Question: 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}} $
Options:
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
Show Answer
Answer:
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( \frac{40-3y}{4} )-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 $ .
BETA
