42. Approximating values for quick computation problems

42. Approximating values for quick computation problems

Approximating values for quick computation problems involve estimating numerical results to simplify calculations, often when exact computations are complex or unnecessary. These problems typically use techniques like rounding, linear approximations, percentage estimates, or simplified fractions to obtain values that are "close enough" for practical purposes. They are common in real-world scenarios like finance, engineering, statistics, or everyday calculations where precision is less critical than speed.

Below are 10 approximating values for quick computation problems with step-by-step solutions and detailed explanations. Each problem includes the problem statement, solution procedure, verification (comparing with exact values), and an explanation of the concept. The problems are diverse, covering scenarios like arithmetic calculations, percentages, roots, and financial estimates, ensuring they are distinct from previous problem types (e.g., surds, sequences, or age problems). They focus on numerical approximations and avoid unrelated contexts (e.g., alcohol or mixing solutions).

Problem 1

Approximate the value of

47 \times 23

for quick computation.

Step-by-Step Solution:

Round the numbers: To simplify, round 47 to 50 and 23 to 20.
Multiply the rounded numbers:
50 \times 20 = 1,000
Adjust for rounding:
- 47 is 3 less than 50, and 23 is 3 less than 20.
- The error is approximately the cross terms:
  50 \times 3 + 20 \times 3 = 150 + 60 = 210
  .
- Subtract the error:
  1,000 - 210 = 790
  .
- Alternatively, compute directly with one rounded number:
  50 \times 23 = 1,150
  , then adjust for 47 being 3 less:
  1,150 - 3 \times 23 = 1,150 - 69 = 1,081
  .
Choose the simpler approximation: For quick computation,
50 \times 20 = 1,000
is often sufficient, but ( 1,081 ) is closer.
Verify: Exact value:
47 \times 23 = 1,081
.
- Approximation (1,000): Error =
  |1,081 - 1,000| = 81
  , about 7.5% off.
- Adjusted approximation (1,081): Exact match.
Answer: Approximately 1,000 (simple) or 1,081 (adjusted).

Explanation: Rounding numbers to multiples of 10 simplifies multiplication, especially mentally. Adjusting for the rounding error improves accuracy but may defeat the "quick" purpose. Here,

50 \times 20 = 1,000

is a fast estimate, while the adjustment aligns with the exact value, showing a trade-off between speed and precision.

Problem 2

Estimate the total cost of items priced at $19.75, $8.20, and $12.95.

Step-by-Step Solution:

Round the prices:
- $19.75 \approx $20
- $8.20 \approx $8
- $12.95 \approx $13
Add the rounded values:
20 + 8 + 13 = 41
Adjust for rounding:
- $19.75 is $0.25 less than $20: error = -$0.25
- $8.20 is $0.20 more than $8: error = +$0.20
- $12.95 is $0.05 less than $13: error = -$0.05
- Net error:
  -0.25 + 0.20 - 0.05 = -0.10
- Adjust:
  41 - 0.10 = 40.90
Verify: Exact sum:
19.75 + 8.20 + 12.95 = 40.90
.
- Simple approximation ($41): Error =
  |40.90 - 41| = 0.10
  , about 0.24% off.
- Adjusted approximation ($40.90): Exact match.
Answer: Approximately $41 (simple) or $40.90 (adjusted).

Explanation: Rounding to the nearest dollar simplifies addition for quick mental computation. Adjusting for the rounding errors provides a more precise estimate but requires additional steps. The simple approximation is close enough for most practical purposes, like budgeting.

Problem 3

Approximate

\sqrt{52}

for quick computation.

Step-by-Step Solution:

Find the nearest perfect square: The closest perfect square to 52 is
7^2 = 49
, since
8^2 = 64
is too far.
Use a linear approximation: For
\sqrt{x}
near a known square
a^2
, approximate:
\sqrt{x} \approx a + \frac{x - a^2}{2a}
Here,
x = 52
,
a = 7
,
a^2 = 49
:
\sqrt{52} \approx 7 + \frac{52 - 49}{2 \cdot 7} = 7 + \frac{3}{14} \approx 7 + 0.214 \approx 7.214
Simplify for quick mental math: Alternatively, estimate
\frac{3}{14} \approx 0.2
, so:
7 + 0.2 = 7.2
Verify: Exact value:
\sqrt{52} \approx 7.211
.
- Approximation (7.2): Error =
  |7.211 - 7.2| \approx 0.011
  , about 0.15% off.
- Precise approximation (7.214): Error =
  |7.211 - 7.214| \approx 0.003
  , very accurate.
Answer: Approximately 7.2 (simple) or 7.214 (adjusted).

