Paper JAMB 1997 General Maths Objective
Objective paper · 46 questions · partial
JAMB 1997 · UME Topics include Number bases, Approximation & error, Quadratics & their graphs, Indices & standard form, Logarithms, Surds.
Our copy of this paper is missing questions 9, 20, 35, 39.
Sit this paper Answer every question in order, timed if you like (suggested 30 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 37 38 40 41 42 43 44 45 46 47 48 49 50 If ( 1 P 03 ) 4 = 115 10 (1P03)_4 = 115_{10} ( 1 P 03 ) 4 = 11 5 10 P P P
Worked solution (try it first) The place values in base four are 64, 16, 4 and 1, so
1 P 03 4 = 64 + 16 P + 0 + 3 1P03_4 = 64 + 16P + 0 + 3 1 P 0 3 4 = 64 + 16 P + 0 + 3 .
Set it equal to 115:
67 + 16 P = 115 67 + 16P = 115 67 + 16 P = 115 , so
16 P = 48 16P = 48 16 P = 48 .
So
P = 3 P = 3 P = 3 , option D.
Watch out
P P P 4 2 4^2 4 2 4 P = 48 4P = 48 4 P = 48 P = 12 P = 12 P = 12 Report a problem with this question
Evaluate 64.764 2 − 35.236 2 64.764^2 - 35.236^2 64.76 4 2 − 35.23 6 2
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 ) .
Here
a − b = 64.764 − 35.236 = 29.528 a - b = 64.764 - 35.236 = 29.528 a − b = 64.764 − 35.236 = 29.528 and
a + b = 100 a + b = 100 a + b = 100 , so the value is
29.528 × 100 = 2952.8 29.528 \times 100 = 2952.8 29.528 × 100 = 2952.8 .
To 3 significant figures, the fourth figure is 2, so round down: 2950, option B.
Watch out
2952.8 rounds to 2950, because the figure after the 5 is 2. Rounding up gives 2960 (option A). Report a problem with this question
Find the value of ( 0.006 ) 3 + ( 0.004 ) 3 (0.006)^3 + (0.004)^3 ( 0.006 ) 3 + ( 0.004 ) 3
A 2.8 × 10 − 9 2.8 \times 10^{-9} 2.8 × 1 0 − 9 B 2.8 × 10 − 8 2.8 \times 10^{-8} 2.8 × 1 0 − 8 C 2.8 × 10 − 7 2.8 \times 10^{-7} 2.8 × 1 0 − 7 D 2.8 × 10 − 6 2.8 \times 10^{-6} 2.8 × 1 0 − 6
Worked solution (try it first) Cube each number:
( 0.006 ) 3 = 0.000000216 (0.006)^3 = 0.000000216 ( 0.006 ) 3 = 0.000000216 = 2.16 × 10 − 7 = 2.16 \times 10^{-7} = 2.16 × 1 0 − 7 , because
6 3 = 216 6^3 = 216 6 3 = 216 and the 3 decimal places become 9.
Likewise
( 0.004 ) 3 = 0.000000064 (0.004)^3 = 0.000000064 ( 0.004 ) 3 = 0.000000064 = 0.64 × 10 − 7 = 0.64 \times 10^{-7} = 0.64 × 1 0 − 7 , because
4 3 = 64 4^3 = 64 4 3 = 64 .
Add with the same power of 10:
( 2.16 + 0.64 ) × 10 − 7 = 2.8 × 10 − 7 (2.16 + 0.64) \times 10^{-7} = 2.8 \times 10^{-7} ( 2.16 + 0.64 ) × 1 0 − 7 = 2.8 × 1 0 − 7 , option C.
Watch out
Cubing triples the decimal places: 0.006 has 3, so ( 0.006 ) 3 (0.006)^3 ( 0.006 ) 3 2.8 × 10 − 6 2.8 \times 10^{-6} 2.8 × 1 0 − 6 Report a problem with this question
Given that log a 2 = 0.693 \log_a 2 = 0.693 log a 2 = 0.693 log a 3 = 1.097 \log_a 3 = 1.097 log a 3 = 1.097 log a 13.5 \log_a 13.5 log a 13.5
Worked solution (try it first) Write 13.5 using 2 and 3:
13.5 = 27 2 = 3 3 2 13.5 = \frac{27}{2} = \frac{3^3}{2} 13.5 = 2 27 = 2 3 3 .
So
log a 13.5 = 3 log a 3 − log a 2 \log_a 13.5 = 3\log_a 3 - \log_a 2 log a 13.5 = 3 log a 3 − log a 2 .
That is
3 ( 1.097 ) − 0.693 = 3.291 − 0.693 = 2.598 3(1.097) - 0.693 = 3.291 - 0.693 = 2.598 3 ( 1.097 ) − 0.693 = 3.291 − 0.693 = 2.598 , option C.
Watch out
Dividing by 2 means subtracting log a 2 \log_a 2 log a 2 3.291 + 0.693 = 3.984 3.291 + 0.693 = 3.984 3.291 + 0.693 = 3.984 Also set as JAMB 2015 · UTME · Q28
Report a problem with this question
Simplify log 2 96 − 2 log 2 6 \log_2 96 - 2\log_2 6 log 2 96 − 2 log 2 6
A − log 2 3 -\log_2 3 − log 2 3 B 3 − log 2 3 3 - \log_2 3 3 − log 2 3 C log 2 3 − 3 \log_2 3 - 3 log 2 3 − 3 D log 2 3 − 2 \log_2 3 - 2 log 2 3 − 2
Worked solution (try it first) Move the 2 up as a power:
2 log 2 6 = log 2 36 2\log_2 6 = \log_2 36 2 log 2 6 = log 2 36 , so the expression is
log 2 96 36 = log 2 8 3 \log_2 \frac{96}{36} = \log_2 \frac83 log 2 36 96 = log 2 3 8 .
Split the fraction:
log 2 8 3 = log 2 8 − log 2 3 \log_2 \frac83 = \log_2 8 - \log_2 3 log 2 3 8 = log 2 8 − log 2 3 .
log 2 8 = 3 \log_2 8 = 3 log 2 8 = 3 , so the answer is
3 − log 2 3 3 - \log_2 3 3 − log 2 3 , option B.
