Full Mock Test (40 Questions)

Solution

Numerator: 96 ⇒ \(\Large \frac{37.5}{100}\normalsize \: \times \: 96 \\ = \normalsize 36\)

Decreased by 36 ⇒ 96 - 36

∴ New numerator = 60

Denominator: 50 ⇒ \(\Large \frac{20 }{100}\normalsize \: \times \: 50 \\ = \normalsize 10\)

Decreased by 10 ⇒ 50 - 10 = 40

New Denominator = 40

New fraction = \(\Large \frac{60}{40} \\ = \normalsize 1.5\)

Solution:

Step1: ⇒ \(\normalsize \left( 0.0008\right)^{\frac{1}{3}} \: \times \: \left( 0.16\right)^{\frac{3}{2}}\)  

can be written as:

⇒ \(\left ( \Large \frac {8}{1000} \right)^{\frac{1}{3}} \; \times \; \left ( \Large \frac {16}{100} \right)^{\frac{3}{2}}\)

⇒ \(\left ( \Large \frac {2^3}{10^3} \right)^{\frac{1}{3}} \; \times \; \left ( \Large \frac {4^2}{10^2} \right)^{\frac{3}{2}}\)

Step 2: Simplifying further, we have

⇒ \(\Large \frac{2^{3 \; \times \; \frac{1}{3}}}{10^{3 \; \times \; \frac{1}{3}}} \; \times \; \frac{4^{2 \; \times \; \frac{3}{2}}}{10^{2 \; \times \; \frac{3}{2}}}\)

= \(\Large \frac{2}{10} \; \times \; \frac{4^3}{10^3}\)

Step 3: This will give:

⇒ \(\Large \frac{2 \; \times \; 64}{10,000} \\ = \Large \frac{128}{10,000}\)

Step 4: Divide by 4

⇒ \(\Large \frac{32}{2500} \\ = \Large \frac{8}{625}\)

Solution

⇒\(\left( \Large\frac{3}{4}\normalsize \; of \; \Large\frac{4}{9} \normalsize \:÷ \: 9 \Large \frac{1}{2}\right)\normalsize \: ÷ \:1\Large \frac{5}{19}\)

Applying the rule of BODMAS carefully, we have:

= \(\left(\Large \frac{3}{4}\normalsize\: \times \:\Large\frac{4}{9} \normalsize \:÷\: \Large\frac{19}{2}\right ) \normalsize\: ÷ \:\Large\frac{24}{19}\)

= \(\left(\Large \frac{3}{9}\normalsize \:÷ \:\Large\frac{19}{2}\right)\normalsize \: ÷ \:\Large\frac{24}{19}\)

= \(\left(\Large \frac{3}{9} \normalsize \: \times \:\Large\frac{2}{19}\right)\normalsize \: ÷\: \Large \frac{24}{19}\)

= \(\left(\Large \frac{1}{3} \normalsize \:\times\: \Large\frac{2}{19}\right) \normalsize\:÷\: \Large \frac{24}{19}\)

= \(\Large \frac{1}{3} \normalsize\:\times\: \Large\frac{2}{19}\normalsize\: \times\: \Large\frac{19}{24}\)

= \(\Large \frac{1}{3}\normalsize \: \times\: \Large \frac{2}{24}\)

= \(\Large \frac{1}{3} \normalsize\: \times \: \Large \frac{1}{12}\)

= \(\Large \frac{1}{36}\)

Solution:

⇒ \(\normalsize X_4 \: + \: 34_5 = 54_6\)

[X × 4⁰] + [3 × 5¹] + [4 × 5⁰] = [5 × 6¹] + [4 × 6⁰]

X + 15 + 4 = 30 + 4

X = 30 - 15 + 4 - 4

X = 15

Solution

Step 1: This can be rewritten as:

⇒ \(\Large \frac{4^{2x}}{4^{3x}} \normalsize = 2\)

⇒ \(\normalsize 4^{2x} \; \div \; 4^{3x} = 2\)

⇒ \(\normalsize 4^{2x \; - \; 3x} = 2\)

Step 2: This can be simplified as:

