Paper JAMB 1998 General Maths Objective
Objective paper · 46 questions · partial
JAMB 1998 · UME Topics include Number bases, Number foundations & fractions, Solid mensuration, Commercial arithmetic, Expressions, formulae & change of subject, Sets & Venn diagrams.
Our copy of this paper is missing questions 3, 4, 5, 40.
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 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 41 42 43 44 45 46 47 48 49 50 If 1011 2 + X 7 = 25 10 1011_2 + X_7 = 25_{10} 101 1 2 + X 7 = 2 5 10 X X X
Worked solution (try it first) Change to base ten:
1011 2 = 8 + 2 + 1 = 11 1011_2 = 8 + 2 + 1 = 11 101 1 2 = 8 + 2 + 1 = 11 .
So
X 7 = 25 − 11 = 14 X_7 = 25 - 11 = 14 X 7 = 25 − 11 = 14 in base ten.
Write 14 in base seven:
14 = 2 × 7 + 0 14 = 2 \times 7 + 0 14 = 2 × 7 + 0 , so
X = 20 X = 20 X = 20 , option B.
Watch out
14 is the base-ten value. X X X Report a problem with this question
Evaluate [ 1 0.03 ÷ 1 0.024 ] − 1 \left[\frac{1}{0.03} \div \frac{1}{0.024}\right]^{-1} [ 0.03 1 ÷ 0.024 1 ] − 1
Worked solution (try it first) Dividing by
1 0.024 \frac{1}{0.024} 0.024 1 is the same as multiplying by 0.024:
1 0.03 × 0.024 = 0.024 0.03 \frac{1}{0.03} \times 0.024 = \frac{0.024}{0.03} 0.03 1 × 0.024 = 0.03 0.024 .
Multiply top and bottom by 1000:
24 30 = 0.8 \frac{24}{30} = 0.8 30 24 = 0.8 .
The power
− 1 -1 − 1 means the reciprocal:
1 0.8 = 1.25 \frac{1}{0.8} = 1.25 0.8 1 = 1.25 , option B.
Watch out
Don't stop at 0.8: the power − 1 -1 − 1 Report a problem with this question
A market woman sells oil in cylindrical tins 10 cm deep and 6 cm in diameter at ₦15.00 each. If she bought a full cylindrical jug 18 cm deep and 10 cm in diameter for ₦50.00, how much did she make by selling all the oil?
A ₦62.50 B ₦35.00 C ₦31.00 D ₦25.00
Worked solution (try it first) Tin: radius 3 cm, so its volume is
π × 3 2 × 10 = 90 π cm 3 \pi \times 3^2 \times 10 = 90\pi\text{ cm}^3 π × 3 2 × 10 = 90 π cm 3 .
Jug: radius 5 cm, so its volume is
π × 5 2 × 18 = 450 π cm 3 \pi \times 5^2 \times 18 = 450\pi\text{ cm}^3 π × 5 2 × 18 = 450 π cm 3 .
That fills
450 π ÷ 90 π = 5 450\pi \div 90\pi = 5 450 π ÷ 90 π = 5 tins.
She takes in
5 × 15 = 75 5 \times 15 = 75 5 × 15 = 75 naira.
Profit:
75 − 50 = 25 75 - 50 = 25 75 − 50 = 25 , so ₦25.00, option D.
Watch out
Square the radii when comparing volumes. With 5 × 18 5 \times 18 5 × 18 3 × 10 3 \times 10 3 × 10 Report a problem with this question
A man is paid r r r q q q
A r ( 35 + q ) r(35 + q) r ( 35 + q ) B q ( 35 r − q ) q(35r - q) q ( 35 r − q ) C q ( 35 r + r ) q(35r + r) q ( 35 r + r ) D r ( 35 r − q ) r(35r - q) r ( 35 r − q )
Worked solution (try it first) The 35 hours include the
q q q overtime hours, so the normal hours are
35 − q 35 - q 35 − q , paid
r ( 35 − q ) r(35 - q) r ( 35 − q ) .
Overtime is at double rate:
q q q hours at
2 r 2r 2 r gives
2 q r 2qr 2 q r .
Total:
35 r − q r + 2 q r = 35 r + q r = r ( 35 + q ) 35r - qr + 2qr = 35r + qr = r(35 + q) 35 r − q r + 2 q r = 35 r + q r = r ( 35 + q ) , option A.
Watch out
The 35-hour week already includes the overtime. Paying 35 normal hours and then q q q 2 r 2r 2 r 35 r + 2 q r 35r + 2qr 35 r + 2 q r Report a problem with this question
Given the universal set U = { 1 , 2 , 3 , 4 , 5 , 6 } U = \{1, 2, 3, 4, 5, 6\} U = { 1 , 2 , 3 , 4 , 5 , 6 } P = { 1 , 2 , 3 , 4 } P = \{1, 2, 3, 4\} P = { 1 , 2 , 3 , 4 } Q = { 3 , 4 , 5 } Q = \{3, 4, 5\} Q = { 3 , 4 , 5 } R = { 2 , 4 , 6 } R = \{2, 4, 6\} R = { 2 , 4 , 6 } P ∪ ( Q ∩ R ) P \cup (Q \cap R) P ∪ ( Q ∩ R )
A { 4 } \{4\} { 4 } B { 1 , 2 , 3 , 4 } \{1, 2, 3, 4\} { 1 , 2 , 3 , 4 } C { 1 , 2 , 3 , 5 , 6 } \{1, 2, 3, 5, 6\} { 1 , 2 , 3 , 5 , 6 } D { 1 , 2 , 3 , 4 , 5 , 6 } \{1, 2, 3, 4, 5, 6\} { 1 , 2 , 3 , 4 , 5 , 6 }
Worked solution (try it first) Do the bracket first:
Q ∩ R Q \cap R Q ∩ R is the elements in both
Q Q Q and
R R R , which is
{ 4 } \{4\} { 4 } .
Then
P ∪ { 4 } P \cup \{4\} P ∪ { 4 } is every element of
P P P together with 4.
Since 4 is already in
P P P , this is
{ 1 , 2 , 3 , 4 } \{1, 2, 3, 4\} { 1 , 2 , 3 , 4 } , option B.
Watch out
{ 4 } \{4\} { 4 } Q ∩ R Q \cap R Q ∩ R P P P Report a problem with this question
In the Venn diagram, the shaded region is
A ( P ∩ Q ) ∪ R (P \cap Q) \cup R ( P ∩ Q ) ∪ R B ( P ∩ Q ) ∩ R (P \cap Q) \cap R ( P ∩ Q ) ∩ R C ( P ∩ Q ′ ) ∪ R (P \cap Q') \cup R ( P ∩ Q ′ ) ∪ R D ( P ∩ Q ′ ) ∩ R (P \cap Q') \cap R ( P ∩ Q ′ ) ∩ R
Worked solution (try it first) The shaded region lies inside both
P P P and
R R R .
It lies outside
Q Q Q , so it is in
P ∩ Q ′ P \cap Q' P ∩ Q ′ as well as in
R R R .
So it is
( P ∩ Q ′ ) ∩ R (P \cap Q') \cap R ( P ∩ Q ′ ) ∩ R , option D.
Watch out
∪ R \cup R ∪ R R R R R R R P P P ∩ R \cap R ∩ R Report a problem with this question
When the expression p m 2 + q m + 1 pm^2 + qm + 1 p m 2 + q m + 1 ( m − 1 ) (m - 1) ( m − 1 ) ( m + 1 ) (m + 1) ( m + 1 ) p p p q q q
A 2 , − 1 2, -1 2 , − 1 B − 1 , 2 -1, 2 − 1 , 2 C 3 , − 2 3, -2 3 , − 2 D − 2 , 3 -2, 3 − 2 , 3
Worked solution (try it first) By the remainder theorem, dividing by
m − 1 m - 1 m − 1 leaves the value at
m = 1 m = 1 m = 1 :
p + q + 1 = 2 p + q + 1 = 2 p + q + 1 = 2 , so
p + q = 1 p + q = 1 p + q = 1 .
Dividing by
m + 1 m + 1 m + 1 leaves the value at
m = − 1 m = -1 m = − 1 :
p − q + 1 = 4 p - q + 1 = 4 p − q + 1 = 4 , so
p − q = 3 p - q = 3 p − q = 3 .
