Approximation is the chapter that rewards the student who is willing to be slightly wrong. That sounds strange, so read it again. The paper is not asking for the exact answer, it is asking which option is closest, and the options are usually far apart. So the moment you start calculating to two decimal places, you have already lost. Round first, calculate second. 19.98 is 20, 4.02 is 4, 34.9% is one third, and none of that will cost you the mark. The only care you need is to round in opposite directions where you can, so that one error cancels the other. Look at how far apart the options are before you begin, because that tells you how rough you are allowed to be. Do these questions with the rounding written down and you will finish a full set in the time you used to spend on five sums.
Approximation Rules and Shortcuts
Look at the options first
The gap between the options tells you how rough you are allowed to be. Wide options, round freely. Close options, round carefully.
Rounding
5 or more rounds up, less than 5 rounds down.
Round in opposite directions where you can, so one error cancels the other. In 19.8 × 5.2, take 20 × 5 = 100, and the true value is 102.96.
BODMAS still applies
Round the numbers, never the order of operations. Most wrong answers here come from breaking BODMAS, not from bad rounding.
The fraction values you must know
50% = 1/2, 33.33% = 1/3, 25% = 1/4, 20% = 1/5, 16.67% = 1/6, 14.28% = 1/7, 12.5% = 1/8, 11.11% = 1/9, 10% = 1/10, 9.09% = 1/11, 8.33% = 1/12, 6.25% = 1/16, 5% = 1/20
So 34.02% of 1200 is roughly one third of 1200 = 400
Percentage shortcuts
x% of y = y% of x, so 16% of 25 is the same as 25% of 16, which is 4.
Break it up. 35% of 240 = 25% of 240 plus 10% of 240 = 60 + 24 = 84.
Root approximation
√N is roughly a + (N − a²) ÷ (2a), so √51 is roughly 7 + 2 ÷ 14 = 7.14
³√N is roughly a + (N − a³) ÷ (3a²), so ³√130 is roughly 5 + 5 ÷ 75 = 5.07
Squares near a round number
a² − b² = (a + b) (a − b), so 4.99² − 5.01² = 10 × (−0.02) = −0.2
Never square them separately when two squares are being subtracted.
Successive percentage change
Net change of a% and b% = a + b + (a × b ÷ 100) percent
A rise of 20% followed by a fall of 20% is a net fall of 4%.
The method
Scan the options, round every number, solve by BODMAS, mark the closest option and move on. Do not go back and check with the exact figures, that is the whole point of the chapter.
60 Approximation Aptitude Questions and Answers (Solved MCQs)
Question 1. What is the approximate value of 487.92 + 613.18 – 298.76 when each number is rounded to the nearest integer?
a) 800
b) 802
c) 804
d) 806
Answer:
b) 802 — Rounding each number to the nearest integer gives 488 + 613 – 299 = 802.
Question 2. What is the approximate value of 39.87 × 20.14 when both numbers are rounded to the nearest integer?
a) 780
b) 790
c) 800
d) 820
Answer:
c) 800 — Rounding to the nearest integer gives 39.87 ≈ 40 and 20.14 ≈ 20. Therefore, the approximate value = 40 × 20 = 800.
Question 3. What is the approximate value of 784.26 ÷ 19.82 when the numbers are rounded to the nearest integers?
a) 38
b) 39
c) 40
d) 42
Answer:
b) 39 — Rounding to the nearest integers gives 784.26 ≈ 784 and 19.82 ≈ 20. Therefore, 784 ÷ 20 = 39.2, which is approximately 39.
Question 4. Estimate 6,483 + 2,719 – 1,264 by rounding each number to the nearest hundred.
a) 7,800
b) 7,900
c) 8,000
d) 8,100
Answer:
b) 7,900 — Rounding to the nearest hundred gives 6,483 ≈ 6,500, 2,719 ≈ 2,700 and 1,264 ≈ 1,300. Thus, 6,500 + 2,700 – 1,300 = 7,900.
Question 5. What is the approximate value of 51.86 × 9.92 + 184.37 when every number is rounded to the nearest integer?
a) 684
b) 694
c) 704
d) 714
Answer:
c) 704 — Rounding to the nearest integer gives 52 × 10 + 184 = 520 + 184 = 704.
