A trader bought goats for ₦4,000 each. He sold them for ₦180,000 at a loss of 25 % 25\% 25%
Worked solution (try it first) He sold at a
25 % 25\% 25% loss, so ₦180,000 is
75 % 75\% 75% of the cost: cost
= 180 000 ÷ 0.75 = = 180\,000 \div 0.75 = = 180 000 ÷ 0.75 = ₦240,000.
Each goat cost ₦4,000, so the number of goats is
240 000 ÷ 4000 = 60 240\,000 \div 4000 = 60 240 000 ÷ 4000 = 60 .
So he bought 60 goats, option D.
Watch out
₦180,000 is what he sold them for, not what he paid. Dividing it by ₦4,000 gives 45 (option B); find the cost price first. Also set as JAMB 2017 · UTME · Q21
Report a problem with this question
Simplify ( 0.7 + 70 ) 2 (\sqrt{0.7} + \sqrt{70})^2 ( 0.7 + 70 ) 2
Worked solution (try it first) Expand:
( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 , so this is
0.7 + 2 0.7 × 70 + 70 0.7 + 2\sqrt{0.7 \times 70} + 70 0.7 + 2 0.7 × 70 + 70 .
0.7 × 70 = 49 0.7 \times 70 = 49 0.7 × 70 = 49 , so the middle term is
2 49 = 14 2\sqrt{49} = 14 2 49 = 14 .
Add:
0.7 + 14 + 70 = 84.7 0.7 + 14 + 70 = 84.7 0.7 + 14 + 70 = 84.7 , option C.
Watch out
Don't forget the middle term 2 a b 2ab 2 ab 0.7 + 70 = 70.7 0.7 + 70 = 70.7 0.7 + 70 = 70.7 Report a problem with this question
Evaluate 0.21 × 0.072 × 0.0054 0.006 × 1.68 × 0.063 \dfrac{0.21 \times 0.072 \times 0.0054}{0.006 \times 1.68 \times 0.063} 0.006 × 1.68 × 0.063 0.21 × 0.072 × 0.0054
A 0.1286 B 0.1285 C 0.01286 D 0.01285
Worked solution (try it first) Cancel before multiplying:
0.21 0.063 = 10 3 \frac{0.21}{0.063} = \frac{10}{3} 0.063 0.21 = 3 10 ,
0.072 1.68 = 3 70 \frac{0.072}{1.68} = \frac{3}{70} 1.68 0.072 = 70 3 and
0.0054 0.006 = 0.9 \frac{0.0054}{0.006} = 0.9 0.006 0.0054 = 0.9 .
Multiply:
10 3 × 3 70 = 1 7 \frac{10}{3} \times \frac{3}{70} = \frac17 3 10 × 70 3 = 7 1 , and
1 7 × 0.9 = 9 70 \frac17 \times 0.9 = \frac{9}{70} 7 1 × 0.9 = 70 9 .
9 70 = 0.128571 … \frac{9}{70} = 0.128571\ldots 70 9 = 0.128571 … .
The fifth significant figure is 7, so round up: 0.1286, option A.
Watch out
Look at the next digit when rounding: 0.12857 has 7 after the 5, so it rounds up to 0.1286. Cutting off without rounding gives 0.1285 (option B). Report a problem with this question
In a school, 220 students offer Biology or Mathematics or both. 125 offer Biology and 110 offer Mathematics. How many offer Biology but not Mathematics?
Worked solution (try it first) Both subjects: add the two subjects and take away the total,
125 + 110 − 220 = 15 125 + 110 - 220 = 15 125 + 110 − 220 = 15 .
Biology but not Mathematics:
125 − 15 = 110 125 - 15 = 110 125 − 15 = 110 , option B.
Watch out
125 (option A) is everyone who offers Biology, including the 15 who also offer Mathematics. Take them away. Report a problem with this question
Simplify 52.4 − 5.7 − 3.45 − 1.75 52.4 - 5.7 - 3.45 - 1.75 52.4 − 5.7 − 3.45 − 1.75
Worked solution (try it first) Add the three numbers being taken away, lining up the decimal points:
5.7 + 3.45 + 1.75 = 10.9 5.7 + 3.45 + 1.75 = 10.9 5.7 + 3.45 + 1.75 = 10.9 .
Then
52.4 − 10.9 = 41.5 52.4 - 10.9 = 41.5 52.4 − 10.9 = 41.5 , option C.
Watch out
Line up the decimal points: write 5.7 as 5.70 and 52.4 as 52.40 before working with 3.45 and 1.75, or the hundredths get added into the wrong column. Report a problem with this question
Without using tables, evaluate ( 343 ) 1 3 × ( 0.14 ) − 1 × ( 25 ) − 1 2 (343)^{\frac13} \times (0.14)^{-1} \times (25)^{-\frac12} ( 343 ) 3 1 × ( 0.14 ) − 1 × ( 25 ) − 2 1
Worked solution (try it first) ( 343 ) 1 3 (343)^{\frac13} ( 343 ) 3 1 is the cube root of 343, which is 7.
( 0.14 ) − 1 (0.14)^{-1} ( 0.14 ) − 1 is the reciprocal of 0.14:
1 0.14 = 100 14 \frac{1}{0.14} = \frac{100}{14} 0.14 1 = 14 100 ( 25 ) − 1 2 = 1 25 (25)^{-\frac12} = \frac{1}{\sqrt{25}} ( 25 ) − 2 1 = 25 1 Multiply:
7 × 50 7 × 1 5 = 10 7 \times \frac{50}{7} \times \frac15 = 10 7 × 7 50 × 5 1 = 10 , option C.
Watch out
A negative index means the reciprocal, not a negative number: ( 25 ) − 1 2 = 1 5 (25)^{-\frac12} = \frac15 ( 25 ) − 2 1 = 5 1 − 5 -5 − 5 1 5 \frac15 5 1 Report a problem with this question
In the diagram are two concentric circles of radii r r r R R R O O O r = 2 5 R r = \frac25R r = 5 2 R π \pi π R R R
R r O
A 9 25 π R 2 \frac9{25}\pi R^2 25 9 π R 2 B 5 9 π R 2 \frac59\pi R^2 9 5 π R 2 C 21 25 π R 2 \frac{21}{25}\pi R^2 25 21 π R 2 D 21 23 π R 2 \frac{21}{23}\pi R^2 23 21 π R 2
Worked solution (try it first) The ring is the big circle minus the small one:
π R 2 − π r 2 \pi R^2 - \pi r^2 π R 2 − π r 2 .
With
r = 2 5 R r = \frac25R r = 5 2 R ,
r 2 = 4 25 R 2 r^2 = \frac4{25}R^2 r 2 = 25 4 R 2 .
So the ring is
π R 2 − 4 25 π R 2 = 21 25 π R 2 \pi R^2 - \frac4{25}\pi R^2 = \frac{21}{25}\pi R^2 π R 2 − 25 4 π R 2 = 25 21 π R 2 , option C.
