There are 42 students in a group. If each student is either

I got C too. solution F=8 and S=34 is the only possibility from S1 and S2

ST 1 - insufficent
st 2 - in suff
combing 1 & 2
we get only one value > 7 which is 8 that satisfy answer

hence c

f + s =42.

1. Insufficient
s > 4f
s > 4(42-s)
s > 33.6
s >=34
s could be 34 ,35....

2. Insufficient.

we dont know anthing about s.

together , Sufficient
as s = 34 , f = 8 is the only combination that satisfies 1 and 2.

Answer is C.

Statement 1 - Insufficient... Many answers are possible
Statement 2 - Nothing can be derived from this

Both together - 35 & 7 is an option but from statement 2, more than 7
34 and 8 is possible (more than 4:1 and also 8)
33 and 9 is not possible as statement 1 is not satisfied

So only one answer when two statements considered together


Answer C

x + y = 42

(1)

x > 4y


42 -y > 4y

y < 8.4

So, y = 8,7 etc.

x = 34, 35 etc.

Not Sufficient

(2)

y > 7

Not Sufficient

(1) and (2)

y = 8, x = 34

Answer - C

The translation of (1) killed me in this problem.
I thought it meant 4s > f. Can someone help me translate it to s > 4f?

The statement clearly states that "more than four times as many seniors as it has freshmen."
Lets forget more than part here and focus on four times as many seniors as it has freshmen.
This means S=4F
now more than part . incorporate > in place of =.
So S > 4F

C.

Statement 1:
s>4F

>4F+F = 42
F & S should be an integers.
So I found factors of 42 fitting to solve the equation .
Factors of 42 = 1,2,3,6,7,14,21,42.
So 5F + F = 42 => F=7 & S = 35
6F+F= 42 => F = 6 & S=36. and so on for other values. => Not sufficient.

Statement 2:
F>7.
Many values satisfy this condition such as F = 8 & S = 34; F=9 & S = 33 & so on =>Insufficient

Combine both :
F>7 and S>4F . F = 8 & S = 34

Hence C

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There are 42 students in a group. If each student is either a freshmen or a senior, how many of the students are seniors?

(1) The group has more than four times as many seniors as it has freshmen.

(2) The group has more than 7 freshmen.


In the original condition, there are 2 variables(f,s) and 1 equation(f+s=42), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), in s>4f, value of s is not unique and not sufficient.
For 2), in f>7, value of s is also not unique and not sufficient. When 1) & 2), they become s>4f and f>7 → s+f>4f+7, 42>4f+7, 35>4f → 35/4=8.75>f. Since f>7, in 8.75>f>7, f=8, s=34, which is unique and sufficient.
Therefore, the answer is C.


--> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

Hi Bunuel , unable to figure out where the 33.6 is coming from ? can you please help ? ive tried to figure it out for the past 30 minutes

\(S>4*(42-S)\)

\(S>4*42-4S\)

\(5S>4*42\)

\(S>\frac{4*42}{5}\)

\(S>33.6\)

Hope it helps.