Question 6. Estimate 8,749 ÷ 291 by rounding the dividend to the nearest thousand and the divisor to the nearest hundred.
a) 20
b) 25
c) 30
d) 35
Answer:
c) 30 — Rounding as directed gives 8,749 ≈ 9,000 and 291 ≈ 300. Therefore, 9,000 ÷ 300 = 30.
Question 7. What is the approximate value of 18.72% of 598 when 18.72% is rounded to 19% and 598 is rounded to 600?
a) 108
b) 110
c) 114
d) 120
Answer:
c) 114 — Using the stated approximations, the required value = 19% of 600 = (19/100) × 600 = 114.
Question 8. Estimate 297.8 × 14.9 ÷ 5.1 by rounding every number to the nearest integer.
a) 846
b) 876
c) 894
d) 910
Answer:
c) 894 — Rounding gives 297.8 ≈ 298, 14.9 ≈ 15 and 5.1 ≈ 5. Thus, (298 × 15) ÷ 5 = 298 × 3 = 894.
Question 9. A shop sold 1,487 notebooks in January, 2,536 in February and 1,962 in March. Approximately how many notebooks were sold altogether when each quantity is rounded to the nearest hundred?
a) 5,800
b) 5,900
c) 6,000
d) 6,100
Answer:
c) 6,000 — Rounding to the nearest hundred gives 1,487 ≈ 1,500, 2,536 ≈ 2,500 and 1,962 ≈ 2,000. Total ≈ 1,500 + 2,500 + 2,000 = 6,000.
Question 10. What is the approximate value of (62.8 × 17.2) – 348.6 when every number is rounded to the nearest integer?
a) 702
b) 712
c) 722
d) 732
Answer:
c) 722 — Rounding gives 62.8 ≈ 63, 17.2 ≈ 17 and 348.6 ≈ 349. Therefore, (63 × 17) – 349 = 1,071 – 349 = 722.
Question 11. Estimate 3,746 × 48 by rounding 3,746 to the nearest hundred and 48 to the nearest ten.
a) 1,75,000
b) 1,80,000
c) 1,85,000
d) 1,90,000
Answer:
c) 1,85,000 — Rounding gives 3,746 ≈ 3,700 and 48 ≈ 50. Therefore, the estimated product = 3,700 × 50 = 1,85,000.
Question 12. What is the approximate value of 999.4 + 501.7 ÷ 10.2 when all numbers are rounded to the nearest integer?
a) 1,039
b) 1,049
c) 1,059
d) 1,069
Answer:
b) 1,049 — Rounding gives 999.4 ≈ 999, 501.7 ≈ 502 and 10.2 ≈ 10. Applying division first, 999 + (502 ÷ 10) = 999 + 50.2 = 1,049.2 ≈ 1,049.
Question 13. Estimate the difference between 74,682 and 28,349 by rounding both numbers to the nearest thousand.
a) 45,000
b) 46,000
c) 47,000
d) 48,000
Answer:
c) 47,000 — Rounding to the nearest thousand gives 74,682 ≈ 75,000 and 28,349 ≈ 28,000. Therefore, the estimated difference = 75,000 – 28,000 = 47,000.
Question 14. What is the approximate value of (24.7 × 40.3) ÷ 8.2 when each number is rounded to the nearest integer?
a) 115
b) 120
c) 125
d) 130
Answer:
c) 125 — Rounding gives 24.7 ≈ 25, 40.3 ≈ 40 and 8.2 ≈ 8. Therefore, (25 × 40) ÷ 8 = 1,000 ÷ 8 = 125.
Question 15. Estimate 4,892 + 3,167 – 2,438 + 1,751 by rounding each number to the nearest hundred.
a) 7,200
b) 7,300
c) 7,400
d) 7,500
Answer:
b) 7,300 — Rounding to the nearest hundred gives 4,892 ≈ 4,900, 3,167 ≈ 3,200, 2,438 ≈ 2,400 and 1,751 ≈ 1,800. Thus, 4,900 + 3,200 – 2,400 + 1,800 = 7,500. Therefore, the correct answer is d) 7,500.