⇒ \(\normalsize 4^{-x} = 2\)

Step 3: Writing this in index form in base 2

⇒ \(\normalsize (2^2)^{-x} = 2^1\)

⇒ \(\normalsize (2)^{-2x} = 2^1\)

Step 4: Dividing through by 2

⇒ \(\Large \frac {2^{-2x}}{2} =\frac {2}{2}\)

-2x = 1

x = \(\Large \frac{-1}{2}\)

Solution

⇒ \(\Large \frac{2\frac{2}{3} \: \times \: 1 \frac{1}{2}}{4 \frac{4}{5}}\)

⇒ \(\Large \frac{\frac{8}{3} \: \times \: \frac{3}{2}}{ \frac{24}{5}}\)

⇒ \(\Large \frac{8}{3} \: \times \: \frac{3}{2} \: \div \: \frac{24}{5}\)

⇒ \(\Large \frac{8}{3} \: \times \: \frac{3}{2} \: \times\: \frac{5}{24}\)

⇒ \(\Large \frac{5}{6}\)

Solution: Given that log 7.5 = 0.8751 to evaluate 2log 75 + log 750

Step 1: First simplify the expression re-writing it as 2log (7.5 × 10) + log (7.5 × 100)

Step 2: Simplify as:

= 2(log 7.5 + log 10) + log 7.5 + log 100

but log 100 = \(\normalsize \log 10^2 = 2 \log 10\)

= 2(log 7.5 + log 10) + log 7.5 + 2 log 10

Step 3: Collecting like terms

2 log 7.5 + log 7.5 + 2 log 10 + 2 log 10

3 log 7.5 + 4 log 10

Step 4:  Note: log 10 = 1

= 3 log 7.5 + 4

Step 5: Substitute the value of log 7.5, we have

⇛ 3(0.8751) + 4

⇛ 2.6253 + 4

⇛ 6. 6253

Solution

Using the expansion method convert (10110.011)₂ to base 10

(1 × 2⁴) + (0 × 2³) + (1 × 2²) + (1 × 2¹) + (0 × 2⁰) + (0 × 2^-1) + (1 × 2^-2)  + (1 × 2^-3)

(1 × 16) + (0 ) + (1 × 4) + (1 × 2) + (0) + (0 ) + (1 × ¹/₄) + + (1 × ¹/₈)

16 + 4 + 2 + (0.25 + 0.125)

22 + 0.375

= 22.375₁₀

Solution: Step 1: Transform the word problem into a mathematical statement

⇛ \(\normalsize y \propto w^2\)

Step 2: Introduce a constant k, we have

y = w²k

Step 3: Find the value of the constant k, by substituting

y = 8, w = 2

8 = 2²k

8 = 4k

k = \(\Large \frac{8}{4}\)

k = 2

Step 4: Substitute the value of k to get the relationship between y and w.

y = 2w²

Step 5: Find y by substituting w = 3 in the formula in step 4

⇛ y = 2 × 3²

y = 2 × 9

y = 18

Solution:

-3, 9, -27

⇒ \(\normalsize T_n = (-3)^n\)

⇒ \(\normalsize T_1 = -3\)

⇒ \(\normalsize T_2 = (-3)^2 = 9\)

⇒ \(\normalsize T_3 = (-3)^3 = -27\)

∴ \(\normalsize T_n = (-3)^n\)

Solution

27x³ - 8y³ = (3x - 2y)³

But 9x² + 6xy + 4y² = (3x +2y)²

So, 27x³ - 8y³ = (3x - 2y)(3x - 2y)²

Hence the other factor is 3x - 2y

Solution:

Given \(\normalsize f(x+3)=x^2+2x+7\).

