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For every positive number x,
f(x)= 1/4 multiplied by greatest integer less than or equal to 4x. If an and b are positive numbers such that f(f(a)f(b))=1.75, which of the following cannot be the value of a?
A. 0.3
B. 0.8
C. 1.3
D. 2.3
E. 3.8
f(x)= 1/4 multiplied by greatest integer less than or equal to 4x. If an and b are positive numbers such that f(f(a)f(b))=1.75, which of the following cannot be the value of a?
A. 0.3
B. 0.8
C. 1.3
D. 2.3
E. 3.8
22 Feb 2026, 04:54
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The key concept here is Custom/Special Functions — specifically the floor function (greatest integer ≤ n). The entire problem hinges on recognizing what outputs this function can actually produce.
Step 1 — Understand what f(x) does.
f(x) = (1/4) × ⌊4x⌋, where ⌊4x⌋ means "round 4x down to the nearest whole number, then divide by 4."
In plain terms: f(x) always rounds x DOWN to the nearest multiple of 0.25. For example:
f(0.3) = (1/4) × ⌊1.2⌋ = (1/4)(1) = 0.25
f(0.8) = (1/4) × ⌊3.2⌋ = (1/4)(3) = 0.75
f(1.3) = (1/4) × ⌊5.2⌋ = (1/4)(5) = 1.25
Every output of f is a multiple of 0.25. This is the core insight.
Step 2 — Simplify f(f(a)).
Since f(a) is already a multiple of 0.25, write f(a) = k/4 for some positive integer k. Then:
f(f(a)) = (1/4) × ⌊4 × (k/4)⌋ = (1/4) × ⌊k⌋ = k/4 = f(a)
So f(f(a)) = f(a) for all positive x. The double application collapses to the same value — a simplification most people miss completely.
Step 3 — Rewrite the condition.
f(f(a)) × f(b) = 1.75 becomes simply: f(a) × f(b) = 7/4
Both outputs must be positive multiples of 0.25 (of the form n/4). For the product to equal 7/4, we need f(a) = 7/n, and for that to itself be a multiple of 1/4, n must divide 28 evenly. So valid values of f(a) are: 0.25, 0.50, 1.00, 1.75, 3.50, 7.00.
Step 4 — Check each answer choice.
A. a = 0.3 → f(a) = 0.25 ✓ (valid — f(b) = 7, achievable when b ∈ [7, 7.25))
B. a = 0.8 → f(a) = 0.75 ✗ (not in valid list — f(b) would need to equal 7/3 ≈ 2.333, which is not a multiple of 0.25 and therefore unreachable)
C. a = 1.3 → f(a) = 1.25 ✗
D. a = 2.3 → f(a) = 2.25 ✗
E. a = 3.8 → f(a) = 3.75 ✗
Answer: B
Common trap: Students evaluate f(a) numerically for each choice but forget to check whether the resulting f(b) requirement is achievable. When f(a) = 0.75, the equation demands f(b) = 7/3 — but since f only outputs multiples of 0.25, that value is simply not reachable for any b.
Takeaway: With floor-type custom functions, identify the output set first. Here, knowing every output is a multiple of 0.25 immediately constrains which values of f(a) are valid — and the answer falls out cleanly.
— Kavya | 725 (Q90, V85, DI79) | GMAT Focus Edition
Step 1 — Understand what f(x) does.
f(x) = (1/4) × ⌊4x⌋, where ⌊4x⌋ means "round 4x down to the nearest whole number, then divide by 4."
In plain terms: f(x) always rounds x DOWN to the nearest multiple of 0.25. For example:
f(0.3) = (1/4) × ⌊1.2⌋ = (1/4)(1) = 0.25
f(0.8) = (1/4) × ⌊3.2⌋ = (1/4)(3) = 0.75
f(1.3) = (1/4) × ⌊5.2⌋ = (1/4)(5) = 1.25
Every output of f is a multiple of 0.25. This is the core insight.
Step 2 — Simplify f(f(a)).
Since f(a) is already a multiple of 0.25, write f(a) = k/4 for some positive integer k. Then:
f(f(a)) = (1/4) × ⌊4 × (k/4)⌋ = (1/4) × ⌊k⌋ = k/4 = f(a)
So f(f(a)) = f(a) for all positive x. The double application collapses to the same value — a simplification most people miss completely.
Step 3 — Rewrite the condition.
f(f(a)) × f(b) = 1.75 becomes simply: f(a) × f(b) = 7/4
Both outputs must be positive multiples of 0.25 (of the form n/4). For the product to equal 7/4, we need f(a) = 7/n, and for that to itself be a multiple of 1/4, n must divide 28 evenly. So valid values of f(a) are: 0.25, 0.50, 1.00, 1.75, 3.50, 7.00.
Step 4 — Check each answer choice.
A. a = 0.3 → f(a) = 0.25 ✓ (valid — f(b) = 7, achievable when b ∈ [7, 7.25))
B. a = 0.8 → f(a) = 0.75 ✗ (not in valid list — f(b) would need to equal 7/3 ≈ 2.333, which is not a multiple of 0.25 and therefore unreachable)
C. a = 1.3 → f(a) = 1.25 ✗
D. a = 2.3 → f(a) = 2.25 ✗
E. a = 3.8 → f(a) = 3.75 ✗
Answer: B
Common trap: Students evaluate f(a) numerically for each choice but forget to check whether the resulting f(b) requirement is achievable. When f(a) = 0.75, the equation demands f(b) = 7/3 — but since f only outputs multiples of 0.25, that value is simply not reachable for any b.
Takeaway: With floor-type custom functions, identify the output set first. Here, knowing every output is a multiple of 0.25 immediately constrains which values of f(a) are valid — and the answer falls out cleanly.
— Kavya | 725 (Q90, V85, DI79) | GMAT Focus Edition
14 Jun 2026, 10:25
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f(f(a).f(b)) is not same as f(f(a) . f(b)..
f(f(a).f(b)) = f(a).f(b).. according to previous steps...
we can write
1/4. [4. f(a).f(b)] = 1.75
1.75=< f(a).f(b)<2
f(f(a).f(b)) = f(a).f(b).. according to previous steps...
we can write
1/4. [4. f(a).f(b)] = 1.75
1.75=< f(a).f(b)<2
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