Question 1

Solve $1\frac{1}{8} + 1\frac{6}{7} + 3\frac{3}{5} = x$

Accepted Answer

Concept & Strategy When adding multiple mixed fractions, the most efficient method is to split the numbers into their integer components and their fractional components. Process them separately before recombining. $$A\frac{b}{c} + X\frac{y}{z} = (A + X) + \left(\frac{b}{c} + \frac{y}{z}ight)$$ Step-by-Step Solution * Separate the integer and fraction parts from the given equation: $$(1 + 1 + 3) + \left(\frac{1}{8} + \frac{6}{7} + \frac{3}{5}ight)$$ * Sum the integer parts: $$1 + 1 + 3 = 5$$ * Find the Least Common Multiple (LCM) for the denominators 8, 7, and 5 to add the fractions. Since they share no common factors, multiply them: $$LCM = 8 	imes 7 	imes 5 = 280$$ * Convert each fraction to have the common denominator of 280: $$\frac{1 	imes 35}{8 	imes 35} = \frac{35}{280}$$ $$\frac{6 	imes 40}{7 	imes 40} = \frac{240}{280}$$ $$\frac{3 	imes 56}{5 	imes 56} = \frac{168}{280}$$ * Add the numerators of the converted fractions together: $$\frac{35 + 240 + 168}{280} = \frac{443}{280}$$ * Convert the improper fraction back into a mixed fraction: $$\frac{443}{280} = 1\frac{163}{280}$$ * Recombine the integer sum and the new mixed fraction: $$5 + 1\frac{163}{280} = 6\frac{163}{280}$$ Exam Strategy & Shortcut Instead of immediately jumping to find the massive LCM of 280, add the whole numbers first to get 5. Glancing at the fractions $\frac{6}{7}$ (which is nearly 1) and $\frac{3}{5}$ (more than half), you know their sum will add at least 1 more to your total integer. This means the final integer part must be 6. Looking at the options, (b) is the only choice starting with 6, allowing you to bypass the heavy fraction math entirely. Common Pitfall A frequent error is converting all mixed fractions into improper fractions right at the start (e.g., turning $3\frac{3}{5}$ into $\frac{18}{5}$). This results in very large, unwieldy numerators that must be multiplied for the common denominator, drastically increasing calculation time and the likelihood of arithmetic mistakes. Final Answer Therefore, the correct answer is **$6\frac{163}{280}$**.

Question 2

Solve $2.5 	imes 4.8 + 7.2 	imes 1.5 - 1.2 	imes 14 = x$

Accepted Answer

### Concept & Logic The problem requires applying the standard order of operations (BODMAS/PEMDAS) to decimal arithmetic. Multiplication must be performed before addition and subtraction. ### Step-by-Step Solution * **Step 1:** Isolate and perform the multiplication operations first. * $2.5 	imes 4.8 = 12$ * $7.2 	imes 1.5 = 10.8$ * $1.2 	imes 14 = 16.8$ * **Step 2:** Substitute these products back into the original equation. * Equation becomes: $12 + 10.8 - 16.8$ * **Step 3:** Perform addition and subtraction from left to right. * $22.8 - 16.8 = 6$ ### Exam Strategy & Shortcut To multiply decimals quickly, temporarily ignore the decimal points, perform integer multiplication, and then place the decimal back based on the total decimal places. For example, for $2.5 	imes 4.8$, calculate $25 	imes 48$. An easy way to do $25 	imes 48$ is $100/4 	imes 48 = 1200$. Since there are two decimal places in total ($2.5$ and $4.8$), shift the decimal two spots to the left: $12.00$. ### Common Pitfall Students often make alignment errors when adding or subtracting the final terms, such as subtracting $16.8$ from $10.8$ incorrectly or misplacing the decimal point during the initial multiplication steps. ### Final Answer Therefore, the correct answer is 6.

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