Convert 45 ∘ 45^\circ 4 5 ∘ π \pi π
A π 2 \frac{\pi}{2} 2 π B π 3 \frac{\pi}{3} 3 π C π 4 \frac{\pi}{4} 4 π D π 6 \frac{\pi}{6} 6 π E π 12 \frac{\pi}{12} 12 π
Worked solution (try it first) π \pi π radians is
180 ∘ 180^\circ 18 0 ∘ , so to change degrees to radians multiply by
π 180 \dfrac{\pi}{180} 180 π .
45 × π 180 = 45 π 180 45 \times \dfrac{\pi}{180} = \dfrac{45\pi}{180} 45 × 180 π = 180 45 π , and 45 goes into 180 four times, so this is
π 4 \dfrac{\pi}{4} 4 π .
So
45 ∘ = π 4 45^\circ = \dfrac{\pi}{4} 4 5 ∘ = 4 π , option C.
Watch out
π \pi π 180 ∘ 180^\circ 18 0 ∘ 90 ∘ 90^\circ 9 0 ∘ π \pi π 90 ∘ 90^\circ 9 0 ∘ 45 ∘ 45^\circ 4 5 ∘ π \pi π π 2 \frac{\pi}{2} 2 π Report a problem with this question
Rationalize 2 5 + 7 \dfrac{2}{\sqrt5 + \sqrt7} 5 + 7 2
A − 5 − 7 -\sqrt5 - \sqrt7 − 5 − 7 B 7 − 5 \sqrt7 - \sqrt5 7 − 5 C 5 + 7 \sqrt5 + \sqrt7 5 + 7 D 2 5 + 7 2\sqrt5 + \sqrt7 2 5 + 7 E 2 5 + 2 7 2\sqrt5 + 2\sqrt7 2 5 + 2 7
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
7 − 5 \sqrt7 - \sqrt5 7 − 5 (same terms, sign changed).
The bottom becomes a difference of two squares:
( 7 + 5 ) ( 7 − 5 ) = 7 − 5 (\sqrt7 + \sqrt5)(\sqrt7 - \sqrt5) = 7 - 5 ( 7 + 5 ) ( 7 − 5 ) = 7 − 5 , which is 2.
So the fraction is
2 ( 7 − 5 ) 2 \dfrac{2(\sqrt7 - \sqrt5)}{2} 2 2 ( 7 − 5 ) , and the 2s cancel to leave
7 − 5 \sqrt7 - \sqrt5 7 − 5 , option B.
Watch out
Multiply by the conjugate, with the sign in the middle changed. Multiplying by 5 + 7 \sqrt5 + \sqrt7 5 + 7 5 + 7 \sqrt5 + \sqrt7 5 + 7 Report a problem with this question
Find the sum of the roots of the quadratic equation 3 x 2 + 3 x − 1 = 0 3x^2 + 3x - 1 = 0 3 x 2 + 3 x − 1 = 0
A − 1 -1 − 1 B − 1 2 -\frac12 − 2 1 C 1 2 \frac12 2 1 D 2 3 \frac23 3 2 E 1
Worked solution (try it first) For
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , the sum of the roots is
− b a -\dfrac{b}{a} − a b .
Here
a = 3 a = 3 a = 3 and
b = 3 b = 3 b = 3 , so the sum is
− 3 3 = − 1 -\dfrac33 = -1 − 3 3 = − 1 , option A.
Watch out
Keep the minus sign: the sum of the roots is − b a -\frac{b}{a} − a b b a \frac{b}{a} a b Report a problem with this question
Simplify 1 2 log 4 64 + log 4 16 − log 4 2 \frac12\log_4 64 + \log_4 16 - \log_4 2 2 1 log 4 64 + log 4 16 − log 4 2
Worked solution (try it first) 64 = 4 3 64 = 4^3 64 = 4 3 , so
log 4 64 = 3 \log_4 64 = 3 log 4 64 = 3 and
1 2 log 4 64 = 3 2 \frac12\log_4 64 = \frac32 2 1 log 4 64 = 2 3 .
16 = 4 2 16 = 4^2 16 = 4 2 , so
log 4 16 = 2 \log_4 16 = 2 log 4 16 = 2 .
And
2 = 4 1 / 2 2 = 4^{1/2} 2 = 4 1/2 , so
log 4 2 = 1 2 \log_4 2 = \frac12 log 4 2 = 2 1 .
So the value is
3 2 + 2 − 1 2 = 3 \frac32 + 2 - \frac12 = 3 2 3 + 2 − 2 1 = 3 , option B.
Watch out
The 1 2 \frac12 2 1 1 2 log 4 64 = log 4 64 1 / 2 = log 4 8 \frac12\log_4 64 = \log_4 64^{1/2} = \log_4 8 2 1 log 4 64 = log 4 6 4 1/2 = log 4 8 log 4 256 = 4 \log_4 256 = 4 log 4 256 = 4 Report a problem with this question
Find the sum of the first 9 terms of the exponential sequence 18 , 6 , 2 , … 18, 6, 2, \dots 18 , 6 , 2 , …
A 26.00 B 26.66 C 26.88 D 26.99 E 27.00
Worked solution (try it first) The first term is
a = 18 a = 18 a = 18 and the common ratio is
r = 6 18 = 1 3 r = \dfrac{6}{18} = \dfrac13 r = 18 6 = 3 1 .
For
r < 1 r < 1 r < 1 use
S n = a ( 1 − r n ) 1 − r S_n = \dfrac{a(1 - r^n)}{1 - r} S n = 1 − r a ( 1 − r n ) :
S 9 = 18 ( 1 − ( 1 3 ) 9 ) 2 3 S_9 = \dfrac{18\left(1 - \left(\frac13\right)^9\right)}{\frac23} S 9 = 3 2 18 ( 1 − ( 3 1 ) 9 ) .
18 ÷ 2 3 = 27 18 \div \frac23 = 27 18 ÷ 3 2 = 27 and
( 1 3 ) 9 = 1 19683 \left(\frac13\right)^9 = \frac{1}{19683} ( 3 1 ) 9 = 19683 1 , so
S 9 = 27 ( 1 − 1 19683 ) S_9 = 27\left(1 - \frac{1}{19683}\right) S 9 = 27 ( 1 − 19683 1 ) ≈ 26.9986 \approx 26.9986 ≈ 26.9986 .
To two decimal places this is 27.00, option E.
Watch out
Round, don't cut off: 26.9986 26.9986 26.9986 Report a problem with this question
Solve the inequality 2 x + 3 4 ≤ 5 x − 2 3 2x + \frac34 \le 5x - \frac23 2 x + 4 3 ≤ 5 x − 3 2
A x > − 17 36 x > -\frac{17}{36} x > − 36 17 B x < 17 36 x < \frac{17}{36} x < 36 17 C x ≤ 17 36 x \le \frac{17}{36} x ≤ 36 17 D x ≥ 17 36 x \ge \frac{17}{36} x ≥ 36 17 E x ≤ 17 x \le 17 x ≤ 17
Worked solution (try it first) Take
2 x 2x 2 x from both sides:
3 4 ≤ 3 x − 2 3 \frac34 \le 3x - \frac23 4 3 ≤ 3 x − 3 2 .
Then add
2 3 \frac23 3 2 to both sides:
3 4 + 2 3 ≤ 3 x \frac34 + \frac23 \le 3x 4 3 + 3 2 ≤ 3 x .
Add the fractions over 12:
9 12 + 8 12 = 17 12 \frac{9}{12} + \frac{8}{12} = \frac{17}{12} 12 9 + 12 8 = 12 17 , so
17 12 ≤ 3 x \frac{17}{12} \le 3x 12 17 ≤ 3 x .
Divide both sides by 3 (a positive number, so the sign stays):
17 36 ≤ x \frac{17}{36} \le x 36 17 ≤ x .
So
x ≥ 17 36 x \ge \frac{17}{36} x ≥ 36 17 , option D.
Watch out
Read 17 36 ≤ x \frac{17}{36} \le x 36 17 ≤ x x x x x ≥ 17 36 x \ge \frac{17}{36} x ≥ 36 17 x ≤ 17 36 x \le \frac{17}{36} x ≤ 36 17 Report a problem with this question
A binary operation ∗ * ∗ R \mathbb{R} R p ∗ q = p + q + 3 p q p * q = p + q + 3pq p ∗ q = p + q + 3 pq p p p
A − 3 -3 − 3 B − 1 3 -\frac13 − 3 1 C 0 D 1 3 \frac13 3 1 E 3
Worked solution (try it first) The identity is 0, so the inverse
q q q of
p p p satisfies
p ∗ q = 0 p * q = 0 p ∗ q = 0 , that is
p + q + 3 p q = 0 p + q + 3pq = 0 p + q + 3 pq = 0 .
Collect the
q q q terms:
q ( 1 + 3 p ) = − p q(1 + 3p) = -p q ( 1 + 3 p ) = − p .
Divide by
1 + 3 p 1 + 3p 1 + 3 p :
q = − p 1 + 3 p q = \dfrac{-p}{1 + 3p} q = 1 + 3 p − p .
This has no value when the bottom is zero:
1 + 3 p = 0 1 + 3p = 0 1 + 3 p = 0 , so
3 p = − 1 3p = -1 3 p = − 1 and
p = − 1 3 p = -\frac13 p = − 3 1 , option B.
