PROBLEMS ON MIXTURE AND INVEST

Problem 1 :

On a vocation, Elwyn averaged 50 mph travelling from Denver to Minneapolis. Returning by a different route that covered the same number of miles he averaged 55 mph. What is the distance between the two cities if the total travelling time was 32 hours ?

Solution :

Time = Distance / speed

Let x be the distance in between Denver and Minneapolis. 

Time taken to reach Minneapolis from Denver be t₁.

t₁ = x/50

Let t₁be the time taken for returning.

t₂ = x/55

t₁ + t₂ = x/50 + x/55

32 = (11x + 10x)/550

32 = 21x/550

x = (32 ⋅ 550) / 21

x = 838.09

So, the distance to be covered is 838.09 miles. 

Problem 2 :

Terry earned $48000 from royalties on her cookbook. She paid 28% income tax of these royalties. The balance was invested in two ways, some of it at 3.25% and some of it at 1.75%. The investment produced $904.80 per year. Find the amount invested at each rate.

Solution :

The net amount get by her excluding the tax

= (100 - 28)% of 48000

= 72% of 48000

= 0.72 (48000)

= 34560

• x is the amount she is investing in the first scheme.
• Then 34560 - x is the amount she is investing in the second scheme.

So, he is investing $20000 in the first scheme and in the second scheme he is investing $14560.

Problem 3 :

A chemist wishes to obtain a 20% of alcohol solution, however she only has 10% alcohol and 30% of alcohol on hand. How much should she add to the 5 gallons of 10% alcohol to obtain 20% mixture ?

Solution :

Quantity of 10% alcohol = 5

Quantity of 20% alcohol = x

10% of 5 + 30% of x = 20% of (x + 5)

Problem 4 :

The sides of the triangular plot of land have a perimeter of 230 meters. If the largest side is twice as large as the smallest and middle side is 20 less than the largest. What is the lengths of each side?

Solution :

Let x be the smaller side

Largest side = 2x

Middle side = 2x - 20

Perimeter of the triangle = 230

x + 2x + 2x - 20 = 230

5x - 20 = 230

5x = 250

x = 50

2x = 100

2x - 20 = 80

So, the side lengths are 50 m, 100 m, 80 m.

Problem 5 :

You have 20 gallons of 45% of antifreeze solution. How many gallons of 57% of antifreeze solution need to be added to make a 51% of antifreeze solution ?

Solution :

Let x be the quantity of 57% of solution.

45% of 20 + 57% of x = 51% of (x + 20)

0.45(20) + 0.57x = 0.51(x + 20)

9 + 0.57x = 0.51x + 10.2

0.57x - 0.51x = 10.2 - 9

0.06x = 1.2

x = 1.2/0.06

x = 20 gallons