A can complete a work in 12 days. B is 60% more efficient than A. The number of days taken by B to finish the same work is<\/p>\n
[amp_mcq option1=”6″ option2=”6\u00bd” option3=”7\u00bd” option4=”8″ correct=”option3″]<\/p>\n
B is 60% more efficient than A.
\nB’s efficiency = A’s efficiency + 60% of A’s efficiency
\nB’s efficiency = A’s efficiency * (1 + 60\/100) = A’s efficiency * (1 + 0.60) = 1.60 * A’s efficiency.
\nB’s efficiency = 1.60 * (W \/ 12) units\/day.
\nB’s efficiency = (1.6 \/ 12) * W units\/day
\nB’s efficiency = (16 \/ 120) * W units\/day
\nB’s efficiency = (2 \/ 15) * W units\/day.<\/p>\n
Let the number of days taken by B to finish the same work be D days.
\nWork done by B = B’s efficiency * Number of days
\nW = (2W \/ 15) * D<\/p>\n
To find D, we can divide both sides by W (assuming W > 0, which is true for work): A can complete a work in 12 days. B is 60% more efficient than A. The number of days taken by B to finish the same work is [amp_mcq option1=”6″ option2=”6\u00bd” option3=”7\u00bd” option4=”8″ correct=”option3″] This question was previously asked in UPSC CISF-AC-EXE – 2019 Download PDFAttempt Online Let the total amount of work to be … <\/p>\n
\n1 = (2 \/ 15) * D
\nD = 15 \/ 2
\nD = 7.5 days.
\n7.5 days is equal to 7\u00bd days.
\n<\/section>\n
\n– If A takes $T_A$ days, A’s efficiency is proportional to $1\/T_A$.
\n– If B is X% more efficient than A, B’s efficiency = $(1 + X\/100) \\times$ A’s efficiency.
\n<\/section>\n
\nRatio of efficiencies (A:B) = 100 : 160 = 10 : 16 = 5 : 8.
\nRatio of time taken (A:B) = 1\/Efficiency ratio (B:A) = 1\/(160:100) = 160:100 reciprocal = 100:160 = 5:8.
\nSo, Time taken by A \/ Time taken by B = 8 \/ 5.
\n12 \/ D = 8 \/ 5
\n8 * D = 12 * 5
\nD = (12 * 5) \/ 8 = 60 \/ 8 = 15 \/ 2 = 7.5 days.
\nThis approach using ratios confirms the result.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"