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$(a){\text{ }} - 19$

$(b){\text{ }} - 18$

$(c){\text{ }}5$

$(d){\text{ }}55$

Answer

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157.2k+ views

Hint: Factorise the given numbers carefully. Also keep in mind that the value of $y$ can be negative as well.

We have the given numbers as:

$210$ and $55$

Now, these can be factorized as

$210 = 2 \times 3 \times 5 \times 7$

$55 = 5 \times 11$

This shows that the HCF of $210$ and $55$ comes out to be $5$.

Since, it is given in the question that the HCF of $210$ and $55$ is of the form \[(210)(5) + 55y\]

That is,

\[5 = (210)(5) + 55y\]

On further evaluation, we get

\[ \Rightarrow 1050 + 55y = 5\]

\[ \Rightarrow 55y = - 1050 + 5\]

\[ \Rightarrow 55y = - 1045\]

\[ \Rightarrow y = \dfrac{{ - 1045}}{{55}}\]

$\therefore y = - 19$

So, the required solution is $(a){\text{ }} - 19$.

Note: To solve these types of questions, find out the HCF of the given numbers and substitute it in the given equation to obtain the required solution.

We have the given numbers as:

$210$ and $55$

Now, these can be factorized as

$210 = 2 \times 3 \times 5 \times 7$

$55 = 5 \times 11$

This shows that the HCF of $210$ and $55$ comes out to be $5$.

Since, it is given in the question that the HCF of $210$ and $55$ is of the form \[(210)(5) + 55y\]

That is,

\[5 = (210)(5) + 55y\]

On further evaluation, we get

\[ \Rightarrow 1050 + 55y = 5\]

\[ \Rightarrow 55y = - 1050 + 5\]

\[ \Rightarrow 55y = - 1045\]

\[ \Rightarrow y = \dfrac{{ - 1045}}{{55}}\]

$\therefore y = - 19$

So, the required solution is $(a){\text{ }} - 19$.

Note: To solve these types of questions, find out the HCF of the given numbers and substitute it in the given equation to obtain the required solution.