Mathematical Reasoning and Aptitude Questions and Answers

Answer: A

The true statement is there is \(m \in X\) such that \(f(m) = m\).

Consider \(g(m) = f(m) - m\). Since f maps into X, \(f(1) \ge 1\), so \(g(1) \ge 0\), and \(f(50) \le 50\), so \(g(50) \le 0\). Take the largest m with \(f(m) \ge m\). If \(f(m) \gt m\), then \(f(m) \ge m+1\), and because f is order-preserving \(f(m+1) \ge f(m) \ge m+1\), contradicting the choice of m. So \(f(m) = m\): every order-preserving map of a finite chain into itself has a fixed point.

The others fail. \(f(m) = m - 1\) for every m would give \(f(1) = 0\), outside X. \(f(m) = m + 1\) would give \(f(50) = 51\), outside X. And the identity map is order-preserving with \(f(m) = m\), so \(f(m) = \frac{1}{2}m\) is not forced for even m.

Answer: B

The pair is (34, 34).

Each X becomes Y and each Y becomes XY. So if a word has x letters X and y letters Y, the next word has y letters X (one from each Y) and x + y letters Y (one from each X and one from each Y).

\[ \begin{aligned} x_{n+1} &= y_{n} \\ y_{n+1} &= x_{n} + y_{n} \end{aligned} \]

Starting from \(w_{0}\) = X, with (x, y) = (1, 0), the counts of Y run 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 for \(w_{0}\) to \(w_{10}\), the Fibonacci numbers. So m, the number of Y in \(w_{9}\), is 34, and l, the number of X in \(w_{10}\), equals the number of Y in \(w_{9}\), also 34.

(21, 34) and (34, 21) come from slipping one step in either count, and (55, 34) takes the Y count of \(w_{10}\) for l.

Answer: D

There will be 89 female rabbits.

Each month, adults (A) stay and each produces one juvenile (J), and each juvenile becomes an adult. So the new adults are the old adults plus the old juveniles, and the new juveniles equal the old adults:

\[ \begin{aligned} A_{n+1} &= A_{n} + J_{n} \\ J_{n+1} &= A_{n} \end{aligned} \]

Starting with one juvenile, the totals month by month are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 for months 0 to 10, the Fibonacci sequence, because each total is the sum of the two before it. After 10 months there are 89 female rabbits.

71, 79 and 81 do not belong to the sequence; 89 is the total after exactly ten steps.

Answer: D

The true statement is \(x_{n} \lt 2\) for every n.

\(x_{n}\) is the sum of the first n terms of a geometric series with first term 1 and ratio \(\frac{1}{2}\):

\[ \begin{aligned} x_{n} &= \frac{1 - (1/2)^{n}}{1 - 1/2} \\ &= 2 - \frac{1}{2^{n-1}} \end{aligned} \]

Since \(\frac{1}{2^{n-1}} \gt 0\) for every n, \(x_{n}\) is always less than 2, though it gets as close to 2 as we like.

So \(x_{n}\) never equals 2 and never exceeds 2, which rules out the options '\(x_{n} = 2\) for some n' and '\(2 \lt x_{n} \lt 3\)'. And \(x_{4} = 1.875\), already above 1.8, so '\(x_{n} \le 1.8\) for every n' is false.

Answer: C

8000 has 28 positive divisors.

Write 8000 as a product of prime powers, then add 1 to each exponent and multiply:

\[ \begin{aligned} 8000 &= 8 \times 1000 = 2^{3} \times 2^{3} \times 5^{3} \\ &= 2^{6} \times 5^{3} \\ d(8000) &= (6+1)(3+1) = 28 \end{aligned} \]

Each divisor is \(2^{a}5^{b}\) with a from 0 to 6 (7 choices) and b from 0 to 3 (4 choices). 25, 26 and 18 do not match this count; 18, for instance, would need exponents such as 5 and 2, which do not give 8000.

Answer: C

The next term is 73.

All the terms are primes. Counting primes in order (2, 3, 5, 7, 11, 13, 17, ...), 17 is the 7th prime, 23 the 9th, 37 the 12th and 53 the 16th. The positions go up by 2, 3 and 4, so the next step is 5, giving position 21. \[ 16 + 5 = 21 \] The 21st prime is 73.

The other options are primes too but sit at the wrong places: 59 is the 17th, 61 the 18th and 71 the 20th. That is why the series cannot be settled by the differences of the terms (6, 14, 16), which follow no simple rule.

Answer: B

The missing pair is TV.

Each pair is a letter and the letter two places after it: E-G, I-K, N-P. The first letters advance by growing steps, counting positions in the alphabet (A = 1):

\[ \begin{aligned} E(5) &\to I(9): +4 \\ I(9) &\to N(14): +5 \\ N(14) &\to T(20): +6 \\ T(20) &\to A(27 = 1): +7 \end{aligned} \]

So the missing pair starts with T and is TV, and the series then wraps round the alphabet from T past Z to A, giving AC as printed. RT, SU and SV do not fit the +6 step or the two-letter gap.

Answer: A

SANGLI is coded as 2021581310.

Each letter is replaced by its position in the alphabet plus 1. For CHENAI: C(3) gives 4, H(8) gives 9, E(5) gives 6, N(14) gives 15, A(1) gives 2 and I(9) gives 10, which written together is 4 9 6 15 2 10, the code 49615210.

For SANGLI: S(19) gives 20, A(1) gives 2, N(14) gives 15, G(7) gives 8, L(12) gives 13 and I(9) gives 10, which gives 20 2 15 8 13 10, that is 2021581310.

The other codes use the plain positions for some letters or miscount N, G or L, so they break the +1 rule.

Answer: A

The ascending order is Football, Tennisball, Basket-ball, Handball.

The statements give three comparisons: football is lighter than tennis-ball; basketball is heavier than tennis-ball; and basketball is lighter than handball. Putting them together from lightest to heaviest:

Football, then Tennisball, then Basketball, then Handball.

The order beginning with Basketball reverses most of these facts. The order ending with Basketball after Handball contradicts 'basketball is lighter than handball'. The order beginning with Tennisball contradicts 'football is lighter than tennis-ball'.

Answer: C

The mode is 26.

For a moderately skewed distribution, Karl Pearson's empirical relation links the three averages:

\[ \begin{aligned} \text{Mode} &= 3\,\text{Median} - 2\,\text{Mean} \\ &= 3(22) - 2(20) \\ &= 66 - 40 = 26 \end{aligned} \]

21 is the average of the mean and median, and 16 comes from reversing the roles (3 x 20 - 2 x 22). 11 does not come from the formula at all.