Mrs. Sharma has a few chocolate candies which she wants to distribute

Question

Mrs. Sharma has a few chocolate candies which she wants to distribute among her five grand-daughters. If she distributes the candies among them in the ratio of 1 : 4 : 5 : 7 : 9, then 1 candy is left out. If she distributes the candies in the ratio of 2 : 3 : 3 : 4 : 5, then 3 candies are left out. However, if she distributes the candies among them in the ratio of 1 : 2 : 3 : 7 : 8, then no candies are left over. What is the minimum number of candies that any of the five grand-daughters can receive in any of the three cases?

A. 3
B. 4
C. 5
D. 6
E. 12

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Official Answer

B

A. 3
B. 4
C. 5
D. 6
E. 12

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harsha3699 · Answer

To answer the above, first we need to find the total no. of chocolates. in the third case we get a clue that the total no. of chocolates is a multiple of 21. This is by adding all the parts of the given ratio. Next, the same multiple of 21 must be equated to case 1+1, since in the first case 1 chocolate was left behind. Similarly, the multiple of 21 must be equated to case 2+3, subsequently we get the total no. of chocolates as 105. 
Plugging this value in different cases, would give us the answer. 
Answer is B.

Gprabhumir · Answer

I dont understand this solution, why should the multiple of 21 be equated to case 2+3?

Regor60 · Answer

N=26K+1=17L+3=21M

So

M=(17/21)L+3/21

Since M is an integer we need to pick up 18/21 with the multiplication by L.

If L=5 we have 85/21 leaving 1/21 remainin after the denominator is multiplied by 4, not enough.

If L=6, then the numerator is 102, which with 4*21=84 is 18 less than 102, which gives us 18/21 we need.

So N=17*6+3 = 105

In the first scenario 1/26 of 105-1=104, or 4 is the least.

In the second scenario 2/17 of 105-3=102, or 12 is given.

In the final scenario 1/21 of 105 or 5 is given.

So 4 is the smallest amount.

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sj296 · Answer

The number of chocolates should be equal.

Distribution 1:Chocolates were distributed in 1 : 4 : 5 : 7 : 9

So, total chocolates = K+4K+5K+7K+9K+1 = 26K + 1 ..................(1)

Distribution 2: Chocolates were distributed in 2 : 3 : 3 : 4 : 5

So, total chocolates = 2p+3p+3p+4p+5p+3 = 17p+3.....................(2)

Distribution 3: Chocolates were distributed in 1 : 2 : 3 : 7 : 8

So, total chocolates = q+2q+3q+7q+8q = 21q...............................(3)

Now, we need to find the values of K, p, and q such that (1), (2) and (3) are equal.

From trial and error, if q = 5; K= 4, and p = 6

So, minimum number of candies = K = 4

Option B

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Mrs. Sharma has a few chocolate candies which she wants to distribute

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Mrs. Sharma has a few chocolate candies which she wants to distribute

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