SAT MATH PROBLEMS ON RATES

Problem 1 :

A faulty register records only $5 for every $8 deposit. If the register shows a total balance of $40, what is the actual balance?

Solution :

The register records $5 for every deposit and the total balance in the register is $40.

Number of deposits :

= ⁴⁰⁄₅

= 8

There are 8 deposits in all.

The original value of each deposit is $8.

Then, the actual balance is

= no. of deposits x original value of each deposit

= 8 x $8

= $64

Problem 2 :

During a trip, Jonathan had driven a total of 75 miles by 6:20 PM and a total of 85 miles by 6:40 PM. He drove at the same rate for the entire trip. At what time had he driven a total of 140 miles?

Solution :

Total distance covered at 6:20 PM = 75 miles.

Total distance covered at 6:40 PM = 85 miles.

Distance covered between 6:20 PM and 6:40 PM :

= 85 - 75

= 10 miles

Time gap between 6:20 PM and 6:40 PM is 20 minutes.

So, distance covered in 20 minutes = 10 miles.

Distance covered in one hour :

= 3 x 10 miles

= 30 miles

In other words, the speed is 30 miles per hour.

Distance covered at 6:40 PM :

= 85 miles

Distance covered at 7:00 PM :

= 85 + 10

= 95 miles

Distance covered at 8:00 PM :

= 95 + 30

= 125 miles

Distance covered at 8:20 PM :

= 125 + 10

= 135 miles

Distance covered at 8:30 PM :

= 125 + 5

= 140 miles

Jonathan had driven a total of 140 miles at 8:30 PM.

Problem 3 :

A train travelling at an average speed of 80 miles per hour takes 8 hours to complete a given trip. How much time would it take the train to complete the same trip if it traveled at an average speed of 120 miles per hour?

Solution :

Formula to find the distance :

= speed x time

Distance traveled by the train in 8 hours at the rate of 80 miles per hour :

= 80 x 8

= 640 miles

Formula to find the time :

= distance/speed

Time taken by the train to complete the same trip (640 miles) at an average speed of 120 miles per hour :

= 640/120

= 5⅓ hours

= 5 hours + ⅓ hours

= 5 hours + ⅓ x 60 minutes

= 5 hours + 20 minutes

= 5 hours 20 minutes

Problem 4 :

At a certain pizza restaurant, 8 ounces of cheese is enough for 2/3 of a pizza. Given that there are 16 ounces in a pound, how many pizzas can be produced with 12 pounds of cheese?

Solution :

Given : 8 ounces of cheese is enough for 2/3 of a pizza.

Let x be the number of ounces of cheese required for one pizza.

Then, we have

x : 1 = 8 : ⅔

x ÷ 1 = 8 ÷ ⅔

ˣ⁄₁ = 8 ⋅ ³⁄₂

x = 12

So, 12 ounces cheese required for 1 pizza.

Convert 12 pounds to ounces :

12 pounds = 12 x 16 ounces

= 192 ounces

Number of pizzas can be produced with 192 ounces (= 12 pounds) of cheese :

= ¹⁹²⁄₁₂

= 16

Problem 5 :

On Friday, Janice read x pages every 30 minutes for 4 hours, and Kim read y pages every 15 minutes for 5 hours. Which of the following represents the total number of pages read by Janice and Kim on Friday?

Solution :

Given : Janice read x pages every 30 minutes for 4 hours.

30 minutes ----> x pages

1 hour ----> 2x pages

4 hours ----> 4(2x) = 8x pages

Given : Kim read y pages every 15 minutes for 5 hours.

15 minutes ----> y pages

1 hour ----> 4y pages

5 hours ----> 5(4y) = 20y pages

Total number of pages read by Janice and Kim on Friday :

= 8x + 20y

Problem 6 :

A salad dressing supplier ensures that there are 7.5 ounces of salad dressing in every container. How many container would be required to hold 150 ounces of salad dressing?

Solution :

Given : One container can hold 7.5 ounces of salad dressing.

Number of containers required to hold 150 ounces of salad dressing :

= 20

Problem 7 :

A small shop can manufacture 4 windows every 5 hours. At that rate, how long will it take to manufacture 7 windows?

Solution :

Given : 5 hours required to manufacture 4 windows.

Let x be the number of hours required to manufacture 7 windows.

x : 7 = 5 : 4

ˣ⁄₇ = ⁵⁄₄

Multiply both sides by 7.

x = ³⁵⁄₄

= 8¾ hours

= 8 hours + ¾ hours

= 8 hours + ¾ x 60 minutes

= 8 hours + 45 minutes

= 8 hours 45 minutes

It will take 8 hours 45 minutes to manufacture 7 windows.

Problem 8 :

One calorie is equivalent to 4.184 joules of energy. One calorie also represents the amount of energy required to raise the temperature of 1 gram of water by 1° Celsius. Based on this information, raising the temperature of 1 gram of water by 120° Fahrenheit would require how many joules of energy, to the nearest hundredth?

(1° Celsius = 33.8° Fahrenheit)

Solution :

Convert 120° Fahrenheit to Celsius.

120° Fahrenheit = 120/33.8 Celsius

120° Fahrenheit ≈ 3.55° Celsius

Given : One calorie of energy required to raise the temperature of 1 gram of water by 1° Celsius.

Let x be the number of calories of energy required to raise the temperature of 1 gram of water by 3.55° Celsius.

x : 3.55 = 1 : 1

Multiply both sides by 3.55.

x = 3.55

3.55 calories of energy required to raise the temperature of 1 gram of water by 3.55° Celsius.

Convert 3.55 calories to joules.

3.55 calories = 3.55 x 4.184 joules

≈ 14.85 joules

Therefore, 14.85 joules of energy required to raise the temperature of 1 gram of water by 3.55° Celsius or 120° Fahrenheit.

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