Paper JAMB 1989 General Maths Objective
Objective paper · 40 questions · partial
JAMB 1989 · UME Topics include Indices & standard form, Number foundations & fractions, Approximation & error, Commercial arithmetic, Solid mensuration, Surds.
Our copy of this paper is missing questions 1, 4, 13, 14, 16, 27, 29, 35, 39, 47.
Sit this paper Answer every question in order, timed if you like (suggested 25 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
2 3 5 6 7 8 9 10 11 12 15 17 18 19 20 21 22 23 24 25 26 28 30 31 32 33 34 36 37 38 40 41 42 43 44 45 46 48 49 50 Evaluate 2,700,000 × 0.03 ÷ 18,000 2{,}700{,}000 \times 0.03 \div 18{,}000 2 , 700 , 000 × 0.03 ÷ 18 , 000
A 4.5 × 10 0 4.5 \times 10^0 4.5 × 1 0 0 B 4.5 × 10 1 4.5 \times 10^1 4.5 × 1 0 1 C 4.5 × 10 2 4.5 \times 10^2 4.5 × 1 0 2 D 4.5 × 10 3 4.5 \times 10^3 4.5 × 1 0 3
Worked solution (try it first) Multiply first:
2 700 000 × 0.03 2\,700\,000 \times 0.03 2 700 000 × 0.03 is 3 hundredths of 2 700 000, which is
81 000 81\,000 81 000 .
Divide:
81 000 ÷ 18 000 = 81 ÷ 18 = 4.5 81\,000 \div 18\,000 = 81 \div 18 = 4.5 81 000 ÷ 18 000 = 81 ÷ 18 = 4.5 .
In standard form
4.5 = 4.5 × 10 0 4.5 = 4.5 \times 10^0 4.5 = 4.5 × 1 0 0 , option A.
Watch out
0.03 is three hundredths, not three tenths. Using 0.3 gives 810 000 ÷ 18 000 = 45 810\,000 \div 18\,000 = 45 810 000 ÷ 18 000 = 45 4.5 × 10 1 4.5 \times 10^1 4.5 × 1 0 1 Report a problem with this question
The prime factors of 2,520 are
A 2, 9, 5 B 2, 9, 7 C 2, 3, 5, 7 D 2, 3, 7, 9
Worked solution (try it first) Divide by primes:
2520 = 2 3 × 315 2520 = 2^3 \times 315 2520 = 2 3 × 315 , and
315 = 3 2 × 35 315 = 3^2 \times 35 315 = 3 2 × 35 = 3 2 × 5 × 7 = 3^2 \times 5 \times 7 = 3 2 × 5 × 7 .
So
2520 = 2 3 × 3 2 × 5 × 7 2520 = 2^3 \times 3^2 \times 5 \times 7 2520 = 2 3 × 3 2 × 5 × 7 , and the prime factors are 2, 3, 5 and 7, option C.
Watch out
9 is a factor of 2520 but not a prime factor, because 9 = 3 × 3 9 = 3 \times 3 9 = 3 × 3 Report a problem with this question
Simplify ( 64 r − 6 3 ) 1 2 \left(\sqrt[3]{64r^{-6}}\right)^{\frac12} ( 3 64 r − 6 ) 2 1
A r r r B 2 r 2r 2 r C 1 2 r \frac{1}{2r} 2 r 1 D 2 r \frac2r r 2
Worked solution (try it first) Cube root first:
64 3 = 4 \sqrt[3]{64} = 4 3 64 = 4 and
r − 6 3 = r − 2 \sqrt[3]{r^{-6}} = r^{-2} 3 r − 6 = r − 2 , so the inside is
4 r − 2 4r^{-2} 4 r − 2 .
The power
1 2 \frac12 2 1 is a square root:
( 4 r − 2 ) 1 2 = 2 r − 1 (4r^{-2})^{\frac12} = 2r^{-1} ( 4 r − 2 ) 2 1 = 2 r − 1 .
A negative index means one over:
2 r − 1 = 2 r 2r^{-1} = \frac2r 2 r − 1 = r 2 , option D.
Watch out
Keep the negative index: r − 1 = 1 r r^{-1} = \frac1r r − 1 = r 1 2 r \frac2r r 2 2 r 2r 2 r Report a problem with this question
What is the difference between 0.007685 correct to three significant figures and 0.007685 correct to four decimal places?
A 10 − 5 10^{-5} 1 0 − 5 B 7 × 10 − 4 7 \times 10^{-4} 7 × 1 0 − 4 C 8 × 10 − 5 8 \times 10^{-5} 8 × 1 0 − 5 D 10 − 6 10^{-6} 1 0 − 6
Worked solution (try it first) To 3 significant figures: the zeros after the point don't count, so keep 7, 6, 8 and look at the 5.
Round up: 0.00769.
To 4 decimal places: keep 0.0076 and look at the fifth decimal, 8.
Round up: 0.0077.
The difference is
0.0077 − 0.00769 = 0.00001 = 10 − 5 0.0077 - 0.00769 = 0.00001 = 10^{-5} 0.0077 − 0.00769 = 0.00001 = 1 0 − 5 , option A.
Watch out
Significant figures start at the first non-zero digit, 7, but decimal places count from the point. Mixing them up (for example taking 3 s.f. as 0.008) gives a difference that matches no option. Report a problem with this question
If a : b = 5 : 8 a : b = 5 : 8 a : b = 5 : 8 x : y = 25 : 16 x : y = 25 : 16 x : y = 25 : 16 a x : b y \frac ax : \frac by x a : y b
A 125 : 128 125 : 128 125 : 128 B 3 : 5 3 : 5 3 : 5 C 3 : 4 3 : 4 3 : 4 D 2 : 5 2 : 5 2 : 5
Worked solution (try it first) Write the ratio as a fraction and divide:
a x ÷ b y = a x × y b \dfrac ax \div \dfrac by = \dfrac ax \times \dfrac yb x a ÷ y b = x a × b y = a b × y x = \dfrac ab \times \dfrac yx = b a × x y .
Put in
a b = 5 8 \frac ab = \frac58 b a = 8 5 and
y x = 16 25 \frac yx = \frac{16}{25} x y = 25 16 :
5 8 × 16 25 = 80 200 \frac58 \times \frac{16}{25} = \frac{80}{200} 8 5 × 25 16 = 200 80 So
a x : b y = 2 : 5 \frac ax : \frac by = 2 : 5 x a : y b = 2 : 5 , option D.
Watch out
x x x 5 × 25 : 8 × 16 5 \times 25 : 8 \times 16 5 × 25 : 8 × 16 125 : 128 125 : 128 125 : 128 Report a problem with this question
Oke deposited ₦800.00 in the bank at the rate of 12 1 2 % 12\frac12\% 12 2 1 %
Worked solution (try it first) The amount is
1 1 2 1\frac12 1 2 1 times ₦800, which is ₦1,200, so the interest is
1200 − 800 = 1200 - 800 = 1200 − 800 = ₦400.
