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04 Feb 2026, 22:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
48% (02:08) correct
52%
(02:07)
wrong
based on 176
sessions
History
Date
Time
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Which of the following is the closest approximation to \(\frac{√(73) * √(239)}{√(7.2) + √(15.7)}\)?
A. 10
B. 15
C. 20
D. 25
E. 30
A. 10
B. 15
C. 20
D. 25
E. 30
05 Feb 2026, 00:49
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Approximate to all to next higher squares - 73 to 81, 239 to 256, 7.2 to 9 and 15.7 to 16.
Thus we would get (9 x 16) /(3 +4) = 144/7 = 20.5
Ans is C
Thus we would get (9 x 16) /(3 +4) = 144/7 = 20.5
Ans is C
nemoexplicabo
05 Feb 2026, 08:07
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This is a fantastic problem for practicing estimation and approximation — skills that are absolutely critical for saving time on the GMAT, especially on questions involving square roots.
The secret here is to approximate each square root to nearby perfect squares, then simplify the expression. Let me show you how:
Step 1: Approximate the square roots in the numerator
√73 ≈ √64 = 8 or √81 = 9 → √73 is between 8 and 9, closer to 8.5
√239 ≈ √225 = 15 or √256 = 16 → √239 is between 15 and 16, closer to 15.5
For cleaner mental math, I'll use √73 ≈ 8.5 and √239 ≈ 15.5
Numerator ≈ 8.5 × 15.5
To multiply these quickly:
8.5 × 15.5 = 8.5 × 15 + 8.5 × 0.5
= 127.5 + 4.25
≈ 131.75
Or think of it as: (8.5)(15.5) ≈ (9)(16) = 144 (slight overestimate)
So the numerator is around 130-135.
Step 2: Approximate the square roots in the denominator
√7.2 ≈ √9 = 3, but 7.2 is closer to √4 = 2 → Actually, √7.2 ≈ 2.7
√15.7 ≈ √16 = 4
Denominator ≈ 2.7 + 4 = 6.7
Step 3: Divide
Final calculation ≈ 132 ÷ 6.7 ≈ 132 ÷ 7 ≈ 19-20
Looking at the answer choices, 20 is the closest.
Answer: C
Common traps to avoid:
A big mistake is trying to calculate the exact values — that will eat up way too much time and isn't necessary on approximation questions. Another pitfall is approximating too roughly. For instance, if you approximate √73 as 8 and √239 as 15, you get 120 in the numerator, which might lead you to answer B (15) instead of C (20). The key is finding the balance between speed and accuracy.
Key takeaway: Approximation questions reward smart rounding, not perfect precision. When you see square roots with non-perfect squares, immediately think about the nearest perfect squares above and below. Practice estimating products and quotients mentally — it's a skill that will save you minutes on test day!
The secret here is to approximate each square root to nearby perfect squares, then simplify the expression. Let me show you how:
Step 1: Approximate the square roots in the numerator
√73 ≈ √64 = 8 or √81 = 9 → √73 is between 8 and 9, closer to 8.5
√239 ≈ √225 = 15 or √256 = 16 → √239 is between 15 and 16, closer to 15.5
For cleaner mental math, I'll use √73 ≈ 8.5 and √239 ≈ 15.5
Numerator ≈ 8.5 × 15.5
To multiply these quickly:
8.5 × 15.5 = 8.5 × 15 + 8.5 × 0.5
= 127.5 + 4.25
≈ 131.75
Or think of it as: (8.5)(15.5) ≈ (9)(16) = 144 (slight overestimate)
So the numerator is around 130-135.
Step 2: Approximate the square roots in the denominator
√7.2 ≈ √9 = 3, but 7.2 is closer to √4 = 2 → Actually, √7.2 ≈ 2.7
√15.7 ≈ √16 = 4
Denominator ≈ 2.7 + 4 = 6.7
Step 3: Divide
Final calculation ≈ 132 ÷ 6.7 ≈ 132 ÷ 7 ≈ 19-20
Looking at the answer choices, 20 is the closest.
Answer: C
Common traps to avoid:
A big mistake is trying to calculate the exact values — that will eat up way too much time and isn't necessary on approximation questions. Another pitfall is approximating too roughly. For instance, if you approximate √73 as 8 and √239 as 15, you get 120 in the numerator, which might lead you to answer B (15) instead of C (20). The key is finding the balance between speed and accuracy.
Key takeaway: Approximation questions reward smart rounding, not perfect precision. When you see square roots with non-perfect squares, immediately think about the nearest perfect squares above and below. Practice estimating products and quotients mentally — it's a skill that will save you minutes on test day!