Add the equations:
2 p = 4 2p = 4 2 p = 4 , so
p = 2 p = 2 p = 2 and
q = − 1 q = -1 q = − 1 , option A.
Watch out
"Respectively" means give p p p q q q − 1 , 2 -1, 2 − 1 , 2 Also set as JAMB 2016 · UTME · Q27
Report a problem with this question
Factorize r 2 − r ( 2 p + q ) + 2 p q r^2 - r(2p + q) + 2pq r 2 − r ( 2 p + q ) + 2 pq
A ( r − 2 q ) ( 2 r − p ) (r - 2q)(2r - p) ( r − 2 q ) ( 2 r − p ) B ( r − q ) ( r + p ) (r - q)(r + p) ( r − q ) ( r + p ) C ( r − q ) ( r − 2 p ) (r - q)(r - 2p) ( r − q ) ( r − 2 p ) D ( 2 r − q ) ( r + p ) (2r - q)(r + p) ( 2 r − q ) ( r + p )
Worked solution (try it first) Expand the bracket:
r 2 − 2 p r − q r + 2 p q r^2 - 2pr - qr + 2pq r 2 − 2 p r − q r + 2 pq .
Group in pairs and take out common factors:
r ( r − 2 p ) − q ( r − 2 p ) r(r - 2p) - q(r - 2p) r ( r − 2 p ) − q ( r − 2 p ) .
Take out the common bracket:
( r − q ) ( r − 2 p ) (r - q)(r - 2p) ( r − q ) ( r − 2 p ) , option C.
Watch out
Expand to check. Option A, ( r − 2 q ) ( 2 r − p ) (r - 2q)(2r - p) ( r − 2 q ) ( 2 r − p ) 2 r 2 2r^2 2 r 2 r 2 r^2 r 2 Report a problem with this question
Solve the equation x − x − 2 − 1 = 0 \sqrt x - \sqrt{x - 2} - 1 = 0 x − x − 2 − 1 = 0
A 3 2 \frac32 2 3 B 2 3 \frac23 3 2 C 4 9 \frac49 9 4 D 9 4 \frac94 4 9
Worked solution (try it first) Isolate one root:
x = 1 + x − 2 \sqrt x = 1 + \sqrt{x - 2} x = 1 + x − 2 .
Square both sides:
x = 1 + 2 x − 2 + x − 2 x = 1 + 2\sqrt{x - 2} + x - 2 x = 1 + 2 x − 2 + x − 2 .
The
x x x terms cancel:
2 x − 2 = 1 2\sqrt{x - 2} = 1 2 x − 2 = 1 , so
x − 2 = 1 2 \sqrt{x - 2} = \frac12 x − 2 = 2 1 .
Square again:
x − 2 = 1 4 x - 2 = \frac14 x − 2 = 4 1 , so
x = 9 4 x = \frac94 x = 4 9 , option D.
Watch out
3 2 \frac32 2 3 x \sqrt x x x x x 9 4 − 1 4 − 1 = 3 2 − 1 2 − 1 = 0 \sqrt{\frac94} - \sqrt{\frac14} - 1 = \frac32 - \frac12 - 1 = 0 4 9 − 4 1 − 1 = 2 3 − 2 1 − 1 = 0 Report a problem with this question
Find the range of values of m m m 3 x 2 − 3 m x + ( m 2 − m − 3 ) = 0 3x^2 - 3mx + (m^2 - m - 3) = 0 3 x 2 − 3 m x + ( m 2 − m − 3 ) = 0
A − 1 < m < 7 -1 < m < 7 − 1 < m < 7 B − 2 < m < 6 -2 < m < 6 − 2 < m < 6 C − 3 < m < 9 -3 < m < 9 − 3 < m < 9 D − 4 < m < 8 -4 < m < 8 − 4 < m < 8
Worked solution (try it first) Real roots need
b 2 − 4 a c ≥ 0 b^2 - 4ac \ge 0 b 2 − 4 a c ≥ 0 :
9 m 2 − 12 ( m 2 − m − 3 ) ≥ 0 9m^2 - 12(m^2 - m - 3) \ge 0 9 m 2 − 12 ( m 2 − m − 3 ) ≥ 0 .
Simplify:
− 3 m 2 + 12 m + 36 ≥ 0 -3m^2 + 12m + 36 \ge 0 − 3 m 2 + 12 m + 36 ≥ 0 .
Divide by
− 3 -3 − 3 and flip the sign:
m 2 − 4 m − 12 ≤ 0 m^2 - 4m - 12 \le 0 m 2 − 4 m − 12 ≤ 0 .
Factorise:
( m − 6 ) ( m + 2 ) ≤ 0 (m - 6)(m + 2) \le 0 ( m − 6 ) ( m + 2 ) ≤ 0 , which holds between the roots.
So
m m m lies between
− 2 -2 − 2 and 6, option B.
Watch out
Dividing an inequality by a negative number flips it. Keeping ≥ \ge ≥ m ≤ − 2 m \le -2 m ≤ − 2 m ≥ 6 m \ge 6 m ≥ 6 Report a problem with this question
Make a x \frac ax x a x + a x − a = m \dfrac{x + a}{x - a} = m x − a x + a = m
A m − 1 m + 1 \dfrac{m - 1}{m + 1} m + 1 m − 1 B 1 + m 1 − m \dfrac{1 + m}{1 - m} 1 − m 1 + m C 1 − m 1 + m \dfrac{1 - m}{1 + m} 1 + m 1 − m D m + 1 m − 1 \dfrac{m + 1}{m - 1} m − 1 m + 1
Worked solution (try it first) Divide the top and bottom of the left side by
x x x :
1 + a x 1 − a x = m \dfrac{1 + \frac ax}{1 - \frac ax} = m 1 − x a 1 + x a = m .
Multiply out:
1 + a x = m − m a x 1 + \frac ax = m - m\frac ax 1 + x a = m − m x a , so
a x ( 1 + m ) = m − 1 \frac ax(1 + m) = m - 1 x a ( 1 + m ) = m − 1 .
Divide by
1 + m 1 + m 1 + m :
a x = m − 1 m + 1 \frac ax = \dfrac{m - 1}{m + 1} x a = m + 1 m − 1 , option A.
Watch out
The subject is a x \frac ax x a x a \frac xa a x m + 1 m − 1 \frac{m + 1}{m - 1} m − 1 m + 1 x a \frac xa a x Report a problem with this question
Divide 2 x 3 + 11 x 2 + 17 x + 6 2x^3 + 11x^2 + 17x + 6 2 x 3 + 11 x 2 + 17 x + 6 2 x + 1 2x + 1 2 x + 1
A x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6 B 2 x 2 + 5 x + 6 2x^2 + 5x + 6 2 x 2 + 5 x + 6 C 2 x 2 − 5 x + 6 2x^2 - 5x + 6 2 x 2 − 5 x + 6 D x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6
Worked solution (try it first) Long division:
2 x 3 ÷ 2 x = x 2 2x^3 \div 2x = x^2 2 x 3 ÷ 2 x = x 2 .
Subtract
x 2 ( 2 x + 1 ) = 2 x 3 + x 2 x^2(2x + 1) = 2x^3 + x^2 x 2 ( 2 x + 1 ) = 2 x 3 + x 2 to leave
10 x 2 + 17 x + 6 10x^2 + 17x + 6 10 x 2 + 17 x + 6 .
10 x 2 ÷ 2 x = 5 x 10x^2 \div 2x = 5x 10 x 2 ÷ 2 x = 5 x .
Subtract
5 x ( 2 x + 1 ) = 10 x 2 + 5 x 5x(2x + 1) = 10x^2 + 5x 5 x ( 2 x + 1 ) = 10 x 2 + 5 x to leave
12 x + 6 12x + 6 12 x + 6 .
12 x ÷ 2 x = 6 12x \div 2x = 6 12 x ÷ 2 x = 6 , and
6 ( 2 x + 1 ) = 12 x + 6 6(2x + 1) = 12x + 6 6 ( 2 x + 1 ) = 12 x + 6 leaves 0.
So the quotient is
x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6 , option A.
