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15 Aug 2024, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:00) correct
36%
(02:33)
wrong
based on 181
sessions
History
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Time
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Not Attempted Yet
In a colony, there are 75 families, each having at most 3 breeds of dogs: pugs, beagles or boxers. Of these families, 40 have exactly one breed. Is there any family in the colony that has beagles and boxers but not pugs? Note: Each family has atleast one breed of dog.
(1) 7 families have all the three breeds of dogs.
(2) 10 families have pugs and exactly one of the other two breeds of dogs.
(1) 7 families have all the three breeds of dogs.
(2) 10 families have pugs and exactly one of the other two breeds of dogs.
15 Aug 2024, 18:52
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Answer should be C
X family only pugs
y family only have beagles
X family only have boxers
a .. Family having pugs and eagle
b.. Family having eagles and boxers
c.. Family having pugs and boxers
T.. family having boxers, pugs and eagles
Out of the 75 family, 40 have only one breed
X+y+z=40
We have to find is there is a family having eagle and boxers but not pugs thus b
Statement 1
7 family have all the three breeds of dogs
Means T=7
We have 40+7+a+b+c=75
Insufficient
Statement 2
10 families have pugs and exactly one of the other two breeds of dogs
a+c=10
40+10+T+b=75
Insufficient
1+2
40+10+7+b=75 —-> b=18
Posted from my mobile device
X family only pugs
y family only have beagles
X family only have boxers
a .. Family having pugs and eagle
b.. Family having eagles and boxers
c.. Family having pugs and boxers
T.. family having boxers, pugs and eagles
Out of the 75 family, 40 have only one breed
X+y+z=40
We have to find is there is a family having eagle and boxers but not pugs thus b
Statement 1
7 family have all the three breeds of dogs
Means T=7
We have 40+7+a+b+c=75
Insufficient
Statement 2
10 families have pugs and exactly one of the other two breeds of dogs
a+c=10
40+10+T+b=75
Insufficient
1+2
40+10+7+b=75 —-> b=18
Posted from my mobile device
15 Aug 2024, 20:50
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Since no of families having exactly one of the breeds is given:
\(\text{Total = exactly one breed + exactly two breeds + exactly three breeds}\)
\(\text{75 = 40 + exactly two breeds + exactly three breeds}\)
Statement 1: 7 families have all the three breeds of dogs.
\(\text{75 = 40 + exactly two breeds + 7}\)
Not Sufficient
Statement 2: 10 families have pugs and exactly one of the other two breeds of dogs.
\(\text{exactly two breeds = Pugs & Beagles + Pugs & Boxers + Beagles & Boxers}\)
\(\text{exactly two breeds = 10 + Beagles & Boxers}\)
\(\text{75 = 40 + 10 + Beagles & Boxers + exactly three breeds}\)
Not Sufficient
Combined:
\(\text{75 = 40 + 10 + Beagles & Boxers + 7}\)
Sufficient
Answer is C
\(\text{Total = exactly one breed + exactly two breeds + exactly three breeds}\)
\(\text{75 = 40 + exactly two breeds + exactly three breeds}\)
Statement 1: 7 families have all the three breeds of dogs.
\(\text{75 = 40 + exactly two breeds + 7}\)
Not Sufficient
Statement 2: 10 families have pugs and exactly one of the other two breeds of dogs.
\(\text{exactly two breeds = Pugs & Beagles + Pugs & Boxers + Beagles & Boxers}\)
\(\text{exactly two breeds = 10 + Beagles & Boxers}\)
\(\text{75 = 40 + 10 + Beagles & Boxers + exactly three breeds}\)
Not Sufficient
Combined:
\(\text{75 = 40 + 10 + Beagles & Boxers + 7}\)
Sufficient
Answer is C