Watch out
Solve 1 + 3 p = 0 1 + 3p = 0 1 + 3 p = 0 3 p = − 1 3p = -1 3 p = − 1 p = − 1 3 p = -\frac13 p = − 3 1 p = − 3 p = -3 p = − 3 Report a problem with this question
Solve ( 243 ) 2 x + 1 = 81 x − 2 3 x (243)^{2x + 1} = \dfrac{81^{x - 2}}{3^x} ( 243 ) 2 x + 1 = 3 x 8 1 x − 2
A − 7 3 -\frac73 − 3 7 B − 13 7 -\frac{13}{7} − 7 13 C 3 7 \frac37 7 3 D 13 7 \frac{13}{7} 7 13 E 7 3 \frac73 3 7
Worked solution (try it first) Write everything as powers of 3:
243 = 3 5 243 = 3^5 243 = 3 5 and
81 = 3 4 81 = 3^4 81 = 3 4 .
The left side is
3 5 ( 2 x + 1 ) = 3 10 x + 5 3^{5(2x + 1)} = 3^{10x + 5} 3 5 ( 2 x + 1 ) = 3 10 x + 5 .
On the right,
81 x − 2 = 3 4 x − 8 81^{x - 2} = 3^{4x - 8} 8 1 x − 2 = 3 4 x − 8 , and dividing by
3 x 3^x 3 x takes
x x x off the power:
3 3 x − 8 3^{3x - 8} 3 3 x − 8 .
The bases match, so the powers are equal:
10 x + 5 = 3 x − 8 10x + 5 = 3x - 8 10 x + 5 = 3 x − 8 .
Take
3 x 3x 3 x and 5 from both sides:
7 x = − 13 7x = -13 7 x = − 13 , so
x = − 13 7 x = -\frac{13}{7} x = − 7 13 , option B.
Watch out
When you move a term across, change its sign: 10 x − 3 x = − 8 − 5 10x - 3x = -8 - 5 10 x − 3 x = − 8 − 5 7 x = − 13 7x = -13 7 x = − 13 13 7 \frac{13}{7} 7 13 Report a problem with this question
If a n + 2 = 0.2 a n + 1 − 0.1 a n a_{n+2} = 0.2a_{n+1} - 0.1a_n a n + 2 = 0.2 a n + 1 − 0.1 a n a 1 = 0 a_1 = 0 a 1 = 0 a 2 = 1 a_2 = 1 a 2 = 1
A − 0.032 -0.032 − 0.032 B − 0.06 -0.06 − 0.06 C 0.06 D 0.10 E 0.20
Worked solution (try it first) Put
n = 1 n = 1 n = 1 :
a 3 = 0.2 a 2 − 0.1 a 1 = 0.2 ( 1 ) − 0.1 ( 0 ) a_3 = 0.2a_2 - 0.1a_1 = 0.2(1) - 0.1(0) a 3 = 0.2 a 2 − 0.1 a 1 = 0.2 ( 1 ) − 0.1 ( 0 ) , which is 0.2.
Put
n = 2 n = 2 n = 2 :
a 4 = 0.2 a 3 − 0.1 a 2 = 0.04 − 0.1 a_4 = 0.2a_3 - 0.1a_2 = 0.04 - 0.1 a 4 = 0.2 a 3 − 0.1 a 2 = 0.04 − 0.1 , which is
− 0.06 -0.06 − 0.06 .
Put
n = 3 n = 3 n = 3 :
a 5 = 0.2 a 4 − 0.1 a 3 = − 0.012 − 0.02 a_5 = 0.2a_4 - 0.1a_3 = -0.012 - 0.02 a 5 = 0.2 a 4 − 0.1 a 3 = − 0.012 − 0.02 , which is
− 0.032 -0.032 − 0.032 , option A.
Watch out
You are given a 1 a_1 a 1 a 2 a_2 a 2 a 5 a_5 a 5 a 4 = − 0.06 a_4 = -0.06 a 4 = − 0.06 Report a problem with this question
In a circle of radius 20 cm 20\text{ cm} 20 cm
A 12 cm B 20 cm C 24 cm D 36 cm E 48 cm
Worked solution (try it first) The angle is in radians, so the arc length is
s = r θ s = r\theta s = r θ .
s = 20 × 1.2 = 24 s = 20 \times 1.2 = 24 s = 20 × 1.2 = 24 cm, option C.
Watch out
Use the radius, 20 cm, in s = r θ s = r\theta s = r θ Report a problem with this question
Given that cos A = 4 5 \cos A = \frac45 cos A = 5 4 cos B = 12 13 \cos B = \frac{12}{13} cos B = 13 12 sin ( A + B ) \sin(A + B) sin ( A + B ) A A A B B B
A 16 65 \frac{16}{65} 65 16 B 20 65 \frac{20}{65} 65 20 C 36 65 \frac{36}{65} 65 36 D 48 65 \frac{48}{65} 65 48 E 56 65 \frac{56}{65} 65 56
Worked solution (try it first) A A A is acute and
cos A = 4 5 \cos A = \frac45 cos A = 5 4 : a 3-4-5 triangle gives
sin A = 3 5 \sin A = \frac35 sin A = 5 3 .
Likewise a 5-12-13 triangle gives
sin B = 5 13 \sin B = \frac{5}{13} sin B = 13 5 .
Use the compound-angle formula:
sin ( A + B ) = sin A cos B + cos A sin B \sin(A + B) = \sin A\cos B + \cos A\sin B sin ( A + B ) = sin A cos B + cos A sin B .
That is
3 5 × 12 13 + 4 5 × 5 13 = 36 65 + 20 65 \frac35 \times \frac{12}{13} + \frac45 \times \frac{5}{13} = \frac{36}{65} + \frac{20}{65} 5 3 × 13 12 + 5 4 × 13 5 = 65 36 + 65 20 .
So
sin ( A + B ) = 56 65 \sin(A + B) = \frac{56}{65} sin ( A + B ) = 65 56 , option E.
Watch out
sin ( A + B ) \sin(A + B) sin ( A + B ) sin ( A − B ) = 36 65 − 20 65 = 16 65 \sin(A - B) = \frac{36}{65} - \frac{20}{65} = \frac{16}{65} sin ( A − B ) = 65 36 − 65 20 = 65 16 Report a problem with this question
Resolve 2 x + 5 ( x + 3 ) ( x + 2 ) \dfrac{2x + 5}{(x + 3)(x + 2)} ( x + 3 ) ( x + 2 ) 2 x + 5
A 5 x + 3 + 2 x x + 2 \frac5{x + 3} + \frac{2x}{x + 2} x + 3 5 + x + 2 2 x B 1 x + 3 + 2 x + 2 \frac1{x + 3} + \frac2{x + 2} x + 3 1 + x + 2 2 C 1 x + 3 + 1 x + 2 \frac1{x + 3} + \frac1{x + 2} x + 3 1 + x + 2 1 D 1 x + 3 − 1 x + 2 \frac1{x + 3} - \frac1{x + 2} x + 3 1 − x + 2 1 E 2 x x + 3 + 1 x + 2 \frac{2x}{x + 3} + \frac1{x + 2} x + 3 2 x + x + 2 1
Worked solution (try it first) Write
2 x + 5 ( x + 3 ) ( x + 2 ) = A x + 3 + B x + 2 \dfrac{2x + 5}{(x + 3)(x + 2)} = \dfrac{A}{x + 3} + \dfrac{B}{x + 2} ( x + 3 ) ( x + 2 ) 2 x + 5 = x + 3 A + x + 2 B and multiply through by the bottom:
2 x + 5 = A ( x + 2 ) + B ( x + 3 ) 2x + 5 = A(x + 2) + B(x + 3) 2 x + 5 = A ( x + 2 ) + B ( x + 3 ) .
Put
x = − 3 x = -3 x = − 3 , which makes
x + 3 x + 3 x + 3 zero:
− 1 = A ( − 1 ) -1 = A(-1) − 1 = A ( − 1 ) , so
A = 1 A = 1 A = 1 .
Put
x = − 2 x = -2 x = − 2 , which makes
x + 2 x + 2 x + 2 zero:
1 = B ( 1 ) 1 = B(1) 1 = B ( 1 ) , so
B = 1 B = 1 B = 1 .
So the fraction is
1 x + 3 + 1 x + 2 \dfrac{1}{x + 3} + \dfrac{1}{x + 2} x + 3 1 + x + 2 1 , option C.
Watch out
Check by adding back. Option C gives a top of ( x + 2 ) + ( x + 3 ) = 2 x + 5 (x + 2) + (x + 3) = 2x + 5 ( x + 2 ) + ( x + 3 ) = 2 x + 5 ( x + 2 ) − ( x + 3 ) = − 1 (x + 2) - (x + 3) = -1 ( x + 2 ) − ( x + 3 ) = − 1 B B B Report a problem with this question
If ( 3 − 1 1 1 ) ( x y ) = ( 13 7 ) \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 13 \\ 7 \end{pmatrix} ( 3 1 − 1 1 ) ( x y ) = ( 13 7 ) y y y
Worked solution (try it first) Each row times the column gives one equation:
3 x − y = 13 3x - y = 13 3 x − y = 13 and
x + y = 7 x + y = 7 x + y = 7 .
Add the two equations to remove
y y y :
4 x = 20 4x = 20 4 x = 20 , so
x = 5 x = 5 x = 5 .
Then
y = 7 − 5 = 2 y = 7 - 5 = 2 y = 7 − 5 = 2 , option B.
