Paper JAMB 2011 General Maths Objective
Objective paper · 45 questions · partial
JAMB 2011 · UTME Topics include Number bases, Number foundations & fractions, Commercial arithmetic, Indices & standard form, Logarithms, Surds.
Our copy of this paper is missing questions 1, 24, 41, 44, 46.
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.
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 42 43 45 47 48 49 50 If 2 q 3 5 = 77 8 2q3_5 = 77_8 2 q 3 5 = 7 7 8 q q q
Worked solution (try it first) Change the right side to base ten:
77 8 = 7 × 8 + 7 = 63 77_8 = 7 \times 8 + 7 = 63 7 7 8 = 7 × 8 + 7 = 63 .
In base five,
2 q 3 5 = 50 + 5 q + 3 = 53 + 5 q 2q3_5 = 50 + 5q + 3 = 53 + 5q 2 q 3 5 = 50 + 5 q + 3 = 53 + 5 q .
Set
53 + 5 q = 63 53 + 5q = 63 53 + 5 q = 63 :
5 q = 10 5q = 10 5 q = 10 , so
q = 2 q = 2 q = 2 , option A.
Watch out
77 8 77_8 7 7 8 5 q = 24 5q = 24 5 q = 24 Report a problem with this question
Simplify 3 2 3 × 5 6 × 2 3 11 15 × 3 4 × 2 27 \dfrac{3\frac23 \times \frac56 \times \frac23}{\frac{11}{15} \times \frac34 \times \frac{2}{27}} 15 11 × 4 3 × 27 2 3 3 2 × 6 5 × 3 2
A 5 2 3 5\frac23 5 3 2 B 30 30 30 C 4 1 3 4\frac13 4 3 1 D 50 50 50
Worked solution (try it first) Top:
3 2 3 = 11 3 3\frac23 = \frac{11}{3} 3 3 2 = 3 11 , so the top is
11 3 × 5 6 × 2 3 = 110 54 \frac{11}{3} \times \frac56 \times \frac23 = \frac{110}{54} 3 11 × 6 5 × 3 2 = 54 110 = 55 27 = \frac{55}{27} = 27 55 .
Bottom:
11 15 × 3 4 × 2 27 = 66 1620 \frac{11}{15} \times \frac34 \times \frac{2}{27} = \frac{66}{1620} 15 11 × 4 3 × 27 2 = 1620 66 = 11 270 = \frac{11}{270} = 270 11 .
Divide by flipping the bottom:
55 27 × 270 11 \frac{55}{27} \times \frac{270}{11} 27 55 × 11 270 .
Cancel 11 into 55 (5) and 27 into 270 (10):
5 × 10 = 50 5 \times 10 = 50 5 × 10 = 50 , option D.
Watch out
Change 3 2 3 3\frac23 3 3 2 11 3 \frac{11}{3} 3 11 3 × 2 3 = 2 3 \times \frac23 = 2 3 × 3 2 = 2 Report a problem with this question
A man invested ₦5,000 for 9 months at 4 % 4\% 4%
Worked solution (try it first) Write the time in years: 9 months is
9 12 = 3 4 \frac{9}{12} = \frac34 12 9 = 4 3 year.
Use
I = P R T 100 I = \dfrac{PRT}{100} I = 100 P R T :
I = 5000 × 4 100 × 3 4 I = 5000 \times \dfrac{4}{100} \times \dfrac34 I = 5000 × 100 4 × 4 3 .
That is
200 × 3 4 = 150 200 \times \frac34 = 150 200 × 4 3 = 150 .
The interest is ₦150, option A.
Watch out
The rate is per year, so the time must be in years. Using T = 9 T = 9 T = 9 3 4 \frac34 4 3 Also set as JAMB 2017 · UTME · Q26
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If the numbers M M M N N N Q Q Q 5 : 4 : 3 5 : 4 : 3 5 : 4 : 3 2 N − Q M \dfrac{2N - Q}{M} M 2 N − Q
Worked solution (try it first) Write the numbers in terms of one part
k k k :
M = 5 k M = 5k M = 5 k ,
N = 4 k N = 4k N = 4 k ,
Q = 3 k Q = 3k Q = 3 k .
Then
2 N − Q = 8 k − 3 k = 5 k 2N - Q = 8k - 3k = 5k 2 N − Q = 8 k − 3 k = 5 k .
So
2 N − Q M = 5 k 5 k = 1 \dfrac{2N - Q}{M} = \dfrac{5k}{5k} = 1 M 2 N − Q = 5 k 5 k = 1 , option C.
Watch out
2 N 2N 2 N N N N 8 k 8k 8 k N N N 4 k − 3 k 5 k = 1 5 \frac{4k - 3k}{5k} = \frac15 5 k 4 k − 3 k = 5 1 Report a problem with this question
Simplify ( 16 81 ) 1 4 ÷ ( 9 16 ) − 1 2 \left(\dfrac{16}{81}\right)^{\frac14} \div \left(\dfrac{9}{16}\right)^{-\frac12} ( 81 16 ) 4 1 ÷ ( 16 9 ) − 2 1
A 2 3 \frac23 3 2 B 1 2 \frac12 2 1 C 8 9 \frac89 9 8 D 1 3 \frac13 3 1
Worked solution (try it first) Fourth roots:
( 16 81 ) 1 4 = 2 3 \left(\frac{16}{81}\right)^{\frac14} = \frac23 ( 81 16 ) 4 1 = 3 2 , since
2 4 = 16 2^4 = 16 2 4 = 16 and
3 4 = 81 3^4 = 81 3 4 = 81 .
The negative index flips the fraction:
( 9 16 ) − 1 2 = ( 16 9 ) 1 2 \left(\frac{9}{16}\right)^{-\frac12} = \left(\frac{16}{9}\right)^{\frac12} ( 16 9 ) − 2 1 = ( 9 16 ) 2 1 Divide by multiplying by the reciprocal:
2 3 × 3 4 = 1 2 \frac23 \times \frac34 = \frac12 3 2 × 4 3 = 2 1 , option B.
