A machine manufactures notebooks in a series of five colors: red, blue, black, white and yellow. After producing a notebook of one color from that series, it produces a notebook of the next color. Once five are produced, the machine repeats the same pattern. If the machine began a day producing a red notebook and completed the day by producing a black notebook, how many notebooks could have been produced that day?
A. 27
B. 34
C. 50
D. 61
E. 78
The OA is E.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Hi LUANDATO,
Let's take a look at your question.
The question states:
A machine manufactures notebooks in a series of five colors: red, blue, black, white and yellow.
If the machine starts its production from a red notebook and ends at a black note book, then the sequence must be like:
RED, BLUE, BLACK, WHITE, YELLOW, RED, BLUE, BLACK, WHITE, YELLOW, RED, BLUE, BLACK, WHITE, ..., BLACK
The sequence will be like this and we need to find how many notebooks were printed during the day.
If we look at the sequence, we can see that,
The minimum number of books that can be printed will be 3 from RED to BLACK. (RED, BLUE, BLACK)
or the number of books could be 8. (RED, BLUE, BLACK, WHITE, YELLOW, RED, BLUE, BLACK)
or 13 if one more cycle is completed. (RED, BLUE, BLACK, WHITE, YELLOW, RED, BLUE, BLACK, WHITE, YELLOW, RED, BLUE, BLACK)
So the sequence of the number of books that can possibly be printed during the day is:
3, 8, 13, ...
This is an arithmetic sequence with:
$$a_1=3$$
$$d=5$$
The maximum number of books that can be manufactured will be from one of the choices given. So we will check each option to check which could be the nth term of this sequence using the formula of arithmetic sequence:
$$a_n=a_1+\left(n-1\right)d$$
$$a_n=3+\left(n-1\right)5$$
$$a_n=3+5n-5$$
$$a_n=5n-2$$
Let's check Option
A.
$$a_n=27$$
Plugin the value in the formula:
$$a_n=5n-2$$
$$27=5n-2$$
$$29=5n$$
$$n=\frac{29}{5}$$
It results into a decimal number, which is not a valid value for the number of terms. n should be a whole number.
So we will discard this.
Let's now check Option
B.
$$a_n=34$$
Plugin the value in the formula:
$$a_n=5n-2$$
$$34=5n-2$$
$$36=5n$$
$$n=\frac{36}{5}$$
It results into a decimal number, which is not a valid value for the number of terms. n should be a whole number.
So we will discard this.
Let's now check Option
C.
$$a_n=50$$
Plugin the value in the formula:
$$a_n=5n-2$$
$$50=5n-2$$
$$52=5n$$
$$n=\frac{52}{5}$$
It results into a decimal number, which is not a valid value for the number of terms. n should be a whole number.
So we will discard this too.
Let's now check Option
D.
$$a_n=61$$
Plugin the value in the formula:
$$a_n=5n-2$$
$$61=5n-2$$
$$63=5n$$
$$n=\frac{63}{5}$$
It results into a decimal number, which is not a valid value for the number of terms. n should be a whole number.
So we will discard this as well.
Let's now check Option
E.
$$a_n=78$$
Plugin the value in the formula:
$$a_n=5n-2$$
$$78=5n-2$$
$$80=5n$$
$$n=\frac{80}{5}$$
$$n=16$$
n is a whole number now. Which shows that the number of books manufactured during the day is 78.
Therefore, Option
E is correct.
Hope it helps.
I am available if you'd like any follow up.
