After being dropped, a ball bounces to a height of 16 meters on its first bounce. Each subsequent bounce reaches...
GMAT Advanced Math : (Adv_Math) Questions
After being dropped, a ball bounces to a height of 16 meters on its first bounce. Each subsequent bounce reaches \(\frac{3}{4}\) of the height of the previous bounce. If \(\mathrm{h}\) represents the height of the \(\mathrm{n}^{\mathrm{th}}\) bounce, which equation gives \(\mathrm{h}\) in terms of \(\mathrm{n}\)?
\(16 \left( \frac{4}{3} \right)^{\mathrm{n}-1}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}+1}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}-1}\)
1. TRANSLATE the problem information
- Given information:
- First bounce height: 16 meters
- Each bounce reaches 3/4 of the previous height
- Need formula for height h of the nth bounce
2. INFER the mathematical pattern
- This describes a geometric sequence because each term is found by multiplying the previous term by the same ratio (3/4)
- In geometric sequences: each new value = previous value × constant ratio
3. INFER the sequence structure
- First term \(\mathrm{(a_1)}\) = 16
- Common ratio \(\mathrm{(r)}\) = \(\mathrm{\frac{3}{4}}\)
- The standard formula for the nth term of a geometric sequence is: \(\mathrm{a_n = a_1 \times r^{(n-1)}}\)
4. SIMPLIFY to get the final formula
- Substituting our values: \(\mathrm{h = 16 \times (\frac{3}{4})^{(n-1)}}\)
- This matches answer choice D
5. Verify by testing n = 1
- \(\mathrm{h_1 = 16 \times (\frac{3}{4})^{(1-1)}}\)
- \(\mathrm{= 16 \times (\frac{3}{4})^0}\)
- \(\mathrm{= 16 \times 1}\)
- \(\mathrm{= 16}\) ✓
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students recognize the decreasing pattern but apply the geometric sequence formula incorrectly, using the exponent n instead of (n-1).
They think: "The height after n bounces should be \(\mathrm{16 \times (\frac{3}{4})^n}\)" without recognizing that the first bounce corresponds to n=1, which should give the original height of 16, not \(\mathrm{16 \times \frac{3}{4}}\).
This leads them to select Choice C \(\mathrm{(16(\frac{3}{4})^n)}\).
Second Most Common Error:
Poor TRANSLATE reasoning: Students misinterpret "reaches 3/4 of the previous bounce" as meaning the ball somehow gains height, confusing the fraction.
They incorrectly use 4/3 as the ratio, thinking each bounce gets higher.
This may lead them to select Choice A \(\mathrm{(16(\frac{4}{3})^{(n-1)})}\).
The Bottom Line:
The key challenge is recognizing that while this is a geometric sequence, the indexing matters crucially - the "first bounce" corresponds to n=1 in the formula, requiring the (n-1) exponent to make the math work correctly.
\(16 \left( \frac{4}{3} \right)^{\mathrm{n}-1}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}+1}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}}\)
\(16 \left( \frac{3}{4} \right)^{\mathrm{n}-1}\)