Watch out
Flip the fraction for the negative index: ( 9 16 ) − 1 2 = 4 3 \left(\frac{9}{16}\right)^{-\frac12} = \frac43 ( 16 9 ) − 2 1 = 3 4 3 4 \frac34 4 3 3 4 \frac34 4 3 2 3 ÷ 3 4 = 8 9 \frac23 \div \frac34 = \frac89 3 2 ÷ 4 3 = 9 8 Report a problem with this question
If log 3 18 + log 3 3 − log 3 x = 3 \log_3 18 + \log_3 3 - \log_3 x = 3 log 3 18 + log 3 3 − log 3 x = 3 x x x
Worked solution (try it first) Combine the logs: adding multiplies and subtracting divides, so the left side is
log 3 18 × 3 x = log 3 54 x \log_3 \frac{18 \times 3}{x} = \log_3 \frac{54}{x} log 3 x 18 × 3 = log 3 x 54 .
Change to index form:
54 x = 3 3 = 27 \frac{54}{x} = 3^3 = 27 x 54 = 3 3 = 27 .
So
x = 54 27 = 2 x = \frac{54}{27} = 2 x = 27 54 = 2 , option B.
Watch out
log 3 N = 3 \log_3 N = 3 log 3 N = 3 N = 3 3 = 27 N = 3^3 = 27 N = 3 3 = 27 x = 6 x = 6 x = 6 Report a problem with this question
Rationalize 2 − 5 3 − 5 \dfrac{2 - \sqrt5}{3 - \sqrt5} 3 − 5 2 − 5
A 1 − 5 2 \dfrac{1 - \sqrt5}{2} 2 1 − 5 B 1 − 5 4 \dfrac{1 - \sqrt5}{4} 4 1 − 5 C 5 − 1 2 \dfrac{\sqrt5 - 1}{2} 2 5 − 1 D 1 + 5 4 \dfrac{1 + \sqrt5}{4} 4 1 + 5
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
3 + 5 3 + \sqrt5 3 + 5 .
Bottom:
( 3 − 5 ) ( 3 + 5 ) = 9 − 5 (3 - \sqrt5)(3 + \sqrt5) = 9 - 5 ( 3 − 5 ) ( 3 + 5 ) = 9 − 5 , which is 4.
Top:
( 2 − 5 ) ( 3 + 5 ) = 6 + 2 5 − 3 5 − 5 (2 - \sqrt5)(3 + \sqrt5) = 6 + 2\sqrt5 - 3\sqrt5 - 5 ( 2 − 5 ) ( 3 + 5 ) = 6 + 2 5 − 3 5 − 5 , which is
1 − 5 1 - \sqrt5 1 − 5 .
So the value is
1 − 5 4 \dfrac{1 - \sqrt5}{4} 4 1 − 5 , option B.
Watch out
Watch the signs in the top: − 5 × 5 = − 5 -\sqrt5 \times \sqrt5 = -5 − 5 × 5 = − 5 2 5 − 3 5 = − 5 2\sqrt5 - 3\sqrt5 = -\sqrt5 2 5 − 3 5 = − 5 + 5 +\sqrt5 + 5 1 + 5 4 \frac{1 + \sqrt5}{4} 4 1 + 5 Also set as JAMB 2017 · UTME · Q27
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Simplify ( 2 + 1 3 ) ( 2 − 1 3 ) \left(\sqrt2 + \dfrac{1}{\sqrt3}\right)\left(\sqrt2 - \dfrac{1}{\sqrt3}\right) ( 2 + 3 1 ) ( 2 − 3 1 )
A 7 3 \frac73 3 7 B 5 3 \frac53 3 5 C 5 2 \frac52 2 5 D 3 2 \frac32 2 3
Worked solution (try it first) The brackets are
( a + b ) ( a − b ) (a + b)(a - b) ( a + b ) ( a − b ) , a difference of two squares:
a 2 − b 2 a^2 - b^2 a 2 − b 2 .
Here
a 2 = ( 2 ) 2 = 2 a^2 = (\sqrt2)^2 = 2 a 2 = ( 2 ) 2 = 2 and
b 2 = ( 1 3 ) 2 b^2 = \left(\frac{1}{\sqrt3}\right)^2 b 2 = ( 3 1 ) 2 So the value is
2 − 1 3 = 5 3 2 - \frac13 = \frac53 2 − 3 1 = 3 5 , option B.
Watch out
( a + b ) ( a − b ) (a + b)(a - b) ( a + b ) ( a − b ) a 2 a^2 a 2 b 2 b^2 b 2 2 + 1 3 = 7 3 2 + \frac13 = \frac73 2 + 3 1 = 3 7 Report a problem with this question
From the Venn diagram, the complement of the set P ∩ Q P \cap Q P ∩ Q
A { a , b , d , e } \{a, b, d, e\} { a , b , d , e } B { b , d } \{b, d\} { b , d } C { a , e } \{a, e\} { a , e } D { c } \{c\} { c }
Worked solution (try it first) P ∩ Q P \cap Q P ∩ Q is the overlap of the circles, which holds only
c c c .
So
P ∩ Q = { c } P \cap Q = \{c\} P ∩ Q = { c } .
Its complement is every other element of
U U U :
{ a , b , d , e } \{a, b, d, e\} { a , b , d , e } , option A.
Watch out
{ a , e } \{a, e\} { a , e } P ∪ Q P \cup Q P ∪ Q P ∩ Q P \cap Q P ∩ Q b b b d d d Report a problem with this question
Rahila has 7 different posters to be hung in her bedroom, living room and kitchen. Assuming she plans to place a poster in each of the 3 rooms, how many choices does she have?
Worked solution (try it first) The rooms are different, so the order matters: this is
7 P 3 ^7P_3 7 P 3 .
The bedroom can have any of the 7 posters, the living room any of the 6 left, and the kitchen any of the 5 left.
So she has
7 × 6 × 5 = 210 7 \times 6 \times 5 = 210 7 × 6 × 5 = 210 choices, option D.
Watch out
Which poster goes in which room matters. 7 C 3 = 35 ^7C_3 = 35 7 C 3 = 35 7 2 = 49 7^2 = 49 7 2 = 49 Report a problem with this question
Make R R R T = K R 2 + M 3 T = \dfrac{KR^2 + M}{3} T = 3 K R 2 + M
A 3 T − K M \sqrt{\dfrac{3T - K}{M}} M 3 T − K B 3 T + M K \sqrt{\dfrac{3T + M}{K}} K 3 T + M C 3 T + K M \sqrt{\dfrac{3T + K}{M}} M 3 T + K D 3 T − M K \sqrt{\dfrac{3T - M}{K}} K 3 T − M
Worked solution (try it first) Multiply both sides by 3:
3 T = K R 2 + M 3T = KR^2 + M 3 T = K R 2 + M .
Subtract
M M M :
K R 2 = 3 T − M KR^2 = 3T - M K R 2 = 3 T − M , then divide by
K K K :
R 2 = 3 T − M K R^2 = \frac{3T - M}{K} R 2 = K 3 T − M .
