Quantitative Aptitude Practice Test

Step-by-step Solution:
1. Convert speed from km/hr to m/s: 54 × (5/18) = 15 m/s.
2. Total distance covered in 30 seconds = Speed × Time = 15 × 30 = 450 meters.
3. Total distance = Length of train (L) + Length of bridge (250m).
4. L = 450 - 250 = 200 meters.

Work Formula: \(M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2\)
12 × 18 × 8 = 16 × D_2 × 9
1,728 = 144 × D_2
D_2 = 1,728 / 144 = 12 days.

Combined 1-minute work = 1/20 + 1/30 = 5/60 = 1/12. Therefore time taken = 12 minutes.

Step 1: Calculate Speed of the Train
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{240\text{ m}}{24\text{ s}} = 10\text{ m/s}$$

Step 2: Calculate Total Distance to Cross the Platform
$$\text{Total Distance} = \text{Train Length} + \text{Platform Length} = 240 + 650 = 890\text{ m}$$

Step 3: Time Taken
$$\text{Time} = \frac{\text{Total Distance}}{\text{Speed}} = \frac{890\text{ m}}{10\text{ m/s}} = 89\text{ seconds}$$

Work rate per day:
A's 1-day work = $\frac{1}{12}$, B's 1-day work = $\frac{1}{15}$.

Combined 1-day work = $\frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} = \frac{3}{20}$.

Work completed in 5 days together = $5 \times \frac{3}{20} = \frac{15}{20} = \frac{3}{4}$.

Remaining work = $1 - \frac{3}{4} = \frac{1}{4}$.

Time taken by A alone = $\frac{\text{Remaining Work}}{\text{A's rate}} = \frac{1/4}{1/12} = \frac{12}{4} = 3\text{ days}$.

Net portion filled in 1 minute = $\frac{1}{20} + \frac{1}{30} - \frac{1}{15}$.

LCM of 20, 30, and 15 is 60.

Net rate = $\frac{3 + 2 - 4}{60} = \frac{1}{60}$.

Thus, the ciste will be filled completely in $60\text{ minutes}$ (1 hour).

$$\text{Simple Interest (SI)} = \text{Amount} - \text{Principal} = 15,500 - 12,500 = ₹3,000$$

Using formula: $$\text{SI} = \frac{P \times R \times T}{100}$$

$$3000 = \frac{12500 \times R \times 4}{100} = 500 \times R$$

$$R = \frac{3000}{500} = 6\%$$

Let Cost Price ($CP$) = ₹100.

Marked Price ($MP$) = $100 + 25 = ₹125$.

Discount = $10\% \text{ of } 125 = ₹12.50$.

Selling Price ($SP$) = $125 - 12.50 = ₹112.50$.

Net Gain % = $112.50 - 100 = 12.5\%$.

Word: 'LEADING' (7 letters: L, E, A, D, I, N, G).

Vowels: E, A, I (3 vowels). Consonants: L, D, N, G (4 consonants).

Treat the 3 vowels as one single unit (EAI). We now have 4 consonants + 1 vowel unit = 5 units.

Number of arrangements of 5 units = $5! = 120$.

The 3 vowels can be arranged amongst themselves in $3! = 6$ ways.

Total arrangements = $120 \times 6 = 720$.

Total possible outcomes = $6 \times 6 = 36$.

Pairs with sum greater than 9 (i.e., sum = 10, 11, or 12):
Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
Sum = 11: (5,6), (6,5) → 2 outcomes
Sum = 12: (6,6) → 1 outcome

Total favorable outcomes = $3 + 2 + 1 = 6$.

$$\text{Probability} = \frac{6}{36} = \frac{1}{6}$$

$$\text{Speed downstream} = \text{Speed in still water} + \text{Speed of stream} = 13 + 4 = 17\text{ km/hr}$$

$$\text{Time taken} = \frac{\text{Distance}}{\text{Speed downstream}} = \frac{68}{17} = 4\text{ hours}$$

Under compound interest: $A = P(1 + r)^t$.

Given: $2P = P(1 + r)^5 \implies (1 + r)^5 = 2$.

To become 8 times: $8P = P(1 + r)^T \implies (1 + r)^T = 8 = 2^3$.

Substitute $2 = (1 + r)^5$: $(1 + r)^T = ((1 + r)^5)^3 = (1 + r)^{15}$.

Therefore, $T = 15\text{ years}$.

When two items are sold at the same selling price, one at a gain of $x\%$ and another at a loss of $x\%$, there is always an overall loss given by:

$$\text{Loss \%} = \frac{x^2}{100}\% = \frac{10^2}{100}\% = \frac{100}{100}\% = 1\%\text{ loss}$$

Total age of 30 students = $30 \times 14 = 420\text{ years}$.

When teacher is included, total persons = $31$, new average = $14 + 1 = 15\text{ years}$.

New total age = $31 \times 15 = 465\text{ years}$.

Teacher's age = $465 - 420 = 45\text{ years}$.

Total balls = $5 + 4 + 3 = 12\text{ balls}$.

Probability of first ball being red = $\frac{5}{12}$.

Probability of second ball being red (without replacement) = $\frac{4}{11}$.

Combined probability = $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$.