ∘ = 68 ∘ 180^\circ - 112^\circ = 68^\circ 18 0 ∘ − 11 2 ∘ = 6 8 ∘ Report a problem with this question
An arc of a circle of radius 6 cm is 8 cm long. Find the area of the sector.
A 5 1 3 cm 2 5\frac13\text{ cm}^2 5 3 1 cm 2 B 24 cm 2 24\text{ cm}^2 24 cm 2 C 36 cm 2 36\text{ cm}^2 36 cm 2 D 48 cm 2 48\text{ cm}^2 48 cm 2
Worked solution (try it first) The area of a sector is
1 2 × \frac12 \times 2 1 × radius
× \times × arc length (the arc plays the part of the base of a thin triangle).
So the area is
1 2 × 6 × 8 = 24 cm 2 \frac12 \times 6 \times 8 = 24\text{ cm}^2 2 1 × 6 × 8 = 24 cm 2 , option B.
Watch out
Don't drop the 1 2 \frac12 2 1 6 × 8 = 48 cm 2 6 \times 8 = 48\text{ cm}^2 6 × 8 = 48 cm 2 Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z X Y = 4 XY = 4 X Y = 4 X Z = 3 XZ = 3 X Z = 3 Y Z = 6 YZ = 6 Y Z = 6 Z Z Z
A 3 4 \frac34 4 3 B 29 36 \frac{29}{36} 36 29 C 2 3 \frac23 3 2 D 1 2 \frac12 2 1
Worked solution (try it first) Angle
Z Z Z is between
X Z = 3 XZ = 3 X Z = 3 and
Y Z = 6 YZ = 6 Y Z = 6 .
The side facing it is
X Y = 4 XY = 4 X Y = 4 .
Cosine rule:
cos Z = 3 2 + 6 2 − 4 2 2 × 3 × 6 \cos Z = \dfrac{3^2 + 6^2 - 4^2}{2 \times 3 \times 6} cos Z = 2 × 3 × 6 3 2 + 6 2 − 4 2 , which is
9 + 36 − 16 36 \dfrac{9 + 36 - 16}{36} 36 9 + 36 − 16 .
So
cos Z = 29 36 \cos Z = \frac{29}{36} cos Z = 36 29 , option B.
Watch out
Subtract the square of the side opposite Z Z Z X Y = 4 XY = 4 X Y = 4 Z Z Z 6 2 6^2 6 2 9 + 16 − 36 24 = − 11 24 \frac{9 + 16 - 36}{24} = -\frac{11}{24} 24 9 + 16 − 36 = − 24 11 Report a problem with this question
In the figure, △ P Q T \triangle PQT △ P QT P Q = Q T PQ = QT P Q = QT S S S P T PT P T ∠ S R Q = 35 ∘ \angle SRQ = 35^\circ ∠ S R Q = 3 5 ∘ ∠ T P Q = 20 ∘ \angle TPQ = 20^\circ ∠ T P Q = 2 0 ∘ P Q R PQR P QR ∠ T S R \angle TSR ∠ T S R
A 20 ∘ 20^\circ 2 0 ∘ B 55 ∘ 55^\circ 5 5 ∘ C 75 ∘ 75^\circ 7 5 ∘ D 140 ∘ 140^\circ 14 0 ∘
Worked solution (try it first) Look at triangle
P S R PSR P S R : its angle at
P P P is
20 ∘ 20^\circ 2 0 ∘ and its angle at
R R R is
35 ∘ 35^\circ 3 5 ∘ .
∠ T S R \angle TSR ∠ T S R is the exterior angle of triangle
P S R PSR P S R at
S S S , because
P S T PST P S T is a straight line.
An exterior angle equals the sum of the two interior opposite angles:
∠ T S R = 20 ∘ + 35 ∘ \angle TSR = 20^\circ + 35^\circ ∠ T S R = 2 0 ∘ + 3 5 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option B.
Watch out
You don't need P Q = Q T PQ = QT P Q = QT ∠ P Q T = 180 ∘ − 2 × 20 ∘ = 140 ∘ \angle PQT = 180^\circ - 2 \times 20^\circ = 140^\circ ∠ P QT = 18 0 ∘ − 2 × 2 0 ∘ = 14 0 ∘ Q Q Q S S S Report a problem with this question
Find the total surface area of a solid cone of radius 2 3 2\sqrt3 2 3 4 3 4\sqrt3 4 3
A 8 3 π 8\sqrt3\,\pi 8 3 π B 24 π 24\pi 24 π C 15 3 π 15\sqrt3\,\pi 15 3 π D 36 π 36\pi 36 π
Worked solution (try it first) A solid cone has a curved surface
π r l \pi rl π r l and a circular base
π r 2 \pi r^2 π r 2 , so the total is
π r ( r + l ) \pi r(r + l) π r ( r + l ) .
