Number Series & Progressions: previous-year questions

Answer: (B) 36

24, 72 and 168 are all multiples of 24 ($24 \times 1$, $24 \times 3$, $24 \times 7$), but 36 is not ($36 \div 24 = 1.5$), so 36 is the odd one out. A second test gives the same answer: 36 is the only perfect square ($6^2$) in the list. In odd-number-out questions, check in this order: divisibility by a common factor, squares and cubes, prime or composite, and sum of digits, then pick the option that fails a property shared by the other three.

Answer: (B) 30

The terms are products of consecutive numbers, $n(n+1)$: $1 \times 2 = 2$, $2 \times 3 = 6$, $3 \times 4 = 12$, $4 \times 5 = 20$, so the next is $5 \times 6 = 30$. The differences confirm it: 4, 6, 8, and therefore 10 next ($20 + 10 = 30$). Such numbers ($n^2 + n$) are called pronic or oblong numbers, and the series continues 42, 56, 72. In any number series, first write down the differences; if they form a simple pattern (here, consecutive even numbers) the answer follows at once.

Answer: (D) 6

The rows form an arithmetic progression with common difference 3. If the front (largest) row has $a$ children and there are $n$ rows, the total is $na - \frac{3n(n-1)}{2} = 630$, so $a = \frac{630}{n} + \frac{3(n-1)}{2}$, which must be a whole number. For $n = 3$, $a = 213$; for $n = 4$, $a = 162$; for $n = 5$, $a = 132$; but for $n = 6$, $a = 105 + 7.5 = 112.5$, which is impossible. Check for $n = 5$: $132 + 129 + 126 + 123 + 120 = 630$. Shortcut: with an odd number of rows the middle row equals the average ($\frac{630}{n}$), so $n$ need only divide 630; with an even number of rows the average must end in .5, which fails for 6 ($\frac{630}{6} = 105$).

Answer: (B) 47

Here a = 5 and d = 7, so the 7th term is a + 6d = 5 + 42 = 47. 40 is the 6th term and 33 the 5th term.

Answer: (C) 550

The sum is $10 \times (1 + 2 + \cdots + 10) = 10 \times 55 = 550$. 450 stops at $i = 9$, and 650 adds an extra 100.

Answer: (C) 72

The terms are $2n^2$: 2 × 1, 2 × 4, 2 × 9, 2 × 16, 2 × 25, so the next is 2 × 36 = 72. The differences 6, 10, 14, 18 also grow by 4, giving 50 + 22 = 72.

Answer: (D) 55

From a + 4d = 41 and a + 10d = 20, 6d = −21 so d = −3·5. Then a = 41 − 4 × (−3·5) = 41 + 14 = 55. 45 and 35 come from using the wrong number of steps or the wrong sign of d.

Answer: (B) 1707

Each term is 4 times the previous one minus 1: 2 × 4 − 1 = 7, 7 × 4 − 1 = 27, 27 × 4 − 1 = 107 and 107 × 4 − 1 = 427. So the next term is 427 × 4 − 1 = 1707. The other options do not follow this rule.

Answer: (D) 760

The rule is next term = term × n + $n^2$ for n = 1, 2, 3…: 1 × 1 + 1 = 2, 2 × 2 + 4 = 8, 8 × 3 + 9 = 33, 33 × 4 + 16 = 148. The next term should be 148 × 5 + 25 = 765, and 765 × 6 + 36 = 4626 confirms it, so 760 is wrong. 8, 33 and 148 all fit the rule.

Answer: (B) 386

Each term is the previous one divided by 4: 24576 ÷ 4 = 6144, 6144 ÷ 4 = 1536, and 1536 ÷ 4 = 384, not 386. Then 384 ÷ 4 = 96 and 96 ÷ 4 = 24 fit, so 386 is the odd number.

Answer: (B) 40

Two series alternate: 10, 13, 16, 19 (adding 3) and 5, 10, 20, ? (doubling). The missing term belongs to the second series, so it is 20 × 2 = 40. 22 would be the next term of the first series.