Explanation: Approximating square roots using the nearest perfect square and a linear correction (derived from the derivative of

\sqrt{x}

) provides a quick estimate. The simple 7.2 is fast for mental math, while the adjusted value is closer, balancing speed and accuracy.

Problem 4

Estimate the value of

\frac{97}{31}

for quick computation.

Step-by-Step Solution:

Round the numbers: Approximate 97 as 100 and 31 as 30.
Divide the rounded numbers:
\frac{100}{30} = \frac{10}{3} \approx 3.333
Adjust for rounding:
- 97 is 3 less than 100, and 31 is 1 less than 30.
- Use the approximation formula for division: For
  \frac{a}{b} \approx \frac{a'}{b'}
  , the error is:
  \frac{a}{b} \approx \frac{a'}{b'} \cdot \frac{1 - \frac{a'-a}{a'}}{1 - \frac{b'-b}{b'}}
  Here,
  a = 97
  ,
  a' = 100
  ,
  b = 31
  ,
  b' = 30
  :
  \frac{a'-a}{a'} = \frac{100-97}{100} = 0.03, \quad \frac{b'-b}{b'} = \frac{30-31}{30} = -\frac{1}{30} \approx -0.0333
  \frac{1 - 0.03}{1 - (-0.0333)} = \frac{0.97}{1.0333} \approx 0.938
  \frac{100}{30} \cdot 0.938 \approx 3.333 \cdot 0.938 \approx 3.126
Simplify for quick math: Alternatively, compute
\frac{100}{31} \approx 3.226
, then adjust for 97 being 3 less:
\frac{97}{31} \approx 3.226 - \frac{3}{31} \approx 3.226 - 0.097 \approx 3.129
Verify: Exact value:
\frac{97}{31} \approx 3.129
.
- Simple approximation (3.333): Error =
  |3.129 - 3.333| \approx 0.204
  , about 6.5% off.
- Adjusted approximation (3.129): Very close to exact.
Answer: Approximately 3.33 (simple) or 3.13 (adjusted).

Explanation: Rounding simplifies division, but the denominator’s proximity to 30 introduces some error. The adjustment accounts for relative changes in numerator and denominator, improving accuracy. The simple estimate (3.33) is quick for mental math, while the adjusted value is nearly exact.

Problem 5

A store offers a 19% discount on a $152 item. Approximate the discounted price.

Step-by-Step Solution:

Round the percentage and price: Approximate 19% as 20% and $152 as $150.
Calculate the discount:
20\% \text{ of } 150 = 0.2 \times 150 = 30
Find the discounted price:
150 - 30 = 120
Adjust for rounding:
- 19% is 1% less than 20%, so the discount is slightly less:
  1\% \text{ of } 150 = 1.5
  , so add $1.50:
  120 + 1.5 = 121.5
  .
- $152 is $2 more than $150, so the original price increases the discount:
  20\% \text{ of } 2 = 0.4
  . Adjust:
  121.5 + 0.4 = 121.9
  .
Verify: Exact discount:
0.19 \times 152 = 28.88
.
- Discounted price:
  152 - 28.88 = 123.12
  .
- Simple approximation ($120): Error =
  |123.12 - 120| \approx 3.12
  , about 2.5% off.
- Adjusted approximation ($121.9): Error =
  |123.12 - 121.9| \approx 1.22
  , about 1% off.
Answer: Approximately $120 (simple) or $121.90 (adjusted).

Explanation: Rounding the percentage to 20% and price to $150 simplifies the calculation. Adjustments account for the slight differences, improving accuracy. The simple estimate is fast for quick budgeting, while the adjusted value is closer to the exact price.

