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Sumit jain

· started a discussion

· 1 Months ago

Question display (Question: 113227.0 )

Question and its is not shown in proper language,
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Question:

A man invested \(\cfrac{1}{3}\) of his capital at 7%, \(\cfrac{1}{4}\) at 8% and the remaining at 10%. If his annual income is Rs. 561, then the capital will be :

Options:
A)
\(\unicode{x20B9} \)5600
B) \(\unicode{x20B9} \)6000
C) \(\unicode{x20B9} \)5000
D) \(\unicode{x20B9} \)6600
Solution:
Ans: (d) Let the capital be \(\unicode{x20B9} \) x
From the question
\(\cfrac{1}{3}\)x of 7% + \(\cfrac{1}{4}\)x of 8% + \(\left [ 1-\left ( \cfrac {1}{3}+\cfrac {1}{4} \right ) \right ]\)x of 10% = \(\unicode{x20B9} \)561
or \(\cfrac{\cfrac{1}{3}x×7}{100}+\cfrac{\cfrac{1}{4}x×8}{100}+\cfrac{\cfrac{5}{12}x × 10}{100}\)  = \(\unicode{x20B9} \)561
or \(\cfrac{7}{3}x+2x+\cfrac{25}{6}\) = 561 × 100
or x = \(\cfrac{561×100×6}{51}\)    = \(\unicode{x20B9} \)6600

Knowledge Expert

· commented

· 1 Months ago

Dear Student
Let the total capital be "C"
1. Investing 7% of (C/3)
2. investing 8% of (C/4)
3. Investing 10% of remaining C i.e. apart from previously invested C/3 and C/4.
7/100*C/3+8/100*C/4+10/100*(1-7C/12)
Here 7C/12 is C/3+C/4
and (1-7C/12) is amount invested apart from C/3+C/4
I hope now both the language of question and solution is clear to you.
Thanks
Regards
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