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A collection of nickels and dimes is worth $2.95. There are 47 coins in all. How many nickels are there?

A.12

B.18

C.29

D.35

Verified Answer :

apologiabiology

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

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

\(n+d=47~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05n+0.10d=2.95\\\\\Rightarrow 5n+10d=295\\\\\Rightarrow n+2d=59~~~~~~~~~~~~~~~~~~~~~~~~(ii)\)

Subtracting equation (i) from equation (ii), we get

\((n+2d)-(n+d)=59-47\\\\\Rightarrow d=12,\)

and from equation (i), we get

\(n+12=47\\\\\Rightarrow n=47-12\\\\\Rightarrow n=35.\)

therefore, there are 35 nickels in the collection.

Option (D) is correct.

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