Watch out
Square each radius before you subtract. Squaring the difference, ( R − r ) 2 = ( 3 5 R ) 2 (R - r)^2 = \left(\frac35R\right)^2 ( R − r ) 2 = ( 5 3 R ) 2 9 25 π R 2 \frac{9}{25}\pi R^2 25 9 π R 2 Report a problem with this question
Find the value of k k k 2 y − k x + 4 = 0 2y - kx + 4 = 0 2 y − k x + 4 = 0 y + 1 4 x − 7 = 0 y + \frac14x - 7 = 0 y + 4 1 x − 7 = 0
Worked solution (try it first) Rearrange
2 y − k x + 4 = 0 2y - kx + 4 = 0 2 y − k x + 4 = 0 :
2 y = k x − 4 2y = kx - 4 2 y = k x − 4 , so
y = k 2 x − 2 y = \frac k2x - 2 y = 2 k x − 2 and the gradient is
k 2 \frac k2 2 k .
Rearrange
y + 1 4 x − 7 = 0 y + \frac14x - 7 = 0 y + 4 1 x − 7 = 0 :
y = − 1 4 x + 7 y = -\frac14x + 7 y = − 4 1 x + 7 , so the gradient is
− 1 4 -\frac14 − 4 1 .
Perpendicular gradients multiply to
− 1 -1 − 1 :
k 2 × ( − 1 4 ) = − 1 \frac k2 \times \left(-\frac14\right) = -1 2 k × ( − 4 1 ) = − 1 , so
k 8 = 1 \frac k8 = 1 8 k = 1 .
So
k = 8 k = 8 k = 8 , option D.
Watch out
Moving − k x -kx − k x + k x +kx + k x + k 2 +\frac k2 + 2 k − k 2 -\frac k2 − 2 k k = − 8 k = -8 k = − 8 Report a problem with this question
A bucket is 12 cm in diameter at the top, 8 cm in diameter at the bottom and 4 cm deep. Calculate its volume.
A 144 π cm 3 144\pi\text{ cm}^3 144 π cm 3 B 304 π 3 cm 3 \frac{304\pi}{3}\text{ cm}^3 3 304 π cm 3 C 72 π cm 3 72\pi\text{ cm}^3 72 π cm 3 D 128 π 3 cm 3 \frac{128\pi}{3}\text{ cm}^3 3 128 π cm 3
Worked solution (try it first) The bucket is a frustum with radii
R = 6 R = 6 R = 6 cm and
r = 4 r = 4 r = 4 cm and height 4 cm.
Volume of a frustum:
π h 3 ( R 2 + R r + r 2 ) = 4 π 3 ( 36 + 24 + 16 ) \frac{\pi h}{3}(R^2 + Rr + r^2) = \frac{4\pi}{3}(36 + 24 + 16) 3 π h ( R 2 + R r + r 2 ) = 3 4 π ( 36 + 24 + 16 ) .
That is
4 π 3 × 76 = 304 π 3 cm 3 \frac{4\pi}{3} \times 76 = \frac{304\pi}{3}\text{ cm}^3 3 4 π × 76 = 3 304 π cm 3 , option B.
Watch out
The bucket narrows, so it is not a cylinder. Treating it as one with the top radius gives π × 36 × 4 = 144 π cm 3 \pi \times 36 \times 4 = 144\pi\text{ cm}^3 π × 36 × 4 = 144 π cm 3 Report a problem with this question
In the diagram, X Z XZ X Z X Y Z XYZ X Y Z O O O 15 2 \frac{15}{2} 2 15 X Y = 12 XY = 12 X Y = 12 X Y Z XYZ X Y Z
A 75 cm 2 75\text{ cm}^2 75 cm 2 B 54 cm 2 54\text{ cm}^2 54 cm 2 C 45 cm 2 45\text{ cm}^2 45 cm 2 D 27 cm 2 27\text{ cm}^2 27 cm 2
Worked solution (try it first) X Z XZ X Z is a diameter, so
∠ X Y Z = 90 ∘ \angle XYZ = 90^\circ ∠ X Y Z = 9 0 ∘ (angle in a semicircle).
The diameter is
2 × 15 2 = 15 2 \times \frac{15}{2} = 15 2 × 2 15 = 15 cm.
Pythagoras:
Y Z 2 = 15 2 − 12 2 = 81 YZ^2 = 15^2 - 12^2 = 81 Y Z 2 = 1 5 2 − 1 2 2 = 81 , so
Y Z = 9 YZ = 9 Y Z = 9 cm.
The two sides at the right angle are the base and height: area
= 1 2 × 12 × 9 = \frac12 \times 12 \times 9 = 2 1 × 12 × 9 = 54 cm 2 = 54\text{ cm}^2 = 54 cm 2 , option B.
Watch out
Use the two shorter sides, 12 and 9, which meet at the right angle at Y Y Y Report a problem with this question
Find the coordinates of the midpoint of the x x x y y y 2 y = 4 x − 8 2y = 4x - 8 2 y = 4 x − 8
A ( − 1 , − 2 ) (-1, -2) ( − 1 , − 2 ) B ( 1 , 2 ) (1, 2) ( 1 , 2 ) C ( 2 , 0 ) (2, 0) ( 2 , 0 ) D ( 1 , − 2 ) (1, -2) ( 1 , − 2 )
Worked solution (try it first) Divide by 2:
y = 2 x − 4 y = 2x - 4 y = 2 x − 4 .
On the
x x x -axis
y = 0 y = 0 y = 0 , so
x = 2 x = 2 x = 2 : the point
( 2 , 0 ) (2, 0) ( 2 , 0 ) .
On the
y y y -axis
x = 0 x = 0 x = 0 , so
y = − 4 y = -4 y = − 4 : the point
( 0 , − 4 ) (0, -4) ( 0 , − 4 ) .
Average the two points:
( 2 + 0 2 , 0 − 4 2 ) = ( 1 , − 2 ) \left(\frac{2 + 0}{2}, \frac{0 - 4}{2}\right) = (1, -2) ( 2 2 + 0 , 2 0 − 4 ) = ( 1 , − 2 ) , option D.
Watch out
The y y y − 4 -4 − 4 ( 1 , 2 ) (1, 2) ( 1 , 2 ) Report a problem with this question
A chord of a circle subtends an angle of 120 ∘ 120^\circ 12 0 ∘ 4 3 4\sqrt3 4 3
A 32 π cm 2 32\pi\text{ cm}^2 32 π cm 2 B 16 π cm 2 16\pi\text{ cm}^2 16 π cm 2 C 8 π cm 2 8\pi\text{ cm}^2 8 π cm 2 D 4 π cm 2 4\pi\text{ cm}^2 4 π cm 2
Worked solution (try it first) The radius is half of
4 3 4\sqrt3 4 3 , so
r = 2 3 r = 2\sqrt3 r = 2 3 cm and
r 2 = 12 r^2 = 12 r 2 = 12 .
