Is 300 Possible?

It is possible to find out the number of correct answers.

Let the number of correct answers be $x$ and the number of wrong answers be $y$ ,

There are totally $90$ questions and the student attempted every single question:

$\Rightarrow x+y=90$ --------------------------------- $(1)$

The total score is $300$ , $+4$ for a correct answer and $-1$ for an incorrect attempt:

$\Rightarrow 4x-y=300$ ------------------------------ $(2)$

$(2)+(1)$ , we get:

$5x=390$

$x=78$

Therefore, the number of correct answers be $\boxed{78}$ and the number of wrong answers be $90-78=12$ .

If the target of the student is $300$ marks to be obtained then student has $60$ marks less to secure full marks $360$ .

Aforementioned supplied conditions clearly speaks that $4$ marks for each question in case it is correctly done however, in case it is done incorrectly then student has a lose of $4$ marks for each question and $-1$ being incorrect.

Noting total marks reduced per question is $5$ and total marks reduced for some(unknown) questions is $60$ . Thus total number of questions which are made mistake are $\begin{aligned} \frac{\text{total reduced marks}}{\text{total marks reduced per question}}=\frac{60}{5} = 12\end{aligned}$ and correctly done are $90-12=78$ . The answer therefore is $\text{yes}$ .

Let $x$ be the number of correct answers and $y$ the number of incorrect answers.

$S = 4x - y = 5x - (x + y) = 5x - 90 = 300; \\ x = 390/5 = 78.$

By increasing the number of correct answers by 1, the student's score is increased by 5, and similarly is reduced by 5 if the number of correct answers is reduced by 1. This means that the student's score must be divisible by 5.

300 is divisible by 5 and in the range of possible scores (between 0 and 450), so the answer is $\boxed{yes}$

A quick intuitive "pattern-spotting" solution.

Marks if the student got:

• 0 right: $(0)\times(4)-(90)\times(1)= -90 marks$

• 1 right: $(1)\times(4)-(89)\times(1)= -85 marks$

• 2 right: $(2)\times(4)-(88)\times(1)= -80 marks$

...

• 90 right $(90)\times(4)-(0)\times(1)= 360 marks$

The pattern is easy to spot and intuitive. As the number of right answers increases, the student's mark increases by 5 each time.

Because the first term (-90 marks) is a multiple of 5, and the mark always increases by 5, the mark will always be a multiple of five.

Therefore, the student's mark can (only) be any multiple of 5 from -90 to 360.

Since 300 is a multiple of 5 between these two values, it is possible to "get a total score of 300." The answer is [Yes]

Let number of correct answers be $x$ and number of wrong answers be $y$ .

The equation becomes

$4x-y$ $=$ $300$

Now let us compute the g.c.d of the coefficient of $x$ and $y$ term [ $4$ and $1$ ]

We find it is $1$ .

We know if RHS is divisible by g.c.d of the coefficients of the terms of a linear equation in two variables, then the equation will have integral solutions.

Hence such a configuration is indeed possible.

We can get the equations

$4x - y = 300$ , so $y = 4x - 300$

and

$x + y = 90$ , so $y = 90 - x$

where x is the number of correct items, and y is the number of incorrect ones

Then $4x - 300 = 90 - x$ , through transitivity

Add $x + 300$ to both sides,

$5x = 390$

Divide both sides by 5,

$x = 78$

and get y,

$y = 90 - (78)$

$y = 12$

Therefore, the student can get a score of 300 when he got exactly 78 items correct and 12 items incorrect.

The highest score is 360 so it is possible to get a $total$ score of 300.

Pretty easy: only incorrect attempts get -1, so no attempt gets nothing, 75 questions answered correctly and 15 questions without answers leaves you with 300 points.

I have a follow-up question:

Given that there are 90 questions, and each question answered correctly adds 4 to the final score and an incorrect answer deducts 1, but ALSO given that two of the 90 questions are worth 7 for a correct answer or -2 for an incorrect answer, how do you determine if a certain score is possible?

Yes simply because if they get 100% right that is 90x4=360 which is >300

X+Y=90--------(1) 4*X-Y=300----(2)

5x=390 ∴x=79 x+y=90 ∴y=12

Maximum points is 360. If you attemt every question, you lose 5 points for each missed one. So you can hit 300 because 60 is multiple of 5.

it is a system of equations. Let's make x the number of questions right and y the number of questions wrong. For each question wrong, 1 point is taken away, and for each question right 4 points are added to the score. Now, the points added for the number of questions right and the points subtracted for the number of questions wrong have to come out to 300:

$4x-1y=300$

In the second equations, as x is the number of questions right and y is the number of questions wrong, the total number of questions need to come out to 90:

$x+y=90$

now to solve the equations, the y's cancel out and you add the x's as they have a common factor of x and the integers as they have a common factor of 1. then you get the equation:

$5x=390$

then you divide by 5 on both sides:

$x=78$

then you replace the x in the second equation and you get that $y=12$ so yes it is, in fact, possible to get 300 points exactly.

Highest possible score is when all answers are correct: $4 \times 90 = 360$ . In order to get exactly 300, points, we need 60 points less.

