A gasoline problem in words?

Question

In a family there are two cars. Car A gets 35 miles per gallon
Car B gets 15 miles per gallon
Combines the two cars drive 1325 miles in a week
total gas consumption of 55 gallons
How many gallons were consumed by each of the two cars that week?
CAR A=
CARB=

Answer 1

What a lot of equations and algebra the other answerers want to use. Who needs it?

If car B used all the gas, and car A none, the distance would be 55 * 15 = 825 miles. That's too little by 500 miles.

Every gallon used by car A instead of car B gives you 20 extra miles. How many of those 20s to make the 500 we need? Looks like 25 to me. So 25 gallons for car A, and 55 less 25 = 30 gallons for car B.

Answer 2

This is a system of equations where

35A + 15B = 1325 and A + B = 55

Use the second equation to solve for A or B; I'll solve for A.

A + B = 55
- B - B
A = 55 - B

Now substitute 55 - B in for A in the other equation.

35(55 - B) + 15B = 1325
1925 - 35B + 15B = 1325 (distribute)
1925 - 20B = 1325 (combine like terms)
-1925 -1925
-20B = -600
/-20 /-20 (divide both sides by -20)

B = 30

Car B used 30 gallons and car A used the remaining 25 gallons.

Of course you should check your answer.
35(25) + 15(30)
875 + 450
1325, The answer checks.

Car A = 25, Car B = 30

Answer 3

let a =miles by car A & b=miles by car B
a+b=1325
a=1325-b

a/35+b/15=55
3a+7b=55*3*5*7=5775
3(1325-b)+7b=5775
3975+7b-3b=5775
4b=5775-3975=1800
b=450
a=1325-b=1325-450=875

check
450/15=30 gal
875/35=25 gal
30+25=55 gal
450+875 1325 miles
Car A 875 miles
Car C 450 miles

Answer 4

x gals + y gals = 55 gals

x gals (35 miles / gal) + y gals(15 miles / gal ) = 1325 miles

So, multiply the first equation by -15.

-15x + -15y = -825

35x +15y = 1325

Add them together

20x = 500

x = 25 gallons

y = 30 gallons

Regards,

Mysstere

Answer 5

35A + 15B = 1325
A + B = 55

Solve these two equations simultaneously and you will have the answer.