Here
r + l = 2 3 + 4 3 = 6 3 r + l = 2\sqrt3 + 4\sqrt3 = 6\sqrt3 r + l = 2 3 + 4 3 = 6 3 .
So the area is
π × 2 3 × 6 3 = 12 × 3 × π \pi \times 2\sqrt3 \times 6\sqrt3 = 12 \times 3 \times \pi π × 2 3 × 6 3 = 12 × 3 × π = 36 π = 36\pi = 36 π cm², option D.
Watch out
A solid cone has a base as well. The curved surface alone is π × 2 3 × 4 3 = 24 π \pi \times 2\sqrt3 \times 4\sqrt3 = 24\pi π × 2 3 × 4 3 = 24 π Report a problem with this question
If U U U V V V W W W ∠ U W V \angle UWV ∠ U W V W W W
A The perpendicular bisector of U V UV U V B A circle with U V UV U V C A line parallel to the line U V UV U V D A circle with the line U V UV U V
Worked solution (try it first) The angle in a semicircle is a right angle, and the converse holds too: if
∠ U W V = 90 ∘ \angle UWV = 90^\circ ∠ U W V = 9 0 ∘ , then
W W W lies on the circle with
U V UV U V as diameter.
As
W W W moves round (on both sides of
U V UV U V ), it traces that whole circle.
So the locus is a circle with
U V UV U V as diameter, option D.
Watch out
U V UV U V U V UV U V U U U V V V W W W 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
In the figure, P Q ∥ S T PQ \parallel ST P Q ∥ S T R S ∥ U V RS \parallel UV R S ∥ U V ∠ P Q R = 35 ∘ \angle PQR = 35^\circ ∠ P QR = 3 5 ∘ ∠ Q R S = 65 ∘ \angle QRS = 65^\circ ∠ QR S = 6 5 ∘ ∠ S T V \angle STV ∠ S T V
A 30 ∘ 30^\circ 3 0 ∘ B 35 ∘ 35^\circ 3 5 ∘ C 55 ∘ 55^\circ 5 5 ∘ D 65 ∘ 65^\circ 6 5 ∘
Worked solution (try it first) Produce
S R SR S R back beyond
R R R to meet
P Q PQ P Q at a point
X X X .
Angles on a straight line:
∠ X R Q = 180 ∘ − 65 ∘ \angle XRQ = 180^\circ - 65^\circ ∠ X R Q = 18 0 ∘ − 6 5 ∘ The angles of triangle
X Q R XQR X QR add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ Q X R = 180 ∘ − 115 ∘ − 35 ∘ \angle QXR = 180^\circ - 115^\circ - 35^\circ ∠ QX R = 18 0 ∘ − 11 5 ∘ − 3 5 ∘ This is the angle between the lines
P Q PQ P Q and
R S RS R S .
S T ∥ P Q ST \parallel PQ S T ∥ P Q and
T V ∥ R S TV \parallel RS T V ∥ R S , so
∠ S T V \angle STV ∠ S T V is the same angle between the same two directions:
∠ S T V = 30 ∘ \angle STV = 30^\circ ∠ S T V = 3 0 ∘ , option A.
Watch out
Don't copy a given angle: 35 ∘ 35^\circ 3 5 ∘ 65 ∘ 65^\circ 6 5 ∘ Q Q Q R R R 65 ∘ − 35 ∘ = 30 ∘ 65^\circ - 35^\circ = 30^\circ 6 5 ∘ − 3 5 ∘ = 3 0 ∘ Report a problem with this question
An open rectangular box externally measures 4 m × 3 m × 4 m 4\text{ m} \times 3\text{ m} \times 4\text{ m} 4 m × 3 m × 4 m
A ₦96.00 B ₦112.00 C ₦136.00 D ₦160.00
Worked solution (try it first) The box is open, so paint the base and the four sides but not the top.
Base:
4 × 3 = 12 m 2 4 \times 3 = 12\text{ m}^2 4 × 3 = 12 m 2 .
Sides:
2 ( 4 × 4 ) + 2 ( 3 × 4 ) = 32 + 24 2(4 \times 4) + 2(3 \times 4) = 32 + 24 2 ( 4 × 4 ) + 2 ( 3 × 4 ) = 32 + 24 = 56 m 2 = 56\text{ m}^2 = 56 m 2 .