Take the square root:
R = 3 T − M K R = \sqrt{\dfrac{3T - M}{K}} R = K 3 T − M , option D.
Watch out
M M M − M -M − M + M +M + M Report a problem with this question
Find the remainder when x 3 − 2 x 2 + 3 x − 3 x^3 - 2x^2 + 3x - 3 x 3 − 2 x 2 + 3 x − 3 x 2 + 1 x^2 + 1 x 2 + 1
A 2 x − 1 2x - 1 2 x − 1 B x + 3 x + 3 x + 3 C 2 x + 1 2x + 1 2 x + 1 D x − 3 x - 3 x − 3
Worked solution (try it first) Long division:
x 3 ÷ x 2 = x x^3 \div x^2 = x x 3 ÷ x 2 = x .
Subtract
x ( x 2 + 1 ) = x 3 + x x(x^2 + 1) = x^3 + x x ( x 2 + 1 ) = x 3 + x to leave
− 2 x 2 + 2 x − 3 -2x^2 + 2x - 3 − 2 x 2 + 2 x − 3 .
− 2 x 2 ÷ x 2 = − 2 -2x^2 \div x^2 = -2 − 2 x 2 ÷ x 2 = − 2 .
Subtract
− 2 ( x 2 + 1 ) = − 2 x 2 − 2 -2(x^2 + 1) = -2x^2 - 2 − 2 ( x 2 + 1 ) = − 2 x 2 − 2 to leave
2 x − 1 2x - 1 2 x − 1 .
2 x − 1 2x - 1 2 x − 1 has a lower power than
x 2 + 1 x^2 + 1 x 2 + 1 , so it is the remainder, option A.
Watch out
Subtracting − 2 -2 − 2 − 3 -3 − 3 − 3 + 2 = − 1 -3 + 2 = -1 − 3 + 2 = − 1 2 x + 1 2x + 1 2 x + 1 Also set as JAMB 2017 · UTME · Q39
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Factorize completely 9 y 2 − 16 x 2 9y^2 - 16x^2 9 y 2 − 16 x 2
A ( 3 y − 2 x ) ( 3 y + 4 x ) (3y - 2x)(3y + 4x) ( 3 y − 2 x ) ( 3 y + 4 x ) B ( 3 y + 4 x ) ( 3 y + 4 x ) (3y + 4x)(3y + 4x) ( 3 y + 4 x ) ( 3 y + 4 x ) C ( 3 y + 2 x ) ( 3 y − 4 x ) (3y + 2x)(3y - 4x) ( 3 y + 2 x ) ( 3 y − 4 x ) D ( 3 y − 4 x ) ( 3 y + 4 x ) (3y - 4x)(3y + 4x) ( 3 y − 4 x ) ( 3 y + 4 x )
Worked solution (try it first) Both terms are squares:
9 y 2 = ( 3 y ) 2 9y^2 = (3y)^2 9 y 2 = ( 3 y ) 2 and
16 x 2 = ( 4 x ) 2 16x^2 = (4x)^2 16 x 2 = ( 4 x ) 2 .
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 ) .
So
9 y 2 − 16 x 2 = ( 3 y − 4 x ) ( 3 y + 4 x ) 9y^2 - 16x^2 = (3y - 4x)(3y + 4x) 9 y 2 − 16 x 2 = ( 3 y − 4 x ) ( 3 y + 4 x ) , option D.
Watch out
The square root of 16 x 2 16x^2 16 x 2 4 x 4x 4 x 2 x 2x 2 x 4 x 2 4x^2 4 x 2 Also set as JAMB 2017 · UTME · Q28
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Solve for x x x y y y − 2 x − 5 y = 3 -2x - 5y = 3 − 2 x − 5 y = 3 x + 3 y = 0 x + 3y = 0 x + 3 y = 0
A − 3 , − 9 -3, -9 − 3 , − 9 B 9 , − 3 9, -3 9 , − 3 C − 9 , 3 -9, 3 − 9 , 3 D 3 , − 9 3, -9 3 , − 9
Worked solution (try it first) Make
x x x the subject of the simpler equation:
x = − 3 y x = -3y x = − 3 y .
Substitute into the first:
− 2 ( − 3 y ) − 5 y = 3 -2(-3y) - 5y = 3 − 2 ( − 3 y ) − 5 y = 3 , so
6 y − 5 y = 3 6y - 5y = 3 6 y − 5 y = 3 and
y = 3 y = 3 y = 3 .
Then
x = − 3 × 3 = − 9 x = -3 \times 3 = -9 x = − 3 × 3 = − 9 .
So
x x x and
y y y are
− 9 , 3 -9, 3 − 9 , 3 , option C.
Watch out
Check both equations. Option B, 9 , − 3 9, -3 9 , − 3 x + 3 y = 0 x + 3y = 0 x + 3 y = 0 − 2 ( 9 ) − 5 ( − 3 ) = − 3 -2(9) - 5(-3) = -3 − 2 ( 9 ) − 5 ( − 3 ) = − 3 Report a problem with this question
If x x x y y y x = 81 x = 81 x = 81 y = 9 y = 9 y = 9 x x x y = 1 7 9 y = 1\frac79 y = 1 9 7
A 20 1 4 20\frac14 20 4 1 B 27 27 27 C 2 1 4 2\frac14 2 4 1 D 36 36 36
Worked solution (try it first) Put in
x = 81 x = 81 x = 81 ,
y = 9 y = 9 y = 9 :
81 = 3 k 81 = 3k 81 = 3 k , so
k = 27 k = 27 k = 27 .
Change the mixed number:
1 7 9 = 16 9 1\frac79 = \frac{16}{9} 1 9 7 = 9 16 , so
y = 4 3 \sqrt y = \frac43 y = 3 4 .
So
x = 27 × 4 3 = 36 x = 27 \times \frac43 = 36 x = 27 × 3 4 = 36 , option D.
Watch out
16 9 = 4 3 \sqrt{\frac{16}{9}} = \frac43 9 16 = 3 4 3 4 \frac34 4 3 27 × 3 4 = 20 1 4 27 \times \frac34 = 20\frac14 27 × 4 3 = 20 4 1 Report a problem with this question
T T T R R R R = 3 R = 3 R = 3 T = 2 81 T = \frac{2}{81} T = 81 2 T T T R = 2 R = 2 R = 2
A 1 18 \frac{1}{18} 18 1 B 1 12 \frac{1}{12} 12 1 C 1 24 \frac{1}{24} 24 1 D 1 6 \frac16 6 1
Worked solution (try it first) T = k R 3 T = \dfrac{k}{R^3} T = R 3 k , so
k = T R 3 k = TR^3 k = T R 3 .