One year's interest at
12 1 2 % 12\frac12\% 12 2 1 % is
0.125 × 800 = 0.125 \times 800 = 0.125 × 800 = ₦100.
So it takes
400 ÷ 100 = 4 400 \div 100 = 4 400 ÷ 100 = 4 years, option B.
Watch out
The interest is only half the principal, ₦400. Taking the interest as the whole ₦800 gives 8 years (option D). Report a problem with this question
If the surface area of a sphere is increased by 44 % 44\% 44%
Worked solution (try it first) Surface area goes with the square of the diameter.
A 44% increase multiplies the area by 1.44.
So the diameter is multiplied by
1.44 = 1.2 \sqrt{1.44} = 1.2 1.44 = 1.2 .
That is a 20% increase, option D.
Watch out
Take the square root of the scale factor, not half the percentage: 1.44 = 1.2 \sqrt{1.44} = 1.2 1.44 = 1.2 Report a problem with this question
Simplify 4 − 1 2 − 3 4 - \dfrac{1}{2 - \sqrt3} 4 − 2 − 3 1
A 2 3 2\sqrt3 2 3 B 2 + 3 2 + \sqrt3 2 + 3 C − 2 + 3 -2 + \sqrt3 − 2 + 3 D 2 − 3 2 - \sqrt3 2 − 3
Worked solution (try it first) Rationalise the fraction by multiplying the top and bottom by
2 + 3 2 + \sqrt3 2 + 3 .
The bottom becomes
4 − 3 = 1 4 - 3 = 1 4 − 3 = 1 , so
1 2 − 3 = 2 + 3 \dfrac{1}{2 - \sqrt3} = 2 + \sqrt3 2 − 3 1 = 2 + 3 .
Subtract from 4:
4 − ( 2 + 3 ) = 2 − 3 4 - (2 + \sqrt3) = 2 - \sqrt3 4 − ( 2 + 3 ) = 2 − 3 , option D.
Watch out
2 + 3 2 + \sqrt3 2 + 3 Similar: JAMB 2004 · UME · Q7
Report a problem with this question
Find p p p q q q log 3 p + 3 log 3 q = 3 \log_3 p + 3\log_3 q = 3 log 3 p + 3 log 3 q = 3
A ( 3 q ) 3 \left(\frac3q\right)^3 ( q 3 ) 3 B ( q 3 ) 1 / 3 \left(\frac q3\right)^{1/3} ( 3 q ) 1/3 C ( q 3 ) 3 \left(\frac q3\right)^3 ( 3 q ) 3 D ( 3 q ) 1 / 3 \left(\frac3q\right)^{1/3} ( q 3 ) 1/3
Worked solution (try it first) Move the 3 up as a power:
3 log 3 q = log 3 q 3 3\log_3 q = \log_3 q^3 3 log 3 q = log 3 q 3 , so the left side is
log 3 ( p q 3 ) = 3 \log_3 (pq^3) = 3 log 3 ( p q 3 ) = 3 .
Change to index form:
p q 3 = 3 3 = 27 pq^3 = 3^3 = 27 p q 3 = 3 3 = 27 .
So
p = 27 q 3 p = \frac{27}{q^3} p = q 3 27 = ( 3 q ) 3 = \left(\frac3q\right)^3 = ( q 3 ) 3 , option A.
Watch out
Divide 27 by q 3 q^3 q 3 q q q ( q 3 ) 3 \left(\frac q3\right)^3 ( 3 q ) 3 Report a problem with this question
What are the values of y y y 9 y − 4 ( 3 y ) + 3 = 0 9^y - 4(3^y) + 3 = 0 9 y − 4 ( 3 y ) + 3 = 0
A − 1 -1 − 1 B − 1 -1 − 1 C 1 and 3 D 0 and 1
Worked solution (try it first) Then
9 y = ( 3 2 ) y = u 2 9^y = (3^2)^y = u^2 9 y = ( 3 2 ) y = u 2 , and the equation is
u 2 − 4 u + 3 = 0 u^2 - 4u + 3 = 0 u 2 − 4 u + 3 = 0 .
Factorise:
( u − 1 ) ( u − 3 ) = 0 (u - 1)(u - 3) = 0 ( u − 1 ) ( u − 3 ) = 0 , so
u = 1 u = 1 u = 1 or
u = 3 u = 3 u = 3 .
3 y = 1 = 3 0 3^y = 1 = 3^0 3 y = 1 = 3 0 gives
y = 0 y = 0 y = 0 , and
3 y = 3 = 3 1 3^y = 3 = 3^1 3 y = 3 = 3 1 gives
y = 1 y = 1 y = 1 .
So
y = 0 y = 0 y = 0 and 1, option D.
Watch out
1 and 3 are the values of u = 3 y u = 3^y u = 3 y y y y 3 0 = 1 3^0 = 1 3 0 = 1 3 1 = 3 3^1 = 3 3 1 = 3 Report a problem with this question
The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is ₦74.00 for 20 students and ₦96.00 for 30 students, find the cost for 15 students.
A ₦68.50 B ₦63.00 C ₦60.00 D ₦52.00
Worked solution (try it first) Partly constant and partly varying as the number
n n n :
C = a + b n C = a + bn C = a + bn .
a + 20 b = 74 a + 20b = 74 a + 20 b = 74 and
a + 30 b = 96 a + 30b = 96 a + 30 b = 96 .
Subtract:
10 b = 22 10b = 22 10 b = 22 , so
b = 2.2 b = 2.2 b = 2.2 .
Then
20 b = 44 20b = 44 20 b = 44 , so
a = 74 − 44 = 30 a = 74 - 44 = 30 a = 74 − 44 = 30 .
For 15 students:
2.2 × 15 = 33 2.2 \times 15 = 33 2.2 × 15 = 33 , so
C = 30 + 33 = 63 C = 30 + 33 = 63 C = 30 + 33 = 63 , which is ₦63.00, option B.
Watch out
The extra ₦22 pays for 10 students, so 5 fewer than 20 saves only ₦11. Taking off the full ₦22 gives ₦52.00 (option D). Report a problem with this question
Find the positive number x x x 2 x 3 − x 2 − 2 x = 1 2^{x^3 - x^2 - 2x} = 1 2 x 3 − x 2 − 2 x = 1
Worked solution (try it first) 2 0 = 1 2^0 = 1 2 0 = 1 , and no other power of 2 equals 1, so the index must be 0:
x 3 − x 2 − 2 x = 0 x^3 - x^2 - 2x = 0 x 3 − x 2 − 2 x = 0 .
Factorise:
x ( x 2 − x − 2 ) = x ( x − 2 ) ( x + 1 ) = 0 x(x^2 - x - 2) = x(x - 2)(x + 1) = 0 x ( x 2 − x − 2 ) = x ( x − 2 ) ( x + 1 ) = 0 , so
x = 0 x = 0 x = 0 , 2 or
− 1 -1 − 1 .
