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Question Numbers: 150-152Three schools namely A, B, and C have different numbers of boys and girls. In School A, if the number of boys were increased by 5 and the number of girls were reduced by 5, then the girls would be 20% of the boys in that school. The total number of students in School B is 10 less than twice the number of boys in School A. In School B, boys constitute 60% of the total students. The total number of students in School C is half of the total number of students in School B. The boys in School C are one-third of the boys in School B, while the girls in School C are 9 fewer than the number of boys in School A.
Solution:
Concept:Use the given relationships between Schools A, B, and C to find the initial strength, then apply the transfer and calculate the required percentage.
Explanation:Let
Ab​,
Ag​ be the boys and girls in School A;
Bb​,
Bg​ in School B; and
Cb​,
Cg​ in School C.
Given:
(Ag​−5)=20% of
(Ab​+5)⇒Ag​−5=51​(Ab​+5)⇒5Ag​−25=Ab​+5⇒Ab​=5Ag​−30 ... (i)
Also, Total B
=2Ab​−10,
Bb​=60% of Total B, Total C
=2Total B​,
Cb​=3Bb​​, and
Cg​=Ab​−9.
Since Total C
=Cb​+Cg​:
⇒2Total B​=3Bb​​+(Ab​−9)Substituting
Bb​=0.6×Total B:
⇒0.5×Total B=0.2×Total B+Ab​−9⇒0.3×Total B=Ab​−9Using Total B
=2Ab​−10:
⇒0.3(2Ab​−10)=Ab​−9⇒0.6Ab​−3=Ab​−9⇒0.4Ab​=6⇒Ab​=15From (i):
15=5Ag​−30⇒Ag​=9Total A
=15+9=24Total B
=2(15)−10=20Bb​=60% of
20=12, so
Bg​=8Total C
=220​=10Cb​=312​=4, and
Cg​=15−9=6Boys transferred from B to C
=25% of
12=3New boys in C
=4+3=7New girls in C
=6+2=8New total strength of C
=7+8=15Required percentage
=157​×100=46.66%Answer:46.66%, which is Option B.
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