Question 16. What is the approximate value of 7/13 of 1,297 when 1,297 is rounded to the nearest convenient number?
a) 650
b) 675
c) 700
d) 725
Answer:
c) 700 — Rounding 1,297 to the nearby number 1,300 gives (7/13) × 1,300 = 7 × 100 = 700.
Question 17. What is the approximate value of 24.8% of 1,598?
a) 380
b) 400
c) 420
d) 440
Answer:
b) 400 — Approximate 24.8% as 25% and 1,598 as 1,600. Therefore, the required value ≈ 25% of 1,600 = 400.
Question 18. What is the approximate value of √1,587?
a) 38
b) 39
c) 40
d) 41
Answer:
c) 40 — The nearest convenient perfect square to 1,587 is 1,600. Therefore, √1,587 ≈ √1,600 = 40.
Question 19. What is the approximate value of ∛4,910?
a) 15
b) 16
c) 17
d) 18
Answer:
c) 17 — The nearest perfect cube to 4,910 is 4,913, and 17³ = 4,913. Therefore, ∛4,910 ≈ 17.
Question 20. Estimate (3/8 of 2,392) + (5/12 of 1,197) by using nearby convenient numbers.
a) 1,300
b) 1,350
c) 1,400
d) 1,450
Answer:
c) 1,400 — Approximate 2,392 as 2,400 and 1,197 as 1,200. Thus, (3/8 × 2,400) + (5/12 × 1,200) = 900 + 500 = 1,400.
Question 21. What is the approximate value of 39.96% of 749.8 – 19.98% of 250.2?
a) 225
b) 240
c) 250
d) 275
Answer:
c) 250 — Approximate 39.96% as 40%, 749.8 as 750, 19.98% as 20% and 250.2 as 250. Therefore, 40% of 750 – 20% of 250 = 300 – 50 = 250.
Question 22. What is the approximate value of √624.7 + ∛728.5?
a) 32
b) 33
c) 34
d) 35
Answer:
c) 34 — Approximate 624.7 as 625 and 728.5 as 729. Therefore, √625 + ∛729 = 25 + 9 = 34.
Question 23. Estimate (5/9 of 1,798) ÷ (2/3 of 749) by using nearby convenient numbers.
a) 1.5
b) 2
c) 2.5
d) 3
Answer:
b) 2 — Approximate 1,798 as 1,800 and 749 as 750. The numerator becomes 5/9 × 1,800 = 1,000, while the denominator becomes 2/3 × 750 = 500. Hence, the approximate value = 1,000 ÷ 500 = 2.
Question 24. If x% of 798 is approximately 240, what is the approximate value of x?
a) 25
b) 30
c) 32
d) 35
Answer:
b) 30 — Approximate 798 as 800. Then x% of 800 ≈ 240, so x = (240/800) × 100 = 30.
Question 25. What is the approximate value of 14.92% of 2,008 + 24.96% of 1,196?
a) 550
b) 575
c) 600
d) 625
Answer:
c) 600 — Approximate 14.92% as 15%, 2,008 as 2,000, 24.96% as 25% and 1,196 as 1,200. Thus, 15% of 2,000 + 25% of 1,200 = 300 + 300 = 600.
Question 26. What is the approximate value of √2,024 ÷ √80.8?
a) 4
b) 5
c) 6
d) 7
Answer:
b) 5 — Approximate 2,024 as 2,025 and 80.8 as 81. Therefore, √2,025 ÷ √81 = 45 ÷ 9 = 5.
Question 27. Estimate (7/16 of 3,198) – (3/8 of 1,604) by using nearby convenient numbers.
a) 750
b) 800
c) 850
d) 900
Answer:
b) 800 — Approximate 3,198 as 3,200 and 1,604 as 1,600. Thus, (7/16 × 3,200) – (3/8 × 1,600) = 1,400 – 600 = 800.
Question 28. What is the approximate value of ∛12,160?
a) 21
b) 22
c) 23
d) 24
Answer:
c) 23 — The nearby perfect cube is 12,167, and 23³ = 12,167. Therefore, ∛12,160 ≈ 23.
Question 29. What is the approximate value of (62.48% of 959.6) ÷ (12.49% of 799.8)?
a) 4
b) 5
c) 6
d) 8
Answer:
c) 6 — Approximate 62.48% as 62.5%, 959.6 as 960, 12.49% as 12.5% and 799.8 as 800. The numerator is 62.5% of 960 = 600, and the denominator is 12.5% of 800 = 100. Therefore, the approximate value = 600 ÷ 100 = 6.