Watch out
The log of a fraction is the log of the top minus the log of the bottom, so log 2 8 3 = 3 − log 2 3 \log_2 \frac83 = 3 - \log_2 3 log 2 3 8 = 3 − log 2 3 log 2 3 − 3 \log_2 3 - 3 log 2 3 − 3 log 2 3 8 \log_2 \frac38 log 2 8 3 Report a problem with this question
If 8 x 2 = 2 3 8 × 4 3 4 8^{\frac x2} = 2^{\frac38} \times 4^{\frac34} 8 2 x = 2 8 3 × 4 4 3 x x x
A 3 8 \frac38 8 3 B 3 4 \frac34 4 3 C 4 5 \frac45 5 4 D 5 4 \frac54 4 5
Worked solution (try it first) Write everything as a power of 2:
8 x 2 = ( 2 3 ) x 2 8^{\frac x2} = (2^3)^{\frac x2} 8 2 x = ( 2 3 ) 2 x = 2 3 x 2 = 2^{\frac{3x}{2}} = 2 2 3 x and
4 3 4 = ( 2 2 ) 3 4 4^{\frac34} = (2^2)^{\frac34} 4 4 3 = ( 2 2 ) 4 3 Right side: add the powers,
3 8 + 3 2 = 3 8 + 12 8 \frac38 + \frac32 = \frac38 + \frac{12}{8} 8 3 + 2 3 = 8 3 + 8 12 Equate the powers:
3 x 2 = 15 8 \frac{3x}{2} = \frac{15}{8} 2 3 x = 8 15 , so
x = 15 8 × 2 3 = 5 4 x = \frac{15}{8} \times \frac23 = \frac54 x = 8 15 × 3 2 = 4 5 , option D.
Watch out
Change 8 and 4 to base 2 first. Equating x 2 \frac x2 2 x 3 8 + 3 4 \frac38 + \frac34 8 3 + 4 3 x = 9 4 x = \frac94 x = 4 9 Report a problem with this question
Simplify 2 3 + 3 5 3 5 − 2 3 \dfrac{2\sqrt3 + 3\sqrt5}{3\sqrt5 - 2\sqrt3} 3 5 − 2 3 2 3 + 3 5
A 19 + 4 15 11 \frac{19 + 4\sqrt{15}}{11} 11 19 + 4 15 B 19 + 4 15 19 \frac{19 + 4\sqrt{15}}{19} 19 19 + 4 15 C 19 + 2 15 11 \frac{19 + 2\sqrt{15}}{11} 11 19 + 2 15 D 19 + 2 15 19 \frac{19 + 2\sqrt{15}}{19} 19 19 + 2 15
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
3 5 + 2 3 3\sqrt5 + 2\sqrt3 3 5 + 2 3 .
Bottom:
( 3 5 ) 2 − ( 2 3 ) 2 = 45 − 12 (3\sqrt5)^2 - (2\sqrt3)^2 = 45 - 12 ( 3 5 ) 2 − ( 2 3 ) 2 = 45 − 12 , which is 33.
Top:
( 2 3 + 3 5 ) 2 = 12 + 12 15 + 45 (2\sqrt3 + 3\sqrt5)^2 = 12 + 12\sqrt{15} + 45 ( 2 3 + 3 5 ) 2 = 12 + 12 15 + 45 , which is
57 + 12 15 57 + 12\sqrt{15} 57 + 12 15 .
Divide top and bottom by 3:
19 + 4 15 11 \dfrac{19 + 4\sqrt{15}}{11} 11 19 + 4 15 , option A.
Watch out
The middle term of the square is 2 × 2 3 × 3 5 = 12 15 2 \times 2\sqrt3 \times 3\sqrt5 = 12\sqrt{15} 2 × 2 3 × 3 5 = 12 15 6 15 6\sqrt{15} 6 15 19 + 2 15 11 \frac{19 + 2\sqrt{15}}{11} 11 19 + 2 15 Report a problem with this question
Find the simple interest rate per cent per annum at which ₦1000 accumulates to ₦1240 in 3 years.
Worked solution (try it first) The interest is the amount minus the principal:
1240 − 1000 = 1240 - 1000 = 1240 − 1000 = ₦240.
Use
R = 100 I P T R = \dfrac{100I}{PT} R = P T 100 I with
I = 240 I = 240 I = 240 ,
P = 1000 P = 1000 P = 1000 and
T = 3 T = 3 T = 3 :
R = 24 000 3000 = 8 R = \dfrac{24\,000}{3000} = 8 R = 3000 24 000 = 8 .
So the rate is
8 % 8\% 8% per annum, option B.
Watch out
Divide by the 3 years: ₦240 is 24 % 24\% 24% 8 % 8\% 8% Report a problem with this question
A survey of 100 students in an institution shows that 80 students speak Hausa and 20 speak Igbo, while only 9 speak both languages. How many students speak neither Hausa nor Igbo?
Worked solution (try it first) At least one language:
n ( H ∪ I ) = 80 + 20 − 9 = 91 n(H \cup I) = 80 + 20 - 9 = 91 n ( H ∪ I ) = 80 + 20 − 9 = 91 .
Neither language:
100 − 91 = 9 100 - 91 = 9 100 − 91 = 9 , option B.
Watch out
Subtract the 9 who speak both from the sum first. 100 − 80 − 20 = 0 100 - 80 - 20 = 0 100 − 80 − 20 = 0 Report a problem with this question
If the function f ( x ) = x 3 + 2 x 2 + q x − 6 f(x) = x^3 + 2x^2 + qx - 6 f ( x ) = x 3 + 2 x 2 + q x − 6 x + 1 x + 1 x + 1 q q q
Worked solution (try it first) Divisible by
x + 1 x + 1 x + 1 means
f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 , by the factor theorem.
f ( − 1 ) = − 1 + 2 − q − 6 = − 5 − q f(-1) = -1 + 2 - q - 6 = -5 - q f ( − 1 ) = − 1 + 2 − q − 6 = − 5 − q .
Set it to 0:
q = − 5 q = -5 q = − 5 , option A.
Watch out
q × ( − 1 ) = − q q \times (-1) = -q q × ( − 1 ) = − q − 5 − q = 0 -5 - q = 0 − 5 − q = 0 q = − 5 q = -5 q = − 5 q = 5 q = 5 q = 5 Also set as JAMB 2015 · UTME · Q29
Report a problem with this question
Solve the simultaneous equations 2 x − 3 y = 2 \frac2x - \frac3y = 2 x 2 − y 3 = 2 4 x + 3 y = 10 \frac4x + \frac3y = 10 x 4 + y 3 = 10
A x = 3 2 , y = 1 2 x = \frac32, y = \frac12 x = 2 3 , y = 2 1 B x = 1 2 , y = 3 2 x = \frac12, y = \frac32 x = 2 1 , y = 2 3 C x = − 1 2 , y = − 3 2 x = -\frac12, y = -\frac32 x = − 2 1 , y = − 2 3 D x = 1 2 , y = − 3 2 x = \frac12, y = -\frac32 x = 2 1 , y = − 2 3
Worked solution (try it first) Add the two equations so the
3 y \frac3y y 3 terms cancel:
6 x = 12 \frac6x = 12 x 6 = 12 .
Multiply both sides by
x x x and divide by 12:
x = 6 12 = 1 2 x = \frac{6}{12} = \frac12 x = 12 6 = 2 1 .
Put
x = 1 2 x = \frac12 x = 2 1 into the first equation:
2 x = 4 \frac2x = 4 x 2 = 4 , so
4 − 3 y = 2 4 - \frac3y = 2 4 − y 3 = 2 and
3 y = 2 \frac3y = 2 y 3 = 2 .