To find f(5), set x + 3 = 5

i.e. x = 2

Therefore,

⇒ \(\normalsize f(5)=2^2+2(2)+7 \\ = \normalsize 4+4+7 \\ = \normalsize 15\)

Solution

-6 ≤ 4 - 2x < 5 - x split inequalities into two and solve each part as follows: 1st part

-6 ≤ 4 - 2x = - 6 - 4 ≤ -2x

-10 ≤ -2x

this can also be written as

-2x ≥ -10

divide both sides by -2

⇒ \(\Large \frac{-2x}{-2} \normalsize \leq \: \Large \frac{-10}{-2}\)

x ≤ 5

∴  x ≤ 5 or 5 ≥ x

2nd part

4 - 2x < 5 - x -2x + x < 5 - 4 -x - 1

giving x > -1 or -1 < x Combining both results gives -1 < x ≤ 5

Solution: Step 1: using graphical method determine the x – intercepts
(x – 3) (x + 2) = 0
x = 3 or – 2

Step 2: sketch the graph of y = (x – 3) (x +2)

Step 3: compare (x – 3) (x + 2) < 0 with y = (x – 3) (x + 2) ⇒y < 0

Step 4: The solution therefore, is the range: - 2 < x < 3

Solution:

The series \(\normalsize 1 + \sin 60^\circ + (\sin 60^\circ)^2 + \cdots\)

is a G.P. with

⇒ \(\normalsize a=1\)

and \(\normalsize r=\sin 60^\circ = \Large \frac{\sqrt3}{2}\)

and \(\normalsize |r|<1\)

Sum to infinity:

⇒ \(\normalsize S_\infty= \Large \frac{a}{1-r} \\ = \Large \frac{1}{1-\frac{\sqrt3}{2}}\)

⇒ \(\normalsize S_\infty= \Large \frac{1}{\frac{2-\sqrt3}{2}}\\ = \Large\frac{2}{2-\sqrt3}\)

Rationalize:

⇒ \(\normalsize S_\infty = \Large\frac{2(2+\sqrt3)}{(2-\sqrt3)(2+\sqrt3)}\\ = \Large\frac{2(2+\sqrt3)}{4-3}\\= \normalsize 2(2+\sqrt3)\)

Solution:

Step1: the number of litres that can be bought with ₦900 is

⇒ \(\Large \frac{900}{63} = \normalsize 14.285 \; litres\)

Step 2: The distance that can be covered with ₦900 worth of petrol is:

14. 285 × 14

= 200 km

Solution √10 - √10 + 2√5 - √5

To find the common ration of the G.P

Step 1:  Divide the 2nd term with the first term

= \(\Large\frac{\sqrt{10} \: +\: 2\sqrt{5}}{\sqrt{10} \: +\: \sqrt{5}}\)

Step 2: Now rationalize the denominator by multiplying and dividing the fraction with the conjugate of the denominator.

(i.e. + √10 + √5)

Conjugate of the denominator = √10 - √5

= \(\Large\frac{\sqrt{10} \: +\: 2\sqrt{5}}{\sqrt{10} \: +\: \sqrt{5}} \: \times \:\frac{\sqrt{10} \: -\: \sqrt{5}}{\sqrt{10} \: -\: \sqrt{5}}\)

Step 3: Then simplify the expression

= \(\Large\frac{(\sqrt{10})(\sqrt{10})\; + \;(\sqrt{10})(-\sqrt{5}) \\ \;+ \;(2\sqrt{5})(\sqrt{10}) \;+ \;(2\sqrt{5})(-\sqrt{5})}{(\sqrt{10})^2\; -\; (\sqrt{5})^2}\)

= \(\Large\frac{10 \;-\; \sqrt{50}\; + \;2\sqrt{50} \;- \;10}{10\; -\; 5}\)

= \(\Large\frac{\sqrt{50}}{5}\)

= \(\Large\frac{\sqrt{25 \;\times \;2}}{5}\)

= \(\Large\frac{5\sqrt{2}}{5}\)

= \(\normalsize \sqrt{2}\)

Solution

⇒ |x - 2| < 3

This implies

⇒ -(x - 2) < 3 .... or .... +(x - 2) < 3

⇒ -x + 2 < 3 .... or .... x - 2 < 3

⇒ -x < 3 - 2 .... or .... x < 3 + 2

⇒ x > -1 .... or .... x < 5

Combining the two inequalities

⇒ -1 < x < 5

Solution

⇒ \(\normalsize y \propto \sqrt{n}\)