Watch out
Divide the leading terms: 2 x 3 ÷ 2 x = x 2 2x^3 \div 2x = x^2 2 x 3 ÷ 2 x = x 2 2 x 2 2x^2 2 x 2 Also set as JAMB 2016 · UTME · Q28
Report a problem with this question
Express 11 x + 2 6 x 2 − x − 1 \dfrac{11x + 2}{6x^2 - x - 1} 6 x 2 − x − 1 11 x + 2
A 1 3 x − 1 + 3 2 x + 1 \frac{1}{3x - 1} + \frac{3}{2x + 1} 3 x − 1 1 + 2 x + 1 3 B 3 3 x + 1 − 1 2 x − 1 \frac{3}{3x + 1} - \frac{1}{2x - 1} 3 x + 1 3 − 2 x − 1 1 C 3 3 x − 1 − 1 2 x + 1 \frac{3}{3x - 1} - \frac{1}{2x + 1} 3 x − 1 3 − 2 x + 1 1 D 1 3 x + 1 + 3 2 x − 1 \frac{1}{3x + 1} + \frac{3}{2x - 1} 3 x + 1 1 + 2 x − 1 3
Worked solution (try it first) Factorise the bottom:
6 x 2 − x − 1 = ( 3 x + 1 ) ( 2 x − 1 ) 6x^2 - x - 1 = (3x + 1)(2x - 1) 6 x 2 − x − 1 = ( 3 x + 1 ) ( 2 x − 1 ) .
Write
11 x + 2 = A ( 2 x − 1 ) + B ( 3 x + 1 ) 11x + 2 = A(2x - 1) + B(3x + 1) 11 x + 2 = A ( 2 x − 1 ) + B ( 3 x + 1 ) .
Put
x = − 1 3 x = -\frac13 x = − 3 1 :
− 5 3 = − 5 3 A -\frac53 = -\frac53A − 3 5 = − 3 5 A , so
A = 1 A = 1 A = 1 .
Put
x = 1 2 x = \frac12 x = 2 1 :
15 2 = 5 2 B \frac{15}{2} = \frac52B 2 15 = 2 5 B , so
B = 3 B = 3 B = 3 .
So the answer is
1 3 x + 1 + 3 2 x − 1 \frac{1}{3x + 1} + \frac{3}{2x - 1} 3 x + 1 1 + 2 x − 1 3 , option D.
Watch out
Check the factors by expanding: ( 3 x + 1 ) ( 2 x − 1 ) = 6 x 2 − x − 1 (3x + 1)(2x - 1) = 6x^2 - x - 1 ( 3 x + 1 ) ( 2 x − 1 ) = 6 x 2 − x − 1 ( 3 x − 1 ) ( 2 x + 1 ) (3x - 1)(2x + 1) ( 3 x − 1 ) ( 2 x + 1 ) 6 x 2 + x − 1 6x^2 + x - 1 6 x 2 + x − 1 Report a problem with this question
If x x x 1 3 x + 1 2 > 1 4 x \frac{1}{3x} + \frac12 > \frac{1}{4x} 3 x 1 + 2 1 > 4 x 1
A x > − 1 6 x > -\frac16 x > − 6 1 B x > 0 x > 0 x > 0 C 0 < x < 4 0 < x < 4 0 < x < 4 D 0 < x < 1 6 0 < x < \frac16 0 < x < 6 1
Worked solution (try it first) Multiply every term by
12 x 12x 12 x , which is positive, so the sign stays:
4 + 6 x > 3 4 + 6x > 3 4 + 6 x > 3 .
Subtract 4 from both sides:
6 x > − 1 6x > -1 6 x > − 1 , so
x > − 1 6 x > -\frac16 x > − 6 1 .
Every positive
x x x satisfies
x > − 1 6 x > -\frac16 x > − 6 1 , so the answer is all
x > 0 x > 0 x > 0 , option B.
Watch out
Option A, x > − 1 6 x > -\frac16 x > − 6 1 x x x x = 0 x = 0 x = 0 1 3 x \frac{1}{3x} 3 x 1 x > 0 x > 0 x > 0 Report a problem with this question
The shaded area in the graph represents
A x ≥ 0 , 3 y + 2 x ≥ 6 x \ge 0, 3y + 2x \ge 6 x ≥ 0 , 3 y + 2 x ≥ 6 B x ≥ 0 , y ≥ 3 , 3 x + 2 y ≥ 6 x \ge 0, y \ge 3, 3x + 2y \ge 6 x ≥ 0 , y ≥ 3 , 3 x + 2 y ≥ 6 C x ≥ 2 , y ≥ 0 , 3 x + 2 y ≤ 6 x \ge 2, y \ge 0, 3x + 2y \le 6 x ≥ 2 , y ≥ 0 , 3 x + 2 y ≤ 6 D x ≥ 0 , y ≥ 0 , 3 x + 2 y ≥ 6 x \ge 0, y \ge 0, 3x + 2y \ge 6 x ≥ 0 , y ≥ 0 , 3 x + 2 y ≥ 6
Worked solution (try it first) The line through
( 0 , 3 ) (0, 3) ( 0 , 3 ) and
( 2 , 0 ) (2, 0) ( 2 , 0 ) has gradient
0 − 3 2 − 0 = − 3 2 \frac{0 - 3}{2 - 0} = -\frac32 2 − 0 0 − 3 = − 2 3 and
y y y -intercept 3, so
y = − 3 2 x + 3 y = -\frac32x + 3 y = − 2 3 x + 3 .
Multiply by 2 and rearrange:
3 x + 2 y = 6 3x + 2y = 6 3 x + 2 y = 6 .
Test the origin:
0 ≥ 6 0 \ge 6 0 ≥ 6 is false, and the origin is not shaded, so the region is
3 x + 2 y ≥ 6 3x + 2y \ge 6 3 x + 2 y ≥ 6 .
The shading is in the first quadrant, so
x ≥ 0 x \ge 0 x ≥ 0 and
y ≥ 0 y \ge 0 y ≥ 0 too: option D.
Watch out
Check the boundary with a point: ( 2 , 0 ) (2, 0) ( 2 , 0 ) 3 ( 2 ) + 2 ( 0 ) = 6 3(2) + 2(0) = 6 3 ( 2 ) + 2 ( 0 ) = 6 3 y + 2 x 3y + 2x 3 y + 2 x 4 4 4 ( 3 , 0 ) (3, 0) ( 3 , 0 ) Report a problem with this question
If p + 1 p + 1 p + 1 2 p − 10 2p - 10 2 p − 10 1 − 4 p 2 1 - 4p^2 1 − 4 p 2 p p p
A − 4 , 2 -4, 2 − 4 , 2 B − 2 , 4 11 -2, \frac4{11} − 2 , 11 4 C − 11 4 , 2 -\frac{11}{4}, 2 − 4 11 , 2 D 5 , − 3 5, -3 5 , − 3
Worked solution (try it first) In an A.P. the middle term is the average of its neighbours, so
2 ( 2 p − 10 ) = ( p + 1 ) + ( 1 − 4 p 2 ) 2(2p - 10) = (p + 1) + (1 - 4p^2) 2 ( 2 p − 10 ) = ( p + 1 ) + ( 1 − 4 p 2 ) .
Tidy up:
4 p − 20 = 2 + p − 4 p 2 4p - 20 = 2 + p - 4p^2 4 p − 20 = 2 + p − 4 p 2 , so
4 p 2 + 3 p − 22 = 0 4p^2 + 3p - 22 = 0 4 p 2 + 3 p − 22 = 0 .
Factorise:
( 4 p + 11 ) ( p − 2 ) = 0 (4p + 11)(p - 2) = 0 ( 4 p + 11 ) ( p − 2 ) = 0 .
So
p = − 11 4 p = -\frac{11}{4} p = − 4 11 or
p = 2 p = 2 p = 2 , option C.
Watch out
Double the middle term: 2 × ( 2 p − 10 ) = 4 p − 20 2 \times (2p - 10) = 4p - 20 2 × ( 2 p − 10 ) = 4 p − 20 2 p 2p 2 p 4 p − 10 4p - 10 4 p − 10 Report a problem with this question
The sum of the first three terms of a geometric progression is half its sum to infinity. Find the positive common ratio of the progression.
A 1 4 \frac14 4 1 B 1 2 \frac12 2 1 C 1 3 3 \frac{1}{\sqrt[3]{3}} 3 3 1 D 1 2 3 \frac{1}{\sqrt[3]{2}} 3 2 1
Worked solution (try it first) The sum of three terms is
a ( 1 − r 3 ) 1 − r \dfrac{a(1 - r^3)}{1 - r} 1 − r a ( 1 − r 3 ) and the sum to infinity is
a 1 − r \dfrac{a}{1 - r} 1 − r a .