So
y = 3 2 y = \frac32 y = 2 3 , and the answer is
x = 1 2 x = \frac12 x = 2 1 ,
y = 3 2 y = \frac32 y = 2 3 , option B.
Watch out
Keep track of which letter each value belongs to. Option A swaps them: x = 3 2 x = \frac32 x = 2 3 2 x = 4 3 \frac2x = \frac43 x 2 = 3 4 Report a problem with this question
Find the minimum value of x 2 − 3 x + 2 x^2 - 3x + 2 x 2 − 3 x + 2 x x x
A − 1 4 -\frac14 − 4 1 B − 1 2 -\frac12 − 2 1 C 1 4 \frac14 4 1 D 1 2 \frac12 2 1
Worked solution (try it first) Complete the square: half of
− 3 -3 − 3 is
− 3 2 -\frac32 − 2 3 , so
x 2 − 3 x = ( x − 3 2 ) 2 − 9 4 x^2 - 3x = \left(x - \frac32\right)^2 - \frac94 x 2 − 3 x = ( x − 2 3 ) 2 − 4 9 .
So
x 2 − 3 x + 2 = ( x − 3 2 ) 2 − 9 4 + 2 x^2 - 3x + 2 = \left(x - \frac32\right)^2 - \frac94 + 2 x 2 − 3 x + 2 = ( x − 2 3 ) 2 − 4 9 + 2 = ( x − 3 2 ) 2 − 1 4 = \left(x - \frac32\right)^2 - \frac14 = ( x − 2 3 ) 2 − 4 1 .
A square is never negative, so the minimum value is
− 1 4 -\frac14 − 4 1 , option A.
Watch out
The constant is 2 − 9 4 = − 1 4 2 - \frac94 = -\frac14 2 − 4 9 = − 4 1 9 4 − 2 \frac94 - 2 4 9 − 2 1 4 \frac14 4 1 Report a problem with this question
Make f f f t = v 1 f + 1 g t = \sqrt{\dfrac{v}{\frac1f + \frac1g}} t = f 1 + g 1 v
A g v − t 2 g t 2 \dfrac{gv - t^2}{gt^2} g t 2 g v − t 2 B g t 2 g v − t 2 \dfrac{gt^2}{gv - t^2} g v − t 2 g t 2 C v t 1 / 2 − 1 g \dfrac{v}{t^{1/2}} - \frac1g t 1/2 v − g 1 D g v t 2 − g \dfrac{gv}{t^2} - g t 2 g v − g
Worked solution (try it first) Square both sides:
t 2 = v 1 f + 1 g t^2 = \frac{v}{\frac1f + \frac1g} t 2 = f 1 + g 1 v , so
1 f + 1 g = v t 2 \frac1f + \frac1g = \frac{v}{t^2} f 1 + g 1 = t 2 v .
Subtract
1 g \frac1g g 1 :
1 f = v t 2 − 1 g \frac1f = \frac{v}{t^2} - \frac1g f 1 = t 2 v − g 1 = g v − t 2 g t 2 = \frac{gv - t^2}{gt^2} = g t 2 g v − t 2 .
Turn both sides upside down:
f = g t 2 g v − t 2 f = \dfrac{gt^2}{gv - t^2} f = g v − t 2 g t 2 , option B.
Watch out
g v − t 2 g t 2 \frac{gv - t^2}{gt^2} g t 2 g v − t 2 1 f \frac1f f 1 f f f Report a problem with this question
What value of g g g 4 x 2 − 18 x y + g 4x^2 - 18xy + g 4 x 2 − 18 x y + g
A 9 B 9 y 2 4 \frac{9y^2}{4} 4 9 y 2 C 81 y 2 81y^2 81 y 2 D 81 y 2 4 \frac{81y^2}{4} 4 81 y 2
Worked solution (try it first) A perfect square starting
4 x 2 4x^2 4 x 2 has the form
( 2 x − k y ) 2 = 4 x 2 − 4 k x y + k 2 y 2 (2x - ky)^2 = 4x^2 - 4kxy + k^2y^2 ( 2 x − k y ) 2 = 4 x 2 − 4 k x y + k 2 y 2 .
Match the middle terms:
4 k = 18 4k = 18 4 k = 18 , so
k = 9 2 k = \frac92 k = 2 9 .
So
g = k 2 y 2 = 81 y 2 4 g = k^2y^2 = \dfrac{81y^2}{4} g = k 2 y 2 = 4 81 y 2 , option D.
Watch out
The middle term is 2 × 2 x × k y = 4 k x y 2 \times 2x \times ky = 4kxy 2 × 2 x × k y = 4 k x y k = 18 4 k = \frac{18}{4} k = 4 18 k = 9 k = 9 k = 9 81 y 2 81y^2 81 y 2 Also set as JAMB 2015 · UTME · Q30
Report a problem with this question
Find the value of K K K 5 + 2 r ( r + 1 ) ( r − 2 ) \dfrac{5 + 2r}{(r + 1)(r - 2)} ( r + 1 ) ( r − 2 ) 5 + 2 r K r − 2 + L r + 1 \dfrac{K}{r - 2} + \dfrac{L}{r + 1} r − 2 K + r + 1 L K K K L L L
Worked solution (try it first) Multiply through:
5 + 2 r = K ( r + 1 ) + L ( r − 2 ) 5 + 2r = K(r + 1) + L(r - 2) 5 + 2 r = K ( r + 1 ) + L ( r − 2 ) .
Put
r = 2 r = 2 r = 2 so the
L L L term vanishes:
9 = 3 K 9 = 3K 9 = 3 K .
So
K = 3 K = 3 K = 3 , option A.
Watch out
K K K r − 2 r - 2 r − 2 r = 2 r = 2 r = 2 r = − 1 r = -1 r = − 1 L L L 3 = − 3 L 3 = -3L 3 = − 3 L L = − 1 L = -1 L = − 1 Report a problem with this question
Let f ( x ) = 2 x + 4 f(x) = 2x + 4 f ( x ) = 2 x + 4 g ( x ) = 6 x + 7 g(x) = 6x + 7 g ( x ) = 6 x + 7 g ( x ) > 0 g(x) > 0 g ( x ) > 0 f ( x ) g ( x ) < 1 \dfrac{f(x)}{g(x)} < 1 g ( x ) f ( x ) < 1
A x < − 3 4 x < -\frac34 x < − 4 3 B x > − 4 3 x > -\frac43 x > − 3 4 C x > − 3 4 x > -\frac34 x > − 4 3 D x > − 12 x > -12 x > − 12
Worked solution (try it first) g ( x ) > 0 g(x) > 0 g ( x ) > 0 , so you can multiply both sides by
g ( x ) g(x) g ( x ) without reversing the sign:
2 x + 4 < 6 x + 7 2x + 4 < 6x + 7 2 x + 4 < 6 x + 7 .
