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B
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Difficulty:
65%
(hard)
Question Stats:
69% (03:03) correct
31%
(03:21)
wrong
based on 143
sessions
History
Date
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Not Attempted Yet
In an accounting class, \(\frac{2}{5}\) of the students who are not lawyers are engineers, and \(\frac{1}{4}\) of the students are engineers. If none of the 21 lawyers in the class are engineers, how many students are neither lawyers nor engineers?
(A) 14
(B) 21
(C) 28
(D) 42
(E) 56
(A) 14
(B) 21
(C) 28
(D) 42
(E) 56
03 May 2026, 20:35
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Let total students = T
Lawyers = 21 ⇒ non-lawyers = T − 21
Given: 2/5 of non-lawyers are engineers
Engineers = (2/5)(T − 21)
Also given: 1/4 of total are engineers
Engineers = T/4
Equate:
(2/5)(T − 21) = T/4
Multiply by 20:
8(T − 21) = 5T
8T − 168 = 5T
3T = 168 ⇒ T = 56
Engineers = T/4 = 14
Neither lawyers nor engineers = total − lawyers − engineers
= 56 − 21 − 14
=> 21
Answer: B
Lawyers = 21 ⇒ non-lawyers = T − 21
Given: 2/5 of non-lawyers are engineers
Engineers = (2/5)(T − 21)
Also given: 1/4 of total are engineers
Engineers = T/4
Equate:
(2/5)(T − 21) = T/4
Multiply by 20:
8(T − 21) = 5T
8T − 168 = 5T
3T = 168 ⇒ T = 56
Engineers = T/4 = 14
Neither lawyers nor engineers = total − lawyers − engineers
= 56 − 21 − 14
=> 21
Answer: B
05 Jul 2026, 09:47
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kevincan Can you explain the solution? I am not able to figure out why Lawyers who are engineers is 0.
