Revision Test Paper 1. 2 hrs 30 mins Answer all questions. Total 100 MARKS

Question

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Here are the solutions to your questions:

1.

a) Rounding 4.9984 to 2 decimal places:

Identify the second decimal place, which is the second '9'.
Look at the third decimal place, which is '8'.
Since $8 \ge 5$ , round up the second decimal place.
$4.9984 \approx \boxed{5.00}$
b) Rounding 794.3 to 1 significant figure:
Identify the first significant figure, which is '7'.
Look at the next digit, which is '9'.
Since $9 \ge 5$ , round up the first significant figure. Replace subsequent digits with zeros to maintain place value.
$794.3 \approx \boxed{800}$
c) Expressing 0.00231 in standard form:
Move the decimal point to the right until there is one non-zero digit before the decimal point.
The number becomes 2.31.
Count how many places the decimal point was moved.
It was moved 3 places to the right.
Since the original number was less than 1, the power of 10 is negative.
$0.00231 = \boxed{2.31 \times 10^{-3}}$
2.

a) Evaluate $7 \times 8^2 + 8 + 6$ to base 8:
Calculate the value in base 10.
$7 \times 8^2 + 8 + 6 = 7 \times 64 + 8 + 6 = 448 + 8 + 6 = 462_{10}$
Convert $462_{10}$ to base 8.
$462 \div 8 = 57 \text{ remainder } 6$ $57 \div 8 = 7 \text{ remainder } 1$ $7 \div 8 = 0 \text{ remainder } 7$
Read the remainders from bottom to top.
$462_{10} = \boxed{716_8}$
b) Convert $57_{10}$ to base 6:
Repeatedly divide 57 by 6 and record the remainders.
$57 \div 6 = 9 \text{ remainder } 3$ $9 \div 6 = 1 \text{ remainder } 3$ $1 \div 6 = 0 \text{ remainder } 1$
Read the remainders from bottom to top.
$57_{10} = \boxed{133_6}$
c) Add $111_2 + 1_2$ to base 2:
Perform binary addition.
$\begin{array}{c} & 1 & 1 & 1_2 \\ + & & & 1_2 \\ \hline \end{array}$
Add the rightmost digits
$1 + 1 = 10_2$ (0 carry 1).
$\begin{array}{c} & 1 & 1 & \overset{1}{1}_2 \\ + & & & 1_2 \\ \hline & & & 0_2 \end{array}$
Add the next digits with the carry
$1 + 0 + 1 = 10_2$ (0 carry 1).
$\begin{array}{c} & 1 & \overset{1}{1} & \overset{1}{1}_2 \\ + & & & 1_2 \\ \hline & & 0 & 0_2 \end{array}$
Add the next digits with the carry
$1 + 0 + 1 = 10_2$ (0 carry 1).
$\begin{array}{c} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1}_2 \\ + & & & 1_2 \\ \hline 1 & 0 & 0 & 0_2 \end{array}$ $111_2 + 1_2 = \boxed{1000_2}$
3.

a) Express 0217 in 12-hour notation:
0217 is in 24-hour format.
The hour is 02, which is 2 AM. The minutes are 17.
$0217 = \boxed{2:17 \text{ AM}}$
b) (i) For a general kite, state the number of lines of symmetry:
A general kite has one line of symmetry, which is its main diagonal.
$\boxed{1}$
b) (ii) For a general kite, state the order of rotational symmetry:
A general kite has an order of rotational symmetry of 1, meaning it only looks the same after a full $360^\circ$ rotation.
$\boxed{1}$
4.
Given $f(x) = 3x - 1$ :

a) Find $f(-2)$ :
Substitute $x = -2$ into the function.
$f(-2) = 3(-2) - 1$
Calculate the value.
$f(-2) = -6 - 1 = \boxed{-7}$
b) Find $x$ if $f(x) = 5$ :
Set the function equal to 5.
$3x - 1 = 5$
Add 1 to both sides.
$3x = 5 + 1$ $3x = 6$
Divide by 3.
$x = \frac{6}{3}$ $x = \boxed{2}$
5.

