Paper JAMB 1992 General Maths Objective
Objective paper · 38 questions · partial
JAMB 1992 · UME Topics include Number bases, Approximation & error, Plane mensuration, Indices & standard form, Surds, Sets & Venn diagrams.
Our copy of this paper is missing questions 3, 14, 17, 22, 23, 24, 25, 28, 29, 34, 41, 46.
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.
1 2 4 5 6 7 8 9 10 11 12 13 15 16 18 19 20 21 26 27 30 31 32 33 35 36 37 38 39 40 42 43 44 45 47 48 49 50 Find n n n 34 n = 10011 2 34_n = 10011_2 3 4 n = 1001 1 2
Worked solution (try it first) Change the right side to base ten:
10011 2 = 16 + 2 + 1 = 19 10011_2 = 16 + 2 + 1 = 19 1001 1 2 = 16 + 2 + 1 = 19 .
In base
n n n ,
34 n = 3 n + 4 34_n = 3n + 4 3 4 n = 3 n + 4 , so
3 n + 4 = 19 3n + 4 = 19 3 n + 4 = 19 .
Subtract 4 and divide by 3:
n = 5 n = 5 n = 5 , option A.
Watch out
The 3 is in the n n n 34 n = 3 n + 4 34_n = 3n + 4 3 4 n = 3 n + 4 4 n + 3 = 19 4n + 3 = 19 4 n + 3 = 19 n = 4 n = 4 n = 4 Report a problem with this question
The radius of a circle is given as 5 cm subject to an error of 0.1 cm. What is the percentage error in the area of the circle?
A 1 25 \frac1{25} 25 1 B 1 4 \frac14 4 1 C 4 D 25
Worked solution (try it first) The area is
A = π r 2 A = \pi r^2 A = π r 2 , so it depends on the square of the radius.
The relative error in
r r r is
0.1 5 = 0.02 \frac{0.1}{5} = 0.02 5 0.1 = 0.02 , or 2%.
Squaring doubles a small relative error, so the error in the area is about
2 × 2 % = 4 % 2 \times 2\% = 4\% 2 × 2% = 4% .
Check:
π ( 5.1 2 − 5 2 ) π × 5 2 = 1.01 25 \frac{\pi(5.1^2 - 5^2)}{\pi \times 5^2} = \frac{1.01}{25} π × 5 2 π ( 5. 1 2 − 5 2 ) = 25 1.01 = 4.04 % = 4.04\% = 4.04% , which is 4%, option C.
Watch out
The answer is a percentage: 4%. The fraction 4 100 = 1 25 \frac{4}{100} = \frac{1}{25} 100 4 = 25 1 Report a problem with this question
What is the value of x x x 4 2 x 4 3 x = 2 \frac{4^{2x}}{4^{3x}} = 2 4 3 x 4 2 x = 2
A − 2 -2 − 2 B − 1 2 -\frac12 − 2 1 C 1 2 \frac12 2 1 D 2 2 2
Worked solution (try it first) Dividing powers of the same base, subtract the indices:
4 2 x − 3 x = 4 − x 4^{2x - 3x} = 4^{-x} 4 2 x − 3 x = 4 − x , so
4 − x = 2 4^{-x} = 2 4 − x = 2 .
Write 4 as
2 2 2^2 2 2 :
2 − 2 x = 2 1 2^{-2x} = 2^1 2 − 2 x = 2 1 .
Equate the powers:
− 2 x = 1 -2x = 1 − 2 x = 1 , so
x = − 1 2 x = -\frac12 x = − 2 1 , option B.
Watch out
2 x − 3 x = − x 2x - 3x = -x 2 x − 3 x = − x x = 1 2 x = \frac12 x = 2 1 1 2 \frac12 2 1 Report a problem with this question
Simplify ( 1.25 × 10 4 ) × ( 2.0 × 10 − 1 ) 6.25 × 10 5 \dfrac{(1.25 \times 10^4) \times (2.0 \times 10^{-1})}{6.25 \times 10^5} 6.25 × 1 0 5 ( 1.25 × 1 0 4 ) × ( 2.0 × 1 0 − 1 )
A 4.0 × 10 − 3 4.0 \times 10^{-3} 4.0 × 1 0 − 3 B 5.0 × 10 − 2 5.0 \times 10^{-2} 5.0 × 1 0 − 2 C 2.0 × 10 − 1 2.0 \times 10^{-1} 2.0 × 1 0 − 1 D 5.0 × 10 3 5.0 \times 10^3 5.0 × 1 0 3
Worked solution (try it first) Top:
1.25 × 2.0 = 2.5 1.25 \times 2.0 = 2.5 1.25 × 2.0 = 2.5 and
10 4 × 10 − 1 = 10 3 10^4 \times 10^{-1} = 10^3 1 0 4 × 1 0 − 1 = 1 0 3 , so the top is
2.5 × 10 3 2.5 \times 10^3 2.5 × 1 0 3 .
Divide:
2.5 ÷ 6.25 = 0.4 2.5 \div 6.25 = 0.4 2.5 ÷ 6.25 = 0.4 and
10 3 ÷ 10 5 = 10 − 2 10^3 \div 10^5 = 10^{-2} 1 0 3 ÷ 1 0 5 = 1 0 − 2 , giving
0.4 × 10 − 2 0.4 \times 10^{-2} 0.4 × 1 0 − 2 .
0.4 is not between 1 and 10, so write it as
4 × 10 − 1 4 \times 10^{-1} 4 × 1 0 − 1 .
The value is
4.0 × 10 − 3 4.0 \times 10^{-3} 4.0 × 1 0 − 3 , option A.
Watch out
When dividing, subtract the powers: 10 3 ÷ 10 5 = 10 − 2 10^3 \div 10^5 = 10^{-2} 1 0 3 ÷ 1 0 5 = 1 0 − 2 4.0 × 10 − 3 4.0 \times 10^{-3} 4.0 × 1 0 − 3 4.0 × 10 − 2 4.0 \times 10^{-2} 4.0 × 1 0 − 2 Report a problem with this question
Simplify 5 18 − 3 72 + 4 50 5\sqrt{18} - 3\sqrt{72} + 4\sqrt{50} 5 18 − 3 72 + 4 50
A 17 4 17\sqrt4 17 4 B 4 17 4\sqrt{17} 4 17 C 17 2 17\sqrt2 17 2 D 12 4 12\sqrt4 12 4
Worked solution (try it first) Take out square factors:
5 18 = 5 × 3 2 5\sqrt{18} = 5 \times 3\sqrt2 5 18 = 5 × 3 2 , which is
15 2 15\sqrt2 15 2 .