The major sector has angle
360 ∘ − 120 ∘ = 240 ∘ 360^\circ - 120^\circ = 240^\circ 36 0 ∘ − 12 0 ∘ = 24 0 ∘ , which is
2 3 \frac23 3 2 of the circle.
Area
= 2 3 × 12 π = 8 π cm 2 = \frac23 \times 12\pi = 8\pi\text{ cm}^2 = 3 2 × 12 π = 8 π cm 2 , option C.
Watch out
The major sector is the larger part, 240 ∘ 240^\circ 24 0 ∘ 120 ∘ 120^\circ 12 0 ∘ 4 π cm 2 4\pi\text{ cm}^2 4 π cm 2 Also set as JAMB 2013 · UTME · Q29
Report a problem with this question
If tan θ = 4 3 \tan\theta = \frac43 tan θ = 3 4 sin 2 θ − cos 2 θ \sin^2\theta - \cos^2\theta sin 2 θ − cos 2 θ
A 7 25 \frac7{25} 25 7 B 9 25 \frac9{25} 25 9 C 16 25 \frac{16}{25} 25 16 D 24 25 \frac{24}{25} 25 24
Worked solution (try it first) tan θ = opposite adjacent \tan\theta = \frac{\text{opposite}}{\text{adjacent}} tan θ = adjacent opposite Pythagoras gives the hypotenuse
16 + 9 = 5 \sqrt{16 + 9} = 5 16 + 9 = 5 .
So
sin θ = 4 5 \sin\theta = \frac45 sin θ = 5 4 and
cos θ = 3 5 \cos\theta = \frac35 cos θ = 5 3 .
Then
sin 2 θ − cos 2 θ = 16 25 − 9 25 \sin^2\theta - \cos^2\theta = \frac{16}{25} - \frac{9}{25} sin 2 θ − cos 2 θ = 25 16 − 25 9 = 7 25 = \frac{7}{25} = 25 7 , option A.
Watch out
Finish the subtraction: sin 2 θ \sin^2\theta sin 2 θ 16 25 \frac{16}{25} 25 16 Report a problem with this question
In the diagram, P S T PST P S T P Q = Q S = R S PQ = QS = RS P Q = QS = R S ∠ R S T = 72 ∘ \angle RST = 72^\circ ∠ R S T = 7 2 ∘ x x x
A 72 ∘ 72^\circ 7 2 ∘ B 36 ∘ 36^\circ 3 6 ∘ C 24 ∘ 24^\circ 2 4 ∘ D 18 ∘ 18^\circ 1 8 ∘
Worked solution (try it first) P Q = Q S PQ = QS P Q = QS , so
∠ Q S P = ∠ Q P S = x \angle QSP = \angle QPS = x ∠ QS P = ∠ QP S = x .
The exterior angle of triangle
P Q S PQS P QS at
Q Q Q is
∠ S Q R = x + x = 2 x \angle SQR = x + x = 2x ∠ S QR = x + x = 2 x .
Q S = R S QS = RS QS = R S , so
∠ Q R S = ∠ S Q R = 2 x \angle QRS = \angle SQR = 2x ∠ QR S = ∠ S QR = 2 x .
∠ R S T \angle RST ∠ R S T is an exterior angle of triangle
P R S PRS P R S , so it equals
∠ P + ∠ R \angle P + \angle R ∠ P + ∠ R :
x + 2 x = 72 ∘ x + 2x = 72^\circ x + 2 x = 7 2 ∘ .
So
3 x = 72 ∘ 3x = 72^\circ 3 x = 7 2 ∘ and
x = 24 ∘ x = 24^\circ x = 2 4 ∘ , option C.
Watch out
The exterior angle at S S S x + 2 x x + 2x x + 2 x 2 x = 72 ∘ 2x = 72^\circ 2 x = 7 2 ∘ 36 ∘ 36^\circ 3 6 ∘ Also set as JAMB 2017 · UTME · Q23
Report a problem with this question
The locus of a point P P P S S S T T T
A a perpendicular to S T ST S T B a line parallel to S T ST S T C the angle bisector of P S PS P S S T ST S T D the perpendicular bisector of S T ST S T
Worked solution (try it first) Points equidistant from two fixed points
S S S and
T T T lie on the line that cuts
S T ST S T in half at right angles.
So the locus is the perpendicular bisector of
S T ST S T , option D.
Watch out
"A perpendicular to S T ST S T S T ST S T S T ST S T Report a problem with this question
A solid hemisphere has radius 7 cm. Find its total surface area. [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 462 cm 2 462\text{ cm}^2 462 cm 2 B 400 cm 2 400\text{ cm}^2 400 cm 2 C 308 cm 2 308\text{ cm}^2 308 cm 2 D 66 cm 2 66\text{ cm}^2 66 cm 2
Worked solution (try it first) A solid hemisphere has a curved surface, half of
4 π r 2 4\pi r^2 4 π r 2 , which is
2 π r 2 2\pi r^2 2 π r 2 , plus a flat circle
π r 2 \pi r^2 π r 2 .
Together that is
3 π r 2 3\pi r^2 3 π r 2 .
3 × 22 7 × 49 = 3 × 154 3 \times \frac{22}{7} \times 49 = 3 \times 154 3 × 7 22 × 49 = 3 × 154 = 462 cm 2 = 462\text{ cm}^2 = 462 cm 2 , option A.
Watch out
A solid hemisphere has a flat face as well. The curved part alone is 2 × 154 = 308 cm 2 2 \times 154 = 308\text{ cm}^2 2 × 154 = 308 cm 2 Report a problem with this question
In the diagram, ∠ P Q R = 50 ∘ \angle PQR = 50^\circ ∠ P QR = 5 0 ∘ R R R 128 ∘ 128^\circ 12 8 ∘ P Q R PQR P QR
A a scalene triangle B an isosceles triangle C an equilateral triangle D an obtuse-angled triangle
Worked solution (try it first) Angles on a straight line:
∠ Q R P = 180 ∘ − 128 ∘ \angle QRP = 180^\circ - 128^\circ ∠ QR P = 18 0 ∘ − 12 8 ∘ The exterior angle equals the sum of the two opposite interior angles:
∠ Q P R = 128 ∘ − 50 ∘ \angle QPR = 128^\circ - 50^\circ ∠ QP R = 12 8 ∘ − 5 0 ∘ The angles are
50 ∘ 50^\circ 5 0 ∘ ,
52 ∘ 52^\circ 5 2 ∘ and
78 ∘ 78^\circ 7 8 ∘ : all different and all acute.
So the sides are all different, and the triangle is scalene, option A.
Watch out
128 ∘ 128^\circ 12 8 ∘ R R R 52 ∘ 52^\circ 5 2 ∘ Report a problem with this question
The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have?