a) Simplify $(-3m)^4$ :
Apply the power to both the coefficient and the variable.
$(-3m)^4 = (-3)^4 \times m^4$
Calculate $(-3)^4$ .
$(-3)^4 = (-3) \times (-3) \times (-3) \times (-3) = 81$ $(-3m)^4 = \boxed{81m^4}$
b) (i) Simplify $49^{3/2}$ :
Rewrite the fractional exponent.
The denominator (2) indicates a square root, and the numerator (3) indicates cubing.
$49^{3/2} = (\sqrt{49})^3$
Calculate the square root.
$\sqrt{49} = 7$
Cube the result.
$7^3 = 7 \times 7 \times 7 = 343$ $49^{3/2} = \boxed{343}$
b) (ii) Simplify $2^{-3}$ :
Apply the negative exponent rule, which states $a^{-n} = \frac{1}{a^n}$ .
$2^{-3} = \frac{1}{2^3}$
Calculate $2^3$ .
$2^3 = 2 \times 2 \times 2 = 8$ $2^{-3} = \boxed{\frac{1}{8}}$
6.
Solve the system of equations:
1. $3x + 2y = 8$
2. $5x + 3y = 4.5$
Multiply equation (1) by 3 and equation (2) by 2 to eliminate $y$ .
$3 \times (3x + 2y = 8) \implies 9x + 6y = 24 \quad (3)$ $2 \times (5x + 3y = 4.5) \implies 10x + 6y = 9 \quad (4)$
Subtract equation (3) from equation (4).
$(10x + 6y) - (9x + 6y) = 9 - 24$ $10x - 9x + 6y - 6y = -15$ $x = -15$
Substitute $x = -15$ into equation (1).
$3(-15) + 2y = 8$ $-45 + 2y = 8$
Add 45 to both sides.
$2y = 8 + 45$ $2y = 53$
Divide by 2.
$y = \frac{53}{2}$ $y = 26.5$
The solution is $\boxed{x = -15, y = 26.5}$ .

7.
Given $\angle ATB = 74^\circ$ .

a) Calculate $\angle AOT$ :
In a circle, the radius is perpendicular to the tangent at the point of contact.
So, $\angle OAT = 90^\circ$ and $\angle OBT = 90^\circ$ .
Consider quadrilateral OATB.
The sum of angles in a quadrilateral is $360^\circ$ .
$\angle AOB + \angle OAT + \angle OBT + \angle ATB = 360^\circ$ $\angle AOB + 90^\circ + 90^\circ + 74^\circ = 360^\circ$ $\angle AOB + 254^\circ = 360^\circ$ $\angle AOB = 360^\circ - 254^\circ = 106^\circ$
Triangle AOT and BOT are congruent right-angled triangles (OA=OB, OT is common, $\angle OAT = \angle OBT = 90^\circ$ ).
Therefore, OT bisects $\angle AOB$ and $\angle ATB$ .
$\angle AOT = \frac{1}{2} \angle AOB$ $\angle AOT = \frac{1}{2} \times 106^\circ = \boxed{53^\circ}$
b) Calculate $\angle BAT$ :
Tangents from an external point to a circle are equal in length, so $TA = TB$ .
This means $\triangle ATB$ is an isosceles triangle.
$\angle TAB = \angle TBA$
The sum of angles in $\triangle ATB$ is $180^\circ$ .
$\angle TAB + \angle TBA + \angle ATB = 180^\circ$ $2 \angle TAB + 74^\circ = 180^\circ$ $2 \angle TAB = 180^\circ - 74^\circ$ $2 \angle TAB = 106^\circ$ $\angle TAB = \frac{106^\circ}{2} = \boxed{53^\circ}$
c) Calculate $\angle ACB$ :
The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.
Arc AB subtends $\angle AOB$ at the center and $\angle ACB$ at the circumference.
$\angle ACB = \frac{1}{2} \angle AOB$
From part (a), we found $\angle AOB = 106^\circ$ .
$\angle ACB = \frac{1}{2} \times 106^\circ = \boxed{53^\circ}$
8.
Given a shape with a width of 14 cm and a height of 10 cm, with a semicircle removed from the top. $\pi = 22/7$ .

a) Calculate the area of the shaded part:
The shaded part is a rectangle with a semicircle removed.
Area of rectangle = length $\times$ width = $14 \text{ cm} \times 10 \text{ cm} = 140 \text{ cm}^2$ .
The diameter of the semicircle is 14 cm, so the radius $r = \frac{14}{2} = 7 \text{ cm}$ .
Area of semicircle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (7 \text{ cm})^2$ .
$= \frac{1}{2} \times \frac{22}{7} \times 49 \text{ cm}^2 = 11 \times 7 \text{ cm}^2 = 77 \text{ cm}^2$
Area of shaded part = Area of rectangle - Area of semicircle.
$= 140 \text{ cm}^2 - 77 \text{ cm}^2 = \boxed{63 \text{ cm}^2}$
b) Calculate the perimeter of the shaded part:
The perimeter consists of two vertical sides, the bottom side, and the arc length of the semicircle.
Length of two vertical sides = $2 \times 10 \text{ cm} = 20 \text{ cm}$ .
Length of bottom side = $14 \text{ cm}$ .
Arc length of semicircle = $\frac{1}{2} \times \text{circumference of full circle} = \frac{1}{2} \times 2 \pi r = \pi r$ .
$= \frac{22}{7} \times 7 \text{ cm} = 22 \text{ cm}$
Total perimeter = $20 \text{ cm} + 14 \text{ cm} + 22 \text{ cm}$ .
$= \boxed{56 \text{ cm}}$
9.
Given the frequency table: Colour (Black, Blue, White, Green), Frequency (6, 2, 1, 1).

a) Modal Colour:
The mode is the category with the highest frequency.
From the table, Black has the highest frequency of 6.
The modal colour is $\boxed{\text{Black}}$ .

b) Pie chart angle for blue:
Calculate the total frequency.
Total frequency = $6 + 2 + 1 + 1 = 10$ .
Calculate the fraction of blue colour.
Fraction for blue = $\frac{\text{Frequency of Blue}}{\text{Total Frequency}} = \frac{2}{10}$ .
Multiply the fraction by $360^\circ$ to find the angle in the pie chart.
Angle for blue = $\frac{2}{10} \times 360^\circ = \frac{1}{5} \times 360^\circ = \boxed{72^\circ}$ .

