Arithmetical Reasoning Questions and Answers

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.

Answer: C

Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20

Answer: C

Let the number of boys and girls participating in sports be 3x and 2x respectively.

Then, 3x = 15 or x = 5.

So, number of girls participating in sports = 2x = 10.

Number of students not participating in sports = 60 - (15 + 10) = 35.

Let number of boys not participating in sports be y.

Then, number of girls not participating in sports = (35 -y).

Therefore (35 - y) = y + 5 

\(\Rightarrow  2y = 30\) 

\(\Rightarrow  y = 15.\)

So, number of girls not participating in sports = (35 - 15) = 20.

Hence, total number of girls in the class = (10 + 20) = 30

Answer: D

Correct answer: 120 rolls

Cutting a roll into 10 pieces takes 9 cuts, not 10. In 24 minutes he makes \[ 45\times24 = 1080\ \mathrm{cuts}, \qquad \frac{1080}{9} = 120\ \mathrm{rolls}. \] Dividing by 10 instead gives 108, which is offered as a distractor - the same fencepost point as the garden question above.

Answer: C

Correct answer: Data inadequate

The fractions apply to two different groups, so the combined fraction depends on how many girls and boys there are. With equal numbers it would be \[ \tfrac{1}{2}\left(\tfrac{1}{5}+\tfrac{1}{8}\right) = \tfrac{13}{80}, \] but with twice as many girls as boys it is different again, and nothing in the question fixes the ratio. Adding the fractions or their numerators and denominators has no justification at all.

Answer: B

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13

Answer: C

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.

Then, woman's age = (10X + y) years; husband's age = (10y + x) years.

Therefore \((10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)\)

\(\Rightarrow\) (9y-9x) = (1/11)(11y + 11x) = (x + y)

 \(\Rightarrow\) \(10x = 8y\)

 \(\Rightarrow\) \( x = (4/5)y\)

Clearly, y should be a single-digit multiple of 5, which is 5.

So, x = 4, y = 5.

Hence, woman's age = 10x + y = 45 years

Answer: D

Correct answer: None of these

A telephone dial carries the digits 0 to 9, and any product containing a zero is zero. So the answer is 0, which is not among the first three options. The item works by inviting the solver to start multiplying 1 through 9 and never notice the zero - and the three plausible-looking six-figure numbers are there to encourage exactly that.