Likewise
3 72 = 3 × 6 2 = 18 2 3\sqrt{72} = 3 \times 6\sqrt2 = 18\sqrt2 3 72 = 3 × 6 2 = 18 2 and
4 50 = 4 × 5 2 = 20 2 4\sqrt{50} = 4 \times 5\sqrt2 = 20\sqrt2 4 50 = 4 × 5 2 = 20 2 .
Combine the like surds:
15 2 − 18 2 + 20 2 = 17 2 15\sqrt2 - 18\sqrt2 + 20\sqrt2 = 17\sqrt2 15 2 − 18 2 + 20 2 = 17 2 , option C.
Watch out
Surds only combine once the numbers under the roots match, so first write each as a multiple of 2 \sqrt2 2 17 2 17\sqrt2 17 2 17 4 17\sqrt4 17 4 4 17 4\sqrt{17} 4 17 Report a problem with this question
If x = 3 − 3 x = 3 - \sqrt3 x = 3 − 3 x 2 + 36 x 2 x^2 + \dfrac{36}{x^2} x 2 + x 2 36
Worked solution (try it first) Square
x x x :
( 3 − 3 ) 2 = 9 − 6 3 + 3 (3 - \sqrt3)^2 = 9 - 6\sqrt3 + 3 ( 3 − 3 ) 2 = 9 − 6 3 + 3 , which is
12 − 6 3 12 - 6\sqrt3 12 − 6 3 .
Then
36 12 − 6 3 = 6 2 − 3 \dfrac{36}{12 - 6\sqrt3} = \dfrac{6}{2 - \sqrt3} 12 − 6 3 36 = 2 − 3 6 .
Multiply the top and bottom by
2 + 3 2 + \sqrt3 2 + 3 .
The bottom becomes
4 − 3 = 1 4 - 3 = 1 4 − 3 = 1 , so this is
12 + 6 3 12 + 6\sqrt3 12 + 6 3 .
Add:
( 12 − 6 3 ) + ( 12 + 6 3 ) = 24 (12 - 6\sqrt3) + (12 + 6\sqrt3) = 24 ( 12 − 6 3 ) + ( 12 + 6 3 ) = 24 , option C.
Watch out
( 3 − 3 ) 2 (3 - \sqrt3)^2 ( 3 − 3 ) 2 9 − 6 3 + 3 9 - 6\sqrt3 + 3 9 − 6 3 + 3 9 − 3 = 6 9 - 3 = 6 9 − 3 = 6 6 + 6 = 12 6 + 6 = 12 6 + 6 = 12 Report a problem with this question
If x = { all prime factors of 44 } x = \{\text{all prime factors of 44}\} x = { all prime factors of 44 } y = { all prime factors of 60 } y = \{\text{all prime factors of 60}\} y = { all prime factors of 60 } x ∪ y x \cup y x ∪ y x ∩ y x \cap y x ∩ y
A { 2 , 4 , 3 , 5 , 11 } \{2, 4, 3, 5, 11\} { 2 , 4 , 3 , 5 , 11 } { 4 } \{4\} { 4 } B { 4 , 3 , 5 , 11 } \{4, 3, 5, 11\} { 4 , 3 , 5 , 11 } { 3 , 4 } \{3, 4\} { 3 , 4 } C { 2 , 5 , 11 } \{2, 5, 11\} { 2 , 5 , 11 } { 2 } \{2\} { 2 } D { 2 , 3 , 5 , 11 } \{2, 3, 5, 11\} { 2 , 3 , 5 , 11 } { 2 } \{2\} { 2 }
Worked solution (try it first) 44 = 2 2 × 11 44 = 2^2 \times 11 44 = 2 2 × 11 , so its prime factors are
x = { 2 , 11 } x = \{2, 11\} x = { 2 , 11 } .
60 = 2 2 × 3 × 5 60 = 2^2 \times 3 \times 5 60 = 2 2 × 3 × 5 , so its prime factors are
y = { 2 , 3 , 5 } y = \{2, 3, 5\} y = { 2 , 3 , 5 } .
The union takes every element in either set:
x ∪ y = { 2 , 3 , 5 , 11 } x \cup y = \{2, 3, 5, 11\} x ∪ y = { 2 , 3 , 5 , 11 } .
The intersection takes those in both:
x ∩ y = { 2 } x \cap y = \{2\} x ∩ y = { 2 } .
That is option D.
Watch out
4 is a factor of both numbers but it is not prime (4 = 2 × 2 4 = 2 \times 2 4 = 2 × 2 Report a problem with this question
If U = { 0 , 2 , 3 , 6 , 7 , 8 , 9 , 10 } U = \{0, 2, 3, 6, 7, 8, 9, 10\} U = { 0 , 2 , 3 , 6 , 7 , 8 , 9 , 10 } E = { 0 , 4 , 6 , 8 } E = \{0, 4, 6, 8\} E = { 0 , 4 , 6 , 8 } F = { x : x 2 = 2 6 , x is odd } F = \{x : x^2 = 2^6, x \text{ is odd}\} F = { x : x 2 = 2 6 , x is odd } ( E ∩ F ) ′ (E \cap F)' ( E ∩ F ) ′ ′ ' ′
A { 0 } \{0\} { 0 } B U U U C ∅ \varnothing ∅ D F F F
Worked solution (try it first) x 2 = 2 6 = 64 x^2 = 2^6 = 64 x 2 = 2 6 = 64 gives
x = 8 x = 8 x = 8 or
x = − 8 x = -8 x = − 8 .
Neither is odd, so
F = ∅ F = \varnothing F = ∅ .
Nothing is in both
E E E and an empty set, so
E ∩ F = ∅ E \cap F = \varnothing E ∩ F = ∅ .
The complement of the empty set is everything in the universal set:
( E ∩ F ) ′ = U (E \cap F)' = U ( E ∩ F ) ′ = U , option B.
