የሒሳብ መግቢያ ፈተና፣ 2014 ዓ.ም. (2022)

Division with remainder writes a number as: dividend = divisor \(\times\) quotient + remainder.

Ask how many whole times 20 fits into 75.
\(20 \times 3 = 60\) fits, but \(20 \times 4 = 80\) is more than 75.
So the quotient is 3.
The remainder is what is left over: \(75 - 60 = 15\).

Check: \(75 = 20 \times 3 + 15\), and \(15\) is smaller than \(20\), so this is a proper remainder.

The question asks for the quotient and the remainder in that order, so the answer is 3 and 15.

Watch out: option A gives the same two numbers in the reversed order, 15 and 3.

Textbook: Mathematics Grade 9, unit The Number System (page 17).

The exterior angles of any convex polygon always add up to \(360^\circ\).

In a regular polygon all exterior angles are equal, so each one is:
\[\text{exterior angle} = \frac{360^\circ}{n}\]

Here \(n = 10\), so:
\(\dfrac{360^\circ}{10} = 36^\circ\).

Watch out: \(144^\circ\) (option B) is the interior angle of a regular 10 sided polygon, because \(180^\circ - 36^\circ = 144^\circ\).

Textbook: Mathematics Grade 9, unit Regular Polygons (page 181).

To change degrees into radians, multiply by \(\dfrac{\pi}{180^\circ}\), because \(180^\circ = \pi\) radians.

\[120^\circ \times \frac{\pi}{180^\circ} = \frac{120\pi}{180} = \frac{2\pi}{3}\]

So \(120^\circ\) equals \(\dfrac{2\pi}{3}\) radians.

Quick sense check: \(120^\circ\) is two thirds of \(180^\circ\), so it must be two thirds of \(\pi\).

Textbook: Mathematics Grade 10, Unit 4 Trigonometric Functions, section 4.1 Radian Measure of Angle (page 187).

For vectors \(\mathbf{u} = a\mathbf{i} + b\mathbf{j}\) and \(\mathbf{v} = c\mathbf{i} + d\mathbf{j}\), the dot (scalar) product is:
\[\mathbf{u} \cdot \mathbf{v} = ac + bd\]

Here \(\mathbf{u} = \mathbf{i} + 2\mathbf{j}\) and \(\mathbf{v} = -2\mathbf{i} + 4\mathbf{j}\).
Multiply the \(\mathbf{i}\) parts: \((1)(-2) = -2\).
Multiply the \(\mathbf{j}\) parts: \((2)(4) = 8\).
Add them: \(-2 + 8 = 6\).

The dot product is 6. Remember the result is a number, not a vector.

Textbook: Mathematics Grade 11, Unit 5 Vectors (page 251).

If a polynomial changes sign between two points, it must cross zero somewhere between them. So we need \(f(0)\) and \(f(1)\) to have opposite signs.

Compute the two values:
\(f(0) = 0^2 + p(0) + 1 = 1\), which is always positive.
\(f(1) = 1^2 + p(1) + 1 = p + 2\).

Since \(f(0) > 0\), we need \(f(1) < 0\):
\(p + 2 < 0\), so \(p < -2\).

Test the choices: only \(p = -3\) satisfies \(p < -2\).
Check: \(f(1) = -3 + 2 = -1 < 0\), so the sign really changes on \((0, 1)\).

Watch out: \(p = -2\) (option C) gives \(f(1) = 0\), a root at \(x = 1\) itself, not inside the open interval.

Textbook: Mathematics Grade 10, Unit 2 Polynomial Functions, section 2.4 Zeros of polynomial functions (page 90).

A vector has both a size and a direction. A scalar has only a size.

Go through the choices:
Weight is the force of gravity on an object. A force has a size and a direction (downward), so weight is a vector.
Volume of a box is just an amount of space, with no direction.
Speed tells how fast something moves but not which way, so it is a scalar.
The width of a room is just a length.

So only the weight of an object is a vector.

Watch out: speed is a scalar; it becomes the vector velocity only when a direction is attached.

Textbook: Mathematics Grade 11, Unit 5 Vectors, section 5.1 Revision on Vectors and Scalars (page 252).