Question 30. What is the approximate value of (√2,501 × ∛1,002) ÷ 24.9?
a) 15
b) 18
c) 20
d) 25
Answer:
c) 20 — Approximate 2,501 as 2,500, 1,002 as 1,000 and 24.9 as 25. Therefore, (√2,500 × ∛1,000) ÷ 25 = (50 × 10) ÷ 25 = 20.
Question 31. What is the approximate value of [(49.8)² – (20.2)²] ÷ 29.7?
a) 60
b) 65
c) 70
d) 75
Answer:
c) 70 — Approximate 49.8 as 50, 20.2 as 20 and 29.7 as 30. Therefore, [(50)² – (20)²] ÷ 30 = (2,500 – 400) ÷ 30 = 2,100 ÷ 30 = 70.
Question 32. What is the approximate value of (798.4 × 15.2) ÷ (39.9 × 7.6)?
a) 32.5
b) 35
c) 37.5
d) 40
Answer:
c) 37.5 — Approximate 798.4 as 800, 15.2 as 15, 39.9 as 40 and 7.6 as 8. Thus, (800 × 15) ÷ (40 × 8) = 12,000 ÷ 320 = 37.5.
Question 33. What is the approximate value of √3,602 + ∛7,995?
a) 75
b) 78
c) 80
d) 82
Answer:
c) 80 — Approximate 3,602 as 3,600 and 7,995 as 8,000. Therefore, √3,600 + ∛8,000 = 60 + 20 = 80.
Question 34. What is the approximate value of 119.8% of 2,499 – 39.9% of 1,252?
a) 2,400
b) 2,500
c) 2,600
d) 2,700
Answer:
b) 2,500 — Approximate 119.8% as 120%, 2,499 as 2,500, 39.9% as 40% and 1,252 as 1,250. Thus, 120% of 2,500 – 40% of 1,250 = 3,000 – 500 = 2,500.
Question 35. If 24.9% of 1,604 + x% of 798 ≈ 720, what is the approximate value of x?
a) 35
b) 40
c) 45
d) 50
Answer:
b) 40 — Approximate 24.9% as 25%, 1,604 as 1,600 and 798 as 800. Then 25% of 1,600 + x% of 800 = 720. Therefore, 400 + 8x = 720, giving x = 40.
Question 36. What is the approximate value of (√4,902 × 14.9) ÷ ∛3,374?
a) 60
b) 65
c) 70
d) 75
Answer:
c) 70 — Approximate 4,902 as 4,900, 14.9 as 15 and 3,374 as 3,375. Therefore, (√4,900 × 15) ÷ ∛3,375 = (70 × 15) ÷ 15 = 70.
Question 37. Estimate [(7/12 of 2,398) + (11/20 of 2,002)] ÷ 49.8.
a) 45
b) 48
c) 50
d) 55
Answer:
c) 50 — Approximate 2,398 as 2,400, 2,002 as 2,000 and 49.8 as 50. Thus, [(7/12 × 2,400) + (11/20 × 2,000)] ÷ 50 = (1,400 + 1,100) ÷ 50 = 50.
Question 38. What is the approximate value of [(79.9)² – (20.1)²] ÷ 99.8?
a) 55
b) 60
c) 65
d) 70
Answer:
b) 60 — Approximate 79.9 as 80, 20.1 as 20 and 99.8 as 100. Therefore, [(80)² – (20)²] ÷ 100 = (6,400 – 400) ÷ 100 = 60.
Question 39. What is the approximate value of 17.98% of 2,498 + 32.02% of 1,251?
a) 800
b) 825
c) 850
d) 875
Answer:
c) 850 — Approximate 17.98% as 18%, 2,498 as 2,500, 32.02% as 32% and 1,251 as 1,250. Therefore, 18% of 2,500 + 32% of 1,250 = 450 + 400 = 850.