With
R = 3 R = 3 R = 3 :
k = 2 81 × 27 = 2 3 k = \frac{2}{81} \times 27 = \frac23 k = 81 2 × 27 = 3 2 .
When
R = 2 R = 2 R = 2 :
T = 2 / 3 8 = 2 24 T = \dfrac{2/3}{8} = \dfrac{2}{24} T = 8 2/3 = 24 2 , which is
1 12 \dfrac{1}{12} 12 1 , option B.
Watch out
Use the cube both times: 3 3 = 27 3^3 = 27 3 3 = 27 2 3 = 8 2^3 = 8 2 3 = 8 k = 2 9 k = \frac29 k = 9 2 T = 1 18 T = \frac{1}{18} T = 18 1 Report a problem with this question
Which of the following diagrams represents the solution of the inequalities y ≤ x − 2 y \le x - 2 y ≤ x − 2 y ≥ x 2 − 4 y \ge x^2 - 4 y ≥ x 2 − 4
Worked solution (try it first) y ≤ x − 2 y \le x - 2 y ≤ x − 2 is the region on or below the straight line.
y ≥ x 2 − 4 y \ge x^2 - 4 y ≥ x 2 − 4 is the region on or above the parabola, inside the U.
They meet where
x 2 − 4 = x − 2 x^2 - 4 = x - 2 x 2 − 4 = x − 2 , that is
x 2 − x − 2 = 0 x^2 - x - 2 = 0 x 2 − x − 2 = 0 , so
( x + 1 ) ( x − 2 ) = 0 (x + 1)(x - 2) = 0 ( x + 1 ) ( x − 2 ) = 0 and
x = − 1 x = -1 x = − 1 or
x = 2 x = 2 x = 2 .
The solution is the region below the line and above the parabola from
x = − 1 x = -1 x = − 1 to
x = 2 x = 2 x = 2 , which is diagram A.
Watch out
y ≤ x − 2 y \le x - 2 y ≤ x − 2 Report a problem with this question
Solve the inequality − 6 ( x + 3 ) ≤ 4 ( x − 2 ) -6(x + 3) \le 4(x - 2) − 6 ( x + 3 ) ≤ 4 ( x − 2 )
A x ≤ 2 x \le 2 x ≤ 2 B x ≥ − 1 x \ge -1 x ≥ − 1 C x ≥ − 2 x \ge -2 x ≥ − 2 D x ≤ − 1 x \le -1 x ≤ − 1
Worked solution (try it first) Expand both brackets:
− 6 x − 18 ≤ 4 x − 8 -6x - 18 \le 4x - 8 − 6 x − 18 ≤ 4 x − 8 .
Add
6 x 6x 6 x and 8 to both sides:
− 10 ≤ 10 x -10 \le 10x − 10 ≤ 10 x .
Divide by 10:
− 1 ≤ x -1 \le x − 1 ≤ x , that is
x ≥ − 1 x \ge -1 x ≥ − 1 , option B.
Watch out
If you collect the x x x − 10 x ≤ 10 -10x \le 10 − 10 x ≤ 10 − 10 -10 − 10 x ≥ − 1 x \ge -1 x ≥ − 1 x ≤ − 1 x \le -1 x ≤ − 1 Report a problem with this question
Solve the inequality x 2 + 2 x > 15 x^2 + 2x > 15 x 2 + 2 x > 15
A x < − 3 x < -3 x < − 3 x > 5 x > 5 x > 5 B − 5 < x < 3 -5 < x < 3 − 5 < x < 3 C x < 3 x < 3 x < 3 x > 5 x > 5 x > 5 D x > 3 x > 3 x > 3 x < − 5 x < -5 x < − 5
Worked solution (try it first) Bring everything to one side:
x 2 + 2 x − 15 > 0 x^2 + 2x - 15 > 0 x 2 + 2 x − 15 > 0 .
Factorise:
( x + 5 ) ( x − 3 ) > 0 (x + 5)(x - 3) > 0 ( x + 5 ) ( x − 3 ) > 0 , so the roots are
x = − 5 x = -5 x = − 5 and
x = 3 x = 3 x = 3 .
"Greater than 0" means outside the roots:
x < − 5 x < -5 x < − 5 or
x > 3 x > 3 x > 3 , option D.
Watch out
Option B, − 5 < x < 3 -5 < x < 3 − 5 < x < 3 x 2 + 2 x − 15 x^2 + 2x - 15 x 2 + 2 x − 15 x = 0 x = 0 x = 0 0 > 15 0 > 15 0 > 15 Also set as JAMB 2017 · UTME · Q29
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Find the sum of the first 18 terms of the series 3 , 6 , 9 , … 3, 6, 9, \ldots 3 , 6 , 9 , …
Worked solution (try it first) This is an A.P. with
a = 3 a = 3 a = 3 and
d = 3 d = 3 d = 3 .
Use
S n = n 2 ( 2 a + ( n − 1 ) d ) S_n = \frac n2\big(2a + (n - 1)d\big) S n = 2 n ( 2 a + ( n − 1 ) d ) with
n = 18 n = 18 n = 18 : the bracket is
6 + 17 × 3 = 57 6 + 17 \times 3 = 57 6 + 17 × 3 = 57 .
So
S 18 = 9 × 57 = 513 S_{18} = 9 \times 57 = 513 S 18 = 9 × 57 = 513 , option B.
Watch out
The bracket has ( n − 1 ) d = 17 × 3 (n - 1)d = 17 \times 3 ( n − 1 ) d = 17 × 3 18 × 3 18 \times 3 18 × 3 9 × 60 = 540 9 \times 60 = 540 9 × 60 = 540 Report a problem with this question
The second term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms.
Worked solution (try it first) The 2nd term is
a r = 4 ar = 4 a r = 4 and the 4th is
a r 3 = 16 ar^3 = 16 a r 3 = 16 .
Divide:
r 2 = 4 r^2 = 4 r 2 = 4 , so take
r = 2 r = 2 r = 2 and then
a = 2 a = 2 a = 2 .
