Rose has a few identical chocolates, less than 100 but more than 80

Question

Rose has a few identical chocolates, less than 100 but more than 80; she also has a few identical bags. When she keeps 4 chocolates in each bag, she is left with only 3 chocolates in the last bag. When she keeps 5 chocolates in each bag, she is left with only 4 chocolates in the last bag. If she keeps 6 chocolates per bag, what would be the number of chocolates in the last bag?

A. 1
B. 2
C. 3
D. 4
E. 5

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A. 1
B. 2
C. 3
D. 4
E. 5

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SKaur3 · Accepted Answer

The approach suggested by  ManifestDreamMBA  is the one-shot approach for this type of question.

Here is another way to solve this as numbers are quite easy here.

No of Chocolates = C.

Given 80<C<100

C = 5q + 4 - (I)
C = 4p + 3 - (II)

All the numbers satisfying Condition (I) are 84, 89, 94, 99.
Out of them, only 99 satisfy condition II.

So C = 99.

Now, the Question asks for the remainder when C is divided by 6, i.e, 99 mod 6 = 3.

Thus (C) is the correct answer.

ManifestDreamMBA · Answer

Let number of chocolates be n 80 3,7,11,15, 19 ,... n = 5q+4 => 4,9,14, 19 ,... n = 20k+19 This means n = 20*4+19 = 99 If she keeps 6 chocolates per bag, 99/6 = 3 will be left in the last bag

Missinga · Answer

80<Chocolates<100
When she keeps 4 chocolates in each bag, she is left with only 3 chocolates in the last bag... 
C=4a+3..... (3,7,11,15,19)

When she keeps 5 chocolates in each bag, she is left with only 4 chocolates in the last bag... 
C=5a+4..... (4,9,14,19)

Also, C=20n+19 (where,n=4 for 80<C<100) 
C=99

Chocolate remaining after putting 6 in all bags = 99/6 (Reaminder=3)

Answer: C

crimsonfawkes · Answer

Hi,

Could you please explain how you are getting n=4 for 80<c<100. Just not able to figure that part out. Thank you!!

Su1206 · Answer

80 < C < 100

when each bag has 4 Cs, last bag has (remainder) 3C. Thus C can be 83, 87, 91, 95, 99
when each bag has 5 Cs, last bag has (remainder) 4C. Thus C can be 84, 89, 94, 99

See that 99 is common in both lists (i.e satisfies both the conditions)

Thus there are 99 Cs.

When 99 is divided by 6, Quotient is 16 (16*6=96) and remainder is 3

harjas2222 · Answer

let us see

80<x<100

if 4- 3 is left
 5- 4 is left
 6- ?

Thus we will take each factor of 4 and 5 between 80 and 100 and see the relation that satisfies the remainder left

99 suits the best

divide 99 by 6 we get 3

mattjmarchand · Answer

ManifestDreamMBA . I am confused on how you get the 20. Why would it not be the LCM of 4, 5, & 6?

ManifestDreamMBA · Answer

@ mattjmarchand  
The reason it would not be LCM of 4, 5, & 6 is because the number of chocolates is not a multiple of these numbers. Notice, a few chocolates are left behind suggesting it's a remainder problem. Also, answer choice has no 0, indicating you cannot take LCM.

Now given 20, the number where both the conditions are fulfilled for the first time in 19 and then it repeats for every 20 numbers. (20 being the LCM of 4 and 5). Think about it
For 19, divide by 4, you get 3 and divide by 5, you get 4
For 39, divide by 4, you get 3 and divide by 5, you get 4
and the pattern countinues...

Note: This is a technique which can be repeated for other similar problems

Hope this helps!

Rose has a few identical chocolates, less than 100 but more than 80

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Rose has a few identical chocolates, less than 100 but more than 80

Prep Toolkit

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