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This question is from a free aptitude test on naukri.com (image)

This question is based upon the figure shown below.

$$ \begin{array}{|ccc|c|ccc|c|ccc|} \; & 41 & \; & \; & \; & 31 & \; & \; & \; & ? & \; \\ 5 & \; & 6 & \; & 9 & \; & 4 & \; & 8 & \; & 7 \\ 7 & \; & 8 & \; & 2 & \; & 6 & \; & 2 & \; & 6 \\ \end{array}$$

  • a. 51
  • b. 62
  • c. 48
  • d. 53

Question:
Find the missing number on the basis of the pattern of figures shown above.

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    $\begingroup$ Taking a screenshot would be easier to read than a physical picture... ;) I've added a rough rendering of the problem in MathJax, to make it easier to read. $\endgroup$
    – Doorknob
    Commented Jun 25, 2014 at 17:48
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    $\begingroup$ Is this an application that has been used for a long time, or is it from a new book? It wouldn't be the first time I've seen a wrong answer key for a problem in a new book. $\endgroup$
    – Logarr
    Commented Jun 26, 2014 at 2:22
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    $\begingroup$ "I tried a lot." If you tried at least four times, choosing a different answer each time, then you should already know the answer, right? What is it? Might be useful for identifying the pattern. $\endgroup$
    – Kevin
    Commented Jun 26, 2014 at 11:37
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    $\begingroup$ @Kevin: I think by "tried a lot" he meant "I have put a lot of effort" $\endgroup$
    – justhalf
    Commented Jun 26, 2014 at 12:21
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    $\begingroup$ So what is this website? Could you give a link? $\endgroup$
    – klm123
    Commented Jun 26, 2014 at 16:44

10 Answers 10

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I found a pattern, which I personally do not like because it uses two additional constants, but: 1. it is simple, 2. it takes into account all given numbers and 3. it provides expected result. I post this answer because there is not other working pattern was suggested up to now.

Let's name the numbers in pentagon like:

 E
a b  
c d

Then for each pentagon: $E = 10\times(a+b+c-d)-59$.

In this case the answer is 51.

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    $\begingroup$ I'll buy it, but strongly agree it is not a good puzzle unless there is a better answer. $\endgroup$
    – Ross Millikan
    Commented Jun 26, 2014 at 17:13
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This answer somewhat works, but part of the logic seems a bit arbitrary, so maybe I am missing something.

5/7 + 6/8 = 82/56 = 41/28. The numerator 41 matches the top number in the first figure.

9/2 + 4/6 = 62/12 = 31/6. The numerator 31 matches the top number in the second figure.

8/2 + 7/6 = 62/12 = 31/6. 31 isn't a choice for the answer, but 62 is. Why we don't reduce the fraction in this case I do not know.

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    $\begingroup$ Interesting... I got 31 too, but with a different method! With first example, consider: a = 5, b = 6, c = 7, d = 8. With that in mind the solution is to calculate the average of (a * d) and (b * c)... so ((5 * 8) + (6 * 7)) / 2 = (40 + 42) / 2 = 82 / 2 = 41 ...I am not really envolved in maths much anymore, but my solution reminds me of something to do with matrices, just can't put my finger on what it was though! (plus I may be recalling incorrectly) $\endgroup$
    – musefan
    Commented Jun 27, 2014 at 11:56
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    $\begingroup$ @musefan I came up with the same method, but was very disappointed when I realized 31 was not an answer choice. I think you're remembering matrix determinants, but instead of averaging, you'd subtract: det = ad-bc $\endgroup$
    – Doresoom
    Commented Jun 27, 2014 at 22:10
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a) The answer is obviously 51. The topmost number is decremented by 10 every step with an odd number, and incremented by 20 on every step with an even number.

b) The answer is obviously 62. The topmost number is decremented by 10 every step with an odd number, and doubled on every step with an even number.

c) The answer is obviously 48. The topmost number is decremented by 10 every step with an odd number, and incremented by 17 on every step with an even number.

d) The answer is obviously 53. In every third image it is set to the constant 53.

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    $\begingroup$ -1. This answer obviously wasn't meant to help find the intended solution of the problem. $\endgroup$
    – klm123
    Commented Jun 25, 2014 at 22:06
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    $\begingroup$ This problem has no real solution. Every number inserted number fullfills some pattern. Even each unicode letter completes some pattern. Or each complex number. Or any other object you can think of. $\endgroup$
    – Jost
    Commented Jun 25, 2014 at 22:42
  • $\begingroup$ this problem was printed in the book, so there is a solution the author meant. This is the one that must be found. $\endgroup$
    – klm123
    Commented Jun 25, 2014 at 22:44
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    $\begingroup$ @Jost is making a valid point; and though the solution image shows radio buttons rather than check boxes, perhaps that's just bad/manipulative UI design when the correct answer is all four. $\endgroup$
    – JohnLBevan
    Commented Jun 27, 2014 at 18:56
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    $\begingroup$ @JohnLBevan: Although what you're saying is, strictly speaking, true, but given the contest of an aptitude test with such UI indicates that we want an answer which can be accepted. I mean, every question regarding "find the next in the sequence" is prone to this reasoning, but we still find those questions interesting (and useful). So I don't think this answer is suited as an answer here. I agree with Doorknob that flagging to close convey the argument better. $\endgroup$
    – justhalf
    Commented Jun 28, 2014 at 13:31
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Edit: My answer isn't even on the list of options. My bad, scrap this. I totally read the 51 as a 31.

