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Question Numbers: 40-43 Directions: Read the data carefully and answer the following questions.
Three individuals, A, B, and C, were given a set of questions. They attempted some of them. The number of questions attempted by B is 50% of the total questions he was given. A was given 20 fewer questions than B. C was given 130 questions. The combined number of questions attempted by A and C is 145. The total number of questions given to all three is twice the total number of questions they attempted. Additionally, the number of questions attempted by B and C is the same.
Solution:
Concept:Use the given relationships among the number of questions given to A, B, C and the number attempted to find the unattempted count for each, then sum them.
Explanation:Let
GA​,
GB​,
GC​ = questions given to A, B, C.
Let
AA​,
AB​,
AC​ = questions attempted by A, B, C.
From given conditions:
AB​=0.5×GB​ (B attempted 50% of his given)
GA​=GB​−20 (A was given 20 fewer than B)
GC​=130AA​+AC​=145GA​+GB​+GC​=2×(AA​+AB​+AC​) (total given is twice total attempted)
AB​=AC​ (attempted by B and C equal)
Substitute
GA​=GB​−20 and
GC​=130 into the total given equation:
(GB​−20)+GB​+130=2(AA​+AB​+AC​)2GB​+110=2(AA​+AB​+AC​)Divide by 2:
GB​+55=AA​+AB​+AC​Using
AA​+AC​=145 and
AB​=AC​:
GB​+55=145−AC​+AC​+AC​ (since
AA​=145−AC​)
GB​+55=145+AC​ →
GB​=90+AC​Also
AB​=0.5×GB​ and
AB​=AC​, so
AC​=0.5×GB​.
Substitute:
GB​=90+0.5×GB​ →
0.5×GB​=90 →
GB​=180.
Then
AC​=0.5×180=90,
AB​=90,
AA​=145−90=55,
GA​=180−20=160,
GC​=130.
Questions not attempted = Given − Attempted:
A:
160−55=105B:
180−90=90C:
130−90=40Sum =
105+90+40=235.
Answer:235 (Option A)
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