Respuesta :
n+d=47
5n+10d=295
divide 2nd by 5
n+2d=59
we have
n+d=47
n+2d=59
eliminate n
multiply first equaton by -1 and add to other equaton
-n-d=-47
n+2d=59 +
0n+1d=12
d=12
sub back
n+d=47
n+12=47
minus 12 both sides
n=35
35 nickles
D is answer
5n+10d=295
divide 2nd by 5
n+2d=59
we have
n+d=47
n+2d=59
eliminate n
multiply first equaton by -1 and add to other equaton
-n-d=-47
n+2d=59 +
0n+1d=12
d=12
sub back
n+d=47
n+12=47
minus 12 both sides
n=35
35 nickles
D is answer
Answer: The correct option is (D) 35.
Step-by-step explanation: Given that a collection of nickels and dimes is worth $2.95 and there are total 47 coins.
We are to find the number of nickels.
We know that
1 nickel = $ 0.05 and 1 dime = $ 0.10.
Let, 'n' and 'd' represents the number of nickels and dimes in the collection.
Then, according to the given information, we have
[tex]n+d=47~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05n+0.10d=2.95\\\\\Rightarrow 5n+10d=295\\\\\Rightarrow n+2d=59~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Subtracting equation (i) from equation (ii), we get
[tex](n+2d)-(n+d)=59-47\\\\\Rightarrow d=12,[/tex]
and from equation (i), we get
[tex]n+12=47\\\\\Rightarrow n=47-12\\\\\Rightarrow n=35.[/tex]
therefore, there are 35 nickels in the collection.
Option (D) is correct.