Getting Fit

let x be the distance of of the course that calvin walks every day, and let the t be the time. the equation will be $x=vt$ when he is feeling energetic, his speed will be v+2 and the time will be (t/2)+(1/4), thus we get $x=(v+2)(\frac{t}{2}+\frac{1}{4})$ when he is feeling lethargic, his speed will be v-1 and the time will be t+(1/2), thus we get $x=(v-1)(t+\frac{1}{2})$ subtitute the first equation for x value $(v+2)(\frac{t}{2}+\frac{1}{4})=(v-1)(t+\frac{1}{2})$ $\frac{v+2}{v-1}=2$ $4=v$ now we can find value of t $x=vt$ $x=4t$ $4t=(4-1)(t+\frac{1}{2})$ $4t=3t+\frac{3}{2}$ $t=\frac{3}{2}$

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$x=vt$ $x=4\times 1.5$ $x=\boxed{6}$

We let $s$ to be the speed, and $t$ to be the time in minutes. Therefore, we can simplify the problem a little bit by writing the equation below,

$(s+2)(\frac{t}{2}+15)=(s-1)(t+30)$

$\frac{st}{2}+15s+t+30=st+30s-t-30$

$2t+60=\frac{st}{2}+1$

$4t+120=st+30s$

$4(t+30)=s(t+30)$

$s=4$

Now, we have to substitute $s=4$ to both sides and equate one of the sides with $\frac{d}{s}=t\implies \frac{d}{4}=t\implies d=4t$ , such that $d$ is the distance. If we equate the equation above again, it will bring us to nowhere.

$3(t+30)= 4t$

$\implies t=30$

Then, we have to convert $t$ into hours, which means that $t=\frac{1}{2} hrs$ . Substitute both values into $d=4t$ , we get $d=\boxed6$ , which is our desired answer.

Suppose that normally he walks distance $d$ at speed $v$ and in time $t$ , then we are given system $\begin{cases} v + 2 = \frac{d}{\frac{t}{2} + 0.25} \\ v - 1 = \frac{d}{t + 0.5} \\ v = \frac{d}{t} \end{cases}$ Which is very easy to solve to get $(v,d,t) = (4,6,1.5)$ , so the answer is $\boxed{6}$

Let Calvin's normal speed be $x$ km/hr and usual time be $t$ hr. Distance be $d$ km.

$d = xt = (x + 2)(\frac{t}{2}+\frac{1}{4}) = (x-1)(t+\frac{1}{2})$

$xt = (x-1)(t+\frac{1}{2}) \implies x = 2t + 1$

Putting $x = 2t + 1$ in $xt = (x + 2)(\frac{t}{2}+\frac{1}{4})$

we get $(2t + 1)t = (2t + 3)(\frac{2t + 1}{4} ) \implies 4t^2 - 4t - 3 = 0 \implies t = \frac{3}{2} or \frac{-1}{2}$

Finally,

$t = \frac{3}{2}$ hr,

$x= 2\cdot \frac{3}{2}+1=4$ km/hr,

$d = x\cdot d = \boxed6$ km.

Let the of the course distance be ${d}$

the normal speed be ${s}$

and the normal time be ${t}$

we know that ${d}={s}\times{t}.....(1)$

In the first instance ${d}=(s+2)\times(\frac{t}{2}+\frac{1}{4}).....(2)$

In the second instance ${d}=(s-1)\times(t+\frac{t}{2}).....(3)$

Note that the times given in minutes are converter into hours hence 1/4hr and 1/2hr

i.e. $(s-1)(t+\frac{t}{2})=(s+2)(\frac{t}{2}+\frac{1}{4})={st}$

Solving, $(s+2)(\frac{t}{2}+\frac{1}{4})={st}$

i.e. $\frac{st}{2}+\frac{s}{4}+{t}+{1}{2}={st}$

which simplifies to ${2st}-{4t}-{s}={2}.....(4)$

solving, $(s-1)(t+\frac{t}{2})={st}$

i.e. ${st}+\frac{s}{2}-{t}-\frac{1}{2}={st}$

which simplifies to ${s-2t}={1}$

from which ${s}={1+2t}.....(5)$

substituting.....(5) into.....(4) we get

$2t+4{t}^2-4t-1-2t=2$

which simplifies to ${4t^2}-{4t}-{3}={0}$

and $(2t-3)(2t+1)={0}$

so $(2t-3)={0}$

and ${t}=\frac{3}{2}hrs$ as the only positive value of t

But ${s}={1+2t}$

i.e. ${s}={1+2(\frac{3}{2})}$

and ${s}={4km/hr}$

from.....(1) ${d}={s}\times{t}$

i.e. ${d}={4}\times{\frac{3}{2}}$

Hence distance of the course equals $\boxed{6km}$

Distance=velocity Time In All Cases, distance travelled is same 2xy=(x+2)(y+0.25)=(x-1)(2y+0.5) 2xy=xy+0.25x+2y+0.5=2xy+0.5x-2y-0.5 xy-0.25x-2y-0.5=0 0.5x-2y-30=0 Subtracting xy-0.75x=0 x(y-0.75)=0 x!=0 implies y=0.75hrs 0.5x-2y-0.5=0 0.5x-1.5-0.5=0 x=4 Distance=2xy=2 4*0.75=6kms

