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Arun

· started a discussion

· 1 Months ago

Substitution is easier (Question: 108551.0 )

This can be easily solved using substitution.

put a = 2, b = 1
c = 2, d = 1
m = 1, n = 1

Question:
If \(\cfrac{a}{b}=\cfrac{c}{d},\) then \(\cfrac{ma+nc}{mb+nd}\)  is not equal to:
Options:
A)


\(\cfrac{a}{b}\)
B) \(\cfrac{c}{d}\)
C) \(\cfrac{a+c}{b+d}\)
D) \(\cfrac{c-a}{b-d}\)
Solution:
Ans: (d)

Let \(\cfrac{a}{b}=\cfrac{c}{d}=k\)

\(\therefore\) a = bk  and c = dk

\(\cfrac{ma+nc}{mb+nd}=\cfrac{m.bk+n.dk}{mb+nd}=k=\cfrac{a}{b}=\cfrac{c}{d}\) 

\(\therefore\) Options (a) and (b) are correct

\(\because\) \(\cfrac{a+c}{b+d}+\cfrac{bk+dk}{b+d}=k=\cfrac{ma+nc}{nb+nd}\)

Hence, option (c) is also correct

Arun

· commented

· 1 Months ago

Sorry A small correction. Put c = 4 and d = 2 [instead of 2 and 1]
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