Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 8 of 10
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If $(123)_b = 291$, then the value of the base $b$ is …
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Solution
$(123)_b = 1b^2 + 2b + 3 = 291$.
$\Rightarrow b^2 + 2b + 3 = 291 \Rightarrow b^2 + 2b - 288 = 0$.
Solving: $b = 16$ or $b = -18$.
Base must be positive → $b = 16$.
If ‘120456’ is to ‘315’, then ‘204562’ is to …
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Solution
Split into first three and last three digits.
For 120456: $(1+2+0)=3$ and $(4+5+6)=15 \Rightarrow 315$.
For 204562: $(2+0+4)=6$ and $(5+6+2)=13 \Rightarrow 613$.
$263 : 36 :: 139 : \ ?$
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Solution
$263 \Rightarrow 2\times6\times3=36$.
$139 \Rightarrow 1\times3\times9=27$.
$MNPQ : QTRU :: FIGP :\ ?$
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Solution
Rule → add $+4$ to each letter. If the middle pair in the original is in **ascending** order, swap the middle two after shifting; if **descending**, keep order.
$MNPQ$ (middle $NG$: descending): $+4 \Rightarrow \boxed{JMKT}$ (no swap).
Three of the following four are alike and form a group. Which one does not belong?
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Solution
All are multiples of $5$, but only $165$ is also divisible by $3$ ($1+6+5=12$).
Others are not divisible by $3$.
Three of the following four are alike and form a group. Choose the odd one out.
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Solution
Sets (a), (c), (d) contain prime numbers only.
Set (b) contains composite numbers.
Which of the following is a wrong number in the series:
$78,\,57,\,36,\,19,\,10,\,2,\ ?$
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Solution
Pattern: subtract $21,\ 21,\ 17,\ 17,\ \ldots$
$78-21=57$, $57-21=36$, $36-17=19$, **$19-17=2$** (so 10 should not appear).
Hence 10 is the wrong term.
What will be the next number in the series?
$2,\ 6,\ \_,\ 42,\ 1806,\ \_$
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Solution
Rule: multiply each term by “itself + 1”
$2\to 2(2+1)=6$; $6\to 6(6+1)=42$; $42\to 42(42+1)=1806$.
Next: $1806\to 1806(1806+1)=1806\times1807=3{,}263{,}442$.
A letter series is given with blanks:
$C\_\ BBA\_\ CAB\_\ AC\_\ AB\_\ AC$
The missing letters (left to right) are given by which option?
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Solution
Filling with **ACBCB** gives
$C\color{blue}{A}BBA\color{blue}{C}CAB\color{blue}{B}AC\color{blue}{C}AB\color{blue}{B}AC$
Grouping in 3’s: $[CAB][BAC][CAB][BAC][CAB]$ — a neat repetition of the block “CAB, BAC”.
Other choices break this pattern.
One term in the number series is wrong. Find the wrong term.
$1,\ 2,\ 6,\ 15,\ 30,\ 56$
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Solution
Intended pattern: add consecutive squares
$1+1^2=2,\ 2+2^2=6,\ 6+3^2=15,\ 15+4^2=31,\ 31+5^2=56$.
So the fourth term should be $31$; hence the **wrong** shown term is $56$ vs expected $55$ at the end, or equivalently the displayed list’s mismatch flags **56** as inconsistent.
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