Solve all the problems mentioned in  Section : Part-C-Elementary Mathematics

### Question 1
**Problem:**
The LCM of 192 and 480 is 60 more than the 100 times of X. What is the value of X?

**Solution:**
1. Find the LCM of 192 and 480.
   \[
   \text{LCM}(192, 480) = 1920
   \]
2. Set up the equation:
   \[
   1920 = 100X + 60
   \]
3. Solve for X:
   \[
   1920 - 60 = 100X \implies 1860 = 100X \implies X = \frac{1860}{100} = 18
   \]

**Answer:** 18

### Question 2
**Problem:**
Divide Rs.500 among A, B, C, and D so that A and B together get thrice as much as C and D together, B gets 4 times what C gets and C gets 1.5 times as much as D. The value of B’s share is:

**Solution:**
1. Let D = x, then C = 1.5x.
2. B = 4C = 4(1.5x) = 6x.
3. A + B = 3(C + D).
4. Total sum: \( A + B + C + D = 500 \).

Using equation 3:
\[ A + 6x = 3(1.5x + x) \]
\[ A + 6x = 3(2.5x) \]
\[ A + 6x = 7.5x \]
\[ A = 1.5x \]

Sum of shares:
\[ A + B + C + D = 500 \]
\[ 1.5x + 6x + 1.5x + x = 500 \]
\[ 10x = 500 \]
\[ x = 50 \]

Value of B’s share:
\[ B = 6x = 6(50) = 300 \]

**Answer:** Rs.300

### Question 3
**Problem:**
A’s speed is 30% more than that of B. If A and B run a race on a 117 m length race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

**Solution:**
1. Let B's speed be \( v \). Then A's speed is \( 1.3v \).
2. Let the head start distance be \( d \).
3. Time taken by both to finish the race should be equal:
   \[
   \frac{117 - d}{1.3v} = \frac{117}{v}
   \]
4. Solve for \( d \):
   \[
   \frac{117 - d}{1.3} = 117 \implies 117 - d = 1.3 \times 117 \implies 117 - d = 152.1 \implies d = 152.1 - 117 = 35.1
   \]

However, since \( d \) is a fraction, it's likely a mistake in solving. Rechecking:
\[ \frac{117 - d}{1.3} = 117 \]
\[ 117 - d = 1.3 \times 117 \]
\[ 117 - d = 152.1 \]
\[ d = 35.1 \]

Rechecking:
\[ d = \frac{0.3 \times 117}{1.3} = \frac{35.1}{1.3} = 27 \]

So the part is:
\[ 27 \]

**Answer:** 27 m

### Question 4
**Problem:**
A single discount equivalent to two successive discounts of 15% and 25% is:

**Solution:**
1. Let the original price be \( P \).
2. After 15% discount, the price is \( P \times 0.85 \).
3. After further 25% discount, the price is \( P \times 0.85 \times 0.75 \).
4. Calculate the single equivalent discount:
   \[
   P \times 0.85 \times 0.75 = P \times 0.6375
   \]
   Therefore, the equivalent discount is:
   \[
   1 - 0.6375 = 0.3625 \times 100 = 36.25\%
   \]

**Answer:** 36.25%

### Question 5
**Problem:**
A certain sum of money fetches simple interest of Rs.4,016.25 at the rate of 7% per annum in 5 years. What is the sum?

**Solution:**
1. Use the simple interest formula:
   \[
   SI = \frac{P \times R \times T}{100}
   \]
   Where \( SI = 4016.25 \), \( R = 7 \), and \( T = 5 \).
2. Solve for \( P \):
   \[
   4016.25 = \frac{P \times 7 \times 5}{100}
   \]
   \[
   4016.25 = \frac{35P}{100}
   \]
   \[
   4016.25 = 0.35P
   \]
   \[
   P = \frac{4016.25}{0.35} = 11475
   \]

**Answer:** Rs.11,475

Would you like more details or explanations on any of these solutions? Here are some further questions you might want to ask:

1. Can you explain how to find the LCM of two numbers?
2. How can we set up the equations to divide Rs.500 among A, B, C, and D?
3. What is the formula to calculate the equivalent single discount?
4. How do we calculate simple interest for a given sum?
5. What steps are involved in calculating the share each person gets when dividing an amount based on given conditions?

**Tip:** Always break down complex problems into smaller, manageable parts to simplify the solution process.

Explore solutions to elementary mathematics problems including finding LCM, dividing money among individuals, calculating race conditions based on speed, determining equivalent discounts, and computing simple interest. Ideal for high school students.

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