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: 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.

Answer: B

Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....

i.e. numbers of the form (8n + 1).

Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger

Answer: B

Correct answer: 29

Every match eliminates exactly one player, and 29 of the 30 must be eliminated to leave one winner, so 29 matches are needed - no more and no fewer. The argument holds whatever the draw looks like and whoever wins, which is why byes and uneven brackets make no difference to the total.

Answer: D

Correct answer: 100 days

Divide the whole statement through by 100: if 100 cats kill 100 mice in 100 days, then one cat kills one mouse in 100 days. Multiplying back up, 4 cats kill 4 mice in the same 100 days, because the cats and the mice scale together and the time is unchanged. The expected answer of 4 days comes from scaling only one of the two.

Answer: B

Let son's age be x years. Then, father's age = (3x) years.

Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.

So, 3x - 5 = 4 (x - 5) 

\(\Rightarrow  3x - 5 = 4x - 20\) 

\(\Rightarrow  x = 15\)

Answer: C

Correct answer: 4

\( 7p + 5w = 38 \) with \( p \) and \( w \) whole numbers. Working modulo 5, \( 2p \equiv 3 \), so \( p \equiv 4 \); and \( p=9 \) already exceeds 38, so \( p=4 \) is the only possibility, with \( w=2 \). Checking: \( 28 + 10 = 38 \). The solution is unique, so the data are not inadequate.

Answer: C

Clearly, we have :

B-3 = E ...(i)

B + 3 = D ...(ii)

A+B = D + E+10 ...(iii)

B = C + 2 ...(iv)

A+B + C + D + E= 133 ...(v)

From (i) and (ii), we have : 2 B = D + E ...(vi)

From (iii) and (vi), we have : A = B + 10 ...(vii)

Using (iv), (vi) and (vii) in (v), we get:

\((B + 10) + B + (B - 2) + 2B = 133\)

\(\Rightarrow  5B = 125 \)

\(\Rightarrow B = 25\)