Hey guys!

My teacher gave us a problem.

Eventually later on even she could not solve it!

Can you help me!

Here is the question:

H.C.F of \(\frac{a}{b}\) and \(\frac{c}{d}\) is \(\frac{2}{105}\)

L.C.M of \(\frac{a}{b}\) and \(\frac{c}{d}\) is \(\frac{12}{5}\)

H.C.F. of \(\frac{a}{c}\) and \(\frac{b}{d}\) is \(\frac{1}{210}\)

L.C.M of \(\frac{a}{c}\) and \(\frac{b}{d}\) is 60.

Find the value of a + b + c + d

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## Comments

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TopNewestSimply use the property : \[\boxed{L.C.M(A,B) \times H.C.F.(A,B)=A \times B}\]

Go on for the calculation part and you'll get the answer.

If you don't get the values even then, it means required values of \(a,b,c\) and \(d\) don't simply exist.

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Ok! Thank you very much sir!

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Does HCF and LCM being fractions have any meaning?

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Yeah It has. For example LCM of 1/3 and and 7/12 is 7/3 @Janardhanan Sivaramakrishnan

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Have you considered the fact that this problem may be wrong??

I'm getting contradictory conclusions.

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Even I thought so but since my teacher gave me this question I thought it ought to be correct. Are you sure that it is wrong?

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Just ask her/him to re check it.

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Maam said that they are in their simplest form. Could you send me a solution?

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Never think like this. Everyone makes mistakes

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@Raghav Vaidyanathan well I got answer in decimals.

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Using the fact that \(H.C.F(x,y)*L.C.M(x,y)=x*y\) where \(x,y\) are both natural numbers or rational fractions, we can get the following from the given data:

\(\frac { ac }{ bd } =\frac { 8 }{ 7\times 25 } \\ \frac { ab }{ cd } =\frac { 2 }{ 7 } \)

From the above equations, we can get:

\(\frac { a }{ d } =\frac { 4 }{ 35 } \\ \frac { c }{ b } =\frac { 2 }{ 5 } \)

This means that we can take:

\(a=4p, d=35p, c=2q\) and \(b=5q\) where \(p\) and \(q\) are co-prime natural numbers.

\(\frac { a }{ b } =\frac { 4p }{ 5q } ,\frac { c }{ d } =\frac { 2q }{ 5p } \quad \quad ........(1)\\ \frac { a }{ c } =\frac { 2p }{ q } ,\frac { b }{ d } =\frac { q }{ 7p } \quad \quad ........(2)\)

The fractions mentioned in Eq\((2)\) have H.C.F as \(\frac { 1 }{ 2\times 3\times 5\times 7 } \). What this means is that, in Eq\((2)\), one of the fractions

MUSTcontain \(5\) in the denominator(if neither contains \(5\) in the denominator, then the H.C.F will not contain a \(5\) in the denominator).This means that either \(p \) or \(q\) is \(5\). (Keep in mind that \(p\) and \(q\) are co-prime).

Now, if we observe Eq\((1)\), both fractions already have exactly one power of \(5\) in the denominator(without considering value of \(p\) or \(q\)). Observe that the H.C.F of the fractions in Eq\((1)\) also has exactly one power of \(5\) in the denominator. To take H.C.F of two fractions, we take the highest power of each prime divisor from the denominator, and since the H.C.F only has one \(5\) in the denominator, this means that both the fractions in Eq\((1)\) can contain maximum of only one \(5\) in the denominator. But we have already seen that either \(p\) or \(q\) is \(5\). So either way, one of the fractions

WILLhave a multiple \(25\) in the denominator. But if \(25\) was present in the denominator, then the H.C.F must also have a multiple of \(25\) in its denominator. This is a contradiction. Hence, our equations are inconsistent and natural numbers \(a,b,c,d\) do not exist for given conditions.@tanveen dhingra @Kalash Verma

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Thank you @Sandeep Bhardwaj sir Is this fine?

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I think the problem is at this step. Why do \( p \) and \( q \) have to be co - prime?

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They don't have to be, even if we assume that \(p\) and \(q\) are rational numbers, we get the same result. The actual thing that makes a difference is \(p/q\) which is also a rational number. The problem is when we find that \(p/q\) must have and not have a \(5\) in the denominator simultaneously, which is impossible.

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Yeah did the same way.That's why it was coming In decimals I think.

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I don't think a rational solution exists for given data.

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@tanveen dhingra I do not think that it is worth to give a solution by me. My solution is 85 percent the same. Though his more clearer .

The reason I get an decimal is due to that never solved for values of A,B ,C,D instead used some equations to straight away find A,B,C,D. Could have done some mistake!!.Log in to reply

hi

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@Sandeep Bhardwaj @Calvin Lin

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