Question

With uniform speed a car covers the distance in 10 hrs. Had tge speed been increased by 6 km/hr the same distance could have beeb covered in 7 1/3 hrs. What is the distance covered?

Answer 1

♠ ANSWER :

Let the distance covered be x m.

Time taken, t1= 10 hrs.

Speed, S1 = x / 10., S = x / 10

Increment in speed = 6 km / hr.

Time taken, t2 = 7 1/3 = 22/3 hrs.

Speed = x / 22 /3.

S + 6 = 3x / 22.

x / 10 + 6 =3 x / 22

6 = 3x / 22 - x / 10

6 = 15 x - 11 x / 110

4 x / 110 = 6 m.

x = 6 * 110 / 4 m.

x = 660 / 4

x = 165 m.

➡️ Distance covered = 165 m.

Answer 2

Let, the distance covered by the car be $x$ km with the uniform speed of $v$ $km$.

So, $v$ = $\frac{x}{10}$ ---- ($i$)

Acc. to question,$v$ + 6 = $\frac{x}{7\frac{1}{3}}$
or
$v$ = $\frac{x}{7\frac{1}{3}}$ - 6 ---- ($ii$)

From ($i$) and ($ii$), we get

$\frac{x}{10}$ = $\frac{x}{7\frac{1}{3}}$ - 6$\frac{x}{10}$ = $\frac{3x}{22}$ - 6$\frac{x}{10}$ = $\frac{3x - 132}{22}$22 × $\frac{x}{10}$ = 3$x$ - 13222
$x$ = ( 3$x$ - 132 ) × 1022$x$ = 30
$x$ - 132030
$x$ - 22$x$ = 13208$x$ = 1320

Therefore, $x$ = $\frac{1320}{8}$ = 165

Hence, distance travelled by the car = 165$km$
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