Directions : Given below is no. of male & female students in classes A, B & C.…
2019
Directions : Given below is no. of male & female students in classes A, B & C. some data are missing which you have to calculate as per instructions provided.
NOTE:
(i) probability of selecting a boy from class A is 5/12 .
(ii) probability of selecting a boy from all the boys of all classes is 14/19 such that the boy selected is either from class B or class C.
(iii) probability of selecting a boy from class B is equal to probability of selecting a boy from class C.
By what percent total students in class B are more than that of in class A?
- A.
16 ⅔%
- B.
37 ½%
- C.
25%
- D.
33 ⅓%
- E.
None of these
Show answer & explanation
Correct answer: D
Concept: In a single-stage selection, the probability of picking a boy from a class equals (boys in that class) / (total students in that class). Each stated probability is therefore one equation linking the missing counts; build the equations, solve them in order, then compare two class totals using percentage change = (new - base)/base x 100, where the base is the quantity you compare AGAINST.
Application:
From note (i), P(boy from A) = boys_A / total_A = 50 / total_A = 5/12, so total_A = 50 x 12/5 = 120. Hence girls_A = 120 - 50 = 70.
Let boys_B + boys_C = X. Note (ii) says X / (boys_A + X) = 14/19, i.e. X / (50 + X) = 14/19. Cross-multiplying: 19X = 14(50 + X) -> 5X = 700 -> X = 140. So boys_B + boys_C = 140.
Note (iii): P(boy from B) = P(boy from C). With girls_B = 80 and girls_C = 60, write boys_B = b and boys_C = 140 - b. Then b/(b + 80) = (140 - b)/(140 - b + 60). Cross-multiplying gives 200b - b^2 = 60b + 11200 - b^2 -> 140b = 11200 -> b = 80. So boys_B = 80 and boys_C = 60.
Total_B = boys_B + girls_B = 80 + 80 = 160; Total_A = 120 (from step 1).
Required percent = (Total_B - Total_A)/Total_A x 100 = (160 - 120)/120 x 100 = 40/120 x 100 = 100/3 = 33 1/3%.
Cross-check: Verify the two probabilities: P(boy B) = 80/160 = 1/2 and P(boy C) = 60/120 = 1/2, so note (iii) holds. Combined boys from B and C = 140 out of 190 total boys = 14/19, matching note (ii). The class B total exceeds class A by 40 students, and 40 is one-third of 120, confirming 33 1/3%.