Set the first equal to half the second and cancel
a 1 − r \dfrac{a}{1 - r} 1 − r a :
1 − r 3 = 1 2 1 - r^3 = \frac12 1 − r 3 = 2 1 .
So
r 3 = 1 2 r^3 = \frac12 r 3 = 2 1 , and taking the cube root gives
r = 1 2 3 r = \dfrac{1}{\sqrt[3]{2}} r = 3 2 1 , option D.
Watch out
r 3 = 1 2 r^3 = \frac12 r 3 = 2 1 r = 1 2 r = \frac12 r = 2 1 Report a problem with this question
The identity element with respect to the multiplication shown in the table is
⊗ \otimes ⊗ p p p q q q r r r s s s
p p p r r r p p p r r r p p p
q q q p p p q q q r r r s s s
r r r r r r r r r r r r r r r
s s s q q q s s s r r r q q q
Worked solution (try it first) The identity
e e e leaves every element unchanged, so its row and its column must repeat the headings
p , q , r , s p, q, r, s p , q , r , s .
The row for
q q q reads
p , q , r , s p, q, r, s p , q , r , s , and so does the column under
q q q .
So the identity is
q q q , option B.
Watch out
r r r r r r r r r Also set as JAMB 2016 · UTME · Q29
Report a problem with this question
The binary operation ∗ * ∗ x ∗ y = x y − y − x x * y = xy - y - x x ∗ y = x y − y − x x x x y y y x ∗ 3 = 2 ∗ x x * 3 = 2 * x x ∗ 3 = 2 ∗ x x x x
Worked solution (try it first) Use the rule with
y = 3 y = 3 y = 3 :
x ∗ 3 = 3 x − 3 − x = 2 x − 3 x * 3 = 3x - 3 - x = 2x - 3 x ∗ 3 = 3 x − 3 − x = 2 x − 3 .
Use it again with 2 first:
2 ∗ x = 2 x − x − 2 = x − 2 2 * x = 2x - x - 2 = x - 2 2 ∗ x = 2 x − x − 2 = x − 2 .
Set them equal:
2 x − 3 = x − 2 2x - 3 = x - 2 2 x − 3 = x − 2 .
Take
x x x from both sides and add 3:
x = 1 x = 1 x = 1 , option C.
Watch out
Watch the signs when you collect the numbers: 2 x − x = − 2 + 3 2x - x = -2 + 3 2 x − x = − 2 + 3 x = 1 x = 1 x = 1 x = 2 − 3 x = 2 - 3 x = 2 − 3 − 1 -1 − 1 Report a problem with this question
The determinant of the matrix ( x 1 0 1 − x 2 3 1 1 + x 4 ) \begin{pmatrix} x & 1 & 0 \\ 1 - x & 2 & 3 \\ 1 & 1 + x & 4 \end{pmatrix} x 1 − x 1 1 2 1 + x 0 3 4 x x x
A − 3 x 2 − 17 -3x^2 - 17 − 3 x 2 − 17 B − 3 x 2 + 9 x − 1 -3x^2 + 9x - 1 − 3 x 2 + 9 x − 1 C 3 x 2 + 17 3x^2 + 17 3 x 2 + 17 D 3 x 2 − 9 x + 5 3x^2 - 9x + 5 3 x 2 − 9 x + 5
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
x ( 2 × 4 − 3 ( 1 + x ) ) = x ( 5 − 3 x ) x(2 \times 4 - 3(1 + x)) = x(5 - 3x) x ( 2 × 4 − 3 ( 1 + x )) = x ( 5 − 3 x ) , which is
5 x − 3 x 2 5x - 3x^2 5 x − 3 x 2 .
The second term is
− 1 × ( 4 ( 1 − x ) − 3 × 1 ) = − ( 1 − 4 x ) -1 \times (4(1 - x) - 3 \times 1) = -(1 - 4x) − 1 × ( 4 ( 1 − x ) − 3 × 1 ) = − ( 1 − 4 x ) , which is
4 x − 1 4x - 1 4 x − 1 .
The third term is 0, because the entry is 0.
Add them:
5 x − 3 x 2 + 4 x − 1 = − 3 x 2 + 9 x − 1 5x - 3x^2 + 4x - 1 = -3x^2 + 9x - 1 5 x − 3 x 2 + 4 x − 1 = − 3 x 2 + 9 x − 1 , option B.
Watch out
The middle term takes a minus sign: − 1 × ( 1 − 4 x ) = 4 x − 1 -1 \times (1 - 4x) = 4x - 1 − 1 × ( 1 − 4 x ) = 4 x − 1 − 3 x 2 + x + 1 -3x^2 + x + 1 − 3 x 2 + x + 1 Report a problem with this question
Let I = ( 1 0 0 1 ) I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} I = ( 1 0 0 1 ) P = ( 2 3 4 5 ) P = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} P = ( 2 4 3 5 ) Q = ( u 4 + u − 2 v v ) Q = \begin{pmatrix} u & 4 + u \\ -2v & v \end{pmatrix} Q = ( u − 2 v 4 + u v ) 2 × 2 2 \times 2 2 × 2 P Q = I PQ = I P Q = I ( u , v ) (u, v) ( u , v )
A ( − 5 2 , − 1 ) \left(-\frac52, -1\right) ( − 2 5 , − 1 ) B ( − 5 2 , 3 2 ) \left(-\frac52, \frac32\right) ( − 2 5 , 2 3 ) C ( − 5 6 , 1 ) \left(-\frac56, 1\right) ( − 6 5 , 1 ) D ( 5 2 , 2 3 ) \left(\frac52, \frac23\right) ( 2 5 , 3 2 )
Worked solution (try it first) Row 1 of
P P P times column 1 of
Q Q Q must give the 1 in
I I I :
2 u + 3 ( − 2 v ) = 1 2u + 3(-2v) = 1 2 u + 3 ( − 2 v ) = 1 , so
2 u − 6 v = 1 2u - 6v = 1 2 u − 6 v = 1 .
Row 2 of
P P P times column 1 of
Q Q Q must give 0:
4 u − 10 v = 0 4u - 10v = 0 4 u − 10 v = 0 , so
u = 5 2 v u = \frac52 v u = 2 5 v .
Substitute:
5 v − 6 v = 1 5v - 6v = 1 5 v − 6 v = 1 , so
v = − 1 v = -1 v = − 1 and
u = − 5 2 u = -\frac52 u = − 2 5 .
So
( u , v ) = ( − 5 2 , − 1 ) (u, v) = \left(-\frac52, -1\right) ( u , v ) = ( − 2 5 , − 1 ) , option A.
Watch out
Keep the sign of the entry − 2 v -2v − 2 v 3 × ( − 2 v ) = − 6 v 3 \times (-2v) = -6v 3 × ( − 2 v ) = − 6 v + 6 v +6v + 6 v v = 1 v = 1 v = 1 ( − 5 2 , 1 ) \left(-\frac52, 1\right) ( − 2 5 , 1 ) Report a problem with this question
In the diagram, P R PR P R P Q R S PQRS P QR S P S T PST P S T Q R T QRT QR T ∠ S P R = 35 ∘ \angle SPR = 35^\circ ∠ S P R = 3 5 ∘ ∠ P T Q = 30 ∘ \angle PTQ = 30^\circ ∠ P T Q = 3 0 ∘ ∠ Q S R \angle QSR ∠ QS R
A 20 ∘ 20^\circ 2 0 ∘ B 25 ∘ 25^\circ 2 5 ∘ C 30 ∘ 30^\circ 3 0 ∘ D 35 ∘ 35^\circ 3 5 ∘
Worked solution (try it first) P R PR P R is a diameter, so
∠ P S R = 90 ∘ \angle PSR = 90^\circ ∠ P S R = 9 0 ∘ .
P S T PST P S T is a straight line, so
∠ R S T = 90 ∘ \angle RST = 90^\circ ∠ R S T = 9 0 ∘ too.
