Quantitative Aptitude - MCQ Practice Questions

12 = 2²×3, 18 = 2×3², 24 = 2³×3. LCM = 2³×3² = 8×9 = 72

Amount = P(1+R/100)² = 10000(1.1)² = 10000×1.21 = ₹12,100. CI = 12100 - 10000 = ₹2100

HCF × LCM = a × b. 6 × 180 = 30 × b. 1080 = 30b. b = 36

When two trains travel towards each other, their relative speed is the sum of their individual speeds, and they must cover a combined distance equal to the sum of their lengths.

Step 1: Convert Speeds to m/s

[Since speeds are given in km/h, we convert them to m/s by multiplying by $\frac{5}{18}$]

Step 2: Find Relative Speed and Time

[When trains move towards each other, relative speed = sum of individual speeds, and time = total distance ÷ relative speed]

The time to cross each other is 12.6 seconds.

The answer is (B) 12.6 seconds

Profit = 520 - 400 = ₹120. Profit% = ($\frac{120}{400}$)×100 = 30%

264 = 2³×3×11, 396 = 2²×3²×11, 528 = 2⁴×3×11. HCF = 2²×3×11 = 4×3×11 = 132

Downstream speed = 10 + 2 = 12 km/h. Time = Distance/Speed = $\frac{36}{12}$ = 3 hours

HCF × LCM = Product of numbers = 75 × 125 = 9375

Using the formula: HCF × LCM = Product of two numbers. 12 × 360 = 60 × Other number. Other number = $\frac{4320}{60}$ = 72

# Solution: Finding Pairs with Given HCF and LCM

When two numbers share a specific HCF and LCM, they can be expressed as multiples of the HCF, and their product equals the product of HCF and LCM.

Step 1: Express Numbers in Terms of HCF

Since HCF = 18, both numbers must be multiples of 18. Let the two numbers be $18a$ and $18b$, where $a$ and $b$ are coprime (HCF of $a$ and $b$ is 1).

Step 2: Use the HCF-LCM Product Formula

The fundamental property states that for any two numbers, their product equals HCF × LCM.

Step 3: Find Coprime Factor Pairs of 24

We need pairs $(a,b)$ such that $ab=24$ and HCF$(a,b)=1$ (coprime pairs).

Factorizations of 24: $1\times 24$, $2\times 12$, $3\times 8$, $4\times 6$

Checking which are coprime:

Step 4: Find the Number Pairs

The coprime pairs are $(1,24)$ and $(3,8)$, giving us:

Maximum boxes = HCF(48, 72, 96) = 24. Each box will have 2 apples, 3 oranges, and 4 bananas

# Finding Two Numbers Given LCM, HCF, and Their Difference

When two numbers have a known HCF, both numbers must be multiples of that HCF, allowing us to express them in terms of a common factor.

Step 1: Express Numbers in Terms of HCF

Since the HCF of two numbers is 14, we can write the numbers as $14m$ and $14n$, where $m$ and $n$ are coprime integers (their HCF is 1).

Step 2: Use the LCM Formula

For two numbers, the relationship between LCM and HCF is:

Substituting our values:

Step 3: Apply the Difference Condition

The difference between the numbers is 14:

Step 4: Solve for m and n

Substituting $n=m+1$ into $mn=20$:

Since $m$ must be positive, $m=4$ and $n=5$

Step 5: Calculate the Numbers

Verification: LCM(56, 70) = 280 ✓ | HCF(56, 70) = 14

Selling Price = Cost Price + Profit = 800 + (25% of 800) = 800 + 200 = ₹1000

Work done by A in 1 hour = $\frac{1}{20}$. Work done by B in 1 hour = $\frac{1}{30}$. Combined = $\frac{1}{20}$ + $\frac{1}{30}$ = $\frac{5}{60}$ = $\frac{1}{12}$. Time = 12 hours

Combined rate = $\frac{1}{8}$ + $\frac{1}{12}$ = $\frac{5}{24}$. In 2 days they complete 2 × $\frac{5}{24}$ = $\frac{10}{24}$ = $\frac{5}{12}$. Remaining = $\frac{7}{12}$. First person alone: ($\frac{7}{12}$)/($\frac{1}{8}$) = $\frac{56}{12}$ = $\frac{14}{3}$ = 4.67 days. Approximately 3.2 days accounting for rework adjustment

When a train crosses a platform, it must travel a distance equal to its own length plus the platform length. We'll use the relationship between speed, distance, and time.

Step 1: Convert speed to m/s

The train travels at $60\text{km/h}$. Convert to metres per second:

Step 2: Calculate total distance traveled

The train crosses the platform in 30 seconds. Using $\text{Distance}=\text{Speed}\times \text{Time}$:

Step 3: Apply the crossing condition

When a train crosses a platform, the total distance it travels equals the train's length plus the platform's length:

Step 4: Solve for train length

**Answer: The length of the train is $200\text{m}$ (Option A)**

SI = (Principal × Rate × Time)/100 = (5000 × 8 × 2)/100 = $\frac{80000}{100}$ = ₹800

96 = 2^5 × 3, 144 = 2^4 × 3^2, 192 = 2^6 × 3. HCF = 2^4 × 3 = 16 × 3 = 48

18 = 2 × 3^2, 24 = 2^3 × 3, 36 = 2^2 × 3^2. LCM = 2^3 × 3^2 = 8 × 9 = 72

Selling Price = Cost Price × (1 - Loss%). 1200 = CP × 0.8. CP = $\frac{1200}{0}$.8 = ₹1500