Even though this question is old and hasn't seen much action recently, it bothers me that the accepted answer uses additional constants when there is a much simpler solution. It can be stated in one line:

The top number is equal to the average of the products of the diagonal numbers.

This can be expressed as the following equation:

E = ((a * d) + (b * c))/2,

where the pentagon is represented by the variables shown below.

  E
a   b
c   d

For the first pentagon: ((5 * 8) + (6 * 7))/2 = (40 + 42)/2 = 41

For the second: ((9 * 6) + (4 * 2))/2 = (54 + 8)/2 = 31

And for the unknown: ((8 * 6) + (7 * 2))/2 = (48 + 14)/2 = 31

Therefore the answer is 31.

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Each answer is correct (if you can justify it)

I looked for solutions that use only products, additions and subtractions, without any magical constants.
With the numbers

a b
c d

I look for possible solutions of the form 

x1*a + x2*b + x3*c + x4*d +
x5*a*a + x6*b*b + x7*c*c + x8*d*d + 
x9*a*b + x10*a*c + x11*a*d +
x12*b*c + x13*b*d + x14*c*d

Where xN in [-1,0,1]
There are a total of 3**14 = 4782969 possible combinations and only 88 of those work for both of the first two examples. Out of those 88 there is at least one solution for each of possible answers:

+a -b +d +bc + bd +cd =                41/31/51
-a -b +bb -cc -ab -ac +ad +bc +bd =    41/31/62
-a -b +c +bb +cc -ab +ad -bc +bd -cd = 41/31/48
-a +b +d +bb -ab +ad +bc -cd =         41/31/48
+a +b +d -dd +ab +cd =                 41/31/53

None of those stands out for being especially simple or symmetric, but c:48 has twice as many solutions as the others, which tips the scales slightly in its favor.

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(5+7+6+8)*2-11=41

(9+2+4+6)*2-11=31

but

(8+2+7+6)*2-11=35

Not the answer looking for but just can be another soln

First I got the above two then I found this.

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Wild guess here - but it still could be a pattern

41 // 4 + 1 = 5 (answer is on top-left) 31 // 3 + 1 = 4 (answer is on top-right) 51 // 5 + 1 = 6 (answer is on bottom-right)

Maybe the pattern here is that the answer to adding each digit of the topmost number from each image is moving clockwise for each image.

^Not sure if that makes sense... But I tried. Would be nice if you tell us the answer if you found it :D

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    $\begingroup$ Not a bad thought actually, at least this doesn't have to try to explain away a missing factor of 2. Does make you wonder why specifically 51 though (and not 15, 24, 33, 42, or 60) $\endgroup$
    – Dennis Meng
    Commented Jun 25, 2014 at 21:28
  • $\begingroup$ Yeah. I figured maybe the other numbers were there to distract people from the pattern. But there probably is a logical reason to it $\endgroup$
    – Chubonga93
    Commented Jun 26, 2014 at 0:02
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Let's say $41$ is the top number and $a=5,b=6,c=7,d=8$.

$$\begin{align} cb-41 + ad &= 41\\ cb-31 + ad &= 31\\ cb-x + ad &= x\\ \implies 14-x+48 &= x\\ 2x &= 62\\ x&=31 \end{align}$$

therefore answer is $31$.

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How about something a little simpler. Using the variable names for the numbers as above:

$$\begin{array}{ccc} &E\\ a&&b\\ c&&d \end{array}$$

Steps:

  1. Sum the numbers ($a$ through $d$).
  2. If 2nd digit of sum is greater than 5 subtract 1st from 2nd, otherwise add.
  3. Append a 1.

Solutions for $E$:

$$\begin{align} 5+6+7+8 &= 26 &&\rightarrow \text{Subtract:}\ 6-2 = 4 &\rightarrow \text{Append:}\ 41.\\ 9+4+2+6 &= 21 &&\rightarrow \text{Add:}\ 2+1 = 3 &\rightarrow \text{Append:}\ 31.\\ 8+7+2+6 &= 23 &&\rightarrow \text{Add:}\ 2+3 = 5 &\rightarrow \text{Append:}\ 51. \end{align}$$

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I may not be correct but just let me know if I have a mistake in my logic.

My thought is that I consider them as sequence with $15,10,25,10,35$ and so on...

Therefore,

$$41-(5+6+7+8) = 15$$ $$31- (9+4+2+6) = 10$$
$$X - (8+7+2+6) = 25$$

Hence, I think $48$ is my answer.

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    $\begingroup$ Why $25$? Shouldn't it be $5$ so your answer is $28$? $\endgroup$
    – kaine
    Commented Jun 27, 2014 at 14:48
  • $\begingroup$ true.but as the assumed sequnce i gave .... i just put my thought out.and plzz do check the other solution i added to the start of the post. though it isn't in the options. $\endgroup$
    – bvsss
    Commented Jun 27, 2014 at 14:49

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