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Sets Practice Test 1
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© examsnet.com
Question : 21 of 33
Marks:
+1
,
-0
In a statistical investigation of 2,000 familiesof Delhi, it was found that 73 families hadneither a car nor a scooter; 849 families had acar and 287 had a scooter. The number offamilies in that group having both the car andscooter is:
1,320
1,021
1,221
1,300
None of these
Validate
Solution:
👈: Video Solution
Concept:
Use the inclusion-exclusion principle for two sets to find the intersection when the union, and individual set sizes are known.
Explanation:
Let
C
C
C
be the set of families having a car.
Let
S
S
S
be the set of families having a scooter.
Total families
=
2000
= 2000
=
2000
.
Families with neither
=
73
= 73
=
73
.
So, families with at least one vehicle,
n
(
C
∪
S
)
=
2000
−
73
=
1927
n(C \cup S) = 2000 - 73 = 1927
n
(
C
∪
S
)
=
2000
−
73
=
1927
.
Given
n
(
C
)
=
849
n(C) = 849
n
(
C
)
=
849
and
n
(
S
)
=
287
n(S) = 287
n
(
S
)
=
287
.
By inclusion-exclusion,
n
(
C
∪
S
)
=
n
(
C
)
+
n
(
S
)
−
n
(
C
∩
S
)
n(C \cup S) = n(C) + n(S) - n(C \cap S)
n
(
C
∪
S
)
=
n
(
C
)
+
n
(
S
)
−
n
(
C
∩
S
)
.
Substitute values:
1927
=
849
+
287
−
n
(
C
∩
S
)
1927 = 849 + 287 - n(C \cap S)
1927
=
849
+
287
−
n
(
C
∩
S
)
.
So,
1927
=
1136
−
n
(
C
∩
S
)
1927 = 1136 - n(C \cap S)
1927
=
1136
−
n
(
C
∩
S
)
.
Thus,
n
(
C
∩
S
)
=
1136
−
1927
=
−
791
n(C \cap S) = 1136 - 1927 = -791
n
(
C
∩
S
)
=
1136
−
1927
=
−
791
.
A negative number of families is impossible.
Therefore, the given data is inconsistent, and no valid number of families can have both a car and a scooter.
Answer:
None of these.
© examsnet.com
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