A) 6
B) 9
C) 12
D) \[\frac{66}{7}\]
Correct Answer: C
Solution :
a : b : c = 2 : 3 : 4 \[\therefore \]\[\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\](let)\[\Rightarrow \]\[a=2k,\]\[b=3k\]and \[c=4k\] \[\therefore \] \[2a-3b+4c=33\] \[\Rightarrow \] \[2\times 2k-3\times 3k+4\times 4k=33\] \[\Rightarrow \] \[4k-9k+16k=33\] \[\Rightarrow \] \[11k=33\]\[\Rightarrow \]\[k=\frac{33}{11}=3\] \[\therefore \]\[c=4k=4\times 3=12\]
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