Uday can cover X distance with Y speed in Z time . He can cover same distance with Y + 10 = speed in (Z-2) time . He can cover same distance X with Y `-15` speed in (Z + 6) time . What can be found from the given data .
(i) time to cover 200 km with speed Y + 10
(ii) distance covered in (Z + 6) time (Y + 10) speed
(iii) speed by which a tunnel can be crossed in `(Z)/(2)` hour
(iv) Ratio between time to cover distance X with speed Y to time to cover distance (X-5) with speed (Y + 10)

Options

A. Only (ii)
B. Only (ii) and (iii)
C. Only (i) and (iii)
D. Only (i) , (ii) and (iv)

Explanation

To solve the problem step by step, we will derive the equations based on the information given and then find the required values.

### Step 1: Derive the equations from the given information

1. **First Statement**: Uday can cover distance \(X\) with speed \(Y\) in time \(Z\).
- From the formula \( \text{Distance} = \text{Speed} \times \text{Time} \), we have:
\[ X = Y \times Z \quad \text{(Equation 1)} \]

2. **Second Statement**: Uday can cover the same distance \(X\) with speed \(Y + 10\) in time \(Z - 2\).
- Again using the distance formula:
\[ X = (Y + 10) \times (Z - 2) \quad \text{(Equation 2)} \]

3. **Third Statement**: Uday can cover the same distance \(X\) with speed \(Y - 15\) in time \(Z + 6\).
- Using the distance formula:
\[ X = (Y - 15) \times (Z + 6) \quad \text{(Equation 3)} \]

### Step 2: Set up equations based on the derived equations

From Equation 1, we can express \(X\) in terms of \(Y\) and \(Z\):
\[ X = YZ \]

Substituting \(X\) into Equations 2 and 3 gives us:
1. From Equation 2:
\[ YZ = (Y + 10)(Z - 2) \]
Expanding this:
\[ YZ = YZ - 2Y + 10Z - 20 \]
Simplifying:
\[ -2Y + 10Z - 20 = 0 \quad \Rightarrow \quad 2Y - 10Z = -20 \quad \Rightarrow \quad Y - 5Z = -10 \quad \text{(Equation 4)} \]

2. From Equation 3:
\[ YZ = (Y - 15)(Z + 6) \]
Expanding this:
\[ YZ = YZ + 6Y - 15Z - 90 \]
Simplifying:
\[ 6Y - 15Z - 90 = 0 \quad \Rightarrow \quad 2Y - 5Z = 30 \quad \text{(Equation 5)} \]

### Step 3: Solve the system of equations

We now have two equations:
1. \(Y - 5Z = -10\) (Equation 4)
2. \(2Y - 5Z = 30\) (Equation 5)

Subtract Equation 4 from Equation 5:
\[ (2Y - 5Z) - (Y - 5Z) = 30 + 10 \]
This simplifies to:
\[ Y = 40 \]

Now substitute \(Y = 40\) back into Equation 4:
\[ 40 - 5Z = -10 \quad \Rightarrow \quad 5Z = 50 \quad \Rightarrow \quad Z = 10 \]

Now substitute \(Y\) and \(Z\) back into Equation 1 to find \(X\):
\[ X = YZ = 40 \times 10 = 400 \]

### Summary of values found:
- \(Y = 40\) km/h
- \(Z = 10\) hours
- \(X = 400\) km

### Step 4: Answer the questions

1. **Time to cover 200 km with speed \(Y + 10\)**:
\[ \text{Speed} = Y + 10 = 40 + 10 = 50 \text{ km/h} \]
\[ \text{Time} = \frac{200}{50} = 4 \text{ hours} \]

2. **Distance covered in \(Z + 6\) time at \(Y + 10\) speed**:
\[ \text{Time} = Z + 6 = 10 + 6 = 16 \text{ hours} \]
\[ \text{Distance} = \text{Speed} \times \text{Time} = 50 \times 16 = 800 \text{ km} \]

3. **Speed by which a tunnel can be crossed in \(\frac{Z}{2}\) hours**:
- We cannot find this without knowing the length of the tunnel.

4. **Ratio between time to cover distance \(X\) with speed \(Y\) to time to cover distance \(X - 5\) with speed \(Y + 10\)**:
\[ \text{Time with } Y = \frac{X}{Y} = \frac{400}{40} = 10 \text{ hours} \]
\[ \text{Time with } Y + 10 = \frac{X - 5}{Y + 10} = \frac{395}{50} = 7.9 \text{ hours} \]
\[ \text{Ratio} = \frac{10}{7.9} \approx 1.27 \]