Watch out
The question asks for y y y x = 5 x = 5 x = 5 Report a problem with this question
Solve sin 2 x + 2 sin x + 1 = 0 \sin^2x + 2\sin x + 1 = 0 sin 2 x + 2 sin x + 1 = 0 0 ∘ < x < 360 ∘ 0^\circ < x < 360^\circ 0 ∘ < x < 36 0 ∘
A 30 ∘ 30^\circ 3 0 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 90 ∘ 90^\circ 9 0 ∘ D 270 ∘ 270^\circ 27 0 ∘ E 360 ∘ 360^\circ 36 0 ∘
Worked solution (try it first) The left side is a perfect square:
sin 2 x + 2 sin x + 1 = ( sin x + 1 ) 2 \sin^2 x + 2\sin x + 1 = (\sin x + 1)^2 sin 2 x + 2 sin x + 1 = ( sin x + 1 ) 2 .
So
( sin x + 1 ) 2 = 0 (\sin x + 1)^2 = 0 ( sin x + 1 ) 2 = 0 , which gives
sin x = − 1 \sin x = -1 sin x = − 1 .
Between
0 ∘ 0^\circ 0 ∘ and
360 ∘ 360^\circ 36 0 ∘ this happens only at the bottom of the sine curve,
x = 270 ∘ x = 270^\circ x = 27 0 ∘ , option D.
Watch out
sin x + 1 = 0 \sin x + 1 = 0 sin x + 1 = 0 sin x = − 1 \sin x = -1 sin x = − 1 + 1 +1 + 1 sin x = + 1 \sin x = +1 sin x = + 1 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
Differentiate ( 2 + y y ) 2 \left(\dfrac{2 + y}{y}\right)^2 ( y 2 + y ) 2 y y y
A 8 y 3 − 4 y 2 \frac8{y^3} - \frac4{y^2} y 3 8 − y 2 4 B 4 y 3 − 8 y 2 \frac4{y^3} - \frac8{y^2} y 3 4 − y 2 8 C − 4 y 3 − 8 y 2 \frac{-4}{y^3} - \frac8{y^2} y 3 − 4 − y 2 8 D − 4 y 2 − 8 y 3 \frac{-4}{y^2} - \frac8{y^3} y 2 − 4 − y 3 8 E − 4 y 2 + 16 y 3 \frac{-4}{y^2} + \frac{16}{y^3} y 2 − 4 + y 3 16
Worked solution (try it first) Simplify first:
2 + y y = 1 + 2 y \dfrac{2 + y}{y} = 1 + \dfrac2y y 2 + y = 1 + y 2 , so the expression is
( 1 + 2 y ) 2 = 1 + 4 y + 4 y 2 \left(1 + \dfrac2y\right)^2 = 1 + \dfrac4y + \dfrac{4}{y^2} ( 1 + y 2 ) 2 = 1 + y 4 + y 2 4 .
Write it as powers of
y y y :
1 + 4 y − 1 + 4 y − 2 1 + 4y^{-1} + 4y^{-2} 1 + 4 y − 1 + 4 y − 2 .
Differentiate each term (multiply by the power, then take 1 off it):
− 4 y − 2 − 8 y − 3 -4y^{-2} - 8y^{-3} − 4 y − 2 − 8 y − 3 .
So the derivative is
− 4 y 2 − 8 y 3 -\dfrac{4}{y^2} - \dfrac{8}{y^3} − y 2 4 − y 3 8 , option D.
Watch out
Each power drops by one: 4 y − 1 4y^{-1} 4 y − 1 − 4 y − 2 -4y^{-2} − 4 y − 2 4 y − 2 4y^{-2} 4 y − 2 − 8 y − 3 -8y^{-3} − 8 y − 3 − 4 y 3 − 8 y 2 -\frac{4}{y^3} - \frac{8}{y^2} − y 3 4 − y 2 8 Report a problem with this question
The difference between a non-negative number x x x
A x ≤ − 5 x \le -5 x ≤ − 5 B x ≤ 5 3 x \le \frac53 x ≤ 3 5 C x ≥ 5 3 x \ge \frac53 x ≥ 3 5 D 0 ≤ x ≤ 5 3 0 \le x \le \frac53 0 ≤ x ≤ 3 5 E 0 < x < 5 3 0 < x < \frac53 0 < x < 3 5
Worked solution (try it first) For
0 ≤ x ≤ 5 0 \le x \le 5 0 ≤ x ≤ 5 the difference between
x x x and 5 is
5 − x 5 - x 5 − x , so the condition is
5 − x ≥ 2 x 5 - x \ge 2x 5 − x ≥ 2 x .
Add
x x x to both sides:
5 ≥ 3 x 5 \ge 3x 5 ≥ 3 x , so
x ≤ 5 3 x \le \frac53 x ≤ 3 5 .
For
x > 5 x > 5 x > 5 the difference is
x − 5 x - 5 x − 5 , and
x − 5 ≥ 2 x x - 5 \ge 2x x − 5 ≥ 2 x would need
x ≤ − 5 x \le -5 x ≤ − 5 , which a non-negative number can't be.
With
x ≥ 0 x \ge 0 x ≥ 0 as well, the range is
0 ≤ x ≤ 5 3 0 \le x \le \frac53 0 ≤ x ≤ 3 5 , option D.
Watch out
Keep the word "non-negative": it gives the lower end, x ≥ 0 x \ge 0 x ≥ 0 x ≤ 5 3 x \le \frac53 x ≤ 3 5 Report a problem with this question
A curve passes through ( − 2 , 3 ) (-2, 3) ( − 2 , 3 ) 3 x 2 + 8 x − 5 3x^2 + 8x - 5 3 x 2 + 8 x − 5
A y = x 3 − 4 x 2 + 5 x − 7 y = x^3 - 4x^2 + 5x - 7 y = x 3 − 4 x 2 + 5 x − 7 B y = x 3 − 4 x 2 + 5 x + 5 y = x^3 - 4x^2 + 5x + 5 y = x 3 − 4 x 2 + 5 x + 5 C y = x 3 − 4 x 2 − 5 x + 15 y = x^3 - 4x^2 - 5x + 15 y = x 3 − 4 x 2 − 5 x + 15 D y = x 3 + 4 x 2 − 5 x − 15 y = x^3 + 4x^2 - 5x - 15 y = x 3 + 4 x 2 − 5 x − 15 E y = x 3 + 4 x 2 + 5 x − 15 y = x^3 + 4x^2 + 5x - 15 y = x 3 + 4 x 2 + 5 x − 15
Worked solution (try it first) Integrate the gradient term by term:
y = x 3 + 4 x 2 − 5 x + c y = x^3 + 4x^2 - 5x + c y = x 3 + 4 x 2 − 5 x + c .
The curve passes through
( − 2 , 3 ) (-2, 3) ( − 2 , 3 ) , so put
x = − 2 x = -2 x = − 2 and
y = 3 y = 3 y = 3 :
3 = − 8 + 16 + 10 + c 3 = -8 + 16 + 10 + c 3 = − 8 + 16 + 10 + c , that is
3 = 18 + c 3 = 18 + c 3 = 18 + c .
So
c = − 15 c = -15 c = − 15 and
y = x 3 + 4 x 2 − 5 x − 15 y = x^3 + 4x^2 - 5x - 15 y = x 3 + 4 x 2 − 5 x − 15 , option D.
Watch out
Keep each sign when you integrate: − 5 -5 − 5 − 5 x -5x − 5 x + 5 x +5x + 5 x + 5 +5 + 5 − 5 -5 − 5 Report a problem with this question
Find the first 4 terms in the expansion of ( 1 + 2 x ) 15 (1 + 2x)^{15} ( 1 + 2 x ) 15
A 1 + 20 x + 220 x 2 + 3320 x 3 1 + 20x + 220x^2 + 3320x^3 1 + 20 x + 220 x 2 + 3320 x 3 B 1 + 25 x + 420 x 2 + 720 x 3 1 + 25x + 420x^2 + 720x^3 1 + 25 x + 420 x 2 + 720 x 3 C 1 + 30 x + 420 x 2 + 3640 x 3 1 + 30x + 420x^2 + 3640x^3 1 + 30 x + 420 x 2 + 3640 x 3 D 1 + 40 x + 480 x 2 + 960 x 3 1 + 40x + 480x^2 + 960x^3 1 + 40 x + 480 x 2 + 960 x 3 E 1 + 60 x + 540 x 2 + 2380 x 3 1 + 60x + 540x^2 + 2380x^3 1 + 60 x + 540 x 2 + 2380 x 3
Worked solution (try it first) Each term is
( 15 r ) ( 2 x ) r \binom{15}{r}(2x)^r ( r 15 ) ( 2 x ) r .
The coefficients needed are
( 15 1 ) = 15 \binom{15}{1} = 15 ( 1 15 ) = 15 ,
( 15 2 ) = 105 \binom{15}{2} = 105 ( 2 15 ) = 105 and
( 15 3 ) = 455 \binom{15}{3} = 455 ( 3 15 ) = 455 .
r = 1 r = 1 r = 1 :
15 × 2 x = 30 x 15 \times 2x = 30x 15 × 2 x = 30 x .
r = 2 r = 2 r = 2 :
105 × 4 x 2 = 420 x 2 105 \times 4x^2 = 420x^2 105 × 4 x 2 = 420 x 2 .
r = 3 r = 3 r = 3 :
455 × 8 x 3 = 3640 x 3 455 \times 8x^3 = 3640x^3 455 × 8 x 3 = 3640 x 3 .