Problem 6

Approximate the value of

1.03^5

for quick computation.

Step-by-Step Solution:

Use the binomial approximation: For small ( x ),
(1 + x)^n \approx 1 + nx
. Here,
x = 0.03
,
n = 5
:
1.03^5 \approx 1 + 5 \cdot 0.03 = 1 + 0.15 = 1.15
Refine the approximation: Include the second term for better accuracy:
(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2} x^2
\frac{5 \cdot 4}{2} \cdot (0.03)^2 = 10 \cdot 0.0009 = 0.009
1.15 + 0.009 = 1.159
Verify: Exact value:
1.03^5 \approx 1.15927
.
- Simple approximation (1.15): Error =
  |1.15927 - 1.15| \approx 0.00927
  , about 0.8% off.
- Refined approximation (1.159): Error =
  |1.15927 - 1.159| \approx 0.00027
  , very accurate.
Answer: Approximately 1.15 (simple) or 1.159 (refined).

Explanation: The binomial approximation is ideal for quick computations of expressions like

(1 + x)^n

when ( x ) is small. The simple linear term (1 + nx) is fast, while adding the quadratic term improves accuracy, especially for larger ( n ).

Problem 7

Estimate the total distance traveled by a car averaging 47 mph for 3.2 hours.

Step-by-Step Solution:

Round the values: Approximate 47 mph as 50 mph and 3.2 hours as 3 hours.
Calculate the distance:
\text{Distance} = 50 \times 3 = 150 \text{ miles}
Adjust for rounding:
- 47 is 3 less than 50: Error for 3 hours =
  3 \times 3 = 9
  miles less.
- 3.2 is 0.2 more than 3: Additional distance =
  47 \times 0.2 \approx 9.4
  .
- Net adjustment:
  150 - 9 + 9.4 = 150.4
  .
Verify: Exact distance:
47 \times 3.2 = 150.4
.
- Simple approximation (150): Error =
  |150.4 - 150| = 0.4
  , about 0.27% off.
- Adjusted approximation (150.4): Exact match.
Answer: Approximately 150 miles (simple) or 150.4 miles (adjusted).

Explanation: Rounding simplifies the multiplication, making it suitable for quick estimation. Adjustments account for the differences in speed and time, yielding an exact result here. The simple estimate is close enough for most practical purposes, like trip planning.

Problem 8

Approximate the area of a circle with a radius of 7.8 cm.

Step-by-Step Solution:

Round the radius: Approximate 7.8 cm as 8 cm.
Use the area formula:
A = \pi r^2
. Approximate
\pi \approx 3.14
:
A \approx 3.14 \times 8^2 = 3.14 \times 64 \approx 200.96
For quicker mental math, use
\pi \approx 3
:
A \approx 3 \times 64 = 192
Adjust for rounding:
- 7.8 is 0.2 less than 8. Use the approximation for
  r^2
  :
  (8 - 0.2)^2 = 64 - 2 \cdot 8 \cdot 0.2 + 0.2^2 \approx 64 - 3.2 + 0.04 = 60.84
  A \approx 3.14 \times 60.84 \approx 191.04
Verify: Exact area:
\pi \times 7.8^2 \approx 3.14159 \times 60.84 \approx 191.13
.
- Simple approximation (192): Error =
  |191.13 - 192| \approx 0.87
  , about 0.46% off.
- Adjusted approximation (191.04): Error =
  |191.13 - 191.04| \approx 0.09
  , very accurate.
Answer: Approximately 192 cm² (simple) or 191 cm² (adjusted).

Explanation: Rounding the radius and using a simpler

\pi

value (3) speeds up the calculation. The adjustment accounts for the squared effect of the radius difference, improving accuracy. The simple estimate is practical for quick geometric calculations.

Problem 9

Estimate the total revenue from selling 195 units at $9.85 each.

