question_answer

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, \[x>y\].