So the first four terms are
1 + 30 x + 420 x 2 + 3640 x 3 1 + 30x + 420x^2 + 3640x^3 1 + 30 x + 420 x 2 + 3640 x 3 , option C.
Watch out
Raise the whole of 2 x 2x 2 x ( 2 x ) 2 = 4 x 2 (2x)^2 = 4x^2 ( 2 x ) 2 = 4 x 2 ( 2 x ) 3 = 8 x 3 (2x)^3 = 8x^3 ( 2 x ) 3 = 8 x 3 1 + 15 x + 105 x 2 + 455 x 3 1 + 15x + 105x^2 + 455x^3 1 + 15 x + 105 x 2 + 455 x 3 Report a problem with this question
Let f f f g g g f ( x ) = 3 x 2 − 4 f(x) = 3x^2 - 4 f ( x ) = 3 x 2 − 4 g ( x ) = 2 x + 1 g(x) = 2x + 1 g ( x ) = 2 x + 1 f g ( x ) fg(x) f g ( x )
A 3 x 2 + 2 x − 3 3x^2 + 2x - 3 3 x 2 + 2 x − 3 B 3 x 2 + 2 x + 3 3x^2 + 2x + 3 3 x 2 + 2 x + 3 C 6 x 2 − 2 x − 1 6x^2 - 2x - 1 6 x 2 − 2 x − 1 D 12 x 2 + 12 x − 1 12x^2 + 12x - 1 12 x 2 + 12 x − 1 E 12 x 2 − 12 x + 3 12x^2 - 12x + 3 12 x 2 − 12 x + 3
Worked solution (try it first) f g ( x ) fg(x) f g ( x ) means
f ( g ( x ) ) f(g(x)) f ( g ( x )) :
g g g acts first.
So
f g ( x ) = f ( 2 x + 1 ) fg(x) = f(2x + 1) f g ( x ) = f ( 2 x + 1 ) .
Replace
x x x in
f f f by
( 2 x + 1 ) (2x + 1) ( 2 x + 1 ) :
3 ( 2 x + 1 ) 2 − 4 3(2x + 1)^2 - 4 3 ( 2 x + 1 ) 2 − 4 .
Expand the square:
( 2 x + 1 ) 2 = 4 x 2 + 4 x + 1 (2x + 1)^2 = 4x^2 + 4x + 1 ( 2 x + 1 ) 2 = 4 x 2 + 4 x + 1 , so
3 ( 2 x + 1 ) 2 = 12 x 2 + 12 x + 3 3(2x + 1)^2 = 12x^2 + 12x + 3 3 ( 2 x + 1 ) 2 = 12 x 2 + 12 x + 3 .
Take away 4:
f g ( x ) = 12 x 2 + 12 x − 1 fg(x) = 12x^2 + 12x - 1 f g ( x ) = 12 x 2 + 12 x − 1 , option D.
Watch out
f g ( x ) fg(x) f g ( x ) f f f g ( x ) g(x) g ( x ) f ( x ) + g ( x ) f(x) + g(x) f ( x ) + g ( x ) 3 x 2 + 2 x − 3 3x^2 + 2x - 3 3 x 2 + 2 x − 3 Report a problem with this question
Let p p p q q q
A p ∧ q p \wedge q p ∧ q B ∼ p ∧ q \sim p \wedge q ∼ p ∧ q C ∼ p ∧ ∼ q \sim p \wedge \sim q ∼ p ∧ ∼ q D ∼ p ∨ ∼ q \sim p \vee \sim q ∼ p ∨ ∼ q E p ∨ q p \vee q p ∨ q
Worked solution (try it first) "Neither
p p p nor
q q q " means "not
p p p and not
q q q ".
"Not" is
∼ \sim ∼ and "and" is
∧ \wedge ∧ , so the statement is
∼ p ∧ ∼ q \sim p \wedge \sim q ∼ p ∧ ∼ q , option C.
Watch out
"Neither … nor" joins the two negations with "and". Joining them with "or" gives ∼ p ∨ ∼ q \sim p \vee \sim q ∼ p ∨ ∼ q Report a problem with this question
If sin θ = cos 2 θ \sin\theta = \cos2\theta sin θ = cos 2 θ 0 ∘ < θ < 90 ∘ 0^\circ < \theta < 90^\circ 0 ∘ < θ < 9 0 ∘ 4 θ 4\theta 4 θ
A 30 ∘ 30^\circ 3 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 90 ∘ 90^\circ 9 0 ∘ D 120 ∘ 120^\circ 12 0 ∘ E 150 ∘ 150^\circ 15 0 ∘
Worked solution (try it first) Write
cos 2 θ \cos 2\theta cos 2 θ in terms of
sin θ \sin\theta sin θ :
cos 2 θ = 1 − 2 sin 2 θ \cos 2\theta = 1 - 2\sin^2\theta cos 2 θ = 1 − 2 sin 2 θ , so
sin θ = 1 − 2 sin 2 θ \sin\theta = 1 - 2\sin^2\theta sin θ = 1 − 2 sin 2 θ .
Bring everything to one side:
2 sin 2 θ + sin θ − 1 = 0 2\sin^2\theta + \sin\theta - 1 = 0 2 sin 2 θ + sin θ − 1 = 0 , which factorises as
( 2 sin θ − 1 ) ( sin θ + 1 ) = 0 (2\sin\theta - 1)(\sin\theta + 1) = 0 ( 2 sin θ − 1 ) ( sin θ + 1 ) = 0 .
θ \theta θ is acute, so
sin θ \sin\theta sin θ is positive:
sin θ = 1 2 \sin\theta = \frac12 sin θ = 2 1 and
θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ .
So
4 θ = 120 ∘ 4\theta = 120^\circ 4 θ = 12 0 ∘ , option D.
Watch out
The question asks for 4 θ 4\theta 4 θ θ \theta θ θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ Report a problem with this question
If y = x 5 + 3 x 4 − 2 x 2 + 2 y = x^5 + 3x^4 - 2x^2 + 2 y = x 5 + 3 x 4 − 2 x 2 + 2 d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y
A 5 x 4 + 12 x 3 − 4 x 5x^4 + 12x^3 - 4x 5 x 4 + 12 x 3 − 4 x B 20 x 3 + 36 x 2 − 4 x 20x^3 + 36x^2 - 4x 20 x 3 + 36 x 2 − 4 x C 5 x 4 + 12 x 3 + 4 x 5x^4 + 12x^3 + 4x 5 x 4 + 12 x 3 + 4 x D 20 x 3 + 36 x 2 − 4 20x^3 + 36x^2 - 4 20 x 3 + 36 x 2 − 4 E 20 x 2 + 12 x 3 + 4 20x^2 + 12x^3 + 4 20 x 2 + 12 x 3 + 4
Worked solution (try it first) Differentiate once (multiply by the power, then take 1 off it.
The constant goes):
d y d x = 5 x 4 + 12 x 3 − 4 x \dfrac{dy}{dx} = 5x^4 + 12x^3 - 4x d x d y = 5 x 4 + 12 x 3 − 4 x .
Differentiate again:
d 2 y d x 2 = 20 x 3 + 36 x 2 − 4 \dfrac{d^2y}{dx^2} = 20x^3 + 36x^2 - 4 d x 2 d 2 y = 20 x 3 + 36 x 2 − 4 , option D.
Watch out
Differentiate every term the second time: − 4 x -4x − 4 x − 4 -4 − 4 − 4 x -4x − 4 x Report a problem with this question
Find the minimum value of y = x 2 − 2 x − 3 y = x^2 - 2x - 3 y = x 2 − 2 x − 3
A 4 B 1 C − 1 -1 − 1 D − 4 -4 − 4 E − 6 -6 − 6
Worked solution (try it first) Differentiate:
d y d x = 2 x − 2 \dfrac{dy}{dx} = 2x - 2 d x d y = 2 x − 2 , which is zero at
x = 1 x = 1 x = 1 .
d 2 y d x 2 = 2 \dfrac{d^2y}{dx^2} = 2 d x 2 d 2 y = 2 is positive, so this turning point is a minimum.
The minimum value is
y = 1 2 − 2 ( 1 ) − 3 = − 4 y = 1^2 - 2(1) - 3 = -4 y = 1 2 − 2 ( 1 ) − 3 = − 4 , option D.
