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Year 6 Maths Homework Word Problems

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Mixture word problems

Mixture word problems

There are many examples of mixture word problems. The goal is to really help you understand deeply a few.Then, you will be able to tackle similar problems on your own

I offer two solution for the mixture word problems here

A store owner wants to mix cashews and almonds. Cashews cost 2 dollars per pound and almonds cost 5 dollars per pound. He plans to sell 150 pounds of a mixture. How many pounds of each type of nuts should be mixed if the mixture will cost 3 dollars?

Solution #1: Use of one variable leading to a linear equation

Let x be the number of pounds of cashews

So, 150 - x will represent the number of almonds

Since each pound of the mixture cost 3 dollars, 150 pounds will cost 3 × 150 = 450 dollars

Cost of cashews + cost of almonds = 450

2 × x + (150 - x) × 5 = 450

2x + 150 × 5 - x × 5 = 450

2x + 750 - 5x = 450

2x - 5x + 750 = 450

-3x + 750 - 750 = 450 - 750

150 - x = 150 - 100 = 50.

The store owner should mix 100 pounds of cashews with 50 pounds of almonds

Solution #2: Use of two variables leading to a system of linear equations

Let x represent the number of pounds of cashews

Let y be the number of pounds of almonds

Multiply x + y = 150 by -2

-2(x + y) = 150 × -2

Put the two equations together again.

Add the left sides and the right sides to get:

Example 2 and 3 are more challenging mixture word problems

Suppose a car can run on ethanol and gas and you have a 15 gallons tank to fill. You can buy fuel that is either 30 percent ethanol or 80 percent ethanol. How much of each type of fuel should you mix so that the mixture is 40 percent ethanol?

Solution #1: Use of one variable leading to a linear equation

Le x represent number of gallons of gas that contain 30 percent ethanol

Let 15 - x be number of gallons of gas that contain 80 percent ethanol

Since the mixture contains 40 percent ethanol, only 40% of the 15 gallons will be ethanol

40% of 15 = (40/100) times 15 = (40/100) times 15/1 = (40 × 15) / (100 × 1) = 600 / 100 = 6

In order for x gallons of gas to contain 30% ethanol, we must take 30% of x or 30% times x

In order for 15 - x gallons of gas to contain 80% ethanol, we must take 80% of 15 - x or 80% times 15 - x

0.30 × x + 0.80 × (15 - x) = 6

0.30x + 0.80 × 15 - 0.80 × x = 6

0.30x + 12 - 0.80x = 6

0.30x - 0.80x + 12 = 6

So 12 gallons of gass contain 30 percent ethanol and 15 - 12 = 3 gallons contain 80 percent ethanol

Therefore, mix 12 gallons of a 30% ethanol with 3 gallons of an 80% ethanol

Indeed 30% of 12 = 0.30 × 12= 3.6 and 80% of 3 = 0.80 × 3 = 2.4

Solution #2: Use of two variables leading to a system of linear equations.

Le x be number of gallons of gas that contain 30 percent ethanol

Let y represent number of gallons of gas that contain 80 percent ethanol

0.30 × x + 0.80 × y = 0.40 × 15

Use x + y = 15 to get x ( subtract y from both sides) and replace x into the other equation

0.30 × (15 - y) + 0.80y = 6

0.30 × 15 - 0.30 × y + 0.80y = 6

4.5 + 0.80y - 0.30y = 6

4.5 - 4.5 + 0.50y = 6 - 4.5

Again, x = 15 - y = 15 - 3 = 12

So far, you probably noticed that mixture word problems can be quite challenging!

You have 6 liters of water that have 20 percent strawberry juice. How many liters of a 80 percent strawberry juice should be added to the mixture to make 75 percent strawberry juice?

Solution: Use of two variables leading to a system of linear equations

Le x be number of liters of water that contain 20 percent strawberry juice

Let y represent number of liters of water that contain 80 percent strawberry juice

0.20x + 0.80y = 0.75 × 6

Solve by substitution

0.20 × (6 - y) + 0.80y = 4.5

0.20 × 6 - 0.20 × y + 0.80y = 4.5

1.2 + 0.80y - 0.20y = 4.5

1.2 - 1.2 + 0.60y = 4.5 - 1.2

0.60y / 0.60 = 3.3 / 0.60

Therefore, if you want your juice to contain 75% strawberry juice, do the following:

Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice

Your juice shoud taste better now!

The mixture word problems I solved above are typical questions. You may not encounter these word problems a lot in algebra. However, it is good idea to know how to solve these mixture word problems

When solving mixture word problems, I suggest you do it with a system of linear equations

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Maths Word Problem Homework

Thread: Maths Word Problem Homework

I'm stuck on this maths word problem for homework. please help.

The first 100 positive whole numbers are multiplied together to form the number N.
(a) What is the very last right-hand (ie "units") digit of N? Explain your answer.
(b) What digit is in the 19th place from the right? Explain your answer.

The last digit has to be 0.
Since any whole number multiplied by, say, 10, equals zero.

And I think. the 19th may also be.
10,20,30,40,50,60,70,80,90,100 add in a zero, as do 5*2,12*15,22*25,32*35,42*45,52*55,62*65, 72*75, 82*85, & 92*95.

Aye, there's 20 of them. So the 19th place from the right is also a 0.

Apr 27th 2009, 02:54 PM #3

Hi Unenlightened, that was quick thank you very much you are a life saver.

Apr 27th 2009, 02:59 PM #4

Notice that .
has a factor of " />.
So N has 24 'trailing' zeros. (Ends in 24 zeros.)

Apr 28th 2009, 02:48 AM #5

Thanks for replying.

Please would you explain how N=100. I was trying to multiply all the numbers together to get N.

Apr 28th 2009, 03:46 AM #6

I'm stuck on this maths word problem for homework. please help.

The first 100 positive whole numbers are multiplied together to form the number N.
(a) What is the very last right-hand (ie "units") digit of N? Explain your answer.
(b) What digit is in the 19th place from the right? Explain your answer.

Multiplying the first numbers means that you are doing the factorial of . .

(a) Lets say you did and last step was to multiply be to get . If you multiply any number by , what is the last digit?

(b) Try to understand how many zeros that has then you many be able to work out what the 19th value is.

This website talks about , giving it's value and explaining how many zero's it has.

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