pls help me to solve the problem?

Question

In a mathematical contest, three problems, A, B and C were posted. Among the participants there were 25 students who solved at least one problem each. Of all the contestants who didn’t solved problem A, the number who solved B was twice the number who solved C. The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem. Of all students who solved just one problem, half didn’t solve problem A. How many students solved only problem B?

Answer 1

let me try to do this mathematically , ofcourse there are a lot of variables, bear with me. join me on this exciting journey which may lead to a dead end if we arent careful



p students solved A alone

q students solved A and B alone

r students solved A and C alone

s students solved A and B and C

t students solved B alone

u students solved B and C alone

v students solved C alone

we have seperated all possible options here and they are all mutually exclusive. the sum of these should be 25. in other words

p+q+r+s+t+u+v = 25.......................[1]

that is our first equation, now lets see the other clues very carefully

1st HINT : Of all the contestants who didn’t solved problem A, the number who solved B was twice the number who solved C.

which contestants didnt solve A its obviously t+u+v, among these the people who solve B which is t+u will be twice who solved C which is u+v

t+u = 2(u+v)

or

t = u + 2v.......................................[2]

2nd HINT: The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem. pretty simple

p = q+r+s +1..................................[3]

3rd HINT : Of all students who solved just one problem, half didn’t solve problem A.

p = t+v ..........................................[4]

excellent we have 4 equations now we should try to find equations with highest coefficients , this will help us to narrow down the trial method to determine the integer solutions alone.

lets just look at equation [1]

p+q+r+s+t+u+v = 25 substitute eq [3] so we can eliminate 3 variables

2p + t+u+v = 26 use equation [4] to get

3p+u = 26 also p = t+v eq [4] and t = u+2v eq[2] we get p = u+3v lets substitute to get

4u+9v = 26



this is the equation with highest coefficients we could arrive at and this will help us most

we know u and v has integer values, inspection shows that the only two numbers which satisfy these are u =2 and v = 2.

wow , isnt that amazing ? so the problem is solved, we can get the others using this

t = u +2v = 6 the sudents who solved B alone

p = t+v students who solved A alone = 8



hope everything was clear, wish you all the best

let me know

Answer 2

The answer is 6. I did not use much math knowledge for this, just educated guesswork.

To start with, let x be those who solved only A. Then x-1 are those that solved A plus atleast one other problem. Similarly, x is the number who solved only 1 problem other than A since "Of all students who solved just one problem, half didn’t solve problem A."

The sum of all these is 3x-1 which cannot exceed 25. Therefore the max value of x can be 8. So use this as the guess value for now.

So 8 students solved only A. 7 students solved A plus atleast one other problem. Which means 15 solved A and 10 did not.

8 students solved juts 1 problem apart from A and therefore 16 solved just one problem.

This means that 25-16=9 students solved 2 or more problems. As mentioned earlier, 7 of these solved A so 2 solved both B&C. We mentioned that 10 students solved either B or C or both. So 8 students solved either just B or C. Only if 6 solved B alone and 2 solved C alone then we add the 2 who solved both B&C to find that "Of all the contestants who didn’t solved problem A, the number who solved B was twice the number who solved C",

Therefore this problem solution is 6 students solved only problem B, 8 solved only A and 2 solved only C.

Answer 3

I think ans is 25/3....

Let the no of students solved B be x

Then no of students who solved A and C let it be Z=25-x

also, Given that no of students didn't solved A=25/2

and, no of student who solved C=x/2

So, x+x/2=25/2 [since,who didn't solved A is equal to the students who solved C and B]

3x/2=25/2

x=25/3

Hence, the ans is 25/3

Answer 4

Some more clarifications needed.

Answer 5

( 8) Eight