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    Question

    The average of four numbers β€˜a’, β€˜b’, β€˜c’

    and β€˜d’ is 60. The value of β€˜c’ is 16.67% more than the value of β€˜d’. The value of β€˜a’ is 8 more than the value of β€˜c’. The ratio between the values of β€˜b’ and β€˜d’ is 3:2 respectively. Which of the following statements is/are true? (It is assumed that each of the numbers is natural.) (i) The value of β€˜d’ is the multiple of 6. (ii) The value of β€˜a’ is 16 more than the value of β€˜d’. (iii) The value of β€˜c’ is 54.
    A Only (i) and (iii) Correct Answer Incorrect Answer
    B Only (ii) and (iii) Correct Answer Incorrect Answer
    C Only (i) and (ii) Correct Answer Incorrect Answer
    D Only (iii) Correct Answer Incorrect Answer
    E All (i), (ii) and (iii) Correct Answer Incorrect Answer

    Solution

    The average of four numbers β€˜a’, β€˜b’, β€˜c’ and β€˜d’ is 60. a+b+c+d = 60x4 = 240Β  Β  Eq.(i) The value of β€˜c’ is 16.67% more than the value of β€˜d’.Β  Let’s assume β€˜d’ = 6y. c = (100+16.67)% of 6y = 116.67% of 6y = (7/6) of 6yΒ  Β  Β  Β  Β  [we know that 16.67% = (1/6).] = 7y The value of β€˜a’ is 8 more than the value of β€˜c’. a = (7y+8) The ratio between the values of β€˜b’ and β€˜d’ is 3:2 respectively.Β  b = (6y/2)x3 = 9y Put the values of β€˜a’, β€˜b’, β€˜c’ and β€˜d’ in terms of β€˜y’ in Eq.(i). (7y+8)+9y+7y+6y = 240 29y+8 = 240 29y = 240-8 = 232 y = 8 So β€˜a’ = (7x8+8) = (56+8) = 64 b = 9x8 = 72 c = 7x8 = 56 d = 6x8 = 48 (i) The value of β€˜d’ is the multiple of 6. The above given statement is true. Because the value of β€˜d’ is the multiple of 6. (ii) The value of β€˜a’ is 16 more than the value of β€˜d’. The above given statement is true. Because β€˜a’ = d+16. 64 = 48+16 (iii) The value of β€˜c’ is 54. The above given statement is not true. Because β€˜c’ = 56.

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