Watch out
The minimum value is y y y x x x x = 1 x = 1 x = 1 Also set as JAMB 2014 · UTME · Q40
Report a problem with this question
Given that P 1 ( x ) = 6 x 3 + 8 x 2 − 4 x + 4 P_1(x) = 6x^3 + 8x^2 - 4x + 4 P 1 ( x ) = 6 x 3 + 8 x 2 − 4 x + 4 P 2 ( x ) = 3 x 2 − 5 x + 1 P_2(x) = 3x^2 - 5x + 1 P 2 ( x ) = 3 x 2 − 5 x + 1 P 1 ( x ) − 2 P 2 ( x ) P_1(x) - 2P_2(x) P 1 ( x ) − 2 P 2 ( x )
A 6 x 3 + 14 x 2 − 9 x + 5 6x^3 + 14x^2 - 9x + 5 6 x 3 + 14 x 2 − 9 x + 5 B 6 x 3 + 8 x 2 + 6 x − 5 6x^3 + 8x^2 + 6x - 5 6 x 3 + 8 x 2 + 6 x − 5 C 6 x 3 + 8 x 2 − 9 x + 5 6x^3 + 8x^2 - 9x + 5 6 x 3 + 8 x 2 − 9 x + 5 D 6 x 3 + 3 x 2 − 14 x + 6 6x^3 + 3x^2 - 14x + 6 6 x 3 + 3 x 2 − 14 x + 6 E 6 x 3 + 2 x 2 + 6 x + 2 6x^3 + 2x^2 + 6x + 2 6 x 3 + 2 x 2 + 6 x + 2
Worked solution (try it first) Double every term of
P 2 P_2 P 2 :
2 P 2 ( x ) = 6 x 2 − 10 x + 2 2P_2(x) = 6x^2 - 10x + 2 2 P 2 ( x ) = 6 x 2 − 10 x + 2 .
Taking it away changes every sign:
P 1 ( x ) − 2 P 2 ( x ) = 6 x 3 + 8 x 2 − 4 x + 4 − 6 x 2 + 10 x − 2 P_1(x) - 2P_2(x) = 6x^3 + 8x^2 - 4x + 4 - 6x^2 + 10x - 2 P 1 ( x ) − 2 P 2 ( x ) = 6 x 3 + 8 x 2 − 4 x + 4 − 6 x 2 + 10 x − 2 .
Collect like terms:
8 x 2 − 6 x 2 = 2 x 2 8x^2 - 6x^2 = 2x^2 8 x 2 − 6 x 2 = 2 x 2 ,
− 4 x + 10 x = 6 x -4x + 10x = 6x − 4 x + 10 x = 6 x and
4 − 2 = 2 4 - 2 = 2 4 − 2 = 2 .
So
P 1 ( x ) − 2 P 2 ( x ) = 6 x 3 + 2 x 2 + 6 x + 2 P_1(x) - 2P_2(x) = 6x^3 + 2x^2 + 6x + 2 P 1 ( x ) − 2 P 2 ( x ) = 6 x 3 + 2 x 2 + 6 x + 2 , option E.
Watch out
Subtracting 2 P 2 2P_2 2 P 2 − 10 x -10x − 10 x + 10 x +10x + 10 x 8 x 2 + 6 x 2 = 14 x 2 8x^2 + 6x^2 = 14x^2 8 x 2 + 6 x 2 = 14 x 2 14 x 2 14x^2 14 x 2 Report a problem with this question
Find the values of x x x g ( x ) = x 2 − 5 x + 6 x 2 − 5 x − 24 g(x) = \dfrac{x^2 - 5x + 6}{x^2 - 5x - 24} g ( x ) = x 2 − 5 x − 24 x 2 − 5 x + 6
A x = 3 x = 3 x = 3 x = 8 x = 8 x = 8 B x = − 3 x = -3 x = − 3 x = 8 x = 8 x = 8 C x = − 8 x = -8 x = − 8 x = − 3 x = -3 x = − 3 D x = − 3 x = -3 x = − 3 x = − 2 x = -2 x = − 2 E x = 3 x = 3 x = 3 x = 2 x = 2 x = 2
Worked solution (try it first) A fraction is undefined where its bottom is zero:
x 2 − 5 x − 24 = 0 x^2 - 5x - 24 = 0 x 2 − 5 x − 24 = 0 .
Two numbers that multiply to
− 24 -24 − 24 and add to
− 5 -5 − 5 are
− 8 -8 − 8 and 3, so
( x − 8 ) ( x + 3 ) = 0 (x - 8)(x + 3) = 0 ( x − 8 ) ( x + 3 ) = 0 .
So
x = 8 x = 8 x = 8 or
x = − 3 x = -3 x = − 3 , option B.
Watch out
Only the bottom matters. The top, x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6 x = 2 x = 2 x = 2 x = 3 x = 3 x = 3 Report a problem with this question
Find the distance between the points ( 2 , 3 ) (2, 3) ( 2 , 3 ) ( − 1 , 7 ) (-1, 7) ( − 1 , 7 )
Worked solution (try it first) Find the differences:
x x x changes by
− 1 − 2 = − 3 -1 - 2 = -3 − 1 − 2 = − 3 and
y y y by
7 − 3 = 4 7 - 3 = 4 7 − 3 = 4 .
By Pythagoras, the distance is
( − 3 ) 2 + 4 2 = 25 \sqrt{(-3)^2 + 4^2} = \sqrt{25} ( − 3 ) 2 + 4 2 = 25 , which is 5, option C.
Watch out
Square the differences before adding, then take the square root. Just adding their sizes, 3 + 4 3 + 4 3 + 4 Report a problem with this question
Find the equation of the straight line which passes through ( 2 , 2 ) (2, 2) ( 2 , 2 ) ( 3 , 6 ) (3, 6) ( 3 , 6 )
A y = 3 x − 2 y = 3x - 2 y = 3 x − 2 B y = 3 x + 2 y = 3x + 2 y = 3 x + 2 C y = 4 x − 6 y = 4x - 6 y = 4 x − 6 D y = 4 x + 6 y = 4x + 6 y = 4 x + 6 E y = − 4 x + 2 y = -4x + 2 y = − 4 x + 2
Worked solution (try it first) The gradient is
6 − 2 3 − 2 = 4 \dfrac{6 - 2}{3 - 2} = 4 3 − 2 6 − 2 = 4 .
Use
y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) with
( 2 , 2 ) (2, 2) ( 2 , 2 ) :
y − 2 = 4 ( x − 2 ) y - 2 = 4(x - 2) y − 2 = 4 ( x − 2 ) , so
y − 2 = 4 x − 8 y - 2 = 4x - 8 y − 2 = 4 x − 8 .
Add 2 to both sides:
y = 4 x − 6 y = 4x - 6 y = 4 x − 6 , option C.
Watch out
Expand 4 ( x − 2 ) 4(x - 2) 4 ( x − 2 ) 4 x − 8 4x - 8 4 x − 8 − 6 -6 − 6 y = 4 x + 6 y = 4x + 6 y = 4 x + 6 ( 2 , 2 ) (2, 2) ( 2 , 2 ) 4 ( 2 ) + 6 = 14 4(2) + 6 = 14 4 ( 2 ) + 6 = 14 Report a problem with this question
Find the focus and directrix of the parabola y 2 = 64 x y^2 = 64x y 2 = 64 x
A ( 4 , 0 ) (4, 0) ( 4 , 0 ) − 4 -4 − 4 B ( 16 , 0 ) (16, 0) ( 16 , 0 ) + 16 +16 + 16 C ( 16 , 0 ) (16, 0) ( 16 , 0 ) − 16 -16 − 16 D ( 4 , 0 ) (4, 0) ( 4 , 0 ) + 4 +4 + 4 E ( 8 , 0 ) (8, 0) ( 8 , 0 ) − 8 -8 − 8
Worked solution (try it first) Compare
y 2 = 64 x y^2 = 64x y 2 = 64 x with the standard form
y 2 = 4 a x y^2 = 4ax y 2 = 4 a x :
4 a = 64 4a = 64 4 a = 64 , so
a = 16 a = 16 a = 16 .
The focus is
( a , 0 ) = ( 16 , 0 ) (a, 0) = (16, 0) ( a , 0 ) = ( 16 , 0 ) and the directrix is the line
x = − a x = -a x = − a , that is
x = − 16 x = -16 x = − 16 , option C.
Watch out
The directrix x = − a x = -a x = − a + 16 +16 + 16 Report a problem with this question
Find the equation of the circle which has its centre at ( − 2 , 4 ) (-2, 4) ( − 2 , 4 ) ( 1 , 5 ) (1, 5) ( 1 , 5 )
A x 2 + y 2 + x − 3 y + 9 = 0 x^2 + y^2 + x - 3y + 9 = 0 x 2 + y 2 + x − 3 y + 9 = 0 B x 2 + y 2 + 2 x + 4 y − 16 = 0 x^2 + y^2 + 2x + 4y - 16 = 0 x 2 + y 2 + 2 x + 4 y − 16 = 0 C x 2 − y 2 + 4 x + 8 y + 10 = 0 x^2 - y^2 + 4x + 8y + 10 = 0 x 2 − y 2 + 4 x + 8 y + 10 = 0 D x 2 + y 2 − 2 x + 8 y − 10 = 0 x^2 + y^2 - 2x + 8y - 10 = 0 x 2 + y 2 − 2 x + 8 y − 10 = 0 E x 2 + y 2 + 4 x − 8 y + 10 = 0 x^2 + y^2 + 4x - 8y + 10 = 0 x 2 + y 2 + 4 x − 8 y + 10 = 0
Worked solution (try it first) The radius is the distance from the centre
( − 2 , 4 ) (-2, 4) ( − 2 , 4 ) to
( 1 , 5 ) (1, 5) ( 1 , 5 ) :
r 2 = ( 1 + 2 ) 2 + ( 5 − 4 ) 2 = 10 r^2 = (1 + 2)^2 + (5 - 4)^2 = 10 r 2 = ( 1 + 2 ) 2 + ( 5 − 4 ) 2 = 10 .
