
The sum of $4$ consecutive integers is $70$. Then find the greatest among them.
(A) $19$
(B) $23$
(C) $17$
(D) $16$
Answer
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Hint:Assume one of the numbers as some variable and make a sequence of four numbers. Now add them up and equate that to $70$. After solving this equation, you will get the value of one number. Use that to find the greatest.
Complete step-by-step answer:
Let’s try to analyse the question first. It is given that the sum of four consecutive integers is $70$ and we have to find out the greatest of these four numbers. Here consecutive numbers means that these numbers are in a sequence with a difference of $1$. For example: $5,6,7,8$ are four consecutive integers.
So, it’s a sequence of four numbers with a difference of $1$ whose sum is $70$
Now, assume that the first number of this sequence to be some integer$'m'$ .
Since numbers are consecutive, second will be $\left( {m + 1} \right)$, the third will be $\left( {m + 2} \right)$ and the fourth will become $\left( {m + 3} \right)$
We have now those four numbers as: $m,\left( {m + 1} \right),\left( {m + 2} \right),\left( {m + 3} \right)$
Also, according to the provided information: $m + \left( {m + 1} \right) + \left( {m + 2} \right) + \left( {m + 3} \right) = 70$
This is an equation with only one variable and can be easily solved for the value of $'m'$
$ \Rightarrow m + m + 1 + m + 2 + m + 3 = 70 \Rightarrow 4m + 6 = 70$
$ \Rightarrow m = \dfrac{{70 - 6}}{4} = \dfrac{{64}}{4} = 16$
Therefore, we have the value of the first number in our sequence as $m = 16$
Hence, the four numbers will be: $16,\left( {16 + 1} \right),\left( {16 + 2} \right),\left( {16 + 3} \right)$
i.e. $16,17,18,19$
And the required greatest of the four integers is $19$
So, the correct answer is “Option A”.
Note:Do not complicate the problem by assuming four different variables. Solve the equations carefully to avoid any complications. An alternative approach can be to assume four numbers in such a way that it reduces the calculation; like $\left( {m - 3} \right),\left( {m - 2} \right),\left( {m - 1} \right),m$. In this way, after solving the equation for a variable $'m'$, you will straight away get the final answer.
Complete step-by-step answer:
Let’s try to analyse the question first. It is given that the sum of four consecutive integers is $70$ and we have to find out the greatest of these four numbers. Here consecutive numbers means that these numbers are in a sequence with a difference of $1$. For example: $5,6,7,8$ are four consecutive integers.
So, it’s a sequence of four numbers with a difference of $1$ whose sum is $70$
Now, assume that the first number of this sequence to be some integer$'m'$ .
Since numbers are consecutive, second will be $\left( {m + 1} \right)$, the third will be $\left( {m + 2} \right)$ and the fourth will become $\left( {m + 3} \right)$
We have now those four numbers as: $m,\left( {m + 1} \right),\left( {m + 2} \right),\left( {m + 3} \right)$
Also, according to the provided information: $m + \left( {m + 1} \right) + \left( {m + 2} \right) + \left( {m + 3} \right) = 70$
This is an equation with only one variable and can be easily solved for the value of $'m'$
$ \Rightarrow m + m + 1 + m + 2 + m + 3 = 70 \Rightarrow 4m + 6 = 70$
$ \Rightarrow m = \dfrac{{70 - 6}}{4} = \dfrac{{64}}{4} = 16$
Therefore, we have the value of the first number in our sequence as $m = 16$
Hence, the four numbers will be: $16,\left( {16 + 1} \right),\left( {16 + 2} \right),\left( {16 + 3} \right)$
i.e. $16,17,18,19$
And the required greatest of the four integers is $19$
So, the correct answer is “Option A”.
Note:Do not complicate the problem by assuming four different variables. Solve the equations carefully to avoid any complications. An alternative approach can be to assume four numbers in such a way that it reduces the calculation; like $\left( {m - 3} \right),\left( {m - 2} \right),\left( {m - 1} \right),m$. In this way, after solving the equation for a variable $'m'$, you will straight away get the final answer.
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