Question

Vince went on a 3 day hiking trip, he walked 3/4 the distance that he walked the day before. He walked 83.25 kilometers total in the trip. How far did Vince walked on the 1st day of the trip ?

Answer 1

Answer:

Vince walked 36 km on the 1st day of the trip.

Step-by-step explanation:

• Vince went on a 3-day hike trip.

• It is stated that Vince walks 3/4 of the distance that he walks the day before.

• His total traveled distance = 83.25

• Let 'x' be the distance traveled on the first day.

As he walks 3/4 of the distance that he walks the day before.

• so the second-day traveling distance = 3/4x

• And the third-day traveling distance = 3/4 × 3/4x

                                                                     = 9/16x

Given his total traveled distance = 83.25

Therefore, the equation becomes

$x\:+\:\frac{3}{4}x\:+\:\frac{9}{16}x\:\:=\:83.25$

Multiply both sides the 16 (L.C.M)

$x\cdot \:16+\frac{3}{4}x\cdot \:16+\frac{9}{16}x\cdot \:16=83.25\cdot \:16$

simplify

$16x+12x+9x=1332$

$37x=1332$

Divide both sides by 37

$\frac{37x}{37}=\frac{1332}{37}$

$x=36$ km

Therefore, Vince walked 36 km on the 1st day of the trip.