Solving Linear Equations - 3 variables

Let the total number of students who chose science, arts, and commerce streams be $$S$$, $$A$$, and $$C$$, respectively.

We have,

$$S+A+C = 1500$$, such that, $$C = 1500-S-A$$ .....(1)

Also, $$1100S + 1000A + 800C = 15,50,000$$,  which can be simplified by dividing by $$100$$;

$$11S + 10A + 8C = 15500$$ .....(2)

Substituting the value of $$C$$ from equation (1) in equation (2), we have

$$11S + 10A + 8(1500-S-A) = 15500$$

$$\Rightarrow 3S + 2A = 3500$$

Since $$S\leq A$$, the maximum value of $$S$$ (if possible), would occur when $$S=A$$, therefore,

$$5S = 3500$$,  or $$S= 700$$.

Let X be the number o f 1 rupee coin, Y be the 50 paise coin and Z be the 25 paise coin.
$$\text{X+Y+Z} = 131$$ ...... (1)
$$\text{X}+\frac{\text{Y}}{2}$$+$$\frac{\text{Z}}{4} = 65$$ ...... (2)
$$\text{X} + \frac{\text{Y}}{4}$$ +$$\frac{\text{Z}}{2} = 62$$ ...... (3)

(1)-(2) -> $$\frac{Y}{2}+\frac{3Z}{4}=66$$ ...... (4)

(3)-(2) -> $$\frac{Y}{4}-\frac{Z}{4}=3$$

$$\frac{3Y}{4}-\frac{3Z}{4}=9$$ ...... (5)

(4)+(5) -> $$\frac{5Y}{4}=75$$

Y = 60

$$\frac{Y}{2}+\frac{3Z}{4}=66$$

$$30+\frac{3Z}{4}=66$$

$$\frac{3Z}{4}=36$$

Z = 48

X = 131 - 60 - 48 

X = 23

Therefore, the number of 1 rupee coins = 23

X+Y+Z = 131, X+$$\frac{Y}{2}$$+$$\frac{Z}{4}$$ = 65, X + $$\frac{Y}{4}$$ +$$\frac{Z}{2}$$ = 62. So, X = 23