Step-by-Step Solution:

Round the values: Approximate 195 units as 200 and $9.85 as $10.
Calculate the revenue:
200 \times 10 = 2,000
Adjust for rounding:
- 195 is 5 less than 200: Error =
  5 \times 10 = 50
  less.
- $9.85 is $0.15 less than $10: Error =
  195 \times 0.15 \approx 29.25
  less.
- Total adjustment:
  2,000 - 50 - 29.25 = 1,920.75
  .
Verify: Exact revenue:
195 \times 9.85 = 1,920.75
.
- Simple approximation ($2,000): Error =
  |1,920.75 - 2,000| = 79.25
  , about 4.1% off.
- Adjusted approximation ($1,920.75): Exact match.
Answer: Approximately $2,000 (simple) or $1,920.75 (adjusted).

Explanation: Rounding simplifies the multiplication, ideal for quick financial estimates. Adjustments account for both quantity and price differences, yielding the exact value here. The simple estimate is useful for rough budgeting or forecasting.

Problem 10

Approximate the time it takes to travel 485 miles at an average speed of 62 mph.

Step-by-Step Solution:

Round the values: Approximate 485 miles as 500 miles and 62 mph as 60 mph.
Calculate the time:
\text{Time} = \frac{\text{Distance}}{\text{Speed}}
:
\frac{500}{60} \approx 8.333 \text{ hours} \approx 8 \frac{1}{3} \text{ hours}
Adjust for rounding:
- 485 is 15 less than 500: Error =
  \frac{15}{60} = 0.25
  hours less.
- 62 is 2 more than 60: Approximate
  \frac{485}{62} \approx \frac{485}{60} \cdot \frac{60}{62} \approx \frac{485}{60} \cdot 0.967
  .
- Compute:
  \frac{485}{60} \approx 8.083
  , then
  8.083 \cdot 0.967 \approx 7.816
  .
Verify: Exact time:
\frac{485}{62} \approx 7.823
.
- Simple approximation (8.333): Error =
  |7.823 - 8.333| \approx 0.51
  , about 6.5% off.
- Adjusted approximation (7.816): Error =
  |7.823 - 7.816| \approx 0.007
  , very accurate.
Answer: Approximately 8.33 hours (simple) or 7.82 hours (adjusted).

Explanation: Rounding simplifies division, making it easier to estimate travel time. The adjustment accounts for differences in distance and speed, improving accuracy. The simple estimate is quick for planning, while the adjusted value is nearly exact.

General Notes on Approximating Values for Quick Computation Problems

Key Concept: These problems aim to simplify complex calculations by rounding numbers, using approximations (e.g., binomial or linear), or estimating percentages, prioritizing speed over exact precision.
Key Steps:
1. Identify numbers that can be rounded to simpler values (e.g., multiples of 10, whole numbers, or approximate percentages).
2. Perform the calculation with rounded values for quick results.
3. Adjust for rounding errors if higher accuracy is needed, using differences or relative errors.
4. Verify by comparing with the exact value to assess the approximation’s accuracy.
Common Techniques:
- Rounding: Adjust numbers to multiples of 10, 100, or simple fractions (e.g., 19% to 20%).
- Binomial Approximation: For
  (1 + x)^n
  , use
  1 + nx
  when ( x ) is small.
- Linear Approximation: For functions like
  \sqrt{x}
  , use
  f(x) \approx f(a) + f'(a)(x-a)
  .
- Percentage Estimates: Round percentages (e.g., 19% to 20%) for quick calculations.
- Error Adjustment: Account for rounding errors by calculating differences or relative changes.
Challenges:
- Balance speed and accuracy: Simple approximations are fast but may have larger errors.
- Avoid overcomplicating adjustments, as they can defeat the purpose of quick computation.
- Ensure rounding is reasonable (e.g., 47 to 50 is better than 40 for small differences).
- Verify approximations to ensure they are within acceptable error margins for the context.
Applications: These problems apply to finance (budgeting, discounts), engineering (quick estimates), statistics (sample proportions), and everyday scenarios (travel time, shopping), where rough estimates suffice.
Verification: Compare approximations with exact values to quantify errors, ensuring the estimate is practical for the intended use. Percentage errors help assess accuracy.

These problems demonstrate a range of approximation techniques, from simple rounding to binomial and linear approximations, applied to practical scenarios.

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42. Approximating values for quick computation problems

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