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rohan2345
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rohan2345
Mary has $4.6 in nickels and dimes only. If she has fewer dimes than nickels, what is the least possible number of nickels she could have?

(A) 24

(B) 27

(C) 30

(D) 32

(E) 36


We need to find the min value of nickels.

So let no of nickels = x
no of dimes = y

So the equation will be 5x+10y=460

First 3 answer choice will make the dimes more than nickels . So A,B,C are eliminated

D) 32*5 = 160

Nickel = 32 Dime = 30


E) 36*5=180
So Nickel = 36 and Dimes = 28

But we need to calculate min value of Nickel

So D is the correct choice.
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1 Nickel = 5 cents, Let Mary have x nickels
1 Dime = 10 cents, let Mary have y dimes
Total amount = 460 cents

We get the equation 5x + 10y = 460

x + 2y = 92
Note that the coefficients are co-prime
First value that satisfies is x = 2, y= 45
After these the subsequent values of x are obtained by adding 2(coefficient of y in the equation) to the previous value
So, x = 2,4,6,8,.... which is an A.P
Similarly the subsequent values of y are obtained by subtracting 1(coefficient of x in the equation) to the previous value
So, y= 45,44,43,42,.... which is an A.P
Now it is given that x > y

Let it happen for the nth term of each A.P
2n > 45 - (n - 1 )

3n > 46

n > 15.33
i.e 16th term
number of nickels = 32 ; number of dimes = 30
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rohan2345
Mary has $4.6 in nickels and dimes only. If she has fewer dimes than nickels, what is the least possible number of nickels she could have?

(A) 24

(B) 27

(C) 30

(D) 32

(E) 36
To me the simpliest way is the following:

We know:

460 = 5X + 10Y -> 92 = X + 2Y
Y<X

First I calculate what would be the number of coins if x =y -> 92 = 3z -> z = 30.something (so if we had the same amount of nikels and dimes we would need 30+ coins).

I know that X = 92 -2Y (from the first equation), and I sustitute Y with 30 (as X will be more for sure since I got that from above)

X = 92-2*30 = 32.

D
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