3 QUESTION 3 3.1 Write down the LCM of the following numbers, give your answer in exponential form. $$ \begin{array}{l} \mathrm{A}=2^{9} \times 3^{6} \times 5^{4} \times 11^{3} \ \mathrm{~B}=2^{2} \times 3^{7} \times 7^{2} \ \mathrm{C}=2^{7} \times 3^{3} \times 13^{3} \end{array} $$ $$ \qquad $$ (2) 3.2 Use prime factors to determine the HCF of $$ 28 ; 196 $$ and 280 $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ (4) 3.3 Two buses, bus $$ A $$ and bus B, both use the same bus stop. - Bus A runs every 10 minutes. - Bus B runs every 14 minutes. - Both buses are at the bus stop at 11 am. What time will both buses next be at the bus stop? $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ (4) [10] QUESTION 4 4.1 Write the following ratio in its simplest form: $$ 2,1 \text { to } 0,7 \text { to } 0,014 $$ $$ \qquad $$ $$ \qquad $$ (2) PSIS NATHEMATICS GRODE B

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**3.1 Finding the LCM**

We are given:
$$
\begin{aligned}
A &= 2^{9} \times 3^{6} \times 5^{4} \times 11^{3}, \
B &= 2^{2} \times 3^{7} \times 7^{2}, \
C &= 2^{7} \times 3^{3} \times 13^{3}.
\end{aligned}
$$

The LCM is found by taking each prime factor with the highest exponent present in any of the numbers.

- For $$2$$: maximum exponent is $$\max(9,2,7)=9$$.
- For $$3$$: maximum exponent is $$\max(6,7,3)=7$$.
- For $$5$$: appears only in $$A$$ as $$5^4$$.
- For $$7$$: appears only in $$B$$ as $$7^2$$.
- For $$11$$: appears only in $$A$$ as $$11^3$$.
- For $$13$$: appears only in $$C$$ as $$13^3$$.

Thus, the LCM is:
$$
\boxed{2^9 \times 3^7 \times 5^4 \times 7^2 \times 11^3 \times 13^3}.
$$

---

**3.2 Determining the HCF using Prime Factors**

We need to find the HCF of $$28$$, $$196$$, and $$280$$.

First, express each number in its prime factorization:
- $$28 = 2^2 \times 7$$.
- $$196 = 14^2 = (2\times7)^2 = 2^2 \times 7^2$$.
- $$280 = 28 \times 10 = (2^2 \times 7) \times (2 \times 5) = 2^3 \times 5 \times 7$$.

The HCF is obtained by taking the common prime factors with the smallest exponents:
- For $$2$$: the smallest exponent is $$\min(2,2,3)=2$$.
- For $$7$$: the smallest exponent is $$\min(1,2,1)=1$$.

So the HCF is:
$$
\boxed{2^2 \times 7 = 4 \times 7 = 28}.
$$

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**3.3 Bus Timetable Problem**

Bus $$A$$ runs every $$10$$ minutes, and bus $$B$$ runs every $$14$$ minutes. Both are at the bus stop at $$11$$ am. We need to find the next time they are together.

Find the LCM of $$10$$ and $$14$$:
- $$10 = 2 \times 5$$.
- $$14 = 2 \times 7$$.

The LCM is:
$$
2 \times 5 \times 7 = 70 \text{ minutes}.
$$

Convert $$70$$ minutes to hours and minutes:
$$
70 \text{ minutes} = 1 \text{ hour } 10 \text{ minutes}.
$$

Adding $$1$$ hour $$10$$ minutes to $$11:00$$ am gives:
$$
11:00 \text{ am} + 1 \text{ hr }10 \text{ min} = 12:10 \text{ pm}.
$$

Thus, both buses will next be at the bus stop at:
$$
\boxed{12:10 \text{ pm}}.
$$

---

**4.1 Simplifying the Ratio**

We are given the ratio:
$$
2.1 \text{ to } 0.7 \text{ to } 0.014.
$$

First, express each number as a fraction (or eliminate decimals) by multiplying by an appropriate power of $$10$$. Multiply each term by $$1000$$ to remove decimals:
$$
2.1 \times 1000 = 2100, \quad 0.7 \times 1000 = 700, \quad 0.014 \times 1000 = 14.
$$

So the ratio becomes:
$$
2100 : 700 : 14.
$$

Divide each term by $$7$$ (the greatest common divisor for these numbers):
$$
2100 \div 7 = 300,\quad 700 \div 7 = 100,\quad 14 \div 7 = 2.
$$

This yields:
$$
300 : 100 : 2.
$$

Finally, we can divide each term by $$2$$ to simplify further:
$$
300 \div 2 = 150,\quad 100 \div 2 = 50,\quad 2 \div 2 = 1.
$$

Thus, the ratio in its simplest form is:
$$
\boxed{150 : 50 : 1}.
$$
**3.1 LCM** The LCM of the numbers is: $$ 2^9 \times 3^7 \times 5^4 \times 7^2 \times 11^3 \times 13^3 $$ **3.2 HCF** The HCF of $$28$$, $$196$$, and $$280$$ is: $$ 28 $$ **3.3 Bus Schedule** Both buses will next be at the bus stop at: $$ 12:10 \text{ pm} $$ **4.1 Ratio Simplification** The simplified ratio is: $$ 150 : 50 : 1 $$

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