SATHEE: Quantitative Aptitude Ques 1845

Quantitative Aptitude Ques 1845

Question: When x is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of x? [IBPS (PO/MT) CWE 2014]

Options:

A) 8

B) 6

C) 4

D) 5

E) None of these

Show Answer

Answer:

Correct Answer: E

Solution:

Here, $ 9-x:15-x::15-x:27-x $ are in continued proportion.

$ \therefore $ Product of extremes = Product of means $ {{(15-x)}^{2}}=(9-x)(27-x) $

$ \Rightarrow $ $ 225-30x+x^{2}=243+x^{2}-36x $

$ \Rightarrow $ $ 6x=243-225 $
$ \Rightarrow $ $ 6x=18 $
$ \Rightarrow $ $ x=3 $

$ \therefore $ Numbers become, $ 9-x=9-3=6 $ $ 15-x=15-3=12 $ and $ 27-x=27-3=24 $ Hence, 6: 12: 24 are in continued proportion on subtracting 3 from 9, 15 and 27.

Quantitative Aptitude Ques 1844

Quantitative Aptitude Ques 1846

Quantitative Aptitude Ques 1845

Question: When x is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of x? [IBPS (PO/MT) CWE 2014]

Options:

Answer:

Solution:

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