Directions: In these given questions two equations are given. You have to solve both the equations and give answer. [IBPS RRB (Office Assistant) 2014] |
I. \[4x+3y=\,\,{{(1600)}^{1/2}}\] |
II. \[6x-5y={{(484)}^{1/2}}\] |
A) If \[x\le y\]
B) If \[x>y\]
C) If \[x<y\]
D) If \[x\ge y\]
E) If \[x=y\] or relationship cannot be established
Correct Answer: B
Solution :
(b)I. \[4x+3y=40\] | ... (i) |
II. \[6x-5y=22\] | ... (ii) |
On multiply Eq, (i) by 6 and Eq. (ii) by 4 and than subtracting, we get | |
i.e. | |
On putting the value of y in Eq. (i), we get \[4x+3\times 4=40\] | |
\[\Rightarrow \] \[4x=40-12\]\[\Rightarrow \]\[4x=28\] | |
\[\therefore \] \[x=7\] | |
Hence, \[x>y\] | |
Alternate Method | |
\[4x+3y=40\] | |
\[x=\frac{40-3y}{4}\] | |
On putting the value of x in Eq. (ii), we get \[6\,\,(x)-5y=22\] | |
\[\Rightarrow \]\[6\left( \frac{40-3y}{4} \right)-5y=22\] \[\Rightarrow \] \[\frac{120-9y}{2}-5y=22\] | |
\[\Rightarrow \]\[120-9y-10y=44\] \[\Rightarrow \] \[-19y-76\]\[\Rightarrow \]\[y=4\] | |
On putting the value of y in Eq, (i), we get \[4x+3\,\,(4)=40\] | |
\[\Rightarrow \] \[4x=40-12\]\[\Rightarrow \]\[x=\frac{28}{4}\] | |
\[\therefore \] \[x=7\] | |
Hence, \[x>y\]. |
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