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For the first one is it write to assume every student has taken either english or algebra
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Bunuel
In a group of 30 students, 18 are enrolled in an English class and 16 are enrolled in an Algebra class. How many students are enrolled in both an English and Algebra class?

(1) 20 are enrolled in exactly one of these two classes.

(2) 3 are not enrolled in either of these classes.



Are You Up For the Challenge: 700 Level Questions

English class- 18 & Algebra class- 16.

Only english + only algebra + both + none = 30, ........ Equation 1

Also, Only english + only algebra + 2 X both= 18+16 = 34 ........ Equation 2

S1. Only English + Only Algebra = 20. So from equation 2, both can be found. Sufficient

S2. none=3, from the equation 1, Only English + only Algebra + both= 27 ....... equation 3
from the equation 2, Only English + only Algebra + 2 X both= 34 ..........equation 4

Equation 4 - equation 3= both . Sufficient .

Both individuals are sufficient. Hence D

No, it's not correct to assume every student is enrolled in either English or Algebra, and in fact, the solution you quoted explicitly accounts for those not enrolled in either class using the group {None}:

{Only English} + {Only Algebra} + {Both} + {None} = 30

Statement (2) further confirms this explicitly by stating that {None} = 3.
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Bunuel
In a group of 30 students, 18 are enrolled in an English class and 16 are enrolled in an Algebra class. How many students are enrolled in both an English and Algebra class?

(1) 20 are enrolled in exactly one of these two classes.

(2) 3 are not enrolled in either of these classes.
From Main Statement: B = Both; N = Neither; then 16+18+N-B = 30;

B-N=4;

From Statement 1: 20 + B + N = 30; B + N = 10; Solving it with B-N=4; B = 7; N = 3; Sufficient

From Statement 2: N = 3; B - N = 4; Solving it for B = 7; Sufficient

Ans D;
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