If a:b = c:d, and e:f = g:h, then (ae+bf):(ae-bf)=?
A
(e+f)(e−f)
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B
(cg+dh)(cg−dh)
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C
(cg−dh)(cg+dh)
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D
e−fe+f
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Solution
The correct option is B(cg+dh)(cg−dh)
Solve by taking values of a,b,c,d and e,f,g and h independiently of each other a=1, b=2, c=3, d=6 and e=3, f=9, g=4 and h =12 gives (ae+bf):(ae-bf)=21: -15 =−75 Option (b) (cg+dh)(cg−dh)=84−60=−75