Year 12 Physics resistance question

All the resistances are 60 ohms , sorry if it’s too small.

If anyone could verify either of these answers for me it would be very appreciated. Just unsure if the question is very simple or I’m not understanding something , thanks.

The three 60-ohms resistors are arranged in parallel. But your calculation for the effective resistance (or total resistance) for resistors in parallel is incorrect - you forget to take the reciprocal in the last step.

Thank you ! I’ll correct it now but is there anyway you could explain how they are arranged in parallel , I thought it was more likely to be series as it looks like some that we’ve been studying with the wires being the only parallel segments , obviously not though.
Thanks

Its a classic hard (but simple looking) problem. A key thing is to try and redraw it in parallel, but a key part of that is to understand the current flow / potential difference across the resistors. A bit loosely
So there are 3 possible paths to the end, each go through one resistor and the pd across each path or resistor is the same as the pd across the circuit (due to the resistor free top and bottom paths) so parallel etc ... Also, labelling the nodes/path splits/joins as A, B, C, ... might make the paths clearer.
If you redraw the circuit, so going from the original to a move/redraw of resistors 2 and 3 then:

Its equivalent to a block of 2 parallel resistors (2 and 3) which are in parallel with thefirst resisitor so ...
As a final one, if you have n identical resistors, R, in parallel the equivalent resistance is R/n which falls out of the usual formula you mention. Its easy to justify as the pd across them is the same and for each resistor the current splits as i/n, so the equivalent resistance is R/n. Its useful if youre checking against typo mistakes like you did in the OP where you forgot to take the reciprocal at the end. In this case 60/3 = 20

Thank you ! I’ll correct it now but is there anyway you could explain how they are arranged in parallel , I thought it was more likely to be series as it looks like some that we’ve been studying with the wires being the only parallel segments , obviously not though.
Thanks

The circuit is a modified “simpler” version of a "Wheatstone bridge circuit".
There are two ways of seeing the three resistors connected in parallel.
The first way is considering the full version as shown below.

Fig. 1
If we trace the current flow, we can transform the circuit into the below circuit. It is as if we “pull” R₂ and R₄ anti-clockwise with R₅ rotating anti-clockwise. 😊

Fig. 2
Use this above circuit as a guide to see how the 3 resistors in your circuit is connected in parallel. R₁, R₃ and R₅ are the three resistors in your circuit without R₂ and R₄. Can you see why R₁, R₃ and R₅ are connected in parallel when R₂ and R₄ are removed?
Once you have gone through the first way, the second way is simpler. But this time, avoid using current flow and ignore current flow. It is a minefield. ☹
Your circuit is “just” the circuit in Fig 1 without R₂ and R₄.
Nodes A and B are connected by a wire implying that there is no potential drop across the connecting wire or there is no p.d. between A and B, so nodes A and B are at the same potential. We can remove the connecting wire and “join nodes A and B together”.
Similarly, nodes C and D are also at the same potential, so we can “join them together”, too.
Can you now see the three resistors are connected in parallel?