Word problems and solution

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More Math Word Problems
Algebra Word Problems

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using elimination and substitution methods?

Grade 8 Algebra Word Problems
How to solve algebra word problems using systems of linear equations?

Example:
Devon is going to make 3 shelves for her father. He has a piece of lumber 12 feet long.
She wants the top shelf to be half a foot shorter than the middle shelf, and the bottom
shelf to be half a foot shorter than twice the length of the top shelf. How long will
each shelf be if she uses the entire 12 feet of wood?

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Grade 8 Algebra Word Problems — Line Segments

Example:
If JK = 7x + 9, JL = 114 and KL = 9x + 9. Find KL.

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Grade 8 number word problems — common core
How to write word problems into systems of linear equations and solve systems of linear equations
using elimination and substitution methods?

Example 1:
The sum of two numbers is 361 and the difference between the two numbers is 173. What are the
two numbers?

Example 2:
There are 356 Grade 8 students at Euclid’s Middle School. Thirty-four more than four times the
number of girls is equal to half the number of boys. How many boys are in Grade 8 at Euclid’s
Middle School? How many girls?

Example 3:
A family member has some five-dollar bills and one-dollar bills in their wallet. Altogether
she has 18 bills and a total of $62. How many of each bill does she have?

Example 1:
A friend bought 2 boxes of pencils and 8 notebooks for school and it cost him $11. He went back
to the store the same day to buy school supplies for his younger brother. He spent $11.25 on 3
boxes of pencils and 5 notebooks. How much would notebooks cost?

Exercises:

  1. A farm raises cows and chickens. The farmer has a total of 42 animals. One day he counts the
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    have? How many chickens?
  2. The length of a rectangle is 4 times the width. The perimeter of the rectangle is 45 inches.
    What is the area of the rectangle?
  3. The sum of the measures of angles x and y is 127″. If the measure of angle x is 34° more
    than half the measure of angle y, what is the measure of each angle?
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Chapter 6: Polynomials

Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below:

Example Mixture Problem Solution Table

Name Amount Value Equation

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

  • The solution names are 50% (S50), 60% (S60), and 80% (S80).
  • The amounts are S50 = 70 mL, S80, and S60 = 70 mL + S80.
  • The concentrations are S50 = 0.50, S60 = 0.60, and S80 = 0.80.
Name Amount Value Equation
S50 70 mL 0.50 0.50 (70 mL)
S80 S80 0.80 0.80 (S80)
S60 70 mL + S80 0.60 0.60 (70 mL + S80)

The equation derived from this data is 0.50 (70 mL) + 0.80 (S80) = 0.60 (70 mL + S80).

Sally and Terry blended a coffee mix that sells for [latex]$2.50[/latex] by mixing two types of coffee. If they used 40 mL of a coffee that costs [latex]$3.00,[/latex] how much of another coffee costing [latex]$1.50[/latex] did they mix with the first?

Name Amount Value Equation
C1.50 C1.50 [latex]$1.50[/latex] [latex]$1.50[/latex] (C1.50)
C3.00 40 mL [latex]$3.00[/latex] [latex]$3.00[/latex] (40 mL)
C2.50 40 mL + C1.50 [latex]$2.50[/latex] [latex]$2.50[/latex] (40 mL + C1.50)

The equation derived from this data is:

[latex]begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \ hline &&-1.00(C_{1.50})&=&-20&& \ \ &&(C_{1.50})&=&dfrac{-20}{-1}&& \ \ &&C_{1.50}&=&20&& end{array}[/latex]

This means 20 mL of the coffee selling for [latex]$1.50[/latex] is needed for the mix.

