5
$\begingroup$

Question:

Determine the missing number in the right figure on the basis of the numbers arranged in the left figure.

enter image description here

My attempt:

I have tried every thing from squares, cubes, products, sum, differences; alternate application of these methods, etc. but cannot at all come close to the options:

(1) 30   (2) 58   (3) 160   (4) 32

I am usually good at these problems but can't solve this one. Please help!

$\endgroup$

closed as too broad by 2012rcampion, Deusovi, bleh, AJL, Spencerkatty Apr 17 '16 at 0:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

6
$\begingroup$

Perhaps the answer is

(2) $~58$.

Argument:

Call the numbers in the outer ring $a,b,c,d,e$, so that in the left figure $a=1$, $b=2$, $c=3$, $d=4$, $e=5$ and in the right figure $a=10$, $b=9$, $c=8$, $d=7$, $e=6$. Call the central number $x$.

Then the numbers are related as

$x=e*(a+b)-c*d$.

Left figure: $~~~~x=5*(1+2)-3*4=3$
Right figure: $~~x=6*(10+9)-8*7=58$

$\endgroup$
1
$\begingroup$

Perhaps the answer is

(2)  $58$

Explanation:

Call the numbers in the outer ring $a,b,c,d,e$, so that in the left figure $a=1$, $b=2$, $c=3$, $d=4$, $e=5$ and in the right figure $a=10$, $b=9$, $c=8$, $d=7$, $e=6$.  Call the central number $x$.
[Thanks, @Gamow, for proposing a concise notation.]

Then,

$x={a~\times~e}-2$


But this is silly; we could just as well say

(1)  $30$

because

$x=\lceil\frac{a~\times~e}2\rceil$

or

$x=4b-e$


or

(3)  $160$

because

$x=a \times c \times (e-4)$


or

(4)  $32$

because

$x=c \times (d-3)$

$\endgroup$
  • $\begingroup$ Thanks for your answer! But I disagree with you about your thought that this question has many answers. To obtain the central number, we are supposed to use ALL the five numbers encircling it. That's why those five numbers are given. This is how these questions are generally solved. These tricks (of not using all the given numbers) are used when there doesn't exist an answer which can be formed using all the given numbers. $\endgroup$ – Gaurang Tandon Apr 17 '16 at 6:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.