Watch out
Take the complement at the end. E ∩ F = ∅ E \cap F = \varnothing E ∩ F = ∅ ( E ∩ F ) ′ (E \cap F)' ( E ∩ F ) ′ U U U ∅ \varnothing ∅ Report a problem with this question
Make t t t s = u t + 1 2 a t 2 s = ut + \frac12at^2 s = u t + 2 1 a t 2
A 1 a [ u ± u 2 − 2 a s ] \frac1a[u \pm \sqrt{u^2 - 2as}] a 1 [ u ± u 2 − 2 a s ] B 1 a [ − u ± u 2 − 2 a s ] \frac1a[-u \pm \sqrt{u^2 - 2as}] a 1 [ − u ± u 2 − 2 a s ] C 1 a [ u ± u 2 + 2 a s ] \frac1a[u \pm \sqrt{u^2 + 2as}] a 1 [ u ± u 2 + 2 a s ] D 1 a [ − u ± u 2 + 2 a s ] \frac1a[-u \pm \sqrt{u^2 + 2as}] a 1 [ − u ± u 2 + 2 a s ]
Worked solution (try it first) Multiply by 2 and bring everything to one side:
a t 2 + 2 u t − 2 s = 0 at^2 + 2ut - 2s = 0 a t 2 + 2 u t − 2 s = 0 , a quadratic in
t t t .
Use the formula with
a a a ,
b = 2 u b = 2u b = 2 u and
c = − 2 s c = -2s c = − 2 s :
t = − 2 u ± 4 u 2 + 8 a s 2 a t = \dfrac{-2u \pm \sqrt{4u^2 + 8as}}{2a} t = 2 a − 2 u ± 4 u 2 + 8 a s .
Take 4 out of the square root, which gives
2 u 2 + 2 a s 2\sqrt{u^2 + 2as} 2 u 2 + 2 a s , and divide top and bottom by 2.
So
t = 1 a [ − u ± u 2 + 2 a s ] t = \frac1a[-u \pm \sqrt{u^2 + 2as}] t = a 1 [ − u ± u 2 + 2 a s ] , option D.
Watch out
Here c = − 2 s c = -2s c = − 2 s − 4 a c = + 8 a s -4ac = +8as − 4 a c = + 8 a s u 2 + 2 a s u^2 + 2as u 2 + 2 a s − 8 a s -8as − 8 a s Report a problem with this question
Factorize 9 p 2 − q 2 + 6 q r − 9 r 2 9p^2 - q^2 + 6qr - 9r^2 9 p 2 − q 2 + 6 q r − 9 r 2
A ( 3 p − 3 q + r ) ( 3 p − q − 9 r ) (3p - 3q + r)(3p - q - 9r) ( 3 p − 3 q + r ) ( 3 p − q − 9 r ) B ( 6 p − 3 q + 3 r ) ( 3 p − q − 4 r ) (6p - 3q + 3r)(3p - q - 4r) ( 6 p − 3 q + 3 r ) ( 3 p − q − 4 r ) C ( 3 p − q + 3 r ) ( 3 p + q − 3 r ) (3p - q + 3r)(3p + q - 3r) ( 3 p − q + 3 r ) ( 3 p + q − 3 r ) D ( 3 p − q + 3 r ) ( 3 p − q − 3 r ) (3p - q + 3r)(3p - q - 3r) ( 3 p − q + 3 r ) ( 3 p − q − 3 r )
Worked solution (try it first) Group the last three terms:
9 p 2 − ( q 2 − 6 q r + 9 r 2 ) 9p^2 - (q^2 - 6qr + 9r^2) 9 p 2 − ( q 2 − 6 q r + 9 r 2 ) .
The bracket is a perfect square,
( q − 3 r ) 2 (q - 3r)^2 ( q − 3 r ) 2 , so the expression is
( 3 p ) 2 − ( q − 3 r ) 2 (3p)^2 - (q - 3r)^2 ( 3 p ) 2 − ( q − 3 r ) 2 .
Difference of two squares:
[ 3 p − ( q − 3 r ) ] [ 3 p + ( q − 3 r ) ] [3p - (q - 3r)][3p + (q - 3r)] [ 3 p − ( q − 3 r )] [ 3 p + ( q − 3 r )] , which is
( 3 p − q + 3 r ) ( 3 p + q − 3 r ) (3p - q + 3r)(3p + q - 3r) ( 3 p − q + 3 r ) ( 3 p + q − 3 r ) , option C.
Watch out
One bracket subtracts q − 3 r q - 3r q − 3 r q q q 3 r 3r 3 r − q -q − q Report a problem with this question
Solve the equation y − 11 y + 24 = 0 y - 11\sqrt y + 24 = 0 y − 11 y + 24 = 0
A 8, 3 B 64, 9 C 6, 4 D 9, − 8 -8 − 8
Worked solution (try it first) Let
u = y u = \sqrt y u = y , so
y = u 2 y = u^2 y = u 2 and the equation is
u 2 − 11 u + 24 = 0 u^2 - 11u + 24 = 0 u 2 − 11 u + 24 = 0 .
Factorise:
( u − 3 ) ( u − 8 ) = 0 (u - 3)(u - 8) = 0 ( u − 3 ) ( u − 8 ) = 0 , so
y = 3 \sqrt y = 3 y = 3 or
y = 8 \sqrt y = 8 y = 8 .
Square:
y = 9 y = 9 y = 9 or
y = 64 y = 64 y = 64 , option B.
Watch out
8 and 3 (option A) are the values of y \sqrt y y y y y Report a problem with this question
A man invested a sum of ₦280.00 partly at 5 % 5\% 5% 4 % 4\% 4% 5 % 5\% 5%
A ₦14.00 B ₦120.00 C ₦140.00 D ₦160.00
Worked solution (try it first) Let
x x x be the amount at
5 % 5\% 5% .
Then
280 − x 280 - x 280 − x is at
4 % 4\% 4% .
The interest adds up to 12.80:
0.05 x + 0.04 ( 280 − x ) = 12.8 0.05x + 0.04(280 - x) = 12.8 0.05 x + 0.04 ( 280 − x ) = 12.8 , which is
0.01 x + 11.2 = 12.8 0.01x + 11.2 = 12.8 0.01 x + 11.2 = 12.8 .
So
0.01 x = 1.6 0.01x = 1.6 0.01 x = 1.6 and
x = 160 x = 160 x = 160 .
He invested ₦160.00 at
5 % 5\% 5% , option D.
