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AQA GCSE · Question 15.2 · Number
Here are two calculations, A and B.
A: 1.92⁷ + 6.9³
B: 5 × ³√1000350
Use approximations to show that answer to A < answer to B.
Here are two calculations, A and B.
A: 1.92⁷ + 6.9³
B: 5 × ³√1000350
Use approximations to show that answer to A < answer to B.
How to approach this question
1. **Approximate A:** Round each number to the nearest integer before performing the calculations. 1.92 becomes 2, and 6.9 becomes 7. Calculate 2⁷ + 7³.
2. **Approximate B:** Round the number inside the cube root to a nearby number that is easy to cube root, like 1,000,000. Calculate the cube root, then multiply by 5.
3. **Compare:** Compare your two estimated answers to show that the estimate for A is less than the estimate for B.
Full Answer
**Approximation for A:**
1.92 is approximately 2.
6.9 is approximately 7.
So, A ≈ 2⁷ + 7³
2⁷ = 2×2×2×2×2×2×2 = 128
7³ = 7×7×7 = 49×7 = 343
A ≈ 128 + 343 = 471
**Approximation for B:**
1000350 is approximately 1000000.
³√1000000 = 100 (since 100 × 100 × 100 = 1000000).
So, B ≈ 5 × ³√1000000
B ≈ 5 × 100 = 500
**Comparison:**
A ≈ 471 and B ≈ 500.
Since 471 < 500, we have shown that the answer to A is less than the answer to B.
The question requires us to use approximations to compare the values of A and B.
**Calculation A: 1.92⁷ + 6.9³**
- We can approximate 1.92 as 2.
- We can approximate 6.9 as 7.
- So, A ≈ 2⁷ + 7³.
- Let's calculate these powers:
- 2⁷ = 2×2×2×2×2×2×2 = 128.
- 7³ = 7×7×7 = 49×7 = (50×7) - (1×7) = 350 - 7 = 343.
- A ≈ 128 + 343 = 471.
**Calculation B: 5 × ³√1000350**
- We can approximate 1000350 as 1000000, which is a convenient cube number.
- ³√1000000 = 100, because 100³ = 100 × 100 × 100 = 1,000,000.
- So, B ≈ 5 × 100.
- B ≈ 500.
**Comparison**
Our approximation for A is 471.
Our approximation for B is 500.
Since 471 < 500, our approximations show that the answer to A is less than the answer to B.
Common mistakes
✗ Choosing poor approximations that make the calculation difficult or inaccurate (e.g., using 1.9 or 1000300).
✗ Errors in calculating powers (e.g., 2⁷ or 7³).
✗ Errors in calculating the cube root.
✗ Not showing the final comparison clearly.
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