Question:hard

In a class of 40 students who study mathematics, physics and chemistry, the number of students studying mathematics is 2 more than 40% of those studying chemistry, while 20 of the students study physics. Three less than one fifth of the total students in the class study all three subjects. The number of students studying only physics is 2 less than the number of students studying mathematics. The number of students studying only mathematics and only chemistry is 3 and 15 respectively. The number of students studying mathematics as well as physics is the same as the number of students studying mathematics as well as chemistry. How many students are studying both mathematics and chemistry but not all the three subjects?

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Set up variables for the pairwise-only overlaps and use the fact that the two given intersections are equal.
Updated On: Jul 21, 2026
  • 25
  • 15
  • 10
  • 2
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The Correct Option is D

Solution and Explanation

This can also be solved by using the general region formula for three sets, Only-a-subject = Subject total minus its two pairwise totals plus the triple overlap, instead of tracking regions with separate variables.
Step 1: Note the fixed values. All three subjects = 5 (one fifth of 40 is 8, minus 3 is 5). Only Mathematics = 3, Only Chemistry = 15, total = 40.
Step 2: Let the common pairwise total be k. Since Mathematics-Physics total equals Mathematics-Chemistry total, call each of these k.
Step 3: Apply the region formula to Mathematics. \(3 = M - k - k + 5\), so \(M = 2k - 2\).
Step 4: Apply it to Physics. Only Physics = \(20 - k - (P\cap C) + 5\), and this also equals \(M - 2 = 2k - 4\). So \(25 - k - (P\cap C) = 2k - 4\), giving \(P\cap C = 29 - 3k\).
Step 5: Apply it to Chemistry. \(15 = C - k - (29 - 3k) + 5\), so \(C = 39 - 2k\).
Step 6: Apply the percentage condition. \(2k - 2 = 2 + 0.4(39 - 2k)\). This gives \(2k - 2 = 17.6 - 0.8k\), so \(2.8k = 19.6\), giving \(k = 7\).
Step 7: Find the required region. Mathematics-Chemistry only = \(k - 5 = 2\), the same answer as before.\[\boxed{2}\]
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