⇒ \(\normalsize y = k \sqrt{n}\)

y = 4 when n = 4,

⇒ \(\normalsize 4 = k \sqrt{4}\)

⇒ \(\normalsize 4 = k 2\)

⇒ \(\normalsize k = \Large \frac{4}{2}\)

⇒ \(\normalsize k = 2\)

Therefore,

⇒ \(\normalsize y = 2 \sqrt{n}\)

⇒ \(\normalsize y = 2 \sqrt{\Large \frac{16}{9}}\)

⇒ \(\normalsize y = 2 \: \times \: \left( \Large \frac{4}{3} \right)\)

⇒ \(\normalsize y = \Large \frac{8}{3}\)

Solution

To help us understand the question better, let's draw a diagram.

From the diagram we can see that the length of the chord is x +  x = 2x

We can find x using pythagoras theorem

⇒ \(\normalsize hyp^2 = opp^2 \: + \: adj^2\)

where hyp = 5 cm
opp = 3 cm
adj = x cm

∴  \(\normalsize 5^2 = 3^2 \: + \: x^2\)

⇒ \(\normalsize x^2 = 5^2 \: - \: 3^2\)

⇒ \(\normalsize x^2 =25 \: - \: 9\)

⇒ \(\normalsize x^2 = 16\)

take the square root of both sides

⇒ \(\normalsize \sqrt{x^2} = \sqrt{16}\)

⇒ \(\normalsize x = \sqrt{16}\)

⇒ \(\normalsize x = 4 \: cm\)

Therefore, the length of the chord

= x + x

= 4 + 4

= 8 cm

Solution

The graph crosses the x-axis at x = -1 and x = 2

Using  \(\normalsize x^2 \: -\: (\alpha \: + \: \beta)x\: + \: \alpha \beta = 0\)

⇒ \(\normalsize x^2 \: -\: (-1 \: + \: 2)x\: + \: (-1) (2) = 0\)

⇒ \(\normalsize x^2 \: -\:x\: - \: 2 = 0\)

∴ y = \(\normalsize x^2 \: -\:x\: - \: 2\)

Solution

Step 1: Length of an arc = \(\Large \frac{\theta}{360} \; \times \; \normalsize 2 \pi r\)

⇒ \(\normalsize 16 \pi = \Large \frac{80}{360} \; \times \; \normalsize 2 \; \times \; \pi \; \times \; r\)

⇒ \(\normalsize 16 = \Large \frac{8}{36} \; \times \; \normalsize 2 \; \times \; r\)

Step 2: Making r the subject

16 = \(\Large \frac {16r}{36}\)

r = \(\Large \frac {36 \; \times \; 16}{16}\)

r = 36

Explanation:

P, Q and R are x-intercepts (where each curve cuts the x-axis), so y = 0.

For \(\normalsize y=x(2-x)\)

⇒ \(\normalsize 0=x(2-x)\)

⇒ \(\normalsize x=0\ \text{or}\ 2-x=0\)

⇒ \(\normalsize x=0\ \text{or}\ x=2\)

For \(\normalsize y=(x-1)(x-3)\)

⇒ \(\normalsize 0=(x-1)(x-3)\)

⇒ \(\normalsize x=1\ \text{or}\ x=3\)

From the diagram, P is at x = 1, Q is at x = 2, and R is at x = 3

⇒ \(\normalsize P, Q, R = 1, \:2, \:3\)

Solution

In Δ TYX

XY = TY

∴ ∠YTX =47° (base ∠s of ISC ΔTYX)

Since TS//YZ ∠ZTS = ∠YZT  = 34° (alternate angles)

In Δ TZX

∠TXY + ∠TZY + ∠ZTX = 180º (Sum of angles in Δ TZX)

∠ZTX = 47º + xº

∴ 47º + 34º + (47º + xº) = 180º

47º + 34º + 47º + xº = 180º

128º + xº = 180º

xº = 180º - 128º

xº = 52º

Solution: Step 1: State the formula for the perimeter of a sector

i.e \(\Large \frac{\theta}{360} \normalsize \: \times \: 2 \pi r \: + \: 2r\)

