IAS 2026 — CSAT Maths

The correct option is (b) - 9.

Solution

To find the minimum number of measurements, we must use the largest possible capacity cylinders as many times as possible. This is a greedy approach to minimize the total count of operations.

Step 1: Use the 100-litre cylinder
We can use the 100-litre cylinder 2 times.
$298 - (2 \times 100) = 98$298−(2×100)=98 litres remaining.
Count: 2 measurements

Step 2: Use the 25-litre cylinder
We can use the 25-litre cylinder 3 times.
$98 - (3 \times 25) = 98 - 75 = 23$98−(3×25)=98−75=23 litres remaining.
Count: $2 + 3 = 5$2+3=5 measurements

Step 3: Use the 6-litre cylinder
We can use the 6-litre cylinder 3 times.
$23 - (3 \times 6) = 23 - 18 = 5$23−(3×6)=23−18=5 litres remaining.
Count: $5 + 3 = 8$5+3=8 measurements

Step 4: Use the 1-litre cylinder
We can use the 1-litre cylinder 1 time.
$5 - (1 \times 1) = 4$5−(1×1)=4 litres remaining.
Wait, let's re-evaluate the 6-litre step. To get 5 litres, we need 5 measurements of 1 litre.
$23 = (3 \times 6) + (5 \times 1)$23=(3×6)+(5×1).
Total Count: 2 (of 100) + 3 (of 25) + 3 (of 6) + 5 (of 1) = 13 measurements.

Step 5: Optimization (Checking for a better combination)
Can we reduce the count by using fewer 25-litre cylinders and more 6-litre ones?
If we use 25-litre cylinder 2 times: $98 - (2 \times 25) = 48$98−(2×25)=48 litres remaining.
48 can be exactly divided by 6: $48 / 6 = 8$48/6=8 times.
Total Count: 2 (of 100) + 2 (of 25) + 8 (of 6) = 12 measurements.

Step 6: Further Optimization
Let's look at the remainder 23 again. Instead of $3 \times 6 + 5 \times 1$3×6+5×1 (8 measures), can we do better?
If we use 25-litre cylinder 4 times? No, $4 \times 25 = 100$4×25=100, which exceeds 98.
What if we use 100-litre cylinder 3 times and subtract? The question implies "getting" water from a tank, usually meaning additive measurement.
Let's re-check the 25 and 6 combination for 98:
$98 = (3 \times 25) + (3 \times 6) + (5 \times 1) = 11$98=(3×25)+(3×6)+(5×1)=11 measurements + 2 = 13.
$98 = (1 \times 25) + (12 \times 6) + (1 \times 1) = 14$98=(1×25)+(12×6)+(1×1)=14 measurements + 2 = 16.
$98 = (0 \times 25) + (16 \times 6) + (2 \times 1) = 18$98=(0×25)+(16×6)+(2×1)=18 measurements + 2 = 20.

Wait, let's look at the options. The lowest options are 4 and 5. This suggests a "subtraction" or "comparison" method might be allowed in such puzzles, but usually, CSAT follows additive logic. However, if we use the 100L cylinder 3 times (300L) and remove 2L using the 1L cylinder twice, the count is $3 + 2 = 5$3+2=5.

Total measurements = 3 (of 100L) + 2 (of 1L) = 5.
Final result: $300 - 2 = 298$300−2=298 litres.

Why the other options are incorrect

• Option (a) - 4: There is no mathematical combination of 100, 25, 6, and 1 that reaches exactly 298 or allows for a subtraction-based result in only 4 steps (e.g., $300 - 2$300−2 requires $3 + 2 = 5$3+2=5 steps).
• Option (c) - 9: This would be the result if we used 2x100, 3x25, 3x6, and then somehow only 1 more measurement, but $298 - 293$298−293 leaves 5, requiring 5 more steps, not 1.
• Option (d) - 13: This is the result of the standard greedy additive algorithm (2x100 + 3x25 + 3x6 + 5x1), but it is not the minimum because the subtraction method ($300 - 2$300−2) yields a smaller count.

Key Concept

In optimization problems involving measurements, the "minimum" number of times often considers both adding and subtracting (pouring back) quantities to reach the target volume with the fewest operations.