Answer: (B) 21-08-2002

The gaps between successive dates grow by one day each time: 18 July to 25 July is 7 days; 25 July to 2 August is 8 days (July has 31 days); 2 August to 11 August is 9 days. The next gap must be 10 days, and 11 August + 10 days = 21-08-2002. The usual slip is to forget that July has 31 days when counting across the month-end. Months with 31 days: January, March, May, July, August, October and December.

Answer: (B) .02

Each bracket has the form $1-\dfrac{1}{n}=\dfrac{n-1}{n}$, so the product is $\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\cdots\times\dfrac{98}{99}\times\dfrac{99}{100}$. Every numerator cancels with the denominator just before it (a telescoping product), leaving only the first numerator and the last denominator: $\dfrac{2}{100}=0.02$. Note that the product starts at $\dfrac{1}{3}$, not $\dfrac{1}{2}$; if it had started at $\dfrac{1}{2}$ the answer would have been $\dfrac{1}{100}=0.01$, which is the trap in option (A).

Answer: (B) 63

PSC's key takes this as a series of consecutive prime numbers: 53, 59, 61, 67, 71. Since $63 = 7 \times 9$ is not prime, 63 is the wrong term and should be 61. Caution: if you look only at the gaps (6, 4, 4, 4) you may be tempted to mark 53, as some guidebooks do, but 59, 67 and 71 being primes shows the intended logic. Primes between 50 and 75: 53, 59, 61, 67, 71, 73.

Answer: (C) 94

Each term is double the previous term plus 2: $4 \times 2 + 2 = 10$, $10 \times 2 + 2 = 22$, $22 \times 2 + 2 = 46$, so the next is $46 \times 2 + 2 = 94$. Equivalently the differences double each time: 6, 12, 24, 48, and $46 + 48 = 94$. Option D (92) comes from forgetting to add 2.

Answer: (D) 9

Two series are mixed alternately. Terms in odd places: 3, 6, ?, each increasing by 3, so the missing term is 9. Terms in even places: 8, 14, 20, each increasing by 6. When a series rises and falls irregularly like this, always test the alternate terms separately.

Answer: (B) (6, 66, 120)

In (7, 77, 140) the second number is $7\times11=77$ and the third is $7\times20=140$, i.e., the pattern is $(n,\ 11n,\ 20n)$. Only (6, 66, 120) fits: $6\times11=66$ and $6\times20=120$. Check the others: for 9 it should be (9, 99, 180), for 8 it should be (8, 88, 160), and for 4 it should be (4, 44, 80). Each wrong option gets at most one term right to tempt you, so verify both relations.

Answer: (B) 226

The differences between successive terms are perfect cubes: $2-1=1^3$, $10-2=8=2^3$, $37-10=27=3^3$, $101-37=64=4^3$. The next difference is $5^3=125$, so the missing term is $101+125=226$. In other words each term is 1 plus a running sum of cubes: $1+1^3=2$, $1+1^3+2^3=10$, $1+1^3+2^3+3^3=37$, and so on. Tip: when the differences grow quickly, test squares and cubes first.

Answer: (C) (63, 8, 3)

In (8, 3, 2) each number is one less than the square of the next: $3^2-1=8$ and $2^2-1=3$. The same rule holds only for (63, 8, 3): $8^2-1=63$ and $3^2-1=8$. Check the rest: in (168, 15, 6), $15^2-1=224\neq168$; in (95, 24, 5), $24^2-1=575\neq95$ (though $5^2-1=24$ holds); in (15, 6, 5), $6^2-1=35\neq15$. Read such sets from right to left when the left-most number is the largest.

Answer: (D) 1

The numbers decrease by successive prime numbers taken in descending order: $32-13=19$, $19-11=8$, and the next prime below 11 is 7, so $8-7=1$. Hence the missing number is 1. The differences are not simply decreasing odd numbers (13, 11, 9), which would give $-1$, a value not among the options; of the choices, only 1 fits a consistent rule. Tip: when the differences are 13 and 11, test primes as well as odd numbers.