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

Name Amount Value Equation
B24 B24 0.24 0.24 (B24)
B18 42 L − B24 0.18 0.18 (42 L − B24)
B20 42 L 0.20 0.20 (42 L)

The equation derived from this data is:

[latex]begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \ &-&7.56&&&&-7.56 \ hline &&&&0.06(B_{24})&=&0.84 \ \ &&&&B_{24}&=&dfrac{0.84}{0.06} \ \ &&&&B_{24}&=&14 end{array}[/latex]

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

In Natasha’s candy shop, chocolate, which sells for [latex]$4[/latex] a kilogram, is mixed with nuts, which are sold for [latex]$2.50[/latex] a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for [latex]$3.50[/latex] a kilogram. How much of each are used to make 30 kilograms of the mixture?

Name Amount Value Equation
Chocolate C [latex]$4.00[/latex] [latex]$4.00[/latex] (C)
Nuts 30 kg − C [latex]$2.50[/latex] [latex]$2.50[/latex] (30 kg − C)
Mix 30 kg [latex]$3.50[/latex] [latex]$3.50[/latex] (30 kg)

The equation derived from this data is:

[latex]begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \ 4.00(C)&+&75&-&2.50(C)&=&105 \ &-&75&&&&-75 \ hline &&&&1.50(C)&=&30 \ \ &&&&C&=&dfrac{30}{1.50} \ \ &&&&C&=&20 end{array}[/latex]

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

Name Amount Value Equation
Antifreeze (A) A 1.00 1.00 (A)
Water (W) 70 L − A 0.00 0.00 (70 L − A)
65% Solution 70 L 0.65 0.65 (70 L)

The equation derived from this data is:

[latex]begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \ &&1.00A&=&45.5 \ &&A&=&45.5 \ end{array}[/latex]

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

Questions

For questions 1 to 9, write the equations that define the relationship.

  1. A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
  2. How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
  3. You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
  4. How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
  5. How many litres of a blue dye that costs [latex]$1.60[/latex] per litre must be mixed with 18 litres of magenta dye that costs [latex]$2.50[/latex] per litre to make a mixture that costs [latex]$1.90[/latex] per litre?
  6. How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
  7. A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
  8. A 20-gram alloy of platinum that costs [latex]$220[/latex] per gram is mixed with an alloy that costs [latex]$400[/latex] per gram. How many grams of the [latex]$400[/latex] alloy should be used to make an alloy that costs [latex]$300[/latex] per gram?
  9. How many kilograms of tea that cost [latex]$4.20[/latex] per kilogram must be mixed with 12 kilograms of tea that cost [latex]$2.25[/latex] per kilogram to make a mixture that costs [latex]$3.40[/latex] per kilogram?

Solve questions 10 to 21.

  1. How many litres of a solvent that costs [latex]$80[/latex] per litre must be mixed with 6 litres of a solvent that costs [latex]$25[/latex] per litre to make a solvent that costs [latex]$36[/latex] per litre?
  2. How many kilograms of hard candy that cost [latex]$7.50[/latex] per kg must be mixed with 24 kg of jelly beans that cost [latex]$3.25[/latex] per kg to make a mixture that sells for [latex]$4.50[/latex] per kg?
  3. How many kilograms of soil supplement that costs [latex]$7.00[/latex] per kg must be mixed with 20 kg of aluminum nitrate that costs [latex]$3.50[/latex] per kg to make a fertilizer that costs [latex]$4.50[/latex] per kg?
  4. A candy mix sells for [latex]$2.20[/latex] per kg. It contains chocolates worth [latex]$1.80[/latex] per kg and other candy worth [latex]$3.00[/latex] per kg. How much of each are in 15 kg of the mixture?
  5. A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
  6. Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
  7. A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
  8. How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
  9. A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
  10. A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
  11. How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
  12. How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

Answer Key 6.8

In today’s post, we are going to see a selection of Second Grade word problems that usually appear at the end of a Smartick session for children with a second-grade level. You will also get a chance to see the solution for each problem.

Ready? Let’s go!

How To Correctly Solve Any Problem

  • Read the word problem carefully! You should understand what is being asked very clearly.
  • Write down the data.
  • Complete the operations and always write down the results for each.
  • Write the solution very clearly and respond to what the problem asks.

