A trader has three types of oils: 403 litres, 434 litres and 465 litre

Question

A trader has three types of oils: 403 litres, 434 litres and 465 litres, respectively. If he wants to fill them separately in tins of equal capacity, what is the least number of tins required?

(A) 84
(B) 42
(C) 29 
(D) 21
(E) 7

MBAB123 · Accepted Answer

DUnada , you need to find the HCF here. By looking at the numbers I realised that they have a gap of 31, so I thought 31 could be the HCF of these numbers and indeed. 403 = 13*31. After that it was just about adding 13,14 and 15 because I already knew that they have a gap of 31.

I initially thought of finding the HCF using the brut force method like you have done, but then decided not to do it. Pattern recognition helped me here and the only calculation I did on paper was dividing 403 by 31.

DUnada · Answer

Can anyone provide a better explanation, below is how I solved for it.

Prime factors of:
403 -> 13 * 31
434 -> 2 * 7 * 31
465 -> 5 * 3 * 31

So you would need X tins of 31 capacity.
And X is equal to 13 + (2*7) + (5*3).

TarunKumar1234 · Answer

A trader has three types of oils: 403 litres, 434 litres and 465 litres, respectively. If he wants to fill them separately in tins of equal capacity, what is the least number of tins required?

403 = 13* 31
434 = 14* 31
465 = 15* 31

So, equal sized, which is 31 liters will be fine. Now, nos. of tins required = 13 + 14 + 15 = 42.

I think B.

CrackGMat656 · Answer

Brian123 ,  Bunuel , @TarunKumar1234

Please let me know what's wrong in my approach.

The question asked us to find out the minimum numbers of cans required to distribute equally. So if we take a sum of all the volumes given, then the total volume is 1302 litres and clearly this number is divisible by 7.

As the question is asking for least number of cans and 1302/7 = 186, he needs 7 tins of 186 litre

TarunKumar1234 · Answer

Hi LoKeshpat ! Let me try to explain. Question asks, If he wants to fill them separately in tins of equal capacity , what is the least number of tins required? So, we have find a common capacity of tin, which is 31 ltr in this case. And no. of tins = Sum of multipliers of 31 litres in three types of oils: 403 litres, 434 litres and 465 litres. Hope, it is clear.

CrackGMat656 · Answer

Yes its clear to me now. Thanks for the explanation  @TarunKumar1234

Nidzo · Answer

Looking at the numbers the first thing that sticks out is that the numbers grow by 31: 
 465 - 434 = 31
434 - 403 = 31

As the numbers are 31 is a prime number, the HCF between the numbers will be 31, or a multiple of 31. However, looking at the ending of the 3 numbers 40 3 , 43 4 , 46 5  it is impossible for it to be a multiple given that 1 is the only number which goes into  3, 4 and 5 , and the next time that a multiple of  31  will end in  1  is  31*11  which is far too large to work.

Now as we know that the difference between the smallest and medium sized tins is 31, and the difference between the medium and largest tin is 31 we only need to divide the smallest by 31 and then we will easily have the answers for the rest:

403/31 = 13

13+14+15 = 42

Answer B

aniruddhawarang29 · Answer

Hello  TarunKumar1234 ,

Can you pls explain why the left overs after calculating the capacity were added?
I still did not get this part.

TarunKumar1234 · Answer

Hi aniruddhawarang29 ! Question asks "you have to take out all liquid from tins and distribute in a way that, 1) new tins used should be of equal size 2) Min nos. of tins are used. Now, we know, 31 is highest common multiplier of 403 litres, 434 litres and 465 litres. As we know, 403 = 13*31. So for 403 litres to distribute, we need 31 liter capacity tin and nos. of tins required = 13. So, Total nos. of tins = Sum of multiples of 31 in 403 litres, 434 litres and 465 litres. Hope, it is clear.

RajatJ79 · Answer

A trader has three types of oils: 403 litres, 434 litres and 465 litres, respectively. If he wants to fill them separately in tins of equal capacity, what is the least number of tins required?

Solution : If tins of equal capacity are required across all 3 types of oils, it means the GCD of the 3 quantities needs to found out.

Taking the first number for prime factorization: 403 = 13 * 31
This means one among 13 and 31 is also a factor of the other 2 numbers. Turns out the common factor is 31, such that -
403 = 13 * 31
434 = 14 * 31
465 = 15 * 31

Since the tins of equal capacity will be of 31 liters each, total tins required will be: 13 + 14 +15 = 42

(A) 84
(B)   42  
(C) 29 
(D) 21
(E) 7

aniruddhawarang29 · Answer

Hi  TarunKumar1234 !

Thank you so much brother. I was completely messed in my head wrt what we are given (ltrs) and what we needed to find out (Tins) and hence the immature Ques. Thanks brother!

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A trader has three types of oils: 403 litres, 434 litres and 465 litre

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A trader has three types of oils: 403 litres, 434 litres and 465 litre

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