Replica 3 Mathematics Practice and Evaluation Guide

Integer Operations and Temperature Change

To find the change in temperature between two points in time, subtract the initial temperature from the final temperature. The problem states that the temperature of a place was $-3^{\circ}\text{C}$ last night and is now $24^{\circ}\text{C}$. Using the formula for change ($\Delta T = T_{final} - T_{initial}$), the calculation is as follows:

$\Delta T = 24^{\circ}\text{C} - (-3^{\circ}\text{C})$

$\Delta T = 24^{\circ}\text{C} + 3^{\circ}\text{C} = 27^{\circ}\text{C}$

Therefore, the total change in temperature is $27^{\circ}\text{C}$.

Ordering Fractions in Ascending Order

When arranging fractions such as $\frac{9}{10}$, $\frac{4}{6}$, $\frac{3}{5}$, and $\frac{4}{5}$ in ascending order without a calculator, it is helpful to find a common denominator or convert them to decimals.

1. For $\frac{3}{5}$, the decimal equivalent is $0.6$.

2. For $\frac{4}{6}$, which simplifies to $\frac{2}{3}$, the decimal equivalent is approximately $0.667$.

3. For $\frac{4}{5}$, the decimal equivalent is $0.8$.

4. For $\frac{9}{10}$, the decimal equivalent is $0.9$.

Comparing these values (0.6 < 0.667 < 0.8 < 0.9), the ascending order is $\frac{3}{5}$, $\frac{4}{6}$, $\frac{4}{5}$, $\frac{9}{10}$.

Arithmetic Evaluation and BODMAS

The expression given for evaluation is $\frac{1}{3}$ of $42 + 12 (2 \frac{1}{2} - \frac{1}{2})$. To solve this, apply the order of operations (BODMAS/PEMDAS):

First, solve the parenthetical expression: $(2 \frac{1}{2} - \frac{1}{2}) = 2$

Next, handle the "of" (multiplication) operation: $\frac{1}{3} \times 42 = 14$

Then, perform the multiplication within the remaining term: $12 \times 2 = 24$

Finally, add the results together: $14 + 24 = 38$

Decimal Evaluation and Simplification

Evaluate the expression $\frac{1.9 \times 0.41}{570 \times 2.46}$ without using a calculator. This can be simplified by identifying common factors and powers of ten:

Step 1: Simplify $\frac{1.9}{570}$. Note that $1.9 \times 300 = 570$. Thus, $\frac{1.9}{570} = \frac{1}{300}$.

Step 2: Simplify $\frac{0.41}{2.46}$. Multiplying both by $100$ gives $\frac{41}{246}$. Since $41\times 6 = 246$, this simplifies to $\frac{1}{6}$.

Step 3: Combine the simplified fractions: $\frac{1}{300} \times \frac{1}{6} = \frac{1}{1800}$

Converting Recurring Decimals to Fractions

To express a recurring decimal like $0.27$ (where the bar or recurring dots are over both $2$ and $7$) as a fraction in its simplest form, follow these steps:

Let $x = 0.272727\dots$ Then $100x = 27.272727\dots$ Subtracting the first equation from the second: $100x - x = 27.272727\dots - 0.272727\dots$ $99x = 27$ $x = \frac{27}{99}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is $9$: $\frac{27 \div 9}{99 \div 9} = \frac{3}{11}$

Using Mathematical Tables for Roots and Squares

To evaluate $\sqrt{547.9} + (0.1097)^2$ using square and square root tables:

1. Find $\sqrt{547.9}$: Using tables for $\sqrt{5.479 \times 10^2}$, we find $10 \times \sqrt{5.479}$. This result is approximately $23.40726$.

2. Find $(0.1097)^2$: Convert to $(1.097 \times 10^{-1})^2 = 1.203409 \times 10^{-2} = 0.01203409$.

Adding the results: $23.40726 + 0.01203409 = 23.41929409$

Rounding to the precision indicated in the options yields $23.41929$.

Least Common Multiple in Synchronous Events

Three lights flash at intervals of $5$, $6$, and $8\,\text{seconds}$. To determine when they will flash together again, find the Least Common Multiple (LCM) of the three numbers:

• The prime factors of $5$ are $5$.

• The prime factors of $6$ are $2 \times 3$.

• The prime factors of $8$ are $2^3$.

The LCM is calculated by taking the highest power of each prime present: $5^1 \times 3^1 \times 2^3 = 5 \times 3 \times 8 = 120$.

The lights will flash together again in $120\,\text{seconds}$, which is equivalent to $2\,\text{minutes}$.

Volume of Cylindrical Pipe Materials

A cylindrical pipe has an internal diameter of $7\,\text{cm}$ ($r = 3.5\,\text{cm}$), an external diameter of $14\,\text{cm}$ ($R = 7\,\text{cm}$), and a length (height) of $21\,\text{cm}$. To find the volume of material used, subtract the internal volume from the external volume:

$V = \pi (R^2 - r^2) h$

$V = \frac{22}{7} \times (7^2 - 3.5^2) \times 21$

$V = 22 \times 3 \times (49 - 12.25)$

$V = 66 \times 36.75 = 2425.5\,\text{cm}^3$

Rates of Work and Proportionality

If it takes $6$ men $2$ days to dig a trench $20\,\text{m}$ long, we can find how long it take $4$ men to dig the same trench.