Watch out
₦120.00 (option B) is the amount at 4 % 4\% 4% x x x Report a problem with this question
Resolve 3 x 2 + x − 2 \dfrac{3}{x^2 + x - 2} x 2 + x − 2 3
A 1 x − 1 − 1 x + 2 \frac{1}{x - 1} - \frac{1}{x + 2} x − 1 1 − x + 2 1 B 1 x + 2 − 1 x − 1 \frac{1}{x + 2} - \frac{1}{x - 1} x + 2 1 − x − 1 1 C 1 x + 1 − 1 x − 2 \frac{1}{x + 1} - \frac{1}{x - 2} x + 1 1 − x − 2 1 D 1 x − 2 + 1 x + 1 \frac{1}{x - 2} + \frac{1}{x + 1} x − 2 1 + x + 1 1
Worked solution (try it first) Factorise the bottom:
x 2 + x − 2 = ( x − 1 ) ( x + 2 ) x^2 + x - 2 = (x - 1)(x + 2) x 2 + x − 2 = ( x − 1 ) ( x + 2 ) .
Write
3 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 \dfrac{3}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} ( x − 1 ) ( x + 2 ) 3 = x − 1 A + x + 2 B .
Multiply through by
( x − 1 ) ( x + 2 ) (x - 1)(x + 2) ( x − 1 ) ( x + 2 ) :
3 = A ( x + 2 ) + B ( x − 1 ) 3 = A(x + 2) + B(x - 1) 3 = A ( x + 2 ) + B ( x − 1 ) .
Put
x = 1 x = 1 x = 1 :
3 = 3 A 3 = 3A 3 = 3 A , so
A = 1 A = 1 A = 1 .
Put
x = − 2 x = -2 x = − 2 :
3 = − 3 B 3 = -3B 3 = − 3 B , so
B = − 1 B = -1 B = − 1 .
So the partial fractions are
1 x − 1 − 1 x + 2 \dfrac{1}{x - 1} - \dfrac{1}{x + 2} x − 1 1 − x + 2 1 , option A.
Watch out
Watch the order of the subtraction. Option B, 1 x + 2 − 1 x − 1 \frac{1}{x + 2} - \frac{1}{x - 1} x + 2 1 − x − 1 1 − 3 ( x − 1 ) ( x + 2 ) \frac{-3}{(x - 1)(x + 2)} ( x − 1 ) ( x + 2 ) − 3 Report a problem with this question
Find all values of x x x − 11 ≤ 4 − 3 x ≤ 28 -11 \le 4 - 3x \le 28 − 11 ≤ 4 − 3 x ≤ 28
A − 5 ≤ x ≤ 18 -5 \le x \le 18 − 5 ≤ x ≤ 18 B 5 ≤ x ≤ 8 5 \le x \le 8 5 ≤ x ≤ 8 C − 8 ≤ x ≤ 5 -8 \le x \le 5 − 8 ≤ x ≤ 5 D − 5 < x ≤ 8 -5 < x \le 8 − 5 < x ≤ 8
Worked solution (try it first) Subtract 4 from all three parts:
− 15 ≤ − 3 x ≤ 24 -15 \le -3x \le 24 − 15 ≤ − 3 x ≤ 24 .
Divide all three parts by
− 3 -3 − 3 and reverse both signs:
5 ≥ x ≥ − 8 5 \ge x \ge -8 5 ≥ x ≥ − 8 .
Read from the left:
− 8 ≤ x ≤ 5 -8 \le x \le 5 − 8 ≤ x ≤ 5 , option C.
Watch out
Divide by − 3 -3 − 3 − 15 ÷ ( − 3 ) = 5 -15 \div (-3) = 5 − 15 ÷ ( − 3 ) = 5 24 ÷ ( − 3 ) = − 8 24 \div (-3) = -8 24 ÷ ( − 3 ) = − 8 + 3 +3 + 3 − 5 ≤ x ≤ 8 -5 \le x \le 8 − 5 ≤ x ≤ 8 Report a problem with this question
Find the sum to infinity of the series 3 + 2 + 4 3 + 8 9 + 16 27 + … 3 + 2 + \frac43 + \frac89 + \frac{16}{27} + \dots 3 + 2 + 3 4 + 9 8 + 27 16 + …
Worked solution (try it first) This is a G.P. with
a = 3 a = 3 a = 3 and
r = 2 ÷ 3 = 2 3 r = 2 \div 3 = \frac23 r = 2 ÷ 3 = 3 2 .
Since
r r r is between
− 1 -1 − 1 and 1, the sum to infinity is
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a .
So
S ∞ = 3 ÷ 1 3 = 9 S_\infty = 3 \div \frac13 = 9 S ∞ = 3 ÷ 3 1 = 9 , option D.
Watch out
The ratio is second term over first, 2 3 \frac23 3 2 3 2 \frac32 2 3 r = 3 2 r = \frac32 r = 2 3 Report a problem with this question
What is the n n n 2 , 6 , 12 , 20 , … 2, 6, 12, 20, \dots 2 , 6 , 12 , 20 , …
A 4 n − 2 4n - 2 4 n − 2 B 2 ( 3 n − 1 ) 2(3n - 1) 2 ( 3 n − 1 ) C n 2 + n n^2 + n n 2 + n D n 2 + 3 n + 2 n^2 + 3n + 2 n 2 + 3 n + 2
Worked solution (try it first) Write each term as a product of two consecutive whole numbers:
2 = 1 × 2 2 = 1 \times 2 2 = 1 × 2 ,
6 = 2 × 3 6 = 2 \times 3 6 = 2 × 3 ,
12 = 3 × 4 12 = 3 \times 4 12 = 3 × 4 ,
20 = 4 × 5 20 = 4 \times 5 20 = 4 × 5 .
The
n n n th term is
n ( n + 1 ) = n 2 + n n(n + 1) = n^2 + n n ( n + 1 ) = n 2 + n .
Check
n = 3 n = 3 n = 3 :
9 + 3 = 12 9 + 3 = 12 9 + 3 = 12 .
So the answer is option C.
Watch out
Test the first term, not just the pattern: option D, ( n + 1 ) ( n + 2 ) (n + 1)(n + 2) ( n + 1 ) ( n + 2 ) 6 , 12 , 20 , … 6, 12, 20, \dots 6 , 12 , 20 , … Report a problem with this question
For an arithmetic sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms.
Worked solution (try it first) 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
a = 2 a = 2 a = 2 ,
d = 3 d = 3 d = 3 and
n = 11 n = 11 n = 11 .
The bracket is
2 × 2 + 10 × 3 = 4 + 30 = 34 2 \times 2 + 10 \times 3 = 4 + 30 = 34 2 × 2 + 10 × 3 = 4 + 30 = 34 .
