Directions : Given below is the information about wind mills in four different…
2018
Directions : Given below is the information about wind mills in four different villages A, B and C and D. Number of wind mills in villages A, B, C and D are 24, 20, 15 and 12 respectively. Number of electricity units produced in one week by one wind mill when they operate with maximum efficiency in village A, B, C and D is 2 lakh units/week, 80000 units/ week, 1 Lakh units/week and 1.5 Lakh units/week respectively. Number of houses in each village A, B, C and D are 540, 240, 150 and 350 respectively. Total units produced are consumed equally by each house in the village
→ Different number of winds mills are operate in four different weeks In first week number of wind mills are operative in village A, B, C and D are 75%, 50%, 40% and 75% respectively. In second week it is 50%, 75%, 60% and 50% respectively. In third week it is 75%, 100%, 80% and 50% respectively. In fourth week it is 100%, 50%, 60% and 75% respectively.
→ Given below is the three ranges of efficiency of a wind mills (number of unit produced /Week by one mill)
Efficiency Type | Range |
|---|---|
Efficiency 1 | 60% - 70% |
Efficiency 2 | 45% - 55% |
Efficiency 3 | 30% - 40% |
Three wind mills also operate on different levels
→ level 1 : Consider upper limit of range of efficiency
→ level 2 : consider mid of range of efficiency
→ level 3 : consider the lower range of efficiency
Eg. If a wind mill is operative at efficiency 2 then its level 2 efficiency will be = (45+55)/2 = 50%
Its level 1 efficiency will be 55%
Its level 3 efficiency will be 45%
Total units produced in village C in second and fourth week at level 1 of efficiency range 1 is what percent of total units produced in village A in first and fourth week at level 2 of efficiency range 1
- A.
25⁷⁄₁₃%
- B.
23²¹⁄₂₇₃%
- C.
13¹²⁄₁₃%
- D.
22⁵⁄₁₃%
- E.
24⁵⁄₁₃%
Show answer & explanation
Correct answer: B
Concept
Units produced by a set of wind mills = (number of operative mills) x (maximum units one mill makes per week) x (efficiency factor). Within an efficiency RANGE, the working level fixes which factor to use: level 1 takes the UPPER limit of the range, level 2 takes the MID-value, level 3 takes the LOWER limit. Efficiency Range 1 is 60%-70%, so its level 1 = 70% and its level 2 = (60+70)/2 = 65%. To express one quantity as a percent of another, divide the first by the second and multiply by 100.
Application
Numerator - Village C, weeks 2 and 4, level 1 of Range 1 (factor = 70%). Village C has 15 mills, each making 1 lakh = 1,00,000 units/week at maximum efficiency.
Operative mills in week 2 = 60% of 15 = 9; in week 4 = 60% of 15 = 9. Total = 18 mills.
Units = 18 x 1,00,000 x 70% = 18 x 1,00,000 x 0.70 = 12,60,000 units.
Denominator - Village A, weeks 1 and 4, level 2 of Range 1 (factor = 65%). Village A has 24 mills, each making 2 lakh = 2,00,000 units/week at maximum efficiency.
Operative mills in week 1 = 75% of 24 = 18; in week 4 = 100% of 24 = 24. Total = 42 mills.
Units = 42 x 2,00,000 x 65% = 42 x 2,00,000 x 0.65 = 54,60,000 units.
Required percent = (12,60,000 / 54,60,000) x 100. Cancel the common 1,00,000 and the common factor: 126/546 = 3/13, so the value = (3/13) x 100 = 300/13 %.
Cross-check
300/13 = 23 remainder 1, i.e. 23 1/13 %. Writing the same fraction with denominator 273 (= 13 x 21): 1/13 = 21/273, so 23 1/13 % = 23 21/273 %. Reverse check: 54,60,000 x (300/13)/100 = 54,60,000 x 3/13 = 12,60,000, which is exactly the numerator, confirming the result.