Subtract
2 x 2x 2 x and 7 from both sides:
− 3 < 4 x -3 < 4x − 3 < 4 x .
Divide by 4:
x > − 3 4 x > -\frac34 x > − 4 3 , option C.
Watch out
From 4 x > − 3 4x > -3 4 x > − 3 − 3 -3 − 3 − 3 4 -\frac34 − 4 3 − 4 3 -\frac43 − 3 4 Report a problem with this question
Find the range of values of x x x 12 x 2 < x + 1 12x^2 < x + 1 12 x 2 < x + 1
A − 1 4 < x < 1 3 -\frac14 < x < \frac13 − 4 1 < x < 3 1 B 1 4 < x < 1 3 \frac14 < x < \frac13 4 1 < x < 3 1 C − 1 3 < x < 1 4 -\frac13 < x < \frac14 − 3 1 < x < 4 1 D − 1 4 < x < − 1 3 -\frac14 < x < -\frac13 − 4 1 < x < − 3 1
Worked solution (try it first) Bring everything to one side:
12 x 2 − x − 1 < 0 12x^2 - x - 1 < 0 12 x 2 − x − 1 < 0 .
Factorise:
( 4 x + 1 ) ( 3 x − 1 ) < 0 (4x + 1)(3x - 1) < 0 ( 4 x + 1 ) ( 3 x − 1 ) < 0 , so the roots are
x = − 1 4 x = -\frac14 x = − 4 1 and
x = 1 3 x = \frac13 x = 3 1 .
"Less than 0" means between the roots:
− 1 4 < x < 1 3 -\frac14 < x < \frac13 − 4 1 < x < 3 1 , option A.
Watch out
Get each root from its own bracket: 4 x + 1 = 0 4x + 1 = 0 4 x + 1 = 0 − 1 4 -\frac14 − 4 1 3 x − 1 = 0 3x - 1 = 0 3 x − 1 = 0 1 3 \frac13 3 1 − 1 3 < x < 1 4 -\frac13 < x < \frac14 − 3 1 < x < 4 1 Report a problem with this question
S n S_n S n n n n S n = n 2 − 1 S_n = n^2 - 1 S n = n 2 − 1 n n n
A 4 n + 1 4n + 1 4 n + 1 B 4 n − 1 4n - 1 4 n − 1 C 2 n + 1 2n + 1 2 n + 1 D 2 n − 1 2n - 1 2 n − 1
Worked solution (try it first) The
n n n th term is the sum of
n n n terms minus the sum of
n − 1 n - 1 n − 1 terms:
T n = S n − S n − 1 T_n = S_n - S_{n - 1} T n = S n − S n − 1 .
So
T n = ( n 2 − 1 ) − ( ( n − 1 ) 2 − 1 ) T_n = (n^2 - 1) - \big((n - 1)^2 - 1\big) T n = ( n 2 − 1 ) − ( ( n − 1 ) 2 − 1 ) = n 2 − ( n − 1 ) 2 = n^2 - (n - 1)^2 = n 2 − ( n − 1 ) 2 .
Expand:
n 2 − ( n 2 − 2 n + 1 ) = 2 n − 1 n^2 - (n^2 - 2n + 1) = 2n - 1 n 2 − ( n 2 − 2 n + 1 ) = 2 n − 1 , option D.
Watch out
Take S n − S n − 1 S_n - S_{n - 1} S n − S n − 1 S n + 1 − S n S_{n + 1} - S_n S n + 1 − S n ( n + 1 ) 2 − n 2 = 2 n + 1 (n + 1)^2 - n^2 = 2n + 1 ( n + 1 ) 2 − n 2 = 2 n + 1 ( n + 1 ) (n + 1) ( n + 1 ) Report a problem with this question
Two binary operations ∗ * ∗ ⊕ \oplus ⊕ m ∗ n = m n − n − 1 m * n = mn - n - 1 m ∗ n = mn − n − 1 m ⊕ n = m n + n − 2 m \oplus n = mn + n - 2 m ⊕ n = mn + n − 2 m m m n n n 3 ⊕ ( 4 ∗ 5 ) 3 \oplus (4 * 5) 3 ⊕ ( 4 ∗ 5 )
Worked solution (try it first) Work out the bracket with the
∗ * ∗ rule:
4 ∗ 5 = 4 × 5 − 5 − 1 = 14 4 * 5 = 4 \times 5 - 5 - 1 = 14 4 ∗ 5 = 4 × 5 − 5 − 1 = 14 .
Now use the
⊕ \oplus ⊕ rule with
m = 3 m = 3 m = 3 and
n = 14 n = 14 n = 14 :
3 ⊕ 14 = 3 × 14 + 14 − 2 3 \oplus 14 = 3 \times 14 + 14 - 2 3 ⊕ 14 = 3 × 14 + 14 − 2 .
That is
42 + 14 − 2 = 54 42 + 14 - 2 = 54 42 + 14 − 2 = 54 , option C.
Watch out
Finish the ⊕ \oplus ⊕ m n = 42 mn = 42 mn = 42 n = 14 n = 14 n = 14 Report a problem with this question
If x ∗ y = x + y − x y x * y = x + y - xy x ∗ y = x + y − x y x x x ( x ∗ 2 ) + ( x ∗ 3 ) = 68 (x * 2) + (x * 3) = 68 ( x ∗ 2 ) + ( x ∗ 3 ) = 68
A 24 B 22 C − 12 -12 − 12 D − 21 -21 − 21
Worked solution (try it first) Use the rule with
y = 2 y = 2 y = 2 :
x ∗ 2 = x + 2 − 2 x = 2 − x x * 2 = x + 2 - 2x = 2 - x x ∗ 2 = x + 2 − 2 x = 2 − x .
With
y = 3 y = 3 y = 3 :
x ∗ 3 = x + 3 − 3 x = 3 − 2 x x * 3 = x + 3 - 3x = 3 - 2x x ∗ 3 = x + 3 − 3 x = 3 − 2 x .
Add them:
5 − 3 x = 68 5 - 3x = 68 5 − 3 x = 68 .
Take 5 from both sides:
− 3 x = 63 -3x = 63 − 3 x = 63 , so
x = − 21 x = -21 x = − 21 , option D.
Watch out
The x y xy x y ( 2 + 3 x ) + ( 3 + 4 x ) = 68 (2 + 3x) + (3 + 4x) = 68 ( 2 + 3 x ) + ( 3 + 4 x ) = 68 x = 9 x = 9 x = 9 Report a problem with this question
Determine x + y x + y x + y ( 2 − 3 − 1 4 ) ( x y ) = ( − 1 8 ) \begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix} ( 2 − 1 − 3 4 ) ( x y ) = ( − 1 8 )
Worked solution (try it first) Multiply out the left side to get two equations:
2 x − 3 y = − 1 2x - 3y = -1 2 x − 3 y = − 1 and
− x + 4 y = 8 -x + 4y = 8 − x + 4 y = 8 .