Second Grade Word Problems

Let’s practice with a selection of Second Grade word problems from Smartick.

Word Problem 1

Second Grade Word Problems at Smartick session

Solution 1

For this word problem, Smartick has given us the solution and it is up to us to select the correct question.

The answer is: ”They were planting 38 chestnut seeds.” Let’s take a look at each option to see which would work with this answer, we’ll use the process of elimination.

  • How many more chestnut seeds than oak seeds were they planting? This is asking us about the different types of seeds, but the solution is only asking about chestnuts so this question is not what we are looking for.
  • Are fewer chestnut trees or fewer oak trees expected to grow? This answer isn’t the one we are looking for because the answer only gives us information about chestnut seeds.
  • How many more poplar seeds than oak seeds were they planting? Nothing was mentioned in the word problem or the answer about poplar trees.
  • How many chestnut seeds were they planting? Here were are being asked about the number of chestnut seeds that were planted and the solution tells us the amount that was planted. This is the question that we are looking for.

Word Problem 2

 Second Grade Word Problems at Smartick session

Solution 2

This word problem is asking how many tablespoons of oil have been added to a stew, and we know that there were 20 tablespoons more added than the 9 tablespoons that the recipe called for. So we need to know how many tablespoons in total were added to the stew.

9 + 20 = 29

29 tablespoons of oil were added to the stew. 

Word Problem 3

 Second Grade Word Problems at Smartick session

Solution 3

Here, we are being asked to calculate the total number of clocks. To help us solve it, we can create a drawing using the information from the word problem. It tells us that there are 2 clocks on each shelf, and there are 6 shelves.

Word problem for second graders.

Word Problem 4

Word problem for second graders.

Solution 4

First, we are asked which operation we need to use to solve the problem and are provided with the options of addition and subtraction. Let’s read the word problem carefully to see which would work best…

Second grade word problems.

There were 6 oranges before Diego added more, and now there are 10. Therefore, the difference between the amount in the bowl now, and the amount there was before, is the amount that Diego added. If we add the numbers, adding the larger quantity to the lesser quantity, it does not give us the number of oranges Diego added to the bowl. However, if we subtract the smaller quantity from the larger quantity, we should find the difference.

The answer to the first question would be: 10 – 6.

And the solution to the word problem: 10 – 6 = 4.

Diego added 4 oranges. 

I hope you have learned something new from this selection of second grade word problems that appear during our daily Smartick sessions. If you would like to learn more about other primary school math topics, register with Smartick and try it for free.

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  • mathrm{Lauren’s:age:is:half:of:Joe’s:age.:Emma:is:four:years:older:than:Joe.:The:sum:of:Lauren,:Emma,:and:Joe’s:age:is:54.:How:old:is:Joe?}

  • mathrm{Kira:went:for:a:drive:in:her:new:car.:She:drove:for:142.5:miles:at:a:speed:of:57:mph.:For:how:many:hours:did:she:drive?}

  • mathrm{Bob’s:age:is:twice:that:of:Barry’s.:Five:years:ago,:Bob:was:three:times:older:than:Barry.:Find:the:age:of:both.}

  • mathrm{Two:men:who:are:traveling:in:opposite:directions:at:the:rate:of:18:and:22:mph:respectively:started:at:the:same:time:at:the:same:place.:In:how:many:hours:will:they:be:250:apart?}

  • mathrm{If:2:tacos:and:3:drinks:cost:12:and:3:tacos:and:2:drinks:cost:13:how:much:does:a:taco:cost?}

Frequently Asked Questions (FAQ)

  • How do you solve word problems?

  • To solve word problems start by reading the problem carefully and understanding what it’s asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
  • How do you identify word problems in math?

  • Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
  • Is there a calculator that can solve word problems?

  • Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
  • What is an age problem?

  • An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as «x years ago,» «in y years,» or «y years later,» which indicate that the problem is related to time and age.

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