Mathematical Reasoning and Aptitude Questions and Answers

Answer: D

The answer is Y wins whenever n is a power of 2.

Each move joins one new pair of circles, so the game lasts exactly as many moves as there are pairs, whatever the players do:

\[ \binom{n}{2} = \frac{n(n-1)}{2} \]

X makes the odd-numbered moves and Y the even-numbered ones, so Y makes the last move, and wins, exactly when this number is even. When n is a power of 2 with \(n \ge 4\), n is a multiple of 4, so \(n(n-1)/2\) is even and Y wins.

Neither player always wins (n = 6 gives 15 moves, a win for X; n = 4 gives 6, a win for Y). 'X wins whenever n is odd' fails for n = 5, which gives 10 moves and a win for Y.

Answer: A

The statement that must be true is For some i, \(k_i = i\).

The labels form a non-decreasing function from {1, ..., 10} to {1, ..., 10}. Such a function always has a fixed point. Let i be the largest index with \(k_i \ge i\) (it exists, since \(k_1 \ge 1\)). If \(k_i \gt i\), then \(k_{i+1} \ge k_i \ge i + 1\), contradicting the choice of i; so \(k_i = i\).

The other options need not hold. Labelling every chair 1 satisfies the condition and has only one fixed point (\(k_1 = 1\)), so neither 'two fixed points' nor 'every i' is forced, and labels need not be distinct, since repeats are allowed.

Answer: A

The basket held 31 flowers.

Work backwards from the fifth person, doubling each time: the fifth got 1, the fourth 2, the third 4, the second 8 and the first 16.

\[ 16 + 8 + 4 + 2 + 1 = 31 \]

Since no flowers remained, the basket held 31. 32 is \(2^{5}\), one too many, 16 is only the first person's share, and 15 leaves out the first person.

Answer: B

The task will take 9 days.

6 men do as much as 8 children, so 1 man = 4/3 child. Convert everything to children:

\[ \begin{aligned} 7\ \text{men} + 12\ \text{children} &= \tfrac{28}{3} + 12 = \tfrac{64}{3}\ \text{children} \\ \text{work} &= 8 \times 24 = 192\ \text{child-days} \\ \text{days} &= 192 \div \tfrac{64}{3} = 9 \end{aligned} \]

12 days would be the time for 16 children alone, and 10 or 16 days come from mixing up the men-children ratio.

Answer: D

The base is 8 cm.

The area of a parallelogram is base times the corresponding altitude. With base \(b\) and altitude \(2b\):

\[ \begin{aligned} b \times 2b &= 128 \\ b^{2} &= 64 \Rightarrow b = 8\ \text{cm} \end{aligned} \]

The altitude is then 16 cm, and 8 x 16 = 128 checks. 6 cm and 7 cm give too small an area, and 12 cm gives 288 square cm.

Answer: C

Only X and Z represent the same set.

Simplify Z. It contains the factor P, so intersect everything with P. With P present, \(Q \cup P^{C}\) reduces to Q, \(P^{C} \cup R\) reduces to R, and \(P \cup R\) and \(P \cup Q\) both contain P already:

\[ \begin{aligned} Z &= P \cap Q \cap R \cap P \\ &= P \cap Q \cap R = X \end{aligned} \]

Y is different. It contains \(P \cap Q\), which includes elements of P and Q that are not in R, and \(Q \cap R^{C}\), which includes elements of Q outside both P and R. Neither kind of element lies in X. So Y is a larger set than X and Z, which rules out 'all three the same', 'only Y and Z' and 'no two the same'.

Answer: C

X and Z give identical necklaces.

Joining the ends turns each string into a circle, so two strings make the same necklace if one is a rotation of the other. Cut X after its sixth pebble and move the first six to the end:

X = GGBRBR | RBBR becomes RBBR + GGBRBR = RBBRGGBRBR, which is exactly Z.

Y does not match either. Each necklace has one pair of adjacent greens (Y's two greens are its end pebbles, which meet when the ends are joined), so read each circle onward from that pair. X and Z give GG BRBRRBBR, and the same necklace turned over gives GG RBBRRBRB, while Y gives GG BRRBRBBR. Y agrees with neither, so 'X and Y', 'Y and Z' and 'no two identical' are wrong.

Answer: A

The weight of the padded (the boatman) is 74 kg.

Find the total weights before and after he is included:

\[ \begin{aligned} \text{children} &= 25 \times 22 = 550\ \text{kg} \\ \text{with him} &= 26 \times 24 = 624\ \text{kg} \\ \text{his weight} &= 624 - 550 = 74\ \text{kg} \end{aligned} \]

A quick check: he must make up 2 kg for each of the 25 children and also bring his own 24 kg average, \(24 + 25 \times 2 = 74\). 75 comes from forgetting that the group size becomes 26, 71 from an arithmetic slip, and 100 from counting 26 extra kilograms per child.

Answer: B

Dayanand's farm is 60 acres.

Add the fractions used for sugarcane and groundnut and find what is left:

\[ \begin{aligned} \tfrac{1}{3} + \tfrac{1}{4} &= \tfrac{7}{12} \\ 1 - \tfrac{7}{12} &= \tfrac{5}{12} \end{aligned} \]

The remaining 5/12 of the farm is the 25 acres of jowar:

\[ \begin{aligned} \tfrac{5}{12} \times T &= 25 \\ T &= 25 \times \tfrac{12}{5} = 60 \end{aligned} \]

Check: sugarcane 20 acres, groundnut 15 acres, jowar 25 acres, total 60. 50 and 75 do not leave exactly 25 acres, and 120 doubles the answer.

Answer: C

The sum to be invested is Rs. 4200.

With compound interest the amount after n years is \(A = P(1 + r)^{n}\). Here \(A = 5458.32\), r = 14% and n = 2:

\[ \begin{aligned} (1.14)^{2} &= 1.2996 \\ P &= \frac{5458.32}{1.2996} = 4200 \end{aligned} \]

Check forwards: 4200 grows by 588 to 4788 after the first year, and by 670.32 to 5458.32 after the second. 4400 and 4120 give amounts of about 5718 and 5354, and 3300 gives only about 4289, so none of them reaches Rs. 5458.32.