The only positive one is
x = 2 x = 2 x = 2 , option C.
Watch out
Set the index to 0, not 1. Trying x = 1 x = 1 x = 1 1 − 1 − 2 = − 2 1 - 1 - 2 = -2 1 − 1 − 2 = − 2 2 − 2 = 1 4 2^{-2} = \frac14 2 − 2 = 4 1 Report a problem with this question
Simplify 324 − 4 x 2 2 x + 18 \frac{324 - 4x^2}{2x + 18} 2 x + 18 324 − 4 x 2
A 2 ( x − 9 ) 2(x - 9) 2 ( x − 9 ) B 2 ( 9 + x ) 2(9 + x) 2 ( 9 + x ) C 81 − x 2 81 - x^2 81 − x 2 D − 2 ( x − 9 ) -2(x - 9) − 2 ( x − 9 )
Worked solution (try it first) Take out 4 from the top:
324 − 4 x 2 = 4 ( 81 − x 2 ) 324 - 4x^2 = 4(81 - x^2) 324 − 4 x 2 = 4 ( 81 − x 2 ) .
Difference of two squares:
81 − x 2 = ( 9 − x ) ( 9 + x ) 81 - x^2 = (9 - x)(9 + x) 81 − x 2 = ( 9 − x ) ( 9 + x ) .
The bottom is
2 x + 18 = 2 ( x + 9 ) 2x + 18 = 2(x + 9) 2 x + 18 = 2 ( x + 9 ) .
Cancel
2 ( x + 9 ) 2(x + 9) 2 ( x + 9 ) to leave
2 ( 9 − x ) 2(9 - x) 2 ( 9 − x ) .
Since
9 − x = − ( x − 9 ) 9 - x = -(x - 9) 9 − x = − ( x − 9 ) , this is
− 2 ( x − 9 ) -2(x - 9) − 2 ( x − 9 ) , option D.
Watch out
9 − x 9 - x 9 − x x − 9 x - 9 x − 9 2 ( 9 − x ) 2(9 - x) 2 ( 9 − x ) 2 ( x − 9 ) 2(x - 9) 2 ( x − 9 ) Report a problem with this question
Factorize completely y 3 − 4 x y + x y 3 − 4 y y^3 - 4xy + xy^3 - 4y y 3 − 4 x y + x y 3 − 4 y
A ( x + x y ) ( y + 2 ) ( y − 2 ) (x + xy)(y + 2)(y - 2) ( x + x y ) ( y + 2 ) ( y − 2 ) B ( y + x y ) ( y + 2 ) ( y − 2 ) (y + xy)(y + 2)(y - 2) ( y + x y ) ( y + 2 ) ( y − 2 ) C y ( 1 + x ) ( y + 2 ) ( y − 2 ) y(1 + x)(y + 2)(y - 2) y ( 1 + x ) ( y + 2 ) ( y − 2 ) D y ( 1 − x ) ( y + 2 ) ( y − 2 ) y(1 - x)(y + 2)(y - 2) y ( 1 − x ) ( y + 2 ) ( y − 2 )
Worked solution (try it first) Group the terms with
4 y 4y 4 y and
4 x y 4xy 4 x y :
( y 3 − 4 y ) + ( x y 3 − 4 x y ) (y^3 - 4y) + (xy^3 - 4xy) ( y 3 − 4 y ) + ( x y 3 − 4 x y ) .
Take out
y y y from the first group and
x y xy x y from the second:
y ( y 2 − 4 ) + x y ( y 2 − 4 ) y(y^2 - 4) + xy(y^2 - 4) y ( y 2 − 4 ) + x y ( y 2 − 4 ) .
Take out the common bracket and then
y y y :
y ( 1 + x ) ( y 2 − 4 ) y(1 + x)(y^2 - 4) y ( 1 + x ) ( y 2 − 4 ) .
Difference of two squares:
y 2 − 4 = ( y + 2 ) ( y − 2 ) y^2 - 4 = (y + 2)(y - 2) y 2 − 4 = ( y + 2 ) ( y − 2 ) , so you get
y ( 1 + x ) ( y + 2 ) ( y − 2 ) y(1 + x)(y + 2)(y - 2) y ( 1 + x ) ( y + 2 ) ( y − 2 ) , option C.
Watch out
Option B has the same value, but ( y + x y ) (y + xy) ( y + x y ) y y y Report a problem with this question
If one factor of x 3 − 8 − 1 x^3 - 8^{-1} x 3 − 8 − 1 x − 2 − 1 x - 2^{-1} x − 2 − 1
A x 2 + 2 − 1 x − 4 − 1 x^2 + 2^{-1}x - 4^{-1} x 2 + 2 − 1 x − 4 − 1 B x 2 − 2 − 1 x − 4 − 1 x^2 - 2^{-1}x - 4^{-1} x 2 − 2 − 1 x − 4 − 1 C x 2 + 2 − 1 x + 4 − 1 x^2 + 2^{-1}x + 4^{-1} x 2 + 2 − 1 x + 4 − 1 D x 2 + 2 − 1 x − 4 − 1 x^2 + 2^{-1}x - 4^{-1} x 2 + 2 − 1 x − 4 − 1
Worked solution (try it first) Write the terms as cubes:
8 − 1 = 1 8 = ( 1 2 ) 3 8^{-1} = \frac18 = \left(\frac12\right)^3 8 − 1 = 8 1 = ( 2 1 ) 3 , so the expression is
x 3 − ( 1 2 ) 3 x^3 - \left(\frac12\right)^3 x 3 − ( 2 1 ) 3 .
A difference of two cubes factorises as
a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) .
With
a = x a = x a = x and
b = 1 2 b = \frac12 b = 2 1 : the other factor is
x 2 + 1 2 x + 1 4 x^2 + \frac12x + \frac14 x 2 + 2 1 x + 4 1 , which is
x 2 + 2 − 1 x + 4 − 1 x^2 + 2^{-1}x + 4^{-1} x 2 + 2 − 1 x + 4 − 1 , option C.
Watch out
In a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) Report a problem with this question
Factorize 4 a 2 + 12 a b − c 2 + 9 b 2 4a^2 + 12ab - c^2 + 9b^2 4 a 2 + 12 ab − c 2 + 9 b 2
A 4 a ( a − 3 b ) + ( 3 b − c ) 2 4a(a - 3b) + (3b - c)^2 4 a ( a − 3 b ) + ( 3 b − c ) 2 B ( 2 a + 3 b − c ) ( 2 a + 3 b + c ) (2a + 3b - c)(2a + 3b + c) ( 2 a + 3 b − c ) ( 2 a + 3 b + c ) C ( 2 a − 3 b − c ) ( 2 a − 3 b + c ) (2a - 3b - c)(2a - 3b + c) ( 2 a − 3 b − c ) ( 2 a − 3 b + c ) D 4 a ( a − 3 b ) + ( 3 b + c ) 2 4a(a - 3b) + (3b + c)^2 4 a ( a − 3 b ) + ( 3 b + c ) 2
Worked solution (try it first) Group the first, second and last terms:
4 a 2 + 12 a b + 9 b 2 = ( 2 a + 3 b ) 2 4a^2 + 12ab + 9b^2 = (2a + 3b)^2 4 a 2 + 12 ab + 9 b 2 = ( 2 a + 3 b ) 2 .
