Arithmetical Reasoning Questions and Answers

Answer: C

63 games are needed.

Every game eliminates exactly one player, and the tournament ends when exactly one player, the winner, is left undefeated. Starting from 64 players, 63 players must be eliminated, so 63 games are played.

Counting round by round gives the same total: 32 + 16 + 8 + 4 + 2 + 1 = 63.

64 counts one game per player, but the winner never loses. 32 is only the first round, and 127 is the number of games for a draw of 128 players.

Answer: D

99 rounds are needed.

Every round eliminates exactly one participant, and the contest ends when one participant is left. Of the 100 entrants, 99 must be eliminated, so 99 rounds are played whatever the pairing.

The 100 option counts the winner as eliminated, and 101 and 98 are off-by-one slips. The general rule for a single-elimination contest is that n players need n − 1 matches.

Answer: A

He touches the top in 56 seconds.

Each second he gains 5 feet but slips back 4, a net gain of 1 foot. The trap is the last climb: once he is within 5 feet of the top, the next climb takes him there and he does not slip back.

\[ \begin{aligned} \text{height before last climb} &= 60 - 5 = 55\ \text{ft} \\ \text{time to reach 55 ft} &= 55 \times 1 = 55\ \text{s} \\ \text{total} &= 55 + 1 = 56\ \text{s} \end{aligned} \]

60 seconds treats every second as a net gain of 1 foot right to the top, ignoring that the final climb ends the task.

Answer: C

The book has 97 pages.

The pages are labelled with a number followed by a letter: 1A, 1B, ... 1Z, then 2A to 2Z, then 3A to 3Z, then 4A onward. Each number therefore covers 26 pages, one for each letter.

\[ \begin{aligned} 1A \text{ to } 3Z &= 3 \times 26 = 78 \\ 4A \text{ to } 4S &= 19\ (\text{S is the 19th letter}) \\ \text{total} &= 78 + 19 = 97 \end{aligned} \]

104 would be four full sets of 26 (up to 4Z), and 98 and 95 miscount the letters of the last set.

Answer: D

The smallest time difference is 49 minutes.

On a 12-hour clock the only possible times are 1:01, 2:02, ..., 9:09, 10:10, 11:11 and 12:12. Each step from one hour to the next adds one hour and one minute, so most neighbouring pairs are 61 minutes apart.

The exception is the change from 12:12 back to 1:01:

\[ \begin{aligned} 12{:}12 \to 1{:}00 &= 48 \text{ min} \\ 1{:}00 \to 1{:}01 &= 1 \text{ min} \\ \text{total} &= 49 \text{ min} \end{aligned} \]

61 minutes is the usual gap, not the smallest. 60 minutes would need the minute digits to stay the same, which the rule does not allow, and 101 minutes is not a gap between any two such times.

Answer: B

Each class period is 40 minutes long.

From 2:00 P.M. to 4:52 P.M. is 172 minutes. Four periods in a row have only three breaks between them, using 12 minutes.

\[ \begin{aligned} 4p + 3 \times 4 &= 172 \\ 4p &= 160 \\ p &= 40 \end{aligned} \]

Counting four breaks instead of three gives (172 - 16) / 4 = 39, the trap answer. 36 and 45 do not fit the 172-minute block with three 4-minute breaks.

Answer: B

The bookworm travels only 36 mm.

Books standing upright on a shelf with spines facing out have their first page on the right side of the book and their last page on the left. So page 1 of volume I lies against volume II, and the last page of volume III also lies against volume II.

\[ \begin{aligned} \text{distance} &= 3 + 30 + 3 \\ &= 36 \text{ mm} \end{aligned} \]

That is the front cover of volume I, the whole of volume II, and the back cover of volume III. 90 mm assumes the worm crosses all three books, 72 mm counts the pages of all three volumes, and 84 mm counts everything except the two outer covers; each puts page 1 and the last page on the wrong side.

Answer: B

D and E together equal B, C and F.

Everyone starts with 10. Following the transfers:

• F borrows 2 from A: A 8, F 12.
• F gives 5 to C: F 7, C 15.
• C gives 3 to B: C 12, B 13.
• B gives 6 to D: B 7, D 16.
• D passes 1 to E: D 15, E 11.

D + E = 26, and B + C + F = 7 + 12 + 7 = 26. A, B and C give 27, A, B and F 22, A, C and F 27.

The total number of cards stays 60 throughout, which is a quick check on the bookkeeping: 8 + 7 + 12 + 15 + 11 + 7 = 60.

Answer: B

Manick's present age = 12 years, Rahul's present age = 4 years.

Let Manick be twice as old as Rahul after x years from now.

Then, 12 + x = 2 (4 + x) 

\(\Rightarrow  12 + x = 8 + 2x\) 

\(\Rightarrow  x = 4.\)

Hence, Manick's required age = 12 + x = 16 years

Answer: D

Correct answer: 5/9

Take the salary as 4 units, so the tips are 5 and the income is 9. The tips are therefore \[ \frac{5}{9} \] of the income. The trap is 5/4, which is the ratio of tips to salary as given rather than to income - the denominator must be the whole, not the other part.