From the second,
x = 4 y − 8 x = 4y - 8 x = 4 y − 8 .
Substitute into the first:
8 y − 16 − 3 y = − 1 8y - 16 - 3y = -1 8 y − 16 − 3 y = − 1 , so
5 y = 15 5y = 15 5 y = 15 and
y = 3 y = 3 y = 3 .
Then
x = 12 − 8 = 4 x = 12 - 8 = 4 x = 12 − 8 = 4 .
So
x + y = 7 x + y = 7 x + y = 7 , option C.
Watch out
The question asks for x + y x + y x + y x y xy x y Report a problem with this question
Find the non-zero positive value of x x x ∣ x 1 0 1 x 1 0 1 x ∣ = 0 \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} = 0 x 1 0 1 x 1 0 1 x = 0
Worked solution (try it first) Expand along the first row:
x ( x ⋅ x − 1 ⋅ 1 ) − 1 ( 1 ⋅ x − 1 ⋅ 0 ) + 0 x(x \cdot x - 1 \cdot 1) - 1(1 \cdot x - 1 \cdot 0) + 0 x ( x ⋅ x − 1 ⋅ 1 ) − 1 ( 1 ⋅ x − 1 ⋅ 0 ) + 0 .
This is
x 3 − x − x = x 3 − 2 x x^3 - x - x = x^3 - 2x x 3 − x − x = x 3 − 2 x , so
x ( x 2 − 2 ) = 0 x(x^2 - 2) = 0 x ( x 2 − 2 ) = 0 .
So
x = 0 x = 0 x = 0 or
x 2 = 2 x^2 = 2 x 2 = 2 .
The non-zero positive value is
x = 2 x = \sqrt2 x = 2 , option C.
Watch out
Don't drop the middle term: it is − 1 × ( x − 0 ) = − x -1 \times (x - 0) = -x − 1 × ( x − 0 ) = − x x ( x 2 − 1 ) = 0 x(x^2 - 1) = 0 x ( x 2 − 1 ) = 0 x = 1 x = 1 x = 1 Report a problem with this question
Each of the base angles of an isosceles triangle is 58 ∘ 58^\circ 5 8 ∘
A 128 ∘ 128^\circ 12 8 ∘ B 116 ∘ 116^\circ 11 6 ∘ C 64 ∘ 64^\circ 6 4 ∘ D 58 ∘ 58^\circ 5 8 ∘
Worked solution (try it first) The angles of the triangle add up to
180 ∘ 180^\circ 18 0 ∘ , so the apex angle is
180 ∘ − 58 ∘ − 58 ∘ = 64 ∘ 180^\circ - 58^\circ - 58^\circ = 64^\circ 18 0 ∘ − 5 8 ∘ − 5 8 ∘ = 6 4 ∘ .
The base is opposite the apex, so the apex angle is the angle the base makes at the circumference.
The angle at the centre is twice that:
2 × 64 ∘ = 128 ∘ 2 \times 64^\circ = 128^\circ 2 × 6 4 ∘ = 12 8 ∘ , option A.
Watch out
Double the apex angle, which faces the base. Doubling a base angle gives 116 ∘ 116^\circ 11 6 ∘ Report a problem with this question
In the figure, F K ∥ G R FK \parallel GR F K ∥ GR F H = G H FH = GH F H = G H ∠ R F K = 34 ∘ \angle RFK = 34^\circ ∠ R F K = 3 4 ∘ ∠ F G H = 47 ∘ \angle FGH = 47^\circ ∠ F G H = 4 7 ∘ x x x ∠ H F R \angle HFR ∠ H F R
A 42 ∘ 42^\circ 4 2 ∘ B 52 ∘ 52^\circ 5 2 ∘ C 64 ∘ 64^\circ 6 4 ∘ D 72 ∘ 72^\circ 7 2 ∘
Worked solution (try it first) F H = G H FH = GH F H = G H , so triangle
F G H FGH F G H is isosceles with equal angles at
G G G and
F F F :
∠ G F H = 47 ∘ \angle GFH = 47^\circ ∠ GF H = 4 7 ∘ .
F K ∥ G R FK \parallel GR F K ∥ GR , so
∠ G F K \angle GFK ∠ GF K and
∠ F G H \angle FGH ∠ F G H are co-interior:
∠ G F K = 180 ∘ − 47 ∘ \angle GFK = 180^\circ - 47^\circ ∠ GF K = 18 0 ∘ − 4 7 ∘ ∠ G F K \angle GFK ∠ GF K is made up of
47 ∘ + x + 34 ∘ 47^\circ + x + 34^\circ 4 7 ∘ + x + 3 4 ∘ , so
x = 133 ∘ − 81 ∘ = 52 ∘ x = 133^\circ - 81^\circ = 52^\circ x = 13 3 ∘ − 8 1 ∘ = 5 2 ∘ , option B.
Watch out
The equal sides F H FH F H G H GH G H H H H G G G F F F H H H 47 ∘ 47^\circ 4 7 ∘ H H H ∠ G F H = 86 ∘ \angle GFH = 86^\circ ∠ GF H = 8 6 ∘ Report a problem with this question
The figure shows circles of radii 3 cm 3\text{ cm} 3 cm 2 cm 2\text{ cm} 2 cm X X X Y Y Y 25 cm 25\text{ cm} 25 cm X Y XY X Y
A 630 cm \sqrt{630}\text{ cm} 630 cm B 626 cm \sqrt{626}\text{ cm} 626 cm C 615 cm \sqrt{615}\text{ cm} 615 cm D 600 cm \sqrt{600}\text{ cm} 600 cm
Worked solution (try it first) Each radius is perpendicular to the tangent, so the two radii and the tangent form a trapezium with two right angles.
Draw a line from
Y Y Y parallel to the tangent to meet radius
X X X at a right angle.
It is 25 cm long and cuts off
3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 cm of that radius.
X Y XY X Y is the hypotenuse of this right-angled triangle, so by Pythagoras
X Y 2 = 25 2 + 1 2 = 626 XY^2 = 25^2 + 1^2 = 626 X Y 2 = 2 5 2 + 1 2 = 626 .
So
X Y = 626 XY = \sqrt{626} X Y = 626 cm, option B.
Watch out
For a direct common tangent the short side is the difference of the radii, 3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 25 2 25^2 2 5 2 5 2 5^2 5 2 600 \sqrt{600} 600 Report a problem with this question
A chord of a circle of diameter 42 cm subtends an angle of 60 ∘ 60^\circ 6 0 ∘ [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 22 cm B 44 cm C 110 cm D 220 cm
Worked solution (try it first) The circumference is
π d = 22 7 × 42 = 132 \pi d = \frac{22}{7} \times 42 = 132 π d = 7 22 × 42 = 132 cm.