Step 2: Substitute the given parameters into the formula

⇒ \(\Large \frac{48}{360} \normalsize \:\times \: 2 \:\times \: \Large \frac{22}{7} \:\times \: \Large \frac{10.5}{1} \normalsize + 2(10.5)\)

= 8.8 + 21

= 29.8 cm

Solution:

Step 1: Gradient m = \(\Large \frac{y_2 \; - \; y_1}{x_2 \; - \; x_1}\)

m = \(\Large \frac{4 \; - \; 2}{1 \; - \; 3}\)

m = \(\Large \frac{2}{-2}\)

m = -1

Solution

ΔU = ΔS = 50 (Alt. Δ)

ΔUST = 180 - 50

= 130º

Solution

The inequalities are:

⇒ \(\normalsize y \le x - 2\)

⇒ \(\normalsize y \ge x^2 - 4\)

The line \(\normalsize y = x - 2\) represents the boundary of the first inequality.
Since the inequality is \(\normalsize \le\), the solution lies below the line.

The curve \(\normalsize y = x^2 - 4\) is a parabola.
Since the inequality is \(\normalsize \ge\), the solution lies above the parabola.

To find where the graphs intersect:

⇒ \(\normalsize x \: - \: 2 = x^2 \: - \: 4\)

⇒ \(\normalsize x^2 \: - \: x \: - \: 2 = 0\)

⇒ \(\normalsize (x\: - \: 2)(x\: + \: 1)=0\)

So the curves intersect at:

⇒ \(\normalsize x = -1 \quad \text{and} \quad x = 2\)

The required region is therefore the area between the parabola and the straight line from
\(x =\: -1\) to \(x = \: 2\)

This corresponds to diagram A.

Solution

Step 1: Replace 2x + 3 with u

i.e. u = 2x + 3 ____________ (1)

Step 2: Differentiate u with respect to x

⇒ \(\Large \frac {du}{dx} \normalsize = 2\)

Step 3: Make dx the subject:

⇒ \(\normalsize dx = \Large \frac {du}{2}\) _____________ (2)

Step 4: Substitute equations (1) and (2) into the integral we have:-

⇒ \(\normalsize \int (2x + 3)^{\Large \frac{1}{2}} \normalsize \delta x = \normalsize \int (u)^{\Large \frac{1}{2}} \Large \frac{du}{2}\)

Step 5: Integrate the new integral w.r.t. u

\(\Large \frac{1}{2} \normalsize \int u^{\Large \frac{1}{2}} \normalsize du = \Large \frac{1}{2}\left [\frac{u^{\Large \frac{1}{2} + 1} }{\frac{3}{2}}\right] \normalsize + k\)

= \(\Large \frac{1}{2} \normalsize\times \Large \frac{2}{3} \normalsize \left[u ^{\Large \frac{3}{2}}\right] \normalsize + k\)

= \(\Large \frac{1}{3} \normalsize u ^ {\Large \frac{3}{2}} \normalsize + k\)  ________(3)

Step 6: Substitute u = 2x +3 in equation (3)

= \(\Large \frac{1}{3} \normalsize (2x + 3) ^ {\Large \frac{3}{2}} \normalsize + k\)

Solution

If y = x sinx, then

Let u = x and v = sinx

⇒ \(\Large \frac{du}{dx} \normalsize = 1\)

⇒ \(\Large\frac{dv}{dx} \normalsize = cosx\)

Using the product rule,

⇒ \(\Large\frac{dy}{dx} = \normalsize V \Large \frac{du}{dx} + \normalsize U \Large \frac{dv}{dx}\\\)

substituting u and v in the equation (u = x, v = sinx)

= (sin x) × 1 + x cosx

= sinx + x cosx

Solution

Evaluate:

⇒ \(\normalsize \int \sin 2x\,dx\)

Step 1: Let

⇒ \(\normalsize u = 2x\)

Then

⇒ \(\Large \frac{du}{dx} \normalsize= 2\)

⇒ \(\normalsize dx = \Large\frac{du}{2}\)

Step 2: Substitute.

