Ramu had 13 notes in the denominations of Rs. 10, Rs. 50 and Rs. 100. The total value of the notes with him was Rs. 830.

Correct option is (C) 3

Let Ramu had x Rs 10, y Rs 50 & z Rs 100 notes.

\(\therefore\) x+y+z = 13         ______________(1)

\((\because\) Ramu had total 13 notes)

And 10x + 50y + 100z = 830

\(\Rightarrow\) x + 5y + 10z = 83     ______________(2)

\((\because\) Total value of notes with Ramu was Rs 830)

Ramu had more Rs 100 notes than Rs 50 notes.

i.e., z > y         ______________(3)

Since, Ramu has Rs 10, 50 & 100 notes with him and total notes are 13.

\(\therefore\) 0 < x < 13, 0 < y < 13 & 0 < z < 13    ______________(4)

Subtract equation (1) from (2), we get

4y + 9z = 83 - 13

\(\Rightarrow\) 4y + 9z = 70

\(\Rightarrow y=\frac{70-9z}4\)         ______________(5)

With the help of inequilities (3) & (4) and equation (5), we can conclude the value of y & z.

\(\because\) z > y > 0

\(\Rightarrow\) z > 1      \((\because y\geq1\) & z > 1)

(i) If z = 2 then \(y=\frac{52}4=13\)

but y < z

So, it is not possible.

(ii) If z = 3 then \(y=\frac{43}4\)      (Not possible)

(Because notes are never be in fraction)

(iii) If z = 4 then \(y=\frac{34}4=8.5\)   (Not possible)

(iv) If z = 5 then \(y=\frac{25}4=6.25\)   (Not possible)

(v) If z = 6 then \(y=\frac{16}4=4\)

Then x = 13 - y - z

= 13 - 6 - 4

= 13 - 10 = 3

Also, x + 5y + 10z = 3+20+60 = 83     (Satisfied)

\(\therefore\) Ramu has 3 Rs 10, 4 Rs 50 & 6 Rs 100 notes.

\(\therefore\) The number of Rs 10 notes with Ramu is 3.