Let x be the distance in kilometres and v km/h be the usual speed. We can construct two equations x/(v+2) = (1/2)(x/v) = 1/4 and x/(v-1) = x/v +1/2. Doubling the first equation we get 2x/(v+2) = x/v + 1/2. Just nice the right side of the two equations are exactly the same!!! Hence 2x/(v+2) = x/(v-1). Eliminating x we get 2(v-1) = v+2 which gives us v=4. Simple substitution and equation solving tells us that x=6.

let s=distance in km, v=normal speed and t=normal time taken to complete the course. hence, we can write (v+2) (t/2+1/4) = (v-1) (t+1/2) = v*t (here, 15min=1/4hrs and 30min=1/2hrs)

solving these equations we get,

v=4 km/hr and t=3/2=1.5 hrs

so, s=v t=4 1.5= 6 km

Distance(d) = speed(s) * Time(t) d=st= (s+2)(t/2 + ¼) ……..(i)and d=st = (s-1)(t + ½)………..(ii) or, (s+2)(t/2 + ¼) = (s-1)(t + ½) or,(s+2)(2t+1)/4 = (s-1)(2t+1)/2
or, (s+2)(2t+1) = 2(s-1)(2t+1) or, 2st+s+4t+2 = 4st+2s-4t-2 or, 2st-8t+s-4=0 or, 2t(s-4)+1(s-4)=0 or,(s-4)(2t-1)=0 s=4,t=-1/2 putting s=4 in equation(ii),we get t= 3/2 then Distance = 4*(3/2)=6

let distance be d and the speed be s

time will be d\s

* d/(s+2) = t/2 + 1/4 *

ie (d/(s+2) - 1/4 ) X 2 = t

2d/(s+2) - 1/2 = t ---- 1

d/(s-1) = t + 1/2

ie d/(s-1) - 1/2 = t ---- 2

1 and 2 =====> s=4 , d=6

s=vt s=(v+2)(t/2+1/4) s=(v-1)(t+1/2)=vt+v/2-t-1/2 v/2-t-1/2=0 or -2t=1-v...(1) s=2t+v/2+1.....(2) s=vt..........(3) vt-2t-v/2-1=0 or vt+1-v-v/2-1=0 or t=3/2 from 1 v=4 s=3/2*4=6

suppose,distance of the course is x km

speed=y

time=x/y

the equations are,

(x)/(y+2)=(x)/(2y)+(1)/(4)..........(1)

(x)/(y-1)=x/y+1/2.......(2)

Dividing equation (1) By equation (2) & using componendo-divodendo method we get the value of y=4

putting the value of y=4 in the equation (1) we get the distance, x=6 km (Ans)

Let x be the rate. Let t be the time (in hours) so 15 mins = ( $\frac{1}{4}$ ) and 30 minutes is ( $\frac{1}{2}$ ).

• 1st condition : (x + 2)( $\frac{t}{2} + \frac{1}{4}$ ) = d
• 2nd condition : (x - 1)(t + $\frac{1}{2}$ ) = d

  - 3rd condition: xt = d

Equating 1st and 2nd condition for the same d:

• $\frac{(2t + 1)(x + 2)}{4}$ = $\frac{(2t + 1)(x - 1)}{2}$
• Cancel "(2t + 1)"

  - x = 4

Substituting "4" into the values of 1st / 2nd condition and equating with the 3rd.

• $\frac{(2t + 1)(4 - 1)}{2}$ = 4t

  - t = $\frac{3}{2}$

Thus

• d = xt
• d = ( $\frac{3}{2}$ )(4)
• d = $\boxed{6}$

Let us take the speed as $x$ km/hr and time as $t$ min. We know the formula, Distance $=$ Speed $\times$ TIme. From that and according to the problem , we have two equations of the distance travelled(d) as---

$d=(x+2)(\frac{1}{2}t+15)$ ...(i) and $d=(x-1)(t+30)$ ....(ii)

From (i) and (ii), we have,

$(x+2)(\frac{1}{2}t+15)=(x-1)(t+30)$

$\implies \frac{(x+2)(t+30)}{2}=(x-1)(t+30)$

$\implies x+2=2(x-1) \implies x+2=2x-2 \implies x=\boxed{4}$

We also have the standard equation as $d=x\times t$ ....(iii)

From (ii) and (iii), we have,

$(x-1)(t+30)=xt \implies (4-1)(t+30)=4t \implies 3(t+30)=4t \implies 3t+90=4t \implies t=\boxed{90}$

So, Speed $=4$ km/hr and Time $=90$ min $=\frac{90}{60}$ hr $=\frac{3}{2}$ hr.