Question 40. What is the approximate value of (∛15,620 × √1,224.8) ÷ 17.48?
a) 45
b) 48
c) 50
d) 52
Answer:
c) 50 — Approximate 15,620 as 15,625, 1,224.8 as 1,225 and 17.48 as 17.5. Thus, (∛15,625 × √1,225) ÷ 17.5 = (25 × 35) ÷ 17.5 = 50.
Question 41. What is the approximate value of 999.2 ÷ 24.8 + 1,501.5 ÷ 49.9?
a) 65
b) 68
c) 70
d) 72
Answer:
c) 70 — Approximate 999.2 as 1,000, 24.8 as 25, 1,501.5 as 1,500 and 49.9 as 50. Therefore, 1,000 ÷ 25 + 1,500 ÷ 50 = 40 + 30 = 70.
Question 42. If √624.8 + ∛1,727.2 + x ≈ 49, what is the approximate value of x?
a) 10
b) 12
c) 14
d) 16
Answer:
b) 12 — Approximate 624.8 as 625 and 1,727.2 as 1,728. Then 25 + 12 + x = 49. Therefore, x = 49 – 37 = 12.
Question 43. What is the approximate value of [(44.9)² + (19.8)²] ÷ 24.7?
a) 92
b) 95
c) 97
d) 100
Answer:
c) 97 — Approximate 44.9 as 45, 19.8 as 20 and 24.7 as 25. Thus, [(45)² + (20)²] ÷ 25 = (2,025 + 400) ÷ 25 = 2,425 ÷ 25 = 97.
Question 44. Estimate (5/8 of 4,797) – (7/15 of 2,998) by using nearby convenient numbers.
a) 1,500
b) 1,550
c) 1,600
d) 1,650
Answer:
c) 1,600 — Approximate 4,797 as 4,800 and 2,998 as 3,000. Therefore, (5/8 × 4,800) – (7/15 × 3,000) = 3,000 – 1,400 = 1,600.
Question 45. What is the approximate value of (124.8% of 1,599.2) ÷ (39.98% of 1,249.5)?
a) 3
b) 3.5
c) 4
d) 4.5
Answer:
c) 4 — Approximate 124.8% as 125%, 1,599.2 as 1,600, 39.98% as 40% and 1,249.5 as 1,250. The numerator becomes 125% of 1,600 = 2,000, and the denominator becomes 40% of 1,250 = 500. Hence, the approximate value = 2,000 ÷ 500 = 4.
Question 46. What is the approximate value of [(√8,102 + ∛27,005) × 19.8] ÷ 39.7?
a) 50
b) 55
c) 60
d) 65
Answer:
c) 60 — Approximate 8,102 as 8,100, 27,005 as 27,000, 19.8 as 20 and 39.7 as 40. Therefore, [(90 + 30) × 20] ÷ 40 = 2,400 ÷ 40 = 60.
Question 47. What is the approximate value of [149.8% of 2,398 – 24.9% of 1,596] ÷ 39.9?
a) 70
b) 75
c) 80
d) 85
Answer:
c) 80 — Approximate 149.8% as 150%, 2,398 as 2,400, 24.9% as 25%, 1,596 as 1,600 and 39.9 as 40. Thus, [150% of 2,400 – 25% of 1,600] ÷ 40 = (3,600 – 400) ÷ 40 = 80.
Question 48. If (√1,224.6 × x) ÷ ∛7,998 ≈ 87.5, what is the approximate value of x?
a) 40
b) 45
c) 50
d) 55
Answer:
c) 50 — Approximate 1,224.6 as 1,225 and 7,998 as 8,000. Then (35 × x) ÷ 20 = 87.5. Therefore, 35x = 1,750 and x = 50.
Question 49. Estimate [(7/9 of 4,498) + (13/16 of 3,198)] ÷ 74.9.
a) 78.3
b) 80
c) 81.3
d) 83.5
Answer:
c) 81.3 — Approximate 4,498 as 4,500, 3,198 as 3,200 and 74.9 as 75. Thus, [(7/9 × 4,500) + (13/16 × 3,200)] ÷ 75 = (3,500 + 2,600) ÷ 75 = 6,100 ÷ 75 ≈ 81.3.
Question 50. What is the approximate value of [(59.98)² – (39.96)²] ÷ 20.05?
a) 90
b) 95
c) 100
d) 105
Answer:
c) 100 — Approximate 59.98 as 60, 39.96 as 40 and 20.05 as 20. Therefore, [(60)² – (40)²] ÷ 20 = (3,600 – 1,600) ÷ 20 = 100.
