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

## 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. – Sandeep Bhardwaj · 2 years, 5 months ago

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– Tanveen Dhingra · 2 years, 5 months ago

Ok! Thank you very much sir!Log in to reply

Does HCF and LCM being fractions have any meaning? – Janardhanan Sivaramakrishnan · 2 years, 5 months ago

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@Janardhanan Sivaramakrishnan – Kalash Verma · 2 years, 5 months ago

Yeah It has. For example LCM of 1/3 and and 7/12 is 7/3Log in to reply

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

I'm getting contradictory conclusions. – Raghav Vaidyanathan · 2 years, 5 months ago

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– Tanveen Dhingra · 2 years, 5 months ago

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?Log in to reply

– Kalash Verma · 2 years, 5 months ago

Just ask her/him to re check it.Log in to reply

– Tanveen Dhingra · 2 years, 5 months ago

Okay! But she seemed pretty sure! And she was ready to give a solution. But I think it was wrong. So means this question is wrongLog in to reply

– Kalash Verma · 2 years, 5 months ago

Well! If there is some info given about a,b,c,d Like are they co-prime or anything I THINK IT CAN BE SOLVED .I solved it assuming that a,b,c,d are co-prime.Log in to reply

Maam said that they are in their simplest form. Could you send me a solution? – Tanveen Dhingra · 2 years, 5 months ago

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– Kalash Verma · 2 years, 5 months ago

Well could I send it tomorrow. I will recheck and confirm then send it.Log in to reply

– Tanveen Dhingra · 2 years, 5 months ago

Ok! Thank youLog in to reply

– Kalash Verma · 2 years, 5 months ago

Never think like this. Everyone makes mistakesLog in to reply

@Raghav Vaidyanathan well I got answer in decimals. – Kalash Verma · 2 years, 5 months ago

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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 – Raghav Vaidyanathan · 2 years, 5 months ago

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@Sandeep Bhardwaj sir Is this fine? – Tanveen Dhingra · 2 years, 5 months ago

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I think the problem is at this step. Why do \( p \) and \( q \) have to be co - prime? – Siddhartha Srivastava · 2 years, 5 months ago

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– Raghav Vaidyanathan · 2 years, 5 months ago

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.Log in to reply

– Kalash Verma · 2 years, 5 months ago

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

– Raghav Vaidyanathan · 2 years, 5 months ago

I don't think a rational solution exists for given data.Log in to reply

@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 . – Kalash Verma · 2 years, 5 months ago

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 – Dhruv Kaushik · 2 years, 5 months ago

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@Sandeep Bhardwaj @Calvin Lin – Tanveen Dhingra · 2 years, 5 months ago

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