The arc is
60 360 = 1 6 \frac{60}{360} = \frac16 360 60 = 6 1 of it:
132 ÷ 6 = 22 132 \div 6 = 22 132 ÷ 6 = 22 cm, option A.
Watch out
42 cm is the diameter. Using it as the radius in 2 π r 2\pi r 2 π r Report a problem with this question
An arc of a circle subtends an angle of 70 ∘ 70^\circ 7 0 ∘ [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 22 cm 2 22\text{ cm}^2 22 cm 2 B 44 cm 2 44\text{ cm}^2 44 cm 2 C 66 cm 2 66\text{ cm}^2 66 cm 2 D 88 cm 2 88\text{ cm}^2 88 cm 2
Worked solution (try it first) Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 , with
r 2 = 36 r^2 = 36 r 2 = 36 .
So the area is
70 360 × 22 7 × 36 \frac{70}{360} \times \frac{22}{7} \times 36 360 70 × 7 22 × 36 .
Cancel the 7 into 70 and the 36 into 360:
10 × 22 10 = 22 cm 2 \frac{10 \times 22}{10} = 22\text{ cm}^2 10 10 × 22 = 22 cm 2 , option A.
Watch out
6 cm is the radius, not the diameter. Taking the radius as 12 gives r 2 = 144 r^2 = 144 r 2 = 144 88 cm 2 88\text{ cm}^2 88 cm 2 Also set as JAMB 2015 · UTME · Q31
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A prism has a uniform cross-section in the shape of a trapezium with parallel sides 8 cm and 10 cm, 5 cm apart. If the prism is 11 cm long, find its volume.
A 990 cm 3 990\text{ cm}^3 990 cm 3 B 880 cm 3 880\text{ cm}^3 880 cm 3 C 550 cm 3 550\text{ cm}^3 550 cm 3 D 495 cm 3 495\text{ cm}^3 495 cm 3
Worked solution (try it first) Area of the trapezium: half the sum of the parallel sides times the distance between them,
1 2 ( 8 + 10 ) × 5 = 45 cm 2 \frac12(8 + 10) \times 5 = 45\text{ cm}^2 2 1 ( 8 + 10 ) × 5 = 45 cm 2 .
Volume = cross-section × length:
45 × 11 = 495 cm 3 45 \times 11 = 495\text{ cm}^3 45 × 11 = 495 cm 3 , option D.
Watch out
Halve the sum of the parallel sides. Without the 1 2 \frac12 2 1 990 cm 3 990\text{ cm}^3 990 cm 3 Report a problem with this question
A cone with a sector angle of 45 ∘ 45^\circ 4 5 ∘ r r r
A r 16 \frac{r}{16} 16 r B r 8 \frac r8 8 r C r 4 \frac r4 4 r D r 2 \frac r2 2 r
Worked solution (try it first) The arc of the sector becomes the circumference of the cone's base
R R R :
45 360 × 2 π r = 2 π R \frac{45}{360} \times 2\pi r = 2\pi R 360 45 × 2 π r = 2 π R .
Divide both sides by
2 π 2\pi 2 π :
R = 45 360 r = r 8 R = \frac{45}{360}r = \frac r8 R = 360 45 r = 8 r , option B.
Watch out
The arc length uses the full circumference 2 π r 2\pi r 2 π r π r \pi r π r r 16 \frac{r}{16} 16 r Report a problem with this question
A point P P P L L L M M M L M LM L M P P P L M LM L M P P P L L L
Worked solution (try it first) P P P is equidistant from
L L L and
M M M , so it lies on the perpendicular bisector of
L M LM L M .
The foot of the perpendicular from
P P P is the mid-point, 8 cm from
L L L .
In the right-angled triangle,
P L = 10 PL = 10 P L = 10 cm is the hypotenuse and 8 cm is one side.
By Pythagoras, the distance from
P P P to
L M LM L M is
10 2 − 8 2 = 36 = 6 \sqrt{10^2 - 8^2} = \sqrt{36} = 6 1 0 2 − 8 2 = 36 = 6 cm, option D.
Watch out
8 cm (option C) is half of L M LM L M Report a problem with this question
The angle between the positive horizontal axis and a given line is 135 ∘ 135^\circ 13 5 ∘ ( 2 , 3 ) (2, 3) ( 2 , 3 )
A x − y = 1 x - y = 1 x − y = 1 B x + y = 1 x + y = 1 x + y = 1 C x + y = 5 x + y = 5 x + y = 5 D x − y = 5 x - y = 5 x − y = 5
Worked solution (try it first) The gradient is the tangent of the angle with the positive
x x x -axis:
tan 135 ∘ = − 1 \tan 135^\circ = -1 tan 13 5 ∘ = − 1 .
Through
( 2 , 3 ) (2, 3) ( 2 , 3 ) :
y − 3 = − 1 ( x − 2 ) y - 3 = -1(x - 2) y − 3 = − 1 ( x − 2 ) , so
y − 3 = − x + 2 y - 3 = -x + 2 y − 3 = − x + 2 .
Add
x x x and 3 to both sides:
x + y = 5 x + y = 5 x + y = 5 , option C.
Watch out
135 ∘ 135^\circ 13 5 ∘ − 1 -1 − 1 1 1 1 + 1 +1 + 1 y − x = 1 y - x = 1 y − x = 1 Report a problem with this question
Find the distance between the point Q ( 4 , 3 ) Q(4, 3) Q ( 4 , 3 ) 2 x − y = 4 2x - y = 4 2 x − y = 4 x + y = 2 x + y = 2 x + y = 2
A 3 10 3\sqrt{10} 3 10 B 3 5 3\sqrt5 3 5 C 26 \sqrt{26} 26 D 13 \sqrt{13} 13
Worked solution (try it first) Add the two equations to remove
y y y :
3 x = 6 3x = 6 3 x = 6 , so
x = 2 x = 2 x = 2 .
Then
y = 2 − 2 = 0 y = 2 - 2 = 0 y = 2 − 2 = 0 , and the lines meet at
( 2 , 0 ) (2, 0) ( 2 , 0 ) .
From
( 2 , 0 ) (2, 0) ( 2 , 0 ) to
Q ( 4 , 3 ) Q(4, 3) Q ( 4 , 3 ) the changes are
4 − 2 = 2 4 - 2 = 2 4 − 2 = 2 and
3 − 0 = 3 3 - 0 = 3 3 − 0 = 3 .
By Pythagoras the distance is
2 2 + 3 2 = 13 \sqrt{2^2 + 3^2} = \sqrt{13} 2 2 + 3 2 = 13 , option D.