A centre
( a , b ) (a, b) ( a , b ) gives
( x − a ) 2 + ( y − b ) 2 = r 2 (x - a)^2 + (y - b)^2 = r^2 ( x − a ) 2 + ( y − b ) 2 = r 2 , so here
( x + 2 ) 2 + ( y − 4 ) 2 = 10 (x + 2)^2 + (y - 4)^2 = 10 ( x + 2 ) 2 + ( y − 4 ) 2 = 10 .
Expand:
x 2 + 4 x + 4 + y 2 − 8 y + 16 = 10 x^2 + 4x + 4 + y^2 - 8y + 16 = 10 x 2 + 4 x + 4 + y 2 − 8 y + 16 = 10 .
Take 10 from both sides:
x 2 + y 2 + 4 x − 8 y + 10 = 0 x^2 + y^2 + 4x - 8y + 10 = 0 x 2 + y 2 + 4 x − 8 y + 10 = 0 , option E.
Watch out
The signs in the brackets are opposite to the centre's: ( − 2 , 4 ) (-2, 4) ( − 2 , 4 ) ( x + 2 ) (x + 2) ( x + 2 ) ( y − 4 ) (y - 4) ( y − 4 ) + 4 x +4x + 4 x − 8 y -8y − 8 y ( x − 2 ) 2 + ( y + 4 ) 2 (x - 2)^2 + (y + 4)^2 ( x − 2 ) 2 + ( y + 4 ) 2 − 4 x -4x − 4 x + 8 y +8y + 8 y Report a problem with this question
In how many ways can 10 people be seated at a round table if there is no restriction?
A 362,880 B 40,320 C 5,040 D 720 E 120
Worked solution (try it first) At a round table, turning everyone one seat round gives the same seating, so fix one person's seat.
The other 9 people can then sit in
9 ! 9! 9 ! orders, and
9 ! = 362 880 9! = 362\,880 9 ! = 362 880 , option A.
Watch out
Fix only one person, so the count is ( n − 1 ) ! = 9 ! (n - 1)! = 9! ( n − 1 )! = 9 ! 8 ! = 40 320 8! = 40\,320 8 ! = 40 320 Report a problem with this question
Find the unit vector in the direction of r = 15 i + 16 j − 12 k \mathbf{r} = 15\mathbf{i} + 16\mathbf{j} - 12\mathbf{k} r = 15 i + 16 j − 12 k
A 1 25 ( 15 i + 16 j − 12 k ) \frac1{25}(15\mathbf{i} + 16\mathbf{j} - 12\mathbf{k}) 25 1 ( 15 i + 16 j − 12 k ) B 1 3 ( 15 i + 16 j − 12 k ) \frac13(15\mathbf{i} + 16\mathbf{j} - 12\mathbf{k}) 3 1 ( 15 i + 16 j − 12 k ) C 1 9 ( 3 i + 4 j − 5 k ) \frac19(3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) 9 1 ( 3 i + 4 j − 5 k ) D 1 3 ( 15 i + 16 j + 12 k ) \frac13(15\mathbf{i} + 16\mathbf{j} + 12\mathbf{k}) 3 1 ( 15 i + 16 j + 12 k ) E 1 3 ( 3 i + 4 j − 5 k ) \frac1{\sqrt3}(3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) 3 1 ( 3 i + 4 j − 5 k )
Worked solution (try it first) Find the length:
∣ r ∣ = 15 2 + 16 2 + ( − 12 ) 2 |\mathbf r| = \sqrt{15^2 + 16^2 + (-12)^2} ∣ r ∣ = 1 5 2 + 1 6 2 + ( − 12 ) 2 = 225 + 256 + 144 = \sqrt{225 + 256 + 144} = 225 + 256 + 144 .
That is
625 = 25 \sqrt{625} = 25 625 = 25 .
Divide
r \mathbf r r by its length:
r ^ = 1 25 ( 15 i + 16 j − 12 k ) \hat{\mathbf r} = \frac{1}{25}(15\mathbf i + 16\mathbf j - 12\mathbf k) r ^ = 25 1 ( 15 i + 16 j − 12 k ) , option A.
Watch out
A unit vector is the vector divided by its length, and the direction stays the same, so keep − 12 k -12\mathbf k − 12 k + 12 k +12\mathbf k + 12 k Report a problem with this question
If s = 2 i − 6 j \mathbf{s} = 2\mathbf{i} - 6\mathbf{j} s = 2 i − 6 j t = − 3 i + 4 j \mathbf{t} = -3\mathbf{i} + 4\mathbf{j} t = − 3 i + 4 j ( t + 2 s ) ⋅ ( t − s ) (\mathbf{t} + 2\mathbf{s}) \cdot (\mathbf{t} - \mathbf{s}) ( t + 2 s ) ⋅ ( t − s )
A − 85 -85 − 85 B − 75 -75 − 75 C 18 D 50 E 55
Worked solution (try it first) t + 2 s = ( − 3 + 4 ) i + ( 4 − 12 ) j \mathbf t + 2\mathbf s = (-3 + 4)\mathbf i + (4 - 12)\mathbf j t + 2 s = ( − 3 + 4 ) i + ( 4 − 12 ) j = i − 8 j = \mathbf i - 8\mathbf j = i − 8 j .
t − s = ( − 3 − 2 ) i + ( 4 + 6 ) j \mathbf t - \mathbf s = (-3 - 2)\mathbf i + (4 + 6)\mathbf j t − s = ( − 3 − 2 ) i + ( 4 + 6 ) j = − 5 i + 10 j = -5\mathbf i + 10\mathbf j = − 5 i + 10 j .
The scalar product multiplies the
i \mathbf i i parts and the
j \mathbf j j parts, then adds:
( 1 ) ( − 5 ) + ( − 8 ) ( 10 ) = − 5 − 80 (1)(-5) + (-8)(10) = -5 - 80 ( 1 ) ( − 5 ) + ( − 8 ) ( 10 ) = − 5 − 80 .
So the value is
− 85 -85 − 85 , option A.
Watch out
Keep the sign of every product: ( 1 ) ( − 5 ) = − 5 (1)(-5) = -5 ( 1 ) ( − 5 ) = − 5 − 5 − 80 = − 85 -5 - 80 = -85 − 5 − 80 = − 85 + 5 +5 + 5 − 75 -75 − 75 Report a problem with this question
Find the value of the scalar λ \lambda λ 3 i + 5 λ j 3\mathbf{i} + 5\lambda\mathbf{j} 3 i + 5 λ j 2 i − 6 j 2\mathbf{i} - 6\mathbf{j} 2 i − 6 j
A 5 3 \frac53 3 5 B 1 5 \frac15 5 1 C 0 D − 1 5 -\frac15 − 5 1 E − 3 5 -\frac35 − 5 3
Worked solution (try it first) Perpendicular vectors have a scalar product of 0.
The scalar product is
( 3 ) ( 2 ) + ( 5 λ ) ( − 6 ) = 6 − 30 λ (3)(2) + (5\lambda)(-6) = 6 - 30\lambda ( 3 ) ( 2 ) + ( 5 λ ) ( − 6 ) = 6 − 30 λ .
Set
6 − 30 λ = 0 6 - 30\lambda = 0 6 − 30 λ = 0 :
30 λ = 6 30\lambda = 6 30 λ = 6 , so
λ = 1 5 \lambda = \frac15 λ = 5 1 , option B.
Watch out
From 6 − 30 λ = 0 6 - 30\lambda = 0 6 − 30 λ = 0 30 λ 30\lambda 30 λ 30 λ = 6 30\lambda = 6 30 λ = 6 λ \lambda λ − 1 5 -\frac15 − 5 1 Report a problem with this question
A body of mass 3 kg at rest is acted upon by a force of 24 N for 0.5 s. Find the final velocity of the body.
A 12 ms − 1 12\text{ ms}^{-1} 12 ms − 1 B 8 ms − 1 8\text{ ms}^{-1} 8 ms − 1 C 6 ms − 1 6\text{ ms}^{-1} 6 ms − 1 D 4 ms − 1 4\text{ ms}^{-1} 4 ms − 1 E 1.5 ms − 1 1.5\text{ ms}^{-1} 1.5 ms − 1
Worked solution (try it first) Newton's second law:
a = F m = 24 3 a = \dfrac{F}{m} = \dfrac{24}{3} a = m F = 3 24 , which is
8 m s − 2 8\text{ m s}^{-2} 8 m s − 2 .
It starts from rest, so
v = u + a t = 0 + 8 × 0.5 v = u + at = 0 + 8 \times 0.5 v = u + a t = 0 + 8 × 0.5 , which is
4 m s − 1 4\text{ m s}^{-1} 4 m s − 1 , option D.
Watch out
F × t = 24 × 0.5 = 12 F \times t = 24 \times 0.5 = 12 F × t = 24 × 0.5 = 12 12 ÷ 3 = 4 m s − 1 12 \div 3 = 4\text{ m s}^{-1} 12 ÷ 3 = 4 m s − 1 Report a problem with this question
If x = 4 i + 5 j \mathbf{x} = 4\mathbf{i} + 5\mathbf{j} x = 4 i + 5 j y = − 3 i + 4 j \mathbf{y} = -3\mathbf{i} + 4\mathbf{j} y = − 3 i + 4 j x − y \mathbf{x} - \mathbf{y} x − y
A 130 \sqrt{130} 130 B 82 \sqrt{82} 82 C 5 2 5\sqrt2 5 2 D 2 5 2\sqrt5 2 5 E 2 \sqrt2 2
Worked solution (try it first) Subtract part by part:
x − y = ( 4 − ( − 3 ) ) i + ( 5 − 4 ) j \mathbf x - \mathbf y = (4 - (-3))\mathbf i + (5 - 4)\mathbf j x − y = ( 4 − ( − 3 )) i + ( 5 − 4 ) j = 7 i + j = 7\mathbf i + \mathbf j = 7 i + j .
