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cupcakes

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by maihuna » Mon Apr 11, 2011 11:51 am
The ratio of cupcakes to children at a particular birthday party is 104 to 7. Each child at the birthday party
eats exactly x cupcakes (where x is a positive integer) and the adults attending the birthday party do not eat
anything. If the number of cupcakes that remain uneaten is less than the number of children at the birthday
party, what must be true about the number of uneaten cupcakes?
I. It is a multiple of 2. II. It is a multiple of 3. III. It is a multiple of 7.
(A) I only (B) II only (C) III only (D) I and II only (E) I, II and III
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Source: — Problem Solving |

by srcc25anu » Mon Apr 11, 2011 12:04 pm
ck : ch = 104:7
no of cupcakes eaten will be 7k. numbers of cupcakes not eaten will be less than 7
in the above case, lets assume cupcakes = 104 and children = 7
7 children will eat 7 *14 each cupcakes = total 98 cupcakes. uneaten cupcakes = 6 (multiple of 2 and 3)
hence Ans D
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by manpsingh87 » Mon Apr 11, 2011 10:45 pm
maihuna wrote:The ratio of cupcakes to children at a particular birthday party is 104 to 7. Each child at the birthday party
eats exactly x cupcakes (where x is a positive integer) and the adults attending the birthday party do not eat
anything. If the number of cupcakes that remain uneaten is less than the number of children at the birthday
party, what must be true about the number of uneaten cupcakes?
I. It is a multiple of 2. II. It is a multiple of 3. III. It is a multiple of 7.
(A) I only (B) II only (C) III only (D) I and II only (E) I, II and III
let cupcakes = p, no. of children=q,
therefore we have p/q=104/7;
also no. of cupcakes eaten by each child = x;

now consider the condition,
If the number of cupcakes that remain uneaten is less than the number of children at the birthdayparty,
therefore we have (p-xq)<q;
p<q(k+1);
P/q<k+1; we know that p/q=104/7;
therefore 104/7-1<k; 97/7<k; 13.85<k; therefore minimum possible value of k=14;
now consider no. of cupcakes be 104 and children=7;
therefore 7 children will eat 14*7=98 cupcakes , uneaten=6 multiple of 2 and 3 hence D
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