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12 / 30ApproximationMediumQuantitative Aptitude

124.983+216.023+343.053≈?\sqrt[3]{124.98} + \sqrt[3]{216.02} + \sqrt[3]{343.05} \approx ?

  1. 15

  2. 18

  3. 21

  4. 24

Show answer & explanation

Correct answer

18

Explanation

Rounding: 1253=5\sqrt[3]{125} = 5; 2163=6\sqrt[3]{216} = 6; 3433=7\sqrt[3]{343} = 7. Sum =5+6+7=18= 5 + 6 + 7 = 18.

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All 30 Approximation questions

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  11. 11.(19.96)^2 × (9.99)^3 ÷ (40.02)^2 ?
  12. 12.[3]124.98 + [3]216.02 + [3]343.05 ?
  13. 13.34.98\% of (899.96 - 250.04) + 51.02\% of 399.98 ?
  14. 14.(49.97)^2 - (29.98)^219.99 ?
  15. 15.(11.98)^3 + (18.99)^3 - (14.02)^3 ?
  16. 16.62.05\% of √6399.8 - 28.98\% of √2400.2 ?
  17. 17.4899.96 × 29.9998.98 × 15.03 ?
  18. 18.√8999.7 - √3599.9 + √1599.8 ?
  19. 19.(23.98)^2 + (36.99)^2 - (17.01)^2 - (29.02)^2 ?
  20. 20.Estimate the value of 48.7 × 19.23.9 × √63.8 by rounding each number to one significant figure, then calculate the approximate result.
  21. 21.Evaluate 48 × 5.2 + 3.7^2√225 - 1.8 by rounding each term to the nearest integer.
  22. 22.Approximate the value of (7.96)^3 × √48.2(2.01)^5 using suitable rounding of each component.
  23. 23.Using the approximation √10 3.16 and π 3.14 , estimate the value of 3π + 2√10π - √10 .
  24. 24.Simplify (4.99)^2 × (1.98)^3/(2.97)^4 by approximating each base to the nearest integer.
  25. 25.Estimate √399 and [3]124 using nearest perfect powers.
  26. 26.Approximate (9.02)^3.01 by first converting to a base and exponent adjustment.
  27. 27.Using the approximation √2 1.41 , √3 1.73 , and π 3.14 , evaluate 2π + 3√25√3 - π to the nearest tenth.
  28. 28.Approximate 0.97 × 15.2 + 4.8^2/2.1^3 - 0.99 by first rounding each term to one decimal place where appropriate, then compute to the…
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