So
S 11 = 11 2 × 34 = 187 S_{11} = \frac{11}{2} \times 34 = 187 S 11 = 2 11 × 34 = 187 , option B.
Watch out
The bracket has ( n − 1 ) d = 10 × 3 (n - 1)d = 10 \times 3 ( n − 1 ) d = 10 × 3 11 × 3 11 \times 3 11 × 3 11 2 × 37 = 203.5 \frac{11}{2} \times 37 = 203.5 2 11 × 37 = 203.5 Report a problem with this question
If the binary operation ∗ * ∗ m ∗ n = m n + m + n m * n = mn + m + n m ∗ n = mn + m + n m m m n n n
A e = 1 e = 1 e = 1 B e = − 1 e = -1 e = − 1 C e = − 2 e = -2 e = − 2 D e = 0 e = 0 e = 0
Worked solution (try it first) The identity
e e e leaves every
m m m unchanged:
m ∗ e = m m * e = m m ∗ e = m for all
m m m .
Use the rule:
m e + m + e = m me + m + e = m m e + m + e = m .
Take
m m m from both sides:
m e + e = 0 me + e = 0 m e + e = 0 , so
e ( m + 1 ) = 0 e(m + 1) = 0 e ( m + 1 ) = 0 .
This must hold for every
m m m , so
e = 0 e = 0 e = 0 , option D.
Watch out
The identity for this operation is not the identity for ordinary multiplication. Trying e = 1 e = 1 e = 1 m ∗ 1 = 2 m + 1 m * 1 = 2m + 1 m ∗ 1 = 2 m + 1 m m m Report a problem with this question
In the figure, P Q R PQR P QR P Q PQ P Q Q R QR QR Q S QS QS Q Q Q P R PR P R Q S QS QS
A Q S = P S ⋅ S R QS = PS \cdot SR QS = P S ⋅ S R B Q S = P S ⋅ S R QS = \sqrt{PS \cdot SR} QS = P S ⋅ S R C Q S = 2 P S ⋅ S R QS = \sqrt2\sqrt{PS \cdot SR} QS = 2 P S ⋅ S R D Q S = 1 2 P S ⋅ S R QS = \frac{1}{\sqrt2}\sqrt{PS \cdot SR} QS = 2 1 P S ⋅ S R
Worked solution (try it first) P R PR P R is a diameter, so
∠ P Q R = 90 ∘ \angle PQR = 90^\circ ∠ P QR = 9 0 ∘ (angle in a semicircle).
Q S QS QS is perpendicular to
P R PR P R , so triangles
P S Q PSQ P S Q and
Q S R QSR QS R are similar and
P S Q S = Q S S R \dfrac{PS}{QS} = \dfrac{QS}{SR} QS P S = S R QS .
Cross-multiply:
Q S 2 = P S ⋅ S R QS^2 = PS \cdot SR Q S 2 = P S ⋅ S R , so
Q S = P S ⋅ S R QS = \sqrt{PS \cdot SR} QS = P S ⋅ S R , option B.
Watch out
The similar triangles give Q S 2 = P S ⋅ S R QS^2 = PS \cdot SR Q S 2 = P S ⋅ S R Q S = P S ⋅ S R QS = PS \cdot SR QS = P S ⋅ S R Report a problem with this question
Determine the distance on the earth's surface between two towns P P P 60 ∘ 60^\circ 6 0 ∘ 20 ∘ 20^\circ 2 0 ∘ Q Q Q 60 ∘ 60^\circ 6 0 ∘ 25 ∘ 25^\circ 2 5 ∘
A 800 π 9 \frac{800\pi}{9} 9 800 π B 800 3 π 9 \frac{800\sqrt3\pi}{9} 9 800 3 π C 800 π 800\pi 800 π D 800 3 π 800\sqrt3\pi 800 3 π
Worked solution (try it first) P P P is east and
Q Q Q is west, so add the longitudes:
20 ∘ + 25 ∘ = 45 ∘ 20^\circ + 25^\circ = 45^\circ 2 0 ∘ + 2 5 ∘ = 4 5 ∘ .
The radius of the parallel of latitude
60 ∘ 60^\circ 6 0 ∘ is
6400 cos 60 ∘ = 3200 6400\cos60^\circ = 3200 6400 cos 6 0 ∘ = 3200 km.
The arc length is
45 360 × 2 π × 3200 = 1 8 × 6400 π \frac{45}{360} \times 2\pi \times 3200 = \frac18 \times 6400\pi 360 45 × 2 π × 3200 = 8 1 × 6400 π = 800 π = 800\pi = 800 π km, option C.
Watch out
The radius of the parallel uses cos 60 ∘ = 1 2 \cos60^\circ = \frac12 cos 6 0 ∘ = 2 1 sin 60 ∘ = 3 2 \sin60^\circ = \frac{\sqrt3}{2} sin 6 0 ∘ = 2 3 800 3 π 800\sqrt3\pi 800 3 π Report a problem with this question
The diagram shows a circle with centre O O O
A 9 π cm 2 9\pi\text{ cm}^2 9 π cm 2 B 9 ( π − 2 ) cm 2 9(\pi - 2)\text{ cm}^2 9 ( π − 2 ) cm 2 C 18 π cm 2 18\pi\text{ cm}^2 18 π cm 2 D 36 π cm 2 36\pi\text{ cm}^2 36 π cm 2
Worked solution (try it first) The sector is a quarter circle:
1 4 × π × 6 2 = 9 π cm 2 \frac14 \times \pi \times 6^2 = 9\pi\text{ cm}^2 4 1 × π × 6 2 = 9 π cm 2 .
The right-angled triangle on the two radii has area
1 2 × 6 × 6 = 18 cm 2 \frac12 \times 6 \times 6 = 18\text{ cm}^2 2 1 × 6 × 6 = 18 cm 2 .
Segment = sector − triangle:
9 π − 18 = 9 ( π − 2 ) cm 2 9\pi - 18 = 9(\pi - 2)\text{ cm}^2 9 π − 18 = 9 ( π − 2 ) cm 2 , option B.
Watch out
9 π 9\pi 9 π Report a problem with this question
The locus of a point which is equidistant from two given fixed points is the
A perpendicular bisector of the straight line joining them B line parallel to the straight line joining them C transversal to the straight line joining them D bisector of the 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) A point the same distance from two fixed points lies on the line that cuts the segment joining them in half at right angles.
So the locus is the perpendicular bisector of the line joining them, option A.
