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firas92
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19 Aug 2020, 12:05
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:49) correct
41%
(02:51)
wrong
based on 216
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In a survey of readers of books A,B and C, it was found that 14 people read book A, 18 people read book B and 15 people read book C. If 2 people read all the three books, which of the following can be the number of people who read exactly one book?
I. 19
II. 20
III. 21
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
I. 19
II. 20
III. 21
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
19 Aug 2020, 19:18
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Since we are surveying only readers of books, then (A U B U C)' = 0 (Those who read none of the 3 types)
Refer to the diagram. Let x, y z be those who read A and B, B and C and A and C only respectively.
Those who read only A = 14 - (x + z + 2)
Those who read only B = 18 - (x + y + 2)
Those who read only C = 15 - (y + z + 2)
Therefore those who read exactly one = 14 - (x + z + 2) + 18 - (x + y + 2) + 15 - (y + z + 2)
= 47 - 2x - 2y - 2z - 6
= 41 - 2(x + y + z)
2(x + y + z) will be an even number. 41 - an even number will always be odd. Therefore II can never be possible.
To get I:
2(x + y + z) = 41 - 19 = 12
x + y + z = 6. This is possible if x = 2, y = 2 and z = 2
Substituting, we get positive values for Only A, Only B and only C. (None of them should be negative)
To get III:
2(x + y + z) = 41 - 21 = 20
x + y + z = 10. This is possible if x = 4, y = 3 and z = 3
Substituting, we get positive values for Only A, Only B and only C.
Therefore both I and III only are possible
Option D
Arun Kumar
Refer to the diagram. Let x, y z be those who read A and B, B and C and A and C only respectively.
Those who read only A = 14 - (x + z + 2)
Those who read only B = 18 - (x + y + 2)
Those who read only C = 15 - (y + z + 2)
Therefore those who read exactly one = 14 - (x + z + 2) + 18 - (x + y + 2) + 15 - (y + z + 2)
= 47 - 2x - 2y - 2z - 6
= 41 - 2(x + y + z)
2(x + y + z) will be an even number. 41 - an even number will always be odd. Therefore II can never be possible.
To get I:
2(x + y + z) = 41 - 19 = 12
x + y + z = 6. This is possible if x = 2, y = 2 and z = 2
Substituting, we get positive values for Only A, Only B and only C. (None of them should be negative)
To get III:
2(x + y + z) = 41 - 21 = 20
x + y + z = 10. This is possible if x = 4, y = 3 and z = 3
Substituting, we get positive values for Only A, Only B and only C.
Therefore both I and III only are possible
Option D
Arun Kumar
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anish777
Joined: 18 Feb 2021
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20 Jan 2024, 00:32
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Consider 20 as 8+7+5. Now, for A which has 14, we can divide this as 7+5+2, where 7 is common b/w A and B. For B, we can divide this as 7+2+1+8, where 8 is in common b/w B and C. C has 15, 5+2+7, where 5 is common b/w C and A. We can see that 20 is a possibility over here. Can anyone check ?