Triangle
R S T RST R S T :
∠ S R T = 180 ∘ − 90 ∘ − 30 ∘ \angle SRT = 180^\circ - 90^\circ - 30^\circ ∠ S R T = 18 0 ∘ − 9 0 ∘ − 3 0 ∘ Q R T QRT QR T is straight, so
∠ Q R S = 180 ∘ − 60 ∘ \angle QRS = 180^\circ - 60^\circ ∠ QR S = 18 0 ∘ − 6 0 ∘ Opposite angles of cyclic quadrilateral
P Q R S PQRS P QR S add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ Q P S = 60 ∘ \angle QPS = 60^\circ ∠ QP S = 6 0 ∘ and
∠ Q P R = 60 ∘ − 35 ∘ \angle QPR = 60^\circ - 35^\circ ∠ QP R = 6 0 ∘ − 3 5 ∘ ∠ Q S R \angle QSR ∠ QS R and
∠ Q P R \angle QPR ∠ QP R both stand on arc
Q R QR QR , so
∠ Q S R = 25 ∘ \angle QSR = 25^\circ ∠ QS R = 2 5 ∘ , option B.
Watch out
∠ Q S R \angle QSR ∠ QS R Q R QR QR ∠ Q P R \angle QPR ∠ QP R ∠ S P R = 35 ∘ \angle SPR = 35^\circ ∠ S P R = 3 5 ∘ S R SR S R Report a problem with this question
In the diagram, P Q ∥ S T PQ \parallel ST P Q ∥ S T ∠ P Q R = 120 ∘ \angle PQR = 120^\circ ∠ P QR = 12 0 ∘ ∠ R S T = 130 ∘ \angle RST = 130^\circ ∠ R S T = 13 0 ∘ x x x
A 50 ∘ 50^\circ 5 0 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 70 ∘ 70^\circ 7 0 ∘ D 80 ∘ 80^\circ 8 0 ∘
Worked solution (try it first) Draw a line through
R R R parallel to
P Q PQ P Q and
S T ST S T .
∠ P Q R \angle PQR ∠ P QR and the angle between
R Q RQ R Q and that line are co-interior:
180 ∘ − 120 ∘ = 60 ∘ 180^\circ - 120^\circ = 60^\circ 18 0 ∘ − 12 0 ∘ = 6 0 ∘ .
∠ R S T \angle RST ∠ R S T and the angle between
R S RS R S and that line are co-interior:
180 ∘ − 130 ∘ = 50 ∘ 180^\circ - 130^\circ = 50^\circ 18 0 ∘ − 13 0 ∘ = 5 0 ∘ .
The
60 ∘ 60^\circ 6 0 ∘ ,
x x x and
50 ∘ 50^\circ 5 0 ∘ make a straight line at
R R R , so
x = 180 ∘ − 60 ∘ − 50 ∘ x = 180^\circ - 60^\circ - 50^\circ x = 18 0 ∘ − 6 0 ∘ − 5 0 ∘ = 70 ∘ = 70^\circ = 7 0 ∘ , option C.
Watch out
50 ∘ 50^\circ 5 0 ∘ R S RS R S x x x Report a problem with this question
In the figure, P Q S T PQST P QS T T S R TSR T S R T S = 10 TS = 10 T S = 10 S R = 8 SR = 8 S R = 8 △ Q R S \triangle QRS △ QR S 20 cm 2 20\text{ cm}^2 20 cm 2 P Q R T PQRT P QR T
A 35 cm 2 35\text{ cm}^2 35 cm 2 B 65 cm 2 65\text{ cm}^2 65 cm 2 C 70 cm 2 70\text{ cm}^2 70 cm 2 D 140 cm 2 140\text{ cm}^2 140 cm 2
Worked solution (try it first) △ Q R S \triangle QRS △ QR S has base
S R = 8 SR = 8 S R = 8 cm:
1 2 × 8 × h = 20 \frac12 \times 8 \times h = 20 2 1 × 8 × h = 20 , so
h = 5 h = 5 h = 5 cm.
This is also the height of the trapezium.
P Q S T PQST P QS T is a parallelogram, so
P Q = T S = 10 PQ = TS = 10 P Q = T S = 10 cm.
The other parallel side is
T R = 10 + 8 = 18 TR = 10 + 8 = 18 T R = 10 + 8 = 18 cm.
Area of the trapezium
= 1 2 ( 10 + 18 ) × 5 = \frac12(10 + 18) \times 5 = 2 1 ( 10 + 18 ) × 5 = 70 cm 2 = 70\text{ cm}^2 = 70 cm 2 , option C.
Watch out
Keep the 1 2 \frac12 2 1 ( 10 + 18 ) × 5 = 140 cm 2 (10 + 18) \times 5 = 140\text{ cm}^2 ( 10 + 18 ) × 5 = 140 cm 2 Also set as JAMB 2016 · UTME · Q30
Report a problem with this question
T Q TQ T Q X Y T R XYTR X Y T R ∠ Y X T = 32 ∘ \angle YXT = 32^\circ ∠ Y X T = 3 2 ∘ ∠ R T Q = 40 ∘ \angle RTQ = 40^\circ ∠ R T Q = 4 0 ∘ ∠ Y T R \angle YTR ∠ Y T R
A 108 ∘ 108^\circ 10 8 ∘ B 121 ∘ 121^\circ 12 1 ∘ C 140 ∘ 140^\circ 14 0 ∘ D 148 ∘ 148^\circ 14 8 ∘
Worked solution (try it first) The angle between tangent
T Q TQ T Q and chord
T R TR T R equals the angle in the alternate segment, so
∠ T X R = ∠ R T Q = 40 ∘ \angle TXR = \angle RTQ = 40^\circ ∠ T X R = ∠ R T Q = 4 0 ∘ .
So the whole angle at
X X X is
∠ Y X R = 32 ∘ + 40 ∘ \angle YXR = 32^\circ + 40^\circ ∠ Y X R = 3 2 ∘ + 4 0 ∘ Opposite angles of cyclic quadrilateral
X Y T R XYTR X Y T R add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ Y T R = 180 ∘ − 72 ∘ \angle YTR = 180^\circ - 72^\circ ∠ Y T R = 18 0 ∘ − 7 2 ∘ = 108 ∘ = 108^\circ = 10 8 ∘ , option A.
Watch out
∠ Y X R \angle YXR ∠ Y X R X X X 32 ∘ + 40 ∘ 32^\circ + 40^\circ 3 2 ∘ + 4 0 ∘ 40 ∘ 40^\circ 4 0 ∘ 180 ∘ − 40 ∘ = 140 ∘ 180^\circ - 40^\circ = 140^\circ 18 0 ∘ − 4 0 ∘ = 14 0 ∘ Report a problem with this question
A chord of a circle of radius 3 \sqrt3 3 60 ∘ 60^\circ 6 0 ∘
A 3 2 \frac{\sqrt3}{2} 2 3 B 3 2 \frac32 2 3 C 3 \sqrt3 3 D 3 cm
Worked solution (try it first) The angle at the centre is twice the angle at the circumference, so the chord makes
120 ∘ 120^\circ 12 0 ∘ at the centre.
The perpendicular from the centre halves the chord and the
120 ∘ 120^\circ 12 0 ∘ .
So half the chord is
3 sin 60 ∘ = 3 × 3 2 \sqrt3\sin60^\circ = \sqrt3 \times \frac{\sqrt3}{2} 3 sin 6 0 ∘ = 3 × 2 3 So the chord is
2 × 3 2 = 3 2 \times \frac32 = 3 2 × 2 3 = 3 cm, option D.
Watch out
A chord equal to the radius, 3 \sqrt3 3 60 ∘ 60^\circ 6 0 ∘ 60 ∘ 60^\circ 6 0 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question
A cylindrical drum of diameter 56 cm contains 123.2 litres of oil when full. Find the height of the drum in centimetres. [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
Worked solution (try it first) Change litres to cm³: 1 litre is
1000 cm 3 1000\text{ cm}^3 1000 cm 3 , so the drum holds
123 200 cm 3 123\,200\text{ cm}^3 123 200 cm 3 .
The radius is
56 ÷ 2 = 28 56 \div 2 = 28 56 ÷ 2 = 28 cm.
Base area:
22 7 × 28 2 = 2464 cm 2 \frac{22}{7} \times 28^2 = 2464\text{ cm}^2 7 22 × 2 8 2 = 2464 cm 2 .
Volume = base area × height, so
h = 123 200 2464 = 50 h = \dfrac{123\,200}{2464} = 50 h = 2464 123 200 = 50 cm, option D.