Use
S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 ) :
S 5 = 2 ( 2 5 − 1 ) 2 − 1 S_5 = \dfrac{2(2^5 - 1)}{2 - 1} S 5 = 2 − 1 2 ( 2 5 − 1 ) , which is
2 × 31 = 62 2 \times 31 = 62 2 × 31 = 62 .
So the sum is 62, option B.
(With
r = − 2 r = -2 r = − 2 the sum is
− 22 -22 − 22 , which is not an option.)
Watch out
Keep the − 1 -1 − 1 r n − 1 r^n - 1 r n − 1 2 ( 32 − 1 ) = 62 2(32 - 1) = 62 2 ( 32 − 1 ) = 62 2 × 32 = 64 2 \times 32 = 64 2 × 32 = 64 Report a problem with this question
A binary operation ⊕ \oplus ⊕ x ⊕ y = x y + x + y x \oplus y = xy + x + y x ⊕ y = x y + x + y 3 ⊕ ( − 2 3 ) 3 \oplus \left(-\frac23\right) 3 ⊕ ( − 3 2 )
A − 1 2 -\frac12 − 2 1 B 1 3 \frac13 3 1 C − 1 -1 − 1 D 2 2 2
Worked solution (try it first) Put
x = 3 x = 3 x = 3 and
y = − 2 3 y = -\frac23 y = − 3 2 .
The product is
x y = 3 × ( − 2 3 ) = − 2 xy = 3 \times \left(-\frac23\right) = -2 x y = 3 × ( − 3 2 ) = − 2 .
So
3 ⊕ ( − 2 3 ) = − 2 + 3 − 2 3 3 \oplus \left(-\frac23\right) = -2 + 3 - \frac23 3 ⊕ ( − 3 2 ) = − 2 + 3 − 3 2 .
That is
1 − 2 3 = 1 3 1 - \frac23 = \frac13 1 − 3 2 = 3 1 , option B.
Watch out
The product 3 × ( − 2 3 ) 3 \times \left(-\frac23\right) 3 × ( − 3 2 ) + 2 +2 + 2 2 + 3 − 2 3 = 13 3 2 + 3 - \frac23 = \frac{13}{3} 2 + 3 − 3 2 = 3 13 Report a problem with this question
Evaluate ∣ 4 2 − 1 2 3 − 1 − 1 1 3 ∣ \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} 4 2 − 1 2 3 1 − 1 − 1 3
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
4 × ( 3 × 3 − ( − 1 ) ( 1 ) ) = 4 × 10 4 \times (3 \times 3 - (-1)(1)) = 4 \times 10 4 × ( 3 × 3 − ( − 1 ) ( 1 )) = 4 × 10 The second term is
− 2 × ( 2 × 3 − ( − 1 ) ( − 1 ) ) = − 2 × 5 -2 \times (2 \times 3 - (-1)(-1)) = -2 \times 5 − 2 × ( 2 × 3 − ( − 1 ) ( − 1 )) = − 2 × 5 The third term is
− 1 × ( 2 × 1 − 3 × ( − 1 ) ) = − 1 × 5 -1 \times (2 \times 1 - 3 \times (-1)) = -1 \times 5 − 1 × ( 2 × 1 − 3 × ( − 1 )) = − 1 × 5 Add them:
40 − 10 − 5 = 25 40 - 10 - 5 = 25 40 − 10 − 5 = 25 , option A.
Watch out
The middle term takes a minus sign. Adding it instead gives 40 + 10 − 5 = 45 40 + 10 - 5 = 45 40 + 10 − 5 = 45 Report a problem with this question
The inverse of matrix N = ( 2 3 1 4 ) N = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} N = ( 2 1 3 4 )
A 1 5 ( 2 1 3 4 ) \frac15\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} 5 1 ( 2 3 1 4 ) B 1 5 ( 4 − 3 − 1 2 ) \frac15\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} 5 1 ( 4 − 1 − 3 2 ) C 1 5 ( 2 − 1 − 3 4 ) \frac15\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix} 5 1 ( 2 − 3 − 1 4 ) D 1 5 ( 4 1 3 2 ) \frac15\begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} 5 1 ( 4 3 1 2 )
Worked solution (try it first) The determinant is
∣ N ∣ = 2 × 4 − 3 × 1 = 5 |N| = 2 \times 4 - 3 \times 1 = 5 ∣ N ∣ = 2 × 4 − 3 × 1 = 5 .
Swap the two diagonal entries (2 and 4) and change the signs of the other two:
( 4 − 3 − 1 2 ) \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} ( 4 − 1 − 3 2 ) .
Divide by the determinant:
N − 1 = 1 5 ( 4 − 3 − 1 2 ) N^{-1} = \frac15\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} N − 1 = 5 1 ( 4 − 1 − 3 2 ) , option B.
Watch out
Swap the diagonal entries 2 and 4 as well as changing the other signs. Option C keeps 2 and 4 where they were. Report a problem with this question
What is the size of each interior angle of a 12-sided regular polygon?
A 120 ∘ 120^\circ 12 0 ∘ B 150 ∘ 150^\circ 15 0 ∘ C 30 ∘ 30^\circ 3 0 ∘ D 180 ∘ 180^\circ 18 0 ∘
Worked solution (try it first) The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so each exterior angle of a regular 12-sided polygon is
360 ∘ ÷ 12 = 30 ∘ 360^\circ \div 12 = 30^\circ 36 0 ∘ ÷ 12 = 3 0 ∘ .
The interior angle is
180 ∘ − 30 ∘ = 150 ∘ 180^\circ - 30^\circ = 150^\circ 18 0 ∘ − 3 0 ∘ = 15 0 ∘ , option B.
Watch out
30 ∘ 30^\circ 3 0 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
A circle of perimeter 28 cm 28\text{ cm} 28 cm
A 56 cm 2 56\text{ cm}^2 56 cm 2 B 49 cm 2 49\text{ cm}^2 49 cm 2 C 98 cm 2 98\text{ cm}^2 98 cm 2 D 28 cm 2 28\text{ cm}^2 28 cm 2
Worked solution (try it first) The length of wire does not change, so the square's perimeter is 28 cm.
Each side is
28 ÷ 4 = 7 28 \div 4 = 7 28 ÷ 4 = 7 cm, so the area is
7 2 = 49 cm 2 7^2 = 49\text{ cm}^2 7 2 = 49 cm 2 , option B.
