I had a set of 50 sums of "Reasoning Aptitude" out of which two i cannot solve. Please help me.Find the missing number in both.
$Q1.~~~43,56,99,180,99,? $
Options
- 202
- 197
- 55
- 155
- 216
$Q2. ~~~ 29,40,44,52,59,73,?$
Options
- 97
- 83
- 95
- 67
$\begingroup$
$\endgroup$
-
$\begingroup$ What does "sum of Reasoning Aptitude" mean? Is there a mathematical question? $\endgroup$ – Jonas Meyer May 22 '12 at 15:54
-
$\begingroup$ Do you want to find the next term in the sequence? $\endgroup$ – user17762 May 22 '12 at 15:55
-
$\begingroup$ Possibly related: math.stackexchange.com/q/144881 $\endgroup$ – Jonas Meyer May 22 '12 at 15:56
-
$\begingroup$ @Marvis Yeah,the next term in the sequence. i cannot get the correct logic... $\endgroup$ – Ashu May 22 '12 at 15:57
-
$\begingroup$ oeis.org will be of help. If we denote the terms of the second sequence as $a(n)$, then $a(n+1) = a(n) + \text{ sum of digits of }a(n)$. Thanks to OEIS. $\endgroup$ – user17762 May 22 '12 at 15:58
$\begingroup$
$\endgroup$
Just to kind of show the absurdity of this sort of question, here's one potential answer for $Q1$:
From the 99, 180, 99 part of the sequence we deduce the next element in the sequence is not determined by the previous element alone. So, it's probable that the first two elements are somehow arbitrary. From this it's not a far stretch to assume they determine the sequence in some way. Now note that:
$$a_2 - a_1 = 56 - 48 = 8$$
$$\frac{a_3}{(\text{sum of digits of } a_2)} = 9 = (a_2 - a_1) + 1$$
$$\frac{a_4}{(\text{sum of digits of } a_3)} = 10 = (a_2 - a_1) + 2$$
$$\frac{a_5}{(\text{sum of digits of } a_4)} = 11 = (a_2 - a_1) + 3$$
So the next element could be:
$$\frac{x}{(\text{sum of digits of } a_5)} = 12 \implies x = 216$$
Is this the correct answer? I don't know - at the very least it seems reasonable, but there's no reason to think it's clever enough to be the one answer. I bet there's an equally plausible answer yielding $202$ as the continuation, or $55$, or $8$.
-
$\begingroup$ Thanks for prettifying the answer, Marvis. I'm awful with Latex :). $\endgroup$ – michael May 22 '12 at 19:54
-
$\begingroup$
$\endgroup$
There has to be a relation in the entire sequence
The answers are
A1) $43,56,99,180,99,?$
$(4+3)*8=56$
$(5+6)*9=99$
$(9+9)*10=180$
$(1+8+0)*11=99$
A2) If we denote the terms of the second sequence as a(n), then a(n+1)=a(n)+ sum of digits of a(n).
$\begingroup$
$\endgroup$
Answer second (use this) 29+2+9=40 40+4+0=44 44+4+4=52 52+5+2=59 59+5+9=73 73+7+3=83 ans
