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Here's the puzzle

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I am solving daily input-output challenge questions in the gradeup android app in order to take preparation for SBI (State Bank of India) PO (Probationary officer). Here I came across this puzzle which I have almost solved except for a single step where I got stuck and unable to find any logic to proceed in the next stage. The step I am referring here is a way to reach step III from step II with proper justification. For convenience here I use round brackets separated by a comma in lieu of separated square boxes joined side by side to write pair of numbers as shown in the above picture.

Let me first describe the first step.In the first step anyway I have to manage to obtain $(2,2), (2,3)$ and $(1,2)$ from the set of pairs $\{(6,8),(3,5) \}, \{(4,9),(1,3) \}$ and $\{ (1,2),(2,7) \}$ respectively. How do we do that?

I observed that $$\begin{align*} | (6×3) - (8×5) | & = 22. \\ | (4×1) - (9×3) | & = 23. \\ | (1×2) - (2×7) | & = 12. \end{align*}$$ So we are through to step I. How to go from I to II. Well in this step we need to get $(1,7)$ and $(4,3)$ from the set of pairs $\{((2,2),(2,3),(1,2) \}.$ How do we get that?

Again my observation did the trick. Observe that

$$\begin{align*} 2^3 + 2^3 + 1^3 & = 17. \\ 2^3+3^3+2^3 & = 43. \end{align*}$$ So we are through to step II. Now I find no proper logic to go to step III from step II. Although I am able to find the way to reach step IV from step III which seems quite easy to me. Let me illustrate. In order to reach at the step IV from step III I need to obtain the pair $(2,5)$ from the set of pairs $\{(1,6),(6,3) \}.$ Right?

Again by the virtue of my observation I found that

$$\begin{align*} (1+6) + (6 × 3) & = 7 + 18 = 25. \end{align*}$$

So I am done except for finding the way of reaching step III from step II. Would anybody please help me finding that? Any help will be highly appreciated. Thank you very much for your valuable time.

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  • 1
    $\begingroup$ Step III -> IV is more likely 61-36=25. $\endgroup$ – JMP Jul 27 at 6:36
  • $\begingroup$ Right. Thanks for your valuable suggestion @JonMark Perry. $\endgroup$ – math maniac. Jul 27 at 6:37
  • $\begingroup$ Also I have noticed $2^2+2^2+2^2+3^2+1^2+2^2=26=43-17$ and $(2+2)\times(2+3)\times(1+2)=60=17+43$! $\endgroup$ – JMP Jul 27 at 16:31
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First, the answer to the actual question in the test:

Using OP's formula ($ab,cd\to|ac-bd|$)

$63,56\to30-18=12\\63,41\to24-3=21\\26,35\to|6-30|=24$.

The sum is $57$, so the answer is (a).

This makes step 2:

From OP, $ab,cd,ef\to a^3+c^3+e^3,b^3+d^3+f^3$.

$1^3+2^3+1^3=10\\2^3+1^3+4^3=73$

Step 3:

The formula appears to be $ab,cd\to ac^2,bd^2$.

For the example: $1\times4\times4=16, 7\times3\times3=63$.

In the question, this gives $1\times7^2=49,0\times3^2=0$

Final output:

Using $\overline{ab}\to ba$:

Example: $\overline{16}-\overline{63}=61-36=25$.

Test: $\overline{49}-\overline{0}=94-0=94$.

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  • $\begingroup$ answer to the given question is obvious from my calculation. Step III is something where I got stuck. $\endgroup$ – math maniac. Jul 27 at 17:49
  • $\begingroup$ Do you know @JonMark Perry what's the mean time to solve a particular question in this test? $\endgroup$ – math maniac. Jul 27 at 17:50
  • $\begingroup$ @mathmaniac.; answer to second question is length of test in time/length of test in questions. The first is (1) to actually answer the question, and (2) to help me follow the logic through. $\endgroup$ – JMP Jul 27 at 18:34

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