Watch out
28 cm is the perimeter, not the area (option D). Find the side first, 28 ÷ 4 = 7 28 \div 4 = 7 28 ÷ 4 = 7 Also set as JAMB 2017 · UTME · Q30
Report a problem with this question
A chord of a circle of radius 7 cm 7\text{ cm} 7 cm 5 cm 5\text{ cm} 5 cm
A 4 6 cm 4\sqrt6\text{ cm} 4 6 cm B 3 6 cm 3\sqrt6\text{ cm} 3 6 cm C 6 6 cm 6\sqrt6\text{ cm} 6 6 cm D 2 6 cm 2\sqrt6\text{ cm} 2 6 cm
Worked solution (try it first) The perpendicular from the centre to a chord bisects it.
The radius, the 5 cm distance and half the chord make a right-angled triangle with hypotenuse 7.
Pythagoras: half the chord is
7 2 − 5 2 = 24 = 2 6 \sqrt{7^2 - 5^2} = \sqrt{24} = 2\sqrt6 7 2 − 5 2 = 24 = 2 6 cm.
So the chord is
2 × 2 6 = 4 6 2 \times 2\sqrt6 = 4\sqrt6 2 × 2 6 = 4 6 cm, option A.
Watch out
2 6 2\sqrt6 2 6 Report a problem with this question
A solid metal cube of side 3 cm 3\text{ cm} 3 cm 3 cm 3\text{ cm} 3 cm 4 cm 4\text{ cm} 4 cm 5 cm 5\text{ cm} 5 cm
A 48 cm 3 48\text{ cm}^3 48 cm 3 B 33 cm 3 33\text{ cm}^3 33 cm 3 C 60 cm 3 60\text{ cm}^3 60 cm 3 D 27 cm 3 27\text{ cm}^3 27 cm 3
Worked solution (try it first) The tank holds
3 × 4 × 5 = 60 cm 3 3 \times 4 \times 5 = 60\text{ cm}^3 3 × 4 × 5 = 60 cm 3 when empty.
The cube takes up
3 3 = 27 cm 3 3^3 = 27\text{ cm}^3 3 3 = 27 cm 3 of that space.
So the water it can now hold is
60 − 27 = 33 cm 3 60 - 27 = 33\text{ cm}^3 60 − 27 = 33 cm 3 , option B.
Watch out
Take away the space the cube fills. 60 cm 3 60\text{ cm}^3 60 cm 3 Report a problem with this question
The perpendicular bisector of a line X Y XY X Y
A whose distance from X X X Y Y Y B whose distance from Y Y Y X X X C which moves on the line X Y XY X Y D which is equidistant from the points X X X Y Y Y
Worked solution (try it first) The perpendicular bisector of
X Y XY X Y passes through the mid-point of
X Y XY X Y at right angles to it.
Any point on it forms two congruent right-angled triangles with
X X X and
Y Y Y , so it is the same distance from
X X X as from
Y Y Y .
That is option D.
Watch out
Points on the perpendicular bisector are equally far from X X X Y Y Y X Y XY X Y X Y XY X Y Report a problem with this question
The midpoint of P ( x , y ) P(x, y) P ( x , y ) Q ( 8 , 6 ) Q(8, 6) Q ( 8 , 6 ) ( 5 , 8 ) (5, 8) ( 5 , 8 ) x x x y y y
A ( 2 , 10 ) (2, 10) ( 2 , 10 ) B ( 2 , 8 ) (2, 8) ( 2 , 8 ) C ( 2 , 12 ) (2, 12) ( 2 , 12 ) D ( 2 , 6 ) (2, 6) ( 2 , 6 )
Worked solution (try it first) The midpoint is the average of the ends:
x + 8 2 = 5 \frac{x + 8}{2} = 5 2 x + 8 = 5 and
y + 6 2 = 8 \frac{y + 6}{2} = 8 2 y + 6 = 8 .
Double each:
x + 8 = 10 x + 8 = 10 x + 8 = 10 and
y + 6 = 16 y + 6 = 16 y + 6 = 16 .
So
x = 2 x = 2 x = 2 and
y = 10 y = 10 y = 10 , option A.
Watch out
Double the midpoint's coordinate first, then subtract Q Q Q y = 16 − 6 = 10 y = 16 - 6 = 10 y = 16 − 6 = 10 y y y Report a problem with this question
Find the equation of a line perpendicular to the line 2 y = 5 x + 4 2y = 5x + 4 2 y = 5 x + 4 ( 4 , 2 ) (4, 2) ( 4 , 2 )
A 5 y − 2 x − 18 = 0 5y - 2x - 18 = 0 5 y − 2 x − 18 = 0 B 5 y + 2 x − 18 = 0 5y + 2x - 18 = 0 5 y + 2 x − 18 = 0 C 5 y − 2 x + 18 = 0 5y - 2x + 18 = 0 5 y − 2 x + 18 = 0 D 5 y + 2 x − 2 = 0 5y + 2x - 2 = 0 5 y + 2 x − 2 = 0
Worked solution (try it first) 2 y = 5 x + 4 2y = 5x + 4 2 y = 5 x + 4 gives
y = 5 2 x + 2 y = \frac52x + 2 y = 2 5 x + 2 , so its gradient is
5 2 \frac52 2 5 .
The perpendicular gradient is the negative reciprocal,
− 2 5 -\frac25 − 5 2 .
Through
( 4 , 2 ) (4, 2) ( 4 , 2 ) :
y − 2 = − 2 5 ( x − 4 ) y - 2 = -\frac25(x - 4) y − 2 = − 5 2 ( x − 4 ) .
Multiply by 5:
5 y − 10 = − 2 x + 8 5y - 10 = -2x + 8 5 y − 10 = − 2 x + 8 .
Collect terms:
5 y + 2 x − 18 = 0 5y + 2x - 18 = 0 5 y + 2 x − 18 = 0 , option B.
Watch out
− 2 5 × ( − 4 ) = + 8 5 -\frac25 \times (-4) = +\frac85 − 5 2 × ( − 4 ) = + 5 8 − 8 5 -\frac85 − 5 8 5 y + 2 x − 2 = 0 5y + 2x - 2 = 0 5 y + 2 x − 2 = 0 Report a problem with this question
In a right-angled triangle, if tan θ = 3 4 \tan\theta = \frac34 tan θ = 4 3 cos θ − sin θ \cos\theta - \sin\theta cos θ − sin θ
A 2 5 \frac25 5 2 B 3 5 \frac35 5 3 C 1 5 \frac15 5 1 D 4 5 \frac45 5 4
Worked solution (try it first) tan θ = opposite adjacent \tan\theta = \frac{\text{opposite}}{\text{adjacent}} tan θ = adjacent opposite Pythagoras gives the hypotenuse
9 + 16 = 5 \sqrt{9 + 16} = 5 9 + 16 = 5 .
