Arithmetical Reasoning Questions and Answers

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:

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

\(\Rightarrow  5B = 125 \)

\(\Rightarrow B = 25\)

Answer: C

Let money with Ken = x. Then, money with Mac = x + £ 3.

Now, 3x = (x + x + £ 3) + £ 2 

\(\Rightarrow  x = £ 5.\)

Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13

Answer: D

Correct answer: 60

With \( d \) deer at four legs and \( p \) peacocks at two: \[ d + p = 80, \qquad 4d + 2p = 200. \] Halving the second gives \( 2d + p = 100 \), and subtracting the first gives \( d = 20 \), so \( p = 60 \). Check: 80 heads and \( 80 + 120 = 200 \) legs.

Answer: A

Correct answer: 71 years

Each of the three was three years younger, so the total was smaller by \( 3\times3 = 9 \): \[ 80 - 9 = 71\ \mathrm{years}. \] Subtracting only 3 gives 77, which is the usual slip - the three years must be counted once for each person.