Revision Test Paper 1. 2 hrs 30 mins Answer all questions. Total 100 MARKS

Step-by-step solution

Identify the second decimal place, which is the second '9'.

Look at the third decimal place, which is '8'.

Identify the first significant figure, which is '7'.

Look at the next digit, which is '9'.

Move the decimal point to the right until there is one non-zero digit before the decimal point.

Count how many places the decimal point was moved.

Since the original number was less than 1, the power of 10 is negative.

Calculate the value in base 10.

Convert 46210462_{10}46210​ to base 8.

Read the remainders from bottom to top.

Repeatedly divide 57 by 6 and record the remainders.

Read the remainders from bottom to top.

Perform binary addition.

Add the rightmost digits

Add the next digits with the carry

Add the next digits with the carry

0217 is in 24-hour format.

Substitute x=−2x = -2x=−2 into the function.

Calculate the value.

Set the function equal to 5.

Add 1 to both sides.

Divide by 3.

Apply the power to both the coefficient and the variable.

Calculate (−3)4(-3)^4(−3)4.

Rewrite the fractional exponent.

Calculate the square root.

Cube the result.

Apply the negative exponent rule, which states a−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​.

Calculate 232^323.

Multiply equation (1) by 3 and equation (2) by 2 to eliminate yyy.

Subtract equation (3) from equation (4).

Substitute x=−15x = -15x=−15 into equation (1).

Add 45 to both sides.

Divide by 2.

In a circle, the radius is perpendicular to the tangent at the point of contact.

Consider quadrilateral OATB.

Triangle AOT and BOT are congruent right-angled triangles (OA=OB, OT is common, ∠OAT=∠OBT=90∘\angle OAT = \angle OBT = 90^\circ∠OAT=∠OBT=90∘).

Tangents from an external point to a circle are equal in length, so TA=TBTA = TBTA=TB.

This means △ATB\triangle ATB△ATB is an isosceles triangle.

The sum of angles in △ATB\triangle ATB△ATB is 180∘180^\circ180∘.

The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.

Arc AB subtends ∠AOB\angle AOB∠AOB at the center and ∠ACB\angle ACB∠ACB at the circumference.

From part (a), we found ∠AOB=106∘\angle AOB = 106^\circ∠AOB=106∘.

The shaded part is a rectangle with a semicircle removed.

The diameter of the semicircle is 14 cm, so the radius r=142=7 cmr = \frac{14}{2} = 7 \text{ cm}r=214​=7 cm.

Area of shaded part = Area of rectangle - Area of semicircle.

The perimeter consists of two vertical sides, the bottom side, and the arc length of the semicircle.

Arc length of semicircle = 12×circumference of full circle=12×2πr=πr\frac{1}{2} \times \text{circumference of full circle} = \frac{1}{2} \times 2 \pi r = \pi r21​×circumference of full circle=21​×2πr=πr.

Total perimeter = 20 cm+14 cm+22 cm20 \text{ cm} + 14 \text{ cm} + 22 \text{ cm}20 cm+14 cm+22 cm.

The mode is the category with the highest frequency.

From the table, Black has the highest frequency of 6.

Calculate the total frequency.

Calculate the fraction of blue colour.

Multiply the fraction by 360∘360^\circ360∘ to find the angle in the pie chart.

Still stuck on this one?

More Mathematics Questions

Convert $462_{10}$ to base 8.

Substitute $x = -2$ into the function.

Calculate $(-3)^4$ .

Apply the negative exponent rule, which states $a^{-n} = \frac{1}{a^n}$ .

Calculate $2^3$ .

Multiply equation (1) by 3 and equation (2) by 2 to eliminate $y$ .

Substitute $x = -15$ into equation (1).

Triangle AOT and BOT are congruent right-angled triangles (OA=OB, OT is common, $\angle OAT = \angle OBT = 90^\circ$ ).

Tangents from an external point to a circle are equal in length, so $TA = TB$ .

This means $\triangle ATB$ is an isosceles triangle.

The sum of angles in $\triangle ATB$ is $180^\circ$ .

Arc AB subtends $\angle AOB$ at the center and $\angle ACB$ at the circumference.

From part (a), we found $\angle AOB = 106^\circ$ .

The diameter of the semicircle is 14 cm, so the radius $r = \frac{14}{2} = 7 \text{ cm}$ .

Arc length of semicircle = $\frac{1}{2} \times \text{circumference of full circle} = \frac{1}{2} \times 2 \pi r = \pi r$ .

Total perimeter = $20 \text{ cm} + 14 \text{ cm} + 22 \text{ cm}$ .

Multiply the fraction by $360^\circ$ to find the angle in the pie chart.