$\frac{1095}{1168}$ when expressed in simplest form is

### Concept & Strategy To reduce a fraction to its simplest form, we need to find the Highest Common Factor (HCF) of its numerator and denominator, and then divide both by this HCF. Alternatively, we can use the difference method: the HCF of two numbers divides their difference. This helps quickly narrow down potential factors without doing long division. ### Step-by-Step Solution * **Given:** The fraction is $\frac{1095}{1168}$. * **Calculation / Deduction:** 1. Let's find the difference between the denominator and the numerator: $$1168 - 1095 = 73$$ 2. The HCF must be a factor of $73$. Since $73$ is a prime number, the HCF is highly likely to be $73$. 3. Test if both numbers are divisible by $73$: $$1095 \div 73 = 15$$ $$1168 \div 73 = 16$$ 4. Since both divide perfectly, we reduce the fraction by dividing the top and bottom by $73$: $$\frac{1095 \div 73}{1168 \div 73} = \frac{15}{16}$$ ### Exam Strategy & Shortcut **Unit Digit Estimation:** Look at the options. The fraction ends in $5$ on top and $8$ on the bottom. We need a multiplier $k$ such that $k \times \text{Numerator Option}$ ends in $5$, and $k \times \text{Denominator Option}$ ends in $8$. Let's test option (b) $\frac{15}{16}$. If $15 \times k$ ends in $5$, $k$ must be odd (like $3, 5, 7, \dots$). If $16 \times k$ ends in $8$, $k$ could be $3$ ($16 \times 3 = 48$) or $8$ ($16 \times 8 = 128$). Let's test $k$ ending in $3$: $15 \times 3 = 45$ (ends in $5$), $16 \times 3 = 48$ (ends in $8$). Scaling this up, let $k = 73$. $15 \times 73 = 1095$. $16 \times 73 = 1168$. The match is instantaneous. ### Common Pitfall Trying to systematically divide $1095$ and $1168$ by small primes ($2, 3, 5, 7, 11\dots$). Since their only common factor is the large prime $73$, you will waste immense time and likely assume the fraction cannot be reduced. Always use the difference method for large numbers. ### Final Answer **Therefore, the correct answer is $\frac{15}{16}$.**