Watch out
Equidistant from two points gives the perpendicular bisector. The angle bisector idea (option D) belongs to two lines, not two points. Report a problem with this question
What is the perpendicular distance of the point ( 2 , 3 ) (2, 3) ( 2 , 3 ) 2 x − 4 y + 3 = 0 2x - 4y + 3 = 0 2 x − 4 y + 3 = 0
A 5 2 \frac{\sqrt5}{2} 2 5 B − 5 20 -\frac{\sqrt5}{20} − 20 5 C − 5 13 -\frac{5}{\sqrt{13}} − 13 5 D 0
Worked solution (try it first) The distance from
( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) to
a x + b y + c = 0 ax + by + c = 0 a x + b y + c = 0 is
∣ a x 1 + b y 1 + c ∣ a 2 + b 2 \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} a 2 + b 2 ∣ a x 1 + b y 1 + c ∣ .
Put in
( 2 , 3 ) (2, 3) ( 2 , 3 ) : the top is
∣ 2 ( 2 ) − 4 ( 3 ) + 3 ∣ = ∣ 4 − 12 + 3 ∣ = 5 |2(2) - 4(3) + 3| = |4 - 12 + 3| = 5 ∣2 ( 2 ) − 4 ( 3 ) + 3∣ = ∣4 − 12 + 3∣ = 5 .
The bottom is
2 2 + 4 2 = 20 = 2 5 \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt5 2 2 + 4 2 = 20 = 2 5 .
So the distance is
5 2 5 = 5 2 \frac{5}{2\sqrt5} = \frac{\sqrt5}{2} 2 5 5 = 2 5 , option A.
Watch out
Take the absolute value on top: a distance can never be negative, which rules out options B and C at once. Without it the top is − 5 -5 − 5 Report a problem with this question
Find the equation of the line through ( 5 , 7 ) (5, 7) ( 5 , 7 ) 7 x + 5 y = 12 7x + 5y = 12 7 x + 5 y = 12
A 5 x + 7 y = 120 5x + 7y = 120 5 x + 7 y = 120 B 7 x + 5 y = 70 7x + 5y = 70 7 x + 5 y = 70 C x + y = 7 x + y = 7 x + y = 7 D 15 x + 17 y = 90 15x + 17y = 90 15 x + 17 y = 90
Worked solution (try it first) Parallel lines have the same gradient, so the new line keeps the same left-hand side:
7 x + 5 y = k 7x + 5y = k 7 x + 5 y = k .
Put in the point
( 5 , 7 ) (5, 7) ( 5 , 7 ) :
k = 7 ( 5 ) + 5 ( 7 ) = 35 + 35 = 70 k = 7(5) + 5(7) = 35 + 35 = 70 k = 7 ( 5 ) + 5 ( 7 ) = 35 + 35 = 70 .
So the line is
7 x + 5 y = 70 7x + 5y = 70 7 x + 5 y = 70 , option B.
Watch out
Keep the coefficients of x x x y y y − 7 5 -\frac75 − 5 7 − 5 7 -\frac57 − 7 5 Report a problem with this question
In the figure, ∠ X Z Y = 90 ∘ \angle XZY = 90^\circ ∠ X Z Y = 9 0 ∘ X Z = 10 XZ = 10 X Z = 10 ∠ Z X R = 30 ∘ \angle ZXR = 30^\circ ∠ Z X R = 3 0 ∘ ∠ R X Y = 15 ∘ \angle RXY = 15^\circ ∠ R X Y = 1 5 ∘ R Y RY R Y
A 10 B 10 ( 1 − 1 3 ) 10\left(1 - \frac{1}{\sqrt3}\right) 10 ( 1 − 3 1 ) C 10 ( 1 − 3 ) 10(1 - \sqrt3) 10 ( 1 − 3 ) D 10 ( 1 − 1 2 ) 10\left(1 - \frac{1}{\sqrt2}\right) 10 ( 1 − 2 1 )
Worked solution (try it first) The whole angle
Z X Y ZXY Z X Y is
30 ∘ + 15 ∘ = 45 ∘ 30^\circ + 15^\circ = 45^\circ 3 0 ∘ + 1 5 ∘ = 4 5 ∘ .
In the right-angled triangle
X Z Y XZY X Z Y :
Z Y = 10 tan 45 ∘ = 10 ZY = 10\tan45^\circ = 10 Z Y = 10 tan 4 5 ∘ = 10 cm.
In the right-angled triangle
X Z R XZR X Z R :
Z R = 10 tan 30 ∘ = 10 3 ZR = 10\tan30^\circ = \frac{10}{\sqrt3} Z R = 10 tan 3 0 ∘ = 3 10 cm.
R R R lies on
Z Y ZY Z Y , so
R Y = Z Y − Z R RY = ZY - ZR R Y = Z Y − Z R = 10 − 10 3 = 10 - \frac{10}{\sqrt3} = 10 − 3 10 = 10 ( 1 − 1 3 ) = 10\left(1 - \frac{1}{\sqrt3}\right) = 10 ( 1 − 3 1 ) cm, option B.
Watch out
R Y RY R Y Z Y ZY Z Y Z R ZR Z R Z Y = 10 ZY = 10 Z Y = 10 Report a problem with this question
Evaluate lim x → 2 ( x − 2 ) ( x 2 + 3 x − 2 ) x 2 − 4 \displaystyle\lim_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} x → 2 lim x 2 − 4 ( x − 2 ) ( x 2 + 3 x − 2 )
Worked solution (try it first) Putting in
x = 2 x = 2 x = 2 straight away gives
0 0 \frac00 0 0 , so simplify first.
Factorise the bottom as a difference of two squares:
x 2 − 4 = ( x − 2 ) ( x + 2 ) x^2 - 4 = (x - 2)(x + 2) x 2 − 4 = ( x − 2 ) ( x + 2 ) .
Cancel the common factor
x − 2 x - 2 x − 2 : the expression becomes
x 2 + 3 x − 2 x + 2 \dfrac{x^2 + 3x - 2}{x + 2} x + 2 x 2 + 3 x − 2 .
Now put in
x = 2 x = 2 x = 2 :
4 + 6 − 2 4 = 8 4 = 2 \dfrac{4 + 6 - 2}{4} = \dfrac84 = 2 4 4 + 6 − 2 = 4 8 = 2 , option B.