Watch out
Halve the diameter. Using 56 cm as the radius makes the base four times too big and gives 12.5 cm (option A). Report a problem with this question
The locus of all points at a distance 8 cm from a point N N N T T T S S S S S S T T T N N N S T N STN S T N
A 4 3 cm 2 4\sqrt3\text{ cm}^2 4 3 cm 2 B 16 3 cm 2 16\sqrt3\text{ cm}^2 16 3 cm 2 C 32 cm 2 32\text{ cm}^2 32 cm 2 D 64 cm 2 64\text{ cm}^2 64 cm 2
Worked solution (try it first) The locus of points 8 cm from
N N N is a circle of radius 8 cm, so
N T = N S = 8 NT = NS = 8 N T = N S = 8 cm.
S S S is equidistant from
T T T and
N N N , so
S T = S N = 8 ST = SN = 8 S T = S N = 8 cm.
All three sides are 8 cm: the triangle is equilateral.
Use area
= 1 2 a b sin C = \frac12ab\sin C = 2 1 ab sin C with the
60 ∘ 60^\circ 6 0 ∘ angle:
1 2 × 8 × 8 × sin 60 ∘ = 32 × 3 2 \frac12 \times 8 \times 8 \times \sin60^\circ = 32 \times \frac{\sqrt3}{2} 2 1 × 8 × 8 × sin 6 0 ∘ = 32 × 2 3 .
So the area is
16 3 cm 2 16\sqrt3\text{ cm}^2 16 3 cm 2 , option B.
Watch out
The angle between the sides is 60 ∘ 60^\circ 6 0 ∘ 90 ∘ 90^\circ 9 0 ∘ 1 2 × 8 × 8 \frac12 \times 8 \times 8 2 1 × 8 × 8 sin 60 ∘ \sin60^\circ sin 6 0 ∘ Report a problem with this question
If the distance between the points ( x , 3 ) (x, 3) ( x , 3 ) ( − x , 2 ) (-x, 2) ( − x , 2 ) x x x
A 6.0 B 2.5 C 6 \sqrt6 6 D 3 \sqrt3 3
Worked solution (try it first) The changes are
x − ( − x ) = 2 x x - (-x) = 2x x − ( − x ) = 2 x and
3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 .
By the distance formula,
( 2 x ) 2 + 1 2 = 5 2 (2x)^2 + 1^2 = 5^2 ( 2 x ) 2 + 1 2 = 5 2 , so
4 x 2 + 1 = 25 4x^2 + 1 = 25 4 x 2 + 1 = 25 .
Then
4 x 2 = 24 4x^2 = 24 4 x 2 = 24 and
x 2 = 6 x^2 = 6 x 2 = 6 .
Take the positive square root:
x = 6 x = \sqrt6 x = 6 , option C.
Watch out
Finish by taking the square root: x 2 = 6 x^2 = 6 x 2 = 6 x = 6 x = \sqrt6 x = 6 Report a problem with this question
The midpoint of the segment of the line y = 4 x + 3 y = 4x + 3 y = 4 x + 3 x x x y y y
A ( − 3 2 , 3 2 ) \left(-\frac32, \frac32\right) ( − 2 3 , 2 3 ) B ( − 2 3 , 3 2 ) \left(-\frac23, \frac32\right) ( − 3 2 , 2 3 ) C ( 3 8 , 3 2 ) \left(\frac38, \frac32\right) ( 8 3 , 2 3 ) D ( − 3 8 , 3 2 ) \left(-\frac38, \frac32\right) ( − 8 3 , 2 3 )
Worked solution (try it first) It meets the
x x x -axis where
y = 0 y = 0 y = 0 :
4 x + 3 = 0 4x + 3 = 0 4 x + 3 = 0 , so
x = − 3 4 x = -\frac34 x = − 4 3 .
That end is
( − 3 4 , 0 ) \left(-\frac34, 0\right) ( − 4 3 , 0 ) .
It meets the
y y y -axis where
x = 0 x = 0 x = 0 :
y = 3 y = 3 y = 3 .
That end is
( 0 , 3 ) (0, 3) ( 0 , 3 ) .
Average the ends:
( − 3 8 , 3 2 ) \left(-\frac38, \frac32\right) ( − 8 3 , 2 3 ) , option D.
Watch out
Keep the sign: 4 x + 3 = 0 4x + 3 = 0 4 x + 3 = 0 x = − 3 4 x = -\frac34 x = − 4 3 x x x − 3 8 -\frac38 − 8 3 Similar: JAMB 2014 · UTME · Q35
Report a problem with this question
Solve the equation cos x + sin x = 1 cos x − sin x \cos x + \sin x = \dfrac{1}{\cos x - \sin x} cos x + sin x = cos x − sin x 1 0 ≤ x < 2 π 0 \le x < 2\pi 0 ≤ x < 2 π
A π 2 , 3 π 2 \frac\pi2, \frac{3\pi}{2} 2 π , 2 3 π B π 3 , 2 π 3 \frac\pi3, \frac{2\pi}{3} 3 π , 3 2 π C 0 , π 3 0, \frac\pi3 0 , 3 π D 0 , π 0, \pi 0 , π
Worked solution (try it first) Multiply both sides by
cos x − sin x \cos x - \sin x cos x − sin x :
( cos x + sin x ) ( cos x − sin x ) = 1 (\cos x + \sin x)(\cos x - \sin x) = 1 ( cos x + sin x ) ( cos x − sin x ) = 1 , so
cos 2 x − sin 2 x = 1 \cos^2 x - \sin^2 x = 1 cos 2 x − sin 2 x = 1 .
The double-angle formula says
cos 2 x − sin 2 x = cos 2 x \cos^2 x - \sin^2 x = \cos2x cos 2 x − sin 2 x = cos 2 x , so
cos 2 x = 1 \cos2x = 1 cos 2 x = 1 .
For
0 ≤ x < 2 π 0 \le x < 2\pi 0 ≤ x < 2 π ,
2 x 2x 2 x runs from 0 to
4 π 4\pi 4 π , and
cos 2 x = 1 \cos2x = 1 cos 2 x = 1 at
2 x = 0 2x = 0 2 x = 0 or
2 π 2\pi 2 π .
So
x = 0 x = 0 x = 0 or
π \pi π , option D.
(At both,
cos x − sin x \cos x - \sin x cos x − sin x is not zero, so the equation makes sense.)
Watch out
Check the options in the equation before choosing. At x = π 2 x = \frac\pi2 x = 2 π 1 0 − 1 = − 1 \frac{1}{0 - 1} = -1 0 − 1 1 = − 1 Report a problem with this question
In the diagram, Q T R QTR QT R P Q = 15 PQ = 15 P Q = 15 P R = 10 PR = 10 P R = 10 P T = 8 PT = 8 P T = 8 ∠ P Q T = 30 ∘ \angle PQT = 30^\circ ∠ P QT = 3 0 ∘ ∠ P T R \angle PTR ∠ P T R
A 8 15 \frac8{15} 15 8 B 2 3 \frac23 3 2 C 3 4 \frac34 4 3 D 15 16 \frac{15}{16} 16 15
Worked solution (try it first) Work in triangle
P Q T PQT P QT .
P Q = 15 PQ = 15 P Q = 15 faces
∠ P T Q \angle PTQ ∠ P T Q , and
P T = 8 PT = 8 P T = 8 faces the
30 ∘ 30^\circ 3 0 ∘ at
Q Q Q .
Sine rule:
sin ∠ P T Q 15 = sin 30 ∘ 8 \dfrac{\sin\angle PTQ}{15} = \dfrac{\sin30^\circ}{8} 15 sin ∠ P T Q = 8 sin 3 0 ∘ .
So
sin ∠ P T Q = 15 × 1 2 8 \sin\angle PTQ = \dfrac{15 \times \frac12}{8} sin ∠ P T Q = 8 15 × 2 1 = 15 16 = \frac{15}{16} = 16 15 .
Q T R QTR QT R is a straight line, so
∠ P T R = 180 ∘ − ∠ P T Q \angle PTR = 180^\circ - \angle PTQ ∠ P T R = 18 0 ∘ − ∠ P T Q , and
sin ( 180 ∘ − A ) = sin A \sin(180^\circ - A) = \sin A sin ( 18 0 ∘ − A ) = sin A .
So
sin ∠ P T R = 15 16 \sin\angle PTR = \frac{15}{16} sin ∠ P T R = 16 15 , option D.