Worked solution (try it first) 20 right angles is
20 × 90 ∘ = 1800 ∘ 20 \times 90^\circ = 1800^\circ 20 × 9 0 ∘ = 180 0 ∘ .
The interior angles of an
n n n -sided polygon add up to
( n − 2 ) × 180 ∘ (n - 2) \times 180^\circ ( n − 2 ) × 18 0 ∘ , so
( n − 2 ) × 180 = 1800 (n - 2) \times 180 = 1800 ( n − 2 ) × 180 = 1800 and
n − 2 = 10 n - 2 = 10 n − 2 = 10 .
So
n = 12 n = 12 n = 12 , option B.
Watch out
Add the 2 back on: n − 2 = 10 n - 2 = 10 n − 2 = 10 n = 12 n = 12 n = 12 Report a problem with this question
Find the equation of the set of points which are equidistant from the parallel lines x = 1 x = 1 x = 1 x = 7 x = 7 x = 7
A y = 4 y = 4 y = 4 B y = 3 y = 3 y = 3 C x = 3 x = 3 x = 3 D x = 4 x = 4 x = 4
Worked solution (try it first) x = 1 x = 1 x = 1 and
x = 7 x = 7 x = 7 are vertical lines, 6 units apart.
The points equidistant from them lie on the vertical line halfway between.
Halfway between 1 and 7 is
1 + 7 2 = 4 \frac{1 + 7}{2} = 4 2 1 + 7 = 4 .
So the locus is
x = 4 x = 4 x = 4 , option D.
Watch out
3 is half the gap, not the position of the middle line. Add it to 1 to get x = 4 x = 4 x = 4 x = 3 x = 3 x = 3 x = 1 x = 1 x = 1 x = 7 x = 7 x = 7 Report a problem with this question
A hunter 1.6 m tall views a bird on top of a tree at an angle of 45 ∘ 45^\circ 4 5 ∘
A 8.8 m B 9.0 m C 10.4 m D 12.0 m
Worked solution (try it first) The angle is measured at his eye.
The height of the bird above eye level is
10.4 tan 45 ∘ = 10.4 10.4\tan45^\circ = 10.4 10.4 tan 4 5 ∘ = 10.4 m.
Add his height: the tree is
10.4 + 1.6 = 12.0 10.4 + 1.6 = 12.0 10.4 + 1.6 = 12.0 m, option D.
Watch out
10.4 m (option C) is only the part above his eyes. Add the 1.6 m height of the hunter. Report a problem with this question
The mean of a set of six numbers is 60. If the mean of the first five is 50, find the sixth number.
Worked solution (try it first) Total of all six numbers:
6 × 60 = 360 6 \times 60 = 360 6 × 60 = 360 .
Total of the first five:
5 × 50 = 250 5 \times 50 = 250 5 × 50 = 250 .
The sixth number is the difference:
360 − 250 = 110 360 - 250 = 110 360 − 250 = 110 , option A.
Watch out
Work with totals. The sixth number is not 60 + ( 60 − 50 ) = 70 60 + (60 - 50) = 70 60 + ( 60 − 50 ) = 70 60 + 5 × 10 = 110 60 + 5 \times 10 = 110 60 + 5 × 10 = 110 Report a problem with this question
The range of the data k + 2 , k − 3 , k + 4 , k − 2 , k , k − 5 , k + 3 , k − 1 k + 2, k - 3, k + 4, k - 2, k, k - 5, k + 3, k - 1 k + 2 , k − 3 , k + 4 , k − 2 , k , k − 5 , k + 3 , k − 1 k + 6 k + 6 k + 6
Worked solution (try it first) The largest value is
k + 6 k + 6 k + 6 and the smallest is
k − 5 k - 5 k − 5 .
The range is
( k + 6 ) − ( k − 5 ) (k + 6) - (k - 5) ( k + 6 ) − ( k − 5 ) .
The
k k k terms cancel, leaving
6 + 5 = 11 6 + 5 = 11 6 + 5 = 11 , option D.
Watch out
Subtracting k − 5 k - 5 k − 5 6 − ( − 5 ) = 11 6 - (-5) = 11 6 − ( − 5 ) = 11 6 − 5 6 - 5 6 − 5 Report a problem with this question
The distribution shows the number of days a group of 260 students were absent from school in a term. How many students were absent for at least four days?
No. of days
1
2
3
4
5
6
No. of students
20
x x x 50
40
2 x 2x 2 x 60
Worked solution (try it first) The frequencies add up to 260:
20 + x + 50 + 40 + 2 x + 60 = 260 20 + x + 50 + 40 + 2x + 60 = 260 20 + x + 50 + 40 + 2 x + 60 = 260 , so
170 + 3 x = 260 170 + 3x = 260 170 + 3 x = 260 .
So
3 x = 90 3x = 90 3 x = 90 and
x = 30 x = 30 x = 30 .
The 5-day group is
2 x = 60 2x = 60 2 x = 60 students.
At least four days means 4, 5 or 6 days:
40 + 60 + 60 = 160 40 + 60 + 60 = 160 40 + 60 + 60 = 160 students, option C.
Watch out
"At least four" includes exactly 4 days. Counting only 5 and 6 days gives 60 + 60 = 120 60 + 60 = 120 60 + 60 = 120 Report a problem with this question
In a class of 80 students, 30 − x 30 - x 30 − x x x x 40 − x 40 - x 40 − x
Worked solution (try it first) The four regions add up to the class:
( 30 − x ) + x + ( 40 − x ) + 20 = 80 (30 - x) + x + (40 - x) + 20 = 80 ( 30 − x ) + x + ( 40 − x ) + 20 = 80 .
Simplify:
90 − x = 80 90 - x = 80 90 − x = 80 , so
x = 10 x = 10 x = 10 .
Music only:
30 − 10 = 20 30 - 10 = 20 30 − 10 = 20 students.
So the probability is
20 80 = 0.25 \frac{20}{80} = 0.25 80 20 = 0.25 , option B.
Watch out
Find x x x 30 80 ≈ 0.38 \frac{30}{80} \approx 0.38 80 30 ≈ 0.38 Report a problem with this question
Find the mean of the data 7, −3, 4, −2, 5, −9, 4, 8, −6, 12.
Worked solution (try it first) Add the positives:
7 + 4 + 5 + 4 + 8 + 12 = 40 7 + 4 + 5 + 4 + 8 + 12 = 40 7 + 4 + 5 + 4 + 8 + 12 = 40 .
Add the negatives:
− 3 − 2 − 9 − 6 = − 20 -3 - 2 - 9 - 6 = -20 − 3 − 2 − 9 − 6 = − 20 .
So the total is
40 − 20 = 20 40 - 20 = 20 40 − 20 = 20 .
There are 10 values, so the mean is
20 10 = 2 \frac{20}{10} = 2 10 20 = 2 , option B.