⇒ \(\normalsize \int \sin 2x\,dx = \int \sin u \cdot \large \frac{du}{2}\)

⇒ \(\normalsize = \frac{1}{2}\int \sin u\,du\)

Step 3: Integrate.

⇒ \(\normalsize \frac{1}{2}(-\cos u) + k\)

Step 4: Substitute back.

⇒ \(\normalsize -\frac{1}{2}\cos 2x + k\)

Explanation:

⇒ \(\normalsize \int_{-1}^{1}(2x+1)^2dx\)

Let \(\normalsize u = 2x+1\)

⇒ \(\Large \frac{du}{dx}\normalsize = 2\)

⇒ \(\normalsize dx = \Large \frac{du}{2}\)

Change limits:

⇒ when \(\normalsize x=-1,; u=2(-1)+1=-1\)
⇒ when \(\normalsize x=1,; u=2(1)+1=3\)

⇒ \(\normalsize \int_{-1}^{3} u^2 \Large \frac{du}{2}\)

⇒ \(\Large \frac{1}{2}\normalsize \int_{-1}^{3} u^2 du\)

⇒ \(\Large \frac{1}{2}\left[\Large \frac{u^3}{3}\right]_{-1}^{3}\)

⇒ \(\Large \frac{1}{2}\left(\Large \frac{3^3}{3}-\frac{(-1)^3}{3}\right)\)

⇒ \(\Large \frac{1}{2}\left(\Large \frac{27}{3}+\frac{1}{3}\right)\)

⇒ \(\Large \frac{1}{2}\times \frac{28}{3}\)

⇒ \(\Large \frac{14}{3}\)

= \(\normalsize 4\frac{2}{3}\)

Explanation:

⇒ \(\normalsize \bar{x} = \Large \frac{\sum fx}{\sum f} \\ = \Large \frac{24}{8} \\ = \normalsize 3\)

Variance = \(\Large \frac{\sum f (x \: - \: \bar{x})^2}{\sum f} \\ = \Large \frac{18}{8} \\ = \Large \frac{9}{4}\)

Solution:

Step 1: We first arrange the numbers in order of size, i.e.

2, 3, 5, 7, 8, 9, 12, 14

Step 2: Hence, the mean P, is

P = \(\Large \frac{2+3+5+7+8+9+12+14}{8}\)

P = \(\Large \frac{60}{8} = \normalsize 7.5\)

Step 4: Median = Q = \(\Large \frac{8+7}{2} \normalsize = 7.5\)

Hence P + Q = 7.5 + 7.5 = 15

Solution

total frequency = 1 + 4 + 3 + 8 + 2 + 5 = 23

This is a number therefore the median falls between ⇒ \(\Large \frac{23+1}{2}\)

= 12

from the table when the frequency is 12 the number is 3

The range is the highest no - lowest no

= 5 - 0 = 5

The median and range of the data respectively = (3, 5)

Explanation:

Total pupils = 6 + 27 + 7 = 40

Pro(at least 11 yrs old) = \(\Large \frac{27+7}{40} \\ = \Large \frac{34}{40} \\ = \Large \frac{17}{20}\)

Solution

ZOOLOGY has 7 letters in total.

The letter O is repeated 3 times.

Number of distinct arrangements:

⇒ \(\frac{7!}{3!}\)

⇒ \(\Large \frac{7\times6\times5\times4\times3\times2\times1}{3\times2\times1}\)

= 840 ways

Solution

Arrange all the values in ascending order,

2 ,2, 3, 3, 3, 4, 4 ,4, 4, 5, 5, 5, 7, 8, 9, 9

Median = \(\Large \frac{4\: + \:4}{2}\)

Median = 4

Solution

From the graph

Total time = 80 + 160 + 200 + 80 + 128 + 72 = 720 minutes

Solution: Step 1: Determine the sum of the acres
i.e. 2 + 5 + 3 + 11 + 9
= 30

Step 2: The angle of the sector of cassava is:

\(\Large \frac{3}{30} \: \times \: \frac{360}{1} \normalsize = 36^o\)