
Divide 136 into two parts one of which when divided by 5 leaves remainder 2 and the other divided by 8 leaves remainder 3.
Answer
498.6k+ views
Hint:
We can take the two parts as x and 132 – x. Then we can take the quotient of the 1st part as p and the quotient of the part as q when divided with 5 and 8 respectively. Then we can form equations for each part with quotient and remainder. Then we can eliminate x by subtracting the 2 equations. Then we can solve for p and q using the hit and trial method. Then we can substitute these values to find the required parts.
Complete step by step solution:
We can take the 2 parts of 136 as x and 132 – x.
It is given that part when divided by 5 leaves remainder 2. Let p be the quotient when divided with 5. So, we can write the part as,
… (1)
It is given that part when divided by 8 leaves remainder 3. Let p be the quotient when divided with 5. So, we can write the part as,
… (2)
Now, we can add equations (1) and (2) to eliminate x.
On adding 5 on both sides, we get
On simplification, we get
On rearranging, we get
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6.
Let q = 2,
On simplification, we get
On dividing the equation by 5, we get
Hence, we have
On substituting the value of p in equation (1), we get
On simplification, we get
Hence, we have
Now the part is 117. The part is given by,
Therefore, the required parts of the number 136 are 117 and 19.
Note:
We must eliminate x and then solve for p and q by hit and trial method. Even though it is not stated, we must know that p and q are natural numbers by the theorem of division algorithm. The values of p and q may not be unique. Another value of p and q is given by,
We have the equation
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6
Let q = 7,
On simplification, we get
On dividing the equation by 5, we get
So, we have
On substituting the value of p in equation (1), we get
On simplification, we get
So, we have
Now the part is 77. The part is given by,
Now, the 2 parts are 77 and 59.
We can take the two parts as x and 132 – x. Then we can take the quotient of the 1st part as p and the quotient of the
Complete step by step solution:
We can take the 2 parts of 136 as x and 132 – x.
It is given that
It is given that
Now, we can add equations (1) and (2) to eliminate x.
On adding 5 on both sides, we get
On simplification, we get
On rearranging, we get
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6.
Let q = 2,
On simplification, we get
On dividing the equation by 5, we get
Hence, we have
On substituting the value of p in equation (1), we get
On simplification, we get
Hence, we have
Now the
Therefore, the required parts of the number 136 are 117 and 19.
Note:
We must eliminate x and then solve for p and q by hit and trial method. Even though it is not stated, we must know that p and q are natural numbers by the theorem of division algorithm. The values of p and q may not be unique. Another value of p and q is given by,
We have the equation
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6
Let q = 7,
On simplification, we get
On dividing the equation by 5, we get
So, we have
On substituting the value of p in equation (1), we get
On simplification, we get
So, we have
Now the
Now, the 2 parts are 77 and 59.
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