Watch out
0 0 \frac00 0 0 Report a problem with this question
If y = x sin x y = x\sin x y = x sin x d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y
A 2 cos x − x sin x 2\cos x - x\sin x 2 cos x − x sin x B sin x + x cos x \sin x + x\cos x sin x + x cos x C sin x − x cos x \sin x - x\cos x sin x − x cos x D x sin x − 2 cos x x\sin x - 2\cos x x sin x − 2 cos x
Worked solution (try it first) Product rule, with
u = x u = x u = x and
v = sin x v = \sin x v = sin x :
d y d x = 1 ⋅ sin x + x cos x \frac{dy}{dx} = 1 \cdot \sin x + x\cos x d x d y = 1 ⋅ sin x + x cos x = sin x + x cos x = \sin x + x\cos x = sin x + x cos x .
Differentiate again.
sin x \sin x sin x gives
cos x \cos x cos x , and by the product rule
x cos x x\cos x x cos x gives
cos x − x sin x \cos x - x\sin x cos x − x sin x .
Add them:
d 2 y d x 2 = 2 cos x − x sin x \frac{d^2y}{dx^2} = 2\cos x - x\sin x d x 2 d 2 y = 2 cos x − x sin x , option A.
Watch out
The question asks for the second derivative. sin x + x cos x \sin x + x\cos x sin x + x cos x Similar: JAMB 2013 · UTME · Q36
Report a problem with this question
Ice forms on a refrigerator ice-box at the rate of ( 4 − 0.6 t ) (4 - 0.6t) ( 4 − 0.6 t ) t t t
A 19.5 g B 17.0 g C 14.5 g D 12.5 g
Worked solution (try it first) The rate is the derivative of the mass, so integrate it to find the ice formed from
t = 0 t = 0 t = 0 to
t = 5 t = 5 t = 5 .
∫ ( 4 − 0.6 t ) d t = 4 t − 0.3 t 2 \int (4 - 0.6t)\,dt = 4t - 0.3t^2 ∫ ( 4 − 0.6 t ) d t = 4 t − 0.3 t 2 .
At
t = 5 t = 5 t = 5 this is
20 − 7.5 = 12.5 20 - 7.5 = 12.5 20 − 7.5 = 12.5 , and at
t = 0 t = 0 t = 0 it is 0.
So 12.5 g of ice forms.
Add the 2 g already there:
2 + 12.5 = 14.5 2 + 12.5 = 14.5 2 + 12.5 = 14.5 g, option C.
Watch out
The integral gives only the ice formed in the 5 minutes. Forgetting the 2 g that was there at the start gives 12.5 g (option D). Report a problem with this question
Obtain the maximum value of the function f ( x ) = x 3 − 12 x + 11 f(x) = x^3 - 12x + 11 f ( x ) = x 3 − 12 x + 11
Worked solution (try it first) At a turning point
f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 :
3 x 2 − 12 = 0 3x^2 - 12 = 0 3 x 2 − 12 = 0 , so
x 2 = 4 x^2 = 4 x 2 = 4 and
x = 2 x = 2 x = 2 or
x = − 2 x = -2 x = − 2 .
f ′ ′ ( x ) = 6 x f''(x) = 6x f ′′ ( x ) = 6 x .
At
x = − 2 x = -2 x = − 2 ,
f ′ ′ = − 12 < 0 f'' = -12 < 0 f ′′ = − 12 < 0 , so this is the maximum.
At
x = 2 x = 2 x = 2 it is the minimum.
f ( − 2 ) = − 8 + 24 + 11 = 27 f(-2) = -8 + 24 + 11 = 27 f ( − 2 ) = − 8 + 24 + 11 = 27 , so the maximum value is 27, option D.
Watch out
Check which turning point is the maximum. f ( 2 ) = 8 − 24 + 11 = − 5 f(2) = 8 - 24 + 11 = -5 f ( 2 ) = 8 − 24 + 11 = − 5 Report a problem with this question
A student blows a balloon and its volume increases at a rate of π ( 20 − t 2 ) cm 3 s − 1 \pi(20 - t^2)\text{ cm}^3\text{s}^{-1} π ( 20 − t 2 ) cm 3 s − 1 t t t 0 cm 3 0\text{ cm}^3 0 cm 3
A 37.00 π 37.00\pi 37.00 π B 37.33 π 37.33\pi 37.33 π C 40.00 π 40.00\pi 40.00 π D 42.67 π 42.67\pi 42.67 π
Worked solution (try it first) Integrate the rate to get the volume:
V = π ( 20 t − t 3 3 ) + c V = \pi\left(20t - \frac{t^3}{3}\right) + c V = π ( 20 t − 3 t 3 ) + c .
The volume is 0 when
t = 0 t = 0 t = 0 , so
c = 0 c = 0 c = 0 .
At
t = 2 t = 2 t = 2 :
V = π ( 40 − 8 3 ) V = \pi\left(40 - \frac83\right) V = π ( 40 − 3 8 ) ≈ 37.33 π cm 3 \approx 37.33\pi\text{ cm}^3 ≈ 37.33 π cm 3 , option B.
Watch out
Keep the minus sign: − t 2 -t^2 − t 2 − t 3 3 -\frac{t^3}{3} − 3 t 3 8 3 \frac83 3 8 42.67 π 42.67\pi 42.67 π Report a problem with this question
A storekeeper checked his stock of five commodities: F 215, G 113, H 108, K 216, M 68. What angle will commodity H represent on a pie chart?
A 216 ∘ 216^\circ 21 6 ∘ B 108 ∘ 108^\circ 10 8 ∘ C 68 ∘ 68^\circ 6 8 ∘ D 54 ∘ 54^\circ 5 4 ∘
Worked solution (try it first) Total stock:
215 + 113 + 108 + 216 + 68 = 720 215 + 113 + 108 + 216 + 68 = 720 215 + 113 + 108 + 216 + 68 = 720 .
Each item gets
360 ∘ 720 = 0.5 ∘ \frac{360^\circ}{720} = 0.5^\circ 720 36 0 ∘ = 0. 5 ∘ .
Commodity H has 108 items:
108 × 0.5 ∘ = 54 ∘ 108 \times 0.5^\circ = 54^\circ 108 × 0. 5 ∘ = 5 4 ∘ , option D.
Watch out
The number of items is not the angle: 108 ∘ 108^\circ 10 8 ∘ 0.5 ∘ 0.5^\circ 0. 5 ∘ Report a problem with this question
If the mean of the frequency distribution below is 5.2, find y y y
x x x 2
4
6
8
f f f 4
y y y 6
5
Worked solution (try it first) Total frequency:
4 + y + 6 + 5 = 15 + y 4 + y + 6 + 5 = 15 + y 4 + y + 6 + 5 = 15 + y .
