NCERT Solutions · Class 9 Mathematics Two Variables, One Line

A robot's corner points, after each full move, lie on a line with slope \(\displaystyle \text{rise}/\text{run}\).(i) Robot $\displaystyle 1$: through \(\displaystyle (0,0)\), slope \(\displaystyle \tfrac{4}{3}\). Robot $\displaystyle 2$: through \(\displaystyle (10,0)\), slope \(\displaystyle 2\).\[y=\tfrac{4}{3}x, \qquad y=2(x-10) \]\[\tfrac{4}{3}x=2x-20 \Rightarrow \tfrac{2}{3}x=20 \Rightarrow x=30, \quad y=40 \]Both robots stand at this corner, after $\displaystyle 10$ and $\displaystyle 20$ cycles.\[(3\cdot 10,\ 4\cdot 10)=(10+20,\ 2\cdot 20)=(30,40) \]The staircases first touch earlier, when Robot $\displaystyle 1$ ($\displaystyle 7$ cycles, then $\displaystyle 3$ right) meets Robot $\displaystyle 2$ ($\displaystyle 14$ cycles).\[(3\cdot 7+3,\ 4\cdot 7)=(10+14,\ 2\cdot 14)=(24,28) \](ii) Robot $\displaystyle 1$: through \(\displaystyle (3,0)\), slope \(\displaystyle 1\). Robot $\displaystyle 2$: through \(\displaystyle (7,0)\), slope \(\displaystyle -\tfrac{3}{5}\).\[y=x-3, \qquad y=-\tfrac{3}{5}(x-7) \]\[5x-15=-3x+21 \Rightarrow x=\tfrac{9}{2}, \quad y=\tfrac{3}{2} \]This point lies behind Robot $\displaystyle 2$'s start \(\displaystyle (7,0)\), so Robot $\displaystyle 2$ never reaches it. The staircases share only the stretch \(\displaystyle (7,0)\) to \(\displaystyle (8,0)\) on the \(\displaystyle x\)-axis.Answer: (i) Yes: the paths first touch at \(\displaystyle (24,28)\); the lines meet at \(\displaystyle (30,40)\). (ii) Yes: along the \(\displaystyle x\)-axis from \(\displaystyle (7,0)\) to \(\displaystyle (8,0)\); the lines meet at \(\displaystyle (4.5,\,1.5)\), behind Robot $\displaystyle 2$'s start.

Measure time in minutes after $\displaystyle 12$ o'clock: the first reading is \(\displaystyle 120+a\), the second is \(\displaystyle 180+b\).\[(120+a)+15=180+b \Rightarrow a-b=45 \]\[a=6b \]\[6b-b=45 \Rightarrow b=9, \quad a=54 \]Check:\[\text{2:54}+15\text{ min}=\text{3:09} \]Answer: The second time was $\displaystyle 3$:$\displaystyle 09$

Let the tens digit be \(\displaystyle x\) and the units digit \(\displaystyle y\), so the number is \(\displaystyle 10x+y\).\[x+y=15 \]\[(10y+x)-(10x+y)=9 \Rightarrow y-x=1 \]\[2y=16 \Rightarrow y=8, \quad x=7 \]Check:\[7+8=15, \qquad 87-78=9 \]Answer: The number is $\displaystyle 78$

Opposite angles of a cyclic quadrilateral are supplementary.\[\angle A+\angle C=180^\circ \Rightarrow (x+7)+(3y+23)=180 \Rightarrow x+3y=150 \quad (1) \]\[\angle B+\angle D=180^\circ \Rightarrow (y+8)+(4x+12)=180 \Rightarrow 4x+y=160 \quad (2) \]\[(2) \Rightarrow y=160-4x \]\[(1) \Rightarrow x+3(160-4x)=150 \Rightarrow 480-11x=150 \Rightarrow x=30, \quad y=40 \]\[\angle A=37^\circ, \quad \angle B=48^\circ, \quad \angle C=143^\circ, \quad \angle D=132^\circ \]Check:\[37+143=180, \qquad 48+132=180 \]Answer: \(\displaystyle \angle A=37^\circ,\ \angle B=48^\circ,\ \angle C=143^\circ,\ \angle D=132^\circ\)

Let the speed be \(\displaystyle x\) km/h and the time \(\displaystyle y\) hours, so the distance is \(\displaystyle xy\) km.\[(x+6)(y-6)=xy \Rightarrow 6y-6x-36=0 \Rightarrow y-x=6 \]\[(x-4)(y+6)=xy \Rightarrow 6x-4y-24=0 \Rightarrow 3x-2y=12 \]\[3x-2(x+6)=12 \Rightarrow x=24, \quad y=30 \]\[xy=24\times 30=720 \]Check:\[30\times 24=720, \qquad 20\times 36=720 \]Answer: Distance $\displaystyle 720$ km, speed $\displaystyle 24$ km/h

Let the father's age be \(\displaystyle x\) years and the sum of the four children's ages \(\displaystyle y\) years. In $\displaystyle 20$ years each child gains $\displaystyle 20$, so the sum gains \(\displaystyle 4\times 20=80\).\[x=y \]\[y+80=2(x+20) \Rightarrow 2x-y=40 \]\[2x-x=40 \Rightarrow x=40 \]Check:\[40+80=120=2(40+20) \]Answer: The father is $\displaystyle 40$ years old