| Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer. [IBPS (RRB) Grade A 2012] |
| I. \[821{{x}^{2}}-757{{x}^{2}}=256\] |
| II. \[\sqrt{196}{{y}^{3}}-12{{y}^{3}}=16\] |
A) If \[x>y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x\le y\]
E) If \[x=y\] or the relationship cannot be established
Correct Answer: D
Solution :
| I. \[821{{x}^{2}}-757{{x}^{2}}=256\] |
| \[\Rightarrow \]\[64\,\,{{x}^{2}}=256\] |
| \[\Rightarrow \]\[{{x}^{2}}=\frac{256}{64}=4\] |
| \[\therefore \]\[x=\pm \,\,2\] |
| II. \[\sqrt{196}{{y}^{3}}-12{{y}^{3}}=16\] |
| \[\Rightarrow \]\[14{{y}^{3}}-12{{y}^{3}}=16\] |
| \[\Rightarrow \]\[2{{y}^{3}}=16\]\[\Rightarrow \]\[{{y}^{3}}=8\] |
| \[\therefore \]\[y=2\] |
| Hence, \[x\le y\] |
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