Total of
f x fx f x :
8 + 4 y + 36 + 40 = 84 + 4 y 8 + 4y + 36 + 40 = 84 + 4y 8 + 4 y + 36 + 40 = 84 + 4 y .
Mean
= ∑ f x ∑ f = \frac{\sum fx}{\sum f} = ∑ f ∑ f x , so
84 + 4 y 15 + y = 5.2 \frac{84 + 4y}{15 + y} = 5.2 15 + y 84 + 4 y = 5.2 and
84 + 4 y = 78 + 5.2 y 84 + 4y = 78 + 5.2y 84 + 4 y = 78 + 5.2 y .
Collect terms:
6 = 1.2 y 6 = 1.2y 6 = 1.2 y , so
y = 5.0 y = 5.0 y = 5.0 , option C.
Watch out
Put y y y 15 + y 15 + y 15 + y 84 + 4 y = 78 84 + 4y = 78 84 + 4 y = 78 y y y Report a problem with this question
Find the mode and median respectively of the distribution below.
No. of children
0
1
2
3
4
5
6
No. of families
7
11
6
7
7
5
3
Worked solution (try it first) The mode is the number of children with the most families: 1 child, 11 families.
There are
7 + 11 + 6 + 7 + 7 + 5 + 3 = 46 7 + 11 + 6 + 7 + 7 + 5 + 3 = 46 7 + 11 + 6 + 7 + 7 + 5 + 3 = 46 families, so the median is halfway between the 23rd and 24th.
Running totals: 7, 18, 24.
The 19th to 24th families all have 2 children, so the median is 2.
So the mode and median are 1 and 2, option B.
Watch out
Keep the order asked for: mode first, then median. Swapping them gives 2, 1 (option A). Report a problem with this question
If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores.
A 2 3 \frac23 3 2 B 3 2 3 \frac{3}{2\sqrt3} 2 3 3 C 2 3 \sqrt{\frac23} 3 2 D 3 2 \frac{\sqrt3}{2} 2 3
Worked solution (try it first) The mean is
5 + 6 + 7 3 = 6 \frac{5 + 6 + 7}{3} = 6 3 5 + 6 + 7 = 6 .
The deviations from 6 are
− 1 -1 − 1 , 0 and 1.
Their squares are 1, 0 and 1, which add up to 2.
The variance is
2 3 \frac23 3 2 , and the standard deviation is its square root,
2 3 \sqrt{\frac23} 3 2 , option C.
Watch out
2 3 \frac23 3 2 Report a problem with this question
Two fair dice are thrown together. Determine the probability of obtaining a total score of 8.
A 1 12 \frac1{12} 12 1 B 5 36 \frac5{36} 36 5 C 1 8 \frac18 8 1 D 7 36 \frac7{36} 36 7
Worked solution (try it first) Two dice give 36 equally likely outcomes.
A total of 8 comes from
( 2 , 6 ) , ( 3 , 5 ) , ( 4 , 4 ) , ( 5 , 3 ) , ( 6 , 2 ) (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) ( 2 , 6 ) , ( 3 , 5 ) , ( 4 , 4 ) , ( 5 , 3 ) , ( 6 , 2 ) , which is 5 outcomes.
So the probability is
5 36 \frac{5}{36} 36 5 , option B.
Watch out
( 2 , 6 ) (2, 6) ( 2 , 6 ) ( 6 , 2 ) (6, 2) ( 6 , 2 ) 3 36 = 1 12 \frac{3}{36} = \frac{1}{12} 36 3 = 12 1 Report a problem with this question
The probability of an event P P P 3 4 \frac34 4 3 Q Q Q 1 6 \frac16 6 1 P P P Q Q Q 1 12 \frac1{12} 12 1 P P P Q Q Q
A 1 96 \frac1{96} 96 1 B 1 8 \frac18 8 1 C 5 6 \frac56 6 5 D 11 12 \frac{11}{12} 12 11
Worked solution (try it first) For "either or", add the two probabilities and take away the overlap, which is counted twice:
P ( P ∪ Q ) = P ( P ) + P ( Q ) − P ( P ∩ Q ) P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) P ( P ∪ Q ) = P ( P ) + P ( Q ) − P ( P ∩ Q ) .
Over 12:
9 12 + 2 12 − 1 12 = 10 12 \frac{9}{12} + \frac{2}{12} - \frac{1}{12} = \frac{10}{12} 12 9 + 12 2 − 12 1 = 12 10 .
So the probability is
5 6 \frac56 6 5 , option C.
Watch out
Subtract the overlap: 3 4 + 1 6 \frac34 + \frac16 4 3 + 6 1 11 12 \frac{11}{12} 12 11 Report a problem with this question
Five people are to be arranged in a row for a group photograph. How many arrangements are there if a married couple in the group insist on sitting next to each other?
Worked solution (try it first) Tie the couple together and treat them as one unit.
Then there are 4 units to arrange in a row:
4 ! = 24 4! = 24 4 ! = 24 ways.
Inside the unit the couple can swap places:
2 ! = 2 2! = 2 2 ! = 2 ways.
Multiply:
24 × 2 = 48 24 \times 2 = 48 24 × 2 = 48 arrangements, option A.
Watch out
Don't forget that husband and wife can swap within their pair. Stopping at 4 ! = 24 4! = 24 4 ! = 24 Report a problem with this question
A student has 5 courses to take from Mathematics and Physics. There are 4 courses in Mathematics and 3 in Physics which he can choose from at will. In how many ways can he choose his courses so that he takes exactly two courses in Physics?
Worked solution (try it first) Exactly two Physics courses from 3:
3 C 2 = 3 ^3C_2 = 3 3 C 2 = 3 ways.
The other
5 − 2 = 3 5 - 2 = 3 5 − 2 = 3 courses come from Mathematics, 3 from 4:
4 C 3 = 4 ^4C_3 = 4 4 C 3 = 4 ways.
Each Physics choice goes with each Mathematics choice, so multiply:
3 × 4 = 12 3 \times 4 = 12 3 × 4 = 12 , option B.
Watch out
Multiply the two counts, don't add them: 3 + 4 = 7 3 + 4 = 7 3 + 4 = 7 Report a problem with this question