Distance $=$ Speed $\times$ Time $=(4\times \frac{3}{2})$ km $=(2\times 3)$ km $=\boxed{6}$ km

Let the normal speed of Calvin be x km/hr.

Let the time taken by Calvin to walk a fixed course be y hr.

Therefore, distance of the course = xy km

According to first condition:

New speed = (x+2) km/hr

New time taken = ( $\frac{y}{2}$ + $\frac{15}{60}$ ) hr

(x+2) ( $\frac{y}{2}$ + $\frac{15}{60}$ ) = xy

(x+2) ( $\frac{y}{2}$ + $\frac{1}{4}$ ) = xy

(x+2) ( $\frac{2y}{4}$ + $\frac{1}{4}$ ) = xy

(x+2) ( $\frac{2y+1}{4}$ ) = xy

$\frac{(x+2)(2y+1)}{4}$ = xy

(x+2)(2y+1) = 4xy

2xy + 4y + x + 2 = 4xy

x + 4y + 2 = 2xy ..................(i)

According to second condition:

(x-1)(y+ $\frac{1}{2}$ ) = xy

(x-1)(2y+1) = 2xy

2xy - 2y +x -1 = 2xy .................(ii)

2xy - 2y +x -1 = x + 4y +2 [From (i)]

2xy = 6y + 3

x = $\frac{6y+3}{2y}$ .......................(iii)

2xy - 2y +x -1 = 2xy [From (ii)]

x - 2y -1 = 0

$\frac{6y+3}{2y}$ - 2y - 1 = 0

$\frac{6y+3-4y^{2} - 2y}{2y}$ = 0

4y+3- $(4y)^{2}$ = 0

$(4y)^{2}$ - 4y - 3 = 0

$(4y)^{2}$ - 6y + 2y - 3 = 0

2y(2y-3) + 1(2y-3) = 0

(2y-3)(2y+1) = 0

2y - 3 = 0

y = $\frac{3}{2}$ hr

Therefore, x = 4 km/hr

xy = 6 km

Hence the distance of the course that Calvin walks every day is 6km.

Let, usual time he takes to finish the course $x$ hrs with normal speed $y$ km/hrs.

So, the total distance of the course is $xy$ km

So, when he walks faster it takes $( \frac{x}{2} + \frac{15}{60}) \implies ( \frac{x}{2} + \frac{1}{4})$ hrs with speed $(y+2)$ km/hrs and when he walks slower it takes $(x + \frac{30}{60}) \implies (x + \frac{1}{2} )$ hrs with speed $(y-1)$ km/hrs.

In both case the total distance is same $(xy)$ .

We know $speed \times time = distance$

So, we get two equation $\implies ( \frac{x}{2} + \frac{1}{4}) \times (y+2) = xy ........(i)$

And,

$\implies (x + \frac{1}{2}) \times (y-1) = xy ........(ii)$

Solving equations (i) and (ii), we get the normal time he takes $x = 1.5$ hrs. and normal speed $y = 4$ km/hrs.

So, the total distance of his course is $\implies xy \implies 1.5 \times 4 \implies \boxed{6}$ km.

Let s=normal speed, t=normal time and d=distance on solving the following equations for s and t: st = (s+2)(t/2+1/4) = (s-1)(t+1/2) we get, s=4 and t=3/2 hence, d=st=4*(3/2)=6

Let $t$ , $d$ , and $v$ be the time,distance and speed respectively . Using the formula $speed*time=distance$ , we have the following system of equations : $\\\Rightarrow \begin{cases} vt=d \\ (v+2)(\frac{t}{2}+\frac{1}{4})=d \\ (v-1)(t+\frac{1}{2})=d \end{cases} \Rightarrow \begin{cases} vt=d \\ \frac{vt}{2}+\frac{v}{4}+t+1/2=d \\ vt+\frac{v}{2}-t-1/2=d \end{cases} \Rightarrow \begin{cases} vt=d \\ \frac{d}{2}+\frac{v}{4}+t+1/2=d \\ d+\frac{v}{2}-t-1/2=d \end{cases}$

Adding the last two together: $\frac{3d}{2}+\frac{3v}{4}=2d \Rightarrow \frac{3v}{4}=\frac{d}{2} \Rightarrow \frac{3v}{2}=d$ Substituting this in $vt=d \Rightarrow t=\frac{3}{2}$

$\begin{cases} vt=d \\ \frac{d}{2}+\frac{v}{4}+t+1/2=d \\ d+\frac{v}{2}-t-1/2=d \end{cases} \Rightarrow \begin{cases} vt=d \\ 2d+v+4t+2=4d \\ 2d+v-2t-1=2d \end{cases}$

Substracting the last from the second : $6t+3=2d \Rightarrow 9+3=2d \Rightarrow \boxed{d=6}$