Problem 1

Three problems

A

B

, and

C

were given on a mathematics olympiad. All 25 students solved at least one of these problems. The number of students who solved

B

and not

A

is twice the number of students who solved

C

and not

A

. The number of students who solved only

A

is greater by 1 than the number of students who, along with

A

, solved at least one other problem. Among the students who solved only one problem, half solved

A

. How many students solved only

B

Step 2 of 6: Translate the "twice as many" condition

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In plain words

Comparing two "solved X but not A" counts is just comparing sums of Venn-diagram regions, the same translation used for the total in step 1.

b+d=2(c+d)\ \Longrightarrow\ d=b-2c

Detailed analysis

Students who solved $B$ but not $A$ are exactly those in groups $b$ (only B) and $d$ (B and C, not A), so this count is $b+d$ . Likewise the students who solved $C$ but not $A$ number $c+d$ . The condition states $b+d=2(c+d)$ , which rearranges to $d=b-2c$ .