Gmat Math (Words Problems)

Front 1

rate

Back 1

rate x time = distance

Front 2

rate question and answer 1

Back 2

The distance that an object travels is equal to the product of the average speed at which it travels and the amount of time it takes to travel that distance.

Question:
If a car travels at an average speed of 70 kilometers per hour for 4 hours, how many kilometers does it travel?

Solution:
Since rate × time = distance, simply multiply 70 km/hour × 4 hours. Thus, the car travels 280 kilometers in 4 hours.

Front 3

rate question and answer 2

Back 3

To determine the average rate at which an object travels, divide the total distance traveled by the total amount of traveling time.

Question:
On a 400-mile trip, car X traveled half the distance at 40 miles per hour and the other half at 50 miles per hour. What was the average speed of car X?

Solution:
First it is necessary to determine the amount of traveling time. During the first 200 miles, the car traveled at 40 mph; therefore, it took 200/40=5 hours to travel the first 200 miles. During the second 200 miles, the car traveled at 50 mph; therefore, it took 200/50=4 hours to travel the second 200 miles. Thus, the average speed of car X was 400/9=44 4/9 mph. NOT 45 mph.

Front 4

work

Back 4

In a work problem, the rates at which certain persons or machines work alone are usually given, and it is necessary to compute the rate at which they work together (or vice versa). The basic formula for solving work problems is: 1/r+1/s=1/h, where r and s are, for example, the number of hours it takes Rae and Sam, respectively, to complete a job when working alone, and h is the number of hours it takes Rae and Sam to do the job when working together. The reasoning is that in 1 hour Rae does 1/r of the job, Sam does 1/s of the job, and Rae and Sam together do 1/h of the job.

Front 5

Work problem and answer 1

Back 5

Question:
If machine X can produce 1,000 bolts in 4 hours and machine Y can produce 1,000 bolts in 5 hours, in how many hours can machines X and Y, working together at these constant rates, produce 1,000 bolts?

Solution:



Working together, machines X and Y can produce 1,000 bolts in 2 2/9 hours hours.

Front 6

work question and answer 2

Back 6

Question:
If Art and Rita can do a job in 4 hours when working together at their respective constant rates and Art can do the job alone in 6 hours, in how many hours can Rita do the job alone?

Solution:



Working alone, Rita can do the job in 12 hours.

Front 7

mixture

Back 7

In mixture problems, substances with different characteristics are combined, and it is necessary to determine the characteristics of the resulting mixture.

Front 8

mixture question and answer 1

Back 8

Question:
If 6 pounds of nuts that cost $1.20 per pound are mixed with 2 pounds of nuts that cost $1.60 per pound, what is the cost per pound of the mixture?

Solution:
The total cost of the 8 pounds of nuts is 6($1.20) + 2($1.60) = $10.40. The cost per pound is
$10.40/8 = $1.30.

Front 9

Mixture question and answer 2

Back 9

Question:
How many liters of a solution that is 15% salt must be added to 5 liters of a solution that is 8% salt so that the resulting solution is 10% salt?

Solution:
Let n represent the number of liters of the 15% solution. The amount of salt in the 15% solution [0.15n] plus the amount of salt in the 8% solution [(0.08)(5)] must be equal to the amount of salt in the 10% mixture [0.10 (n + 5)]. Therefore:



Two liters of the 15% salt solution must be added to the 8% solution to obtain the 10% solution.

Front 10

interest

Back 10

Interest can be computed in two basic ways. With simple annual interest, the interest is computed on the principal only and is equal to

(principal) × (interest rate) × (time).

If interest is compounded, then interest is computed on the principal as well as on any interest already earned.

Front 11

interest questionand answer 1

Back 11

Question:
If $8,000 is invested at 6% simple annual interest, how much interest is earned after 3 months?

Solution:
Since the annual interest rate is 6%, the interest for 1 year is (0.06)(8,000) = $480. The interest earned in 3 months is
3/12 (480) = $120

Front 12

interest question and answer 2

Back 12

Question:
If $10,000 is invested at 10% annual interest, compounded semiannually, what is the balance after 1 year?

Solution:
The balance after the first 6 months would be 10,000 + (10,000)(0.05) = 10,500.

The balance after one year would be 10,500 + (10,500)(0.05) = $11,025.

The balance after one year can also be expressed as
10,000 (1+ 0.10/2)2

Front 13

discount

Back 13

If a price is discounted by n percent, the price becomes (100 − n) percent of the original price.

Front 14

discount question and answer 1

Back 14

Question:
A certain customer paid $24 for a dress. If that price represented a 25% discount on the original price of the dress, what was the original price of the dress?

Solution:
If p is the original price of the dress, then 0.75p is the discounted price and 0.75p = $24, or p = $32. The original price of the dress was $32.

Front 15

discount question and answer 2

Back 15

Question:
The price of an item is discounted by 20% and then this reduced price is discounted by an additional 30%. These two discounts are equal to an overall discount of what percent?

Solution:
If p is the original price of the item, then 0.8p is the price after the first discount. The price after the second discount is (0.7)(0.8)p = 0.56p. This represents an overall discount of 44% (100% − 56%).

Front 16

profit

Back 16

Gross profit is equal to revenues minus expenses, or selling price minus cost.

Front 17

profit question and answer 1

Back 17

Question:
A certain appliance costs a merchant $30. At what price should the merchant sell the appliance in order to make a gross profit of 50 percent of the cost of the appliance?

Solution:
If s is the selling price of the appliance, then s − 30 = (0.5)(30), or s = $45. The merchant should sell the appliance for $45.

Front 18

sets

Back 18

If S is the set of numbers 1, 2, 3, and 4, you can write S = {1, 2, 3, 4}. Sets can also be represented by Venn diagrams. That is, the relationship among the members of sets can be represented by circles.

Front 19

sets question and answer 1

Back 19

Question:
Each of 25 people is enrolled in history, mathematics, or both. If 20 are enrolled in history and 18 are enrolled in mathematics, how many are enrolled in both history and mathematics?

Solution:
The 25 people can be divided into three sets: those who are enrolled in history only, those who are enrolled in mathematics only, and those who are enrolled in history and mathematics. Thus, a Venn diagram may be drawn as follows, where n is the number of people enrolled in both courses, 20 − n is the number enrolled in history only, and 18 − n is the number enrolled in mathematics only.Since there is a total of 25 people, (20 − n) + n + (18 − n) = 25, or n = 13. Thirteen people are enrolled in both history and mathematics.

Front 20

sets question and answer 2

Back 20

Question:
In a certain production lot, 40% of the toys are red and the remaining toys are green. Half of the toys are small and half are large. If 10% of the toys are red and small, and 40 toys are green and large, how many of the toys are red and large?

Solution:
use graph. Since 20% of the number of toys (n) are green and large, 0.20n = 40 (40 toys are green and large), or n = 200. Therefore, 30% of the 200 toys, or (0.3)(200) = 60, are red and large.

Front 21

measurement question and answer

Back 21

Question:
A train travels at a constant rate of 25 meters per second. How many kilometers does it travel in 5 minutes? (1 kilometer = 1,000 meters)

Solution:
In 1 minute the train travels (25)(60) = 1,500 meters, so in 5 minutes it travels 7,500 meters. Since 1 kilometer = 1,000 meters, 7,500 meters equals 7500/1000 or 7.5 kilometers.