So the expression is
( 2 a + 3 b ) 2 − c 2 (2a + 3b)^2 - c^2 ( 2 a + 3 b ) 2 − c 2 , a difference of two squares.
So it factorises as
( 2 a + 3 b − c ) ( 2 a + 3 b + c ) (2a + 3b - c)(2a + 3b + c) ( 2 a + 3 b − c ) ( 2 a + 3 b + c ) , option B.
Watch out
The middle term + 12 a b +12ab + 12 ab ( 2 a + 3 b ) 2 (2a + 3b)^2 ( 2 a + 3 b ) 2 ( 2 a − 3 b ) 2 (2a - 3b)^2 ( 2 a − 3 b ) 2 − 12 a b -12ab − 12 ab Report a problem with this question
What are K K K L L L 1 2 ( 3 y − 4 x ) 2 = 8 x 2 + K x y + L y 2 \frac12(3y - 4x)^2 = 8x^2 + Kxy + Ly^2 2 1 ( 3 y − 4 x ) 2 = 8 x 2 + K x y + L y 2
A − 12 , 9 2 -12, \frac92 − 12 , 2 9 B − 6 , 9 -6, 9 − 6 , 9 C 6 , 9 6, 9 6 , 9 D 12 , 9 2 12, \frac92 12 , 2 9
Worked solution (try it first) Expand the bracket:
( 3 y − 4 x ) 2 = 9 y 2 − 24 x y + 16 x 2 (3y - 4x)^2 = 9y^2 - 24xy + 16x^2 ( 3 y − 4 x ) 2 = 9 y 2 − 24 x y + 16 x 2 .
Halve every term:
1 2 ( 3 y − 4 x ) 2 = 8 x 2 − 12 x y + 9 2 y 2 \frac12(3y - 4x)^2 = 8x^2 - 12xy + \frac92y^2 2 1 ( 3 y − 4 x ) 2 = 8 x 2 − 12 x y + 2 9 y 2 .
Compare with
8 x 2 + K x y + L y 2 8x^2 + Kxy + Ly^2 8 x 2 + K x y + L y 2 :
K = − 12 K = -12 K = − 12 and
L = 9 2 L = \frac92 L = 2 9 , option A.
Watch out
The middle term is 2 × 3 y × ( − 4 x ) = − 24 x y 2 \times 3y \times (-4x) = -24xy 2 × 3 y × ( − 4 x ) = − 24 x y K = 12 K = 12 K = 12 Report a problem with this question
Solve the pair of equations 2 x − 1 − 3 y − 1 = 4 2x^{-1} - 3y^{-1} = 4 2 x − 1 − 3 y − 1 = 4 4 x − 1 + y − 1 = 1 4x^{-1} + y^{-1} = 1 4 x − 1 + y − 1 = 1 x x x y y y
A − 1 , 2 -1, 2 − 1 , 2 B 1 , 2 1, 2 1 , 2 C 2 , 1 2, 1 2 , 1 D 2 , − 1 2, -1 2 , − 1
Worked solution (try it first) Let
u = 1 x u = \frac1x u = x 1 and
v = 1 y v = \frac1y v = y 1 .
The equations become
2 u − 3 v = 4 2u - 3v = 4 2 u − 3 v = 4 and
4 u + v = 1 4u + v = 1 4 u + v = 1 .
From the second,
v = 1 − 4 u v = 1 - 4u v = 1 − 4 u .
Substitute:
2 u − 3 + 12 u = 4 2u - 3 + 12u = 4 2 u − 3 + 12 u = 4 , so
14 u = 7 14u = 7 14 u = 7 and
u = 1 2 u = \frac12 u = 2 1 .
Then
v = 1 − 2 = − 1 v = 1 - 2 = -1 v = 1 − 2 = − 1 .
Turn back:
x = 1 u = 2 x = \frac1u = 2 x = u 1 = 2 and
y = 1 v = − 1 y = \frac1v = -1 y = v 1 = − 1 .
So
x x x and
y y y are
2 , − 1 2, -1 2 , − 1 , option D.
Watch out
Order matters: "for x x x y y y x x x − 1 , 2 -1, 2 − 1 , 2 4 x − 1 + y − 1 = 1 4x^{-1} + y^{-1} = 1 4 x − 1 + y − 1 = 1 Report a problem with this question
What value of Q Q Q 4 x 2 + 5 x + Q 4x^2 + 5x + Q 4 x 2 + 5 x + Q
A 25 16 \frac{25}{16} 16 25 B 25 64 \frac{25}{64} 64 25 C 5 8 \frac58 8 5 D 5 4 \frac54 4 5
Worked solution (try it first) A complete square with
4 x 2 4x^2 4 x 2 has the form
( 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 .
Match the
x x x terms:
4 k = 5 4k = 5 4 k = 5 , so
k = 5 4 k = \frac54 k = 4 5 .
So
Q = k 2 = 25 16 Q = k^2 = \frac{25}{16} Q = k 2 = 16 25 , option A.
Watch out
If you take out the 4 first, 4 ( x 2 + 5 4 x + Q 4 ) 4\left(x^2 + \frac54x + \frac Q4\right) 4 ( x 2 + 4 5 x + 4 Q ) Q 4 = ( 5 8 ) 2 = 25 64 \frac Q4 = \left(\frac58\right)^2 = \frac{25}{64} 4 Q = ( 8 5 ) 2 = 64 25 Q = 25 16 Q = \frac{25}{16} Q = 16 25 Report a problem with this question
Find the range of values of r r r r a + r b + r c > 1 \frac ra + \frac rb + \frac rc > 1 a r + b r + c r > 1 a a a b b b c c c
A r > a b c b c + a c + a b r > \dfrac{abc}{bc + ac + ab} r > b c + a c + ab ab c B r > a b c r > abc r > ab c C r > 1 a + 1 b + 1 c r > \frac1a + \frac1b + \frac1c r > a 1 + b 1 + c 1 D r > 1 a b c r > \dfrac{1}{abc} r > ab c 1
Worked solution (try it first) Take
r r r out as a common factor:
r ( 1 a + 1 b + 1 c ) > 1 r\left(\frac1a + \frac1b + \frac1c\right) > 1 r ( a 1 + b 1 + c 1 ) > 1 .