Watch out
Keep the minus signs. Adding every number as positive gives 60 and a mean of 6, which is not an option. Report a problem with this question
The probability of a student passing any examination is 2 3 \frac23 3 2
A 1 27 \frac1{27} 27 1 B 8 27 \frac8{27} 27 8 C 4 9 \frac49 9 4 D 2 3 \frac23 3 2
Worked solution (try it first) The chance of failing one examination is
1 − 2 3 = 1 3 1 - \frac23 = \frac13 1 − 3 2 = 3 1 .
Not passing any means failing all three, and the examinations are independent, so multiply:
( 1 3 ) 3 = 1 27 \left(\frac13\right)^3 = \frac{1}{27} ( 3 1 ) 3 = 27 1 , option A.
Watch out
Use the chance of failing, 1 3 \frac13 3 1 8 27 \frac{8}{27} 27 8 Report a problem with this question
How many three-digit numbers can be formed from 32564 without any digit being repeated?
Worked solution (try it first) There are 5 different digits.
The hundreds digit can be any of the 5.
With no repeats, the tens digit has 4 choices left and the units digit 3.
So there are
5 × 4 × 3 = 60 5 \times 4 \times 3 = 60 5 × 4 × 3 = 60 numbers, option C.
Watch out
Order matters: 325 and 523 are different numbers. 5 C 3 = 10 ^5C_3 = 10 5 C 3 = 10 Report a problem with this question
The acres of rice, plantain, cassava, cocoa and palm oil in a certain district are 2, 5, 3, 11 and 9 respectively. What is the angle of the sector for cassava in a pie chart?
A 36 ∘ 36^\circ 3 6 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 108 ∘ 108^\circ 10 8 ∘ D 180 ∘ 180^\circ 18 0 ∘
Worked solution (try it first) Total area:
2 + 5 + 3 + 11 + 9 = 30 2 + 5 + 3 + 11 + 9 = 30 2 + 5 + 3 + 11 + 9 = 30 acres.
Each acre gets
360 ∘ 30 = 12 ∘ \frac{360^\circ}{30} = 12^\circ 30 36 0 ∘ = 1 2 ∘ .
Cassava has 3 acres:
3 × 12 ∘ = 36 ∘ 3 \times 12^\circ = 36^\circ 3 × 1 2 ∘ = 3 6 ∘ , option A.
Watch out
Take the right crop: cassava is the third item, 3 acres. Using plantain's 5 acres gives 60 ∘ 60^\circ 6 0 ∘ Report a problem with this question
Calculate the mean deviation of the numbers 7, 3, 14, 9, 7 and 8.
A 2 1 2 2\frac12 2 2 1 B 2 1 3 2\frac13 2 3 1 C 2 1 6 2\frac16 2 6 1 D 1 1 6 1\frac16 1 6 1
Worked solution (try it first) The numbers add up to 48 and there are 6 of them, so the mean is 8.
The distances from 8 are 1, 5, 6, 1, 1 and 0, which add up to 14.
The mean deviation is
14 6 = 2 1 3 \frac{14}{6} = 2\frac13 6 14 = 2 3 1 , option B.
Watch out
Add the distances carefully: they total 14, not 15. A total of 15 gives 15 6 = 2 1 2 \frac{15}{6} = 2\frac12 6 15 = 2 2 1 Also set as JAMB 2017 · UTME · Q4
Report a problem with this question
Find the maximum value of y y y y = 1 − 2 x − 3 x 2 y = 1 - 2x - 3x^2 y = 1 − 2 x − 3 x 2
A 5 3 \frac53 3 5 B 4 3 \frac43 3 4 C 5 4 \frac54 4 5 D 3 4 \frac34 4 3
Worked solution (try it first) The turning point is at
x = − b 2 a x = -\frac{b}{2a} x = − 2 a b .
With
a = − 3 a = -3 a = − 3 and
b = − 2 b = -2 b = − 2 , that is
x = − 1 3 x = -\frac13 x = − 3 1 .
Substitute:
y = 1 − 2 ( − 1 3 ) − 3 ( 1 9 ) y = 1 - 2\left(-\frac13\right) - 3\left(\frac19\right) y = 1 − 2 ( − 3 1 ) − 3 ( 9 1 ) = 1 + 2 3 − 1 3 = 1 + \frac23 - \frac13 = 1 + 3 2 − 3 1 .
So the maximum value is
4 3 \frac43 3 4 , option B.
Watch out
Don't drop the x 2 x^2 x 2 − 3 ( − 1 3 ) 2 = − 1 3 -3\left(-\frac13\right)^2 = -\frac13 − 3 ( − 3 1 ) 2 = − 3 1 1 + 2 3 = 5 3 1 + \frac23 = \frac53 1 + 3 2 = 3 5 Report a problem with this question
If the 9th term of an A.P. is five times the 5th term, find the relationship between a a a d d d
A a + 2 d = 0 a + 2d = 0 a + 2 d = 0 B a + 3 d = 0 a + 3d = 0 a + 3 d = 0 C 3 a + 5 d = 0 3a + 5d = 0 3 a + 5 d = 0 D 2 a + d = 0 2a + d = 0 2 a + d = 0
Worked solution (try it first) The 9th term is
a + 8 d a + 8d a + 8 d and the 5th term is
a + 4 d a + 4d a + 4 d .
So
a + 8 d = 5 ( a + 4 d ) = 5 a + 20 d a + 8d = 5(a + 4d) = 5a + 20d a + 8 d = 5 ( a + 4 d ) = 5 a + 20 d .
Collect terms:
0 = 4 a + 12 d 0 = 4a + 12d 0 = 4 a + 12 d .
Divide by 4:
a + 3 d = 0 a + 3d = 0 a + 3 d = 0 , option B.
Watch out
Multiply the whole 5th term by 5: 5 ( a + 4 d ) = 5 a + 20 d 5(a + 4d) = 5a + 20d 5 ( a + 4 d ) = 5 a + 20 d a a a a + 8 d = 5 a + 4 d a + 8d = 5a + 4d a + 8 d = 5 a + 4 d a = d a = d a = d Report a problem with this question
The time taken to do a piece of work is inversely proportional to the number of men employed. If it takes 45 men 5 days to do the work, how long will it take 25 men?
A 5 days B 9 days C 12 days D 15 days
Worked solution (try it first) Inverse proportion means (men) × (days) stays the same:
45 × 5 = 225 45 \times 5 = 225 45 × 5 = 225 man-days.
With 25 men:
225 ÷ 25 = 9 225 \div 25 = 9 225 ÷ 25 = 9 days, option B.
Watch out
Fewer men need more days, so the answer must be more than 5. Scaling the other way, 5 × 25 45 5 \times \frac{25}{45} 5 × 45 25 Report a problem with this question
The binary operation ∗ * ∗ p ∗ q = p q + p + q p * q = pq + p + q p ∗ q = pq + p + q 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) 2 ∗ ( 3 ∗ 4 )
Worked solution (try it first) Work out the bracket first:
3 ∗ 4 = 12 + 3 + 4 = 19 3 * 4 = 12 + 3 + 4 = 19 3 ∗ 4 = 12 + 3 + 4 = 19 .