Watch out
No angle here is a right angle, so a plain side ratio such as P T P Q = 8 15 \frac{PT}{PQ} = \frac{8}{15} P Q P T = 15 8 P Q T PQT P QT P R = 10 PR = 10 P R = 10 Report a problem with this question
For what value of x x x 6 sin ( 2 x − 25 ) ∘ 6\sin(2x - 25)^\circ 6 sin ( 2 x − 25 ) ∘ 0 ∘ ≤ x ≤ 180 ∘ 0^\circ \le x \le 180^\circ 0 ∘ ≤ x ≤ 18 0 ∘
A 12 1 2 12\frac12 12 2 1 B 32 1 2 32\frac12 32 2 1 C 57 1 2 57\frac12 57 2 1 D 147 1 2 147\frac12 147 2 1
Worked solution (try it first) sin θ \sin\theta sin θ reaches its maximum value, 1, when
θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ .
So set
2 x − 25 = 90 2x - 25 = 90 2 x − 25 = 90 .
Add 25 to both sides:
2 x = 115 2x = 115 2 x = 115 .
Divide by 2:
x = 57 1 2 x = 57\frac12 x = 57 2 1 , option C.
The next maximum,
2 x − 25 = 450 2x - 25 = 450 2 x − 25 = 450 , gives
x = 237 1 2 x = 237\frac12 x = 237 2 1 , outside the range.
Watch out
Add the 25 when you move it across: 2 x = 90 + 25 = 115 2x = 90 + 25 = 115 2 x = 90 + 25 = 115 2 x = 65 2x = 65 2 x = 65 x = 32 1 2 x = 32\frac12 x = 32 2 1 Report a problem with this question
From the top of a vertical mast 150 m high, two huts on the same ground level are observed, one due east and the other due west of the mast. Their angles of depression are 60 ∘ 60^\circ 6 0 ∘ 45 ∘ 45^\circ 4 5 ∘
A 150 ( 1 + 3 ) 150(1 + \sqrt3) 150 ( 1 + 3 ) B 50 ( 3 + 3 ) 50(3 + \sqrt3) 50 ( 3 + 3 ) C 150 3 150\sqrt3 150 3 D 50 3 \frac{50}{\sqrt3} 3 50
Worked solution (try it first) The huts are on opposite sides of the mast, so the distance between them is the sum of their distances from its foot.
East hut:
150 tan 60 ∘ = 150 3 \frac{150}{\tan60^\circ} = \frac{150}{\sqrt3} t a n 6 0 ∘ 150 = 3 150 West hut:
150 tan 45 ∘ = 150 \frac{150}{\tan45^\circ} = 150 t a n 4 5 ∘ 150 = 150 m.
So the distance is
150 + 50 3 = 50 ( 3 + 3 ) 150 + 50\sqrt3 = 50(3 + \sqrt3) 150 + 50 3 = 50 ( 3 + 3 ) m, option B.
Watch out
Divide the height by tan \tan tan 150 3 + 150 = 150 ( 1 + 3 ) 150\sqrt3 + 150 = 150(1 + \sqrt3) 150 3 + 150 = 150 ( 1 + 3 ) Report a problem with this question
If y = 243 ( 4 x + 5 ) − 2 y = 243(4x + 5)^{-2} y = 243 ( 4 x + 5 ) − 2 d y d x \frac{dy}{dx} d x d y x = 1 x = 1 x = 1
A − 8 3 -\frac83 − 3 8 B 3 8 \frac38 8 3 C 9 8 \frac98 8 9 D − 8 9 -\frac89 − 9 8
Worked solution (try it first) Chain rule: bring down the power
− 2 -2 − 2 and multiply by 4, the derivative of
4 x + 5 4x + 5 4 x + 5 .
So
d y d x = 243 × ( − 2 ) × 4 × ( 4 x + 5 ) − 3 \frac{dy}{dx} = 243 \times (-2) \times 4 \times (4x + 5)^{-3} d x d y = 243 × ( − 2 ) × 4 × ( 4 x + 5 ) − 3 = − 1944 ( 4 x + 5 ) − 3 = -1944(4x + 5)^{-3} = − 1944 ( 4 x + 5 ) − 3 .
At
x = 1 x = 1 x = 1 ,
4 x + 5 = 9 4x + 5 = 9 4 x + 5 = 9 and
9 3 = 729 9^3 = 729 9 3 = 729 .
So
d y d x = − 1944 729 \frac{dy}{dx} = -\frac{1944}{729} d x d y = − 729 1944 = − 8 3 = -\frac83 = − 3 8 , option A.
Watch out
Multiply by the 4 from the inside bracket. Without it you get − 486 729 = − 2 3 -\frac{486}{729} = -\frac23 − 729 486 = − 3 2 Report a problem with this question
Differentiate x cos x \dfrac{x}{\cos x} cos x x x x x
A 1 + x sec x tan x 1 + x\sec x\tan x 1 + x sec x tan x B 1 + sec 2 x 1 + \sec^2x 1 + sec 2 x C cos x + x tan x \cos x + x\tan x cos x + x tan x D sec x + x sec x tan x \sec x + x\sec x\tan x sec x + x sec x tan x
Worked solution (try it first) Write
x cos x = x sec x \dfrac{x}{\cos x} = x\sec x cos x x = x sec x , and recall that
sec x \sec x sec x differentiates to
sec x tan x \sec x\tan x sec x tan x .
Product rule:
1 ⋅ sec x + x ⋅ sec x tan x 1 \cdot \sec x + x \cdot \sec x\tan x 1 ⋅ sec x + x ⋅ sec x tan x .
So the derivative is
sec x + x sec x tan x \sec x + x\sec x\tan x sec x + x sec x tan x , option D.
Watch out
In the product rule the first term is d d x ( x ) × sec x = sec x \frac{d}{dx}(x) \times \sec x = \sec x d x d ( x ) × sec x = sec x sec x \sec x sec x Report a problem with this question
Find the equation of the curve which passes through the point ( 2 , 5 ) (2, 5) ( 2 , 5 ) 6 x − 5 6x - 5 6 x − 5
A 6 x 2 − 5 x + 5 6x^2 - 5x + 5 6 x 2 − 5 x + 5 B 6 x 2 + 5 x + 5 6x^2 + 5x + 5 6 x 2 + 5 x + 5 C 3 x 2 − 5 x − 5 3x^2 - 5x - 5 3 x 2 − 5 x − 5 D 3 x 2 − 5 x + 3 3x^2 - 5x + 3 3 x 2 − 5 x + 3
Worked solution (try it first) Integrate the gradient:
y = 3 x 2 − 5 x + c y = 3x^2 - 5x + c y = 3 x 2 − 5 x + c .
The curve passes through
( 2 , 5 ) (2, 5) ( 2 , 5 ) :
5 = 12 − 10 + c 5 = 12 - 10 + c 5 = 12 − 10 + c , so
c = 3 c = 3 c = 3 .
So
y = 3 x 2 − 5 x + 3 y = 3x^2 - 5x + 3 y = 3 x 2 − 5 x + 3 , option D.
Watch out
6 x 6x 6 x 3 x 2 3x^2 3 x 2 6 x 2 6x^2 6 x 2 Also set as JAMB 2016 · UTME · Q32
Report a problem with this question
If m m m n n n m + 2 n m + 2n m + 2 n
Worked solution (try it first) The numbers add up to 47, so the mean is
m = 47 8 = 5.875 m = \frac{47}{8} = 5.875 m = 8 47 = 5.875 .
In order: 2, 3, 5, 6, 7, 7, 8, 9.
The median is halfway between the 4th and 5th:
n = 6 + 7 2 = 6.5 n = \frac{6 + 7}{2} = 6.5 n = 2 6 + 7 = 6.5 .
So
m + 2 n = 5.875 + 13 = 18.875 m + 2n = 5.875 + 13 = 18.875 m + 2 n = 5.875 + 13 = 18.875 , which is 19 to the nearest whole number, option A.
Watch out
Double the median, not the mean: 2 m + n = 11.75 + 6.5 = 18.25 2m + n = 11.75 + 6.5 = 18.25 2 m + n = 11.75 + 6.5 = 18.25 Report a problem with this question
Estimate the mode of the frequency distribution below.
Average hourly earnings (₦)
5–9
10–14
15–19
20–24
No. of workers
17
32
25
24
Worked solution (try it first) The modal class is 10–14, with 32 workers.
