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Three circuits' insulation resistance in parallel200, 50 and 40 megohms together. What reads?

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The scenario for Q15During a periodic, three circuits are insulation resistance tested. Each reads the same live–live and live–earth: circuit 1 = 200 MΩ, circuit 2 = 50 MΩ, circuit 3 = 40 MΩ.
Q15Calculation

What would you read if all three circuits were tested together?

Answer: 20 MΩ.

Testing circuits together puts their insulation in parallel. Golden rule: the total of resistances in parallel is always lower than the lowest one. The lowest here is 40 MΩ, so anything above that is out straight away, including 290 MΩ, which is just the three added together. If only one option is below the lowest, that's your answer. Otherwise, work it out.

L E 200 MΩcircuit 1 50 MΩcircuit 2 40 MΩcircuit 3 tested together = in parallel 1/R = 1/200 + 1/50 + 1/40 = 0.005 + 0.02 + 0.025 = 0.05 R = 1 ÷ 0.05 = 20 MΩ lower than the lowest ✓ (20 < 40)
Resistances in parallel always give a total lower than the smallest one. Adding them (290 MΩ) is the series answer, and it's wrong here.

Method 1: the x⁻¹ button (quickest)

200 x⁻¹  +  50 x⁻¹  +  40 x⁻¹  =   → 1/20 (S⇔D → 0.05)
Ans x⁻¹  =                        → 20

Don't stop at 0.05! That's 1/R. Press x⁻¹ once more.
Answer: 20 MΩ

Method 2: the long way

1 ÷ 200 = 0.005
1 ÷ 50  = 0.02
1 ÷ 40  = 0.025
0.005 + 0.02 + 0.025 = 0.05
1 ÷ 0.05 = 20 MΩ
Finding x⁻¹
On older Casios it's a labelled key next to sin. On newer Casios it's awkward: you press SHIFT and the key directly under the D-pad, and it isn't clearly labelled. Use the S⇔D key to change a fraction to a decimal.

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