Understanding the Problem: Making the Equation Correct

The question asks us to identify which pair of numbers, when swapped in the given mathematical equation, makes the equation true. We are given an equation with several arithmetic operations and multiple options for pairs of numbers to interchange.

The original equation is:

$(36 \div 2) \times 5 – 15 \times 4 = 20 − 260 \div (5 \times 3 + 2)$

We need to check each option by swapping the specified numbers and then evaluating both sides of the modified equation to see if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

Applying the Order of Operations (BODMAS/PEMDAS)

To correctly evaluate the equation after swapping numbers, we must follow the standard order of operations:

• B/P: Brackets or Parentheses
• O/E: Orders or Exponents
• D/M: Division or Multiplication (from left to right)
• A/S: Addition or Subtraction (from left to right)

Evaluating the Original Equation

Let's first evaluate the original equation to see if it is already correct (it should not be, based on the question asking for a swap).

LHS: $(36 \div 2) \times 5 – 15 \times 4$

• Inside brackets: $36 \div 2 = 18$
• Equation becomes: $18 \times 5 – 15 \times 4$
• Multiplication: $18 \times 5 = 90$, and $15 \times 4 = 60$
• Equation becomes: $90 – 60$
• Subtraction: $90 – 60 = 30$
• So, LHS = $30$.

RHS: $20 − 260 \div (5 \times 3 + 2)$

• Inside brackets: $5 \times 3 = 15$, then $15 + 2 = 17$
• Equation becomes: $20 − 260 \div 17$
• Division: $260 \div 17$ is not a whole number. $\frac{260}{17} \approx 15.29$
• Equation becomes: $20 − \frac{260}{17}$
• Subtraction: $20 − \frac{260}{17} = \frac{20 \times 17 − 260}{17} = \frac{340 − 260}{17} = \frac{80}{17} \approx 4.71$
• So, RHS $\approx 4.71$.

Since $30 \neq \frac{80}{17}$, the original equation is incorrect.

Testing the Options by Swapping Numbers

Option 1: Swapping 4 and 2

If we swap 4 and 2, the equation becomes:

$(36 \div 4) \times 5 – 15 \times 2 = 20 − 260 \div (5 \times 3 + 4)$

LHS: $(36 \div 4) \times 5 – 15 \times 2$

• Inside brackets: $36 \div 4 = 9$
• Equation becomes: $9 \times 5 – 15 \times 2$
• Multiplication: $9 \times 5 = 45$, and $15 \times 2 = 30$
• Equation becomes: $45 – 30$
• Subtraction: $45 – 30 = 15$
• So, LHS = $15$.

RHS: $20 − 260 \div (5 \times 3 + 4)$

• Inside brackets: $5 \times 3 = 15$, then $15 + 4 = 19$
• Equation becomes: $20 − 260 \div 19$
• Division: $\frac{260}{19} \approx 13.68$
• Equation becomes: $20 − \frac{260}{19}$
• Subtraction: $20 − \frac{260}{19} = \frac{20 \times 19 − 260}{19} = \frac{380 − 260}{19} = \frac{120}{19} \approx 6.32$
• So, RHS $\approx 6.32$.

Since $15 \neq \frac{120}{19}$, swapping 4 and 2 does not make the equation correct.

Option 2: Swapping 3 and 5

If we swap 3 and 5, the equation becomes:

$(36 \div 2) \times 3 – 15 \times 4 = 20 − 260 \div (3 \times 5 + 2)$

LHS: $(36 \div 2) \times 3 – 15 \times 4$

• Inside brackets: $36 \div 2 = 18$
• Equation becomes: $18 \times 3 – 15 \times 4$
• Multiplication: $18 \times 3 = 54$, and $15 \times 4 = 60$
• Equation becomes: $54 – 60$
• Subtraction: $54 – 60 = -6$
• So, LHS = $-6$.

RHS: $20 − 260 \div (3 \times 5 + 2)$

• Inside brackets: $3 \times 5 = 15$, then $15 + 2 = 17$
• Equation becomes: $20 − 260 \div 17$
• Division: $\frac{260}{17} \approx 15.29$
• Equation becomes: $20 − \frac{260}{17}$
• Subtraction: $20 − \frac{260}{17} = \frac{340 − 260}{17} = \frac{80}{17} \approx 4.71$
• So, RHS $\approx 4.71$.

Since $-6 \neq \frac{80}{17}$, swapping 3 and 5 does not make the equation correct.

Option 3: Swapping 20 and 36

If we swap 20 and 36, the equation becomes:

$(20 \div 2) \times 5 – 15 \times 4 = 36 − 260 \div (5 \times 3 + 2)$

LHS: $(20 \div 2) \times 5 – 15 \times 4$

• Inside brackets: $20 \div 2 = 10$
• Equation becomes: $10 \times 5 – 15 \times 4$
• Multiplication: $10 \times 5 = 50$, and $15 \times 4 = 60$
• Equation becomes: $50 – 60$
• Subtraction: $50 – 60 = -10$
• So, LHS = $-10$.

RHS: $36 − 260 \div (5 \times 3 + 2)$

• Inside brackets: $5 \times 3 = 15$, then $15 + 2 = 17$
• Equation becomes: $36 − 260 \div 17$
• Division: $\frac{260}{17} \approx 15.29$
• Equation becomes: $36 − \frac{260}{17}$
• Subtraction: $36 − \frac{260}{17} = \frac{36 \times 17 − 260}{17} = \frac{612 − 260}{17} = \frac{352}{17} \approx 20.71$
• So, RHS $\approx 20.71$.

Since $-10 \neq \frac{352}{17}$, swapping 20 and 36 does not make the equation correct.

Option 4: Swapping 2 and 3

If we swap 2 and 3, the equation becomes:

$(36 \div 3) \times 5 – 15 \times 4 = 20 − 260 \div (5 \times 2 + 3)$

LHS: $(36 \div 3) \times 5 – 15 \times 4$

• Inside brackets: $36 \div 3 = 12$
• Equation becomes: $12 \times 5 – 15 \times 4$
• Multiplication: $12 \times 5 = 60$, and $15 \times 4 = 60$
• Equation becomes: $60 – 60$
• Subtraction: $60 – 60 = 0$
• So, LHS = $0$.

RHS: $20 − 260 \div (5 \times 2 + 3)$

• Inside brackets: $5 \times 2 = 10$, then $10 + 3 = 13$
• Equation becomes: $20 − 260 \div 13$
• Division: $260 \div 13 = 20$
• Equation becomes: $20 − 20$
• Subtraction: $20 − 20 = 0$
• So, RHS = $0$.

Since $0 = 0$, swapping 2 and 3 makes the equation correct.

Conclusion

After testing each option, we found that swapping the numbers 2 and 3 makes the given equation correct.

Revision Table: Key Concepts

Additional Information: Strategies for Number Swapping Problems

When solving problems that require swapping numbers or operations to make an equation correct, consider these strategies:

• Understand the Target: Sometimes estimating the rough value of the target side of the equation can give clues about which swap might be effective.
• Analyze Operations: Look at where the numbers are located and the operations involved. Swapping numbers might change division results significantly (e.g., small number dividing large vs. large dividing small) or change signs.
• Systematic Testing: Test each option methodically. Don't guess. Calculate the LHS and RHS for every proposed swap.
• Check Division: Pay special attention to division, as swapping numbers might make a division possible that wasn't before, or vice-versa.
• Be Careful with Brackets: Operations within brackets are performed first. Swapping numbers inside or outside brackets will affect the outcome differently.

These types of problems test your understanding of arithmetic operations and your ability to apply them correctly under different conditions.