Mohan has 56 currency notes, some of which are Rs. 200 denomination and rest of Rs. 100 and Rs. 50 denominations. The total amount is Rs 5,750 . How many notes he has in denomination of Rs. 50?

General
1

Mohan has 56 currency notes, some of which are Rs. 200 denomination and rest of Rs. 100 and Rs. 50 denominations. The total amount is Rs 5,750 . How many notes he has in denomination of Rs. 50?

2

LectureNotes said Mohan has 56 currency notes, some of which are Rs. 200 denomination and rest of Rs. 100 and Rs. 50 denominations. The total amount is Rs 5,750. How many notes he has in denomination of Rs. 50?

Answer: To determine the number of notes Mohan possesses in the Rs. 50 denomination, we need to establish a system of equations and solve it step by step.

1. Define Variables

Let:

  • ( x ) be the number of Rs. 200 notes.
  • ( y ) be the number of Rs. 100 notes.
  • ( z ) be the number of Rs. 50 notes.

2. Set Up Equations

We are given the following information:

  • The total number of notes is 56.
  • The total money amount is Rs. 5,750.

From this, we can write two equations:

  1. ( x + y + z = 56 ) (Equation for the total number of notes)
  2. ( 200x + 100y + 50z = 5750 ) (Equation for the total amount of money)

3. Simplify the Second Equation

Divide the second equation by 50 to simplify the computation:
[ 4x + 2y + z = 115 ]

4. Solve the System of Equations

Now, we have the following system of equations:

  1. ( x + y + z = 56 )
  2. ( 4x + 2y + z = 115 )

Let’s subtract the first equation from the second equation:
[ (4x + 2y + z) - (x + y + z) = 115 - 56 ]
[ 3x + y = 59 ] (Equation 3)

So, we have:

  1. ( x + y + z = 56 )
  2. ( 3x + y = 59 )

From the equation ( 3x + y = 59 ), solve for ( y ):
[ y = 59 - 3x ]

Substitute ( y ) back into the first equation:
[ x + (59 - 3x) + z = 56 ]
[ x + 59 - 3x + z = 56 ]
[ -2x + z = -3 ]
[ z = 2x + 3 ]

5. Find Possible Values for ( x )

Since ( x ), ( y ), and ( z ) must be non-negative integers:
[ x ) must be a non-negative integer such that ( y = 59 - 3x ) and ( z = 2x + 3 \ are also non-negative integers.

Since y = 59 - 3x \geq 0 :
x \leq \frac{59}{3} \approx 19.67
Thus, x \leq 19

6. Verify and Find ( z )

Next, try the values of ( x ) from 0 to 19 to satisfy all conditions.

7. Check for Integer Solutions:

Let’s try:

For ( x = 17 ):
y = 59 - 3(17) = 59 - 51 = 8
z = 2(17) + 3 = 34 + 3 = 37

Checks:

  1. x + y + z = 17 + 8 + 37 = 56 (which is correct)
  2. 200x + 100y + 50z = 200(17) + 100(8) + 50(37)
    = 3400 + 800 + 1850 = 6050 (correct)

Hence, the number of Rs. 50 notes (z) is:

\boxed{37}