Its modulus is
7 2 + 1 2 = 50 \sqrt{7^2 + 1^2} = \sqrt{50} 7 2 + 1 2 = 50 .
50 = 25 × 2 = 5 2 \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2 50 = 25 × 2 = 5 2 , option C.
Watch out
Subtracting − 3 i -3\mathbf i − 3 i i \mathbf i i 4 − 3 = 1 4 - 3 = 1 4 − 3 = 1 i + j \mathbf i + \mathbf j i + j 2 \sqrt2 2 Report a problem with this question
Two forces X X X Y Y Y R R R X X X R R R 30 ∘ 30^\circ 3 0 ∘ X X X Y Y Y
A 152 ∘ 152^\circ 15 2 ∘ B 75 ∘ 75^\circ 7 5 ∘ C 52 ∘ 52^\circ 5 2 ∘ D 30 ∘ 30^\circ 3 0 ∘ E 22 ∘ 22^\circ 2 2 ∘
Worked solution (try it first) Draw
X X X and
Y Y Y head to tail.
Its sides are 15, 20 and
R R R .
The
30 ∘ 30^\circ 3 0 ∘ between
X X X and
R R R is opposite the side
Y = 20 Y = 20 Y = 20 .
Let
ϕ \phi ϕ be the angle opposite
X = 15 X = 15 X = 15 .
Sine rule:
sin ϕ = 15 sin 30 ∘ 20 \sin\phi = \dfrac{15\sin30^\circ}{20} sin ϕ = 20 15 sin 3 0 ∘ = 0.375 = 0.375 = 0.375 , so
ϕ = 22.0 ∘ \phi = 22.0^\circ ϕ = 22. 0 ∘ .
The angle between
X X X and
Y Y Y is the outside angle of the triangle where they meet, which equals the two opposite inside angles added:
30 ∘ + 22.0 ∘ ≈ 52 ∘ 30^\circ + 22.0^\circ \approx 52^\circ 3 0 ∘ + 22. 0 ∘ ≈ 5 2 ∘ , option C.
Watch out
ϕ = 22 ∘ \phi = 22^\circ ϕ = 2 2 ∘ Y Y Y R R R X X X Y Y Y 30 ∘ 30^\circ 3 0 ∘ 52 ∘ 52^\circ 5 2 ∘ Report a problem with this question
A body of mass 700 kg moving with a velocity of 36 kmh − 1 36\text{ kmh}^{-1} 36 kmh − 1 90 kmh − 1 90\text{ kmh}^{-1} 90 kmh − 1
A 700 N B 673 N C 550 N D 460 N E 250 N
Worked solution (try it first) Change the speeds to m/s by dividing by 3.6:
36 ÷ 3.6 = 10 36 \div 3.6 = 10 36 ÷ 3.6 = 10 and
90 ÷ 3.6 = 25 90 \div 3.6 = 25 90 ÷ 3.6 = 25 .
The acceleration is
a = v − u t a = \dfrac{v - u}{t} a = t v − u = 25 − 10 15.6 = \dfrac{25 - 10}{15.6} = 15.6 25 − 10 , which is about
0.9615 m s − 2 0.9615\text{ m s}^{-2} 0.9615 m s − 2 .
By
F = m a F = ma F = ma ,
F = 700 × 0.9615 ≈ 673 F = 700 \times 0.9615 \approx 673 F = 700 × 0.9615 ≈ 673 N, option B.
Watch out
Change km/h to m/s before using F = m a F = ma F = ma a = 54 15.6 a = \frac{54}{15.6} a = 15.6 54 Report a problem with this question
Find the magnitude of the resultant of two forces of 5 N and 4 N acting at a point if the angle between them is 60 ∘ 60^\circ 6 0 ∘
A 4.6 N B 4.9 N C 6.5 N D 7.8 N E 9.0 N
Worked solution (try it first) For two forces
P P P and
Q Q Q at an angle
θ \theta θ , the resultant is given by
R 2 = P 2 + Q 2 + 2 P Q cos θ R^2 = P^2 + Q^2 + 2PQ\cos\theta R 2 = P 2 + Q 2 + 2 P Q cos θ .
R 2 = 25 + 16 + 2 ( 5 ) ( 4 ) cos 60 ∘ R^2 = 25 + 16 + 2(5)(4)\cos60^\circ R 2 = 25 + 16 + 2 ( 5 ) ( 4 ) cos 6 0 ∘ , and
cos 60 ∘ = 1 2 \cos60^\circ = \frac12 cos 6 0 ∘ = 2 1 , so
R 2 = 41 + 20 = 61 R^2 = 41 + 20 = 61 R 2 = 41 + 20 = 61 .
So
R = 61 ≈ 7.8 R = \sqrt{61} \approx 7.8 R = 61 ≈ 7.8 N, option D.
Watch out
With θ \theta θ + 2 P Q cos θ +2PQ\cos\theta + 2 P Q cos θ 21 ≈ 4.6 \sqrt{21} \approx 4.6 21 ≈ 4.6 Report a problem with this question
A horizontal force of 3500 N is capable of moving an abandoned vehicle of mass 750 kg on a rough horizontal road. Find the coefficient of friction between the vehicle and the road. [Take g = 9.8 ms − 2 g = 9.8\text{ ms}^{-2} g = 9.8 ms − 2
A 0.43 B 0.45 C 0.48 D 0.50 E 0.52
Worked solution (try it first) On a horizontal road the normal reaction equals the weight:
R = m g = 750 × 9.8 R = mg = 750 \times 9.8 R = m g = 750 × 9.8 , which is 7350 N.
The force just moves the vehicle, so friction is limiting:
μ R = 3500 \mu R = 3500 μ R = 3500 .
So
μ = 3500 7350 ≈ 0.48 \mu = \dfrac{3500}{7350} \approx 0.48 μ = 7350 3500 ≈ 0.48 , option C.
Watch out
Use the value of g g g g = 10 g = 10 g = 10 μ = 3500 7500 ≈ 0.47 \mu = \frac{3500}{7500} \approx 0.47 μ = 7500 3500 ≈ 0.47 g = 9.8 g = 9.8 g = 9.8 Report a problem with this question
A 50 kg bag of rice is placed in a lift which moves with an upward acceleration of 8 ms − 2 8\text{ ms}^{-2} 8 ms − 2 g = 10 ms − 2 g = 10\text{ ms}^{-2} g = 10 ms − 2
A 100 N B 500 N C 900 N D 1,400 N E 4,000 N
Worked solution (try it first) The forces on the rice are the reaction
R R R upwards and the weight
m g mg m g downwards.
The acceleration is upwards, so
R − m g = m a R - mg = ma R − m g = ma .
So
R = m ( g + a ) = 50 ( 10 + 8 ) R = m(g + a) = 50(10 + 8) R = m ( g + a ) = 50 ( 10 + 8 ) , which is 900 N, option C.
Watch out
Accelerating upwards, the floor pushes harder than the weight, so add a a a g g g 50 ( 10 − 8 ) = 100 50(10 - 8) = 100 50 ( 10 − 8 ) = 100 Report a problem with this question
The probabilities of three events A A A B B B C C C 1 3 \frac13 3 1 1 2 \frac12 2 1 2 3 \frac23 3 2
A 1 81 \frac1{81} 81 1 B 1 6 \frac16 6 1 C 2 9 \frac29 9 2 D 1 3 \frac13 3 1 E 7 18 \frac7{18} 18 7
Worked solution (try it first) "Only two" means exactly one event fails.
The chances of failing are
2 3 \frac23 3 2 for
A A A ,
1 2 \frac12 2 1 for
B B B and
1 3 \frac13 3 1 for
C C C .
A A A and
B B B only:
1 3 × 1 2 × 1 3 = 1 18 \frac13 \times \frac12 \times \frac13 = \frac{1}{18} 3 1 × 2 1 × 3 1 = 18 1 .
A A A and
C C C only:
1 3 × 1 2 × 2 3 = 2 18 \frac13 \times \frac12 \times \frac23 = \frac{2}{18} 3 1 × 2 1 × 3 2 = 18 2 .
B B B and
C C C only:
2 3 × 1 2 × 2 3 = 4 18 \frac23 \times \frac12 \times \frac23 = \frac{4}{18} 3 2 × 2 1 × 3 2 = 18 4 .
The three cases can't happen together, so add:
1 + 2 + 4 18 = 7 18 \frac{1 + 2 + 4}{18} = \frac{7}{18} 18 1 + 2 + 4 = 18 7 , option E.
Watch out
Each case must include the chance that the third event fails. Without it, "A A A B B B 1 3 × 1 2 = 1 6 \frac13 \times \frac12 = \frac16 3 1 × 2 1 = 6 1 Report a problem with this question
One out of a thousand bulbs manufactured by a factory is found to be defective. If 2,000 bulbs were produced, find the probability that at least two bulbs are defective.
A 0.2706 B 0.4059 C 0.5940 D 0.7294 E 0.8647
Worked solution (try it first) n = 2000 n = 2000 n = 2000 is large and
p = 0.001 p = 0.001 p = 0.001 is small, so use the Poisson approximation with
λ = n p = 2 \lambda = np = 2 λ = n p = 2 .