First, calculate the total man-days required for the job: $\text{Total Work} = 6\,\text{men} \times 2\,\text{days} = 12\,\text{man-days}$

To find the time (t) for $4$ men to complete the same $12$-man-day task: $4\,\text{men} \times t = 12\,\text{man-days}$ $t = \frac{12}{4} = 3\,\text{days}$

Chronological Applications of LCM

Three bells ring at intervals of $12$, $16$, and $42\,\text{minutes}$. They ring together at $11:00\,\text{p.m.}$ To find the last time they rang together, find the LCM of the intervals:

• Prime factors of $12 = 2^2 \times 3$

• Prime factors of $16 = 2^4$

• Prime factors of $42 = 2 \times 3 \times 7$

$\text{LCM} = 2^4 \times 3 \times 7 = 16 \times 21 = 336\,\text{minutes}$

Convert $336\,\text{minutes}$ into hours and minutes: $336 \div 60 = 5\,\text{hours with a remainder of } 36\,\text{minutes}$

To find the previous time, subtract $5\,\text{hours } 36\,\text{minutes}$ from $11:00\,\text{p.m.}$: $11:00\,\text{p.m.} - 5:00 = 6:00\,\text{p.m.}$ $6:00\,\text{p.m.} - 36\,\text{minutes} = 5:24\,\text{p.m.}$

Geometry of Regular Polygons

The interior angle ($I$) of a regular polygon is twice the size of its exterior angle ($E$). Because the sum of the interior and exterior angles at any vertex is always $180^{\circ}$, we set up an equation:

$I = 2E$ $I + E = 180^{\circ}$ $2E + E = 180^{\circ}$ $3E = 180^{\circ} \rightarrow E = 60^{\circ}$

The number of sides ($n$) for a regular polygon is found by dividing $360^{\circ}$ by the exterior angle: $n = \frac{360^{\circ}}{60^{\circ}} = 6$

A polygon with $6$ sides is a hexagon.

Number Theory: Remainder and Smallest Multiples

A number $P$, when divided by $16$ and $24$, leaves a remainder of $5$. To find the smallest possible value of $P$, determine the LCM of the divisors and add the remainder:

• Prime factors of $16 = 2^4$

• Prime factors of $24 = 2^3 \times 3$

$\text{LCM}(16, 24) = 2^4 \times 3 = 16 \times 3 = 48$

$P = \text{LCM} + \text{remainder} = 48 + 5 = 53$

Ratio and Price Calculations

The price of a machine increased in the ratio $7:2$. If the original cost was Sh. $90,000$, the current price is calculated by multiplying the original price by the growth factor:

$\text{Current price per machine} = \text{Sh. } 90,000 \times \frac{7}{2} = \text{Sh. } 315,000$

To find the current price of two machines: $\text{Total cost} = \text{Sh. } 315,000 \times 2 = \text{Sh. } 630,000$

Algebraic Age Problems

Tom is twice as old as Mark. In $10\,\text{years}$, Tom will be $1.5$ times Mark's age. Let $M$ be Mark's current age across and $T$ be Tom's current age:

1. $T = 2M$

2. $T + 10 = 1.5(M + 10)$

Substitute the first equation into the second: $2M + 10 = 1.5M + 15$ $2M - 1.5M = 15 - 10$ $0.5M = 5 \rightarrow M = 10$

Since Mark is $10$, Tom is $2 \times 10 = 20$. The sum of their current ages is: $20 + 10 = 30\,\text{years}$

Solving Simultaneous Equations

Consider the system of equations:

1. $2x + 3y = 5$

2. $3x + 2y = 0$

From equation (2), solve for $y$ in terms of $x$: $2y = -3x \rightarrow y = -1.5x$

Substitute this into equation (1): $2x + 3(-1.5x) = 5$ $2x - 4.5x = 5$ $-2.5x = 5 \rightarrow x = -2$

Now find $y$: $y = -1.5(-2) = 3$

The solution is $x = -2, y = 3$.

Algebraic Area of a Rectangle

Find the area of a rectangle with dimensions $2(x + 3)\,\text{cm}$ and $(x + 8)\,\text{cm}$. The area is computed by multiplying length by width:

$\text{Area} = [2(x + 3)] \times (x + 8)$ $\text{Area} = (2x + 6)(x + 8)$

Expand the expression using FOIL (First, Outer, Inner, Last): $\text{Area} = 2x(x) + 2x(8) + 6(x) + 6(8)$ $\text{Area} = 2x^2 + 16x + 6x + 48$ $\text{Area} = 2x^2 + 22x + 48$

The final algebraic expression for the area is $2x^2 + 22x + 48\,\text{cm}^2$.