Arithmetical Reasoning Questions and Answers

Answer: C

A has 40 cards.

Translate the statements. Giving 8 to B: B + 8 = C and A - 8 = C - 3. Taking 6 from C: A + 6 = 2D. Also B + D = 50.

\[ \begin{aligned} C &= A - 5,\quad B = C - 8 = A - 13,\quad D = \tfrac{A + 6}{2} \\ (A - 13) + \tfrac{A + 6}{2} &= 50 \Rightarrow 3A = 120 \Rightarrow A = 40 \end{aligned} \]

Check: B = 27, C = 35, D = 23, and B + D = 50. 27 is B's count, and 30 and 45 do not satisfy the equations.

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: 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

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: A

Correct answer: 8

Let \( n \) be the original number. Each was to pay \( 96/n \) and each of the \( n-4 \) who came paid 4 more: \[ \frac{96}{n-4} - \frac{96}{n} = 4 \;\Rightarrow\; 384 = 4n(n-4) \;\Rightarrow\; n^2 - 4n - 96 = 0, \] giving \( n = 12 \). The question asks how many attended, which is \( 12-4 = 8 \) - and answering 12 is the trap the wording sets.

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

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

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: 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: D

Let number of notes of each denomination be x.

Then, x + 5x + 10x = 480 

\(\Rightarrow  16x = 480\) 

\(\Rightarrow x = 30.\)

Hence, total number of notes = 3x = 90