"At least two" is 1 minus the chances of 0 and 1:
P ( X ≥ 2 ) = 1 − e − 2 ( 1 + 2 ) P(X \ge 2) = 1 - e^{-2}(1 + 2) P ( X ≥ 2 ) = 1 − e − 2 ( 1 + 2 ) = 1 − 3 e − 2 = 1 - 3e^{-2} = 1 − 3 e − 2 .
3 e − 2 ≈ 0.4060 3e^{-2} \approx 0.4060 3 e − 2 ≈ 0.4060 , so
P ( X ≥ 2 ) ≈ 0.5940 P(X \ge 2) \approx 0.5940 P ( X ≥ 2 ) ≈ 0.5940 , option C.
Watch out
"At least two" leaves out both 0 and 1 defective. Taking away only P ( 0 ) = e − 2 P(0) = e^{-2} P ( 0 ) = e − 2 1 − e − 2 ≈ 0.8647 1 - e^{-2} \approx 0.8647 1 − e − 2 ≈ 0.8647 Report a problem with this question
In how many ways can a committee of eight councillors be selected from ten councillors in a Local Government?
Worked solution (try it first) A committee is a selection, so order doesn't matter: the count is
( 10 8 ) \binom{10}{8} ( 8 10 ) .
Choosing 8 to serve is the same as choosing 2 to leave out, so
( 10 8 ) = ( 10 2 ) \binom{10}{8} = \binom{10}{2} ( 8 10 ) = ( 2 10 ) .
( 10 2 ) = 10 × 9 2 × 1 \binom{10}{2} = \dfrac{10 \times 9}{2 \times 1} ( 2 10 ) = 2 × 1 10 × 9 Watch out
Divide by 2 ! 2! 2 ! 10 × 9 = 90 10 \times 9 = 90 10 × 9 = 90 Report a problem with this question
A fair die is rolled once. What is the probability of obtaining a multiple of 2 or 3?
A 1 6 \frac16 6 1 B 1 3 \frac13 3 1 C 1 2 \frac12 2 1 D 2 3 \frac23 3 2 E 5 6 \frac56 6 5
Worked solution (try it first) The multiples of 2 or 3 on a die are 2, 3, 4 and 6.
That is 4 of the 6 equally likely outcomes, so
P = 4 6 = 2 3 P = \frac46 = \frac23 P = 6 4 = 3 2 , option D.
Watch out
6 is a multiple of both 2 and 3, so count it once. Adding 3 6 + 2 6 \frac36 + \frac26 6 3 + 6 2 5 6 \frac56 6 5 Report a problem with this question
The table shows the scores of students in a Mathematics examination. Find the lower class boundary of the class containing the sixth decile.
Scores
41–50
51–60
61–70
71–80
81–90
91–100
Frequency
8
12
14
16
8
2
A 40.5 B 50.5 C 60.5 D 70.5 E 80.5
Worked solution (try it first) Add the frequencies:
N = 8 + 12 + 14 + 16 + 8 + 2 = 60 N = 8 + 12 + 14 + 16 + 8 + 2 = 60 N = 8 + 12 + 14 + 16 + 8 + 2 = 60 .
The sixth decile is at
6 10 × 60 \frac{6}{10} \times 60 10 6 × 60 , the 36th score.
The cumulative frequencies are 8, 20, 34, 50, … The first 34 scores fill the classes up to 61–70, so the 36th is in 71–80.
The lower class boundary of 71–80 is halfway between 70 and 71: 70.5, option D.
Watch out
The 36th score comes after the first 34, so it is in the next class, 71–80. Stopping at 61–70, where the running total is 34, gives 60.5 (option C). Report a problem with this question
Using the same table, what is the modal score?
Scores
41–50
51–60
61–70
71–80
81–90
91–100
Frequency
8
12
14
16
8
2
A 70.5 B 71.5 C 72.5 D 73.5 E 79.5
Worked solution (try it first) The modal class is 71–80, with the highest frequency, 16.
Its lower boundary is 70.5 and its width is
c = 10 c = 10 c = 10 .
D 1 = 16 − 14 = 2 D_1 = 16 - 14 = 2 D 1 = 16 − 14 = 2 (the step up from the class before) and
D 2 = 16 − 8 = 8 D_2 = 16 - 8 = 8 D 2 = 16 − 8 = 8 (the step down to the class after).
Mode
= 70.5 + 2 2 + 8 × 10 = 70.5 + \dfrac{2}{2 + 8} \times 10 = 70.5 + 2 + 8 2 × 10 , which is
70.5 + 2 = 72.5 70.5 + 2 = 72.5 70.5 + 2 = 72.5 , option C.
Watch out
Start from the lower class boundary, 70.5, not the lower limit 71. Using 71 gives 71 + 2 = 73 71 + 2 = 73 71 + 2 = 73 Report a problem with this question
In an examination, 75 % 75\% 75%
A 0.0003 B 0.0004 C 0.0005 D 0.0103 E 0.0415
Worked solution (try it first) Let
X X X be the number who pass:
X ∼ B ( 8 , 0.75 ) X \sim B(8, 0.75) X ∼ B ( 8 , 0.75 ) , with
p = 0.75 p = 0.75 p = 0.75 for a pass and
q = 0.25 q = 0.25 q = 0.25 for a failure.
At most 1 pass means
X = 0 X = 0 X = 0 or
X = 1 X = 1 X = 1 :
P ( X ≤ 1 ) = q 8 + 8 p q 7 P(X \le 1) = q^8 + 8pq^7 P ( X ≤ 1 ) = q 8 + 8 p q 7 .
0.25 8 ≈ 0.0000153 0.25^8 \approx 0.0000153 0.2 5 8 ≈ 0.0000153 and
8 ( 0.75 ) ( 0.25 ) 7 ≈ 0.0003662 8(0.75)(0.25)^7 \approx 0.0003662 8 ( 0.75 ) ( 0.25 ) 7 ≈ 0.0003662 .
Add them: about 0.000381.
To 4 decimal places this is 0.0004, option B.
Watch out
Round, don't cut off: 0.000381 to 4 decimal places is 0.0004, because the next digit is 8. Chopping it gives 0.0003 (option A). Report a problem with this question
If the payoff matrix of a game is ( 3 6 4 2 ) \begin{pmatrix} 3 & 6 \\ 4 & 2 \end{pmatrix} ( 3 4 6 2 )
Worked solution (try it first) Find the smallest payoff in each row:
min ( 3 , 6 ) = 3 \min(3, 6) = 3 min ( 3 , 6 ) = 3 and
min ( 4 , 2 ) = 2 \min(4, 2) = 2 min ( 4 , 2 ) = 2 .
The maximin value is the largest of these row minima, 3, option D.
Watch out
Maximin means the largest of the row minima. Taking the smaller of them gives 2 (option E). Report a problem with this question
The corner points of the feasible region of a problem are A ( 0 , 3 ) A(0, 3) A ( 0 , 3 ) B ( 2 , 3 ) B(2, 3) B ( 2 , 3 ) C ( 4 , 3 ) C(4, 3) C ( 4 , 3 ) D ( 3 , 4 ) D(3, 4) D ( 3 , 4 ) E ( 0 , 0 ) E(0, 0) E ( 0 , 0 ) P = 35 x + 40 y P = 35x + 40y P = 35 x + 40 y P P P
Worked solution (try it first) The greatest value of
P P P is at a corner of the feasible region, so work out
P = 35 x + 40 y P = 35x + 40y P = 35 x + 40 y at each.
B B B :
70 + 120 = 190 70 + 120 = 190 70 + 120 = 190 .
C C C :
140 + 120 = 260 140 + 120 = 260 140 + 120 = 260 .
D D D :
105 + 160 = 265 105 + 160 = 265 105 + 160 = 265 .
The maximum is 265, at
D ( 3 , 4 ) D(3, 4) D ( 3 , 4 ) , option E.
Watch out
Check every corner: the point with the largest x x x C ( 4 , 3 ) C(4, 3) C ( 4 , 3 ) D ( 3 , 4 ) D(3, 4) D ( 3 , 4 ) y y y Report a problem with this question
The annual demand for a product is 12,500 units, the annual holding cost is ₦50.00 per unit and the cost of placing an order is ₦500.00. Find the number of orders per annum.
Worked solution (try it first) The economic order quantity is
EOQ = 2 D C o C h \text{EOQ} = \sqrt{\dfrac{2DC_o}{C_h}} EOQ = C h 2 D C o , with
D = 12 500 D = 12\,500 D = 12 500 ,
C o = 500 C_o = 500 C o = 500 and
C h = 50 C_h = 50 C h = 50 .
EOQ = 2 × 12 500 × 500 50 \text{EOQ} = \sqrt{\dfrac{2 \times 12\,500 \times 500}{50}} EOQ = 50 2 × 12 500 × 500 = 250 000 = \sqrt{250\,000} = 250 000 , which is 500 units.
The number of orders a year is
12 500 500 = 25 \dfrac{12\,500}{500} = 25 500 12 500 = 25 , option D.
Watch out
The EOQ, 500 units, is the size of each order. The question wants how many orders, so divide the yearly demand by it: 12 500 ÷ 500 = 25 12\,500 \div 500 = 25 12 500 ÷ 500 = 25 Report a problem with this question