Then
2 ∗ 19 = 2 × 19 + 2 + 19 = 38 + 21 2 * 19 = 2 \times 19 + 2 + 19 = 38 + 21 2 ∗ 19 = 2 × 19 + 2 + 19 = 38 + 21 .
So
2 ∗ ( 3 ∗ 4 ) = 59 2 * (3 * 4) = 59 2 ∗ ( 3 ∗ 4 ) = 59 , option C.
Watch out
Use the whole rule at the second stage too: 2 ∗ 19 2 * 19 2 ∗ 19 38 + 2 + 19 38 + 2 + 19 38 + 2 + 19 2 × 19 = 38 2 \times 19 = 38 2 × 19 = 38 Report a problem with this question
If − 2 -2 − 2 2 x + 1 − 3 c = 2 c + 3 x − 7 2x + 1 - 3c = 2c + 3x - 7 2 x + 1 − 3 c = 2 c + 3 x − 7 c c c
Worked solution (try it first) Put
x = − 2 x = -2 x = − 2 into both sides.
The left is
2 ( − 2 ) + 1 − 3 c = − 3 − 3 c 2(-2) + 1 - 3c = -3 - 3c 2 ( − 2 ) + 1 − 3 c = − 3 − 3 c .
The right is
2 c + 3 ( − 2 ) − 7 = 2 c − 13 2c + 3(-2) - 7 = 2c - 13 2 c + 3 ( − 2 ) − 7 = 2 c − 13 .
So
− 3 − 3 c = 2 c − 13 -3 - 3c = 2c - 13 − 3 − 3 c = 2 c − 13 .
Add 13 and
3 c 3c 3 c to both sides:
10 = 5 c 10 = 5c 10 = 5 c .
Divide by 5:
c = 2 c = 2 c = 2 , option B.
Watch out
Substitute the negative carefully: 3 x = 3 ( − 2 ) = − 6 3x = 3(-2) = -6 3 x = 3 ( − 2 ) = − 6 2 c − 13 2c - 13 2 c − 13 + 6 +6 + 6 2 c − 1 2c - 1 2 c − 1 c = − 2 5 c = -\frac25 c = − 5 2 Report a problem with this question
If N = ( 3 5 − 4 6 − 3 − 5 − 2 2 1 ) N = \begin{pmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{pmatrix} N = 3 6 − 2 5 − 3 2 − 4 − 5 1 ∣ N ∣ |N| ∣ N ∣
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
3 × ( ( − 3 ) ( 1 ) − ( − 5 ) ( 2 ) ) = 3 × 7 3 \times ((-3)(1) - (-5)(2)) = 3 \times 7 3 × (( − 3 ) ( 1 ) − ( − 5 ) ( 2 )) = 3 × 7 The second term is
− 5 × ( 6 × 1 − ( − 5 ) ( − 2 ) ) = − 5 × ( − 4 ) -5 \times (6 \times 1 - (-5)(-2)) = -5 \times (-4) − 5 × ( 6 × 1 − ( − 5 ) ( − 2 )) = − 5 × ( − 4 ) The third term is
− 4 × ( 6 × 2 − ( − 3 ) ( − 2 ) ) = − 4 × 6 -4 \times (6 \times 2 - (-3)(-2)) = -4 \times 6 − 4 × ( 6 × 2 − ( − 3 ) ( − 2 )) = − 4 × 6 Add them:
21 + 20 − 24 = 17 21 + 20 - 24 = 17 21 + 20 − 24 = 17 , option D.
Watch out
The middle term takes a minus sign, and − 5 × ( − 4 ) = + 20 -5 \times (-4) = +20 − 5 × ( − 4 ) = + 20 + 5 +5 + 5 21 − 20 − 24 = − 23 21 - 20 - 24 = -23 21 − 20 − 24 = − 23 Also set as JAMB 2017 · UTME · Q25
Report a problem with this question
Use the graph to find the values of p p p q q q p x + q y < 4 px + qy < 4 p x + q y < 4
A p = 1 , q = 2 p = 1, q = 2 p = 1 , q = 2 B p = 2 , q = 1 p = 2, q = 1 p = 2 , q = 1 C p = − 1 , q = 2 p = -1, q = 2 p = − 1 , q = 2 D p = 2 , q = − 1 p = 2, q = -1 p = 2 , q = − 1
Worked solution (try it first) The line through
( − 4 , 0 ) (-4, 0) ( − 4 , 0 ) and
( 0 , 2 ) (0, 2) ( 0 , 2 ) has gradient
2 − 0 0 − ( − 4 ) = 1 2 \frac{2 - 0}{0 - (-4)} = \frac12 0 − ( − 4 ) 2 − 0 = 2 1 and
y y y -intercept 2, so
y = 1 2 x + 2 y = \frac12x + 2 y = 2 1 x + 2 .
Multiply by 2 and rearrange:
− x + 2 y = 4 -x + 2y = 4 − x + 2 y = 4 .
The origin is in the shaded region, and
− 0 + 2 ( 0 ) = 0 < 4 -0 + 2(0) = 0 < 4 − 0 + 2 ( 0 ) = 0 < 4 , so the region is
− x + 2 y < 4 -x + 2y < 4 − x + 2 y < 4 .
So
p = − 1 p = -1 p = − 1 and
q = 2 q = 2 q = 2 , option C.
Watch out
Check the line with ( − 4 , 0 ) (-4, 0) ( − 4 , 0 ) − ( − 4 ) + 0 = 4 -(-4) + 0 = 4 − ( − 4 ) + 0 = 4 x + 2 y = 4 x + 2y = 4 x + 2 y = 4 − 4 -4 − 4 ( 4 , 0 ) (4, 0) ( 4 , 0 ) Report a problem with this question
The inverse of the function f ( x ) = 3 x + 4 f(x) = 3x + 4 f ( x ) = 3 x + 4
A 1 3 ( x + 4 ) \frac13(x + 4) 3 1 ( x + 4 ) B 1 4 ( x + 3 ) \frac14(x + 3) 4 1 ( x + 3 ) C 1 5 ( x − 5 ) \frac15(x - 5) 5 1 ( x − 5 ) D 1 3 ( x − 4 ) \frac13(x - 4) 3 1 ( x − 4 )
Worked solution (try it first) Write
y = 3 x + 4 y = 3x + 4 y = 3 x + 4 and make
x x x the subject.
Subtract 4:
y − 4 = 3 x y - 4 = 3x y − 4 = 3 x .
Divide by 3:
x = 1 3 ( y − 4 ) x = \frac13(y - 4) x = 3 1 ( y − 4 ) .