Add the fractions over the common denominator
a b c abc ab c :
1 a + 1 b + 1 c = b c + a c + a b a b c \frac1a + \frac1b + \frac1c = \dfrac{bc + ac + ab}{abc} a 1 + b 1 + c 1 = ab c b c + a c + ab .
This fraction is positive, since
a a a ,
b b b and
c c c are, so dividing by it keeps the sign:
r > a b c b c + a c + a b r > \dfrac{abc}{bc + ac + ab} r > b c + a c + ab ab c , option A.
Watch out
To get r r r 1 a + 1 b + 1 c \frac1a + \frac1b + \frac1c a 1 + b 1 + c 1 a b c b c + a c + a b \dfrac{abc}{bc + ac + ab} b c + a c + ab ab c Report a problem with this question
Express 1 x + 1 − 1 x − 2 \dfrac{1}{x + 1} - \dfrac{1}{x - 2} x + 1 1 − x − 2 1
A − 3 ( x + 1 ) ( 2 − x ) \dfrac{-3}{(x + 1)(2 - x)} ( x + 1 ) ( 2 − x ) − 3 B 3 ( x + 1 ) ( 2 − x ) \dfrac{3}{(x + 1)(2 - x)} ( x + 1 ) ( 2 − x ) 3 C − 1 ( x + 1 ) ( x − 2 ) \dfrac{-1}{(x + 1)(x - 2)} ( x + 1 ) ( x − 2 ) − 1 D 1 ( x + 1 ) ( x − 2 ) \dfrac{1}{(x + 1)(x - 2)} ( x + 1 ) ( x − 2 ) 1
Worked solution (try it first) Over the LCD
( x + 1 ) ( x − 2 ) (x + 1)(x - 2) ( x + 1 ) ( x − 2 ) the top is
( x − 2 ) − ( x + 1 ) = − 3 (x - 2) - (x + 1) = -3 ( x − 2 ) − ( x + 1 ) = − 3 .
So the fraction is
− 3 ( x + 1 ) ( x − 2 ) \frac{-3}{(x + 1)(x - 2)} ( x + 1 ) ( x − 2 ) − 3 .
The options use
2 − x 2 - x 2 − x , and
x − 2 = − ( 2 − x ) x - 2 = -(2 - x) x − 2 = − ( 2 − x ) .
Swapping the bracket cancels the minus:
3 ( x + 1 ) ( 2 − x ) \dfrac{3}{(x + 1)(2 - x)} ( x + 1 ) ( 2 − x ) 3 , option B.
Watch out
Turning x − 2 x - 2 x − 2 2 − x 2 - x 2 − x − 1 -1 − 1 − 3 -3 − 3 Report a problem with this question
On the curve shown, the points at which the gradient of the curve is equal to zero are
A c, d, f, i, l B b, e, g, j, m C a, b, c, d, f, i, j, l D c, d, f, h, i, l
Worked solution (try it first) The gradient is zero where the tangent is horizontal, which happens at the turning points: the peaks and the troughs.
The peaks are
b b b ,
g g g and
m m m and the troughs are
e e e and
j j j .
So the points are
b , e , g , j , m b, e, g, j, m b , e , g , j , m , option B.
Watch out
Gradient zero is not the same as y = 0 y = 0 y = 0 x x x d d d f f f i i i l l l Report a problem with this question
If − 8 , m , n , 19 -8, m, n, 19 − 8 , m , n , 19 ( m , n ) (m, n) ( m , n )
A ( 1 , 10 ) (1, 10) ( 1 , 10 ) B ( 2 , 10 ) (2, 10) ( 2 , 10 ) C ( 3 , 13 ) (3, 13) ( 3 , 13 ) D ( 4 , 16 ) (4, 16) ( 4 , 16 )
Worked solution (try it first) Four terms in an A.P. have three equal gaps between them, so
− 8 + 3 d = 19 -8 + 3d = 19 − 8 + 3 d = 19 .
Then
3 d = 27 3d = 27 3 d = 27 and
d = 9 d = 9 d = 9 .
So
m = − 8 + 9 = 1 m = -8 + 9 = 1 m = − 8 + 9 = 1 and
n = 1 + 9 = 10 n = 1 + 9 = 10 n = 1 + 9 = 10 , option A.
Watch out
Count the gaps, not the terms: from − 8 -8 − 8 Report a problem with this question
M N MN M N M M M M R MR M R M Q MQ M Q R R R Q Q Q N N N ∠ Q M N = 60 ∘ \angle QMN = 60^\circ ∠ QM N = 6 0 ∘ ∠ M N Q = 40 ∘ \angle MNQ = 40^\circ ∠ M N Q = 4 0 ∘ ∠ R M Q \angle RMQ ∠ R M Q
A 120 ∘ 120^\circ 12 0 ∘ B 11 ∘ 11^\circ 1 1 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 20 ∘ 20^\circ 2 0 ∘
Worked solution (try it first) Tangent–chord: the angle between tangent
M N MN M N and chord
M Q MQ M Q equals the angle in the alternate segment, so
∠ M R Q = ∠ Q M N = 60 ∘ \angle MRQ = \angle QMN = 60^\circ ∠ M R Q = ∠ QM N = 6 0 ∘ .
Triangle
M Q N MQN M QN :
∠ M Q N = 180 ∘ − 60 ∘ − 40 ∘ \angle MQN = 180^\circ - 60^\circ - 40^\circ ∠ M QN = 18 0 ∘ − 6 0 ∘ − 4 0 ∘ Angles on the straight line
R Q N RQN R QN give
∠ M Q R = 180 ∘ − 80 ∘ \angle MQR = 180^\circ - 80^\circ ∠ M QR = 18 0 ∘ − 8 0 ∘ Triangle
R M Q RMQ R M Q :
∠ R M Q = 180 ∘ − 60 ∘ − 100 ∘ \angle RMQ = 180^\circ - 60^\circ - 100^\circ ∠ R M Q = 18 0 ∘ − 6 0 ∘ − 10 0 ∘ = 20 ∘ = 20^\circ = 2 0 ∘ , option D.
Watch out
60 ∘ 60^\circ 6 0 ∘ ∠ M R Q \angle MRQ ∠ M R Q R R R ∠ R M Q \angle RMQ ∠ R M Q M M M Report a problem with this question
In the diagram, H K HK H K Q R QR QR P H = 4 PH = 4 P H = 4 H Q = 3 HQ = 3 H Q = 3 K R : P R KR : PR K R : P R
A 7 : 3 7 : 3 7 : 3 B 3 : 7 3 : 7 3 : 7 C 3 : 4 3 : 4 3 : 4 D 4 : 3 4 : 3 4 : 3
Worked solution (try it first) P Q = P H + H Q = 4 + 3 = 7 PQ = PH + HQ = 4 + 3 = 7 P Q = P H + H Q = 4 + 3 = 7 cm.
A line parallel to one side of a triangle divides the other two sides in the same ratio, so
K R : P R = H Q : P Q KR : PR = HQ : PQ K R : P R = H Q : P Q .