Total
12 + 56 = 68 m 2 12 + 56 = 68\text{ m}^2 12 + 56 = 68 m 2 , and the cost is
68 × 2 = 136 68 \times 2 = 136 68 × 2 = 136 , so ₦136.00, option C.
Watch out
Leave out the top of an open box. Painting it as well adds 12 m 2 12\text{ m}^2 12 m 2 Report a problem with this question
Of 900 students admitted to a university in 1979, the distribution by state was: Anambra 185, Imo 135, Kaduna 90, Kwara 110, Ondo 155, Oyo 225. In a pie chart of this distribution, the angle at the centre for Anambra is
A 50 ∘ 50^\circ 5 0 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 74 ∘ 74^\circ 7 4 ∘ D 88 ∘ 88^\circ 8 8 ∘
Worked solution (try it first) The total is 900 students, so each student gets
360 ∘ 900 = 0.4 ∘ \frac{360^\circ}{900} = 0.4^\circ 900 36 0 ∘ = 0. 4 ∘ .
Anambra has 185 students:
185 × 0.4 ∘ = 74 ∘ 185 \times 0.4^\circ = 74^\circ 185 × 0. 4 ∘ = 7 4 ∘ , option C.
Watch out
Divide by the total, 900, not by the number of states. Check the total first: 185 + 135 + 90 + 110 + 155 + 225 = 900 185 + 135 + 90 + 110 + 155 + 225 = 900 185 + 135 + 90 + 110 + 155 + 225 = 900 Report a problem with this question
Find the median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119.
Worked solution (try it first) Put the 10 numbers in order: 89, 100, 108, 119, 120, 130, 131, 131, 141, 161.
With an even count, the median is halfway between the 5th and 6th: 120 and 130.
So the median is
120 + 130 2 = 125 \frac{120 + 130}{2} = 125 2 120 + 130 = 125 , option B.
Watch out
Order the list first. 131 (option A) is the mode, and 120 (option D) is only the 5th value; with 10 numbers you average the 5th and 6th. Report a problem with this question
Find the probability that a number selected at random from 40 to 50 is a prime.
A 3 11 \frac3{11} 11 3 B 5 11 \frac5{11} 11 5 C 3 10 \frac3{10} 10 3 D 4 11 \frac4{11} 11 4
Worked solution (try it first) From 40 to 50 inclusive there are
50 − 40 + 1 = 11 50 - 40 + 1 = 11 50 − 40 + 1 = 11 numbers.
The primes among them are 41, 43 and 47.
So the probability is
3 11 \frac{3}{11} 11 3 , option A.
Watch out
Include both ends: 40 to 50 is 11 numbers, not 50 − 40 = 10 50 - 40 = 10 50 − 40 = 10 3 10 \frac{3}{10} 10 3 Report a problem with this question
A man kept 6 black, 5 brown and 7 purple shirts in a drawer. What is the probability of his picking a purple shirt with his eyes closed?
A 1 7 \frac17 7 1 B 11 18 \frac{11}{18} 18 11 C 7 18 \frac7{18} 18 7 D 7 11 \frac7{11} 11 7
Worked solution (try it first) There are
6 + 5 + 7 = 18 6 + 5 + 7 = 18 6 + 5 + 7 = 18 shirts, each equally likely.
7 of them are purple, so the probability is
7 18 \frac{7}{18} 18 7 , option C.
Watch out
Divide by all the shirts, not by the other shirts: 7 11 \frac{7}{11} 11 7 Report a problem with this question
The table gives the scores of a group of students in a Mathematics test. If the mode is m m m S S S ( S , m ) (S, m) ( S , m )
Score
1
2
3
4
5
6
7
8
Frequency
2
4
7
14
12
6
4
1
A ( 27 , 4 ) (27, 4) ( 27 , 4 ) B ( 14 , 4 ) (14, 4) ( 14 , 4 ) C ( 13 , 4 ) (13, 4) ( 13 , 4 ) D ( 4 , 4 ) (4, 4) ( 4 , 4 )
Worked solution (try it first) The students who scored 4 or less are those with scores 1, 2, 3 and 4:
S = 2 + 4 + 7 + 14 = 27 S = 2 + 4 + 7 + 14 = 27 S = 2 + 4 + 7 + 14 = 27 .
The mode is the score with the highest frequency: score 4, with frequency 14.
So
( S , m ) = ( 27 , 4 ) (S, m) = (27, 4) ( S , m ) = ( 27 , 4 ) , option A.
Watch out
"4 or less" means scores 1 to 4 together, not just the 14 students who scored exactly 4 (option B). Report a problem with this question