Rename
y y y as
x x x :
f − 1 ( x ) = 1 3 ( x − 4 ) f^{-1}(x) = \frac13(x - 4) f − 1 ( x ) = 3 1 ( x − 4 ) , option D.
Watch out
The inverse undoes "add 4" by subtracting 4. Adding 4 gives 1 3 ( x + 4 ) \frac13(x + 4) 3 1 ( x + 4 ) Report a problem with this question
Solve for x x x x 3 − 5 x 2 − x + 5 = 0 x^3 - 5x^2 - x + 5 = 0 x 3 − 5 x 2 − x + 5 = 0
A 1, 1 or 5 B − 1 -1 − 1 − 5 -5 − 5 C 1, 1 or − 5 -5 − 5 D 1, − 1 -1 − 1
Worked solution (try it first) Group the terms in pairs:
x 2 ( x − 5 ) − 1 ( x − 5 ) = 0 x^2(x - 5) - 1(x - 5) = 0 x 2 ( x − 5 ) − 1 ( x − 5 ) = 0 .
Take out the common bracket:
( x − 5 ) ( x 2 − 1 ) = 0 (x - 5)(x^2 - 1) = 0 ( x − 5 ) ( x 2 − 1 ) = 0 , and
x 2 − 1 = ( x − 1 ) ( x + 1 ) x^2 - 1 = (x - 1)(x + 1) x 2 − 1 = ( x − 1 ) ( x + 1 ) .
So
x = 5 x = 5 x = 5 ,
x = 1 x = 1 x = 1 or
x = − 1 x = -1 x = − 1 , option D.
Watch out
x 2 − 1 x^2 - 1 x 2 − 1 ( x − 1 ) ( x + 1 ) (x - 1)(x + 1) ( x − 1 ) ( x + 1 ) ( x − 1 ) 2 (x - 1)^2 ( x − 1 ) 2 1 , 1 1, 1 1 , 1 Report a problem with this question
If P = ( 2 1 − 3 0 ) P = \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix} P = ( 2 − 3 1 0 ) I I I 2 × 2 2 \times 2 2 × 2 P 2 − 2 P + 4 I P^2 - 2P + 4I P 2 − 2 P + 4 I
A ( 2 1 4 1 ) \begin{pmatrix} 2 & 1 \\ 4 & 1 \end{pmatrix} ( 2 4 1 1 ) B ( 1 0 0 1 ) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 1 0 0 1 ) C ( − 3 0 0 − 3 ) \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} ( − 3 0 0 − 3 ) D ( 9 4 12 1 ) \begin{pmatrix} 9 & 4 \\ 12 & 1 \end{pmatrix} ( 9 12 4 1 )
Worked solution (try it first) Square
P P P row by column:
P 2 = ( 4 − 3 2 + 0 − 6 + 0 − 3 + 0 ) P^2 = \begin{pmatrix} 4 - 3 & 2 + 0 \\ -6 + 0 & -3 + 0 \end{pmatrix} P 2 = ( 4 − 3 − 6 + 0 2 + 0 − 3 + 0 ) = ( 1 2 − 6 − 3 ) = \begin{pmatrix} 1 & 2 \\ -6 & -3 \end{pmatrix} = ( 1 − 6 2 − 3 ) .
Subtract
2 P = ( 4 2 − 6 0 ) 2P = \begin{pmatrix} 4 & 2 \\ -6 & 0 \end{pmatrix} 2 P = ( 4 − 6 2 0 ) :
P 2 − 2 P = ( − 3 0 0 − 3 ) P^2 - 2P = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} P 2 − 2 P = ( − 3 0 0 − 3 ) .
Add
4 I 4I 4 I , which adds 4 to each diagonal entry:
( 1 0 0 1 ) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 1 0 0 1 ) , option B.
Watch out
Don't stop before the last term: ( − 3 0 0 − 3 ) \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} ( − 3 0 0 − 3 ) P 2 − 2 P P^2 - 2P P 2 − 2 P + 4 I +4I + 4 I Also set as JAMB 2017 · UTME · Q2
Report a problem with this question
Find the range of values of x x x x + 2 4 − 2 x − 3 3 < 4 \dfrac{x + 2}{4} - \dfrac{2x - 3}{3} < 4 4 x + 2 − 3 2 x − 3 < 4
A x > − 3 x > -3 x > − 3 B x < 4 x < 4 x < 4 C x > − 6 x > -6 x > − 6 D x < 8 x < 8 x < 8
Worked solution (try it first) Multiply every term by 12, the LCM of 4 and 3:
3 ( x + 2 ) − 4 ( 2 x − 3 ) < 48 3(x + 2) - 4(2x - 3) < 48 3 ( x + 2 ) − 4 ( 2 x − 3 ) < 48 .
Expand, taking care with the minus:
3 x + 6 − 8 x + 12 < 48 3x + 6 - 8x + 12 < 48 3 x + 6 − 8 x + 12 < 48 , so
− 5 x + 18 < 48 -5x + 18 < 48 − 5 x + 18 < 48 .
Subtract 18 from both sides:
− 5 x < 30 -5x < 30 − 5 x < 30 .
Divide by
− 5 -5 − 5 and reverse the sign:
x > − 6 x > -6 x > − 6 , option C.
Watch out
The minus in front of the second fraction multiplies the whole top: − 4 ( 2 x − 3 ) = − 8 x + 12 -4(2x - 3) = -8x + 12 − 4 ( 2 x − 3 ) = − 8 x + 12 − 8 x − 12 -8x - 12 − 8 x − 12 − 5 x < 54 -5x < 54 − 5 x < 54 Report a problem with this question
If x x x n \sqrt n n x = 9 x = 9 x = 9 n = 9 n = 9 n = 9 x x x n = 17 9 n = \frac{17}{9} n = 9 17
A 27 B 17 \sqrt{17} 17 C 4 D 3 \sqrt3 3
Worked solution (try it first) Put in
x = 9 x = 9 x = 9 ,
n = 9 n = 9 n = 9 :
9 = 3 k 9 = 3k 9 = 3 k , so
k = 3 k = 3 k = 3 .
When
n = 17 9 n = \frac{17}{9} n = 9 17 :
n = 17 3 \sqrt n = \dfrac{\sqrt{17}}{3} n = 3 17 , because
9 = 3 \sqrt9 = 3 9 = 3 .
So
x = 3 × 17 3 x = 3 \times \dfrac{\sqrt{17}}{3} x = 3 × 3 17 = 17 = \sqrt{17} = 17 , option B.