So
K R : P R = 3 : 7 KR : PR = 3 : 7 K R : P R = 3 : 7 , option B.
Watch out
P R PR P R P Q = 7 PQ = 7 P Q = 7 P H = 4 PH = 4 P H = 4 3 : 4 3 : 4 3 : 4 K R : P K KR : PK K R : P K Report a problem with this question
A regular polygon of ( 2 k + 1 ) (2k + 1) ( 2 k + 1 ) 140 ∘ 140^\circ 14 0 ∘ k k k
A 4 B 4 1 2 4\frac12 4 2 1 C 8 D 8 1 2 8\frac12 8 2 1
Worked solution (try it first) Each exterior angle is
180 ∘ − 140 ∘ = 40 ∘ 180^\circ - 140^\circ = 40^\circ 18 0 ∘ − 14 0 ∘ = 4 0 ∘ , so the polygon has
360 ÷ 40 = 9 360 \div 40 = 9 360 ÷ 40 = 9 sides.
So
2 k + 1 = 9 2k + 1 = 9 2 k + 1 = 9 .
Subtract 1:
2 k = 8 2k = 8 2 k = 8 .
Divide by 2:
k = 4 k = 4 k = 4 , option A.
Watch out
Subtract 1 before halving. Halving 9 gives 4 1 2 4\frac12 4 2 1 k k k Report a problem with this question
In the diagram, P S T PST P S T P Q R PQR P QR P Q = Q S = S R PQ = QS = SR P Q = QS = S R ∠ S P Q = 24 ∘ \angle SPQ = 24^\circ ∠ S P Q = 2 4 ∘ y = ∠ R S T y = \angle RST y = ∠ R S T
A 24 ∘ 24^\circ 2 4 ∘ B 48 ∘ 48^\circ 4 8 ∘ C 72 ∘ 72^\circ 7 2 ∘ D 84 ∘ 84^\circ 8 4 ∘
Worked solution (try it first) P Q = Q S PQ = QS P Q = QS , so
∠ Q S P = ∠ Q P S = 24 ∘ \angle QSP = \angle QPS = 24^\circ ∠ QS P = ∠ QP S = 2 4 ∘ .
The exterior angle of triangle
P Q S PQS P QS at
Q Q Q is
∠ S Q R = 24 ∘ + 24 ∘ \angle SQR = 24^\circ + 24^\circ ∠ S QR = 2 4 ∘ + 2 4 ∘ Q S = S R QS = SR QS = S R , so
∠ S R Q = ∠ S Q R = 48 ∘ \angle SRQ = \angle SQR = 48^\circ ∠ S R Q = ∠ S QR = 4 8 ∘ .
In triangle
Q S R QSR QS R ,
∠ Q S R = 180 ∘ − 48 ∘ − 48 ∘ \angle QSR = 180^\circ - 48^\circ - 48^\circ ∠ QS R = 18 0 ∘ − 4 8 ∘ − 4 8 ∘ P S T PST P S T is a straight line, so
y = 180 ∘ − 24 ∘ − 84 ∘ y = 180^\circ - 24^\circ - 84^\circ y = 18 0 ∘ − 2 4 ∘ − 8 4 ∘ = 72 ∘ = 72^\circ = 7 2 ∘ , option C.
Watch out
84 ∘ 84^\circ 8 4 ∘ ∠ Q S R \angle QSR ∠ QS R y y y S S S 24 ∘ 24^\circ 2 4 ∘ 84 ∘ 84^\circ 8 4 ∘ Report a problem with this question
P Q R S PQRS P QR S P R 2 + Q S 2 = k P Q 2 PR^2 + QS^2 = kPQ^2 P R 2 + Q S 2 = k P Q 2 k k k
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
P R 2 \frac{PR}{2} 2 P R and
Q S 2 \frac{QS}{2} 2 QS .
By Pythagoras,
P Q 2 = P R 2 4 + Q S 2 4 PQ^2 = \frac{PR^2}{4} + \frac{QS^2}{4} P Q 2 = 4 P R 2 + 4 Q S 2 .
Multiply by 4:
P R 2 + Q S 2 = 4 P Q 2 PR^2 + QS^2 = 4PQ^2 P R 2 + Q S 2 = 4 P Q 2 , so
k = 4 k = 4 k = 4 , option D.
Watch out
The legs are half the diagonals. Using the whole diagonals gives P R 2 + Q S 2 = P Q 2 PR^2 + QS^2 = PQ^2 P R 2 + Q S 2 = P Q 2 k = 1 k = 1 k = 1 Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z ∠ Y = ∠ Z = 30 ∘ \angle Y = \angle Z = 30^\circ ∠ Y = ∠ Z = 3 0 ∘ X Z = 3 XZ = 3 X Z = 3 Y Z YZ Y Z
A 3 2 \frac{\sqrt3}{2} 2 3 B 3 3 2 \frac{3\sqrt3}{2} 2 3 3 C 3 3 3\sqrt3 3 3 D 2 3 2\sqrt3 2 3
Worked solution (try it first) The angles add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ X = 180 ∘ − 30 ∘ − 30 ∘ \angle X = 180^\circ - 30^\circ - 30^\circ ∠ X = 18 0 ∘ − 3 0 ∘ − 3 0 ∘ Sine rule:
Y Z YZ Y Z faces
∠ X \angle X ∠ X and
X Z = 3 XZ = 3 X Z = 3 faces
∠ Y \angle Y ∠ Y , so
Y Z sin 120 ∘ = 3 sin 30 ∘ \dfrac{YZ}{\sin120^\circ} = \dfrac{3}{\sin30^\circ} sin 12 0 ∘ Y Z = sin 3 0 ∘ 3 .
So
Y Z = 3 × 3 2 1 2 YZ = \dfrac{3 \times \frac{\sqrt3}{2}}{\frac12} Y Z = 2 1 3 × 2 3 , which is
3 3 3\sqrt3 3 3 cm, option C.
Watch out
Divide by sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 3 sin 120 ∘ = 3 3 2 3\sin120^\circ = \frac{3\sqrt3}{2} 3 sin 12 0 ∘ = 2 3 3 Report a problem with this question
In △ P Q R \triangle PQR △ P QR ∠ Q P R \angle QPR ∠ QP R Q R QR QR S S S P Q PQ P Q V V V ∠ V Q S \angle VQS ∠ V QS P S PS P S T T T ∠ Q P R = 46 ∘ \angle QPR = 46^\circ ∠ QP R = 4 6 ∘ ∠ Q S T = 75 ∘ \angle QST = 75^\circ ∠ QS T = 7 5 ∘ ∠ Q T S \angle QTS ∠ QT S
A 41 ∘ 41^\circ 4 1 ∘ B 52 ∘ 52^\circ 5 2 ∘ C 64 ∘ 64^\circ 6 4 ∘ D 82 ∘ 82^\circ 8 2 ∘
Worked solution (try it first) P S PS P S bisects
∠ Q P R \angle QPR ∠ QP R , so
∠ Q P S = 46 ∘ ÷ 2 = 23 ∘ \angle QPS = 46^\circ \div 2 = 23^\circ ∠ QP S = 4 6 ∘ ÷ 2 = 2 3 ∘ .