Its class boundaries are 9.5 and 14.5, so
L = 9.5 L = 9.5 L = 9.5 and
c = 5 c = 5 c = 5 .
Differences from the neighbours:
Δ 1 = 32 − 17 = 15 \Delta_1 = 32 - 17 = 15 Δ 1 = 32 − 17 = 15 and
Δ 2 = 32 − 25 = 7 \Delta_2 = 32 - 25 = 7 Δ 2 = 32 − 25 = 7 .
Mode
= L + Δ 1 Δ 1 + Δ 2 × c = L + \frac{\Delta_1}{\Delta_1 + \Delta_2} \times c = L + Δ 1 + Δ 2 Δ 1 × c = 9.5 + 15 22 × 5 = 9.5 + \frac{15}{22} \times 5 = 9.5 + 22 15 × 5 , which is
9.5 + 3.41 9.5 + 3.41 9.5 + 3.41 .
So the mode is about ₦12.9, option C.
Watch out
Use the lower class boundary, 9.5, not the lower class limit 10. Starting from 10 gives 13.4 (option D). Report a problem with this question
Find the variance of the numbers K K K K + 1 K + 1 K + 1 K + 2 K + 2 K + 2
A 2 3 \frac23 3 2 B 1 C K + 1 K + 1 K + 1 D ( K + 1 ) 2 (K + 1)^2 ( K + 1 ) 2
Worked solution (try it first) The three numbers add up to
3 K + 3 3K + 3 3 K + 3 , so the mean is
3 K + 3 3 = K + 1 \frac{3K + 3}{3} = K + 1 3 3 K + 3 = K + 1 .
The deviations are
− 1 -1 − 1 , 0 and 1, so the squared deviations add up to 2.
The variance is
2 3 \frac23 3 2 , option A.
It does not depend on
K K K .
Watch out
Divide by the number of values, 3. Dividing by 2 gives 1 (option B). Report a problem with this question
Find the positive value of x x x 1 1 1 x + 1 x + 1 x + 1 2 x + 1 2x + 1 2 x + 1 6 \sqrt6 6
Worked solution (try it first) The three numbers add up to
3 x + 3 3x + 3 3 x + 3 , so the mean is
3 x + 3 3 = x + 1 \frac{3x + 3}{3} = x + 1 3 3 x + 3 = x + 1 .
The deviations are
− x -x − x , 0 and
x x x , so the variance is
x 2 + 0 + x 2 3 = 2 x 2 3 \frac{x^2 + 0 + x^2}{3} = \frac{2x^2}{3} 3 x 2 + 0 + x 2 = 3 2 x 2 .
The standard deviation is
6 \sqrt6 6 , so the variance is 6.
Then
2 x 2 3 = 6 \frac{2x^2}{3} = 6 3 2 x 2 = 6 gives
x 2 = 9 x^2 = 9 x 2 = 9 .
The positive value is
x = 3 x = 3 x = 3 , option C.
Watch out
Square the standard deviation first: the variance is 6, not 6 \sqrt6 6 2 x 2 = 6 2x^2 = 6 2 x 2 = 6 x = 3 x = \sqrt3 x = 3 Report a problem with this question
A bag contains 16 red balls and 20 blue balls only. How many white balls must be added to the bag so that the probability of randomly picking a red ball is 2 5 \frac25 5 2
Worked solution (try it first) Let
w w w white balls be added.
The total becomes
16 + 20 + w = 36 + w 16 + 20 + w = 36 + w 16 + 20 + w = 36 + w .
So
16 36 + w = 2 5 \frac{16}{36 + w} = \frac25 36 + w 16 = 5 2 .
Cross-multiply:
80 = 72 + 2 w 80 = 72 + 2w 80 = 72 + 2 w .
Take 72 from both sides:
2 w = 8 2w = 8 2 w = 8 , so
w = 4 w = 4 w = 4 , option A.
Watch out
The total includes the 20 blue balls. Writing 16 16 + w = 2 5 \frac{16}{16 + w} = \frac25 16 + w 16 = 5 2 w = 24 w = 24 w = 24 Report a problem with this question
The pie chart shows the monthly expenditure of a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7,200?
A ₦1,000 B ₦2,000 C ₦3,000 D ₦4,000
Worked solution (try it first) Transport is marked with a right angle, so it is
90 ∘ 90^\circ 9 0 ∘ .
Housing and School fees share
360 ∘ − 120 ∘ − 90 ∘ = 150 ∘ 360^\circ - 120^\circ - 90^\circ = 150^\circ 36 0 ∘ − 12 0 ∘ − 9 0 ∘ = 15 0 ∘ .
Housing is twice School fees, so split
150 ∘ 150^\circ 15 0 ∘ in the ratio
2 : 1 2 : 1 2 : 1 : Housing gets
2 3 × 150 ∘ = 100 ∘ \frac23 \times 150^\circ = 100^\circ 3 2 × 15 0 ∘ = 10 0 ∘ .
Housing costs
100 360 × 7200 = 2000 \frac{100}{360} \times 7200 = 2000 360 100 × 7200 = 2000 , so the worker spends ₦2,000, option B.
Watch out
Housing is the larger share, 100 ∘ 100^\circ 10 0 ∘ 50 ∘ 50^\circ 5 0 ∘ Report a problem with this question
The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?
A 59.4 % 59.4\% 59.4% B 50.0 % 50.0\% 50.0% C 41.7 % 41.7\% 41.7% D 25.0 % 25.0\% 25.0%
Worked solution (try it first) The pass mark is 5, so the pupils who failed scored 0, 1, 2, 3 or 4.
Read those bars:
1 + 3 + 4 + 7 + 10 = 25 1 + 3 + 4 + 7 + 10 = 25 1 + 3 + 4 + 7 + 10 = 25 pupils failed.
As a percentage of 60:
25 60 × 100 = 41.7 % \frac{25}{60} \times 100 = 41.7\% 60 25 × 100 = 41.7% , option C.
Watch out
A score of 4 is below the pass mark, so include its bar of 10 pupils. Stopping at 3 gives 15 60 = 25 % \frac{15}{60} = 25\% 60 15 = 25% Report a problem with this question
In a recent zonal championship involving 10 teams, teams X X X Y Y Y 2 5 \frac25 5 2 1 3 \frac13 3 1
A 2 15 \frac2{15} 15 2 B 7 15 \frac7{15} 15 7 C 11 15 \frac{11}{15} 15 11 D 13 15 \frac{13}{15} 15 13
Worked solution (try it first) Only one team can win the gold, so the two events can't happen together: they are mutually exclusive.
For mutually exclusive events, "either" means add:
2 5 + 1 3 = 6 15 + 5 15 \frac25 + \frac13 = \frac{6}{15} + \frac{5}{15} 5 2 + 3 1 = 15 6 + 15 5 .
So the probability is
11 15 \frac{11}{15} 15 11 , option C.
Watch out
"Either" means add here, not multiply. Multiplying gives 2 15 \frac{2}{15} 15 2 Report a problem with this question
If x x x y y y { 1 , 2 , 3 , 4 } \{1, 2, 3, 4\} { 1 , 2 , 3 , 4 } x x x y y y
A 5 8 \frac58 8 5 B 5 16 \frac5{16} 16 5 C 1 2 \frac12 2 1 D 3 8 \frac38 8 3
Worked solution (try it first) x x x and
y y y each have 4 values, so there are
4 × 4 = 16 4 \times 4 = 16 4 × 4 = 16 ordered pairs.
Products not greater than 6:
x = 1 x = 1 x = 1 with any
y y y (4 pairs),
x = 2 x = 2 x = 2 with
y = 1 , 2 , 3 y = 1, 2, 3 y = 1 , 2 , 3 (3 pairs),
x = 3 x = 3 x = 3 with
y = 1 , 2 y = 1, 2 y = 1 , 2 (2 pairs) and
( 4 , 1 ) (4, 1) ( 4 , 1 ) .
That is 10 pairs.
So the probability is
10 16 = 5 8 \frac{10}{16} = \frac58 16 10 = 8 5 , option A.
Watch out
"Not greater than 6" includes 6 itself. Leaving out ( 2 , 3 ) (2, 3) ( 2 , 3 ) ( 3 , 2 ) (3, 2) ( 3 , 2 ) 8 16 = 1 2 \frac{8}{16} = \frac12 16 8 = 2 1 Report a problem with this question