Watch out
17 9 \frac{17}{9} 9 17 1 7 9 1\frac79 1 9 7 16 9 \frac{16}{9} 9 16 n = 4 3 \sqrt n = \frac43 n = 3 4 x = 4 x = 4 x = 4 Report a problem with this question
The sum to infinity of the series 1 + 1 3 + 1 9 + 1 27 + … 1 + \frac13 + \frac19 + \frac1{27} + \dots 1 + 3 1 + 9 1 + 27 1 + …
A 3 2 \frac32 2 3 B 5 2 \frac52 2 5 C 10 3 \frac{10}{3} 3 10 D 11 3 \frac{11}{3} 3 11
Worked solution (try it first) This is a G.P. with
a = 1 a = 1 a = 1 and
r = 1 3 r = \frac13 r = 3 1 .
Since
r r r is between
− 1 -1 − 1 and 1,
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a , and
1 − 1 3 = 2 3 1 - \frac13 = \frac23 1 − 3 1 = 3 2 .
So
S ∞ = 1 ÷ 2 3 = 3 2 S_\infty = 1 \div \frac23 = \frac32 S ∞ = 1 ÷ 3 2 = 2 3 , option A.
Watch out
Divide by 1 − r 1 - r 1 − r r r r 1 3 \frac13 3 1 Report a problem with this question
Evaluate ∫ sin 3 x d x \displaystyle\int \sin3x\,dx ∫ sin 3 x d x
A − 2 3 cos 3 x + c -\frac23\cos3x + c − 3 2 cos 3 x + c B − 1 3 cos 3 x + c -\frac13\cos3x + c − 3 1 cos 3 x + c C 1 3 cos 3 x + c \frac13\cos3x + c 3 1 cos 3 x + c D 2 3 cos 3 x + c \frac23\cos3x + c 3 2 cos 3 x + c
Worked solution (try it first) sin \sin sin integrates to
− cos -\cos − cos .
For
sin 3 x \sin3x sin 3 x , also divide by 3, the coefficient of
x x x .
So
∫ sin 3 x d x = − 1 3 cos 3 x + c \int \sin3x\,dx = -\frac13\cos3x + c ∫ sin 3 x d x = − 3 1 cos 3 x + c , option B.
Watch out
Keep the minus sign: the derivative of − cos 3 x -\cos3x − cos 3 x + 3 sin 3 x +3\sin3x + 3 sin 3 x 1 3 cos 3 x \frac13\cos3x 3 1 cos 3 x − sin 3 x -\sin3x − sin 3 x Report a problem with this question
A circle with a radius of 5 cm has its radius increasing at the rate of 0.2 cm s − 1 0.2\text{ cm s}^{-1} 0.2 cm s − 1
A 5 π 5\pi 5 π B 4 π 4\pi 4 π C 2 π 2\pi 2 π D π \pi π
Worked solution (try it first) The area is
A = π r 2 A = \pi r^2 A = π r 2 , so
d A d r = 2 π r \frac{dA}{dr} = 2\pi r d r d A = 2 π r .
Chain rule:
d A d t = d A d r × d r d t \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} d t d A = d r d A × d t d r = 2 π × 5 × 0.2 = 2\pi \times 5 \times 0.2 = 2 π × 5 × 0.2 .
That is
2 π cm 2 s − 1 2\pi\text{ cm}^2\text{s}^{-1} 2 π cm 2 s − 1 , option C.
Watch out
Multiply d r d t \frac{dr}{dt} d t d r d A d r = 2 π r \frac{dA}{dr} = 2\pi r d r d A = 2 π r π r 2 × 0.2 = 5 π \pi r^2 \times 0.2 = 5\pi π r 2 × 0.2 = 5 π Report a problem with this question
If d y d x = 2 x − 3 \frac{dy}{dx} = 2x - 3 d x d y = 2 x − 3 y = 3 y = 3 y = 3 x = 0 x = 0 x = 0 y y y x x x
A x 2 − 3 x x^2 - 3x x 2 − 3 x B x 2 − 3 x + 3 x^2 - 3x + 3 x 2 − 3 x + 3 C 2 x 2 − 3 x 2x^2 - 3x 2 x 2 − 3 x D x 2 − 3 x − 3 x^2 - 3x - 3 x 2 − 3 x − 3
Worked solution (try it first) Integrate:
y = x 2 − 3 x + c y = x^2 - 3x + c y = x 2 − 3 x + c .
Put in
x = 0 x = 0 x = 0 ,
y = 3 y = 3 y = 3 :
3 = 0 − 0 + c 3 = 0 - 0 + c 3 = 0 − 0 + c , so
c = 3 c = 3 c = 3 .
So
y = x 2 − 3 x + 3 y = x^2 - 3x + 3 y = x 2 − 3 x + 3 , option B.
Watch out
Don't forget the constant. x 2 − 3 x x^2 - 3x x 2 − 3 x y = 0 y = 0 y = 0 x = 0 x = 0 x = 0 Also set as JAMB 2017 · UTME · Q1
Report a problem with this question
Find the derivative of y = sin 2 ( 5 x ) y = \sin^2(5x) y = sin 2 ( 5 x ) x x x
A 2 sin 5 x cos 5 x 2\sin5x\cos5x 2 sin 5 x cos 5 x B 5 sin 5 x cos 5 x 5\sin5x\cos5x 5 sin 5 x cos 5 x C 10 sin 5 x cos 5 x 10\sin5x\cos5x 10 sin 5 x cos 5 x D 15 sin 5 x cos 5 x 15\sin5x\cos5x 15 sin 5 x cos 5 x
Worked solution (try it first) Write
y = ( sin 5 x ) 2 y = (\sin5x)^2 y = ( sin 5 x ) 2 .
Chain rule: bring down the 2 to get
2 sin 5 x 2\sin5x 2 sin 5 x , then multiply by the derivative of
sin 5 x \sin5x sin 5 x , which is
5 cos 5 x 5\cos5x 5 cos 5 x .
So
d y d x = 2 sin 5 x × 5 cos 5 x \frac{dy}{dx} = 2\sin5x \times 5\cos5x d x d y = 2 sin 5 x × 5 cos 5 x = 10 sin 5 x cos 5 x = 10\sin5x\cos5x = 10 sin 5 x cos 5 x , option C.
Watch out
sin 5 x \sin5x sin 5 x 5 cos 5 x 5\cos5x 5 cos 5 x cos 5 x \cos5x cos 5 x 2 sin 5 x cos 5 x 2\sin5x\cos5x 2 sin 5 x cos 5 x Report a problem with this question
The slope of the tangent to the curve y = 3 x 2 − 2 x + 5 y = 3x^2 - 2x + 5 y = 3 x 2 − 2 x + 5 ( 1 , 6 ) (1, 6) ( 1 , 6 )
Worked solution (try it first) The slope of the tangent is
d y d x = 6 x − 2 \frac{dy}{dx} = 6x - 2 d x d y = 6 x − 2 .
At
x = 1 x = 1 x = 1 :
6 − 2 = 4 6 - 2 = 4 6 − 2 = 4 , option B.
Watch out
3 x 2 3x^2 3 x 2 6 x 6x 6 x 3 x 3x 3 x 3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 Report a problem with this question