Introduction to Sole Math Problems:
In this topic, we will be discussing few word problems that will help you to solve some math problems. Those are generally a set of problems on the applications of Algebra in day today life activities and few from percentage to calculate the cost price and the selling price. I hope this article will help you in solving few day today life problems. Let us solve few of them on this topic math problem solve. I like to share this What is Experimental Probability with you all through my article.
Examples problems to solve math problems:
Ex 1: Lawrence covers a distance of 15km in 3 hours, partly by walking and partly by running. If he walks at 3km/h and runs at 9km/h, find the distance covered by running.
Sol: Suppose x km denotes the distance covered by running.
Therefore, the distance covered by walking = (15 – x) km
Time taken to cover x km by running = `x/9` hr.
Time taken to cover (15 – x) km by walking = `(15 ** x)/3` hr
Since, Lawrence covers the total distance in 3 hours, we have
x/9 + `(15** x)/3` = 3
[x + 3(15 – x)]/9 = 3
( -2x + 45 ) = 9 × 3 = 27
-2x = 27 – 45 = -18
x = `(-18)/-2` = 9
Therefore the distance covered by running = 9km.
Ex 2: The cost of an article X is 15% less than that of article Y. If their total cost is 2,775 dollars, find the cost of each article.
Sol: Let the cost of the article Y be A dollars.
Then the cost of the article X = (100 – 15) % of A
= 85% of A
Given : A + 85% of A = 2775
`185/100` × A = 2775
A = 2775 ×`100/185`
A = 1500 dollars
Therefore the price of Y is 1500 dollars and the price of X is 2775 – 1500 = 1275 dollars.
Understanding algebra 2 help online for free is always challenging for me but thanks to all math help websites to help me out.
More examples problems to solve math problems:
Ex3. A car can go 30 miles per liter of petrol. The fuels utilized by the car are 12 liters. What is the distance covered by it?
Sol: Given: Mileage = 30 miles/liter, Fuel = 12 liters.
Therefore, D = Mileage x Fuel utilized
= 30 x 12
= 360 miles
Therefore the distanced covered by the car is 360 miles.
Practice problems on math problem solve
1. To cover a distance of 15o miles, a mini truck utilized 15 liters of diesel. What is the mileage of the mini truck?
[ Answer: 10 miles per liter of diesel]
2. A farmer sold a cow and a calf for 760 dollars, thereby making a profit of 25% on the calf and 10% on the cow. By selling them for 767.50, he would have realised a profit of 10% on the calf and 25% on the cow. Find the cost price of each.
[ Answer: Calf = 300 dollars, Cow = 350 dollars]
In this topic, we will be discussing few word problems that will help you to solve some math problems. Those are generally a set of problems on the applications of Algebra in day today life activities and few from percentage to calculate the cost price and the selling price. I hope this article will help you in solving few day today life problems. Let us solve few of them on this topic math problem solve. I like to share this What is Experimental Probability with you all through my article.
Examples problems to solve math problems:
Ex 1: Lawrence covers a distance of 15km in 3 hours, partly by walking and partly by running. If he walks at 3km/h and runs at 9km/h, find the distance covered by running.
Sol: Suppose x km denotes the distance covered by running.
Therefore, the distance covered by walking = (15 – x) km
Time taken to cover x km by running = `x/9` hr.
Time taken to cover (15 – x) km by walking = `(15 ** x)/3` hr
Since, Lawrence covers the total distance in 3 hours, we have
x/9 + `(15** x)/3` = 3
[x + 3(15 – x)]/9 = 3
( -2x + 45 ) = 9 × 3 = 27
-2x = 27 – 45 = -18
x = `(-18)/-2` = 9
Therefore the distance covered by running = 9km.
Ex 2: The cost of an article X is 15% less than that of article Y. If their total cost is 2,775 dollars, find the cost of each article.
Sol: Let the cost of the article Y be A dollars.
Then the cost of the article X = (100 – 15) % of A
= 85% of A
Given : A + 85% of A = 2775
`185/100` × A = 2775
A = 2775 ×`100/185`
A = 1500 dollars
Therefore the price of Y is 1500 dollars and the price of X is 2775 – 1500 = 1275 dollars.
Understanding algebra 2 help online for free is always challenging for me but thanks to all math help websites to help me out.
More examples problems to solve math problems:
Ex3. A car can go 30 miles per liter of petrol. The fuels utilized by the car are 12 liters. What is the distance covered by it?
Sol: Given: Mileage = 30 miles/liter, Fuel = 12 liters.
Therefore, D = Mileage x Fuel utilized
= 30 x 12
= 360 miles
Therefore the distanced covered by the car is 360 miles.
Practice problems on math problem solve
1. To cover a distance of 15o miles, a mini truck utilized 15 liters of diesel. What is the mileage of the mini truck?
[ Answer: 10 miles per liter of diesel]
2. A farmer sold a cow and a calf for 760 dollars, thereby making a profit of 25% on the calf and 10% on the cow. By selling them for 767.50, he would have realised a profit of 10% on the calf and 25% on the cow. Find the cost price of each.
[ Answer: Calf = 300 dollars, Cow = 350 dollars]
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