f.1 manths

let the no. of 5 coins is X

5X + (120-X)*10=810
5X + 1200 - 10X = 810
5X = 390
X = 78

therefore, there are 78 $5 coins and (120-78) 42 $10 coins in the bag.

Let A be the number of $10 coins.
so,the bag have(120-A) $5 coins .
10A+5(120-A)=810
10A+600-5A=810
5A=810-600
A=210/5
A=42
Therefore,there are42 $10 coins and
(120-42) $5 coins
=78 $5 coins

Let x be the no. of $5 coins, then y b the no. of $10 coins.
Two equations
5x+10y=810......1
x+y=120......2
From 2, x=120-y......3
sub 3 into 1, 5(120-y)+10y=810
600-5y+10y=810
5y=810-600
5y=210
y = 42
Sub y=42 into 3
x=120-42
x=78

Therefore the no. of $5 coins and that of $10 coins are 42 and 78 respictively.

If $5 coins have x,$10 coins have y,
x+y=120---------(1)
5x+10y=810-------(2)
(1),
x=120-y
(2),
5(120-y)+10y=810
600-5y+10y=810
600+5y=810
5y=210
y=42
(1),
x=120-y
x=120-42
x=78

so number of $5 coins are 78 and that of $10 coins have 42 in the bag.

Llet the no of $5 and $10 coins be x and y respectively
x+y=120--------(1)
5x+10y=810----(2)]
(2)/5: x+2y=162----(3)
(3)-(1): y=42
x+42=120
x=78
Thetre are 78 $5 and 42$10 coins in the bag.

Let n be number of $5 coins

so, there is (120-n) $10 coins

5n + 10 x (120 - n) = 810
5n + 1200 - 10n = 810
5n = 390
n = 78

Therefore, the number of $5 coins is 78 and the number of $10 coins is 120-78 = 42