Watch out
Subtract the coordinates before squaring. Adding them gives 6 2 + 3 2 = 45 = 3 5 \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt5 6 2 + 3 2 = 45 = 3 5 Report a problem with this question
In a triangle X Y Z XYZ X Y Z ∠ X Y Z = 60 ∘ \angle XYZ = 60^\circ ∠ X Y Z = 6 0 ∘ X Y = 3 XY = 3 X Y = 3 Y Z = 4 YZ = 4 Y Z = 4 X Z XZ X Z
A 23 \sqrt{23} 23 B 13 \sqrt{13} 13 C 2 5 2\sqrt5 2 5 D 2 3 2\sqrt3 2 3
Worked solution (try it first) You have two sides and the angle between them, so use the cosine rule:
X Z 2 = 3 2 + 4 2 − 2 ( 3 ) ( 4 ) cos 60 ∘ XZ^2 = 3^2 + 4^2 - 2(3)(4)\cos60^\circ X Z 2 = 3 2 + 4 2 − 2 ( 3 ) ( 4 ) cos 6 0 ∘ .
cos 60 ∘ = 1 2 \cos60^\circ = \frac12 cos 6 0 ∘ = 2 1 , so the last term is
24 × 1 2 = 12 24 \times \frac12 = 12 24 × 2 1 = 12 and
X Z 2 = 25 − 12 = 13 XZ^2 = 25 - 12 = 13 X Z 2 = 25 − 12 = 13 .
So
X Z = 13 XZ = \sqrt{13} X Z = 13 cm, option B.
Watch out
2 3 = 12 2\sqrt3 = \sqrt{12} 2 3 = 12 2 a b cos C 2ab\cos C 2 ab cos C 3 2 + 4 2 = 25 3^2 + 4^2 = 25 3 2 + 4 2 = 25 Report a problem with this question
In the figure, X Y Z XYZ X Y Z X Y = 5 XY = 5 X Y = 5 X Z = 2 XZ = 2 X Z = 2 X Z XZ X Z E E E ∠ Y Z E = 150 ∘ \angle YZE = 150^\circ ∠ Y Z E = 15 0 ∘ ∠ X Y Z = θ \angle XYZ = \theta ∠ X Y Z = θ sin θ \sin\theta sin θ
A 3 5 \frac35 5 3 B 1 2 \frac12 2 1 C 2 5 \frac25 5 2 D 1 5 \frac15 5 1
Worked solution (try it first) Angles on a straight line add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ X Z Y = 180 ∘ − 150 ∘ \angle XZY = 180^\circ - 150^\circ ∠ X Z Y = 18 0 ∘ − 15 0 ∘ Sine rule:
X Z = 2 XZ = 2 X Z = 2 faces
θ \theta θ and
X Y = 5 XY = 5 X Y = 5 faces the
30 ∘ 30^\circ 3 0 ∘ at
Z Z Z .
So
sin θ 2 = sin 30 ∘ 5 \dfrac{\sin\theta}{2} = \dfrac{\sin30^\circ}{5} 2 sin θ = 5 sin 3 0 ∘ .
So
sin θ = 2 × 1 2 5 \sin\theta = \dfrac{2 \times \frac12}{5} sin θ = 5 2 × 2 1 = 1 5 = \frac15 = 5 1 , option D.
Watch out
The triangle is not right-angled, so sin θ \sin\theta sin θ 2 5 \frac{2}{5} 5 2 sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 Report a problem with this question
Differentiate 6 x 3 − 5 x 2 + 1 3 x 2 \dfrac{6x^3 - 5x^2 + 1}{3x^2} 3 x 2 6 x 3 − 5 x 2 + 1 x x x
A 2 + 2 3 x 3 2 + \frac{2}{3x^3} 2 + 3 x 3 2 B 2 + 1 6 x 2 + \frac{1}{6x} 2 + 6 x 1 C 2 − 2 3 x 3 2 - \frac{2}{3x^3} 2 − 3 x 3 2 D 2 − 1 6 x 2 - \frac{1}{6x} 2 − 6 x 1
Worked solution (try it first) Divide each term by
3 x 2 3x^2 3 x 2 first:
y = 2 x − 5 3 + 1 3 x − 2 y = 2x - \frac53 + \frac13x^{-2} y = 2 x − 3 5 + 3 1 x − 2 .
Differentiate term by term:
2 x 2x 2 x gives 2, the constant gives 0, and
1 3 x − 2 \frac13x^{-2} 3 1 x − 2 gives
− 2 3 x − 3 -\frac23x^{-3} − 3 2 x − 3 .
So
d y d x = 2 − 2 3 x 3 \frac{dy}{dx} = 2 - \frac{2}{3x^3} d x d y = 2 − 3 x 3 2 , option C.
Watch out
Bringing down a negative power makes the term negative: 1 3 × ( − 2 ) = − 2 3 \frac13 \times (-2) = -\frac23 3 1 × ( − 2 ) = − 3 2 Report a problem with this question
Find the gradient of the curve y = 2 x − 1 x y = 2\sqrt x - \frac1x y = 2 x − x 1 x = 1 x = 1 x = 1
Worked solution (try it first) Write the terms as powers:
y = 2 x 1 / 2 − x − 1 y = 2x^{1/2} - x^{-1} y = 2 x 1/2 − x − 1 .
Differentiate:
d y d x = x − 1 / 2 + x − 2 \frac{dy}{dx} = x^{-1/2} + x^{-2} d x d y = x − 1/2 + x − 2 .
At
x = 1 x = 1 x = 1 :
1 + 1 = 2 1 + 1 = 2 1 + 1 = 2 , option C.
Watch out
− x − 1 -x^{-1} − x − 1 + x − 2 +x^{-2} + x − 2 − 1 × − 1 = + 1 -1 \times -1 = +1 − 1 × − 1 = + 1 1 − 1 = 0 1 - 1 = 0 1 − 1 = 0 Report a problem with this question
Integrate 1 x + cos x \frac1x + \cos x x 1 + cos x x x x
A − 1 x 2 + sin x + k -\frac{1}{x^2} + \sin x + k − x 2 1 + sin x + k B ln x + sin x + k \ln x + \sin x + k ln x + sin x + k C ln x − sin x + k \ln x - \sin x + k ln x − sin x + k D − 1 x 2 − sin x + k -\frac{1}{x^2} - \sin x + k − x 2 1 − sin x + k
Worked solution (try it first) 1 x \frac1x x 1 integrates to
ln x \ln x ln x , and
cos x \cos x cos x integrates to
sin x \sin x sin x .
So the integral is
ln x + sin x + k \ln x + \sin x + k ln x + sin x + k , option B.
