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A system in [x] and {x} with x + [y] + {z} = 1.1 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If
x+[y]+{z}=1.1,[x]+{y}+z=2.2,{x}+y+[z]=3.3,x+\left[y\right]+\left\{z\right\} = 1.1 ,\qquad \left[x\right]+\left\{y\right\}+z = 2.2 ,\qquad \left\{x\right\}+y+\left[z\right] = 3.3 ,
 then ([ ⋅ ]\left[\ \cdot\ \right] is G.I.F and { ⋅ }\left\{\ \cdot\ \right\} is fractional part)
Ax+y+z=3.3x+y+z = 3.3correct
By−2x=1y-2x = 1correct
C2(z+1)=5y2\left(z+1\right) = 5ycorrect
D{x}+{y}+{z}=0.3\left\{x\right\}+\left\{y\right\}+\left\{z\right\} = 0.3correct
Solution
Step 1: Every variable appears exactly three times across the system, once whole, once as its integer part, once as its fractional part, and [t]+{t}=t\left[t\right]+\left\{t\right\} = t, so on adding the three equations,
(x+[x]+{x})+([y]+{y}+y)+({z}+z+[z])=1.1+2.2+3.3=6.6\left(x+\left[x\right]+\left\{x\right\}\right)+\left(\left[y\right]+\left\{y\right\}+y\right) +\left(\left\{z\right\}+z+\left[z\right]\right) = 1.1+2.2+3.3 = 6.6
2(x+y+z)=6.6⟹x+y+z=3.3.(4)2\left(x+y+z\right) = 6.6 \quad\Longrightarrow\quad x+y+z = 3.3 . \tag{4}
 So (A) is true. Step 2: Subtracting (1) from (4),
(x+y+z)−(x+[y]+{z})={y}+[z]=3.3−1.1=2.2\left(x+y+z\right)-\left(x+\left[y\right]+\left\{z\right\}\right) = \left\{y\right\}+\left[z\right] = 3.3-1.1 = 2.2
 Here 0≤{y}<10 \le \left\{y\right\}<1 and [z]∈Z\left[z\right] \in \mathbb{Z}, so the only split is
{y}=0.2,[z]=2\left\{y\right\} = 0.2 ,\qquad \left[z\right] = 2
 Subtracting (2) from (4), by the same argument,
{x}+[y]=3.3−2.2=1.1⟹{x}=0.1,[y]=1\left\{x\right\}+\left[y\right] = 3.3-2.2 = 1.1 \quad\Longrightarrow\quad \left\{x\right\} = 0.1 ,\qquad \left[y\right] = 1
 Subtracting (3) from (4),
[x]+{z}=3.3−3.3=0⟹[x]=0,{z}=0\left[x\right]+\left\{z\right\} = 3.3-3.3 = 0 \quad\Longrightarrow\quad \left[x\right] = 0 ,\qquad \left\{z\right\} = 0
 since [x]=−{z}\left[x\right] = -\left\{z\right\} with [x]\left[x\right] an integer and 0≤{z}<10 \le \left\{z\right\}<1 gives [x]∈(−1,0]\left[x\right] \in \left(-1,0\right], i.e. [x]=0\left[x\right] = 0. Step 3:
x=[x]+{x}=0+0.1=0.1x = \left[x\right]+\left\{x\right\} = 0+0.1 = 0.1
y=[y]+{y}=1+0.2=1.2y = \left[y\right]+\left\{y\right\} = 1+0.2 = 1.2
z=[z]+{z}=2+0=2z = \left[z\right]+\left\{z\right\} = 2+0 = 2
 Step 4: Checking the three given equations,
x+[y]+{z}=0.1+1+0=1.1x+\left[y\right]+\left\{z\right\} = 0.1+1+0 = 1.1
[x]+{y}+z=0+0.2+2=2.2\left[x\right]+\left\{y\right\}+z = 0+0.2+2 = 2.2
{x}+y+[z]=0.1+1.2+2=3.3\left\{x\right\}+y+\left[z\right] = 0.1+1.2+2 = 3.3
 Step 5:
(1) x+y+z=0.1+1.2+2=3.3\text{(1)}\ x+y+z = 0.1+1.2+2 = 3.3
(2) y−2x=1.2−0.2=1\text{(2)}\ y-2x = 1.2-0.2 = 1
(3) 2(z+1)=2×3=6,5y=5×1.2=6\text{(3)}\ 2\left(z+1\right) = 2\times3 = 6 ,\qquad 5y = 5\times1.2 = 6
(4) {x}+{y}+{z}=0.1+0.2+0=0.3\text{(4)}\ \left\{x\right\}+\left\{y\right\}+\left\{z\right\} = 0.1+0.2+0 = 0.3
 Answer: (1), (2), (3) and (4)
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