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In the faculty of Reverse-Engineering, 226 second year...

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In the faculty of Reverse-Engineering, 226 second year...

by AAPL » Wed Mar 21, 2018 7:29 am
In the faculty of Reverse-Engineering, 226 second-year students study numeric methods, 423 second-year students study automatic control of airborne vehicles and 134 second-year students study them both. How many students are there in the faculty if the second-year students are approximately 80% of the total?

A. 515
B. 545
C. 618
D. 644
E. 666

The OA is D.

I solved this PS question of the following way,

NM = 226

AV = 423

Both = 134

Total second year students = (226 - 134) + (423 - 134) + 134 = 226 + 423 - 134 = 515

Assume total students = x

Therefore 0.8x = 515

x = 515/0.8 = 5150/8 = 643 approximately. Option D.

Is there another strategic approach to solve this question? Can any experts help, please? Thanks!
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by swerve » Wed Mar 21, 2018 9:25 am
Hi AAPL,

You can try as follow,

The total number of students studying both are 423 + 426 - 134 = 515 (subtracting the 134 since they were included in both the other numbers already).

So 80% of the total is 515, so 100% is approximately 644.

Regards!
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by deloitte247 » Sat Mar 24, 2018 11:49 am
Numeric method = 226 second year students
Automatic control= 423 second year students
Both numeric and automatic = 134 second year students

Therefore, total second year student= (numeric + automatic) - both = [226+423]-134
=649-134= 515 second year students in faculty of reverse engineering. 2nd year student made 80% of the total student.
if 80% =515
then 100%=
$$\frac{\left(515\cdot100\right)}{80}=643.75=644$$
Hence, option D is correct
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