Changing a single question from a correct one to an incorrect one means loss of 5 points, and since 60 is divisible by 5, we know there's whole number of answers we can flip, i.e. it is possible to hit exactly 300.

To get a score = 300 (or more), it is obvious that if the student gets all 90 questions correct, they would score 360 (60x4), so the answer is yes, they can get a score greater than 300.

4a - 1b = 300 and a + b = 90 therefor a = 90 - b

4 (90-b) - b = 300 360 - 4b - b = 300 360 - 5b = 300 5b = 60 b = 12 and a = 90 - 12 = 78

So, yes it is possible when a= 78 and b = 12

I solved this using the “cheating” way known as graphing. Remember that the method doesn’t matter as long as it works. :)

We’re given two equations: 4x-y=300, and x+y=90, where x is the number of correct questions and y is the number of incorrect questions. By graphing these lines, we can see that they intersect at (78,12) - that is, 78 correct, 12 incorrect.

Just pointing out that, at the moment, the percentage of people who got it correct is the number of correct answers required.

4X - 1Y = 300 , X being the correct answers, Y being incorrect and 300 being our goal.

X + Y = 90 , for the total amount of questions is 90

X = 90 - Y

4 (90 - Y) - 1Y = 360 - 5Y = 300

60 = 5Y

Y = 12

X = 90 - 12 = 78

so it is possible to get a score of 300.

so c = correct. i = incorrect. 4c - y = 300. c + y = 90. so we can write y as 90 - c. so if we substitute y for 90 - c we get: 4c -(90 - c) = 300. so 5c - 90 = 300. so 5c = 390. c = 390 / 5 = 78. since we got a whole number then its possible. c = 78. i = 12

x = number of correct answers; whole number; 0 <= x <= 90 4 x -1 (90-x) = 300 5x = 390 is there such x (within bounds) so that sentence is true? yes.

$(90-n)\times 4 - n = 300$

$360 - 4n - n = 300$

$-5n = 300 - 360$

$n = \frac{60}{5} = 12$

$90-12=78$

Thus, 78 correct answers.

$78\times 4 - 12 = 300$

Let x = number of incorrect answers.

Four points per #correct answer (which is total answers minus incorrect answers) minus one point per incorrect answer needs to equal 300.

4(90-x) - x = 300

360 - 4x - x = 300

360 - 5x = 300 60 = 5x x = 12 90 - x = 78

78 correct answers 12 incorrect answers

• Suppose the number of correct answer is $n$ , number of incorrect answer be $90-n$
• For a total of 300 marks, 4n+(-1)(90-n)=300
• Solving it obtain:
• $4n-90+n=300$
• $5n=390$
• $n=78$
• Thus, obtaining $78$ correct answer result in 300 marks

4x + (-1y) = 300 Y=90-x 4x -90 + x = 300 X = 78 Y = 12 78^4=312 12^-1=-12

If the student answers x answers correctly and we assume they score 300. The we get the following equation...

4x - (90 - x) = 300

And if we can show that x is an integer between 0 and 90 then it is possible.

4x - 90 + x = 300

5x = 390

x = 78

So, yes, it is possible. If the student answers 78 correctly and 12 incorrectly.

If x were the number of right answers then the number of wrong answers will be y=90-x

4×x-1×(90-x)=300.....x=78 ; y=12

90 questions all answered correctly gives 360 points. A wrong answer would reduce 5 points from that. 12 wrong answers would take us down to 300 points. That is 78 correct and 12 wrong answers.

$4c - (90 - c) = 300$

$4c - 90 + c = 300$

$5c - 90 = 300$

$5c = 390$

$c = 78$

75 X 4 = 300

For every right answer after that 4 wrong answers must be made to keep the total at 300. This makes groups of five attempts. 90 - 75 = 15 and 5 goes into 15 evenly, making the solution possible.

Approach 1

Suppose $X$ is the total amount of marks that can be obtained in a $90$ -question multiple-choice test where a student earns 4 points for a correct answer and is penalised one point for an incorrect answer, and let $x$ be the number of correct answers out of $90$ . Then there must be $90-x$ incorrect answers, and $X$ must satisfy the following equation:

$X = 4x-(90-x) = 5x-90 = 5(x-18)$

Integer solutions can only be obtained precisely when $X$ is a multiple of $5$ , and since $300$ is a multiple of $5$ we can conclude with our desired result.

Approach 2

Building on from Approach 1, we can simply substitute $X = 300$ and obtain our desired result from there. So, we have that

$5(x-18) = 300 \Rightarrow x-18 = 60 \Rightarrow x=78$

and thus we conclude with our desired result that it is possible to obtain a mark of 300 in this test.

If the total number of questions the student got correct was $c$ , the number of questions they got incorrect would be $90-c$ . The total number of points is $4\times$ the number of correct answers plus $-1\times$ the number of incorrect answers.

$4c-(90-c) = 300$

$5c-90 = 300$

$5c = 390 \rightarrow \boxed{c=78}$

Let x be the number of correct answers and y be the number of wrong answers

x+y=90

x=90-y

Solving

4x-y=300

4(90-y)-y=300

360-4y-y=300

5y=60

y=12

So,it is possible to get 300 marks