Determine the missing number in a grid of numbers

Question

I met this question yesterday while solving an I.Q test. It took me around 1 hour to come up with a solution that I'm not certain is the correct one.

Alright, without further ado, here's the puzzle:

What is the number that you should put instead of the dot?

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****SPOILER ALERT****

MY INCOMPLETE SOLUTION

As you can see, every row is divided into 1 big number on the left, followed by four 1-digit numbers on the right.

So my idea is that if you take the sum of the first and second small numbers, then you take the sum of the third and fourth small numbers, and then you take the sum of the digits of both of the sums, then the result will be equal to the sum of the digits of the big number.

For example, take the third row:

The big number: 153

The four 1-digit numbers: 2, 1, 8, 7

The sum of 1st and 2nd = 2 + 1 = 3

The sum of 3rd and 4th = 8 + 7 = 15

The sum of the digits of both of the sums = 3 + 1 + 5 = 9

The sum of the digits of the big number = 1 + 5 + 3 = 9

However, this works for all rows, except the first one at the top. That's the closest I could get to the solution. Should a person find the complete solution, I'll be extremely grateful.

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NOTE: it is entirely possible that the person who wrote this puzzle is playing "mind-games" with us, and that he included the first row just to deceive us, thus making the question even harder. But let's not make any assumptions here.

It is also worth noting that the choices that came with the puzzle were as follows:

A) 110

B) 128

C) 164

D) 92

Maybe the idea of the question is to choose the most different choice. the sum of the digits of all of these choices is 11, except for 110:

1 + 2 + 8 = 1 + 6 + 4 = 9 + 2 = 11

1 + 1 + 0 = 2

Answer 1

Operating under the assumption that we are being deceived by the puzzle maker, i.e. the first row is a ruse, the answer is fairly simple. Take the sum of the last two columns, multiply them by $10$, then add the sum of the first two columns to that number. In row $2$, for example:

$$5+7\rightarrow 120,9+3=12\rightarrow 120+12=132$$

This means, for our currently unsolved row, we'd get $B$, or $128$.

Can you elaborate why the puzzle maker might want to deceive us by giving us a faulty first row?

I'll update this if a more comprehensive solution comes to mind.

Answer 2

If you assume a linear model $y=\sum_{j=1}^4 \beta_j x_j$ with no intercept and ignore the first row, you get a perfect fit $\beta=(1, 1, 10, 10)$, as others have proposed, yielding 128.

If you assume a linear model $y=\beta_0+\sum_{j=1}^4 \beta_j x_j$ with intercept and include the first row, you get a perfect fit $\beta=(744, 56, -91, 252, 171)/26$, yielding $3340/26=128.462$. Close enough?