| Column - I | Column - II |
| P. If \[\frac{5m}{6}+\frac{3m}{4}=\frac{19}{12},\] then \[m=\] | (i) \[\frac{1}{6}\] |
| Q. If \[2x+\frac{3}{4}=\frac{x}{2}+1,\] then \[x=\] | (ii) \[36\] |
| R. If \[\frac{z}{2}-\frac{3z}{4}+\frac{5z}{6}=21,\] then \[z=\] | (iii) \[\frac{27}{10}\] |
| S. If \[\frac{y}{2}-\frac{1}{5}=\frac{y}{3}+\frac{1}{4},\] then \[y=\] | (iv) 1 |
A) P\[\to \](iii); Q\[\to \](iv); R\[\to \](i); S\[\to \](ii)
B) P\[\to \](iv); Q\[\to \](ii): R\[\to \](iii): S\[\to \](i)
C) P\[\to \](ii); Q\[\to \](i); R\[\to \](iii); S\[\to \](iv)
D) P\[\to \](iv); Q\[\to \](i); R\[\to \](ii); 5\[\to \](iii)
Correct Answer: D
Solution :
P. \[\frac{5m}{6}+\frac{3m}{4}=\frac{19}{12}\] \[\Rightarrow \frac{10m+9m}{12}=\frac{19}{12}\Rightarrow 19m=19\Rightarrow m=1\] \[\text{Q}\text{.}\]\[2x+\frac{3}{4}=\frac{x}{2}+1\Rightarrow \frac{4x-x}{2}=\frac{4-3}{4}\] \[\Rightarrow 4(3x)=2\Rightarrow 12x=2\Rightarrow x=\frac{2}{12}=\frac{1}{6}\] R. \[\frac{z}{2}-\frac{3z}{4}+\frac{5z}{6}=21\] \[\frac{6z-9z+10z}{12}=21\Rightarrow 7z=21\times 12\] \[\Rightarrow z=\frac{21\times 12}{7}\Rightarrow z=36\] S. \[\frac{y}{2}-\frac{1}{5}=\frac{y}{3}+\frac{1}{4}\] \[\Rightarrow \frac{y}{2}-\frac{y}{3}=\frac{1}{4}+\frac{1}{5}\Rightarrow \frac{3y-2y}{6}=\frac{5+4}{20}\] \[\Rightarrow \frac{y}{6}=\frac{9}{20}\Rightarrow y=\frac{9\times 6}{20}=\frac{9\times 3}{10}=\frac{27}{10}\]
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