Here are the solutions to the problems: 10. A school bought 879432 exercise books to share among 11 classes. Determine the number of books that should be added for the classes to get equal shares. Step 1: Divide the total number of books by the number of classes to find the remainder. $$ 879432 \div 11 $$ Using long division or divisibility rule for 11: $879432 = 11 \times 79948 + 4$ The remainder is $4$. Step 2: Determine how many more books are needed to make the total divisible by 11. To get an equal share, the total number of books must be a multiple of 11. Since the remainder is 4, we need $11 - 4 = 7$ more books. The correct option is D. $\boxed{\text{7}}$ 11. What is the value of $\frac{0.5 + 0.2 - 0.25}{0.2}$? Step 1: Calculate the numerator. $$ 0.5 + 0.2 - 0.25 = 0.7 - 0.25 = 0.45 $$ Step 2: Divide the numerator by the denominator. $$ \frac{0.45}{0.2} = \frac{45}{20} = \frac{9}{4} = 2.25 $$ The correct option is D. $\boxed{2.25}$ 12. A certain quadrilateral has the following properties: i) Has two pairs of parallel sides ii) All sides are equal iii) Some angles are equal iv) Diagonals are perpendicular bisectors Which quadrilateral has the properties above? Let's analyze the properties: • Property (i) "Has two pairs of parallel sides" describes a parallelogram. • Property (ii) "All sides are equal" narrows it down to a rhombus or a square. • Property (iv) "Diagonals are perpendicular bisectors" is true for a rhombus and a square. • Property (iii) "Some angles are equal" is true for both a rhombus (opposite angles are equal) and a square (all angles are equal). A square is a special type of rhombus where all angles are $90^\circ$. Since the properties do not explicitly state that all angles are $90^\circ$, but only "some angles are equal", the most general quadrilateral that fits all these properties is a rhombus. The correct option is B. $\boxed{\text{Rhombus}}$ 13. In Mbagathi Hospital, a doctor used a syringe that was 0.00689cm. How many significant figures was the diameter of the syringe? To count significant figures: • Non-zero digits are always significant. • Leading zeros (zeros before non-zero digits) are not significant. In $0.00689$: The leading zeros ($0.00$) are not significant. The digits $6$, $8$, and $9$ are non-zero, so they are significant. Therefore, there are $3$ significant figures. The correct option is D. $\boxed{3}$ 14. A drop of ink was recorded as 0.0000892ml at Kiama School laboratory. Express the drop of ink in standard form. Standard form (scientific notation) is written as $a \times 10^b$, where $1 \le |a| < 10$ and $b$ is an integer. To convert $0.0000892$ to standard form, move the decimal point to the right until there is one non-zero digit before the decimal point. $0.0000892 \rightarrow 8.92$ The decimal point was moved $5$ places to the right, so the exponent $b$ is $-5$. $$ 0.0000892 = 8.92 \times 10^{-5} $$ The correct option is A. $\boxed{8.92 \times 10^{-5}}$ 15. Grade 7 learners where asked to study the figure below. What value did they get for the angle marked $x$? The figure shows two intersecting straight lines. The angle $150^\circ$ and the angle adjacent to it on the straight line form a linear pair, so their sum is $180^\circ$. Let the adjacent angle be $y$. $$ y + 150^\circ = 180^\circ $$ $$ y = 180^\circ - 150^\circ = 30^\circ $$ Angle $x$ is vertically opposite to angle $y$. Vertically opposite angles are equal. $$ x = y = 30^\circ $$ The correct option is D. $\boxed{30^\circ}$ 16. The county government of Kisumu donated 14,568,139 mosquito nets to flood hit areas in Kano Plains. What is the number rounded off to the nearest tens of thousands? Step 1: Identify the tens of thousands digit. The number is $14,568,139$. The tens of thousands digit is $6$. Step 2: Look at the digit to the right of the tens of thousands digit. The digit to the right of $6$ is $8$. Step 3: Round up or down. Since $8 \ge 5$, we round up the tens of thousands digit. $6$ becomes $7$. All digits to the right become $0$. $$ 14,568,139 \approx 14,570,000 $$ The correct option is D. $\boxed{14,570,000}$ 17. Jonathan used a ladder to climb an orange tree. The foot of the ladder was 24m from the tree. The ladder leaned on the tree at a height of 10m. Determine the length of the ladder? This problem forms a right-angled triangle where: • The distance from the foot of the ladder to the tree is one leg ($a = 24 \text{ m}$). • The height the ladder reaches on the tree is the other leg ($b = 10 \text{ m}$). • The length of the ladder is the hypotenuse ($c$). Using the Pythagorean theorem, $a^2 + b^2 = c^2$: $$ 24^2 + 10^2 = c^2 $$ $$ 576 + 100 = c^2 $$ $$ 676 = c^2 $$ $$ c = \sqrt{676} $$ $$ c = 26 \text{ m} $$ The correct option is C. $\boxed{26 \text{ m}}$ 18. Halima paid Sh. 950 for an item after getting a discount of Sh. 50. What percentage discount did she get? Step 1: Calculate the original price of the item. Original price = Price paid + Discount amount $$ \text{Original price} = 950 \text{ Sh} + 50 \text{ Sh} = 1000 \text{ Sh} $$ Step 2: Calculate the percentage discount. $$ \text{Percentage discount} = \frac{\text{Discount amount}}{\text{Original price}} \times 100\% $$ $$ \text{Percentage discount} = \frac{50}{1000} \times 100\% $$ $$ \text{Percentage discount} = \frac{1}{20} \times 100\% $$ $$ \text{Percentage discount} = 5\% $$ The correct option is D. $\boxed{5\%}$ 19. The base and the height of the figure below are equal. Work out the length of the base given that the area of the figure is $40.5 \text{cm}^2$. The figure is a right-angled triangle. Let the base be $b$ and the height be $h$. Given that the base and height are equal, so $b = h$. The area of a triangle is given by the formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$ Given Area $= 40.5 \text{ cm}^2$. Substitute $b$ for height: $$ 40.5 = \frac{1}{2} \times b \times b $$ $$ 40.5 = \frac{1}{2} b^2 $$ Multiply both sides by 2: $$ 40.5 \times 2 = b^2 $$ $$ 81 = b^2 $$ Take the square root of both sides: $$ b = \sqrt{81} $$ $$ b = 9 \text{ cm} $$ The correct option is A. $\boxed{9 \text{ cm}}$ That's 2 down. 3 left today — send the next one.