So
cos θ = 4 5 \cos\theta = \frac45 cos θ = 5 4 and
sin θ = 3 5 \sin\theta = \frac35 sin θ = 5 3 .
Then
cos θ − sin θ = 4 5 − 3 5 \cos\theta - \sin\theta = \frac45 - \frac35 cos θ − sin θ = 5 4 − 5 3 = 1 5 = \frac15 = 5 1 , option C.
Watch out
Sine uses the opposite side (3) and cosine the adjacent side (4). Swapping them gives 3 5 − 4 5 = − 1 5 \frac35 - \frac45 = -\frac15 5 3 − 5 4 = − 5 1 Report a problem with this question
A man walks 100 m 100\text{ m} 100 m X X X Y Y Y 100 m 100\text{ m} 100 m Z Z Z X X X Z Z Z
A 195 ∘ 195^\circ 19 5 ∘ B 135 ∘ 135^\circ 13 5 ∘ C 225 ∘ 225^\circ 22 5 ∘ D 045 ∘ 045^\circ 04 5 ∘
Worked solution (try it first) Z Z Z is 100 m west and 100 m north of
X X X .
So from
Z Z Z ,
X X X is 100 m east and 100 m south.
Equal amounts east and south means
X X X is exactly south-east of
Z Z Z .
South-east is
90 ∘ + 45 ∘ = 135 ∘ 90^\circ + 45^\circ = 135^\circ 9 0 ∘ + 4 5 ∘ = 13 5 ∘ , option B.
Watch out
Stand at Z Z Z X X X 135 ∘ 135^\circ 13 5 ∘ 225 ∘ 225^\circ 22 5 ∘ Report a problem with this question
The derivative of ( 2 x + 1 ) ( 3 x + 1 ) (2x + 1)(3x + 1) ( 2 x + 1 ) ( 3 x + 1 )
A 12 x + 1 12x + 1 12 x + 1 B 6 x + 5 6x + 5 6 x + 5 C 6 x + 1 6x + 1 6 x + 1 D 12 x + 5 12x + 5 12 x + 5
Worked solution (try it first) Expand:
( 2 x + 1 ) ( 3 x + 1 ) = 6 x 2 + 2 x + 3 x + 1 (2x + 1)(3x + 1) = 6x^2 + 2x + 3x + 1 ( 2 x + 1 ) ( 3 x + 1 ) = 6 x 2 + 2 x + 3 x + 1 = 6 x 2 + 5 x + 1 = 6x^2 + 5x + 1 = 6 x 2 + 5 x + 1 .
Differentiate term by term:
12 x + 5 12x + 5 12 x + 5 , option D.
Watch out
6 x 2 6x^2 6 x 2 12 x 12x 12 x 6 x 6x 6 x 6 x + 5 6x + 5 6 x + 5 Report a problem with this question
Find the derivative of sin θ cos θ \dfrac{\sin\theta}{\cos\theta} cos θ sin θ
A sec 2 θ \sec^2\theta sec 2 θ B tan θ cosec θ \tan\theta\,\text{cosec}\,\theta tan θ cosec θ C cosec θ sec θ \text{cosec}\,\theta\sec\theta cosec θ sec θ D cosec 2 θ \text{cosec}^2\theta cosec 2 θ
Worked solution (try it first) sin θ cos θ = tan θ \dfrac{\sin\theta}{\cos\theta} = \tan\theta cos θ sin θ = tan θ .
The derivative of
tan θ \tan\theta tan θ is
sec 2 θ \sec^2\theta sec 2 θ , option A.
Watch out
cosec 2 θ \text{cosec}^2\theta cosec 2 θ cot θ \cot\theta cot θ − cosec 2 θ -\text{cosec}^2\theta − cosec 2 θ tan θ \tan\theta tan θ sec 2 θ \sec^2\theta sec 2 θ Report a problem with this question
Find the value of x x x y = x 3 + x 2 − x + 1 y = x^3 + x^2 - x + 1 y = x 3 + x 2 − x + 1
A 1 3 \frac13 3 1 B − 1 3 -\frac13 − 3 1 C 1 1 1 D − 1 -1 − 1
Worked solution (try it first) At a turning point
d y d x = 3 x 2 + 2 x − 1 = 0 \frac{dy}{dx} = 3x^2 + 2x - 1 = 0 d x d y = 3 x 2 + 2 x − 1 = 0 .
Factorise:
( 3 x − 1 ) ( x + 1 ) = 0 (3x - 1)(x + 1) = 0 ( 3 x − 1 ) ( x + 1 ) = 0 , so
x = 1 3 x = \frac13 x = 3 1 or
x = − 1 x = -1 x = − 1 .
d 2 y d x 2 = 6 x + 2 \frac{d^2y}{dx^2} = 6x + 2 d x 2 d 2 y = 6 x + 2 .
At
x = 1 3 x = \frac13 x = 3 1 it is
4 > 0 4 > 0 4 > 0 (minimum).
At
x = − 1 x = -1 x = − 1 it is
− 4 < 0 -4 < 0 − 4 < 0 (maximum).
So the minimum point is at
x = 1 3 x = \frac13 x = 3 1 , option A.
Watch out
Use the second derivative to tell the two turning points apart. x = − 1 x = -1 x = − 1 − 1 3 -\frac13 − 3 1 Report a problem with this question
Evaluate ∫ 0 1 ( 3 − 2 x ) d x \displaystyle\int_0^1 (3 - 2x)\,dx ∫ 0 1 ( 3 − 2 x ) d x
Worked solution (try it first) Integrate:
[ 3 x − x 2 ] 0 1 \left[3x - x^2\right]_0^1 [ 3 x − x 2 ] 0 1 .
At
x = 1 x = 1 x = 1 :
3 − 1 = 2 3 - 1 = 2 3 − 1 = 2 .
So the integral is 2, option C.
Watch out
Integrate before substituting: 3 is the value of 3 − 2 x 3 - 2x 3 − 2 x x = 0 x = 0 x = 0 Report a problem with this question
Find ∫ cos 4 x d x \displaystyle\int \cos 4x\,dx ∫ cos 4 x d x
A 3 4 sin 4 x + k \frac34\sin 4x + k 4 3 sin 4 x + k B − 1 4 sin 4 x + k -\frac14\sin 4x + k − 4 1 sin 4 x + k C − 3 4 sin 4 x + k -\frac34\sin 4x + k − 4 3 sin 4 x + k D 1 4 sin 4 x + k \frac14\sin 4x + k 4 1 sin 4 x + k
Worked solution (try it first) cos \cos cos integrates to
sin \sin sin .