Watch out
− 1 x 2 -\frac{1}{x^2} − x 2 1 1 x \frac1x x 1 x − 1 x^{-1} x − 1 ln x \ln x ln x Also set as JAMB 2015 · UTME · Q33
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If y = x ( x 4 + x 2 + 1 ) y = x(x^4 + x^2 + 1) y = x ( x 4 + x 2 + 1 ) ∫ − 1 1 y d x \displaystyle\int_{-1}^{1} y\,dx ∫ − 1 1 y d x
A 11 12 \frac{11}{12} 12 11 B 11 16 \frac{11}{16} 16 11 C 5 6 \frac56 6 5 D 0
Worked solution (try it first) Expand:
y = x 5 + x 3 + x y = x^5 + x^3 + x y = x 5 + x 3 + x , so the integral is
[ x 6 6 + x 4 4 + x 2 2 ] − 1 1 \left[\frac{x^6}{6} + \frac{x^4}{4} + \frac{x^2}{2}\right]_{-1}^{1} [ 6 x 6 + 4 x 4 + 2 x 2 ] − 1 1 .
Every power is even, so the value at
x = 1 x = 1 x = 1 and at
x = − 1 x = -1 x = − 1 is the same,
11 12 \frac{11}{12} 12 11 .
Subtract:
11 12 − 11 12 = 0 \frac{11}{12} - \frac{11}{12} = 0 12 11 − 12 11 = 0 , option D.
Watch out
Subtract the value at the lower limit too. Using only the upper limit gives 11 12 \frac{11}{12} 12 11 Report a problem with this question
The pie chart shows the income of a civil servant in a month. If his monthly income is ₦6000, find his monthly basic salary.
Worked solution (try it first) The angles at the centre add up to
360 ∘ 360^\circ 36 0 ∘ , so Basic is
360 ∘ − ( 69 ∘ + 60 ∘ + 61 ∘ + 50 ∘ ) 360^\circ - (69^\circ + 60^\circ + 61^\circ + 50^\circ) 36 0 ∘ − ( 6 9 ∘ + 6 0 ∘ + 6 1 ∘ + 5 0 ∘ ) , which is
120 ∘ 120^\circ 12 0 ∘ .
120 ∘ 120^\circ 12 0 ∘ is
120 360 = 1 3 \frac{120}{360} = \frac13 360 120 = 3 1 of the circle.
So the basic salary is
1 3 × 6000 = 2000 \frac13 \times 6000 = 2000 3 1 × 6000 = 2000 naira: ₦2000, option A.
Watch out
Find the unlabelled Basic angle first by taking the others from 360 ∘ 360^\circ 36 0 ∘ Report a problem with this question
In an examination, the result of a certain school is as shown in the histogram. How many candidates did the school present?
Worked solution (try it first) Each bar's height is how many sat the exam with a score in that class.
Read the bars: 3, 5, 8, 1 and 2.
Add them:
3 + 5 + 8 + 1 + 2 = 19 3 + 5 + 8 + 1 + 2 = 19 3 + 5 + 8 + 1 + 2 = 19 , option D.
Watch out
Read every bar, including the short ones: missing the bar of height 1 gives 18 (option C). Report a problem with this question
Find the median age of the frequency distribution below.
Age
20
25
30
35
40
45
No. of students
3
5
1
1
2
3
Worked solution (try it first) There are
3 + 5 + 1 + 1 + 2 + 3 = 15 3 + 5 + 1 + 1 + 2 + 3 = 15 3 + 5 + 1 + 1 + 2 + 3 = 15 students, so the median is the
15 + 1 2 = 8 \frac{15 + 1}{2} = 8 2 15 + 1 = 8 th age.
Running totals: 3 (age 20), 8 (age 25).
The 4th to 8th students are all 25.
So the median age is 25, option B.
Watch out
The median is the middle student, not the middle of the age column. Using running totals, the 8th student is 25, not 30 or 35. Report a problem with this question
The scores of ten students in a test of 20 marks are 15, 16, 17, 13, 16, 8, 5, 16, 19, 17. What is the modal score?
Worked solution (try it first) Count each score: 16 appears three times, 17 twice, and every other score once.
The mode is the score that occurs most often: 16, option C.
Watch out
The mode is the most frequent score, not the highest score (19, option D). Report a problem with this question
Find the standard deviation of the data − 5 , − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , 4 , 5 -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 − 5 , − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , 4 , 5
A 2 B 3 C 10 \sqrt{10} 10 D 11 \sqrt{11} 11
Worked solution (try it first) The numbers are symmetric about 0, so the mean is 0 and each deviation is the number itself.
∑ x 2 = 2 ( 1 + 4 + 9 + 16 + 25 ) = 110 \sum x^2 = 2(1 + 4 + 9 + 16 + 25) = 110 ∑ x 2 = 2 ( 1 + 4 + 9 + 16 + 25 ) = 110 .
There are 11 numbers, so the variance is
110 11 = 10 \frac{110}{11} = 10 11 110 = 10 and the standard deviation is
10 \sqrt{10} 10 , option C.
Watch out
Count the 0: there are 11 numbers, not 10. Dividing 110 by 10 gives 11 \sqrt{11} 11 Report a problem with this question
Find the difference between the range and the variance of the numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7, where ∑ d 2 = 60 \sum d^2 = 60 ∑ d 2 = 60
Worked solution (try it first) The range is the largest number minus the smallest:
10 − 2 = 8 10 - 2 = 8 10 − 2 = 8 .
The mean is
60 10 = 6 \frac{60}{10} = 6 10 60 = 6 and
∑ d 2 = 60 \sum d^2 = 60 ∑ d 2 = 60 is given, so the variance is
60 10 = 6 \frac{60}{10} = 6 10 60 = 6 .
The difference is
8 − 6 = 2 8 - 6 = 2 8 − 6 = 2 , option A.
Watch out
For the range, use the largest and smallest numbers in the whole list (10 and 2), not the first and last. Using 10 − 4 = 6 10 - 4 = 6 10 − 4 = 6 Report a problem with this question
In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that it is either a grape or a banana?
A 17 30 \frac{17}{30} 30 17 B 11 30 \frac{11}{30} 30 11 C 6 30 \frac6{30} 30 6 D 5 30 \frac5{30} 30 5
Worked solution (try it first) There are
6 + 11 + 13 = 30 6 + 11 + 13 = 30 6 + 11 + 13 = 30 fruits.
A grape or a banana is
6 + 11 = 17 6 + 11 = 17 6 + 11 = 17 fruits, since one fruit can't be both.
So the probability is
17 30 \frac{17}{30} 30 17 , option A.
Watch out
"Either a grape or a banana" includes both kinds. Using only the bananas gives 11 30 \frac{11}{30} 30 11 Report a problem with this question
A number is selected at random between 10 and 20, both numbers inclusive. Find the probability that the number is even.
A 5 11 \frac5{11} 11 5 B 1 2 \frac12 2 1 C 6 11 \frac6{11} 11 6 D 7 10 \frac7{10} 10 7
Worked solution (try it first) From 10 to 20 inclusive there are 11 numbers.
The even ones are 10, 12, 14, 16, 18, 20, which is 6 numbers.
So the probability is
6 11 \frac{6}{11} 11 6 , option C.
Watch out
Both ends are even, so there is one more even number than odd. Assuming half are even gives 1 2 \frac12 2 1 Report a problem with this question