P S T PST P S T is a straight line, so
∠ Q S P = 180 ∘ − 75 ∘ \angle QSP = 180^\circ - 75^\circ ∠ QS P = 18 0 ∘ − 7 5 ∘ ∠ V Q S \angle VQS ∠ V QS is an exterior angle of triangle
P Q S PQS P QS , so
∠ V Q S = 23 ∘ + 105 ∘ \angle VQS = 23^\circ + 105^\circ ∠ V QS = 2 3 ∘ + 10 5 ∘ Q T QT QT bisects it:
∠ V Q T = 64 ∘ \angle VQT = 64^\circ ∠ V QT = 6 4 ∘ .
∠ V Q T \angle VQT ∠ V QT is an exterior angle of triangle
P Q T PQT P QT , so
64 ∘ = ∠ Q P T + ∠ Q T P 64^\circ = \angle QPT + \angle QTP 6 4 ∘ = ∠ QP T + ∠ QT P = 23 ∘ + ∠ Q T S = 23^\circ + \angle QTS = 2 3 ∘ + ∠ QT S .
So
∠ Q T S = 64 ∘ − 23 ∘ \angle QTS = 64^\circ - 23^\circ ∠ QT S = 6 4 ∘ − 2 3 ∘ = 41 ∘ = 41^\circ = 4 1 ∘ , option A.
Watch out
64 ∘ 64^\circ 6 4 ∘ ∠ V Q T \angle VQT ∠ V QT ∠ V Q S \angle VQS ∠ V QS P Q T PQT P QT 23 ∘ 23^\circ 2 3 ∘ P P P Report a problem with this question
Triangle R S T RST R S T S S S R S = x RS = x R S = x S T = y ST = y S T = y R T = z RT = z R T = z x : y = 5 : 12 x : y = 5 : 12 x : y = 5 : 12 z = 52 z = 52 z = 52
A 68 cm B 84 cm C 100 cm D 120 cm
Worked solution (try it first) Write
x = 5 k x = 5k x = 5 k and
y = 12 k y = 12k y = 12 k .
By Pythagoras,
z 2 = 25 k 2 + 144 k 2 = 169 k 2 z^2 = 25k^2 + 144k^2 = 169k^2 z 2 = 25 k 2 + 144 k 2 = 169 k 2 , so
z = 13 k z = 13k z = 13 k .
13 k = 52 13k = 52 13 k = 52 , so
k = 4 k = 4 k = 4 .
The sides are
x = 20 x = 20 x = 20 cm and
y = 48 y = 48 y = 48 cm.
Perimeter
= 20 + 48 + 52 = 120 = 20 + 48 + 52 = 120 = 20 + 48 + 52 = 120 cm, option D.
Watch out
Include the hypotenuse in the perimeter: 20 + 48 = 68 20 + 48 = 68 20 + 48 = 68 z = 52 z = 52 z = 52 Report a problem with this question
The pilot of an aeroplane flying 10 km above the ground towards a landmark sees the landmark at angles of depression of 35 ∘ 35^\circ 3 5 ∘ 55 ∘ 55^\circ 5 5 ∘
A 10 ( sin 35 ∘ − sin 55 ∘ ) 10(\sin35^\circ - \sin55^\circ) 10 ( sin 3 5 ∘ − sin 5 5 ∘ ) B 10 ( cos 35 ∘ − cos 55 ∘ ) 10(\cos35^\circ - \cos55^\circ) 10 ( cos 3 5 ∘ − cos 5 5 ∘ ) C 10 ( tan 35 ∘ − tan 55 ∘ ) 10(\tan35^\circ - \tan55^\circ) 10 ( tan 3 5 ∘ − tan 5 5 ∘ ) D 10 ( cot 35 ∘ − cot 55 ∘ ) 10(\cot35^\circ - \cot55^\circ) 10 ( cot 3 5 ∘ − cot 5 5 ∘ )
Worked solution (try it first) The angle of depression from the plane equals the angle of elevation from the landmark (alternate angles).
So each horizontal distance
d d d has
tan θ = 10 d \tan\theta = \frac{10}{d} tan θ = d 10 .
So
d = 10 tan θ = 10 cot θ d = \frac{10}{\tan\theta} = 10\cot\theta d = t a n θ 10 = 10 cot θ : first
10 cot 35 ∘ 10\cot35^\circ 10 cot 3 5 ∘ , then
10 cot 55 ∘ 10\cot55^\circ 10 cot 5 5 ∘ .
The plane flew the difference:
10 ( cot 35 ∘ − cot 55 ∘ ) 10(\cot35^\circ - \cot55^\circ) 10 ( cot 3 5 ∘ − cot 5 5 ∘ ) , option D.
Watch out
The height is opposite the angle and the distance is adjacent, so the distance is 10 ÷ tan θ 10 \div \tan\theta 10 ÷ tan θ 10 × tan θ 10 \times \tan\theta 10 × tan θ Report a problem with this question
If 4 sin 2 x − 3 = 0 4\sin^2x - 3 = 0 4 sin 2 x − 3 = 0 x x x 0 ∘ < x < 90 ∘ 0^\circ < x < 90^\circ 0 ∘ < x < 9 0 ∘
A 30 ∘ 30^\circ 3 0 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) Add 3 and divide by 4:
sin 2 x = 3 4 \sin^2 x = \frac34 sin 2 x = 4 3 .
Take the square root (positive, since
x x x is acute):
sin x = 3 2 \sin x = \frac{\sqrt3}{2} sin x = 2 3 .
So
x = 60 ∘ x = 60^\circ x = 6 0 ∘ , option C.
Watch out
sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 sin 60 ∘ = 3 2 \sin60^\circ = \frac{\sqrt3}{2} sin 6 0 ∘ = 2 3 30 ∘ 30^\circ 3 0 ∘ 3 2 \frac{\sqrt3}{2} 2 3 Report a problem with this question
A square tile has side 30 cm. How many of these tiles cover a rectangular floor of length 7.2 m and width 4.2 m?
Worked solution (try it first) Work in centimetres: the floor is 720 cm by 420 cm.
Along the length,
720 ÷ 30 = 24 720 \div 30 = 24 720 ÷ 30 = 24 tiles fit.
Along the width,
420 ÷ 30 = 14 420 \div 30 = 14 420 ÷ 30 = 14 tiles.
So
24 × 14 = 336 24 \times 14 = 336 24 × 14 = 336 tiles, option A.