Question 51. What is the approximate value of (∛15,620 + √2,027) ÷ 13.95?
a) 4
b) 5
c) 6
d) 7
Answer:
b) 5 — Approximate 15,620 as 15,625, 2,027 as 2,025 and 13.95 as 14. Therefore, (25 + 45) ÷ 14 = 70 ÷ 14 = 5.
Question 52. If 37.48% of 3,198 + x% of 1,998 ≈ 1,700, what is the approximate value of x?
a) 20
b) 22.5
c) 25
d) 30
Answer:
c) 25 — Approximate 37.48% as 37.5%, 3,198 as 3,200 and 1,998 as 2,000. Then 37.5% of 3,200 + x% of 2,000 = 1,700. This gives 1,200 + 20x = 1,700, so x = 25.
Question 53. What is the approximate value of (√6,402 × ∛2,196) ÷ 25.9?
a) 35
b) 40
c) 45
d) 50
Answer:
b) 40 — Approximate 6,402 as 6,400, 2,196 as 2,197 and 25.9 as 26. Therefore, (80 × 13) ÷ 26 = 1,040 ÷ 26 = 40.
Question 54. What is the approximate value of 249.7% of 799.6 + 74.9% of 1,601 – 39.8% of 1,004?
a) 2,600
b) 2,700
c) 2,800
d) 2,900
Answer:
c) 2,800 — Approximate 249.7% as 250%, 799.6 as 800, 74.9% as 75%, 1,601 as 1,600, 39.8% as 40% and 1,004 as 1,000. Thus, 2,000 + 1,200 – 400 = 2,800.
Question 55. Estimate [(17/24 of 4,798) – (11/18 of 3,597)] ÷ 39.9.
a) 25
b) 30
c) 35
d) 40
Answer:
b) 30 — Approximate 4,798 as 4,800, 3,597 as 3,600 and 39.9 as 40. Thus, [(17/24 × 4,800) – (11/18 × 3,600)] ÷ 40 = (3,400 – 2,200) ÷ 40 = 30.
Question 56. What is the approximate value of √9,803 + √2,502 + ∛999.4?
a) 155
b) 157
c) 159
d) 161
Answer:
c) 159 — Approximate 9,803 as 9,801, 2,502 as 2,500 and 999.4 as 1,000. Therefore, √9,801 + √2,500 + ∛1,000 = 99 + 50 + 10 = 159.
Question 57. What is the approximate value of [(84.9)² – (34.8)²] ÷ 49.8?
a) 110
b) 115
c) 120
d) 125
Answer:
c) 120 — Approximate 84.9 as 85, 34.8 as 35 and 49.8 as 50. Therefore, [(85)² – (35)²] ÷ 50 = (7,225 – 1,225) ÷ 50 = 6,000 ÷ 50 = 120.
Question 58. If (62.4% of 1,598) ÷ (x% of 2,498) ≈ 2, what is the approximate value of x?
a) 15
b) 20
c) 25
d) 30
Answer:
b) 20 — Approximate 62.4% as 62.5%, 1,598 as 1,600 and 2,498 as 2,500. The numerator is 62.5% of 1,600 = 1,000. Thus, 1,000 ÷ (x% of 2,500) = 2, so x% of 2,500 = 500. Therefore, x = 20.
Question 59. What is the approximate value of [(∛64,002 + √3,598) × 24.9] ÷ 49.8?
a) 45
b) 48
c) 50
d) 55
Answer:
c) 50 — Approximate 64,002 as 64,000, 3,598 as 3,600, 24.9 as 25 and 49.8 as 50. Therefore, [(40 + 60) × 25] ÷ 50 = 2,500 ÷ 50 = 50.
Question 60. What is the approximate value of [(199.8)² – (100.2)²] ÷ 149.7?
a) 180
b) 190
c) 200
d) 210
Answer:
c) 200 — Approximate 199.8 as 200, 100.2 as 100 and 149.7 as 150. Therefore, [(200)² – (100)²] ÷ 150 = (40,000 – 10,000) ÷ 150 = 30,000 ÷ 150 = 200.
