6050 = 5000(1+r/100)²; 1.21 = (1+r/100)²; 1+r/100 = 1.1; r = 10%

When a boat travels upstream, it moves against the current, so its effective speed decreases. Use the relationship: \(\text{Upstream Speed} = \text{Speed in still water} - \text{Stream speed}\)

Step 1: Find the upstream speed from distance and time

Given: Distance = 500 km, Time = 20 hours

Step 2: Set up the upstream speed equation

Let \(s\) = speed of the stream (in km/hr)

The boat's speed in still water is 50 km/hr, so:

Step 3: Solve for stream speed

From Steps 1 and 2:

Verification: Upstream speed = \(50 - 25 = 25\) km/hr ✓

Distance covered in 20 hours = \(25 \times 20 = 500\) km ✓

Answer: The speed of the stream is 25 km/hr (Option B)

Relative speed = 60 + 40 = 100 km/h = 100×5/18 = 250/9 m/s; Total distance = 150+100 = 250m; Time = 250/(250/9) = 9 seconds

When three numbers share a common HCF and LCM, we can use the fundamental property that for any number, both the HCF and LCM must divide it appropriately, and the product relationship: HCF × LCM = product of pairs.

Step 1: Verify the given numbers satisfy HCF and LCM conditions

For numbers 45 and 60:

Both must divide the LCM of all three numbers (360) and be divisible by the HCF (15). ✓

Step 2: Use the property relating HCF, LCM, and numbers

For any three numbers \(a\), \(b\), \(c\) with HCF \(h\) and LCM \(l\):

- Each number must be a multiple of HCF: \(a = 15m_1\), \(b = 15m_2\), \(c = 15m_3\)

- The LCM of all three must equal 360

For 45 and 60: \(45 = 15 \times 3\) and \(60 = 15 \times 4\)

Step 3: Find prime factorizations in terms of HCF multiples

The third number \(c\) must satisfy:

- \(\text{HCF}(45, 60, c) = 15 = 3 \times 5\) (so \(c\) must have exactly \(3^1 \times 5^1\) as factors)

- \(\text{LCM}(45, 60, c) = 360 = 2^3 \times 3^2 \times 5\)

Step 4: Determine the third number

From LCM, the third number must contribute the \(2^3\) factor (since 45 and 60 together only have \(2^2\)).

Let \(c = 2^a \times 3^b \times 5^d \times \ldots\)

For HCF to be exactly 15: we need \(b = 1\) and \(d = 1\)

For LCM to be 360: we need \(a = 3\) (the maximum power of 2)

Verification: \(\text{HCF}(45, 60, 120) = 15\) ✓ and \(\text{LCM}(45, 60, 120) = 360\) ✓

Answer: The third number is \(120\) (Option A)

Let CP = x; At 20% profit: SP = 1.2x; At 30% profit: SP = 1.3x; 1.3x - 1.2x = 60; 0.1x = 60; x = 600

For compound interest, the amount formula is \(A = P\left(1 + \frac{r}{100}\right)^n\), where \(P\) is principal, \(r\) is rate p.a., and \(n\) is time in years.

Step 1: Identify given values

Step 2: Apply the compound interest formula

Step 3: Isolate the rate term

Step 4: Take the cube root

Step 5: Solve for rate

Verification: \(8000 \times (1.05)^3 = 8000 \times 1.157625 = 9261\) ✓

Answer: The rate is \(5\% \text{ p.a.}\) (Option A)

A fills in 10 hrs = 10/20 = 1/2 tank; Remaining = 1/2; Combined rate = 1/20 + 1/30 = 5/60 = 1/12; Time for half = (1/2)/(1/12) = 6 hours. Hmm, not in options. Reread: B alone for remaining = (1/2)/(1/30) = 15 hours

To solve this problem, we need to find the original speed, calculate the reduced speed, and then determine the time needed to cover the reduced distance.

Step 1: Calculate Original Speed

The train travels 240 km in 4 hours, so we divide distance by time to find speed.

Step 2: Calculate Reduced Speed

The speed reduces by 25%, so the new speed is 75% of the original speed.

Step 3: Calculate Time for 180 km at Reduced Speed

Using the formula Time = Distance ÷ Speed, we find how long it takes to cover 180 km.

The train will take 4 hours to cover 180 km at the reduced speed.

Original speed of the train:

Speed=

4

240

​

=60 km/h

Speed reduced by 25%:

Reduced speed=60−25% of 60

=60−15=45 km/h

Time to cover 180 km at reduced speed:

Time=

45

180

​

=4 hours

Therefore, the train will take:

Answer: (A) 4 hours

MP = 1.4×CP; SP = 1.4×CP × 0.85 = 1.19×CP; Profit% = 19%

This problem involves two sequential investments: first at simple interest, then the accumulated amount at compound interest.

Step 1: Calculate Simple Interest for first 2 years

Using the Simple Interest formula \(SI = \frac{P \times R \times T}{100}\):

Step 2: Find Amount after 2 years

This amount becomes the principal for the next investment.

Step 3: Apply Compound Interest for next 3 years

Using the Compound Interest formula \(A = P\left(1 + \frac{R}{100}\right)^T\):

Step 4: Calculate final amount

Answer: The total amount after 5 years is ₹18,264.06 (Option A)