Math Problems

Algebra Problems:

Age Problem –

Given-

Jason is 4 times older than Bob at present. 8 years ago Jason was 12 times older. How old are Jason & Bob?

Solution-

x for Bobs age and

4x for Jason

So 4x = x

But 8 years ago Jason was 12 times older so

4x - 8 = 12(x-8)

4x - 8 = 12x - 96

88 = 8x 

x = 11 So Bob is currently 11 years old and Jason is 4(11) or 44 years old.

Given-

Bob is currently twice as old as Steve. Twenty years ago Bob was 6 times as old as Steve. What are their current ages?

Solution-

2x - 20 for Bob

and 

x - 20 for Steve

but, 20 years ago Bob was 6 times older so

2x - 20 = 6(x - 20)

2x - 20 = 6x - 120

4x = 100

x = 100/4

x = 25

So x or Bob is 25 and Steve is twice or 2(25): 50 years old.

Coin Problem –

Given-

Steve had 1 more than twice as many nickels as pennies, twice as many dimes as nickels, 1 more than twice as many 50 cent pieces as dimes, and twice as many quarters as 50 cent pieces. Altogether he has $107.37. How many pennies, nickels, dimes, quarters, and 50 cent pieces does he have?

Solution-

x + 10x + 5 + 40x + 20 + 400x + 250 + 400x + 250 = 10737 =

851x + 525 = 10737 = 

851x = 10737 - 525 =

851x = 10212 =

x = 10212/851

x = 12

So Steve have 12 pennies, 2(12) + 1 or 25 nickels, 4(12) + 2 or 50 dimes, 8(12) + 5 or 101 Fifty cent pieces, and 16(12) + 10 or 202 quarters....

Given-

John has a total of 71 cents of change in his pocket. He has 1 more nickel than he has pennies and 6 times as many dimes as pennies. How many of each coin type does he have?

Solution-

x + 5(x+1) + 10(6x) = 71 =

x + 5x + 5 + 60x = 71 =

66x = 66 =

x = 66/66 = 1

So thers 1 penny, (1 + 1) or 2 nickels and 6(1) 6 dimes 

check, if you add those up 1 + 10 + 60 you get 71 so the answer is correct.

Motion Problem –

Given-

On a round trip of 350 km in each direction, mr. Sinclair averaged 1o km/h more returning than he did going. If the return trip took 4...