I wanted to understand the difference between these two questions and understanding the calc differences. Would appreciate some feedback.

AndyShanky wrote:I wanted to understand the difference between these two questions and understanding the calc differences. Would appreciate some feedback.

First asks you to find "more than one of the three products"

This will include ppl who like 2 products and also like all the three.



second asks you to find "How many students are enrolled in exactly two of the courses"

This will not include ppl in all the three courses.
It only wants you to find ppl in two courses

Hope this helps!!

Could you please confirm the OA of the first question? I seem to get 10% as the answer. Thanks.

Both these questions deal with three overlapping sets. Check out the theory here:
https://www.gmathacks.com/gmat-math/thre ... -sets.html

-BM-

Guys, I do not understand why my logic for 1 st Q is incorrect?

The first one, I myself am getting as 10%

(A u B u C) = 85
(A) = 50
(B) = 30
(C) = 20
(A n B n C) = 5.........all given


all area that is atleast common to 2 =

(A) + (B) + (C) - (A u B u C) - (A n B n C) = 5



the second is right...even i got the answer as 5


(A u B u C) = 70
(A) = 40
(B) = 30
(C) = 35
(A n B n C) = 15.........all given

all area that is atleast common to 2 excluding the area that is common to all 3 =

(A) + (B) + (C) - (A u B u C) - (A n B n C) - (A n B n C) = 10



The difference between the 2 sums is that the second sum asks for 'only the number of students that study exactly 2 subjects', that is, it excludes the area common to all 3.....


If this is insufficient, let me know, i will later help with the help of venn diagrams

I am getting 5 for both 1 and 2. Am I missing any thing ?
shouldn't it be
=50+30+20-85-2*(5)

I am getting 5 for both 1 and 2. Am I missing any thing ?
shouldn't it be
=50+30+20-85-2*(5)

I have seen more variations of these formulas and sometimes get confused also.

Q 1)

Total - None(15% not surveyed) = P(A) + P(B)+P(C) - P(AnB)-P(BnC)-P(CnA) +P(AnBnC)

P(AuBuC) = P(A) + P(B)+P(C) - P(AnB)-P(BnC)-P(CnA) +P(AnBnC)

100-15 = 50+30+20 - P(AnB) - P(BnC) - P(CnA) +5

85 = 50+30+20 - P(AnB) - P(BnC) - P(CnA) +5

P(AnB)-P(BnC)-P(CnA)= 50+30+20-85+5
= 20


The formula for more than 1 (i.e Number of people in two or more sets)

is P(AnB)+P(BnC)+P(CnA) - 2 P(AnBnC)

= 20 - 2(5)
= 10


Q2 :
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.


Total - None= P(A) + P(B)+P(C) - P(AnB)-P(BnC)-P(CnA) +P(AnBnC)

P(AuBuC) = P(A) + P(B)+P(C) - P(AnB)-P(BnC)-P(CnA) +P(AnBnC)

70 - 0 = 40+30+35 - P(AnB)-P(BnC)-P(CnA)+15

P(AnB)+P(BnC)+P(CnA) = 50


Number of people in exactly two of the sets =

P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

= 50 - 3(15)

= 5


Hope I dint mess it up too much!

The OA must be wrong for Q1)

It should be 10 and not 20.

The explanation for the 1st one , the official answer 20 is correct :

n(1) = 50
n(2) = 30
n(3) = 20

What is asked is the % of consumers who liked more than one product, which means :

n(A int B) + n(B int C) + n(A int C)

n(A) + n(B) + n(C) - n(AUBUC) + n(A int B int C)

(derived from the formula of AUBUC)

so , 50 + 30 + 20 - X + 5 = 85 ( as provided in the question )

X ( the value we want ) = 20

Hope this helps !

hahaha))) I got the answer for Q1

1st equation:
a+b+c+d+e+f+g=85
since e=5, a+b+c+d+f+g=80


By addition of
a+b+d+e=50
c+b+e+f=30
g+d+e+f=20
and if e=5
we get 2nd equation:
a+c+g+2b+2d+2f=95

subtract 1st equation from 2nd equation:
b+d+f=15


what percentage of the survey participants liked more than one of the three products?

(b+d+f+e)*100% / 100 = 20%

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

use this .........

AndyShanky wrote:Q1 :
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

OA apparently is 20

We want the % that liked more than one... i.e. that liked either 2 or 3 of the products.

We know that 5% liked all 3, so let's ignore that for now. Taking out the 5% that like all 3, we now have the question:

In a consumer survey, 80% of those surveyed liked either 1 or 2 of three products. 45% of those asked liked product 1, 25% liked product 2, and 15% liked product 3. What % liked exactly 2 of the products?

Well, 45 + 25 + 15 = 85. We only have 80 to go around, so those extra 5% of the people are in exactly 2 groups.

5% like all 3, 5% like exactly 2: 10% like more than 1 product.

If 10% isn't the OA, I'm going out on a limb and saying that the "OA" is wrong!

4meonly wrote:hahaha))) I got the answer for Q1

1st equation:
a+b+c+d+e+f+g=85
since e=5, a+b+c+d+f+g=80


By addition of
a+b+d+e=50
c+b+e+f=30
g+d+e+f=20
and if e=5
we get 2nd equation:
a+c+g+2b+2d+2f=95
subtract 1st equation from 2nd equation:
b+d+f=15

Close but no cigar!

If you add up those 3 equations, you get:

a+c+g+2b+2d+2f+3e=100

Since e=5, you then get:

a+c+g+2b+2d+2f=85

Which is why your answer is 10 more than it should be.

AndyShanky wrote:Q2 :
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

OA apparently is 5

Let's approach this one the same way.

If 15 are in all 3, then we know that:

There are 55 students who take either 1 or 2 of math, english and german. Exactly 25 are in math, 15 in german and 20 in english. How many take exactly 2/3?

M+G+E = 25 + 15 + 20 = 60. However, we only have 55 students to go around, so 5 of them must be double counted. Therefore, 5 students take exactly 2 of the courses.

Note that this method is really just another way to apply the 3 set equation quoted by Ronnie. If you're an equation person, it's a good equation to know. However, like most questions on the GMAT, you can often reason your way to the correct answer even if you don't know the right formula.