Answer-‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ are five friends and th..385315

Q: ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ are five friends and their average age is ____ years.

ForILetthepresentageofAxyearsPresentageofC2xyearsSumofthepresentagesofBandC36272yearsPresentageofB722xyearsPresentageofD722x613yearsAftereightyearstheageofAx8yearsAftereightyearstheageofEx8713yearsPresentageofEx87138yearsSumoftheagesofADandE40572128yearsx722x613x87138128x93626x613x1047812842x6552182x78x6243361284262x1464x146462ThereforeIcannotbetrueForIILetthepresentageofAxyearsPresentageofC2xyearsSumofthepresentagesofBandC32264yearsPresentageofB642xyearsPresentageofD642x613yearsAftereightyearstheageofAx8yearsAftereightyearstheageofEx8713yearsPresentageofEx87138yearsSumoftheagesofADandE36564116yearsx642x613x87138116x83226x613x1047811642x5824182x78x6243361164262x1240x20PresentageofEx8713844yearsPresentageofD642x61352yearsDifference52448yearsThereforeIIcanbetrueForIIILetthepresentageofAxyearsPresentageofE2xyearsSumofthepresentagesofBandC20240yearsPresentageofA402xyearsPresentageofD402x613yearsAftereightyearstheageofAx8yearsAftereightyearstheageofEx8713yearsPresentageofEx87138yearsSumoftheagesofADandE30540110yearsx402x613x87138110x52026x613x1047811042x3640182x78x6243361104262x692x69262AgecannotbenegativeThereforeIIIisfalse

Question

‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ are five friends and their average age is years. At present the age ‘A’ is half of the age of ‘C’ and the ratio of the present ages of ‘B’ to ‘D’ is 6:13. Eight years hence from now, the ratio of the ages of ‘A’ to ‘E’ will become 7:13. If the present average age of ‘B’ and ‘C’ is years, and 'E' is 8 years younger to 'D'. (The age of each person will be in whole years).

The values given in which of the following options will fill the blanks in the same order in which is it given to make the statement true:

I. 40, 36

II. 36, 32

III. 30, 20

A Only II

B Only I

C Only I and III

D Only II and III

E None of these

Solution

For I: Let the present age of ‘A’ = x years Present age of ‘C’ = 2x years Sum of the present ages of ‘B’ and ‘C’ = 36 × 2 = 72 years Present age of ‘B’ = (72 – 2x) years Present age of ‘D’ = [(72 – 2x)/6] × 13 years After eight years, the age of ‘A’ = (x + 8) years After eight years, the age of ‘E’ = [(x + 8)/7] × 13 years Present age of ‘E’ = [(x + 8)/7] × 13 – 8 years Sum of the ages of ‘A’, ‘D’ and ‘E’ = (40 × 5) – 72 = 128 years x + {[(72 – 2x)/6] × 13} + {[(x + 8)/7] × 13} - 8= 128 x + (936 – 26x)/6 + (13x + 104)/7 - 8= 128 42x + 6552 – 182x + 78x + 624 – 336 = 128 × 42 62x = 1464 x = 1464/62 Therefore, I cannot be true. For II: Let the present age of ‘A’ = x years Present age of ‘C’ = 2x years Sum of the present ages of ‘B’ and ‘C’ = 32 × 2 = 64 years Present age of ‘B’ = (64 – 2x) years Present age of ‘D’ = [(64 – 2x)/6] × 13 years After eight years, the age of ‘A’ = (x + 8) years After eight years, the age of ‘E’ = [(x + 8)/7] × 13 years Present age of ‘E’ = [(x + 8)/7] × 13 – 8 years Sum of the ages of ‘A’, ‘D’ and ‘E’ = (36 × 5) – 64 = 116 years x + {[(64 – 2x)/6] × 13} + {[(x + 8)/7] × 13} - 8= 116 x + (832 – 26x)/6 + (13x + 104)/7 - 8= 116 42x + 5824 – 182x + 78x + 624 – 336 = 116 × 42 62x = 1240 x = 20 Present age of ‘E’ = [(x + 8)/7] × 13 – 8 = 44 years Present age of ‘D’ = [(64 – 2x)/6] × 13 = 52 years Difference = 52 – 44 = 8 years Therefore, ‘II’ can be true. For III: Let the present age of ‘A’ = x years Present age of ‘E’ = 2x years Sum of the present ages of ‘B’ and ‘C’ = 20 × 2 = 40 years Present age of ‘A’ = (40 – 2x) years Present age of ‘D’ = [(40 – 2x)/6] × 13 years After eight years, the age of ‘A’ = (x + 8) years After eight years, the age of ‘E’ = [(x + 8)/7] × 13 years Present age of ‘E’ = [(x + 8)/7] × 13 – 8 years Sum of the ages of ‘A’, ‘D’ and ‘E’ = (30 × 5) – 40 = 110 years x + {[(40 – 2x)/6] × 13} + {[(x + 8)/7] × 13} - 8= 110 x + (520 – 26x)/6 + (13x + 104)/7 – 8 = 110 42x + 3640 – 182x + 78x + 624 – 336 = 110 × 42 -62x = 692 x = -692/62 Age cannot be negative. Therefore, III is false.

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