Watch out
Put everything in the same units first. The floor is 30.24 m 2 30.24\text{ m}^2 30.24 m 2 0.3 2 = 0.09 m 2 0.3^2 = 0.09\text{ m}^2 0. 3 2 = 0.09 m 2 0.3 m 2 0.3\text{ m}^2 0.3 m 2 Also set as JAMB 2013 · UTME · Q27
Report a problem with this question
A cylindrical metal pipe 1 m long has an outer diameter of 7.2 cm and an inner diameter of 2.8 cm. Find the volume of metal used for the pipe.
A 440 π cm 3 440\pi\text{ cm}^3 440 π cm 3 B 1,100 π cm 3 1{,}100\pi\text{ cm}^3 1 , 100 π cm 3 C 4,400 π cm 3 4{,}400\pi\text{ cm}^3 4 , 400 π cm 3 D 11,000 π cm 3 11{,}000\pi\text{ cm}^3 11 , 000 π cm 3
Worked solution (try it first) Halve the diameters: outer radius 3.6 cm, inner radius 1.4 cm.
The length is 1 m, which is 100 cm.
The metal is the ring times the length:
π ( 3.6 2 − 1.4 2 ) × 100 = π ( 12.96 − 1.96 ) × 100 \pi(3.6^2 - 1.4^2) \times 100 = \pi(12.96 - 1.96) \times 100 π ( 3. 6 2 − 1. 4 2 ) × 100 = π ( 12.96 − 1.96 ) × 100 .
That is
11 π × 100 = 1100 π cm 3 11\pi \times 100 = 1100\pi\text{ cm}^3 11 π × 100 = 1100 π cm 3 , option B.
Watch out
Use radii, not diameters. With 7.2 7.2 7.2 2.8 2.8 2.8 π ( 51.84 − 7.84 ) × 100 = 4400 π cm 3 \pi(51.84 - 7.84) \times 100 = 4400\pi\text{ cm}^3 π ( 51.84 − 7.84 ) × 100 = 4400 π cm 3 Report a problem with this question
O X Y Z W OXYZW O X Y Z W O X = O Y = O Z = O W = 5 OX = OY = OZ = OW = 5 O X = O Y = O Z = O W = 5 X Y = X W = Y Z = W Z = 6 XY = XW = YZ = WZ = 6 X Y = X W = Y Z = W Z = 6 O T OT O T
A 2 5 2\sqrt5 2 5 B 3 C 4 D 7 \sqrt7 7
Worked solution (try it first) T T T is the centre of the square base.
The base diagonal is
6 2 6\sqrt2 6 2 cm, so
X T XT X T is half of it,
3 2 3\sqrt2 3 2 cm.
In the right-angled triangle
O T X OTX O T X , Pythagoras gives
O T 2 = O X 2 − X T 2 OT^2 = OX^2 - XT^2 O T 2 = O X 2 − X T 2 , which is
25 − 18 = 7 25 - 18 = 7 25 − 18 = 7 .
So
O T = 7 OT = \sqrt7 O T = 7 cm, option D.
Watch out
O X OX O X 3 2 3\sqrt2 3 2 25 − 9 = 4 \sqrt{25 - 9} = 4 25 − 9 = 4 Report a problem with this question
In preparing rice cutlets, a cook used 75 g of rice, 40 g of margarine, 105 g of meat and 20 g of bread crumbs. Find the angle of the sector which represents meat in a pie chart.
A 30 ∘ 30^\circ 3 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 112.5 ∘ 112.5^\circ 112. 5 ∘ D 157.5 ∘ 157.5^\circ 157. 5 ∘
Worked solution (try it first) Total mass:
75 + 40 + 105 + 20 = 240 75 + 40 + 105 + 20 = 240 75 + 40 + 105 + 20 = 240 g.
Meat's share of the circle:
105 240 × 360 ∘ = 157.5 ∘ \frac{105}{240} \times 360^\circ = 157.5^\circ 240 105 × 36 0 ∘ = 157. 5 ∘ , option D.
Watch out
Use the meat, 105 g. 112.5 ∘ 112.5^\circ 112. 5 ∘ 75 240 × 360 ∘ \frac{75}{240} \times 360^\circ 240 75 × 36 0 ∘ Report a problem with this question
In a family of 21 people, the average age is 14 years. If the age of the grandfather is not counted, the average age drops to 12 years. What is the age of the grandfather?
A 35 years B 40 years C 42 years D 54 years
Worked solution (try it first) Total of all 21 ages:
21 × 14 = 294 21 \times 14 = 294 21 × 14 = 294 years.
Total of the other 20 ages:
20 × 12 = 240 20 \times 12 = 240 20 × 12 = 240 years.
The grandfather is the difference:
294 − 240 = 54 294 - 240 = 54 294 − 240 = 54 years, option D.
Watch out
Without the grandfather there are 20 people, not 21. Using 21 × 12 = 252 21 \times 12 = 252 21 × 12 = 252 Report a problem with this question
If n n n m m m ( n , m ) (n, m) ( n , m )
A ( 2.6 , 2.6 ) (2.6, 2.6) ( 2.6 , 2.6 ) B ( 2.5 , 2.6 ) (2.5, 2.6) ( 2.5 , 2.6 ) C ( 2.6 , 2.5 ) (2.6, 2.5) ( 2.6 , 2.5 ) D ( 2.5 , 2.1 ) (2.5, 2.1) ( 2.5 , 2.1 )
Worked solution (try it first) Put the 8 numbers in order: 1.6, 2.1, 2.1, 2.4, 2.6, 2.6, 2.6, 3.7.
The median is halfway between the 4th and 5th:
n = 2.4 + 2.6 2 = 2.5 n = \frac{2.4 + 2.6}{2} = 2.5 n = 2 2.4 + 2.6 = 2.5 .
2.6 occurs three times, more than any other, so
m = 2.6 m = 2.6 m = 2.6 .
So
( n , m ) = ( 2.5 , 2.6 ) (n, m) = (2.5, 2.6) ( n , m ) = ( 2.5 , 2.6 ) , option B.
Watch out
With 8 numbers, average the 4th and 5th. Taking the 5th alone gives 2.6 and ( 2.6 , 2.6 ) (2.6, 2.6) ( 2.6 , 2.6 ) Report a problem with this question
Two numbers are chosen at random from the three numbers 1, 3, 6. Find the probability that the sum of the two is not odd.
A 2 3 \frac23 3 2 B 1 2 \frac12 2 1 C 1 3 \frac13 3 1 D 1 6 \frac16 6 1
Worked solution (try it first) There are three possible pairs:
{ 1 , 3 } \{1, 3\} { 1 , 3 } ,
{ 1 , 6 } \{1, 6\} { 1 , 6 } and
{ 3 , 6 } \{3, 6\} { 3 , 6 } .
Their sums are 4, 7 and 9. "Not odd" means even, and only
1 + 3 = 4 1 + 3 = 4 1 + 3 = 4 is even.
So the probability is
1 3 \frac13 3 1 , option C.
Watch out
2 3 \frac23 3 2 Report a problem with this question