For
cos 4 x \cos4x cos 4 x , also divide by 4, the coefficient of
x x x .
So
∫ cos 4 x d x = 1 4 sin 4 x + k \int\cos4x\,dx = \frac14\sin4x + k ∫ cos 4 x d x = 4 1 sin 4 x + k , option D.
Check:
d d x ( 1 4 sin 4 x ) = cos 4 x \frac{d}{dx}\left(\frac14\sin4x\right) = \cos4x d x d ( 4 1 sin 4 x ) = cos 4 x .
Watch out
cos integrates to + sin +\sin + sin ∫ sin \int\sin ∫ sin − 1 4 sin 4 x -\frac14\sin4x − 4 1 sin 4 x Report a problem with this question
The bar chart shows the distribution of SS2 students in a school. Find the total number of students.
Worked solution (try it first) Each grid step on the vertical axis is 15 students, so read the bars: I is 45, II is 60, III is 30 and IV is 45.
Add them:
45 + 60 + 30 + 45 = 180 45 + 60 + 30 + 45 = 180 45 + 60 + 30 + 45 = 180 students, option A.
Watch out
Use the scale: each grid step is 15 students, not 1. Missing one of the two 45 bars gives 135 (option B). Report a problem with this question
The sum of four consecutive integers is 34. Find the least of these numbers.
Worked solution (try it first) Let the least integer be
n n n .
The four are
n n n ,
n + 1 n + 1 n + 1 ,
n + 2 n + 2 n + 2 and
n + 3 n + 3 n + 3 , with sum
4 n + 6 4n + 6 4 n + 6 .
So
4 n + 6 = 34 4n + 6 = 34 4 n + 6 = 34 .
Take 6 from both sides:
4 n = 28 4n = 28 4 n = 28 .
Divide by 4:
n = 7 n = 7 n = 7 , option A.
(The numbers are 7, 8, 9 and 10.)
Watch out
34 ÷ 4 = 8.5 34 \div 4 = 8.5 34 ÷ 4 = 8.5 8 + 9 + 10 + 11 = 38 8 + 9 + 10 + 11 = 38 8 + 9 + 10 + 11 = 38 Report a problem with this question
Class interval
0–2
3–5
6–8
9–11
Frequency
3
2
5
3
Find the mode of the distribution.
Worked solution (try it first) The modal class is 6–8, with frequency 5.
Its lower class boundary is 5.5 and its width is 3.
Differences from the neighbours:
Δ 1 = 5 − 2 = 3 \Delta_1 = 5 - 2 = 3 Δ 1 = 5 − 2 = 3 and
Δ 2 = 5 − 3 = 2 \Delta_2 = 5 - 3 = 2 Δ 2 = 5 − 3 = 2 .
Mode
= 5.5 + 3 3 + 2 × 3 = 5.5 + \frac{3}{3 + 2} \times 3 = 5.5 + 3 + 2 3 × 3 = 5.5 + 1.8 = 5.5 + 1.8 = 5.5 + 1.8 , which is 7.3.
So the mode is about 7, option D.
Watch out
Start from the class boundary 5.5, not the class limit 6. Using 6 gives 7.8, which rounds to 8 (option B). Report a problem with this question
In how many ways can the letters of the word ELATION be arranged?
A 6 ! 6! 6 ! B 7 ! 7! 7 ! C 5 ! 5! 5 ! D 8 ! 8! 8 !
Worked solution (try it first) ELATION has 7 letters: E, L, A, T, I, O, N, and no letter is repeated.
So there are
7 ! 7! 7 ! arrangements, option B.
Watch out
Count the letters one by one: there are 7, not 6. 6 ! 6! 6 ! Report a problem with this question
In how many ways can five people sit round a circular table?
Worked solution (try it first) Round a table only the positions relative to each other matter, so fix one person's seat.
The other 4 people can then be arranged in
4 ! = 24 4! = 24 4 ! = 24 ways.
So there are
( 5 − 1 ) ! = 24 (5 - 1)! = 24 ( 5 − 1 )! = 24 ways, option A.
Watch out
5 ! = 120 5! = 120 5 ! = 120 Report a problem with this question
Find the probability that a number picked at random from the set { 43 , 44 , 45 , … , 60 } \{43, 44, 45, \ldots, 60\} { 43 , 44 , 45 , … , 60 }
A 2 3 \frac23 3 2 B 1 3 \frac13 3 1 C 2 9 \frac29 9 2 D 7 9 \frac79 9 7
Worked solution (try it first) From 43 to 60 inclusive there are
60 − 43 + 1 = 18 60 - 43 + 1 = 18 60 − 43 + 1 = 18 numbers.
The primes are 43, 47, 53 and 59.
51 is
3 × 17 3 \times 17 3 × 17 and 57 is
3 × 19 3 \times 19 3 × 19 .
So the probability is
4 18 = 2 9 \frac{4}{18} = \frac29 18 4 = 9 2 , option C.
Watch out
51 and 57 look prime but are divisible by 3. Counting them gives 6 18 = 1 3 \frac{6}{18} = \frac13 18 6 = 3 1 Report a problem with this question
In a class of 60 students, 30 offer Physics and 40 offer Chemistry. If a student is picked at random from the class, what is the probability that the student offers both Physics and Chemistry?
A 1 3 \frac13 3 1 B 1 4 \frac14 4 1 C 1 2 \frac12 2 1 D 1 6 \frac16 6 1
Worked solution (try it first) Each student offers at least one of the subjects, so those doing both are counted twice in
30 + 40 = 70 30 + 40 = 70 30 + 40 = 70 .
Both
= 70 − 60 = 10 = 70 - 60 = 10 = 70 − 60 = 10 students.
So the probability is
10 60 = 1 6 \frac{10}{60} = \frac16 60 10 = 6 1 , option D.
Watch out
The overlap is the excess over the class size: 30 + 40 − 60 = 10 30 + 40 - 60 = 10 30 + 40 − 60 = 10 30 60 = 1 2 \frac{30}{60} = \frac12 60 30 = 2 1 Also set as JAMB 2017 · UTME · Q40
Report a problem with this question