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Was it a guess ??? or how do we approach to this kind of problem...

prab wrote:

are there any theories for these kinds of problems? haven't came across any yet, i think hit and trial is the best one!

Which of the following could be the sum of the reciprocals of two different prime numbers? A. 7/13 B. 10/12 C. 11/30 D. 23/50 E. 19/77

Let \(x\) and \(y\) be two different primes, then their sum will be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\). So, the denominator of the reduced fraction must be the product of two different primes: only B, \(\frac{10}{12}=\frac{5}{6}=\frac{5}{2*3}\), and E, \(\frac{19}{77}=\frac{19}{7*11}\), have such denominators and nominator must be the sum of the primes in denominator, thus only B is left: \(\frac{10}{12}=\frac{5}{6}=\frac{2+3}{2*3}\).

Was it a guess ??? or how do we approach to this kind of problem...

prab wrote:

are there any theories for these kinds of problems? haven't came across any yet, i think hit and trial is the best one!

Which of the following could be the sum of the reciprocals of two different prime numbers? A. 7/13 B. 10/12 C. 11/30 D. 23/50 E. 19/77

Let \(x\) and \(y\) be two different primes, then their sum will be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\). So, the denominator of the reduced fraction must be the product of two different primes: only B, \(\frac{10}{12}=\frac{5}{6}=\frac{5}{2*3}\), and E, \(\frac{19}{77}=\frac{19}{7*11}\), have such denominators and nominator must be the sum of the primes in denominator, thus only B is left: \(\frac{10}{12}=\frac{5}{6}=\frac{2+3}{2*3}\).

Answer: B.

Hope it's clear.

Hey Bunnel,

Thank u so much.. for this explanation... very very simple and robust way of explaning things...

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Which of the following could be the sum of the reciprocals of two different prime numbers?

A. 7/13 B. 10/12 C. 11/30 D. 23/50 E. 19/77

This question has to be solved by taking on option at a time. However, we do not need to check all the options here as some can be eliminated by checking the given condition.

We have to find the sum of reciprocals of two different prime numbers. This means that the denominator will be the multiplication of the two different prime numbers.

Checking the options for this.

Option A = 13*1. Does not satisfy the condition of two prime numbers Option B = 10/12 = 5/6, Here 6 = 2*3. This is what we need. To cross check, see that 1/2 + 1/3 = 5/6 = 10/12 Our answer.

Let us check for the others too Option C: 30 = 5*6 or 3*10. Does not satisfy the condition of two prime numbers Option D: 50 = 25*2 or 50*1. Does not satisfy the condition of two prime numbers Option E: 77 = 7*11. This satisfies our condition. Checking for the numerator: 1/11 + 1/7 = 18/77 But option has 19/77. Hence not the correct answer.

Re: Which of the following could be the sum of the reciprocals [#permalink]

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27 Feb 2016, 10:27

what a brain tease... to have an odd number in the numerator, it must be that one of the prime numbers is 2. 2*3=6. 2+2=5 5/6 *2 = 10/12

B.

we can also rewrite: a+b/ab where ab must be either the result of 2 prime numbers multiplied, or we must have a factor as well: abx - where x is the factor. A. 7/13 - ab=13 - can't have such option. B. 10/12 = ab=2*3*2 => so 2 primes multiplied by a factor of 2. numerator must also be the sum of 2 prime numbers multiplied by 2. C. 11/30 = ab=5*3*2 => so either ab is multiplied by 5, 3, or by 2. in this case, a+b as well needs to be multiplied by the same factor. but 11 is not a multiple of 5, 3, or 2. so out. D. 23/50 = 5x2x5 -> 23 is not a factor of 5 or 2. so out. E. 19/77 = 7x11 => so 2 primes. a+b must be equal to 18. so definitely not what we need.

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