I'm Up and Down, and Round and Round — CBSE Class 9 Maths Important Questions

Answer: (a) Straight line.

Every equation of the form $ax+by+c=0$ax+by+c=0 (with $a,b$a,b not both zero) has infinitely many solutions whose plotted points all lie on one straight line.

Answer: (b) $(x,0)$(x,0).

On the x-axis the vertical distance (the y-coordinate) is always $0$0, so points look like $(x,0)$(x,0), e.g. $(3,0)$(3,0) or $(-5,0)$(-5,0).

Answer: (a) $(2,6)$(2,6).

Substitute $x=2$x=2: $y=3\times2=6$y=3×2=6, so $(2,6)$(2,6) satisfies $y=3x$y=3x. The others fail: e.g. for $(6,2)$(6,2), $3\times6=18\neq2$3×6=18≠2.

Answer: (c) Blue.

The cycle length is $4$4. Divide the position by $4$4: $27=4\times6+3$27=4×6+3, remainder $3$3. The $3$3rd item in the cycle (Red, Green, Blue, Yellow) is Blue.

Answer: (a) Both A and R are true and R is the correct explanation of A. Since $x$x stays $3$3 for every $y$y, all points like $(3,0),(3,1),(3,2)$(3,0),(3,1),(3,2) line up vertically, giving a line parallel to the y-axis.

Choose values of $x$x and compute $y$y.

Take $x=0$x=0: $y=2(0)+1=1$y=2(0)+1=1, giving $(0,1)$(0,1).

Take $x=1$x=1: $y=2(1)+1=3$y=2(1)+1=3, giving $(1,3)$(1,3).

So two solutions are $(0,1)$(0,1) and $(1,3)$(1,3).

Meets the x-axis: put $y=0$y=0: $2x=6\Rightarrow x=3$2x=6 x=3, giving $(3,0)$(3,0).

Meets the y-axis: put $x=0$x=0: $y=6$y=6, giving $(0,6)$(0,6).

Substitute each value of $x$x into $y=x+2$y=x+2:

$x=0\Rightarrow y=0+2=2$x=0 y=0+2=2

$x=1\Rightarrow y=1+2=3$x=1 y=1+2=3

$x=2\Rightarrow y=2+2=4$x=2 y=2+2=4

Table:

The three points to plot are $(0,2)$(0,2), $(1,3)$(1,3) and $(2,4)$(2,4); joining them gives the straight line $y=x+2$y=x+2.

The line cuts the y-axis at $(0,1)$(0,1), so the y-intercept is $c=1$c=1.

Slope $m=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{5-1}{2-0}=\dfrac{4}{2}=2$m=change in y/change in x=5-1/2-0=4/2=2.

Therefore the equation is $y=2x+1$y=2x+1.

The repeating block $142857$142857 has length $6$6.

Divide the position by $6$6: $100=6\times16+4$100=6×16+4, remainder $4$4.

The $4$4th digit of the block $1,4,2,8,5,7$1,4,2,8,5,7 is $\mathbf{8}$8.

So the $100$100th digit after the decimal point is $8$8.

Make a table of solutions:

$x=0\Rightarrow y=2(0)-1=-1$x=0 y=2(0)-1=-1

$x=1\Rightarrow y=2(1)-1=1$x=1 y=2(1)-1=1

$x=2\Rightarrow y=2(2)-1=3$x=2 y=2(2)-1=3

Plot $(0,-1)$(0,-1), $(1,1)$(1,1), $(2,3)$(2,3) and join them with a straight line:

Extending the line to $x=3$x=3 gives $y=2(3)-1=5$y=2(3)-1=5. So when $x=3$x=3, $y=5$y=5.

One full turn takes $12$12 minutes, so the motion is periodic with period $12$12.

(a) The rider returns to the bottom after each full turn: first at $t=12$t=12 minutes.

(b) Reduce $30$30 modulo $12$12: $30=12\times2+6$30=12×2+6, remainder $6$6. At $6$6 minutes the rider is half-way round, i.e. at the top. So at $t=30$t=30 min the rider is at the top.

(c) The top is reached at the middle of each turn: $t=6$t=6, then every $12$12 minutes after: $t=6,18,30,42,54$t=6,18,30,42,54 minutes (all within the first $60$60 minutes).

(i) Put $x=2$x=2: $y=15\times2=30$y=15×2=30 km.

(ii) Put $x=0$x=0: $y=0$y=0, so the graph cuts the y-axis at the origin $(0,0)$(0,0).

(iii) Yes. $y=15x$y=15x is a linear equation in two variables, and every such equation has a straight-line graph.

(iv) Put $y=60$y=60: $60=15x\Rightarrow x=\dfrac{60}{15}=4$60=15x x=60/15=4 hours.