Could you guys help me with math problems???

Question

My teacher gave us these problems and told us to answer them.. Since I can't understand any of these, could you guys at least help me answer the problems??

1. Find three consecutive odd integers such that the sum of the last two is 43 more than the first.

2. Sigfried is 14 years old. His age now is two times the age of his brother 4 years ago. How old is Sigfried's brother?

3. How much does Alex have is he has 64 coins and he found that his 50c-coins are double and the number of 25c-coins and the 10c-coins are 8 less than the no. of 25c-coins?

4. Once a week, Ric goes from his hometown in Davao City. The trip usually takes him 4 hours. If the distance between his hometown and Davao is 180 km, how fast should Ric drive if he wants to reach Davao City an hour earlier?

You could just answer or help me with at least one problem so I have an idea of what to answer.. Thank you guys for helping me!!

Answer 1

These are the solutions for you "kababayan".

Problem 1:
Consecutive numbers mean they are in series or like counting numbers.
Examples 1) 11, 12, 13
2) 34, 35, 36
3) 5, 6, 7

The problem says find 3 consecutive numbers such that the sum of the last two is 43 more than the first.

Let x = the first number
x + 1 = the second number
x + 2 = the third number

The sum of the last two numbers is 43 more than the first
So we need to ADD the last two numbers
(x+1) + (x + 2) =
The sum must be 43 more than the first
Since the first number is x, more than 43 means x + 43
So...
(x + 1) + (x+2) = x + 43
Remove the parenthesis
x + 1 + x + 2 = x + 43
2x + 3 = x + 43
Transpose 3 to the right by adding -3 to both sides, and transpose x to the left by adding -x to both sides
2x + 3 - 3 - x = x - x + 43 -3
x = 40 ===> the first number
x + 1 = 40 + 1 = 41 ====> the second number
x + 2 = 40 + 2 = 42 ====> the third number
So the numbers are 40, 41 & 42
To check
(x + 1) + (x + 2) = x + 43
41 + 42 = 40 + 43
83 = 83

Problem 2:
The age of Sigfried is 14 years old. His age is two times the age of his brother 4 years ago. How old is Sigfried's brother?
Let x = age of Sigfried's brother
x - 4 = age of Sigfried's brother 4 years ago. This age is two times that of Sigfried

2 times ( x-4) = 14
2 (x-4) = 14
2x - 8 = 14
Add 8 to both sides to transpose - 8 to the right side
2x - 8 + 8 = 14 + 8
2x = 22
x = 22/2
x = 11 ===> age of Sigfried's brother now
x - 4 = 11 - 4 = 7 ===> age of Sigfried's brother 4 years ago
which is twice the age of Sigfried now.
2 times 7 = 14
14 = 14

Problem 3:
Total number of coins = 64
Let x = number of 25c coins
x - 8 = number of 10c coins
2x = number of 50c coins
Equation:
(x-8) + x + (2x) = 64
x - 8 + x + 2x = 64
4x - 8 = 64
Add 8 to both sides to transpose -8 to the right
4x - 8 + 8 = 64 + 8
4x = 72
x = 72/4
x = 18 ===> number of 25c coins
x - 8 = 18 - 8 = 10 ===> number of 10c coins
2x = 2 (18) = 36 ====> number of 50c coins
10 + 18 + 36 = 64 coins
To get how much Alex has

10 (.10) + 18 (.25) + 36 (.50) =
1.00 + 4.50 + 18.00 = 23.50 ===> amount Alex has

Poblem 4:
Distance from Ric's home to Davao City = 180 km
Time Ric's travel from home to Davao City = 4 hours

Speed = distance / time
Speed = 180 km/ 4 hours
Speed = 45 km/hr ===> normal speed of Ric
To reach Davao City an hour earlier, what should his speed be?

Speed = 180 km/3 hours
Speed = 60 km/h ===> the speed Ric makes to reach Davao City an hour earlier

To check
3 times 60 = 180
3 (60) = 180
180 = 180

I hope you understand....

Answer 2

1. The even integers are represented by 2n
and the odd integers by 2n - 1, for n = 1, 2, 3, ...
Check :
For n = 1, 2 and 3, 2n = 2, 4 and 6.
For n = 1, 2 and 3, 2n - 1 = 1, 3 and 5.
Let your 1st odd integer be 2n - 1. The next one
will be 2 more, that is 2n + 1, and the 3rd one will
be 2n + 3.
The question states :
(2n + 1) + (2n + 3) = (2n - 1) + 43
You should be able to work out n from that, and then work out what 2n -1, 2n + 1 and 2n + 3 is.

2. Sigfried now = 14, Brother now = B
Four years ago, the brother's age was B - 4.
The question states :
14 = 2 * (B - 4)
Now you should be able to work out B.

3. Alex has 64 coins. This is made up of A 50c
coins, B 25c coins and C 10c coins.
So, A + B + C = 64.
The question states (I think) that :
A = 2B and C = B - 8.
We have A and C in terms of B, so replace
their values into the equation A + B + C = 64.
This gives : (2B) + (B) + (B - 8) = 64.
From that, you should be able to work out B.
Then substitute the value for B into A = 2B,
which will give you A.
And do the same for C = B - 8 to find C.
Now that you know A, B and C, you need to work
out A * 50c + B * 25c + C * 10c, which will give
the number of cents, and then convert to dollars.

4. Distance = Speed * Time
Therefore, 180 km = Speed (km/hr) * (4 - 1) hr.
You should be able to work out the speed now.

Answer 3

1. If the first number is x, then the next is x+2 and the third is x+4 (since they're all odd, they're increasing by 2's). The second plus the third, minus the second, equals 43, so (x+2)+(x+4)-x=43 or x+x-x +6 = 43 or x+6 = 43. So x = 37. Therefore, the three numbers are 37, 39, and 41.

2. If S=14 and B = age of brother now, then S = 2(B-4). So 14 = 2B - 8. 22 = 2B. B = 11. His brother is 11 years old now.

3. H = half dollar coins
Q = quarters
D = dimes
H = 2Q, D = Q-8, H+Q+D = 64 --- Substitute so that the last equation only has Q's in it. 2Q + Q + Q - 8 = 64. 4Q = 56. Q = 18. Therefore, H = 36 and D = 10.

4. If he wants to travel 180 km in 3 hours instead of 4, then he wants to go 180 km/3 hours, or 60 km per hour, or 60kph.

Answer 4

I hope no one gives you the answers -- you wouldn't learn any math that way.

You need to set up an equation for each one. The easiest way to do this is to start with a statement, beginning with the word "let", that describes what you are trying to find. You then add more statements that map the word problem to math variables. In the end, you have a single algebraic equation to solve (it should be easy from there). One of my teachers calls this the "Let - Then - And" method, and it works GREAT for problems like these.

Let me give you another problem, to show you how it works.

Find three consecutive mulitples of 5 where the sum of the second and third is 50 more than the first.

First, set it up:
Let x = the smallest of the three integers
Then (x+5) and (x+10) are the other three integers.
And (x+5) + (x+10) = x + 50

Now, you simply solve for x:

2x + 15 = x + 50
x = 35

So, the three integers are 35, 40, and 45.

Lastly, check to make sure everything worked:

40 + 45 = 85, which is 50 more than 35.

Lastly, I see that others answered your questions for you. I *strongly* urge you to go back and solve them independently, so that you know how to do it.

Give someone their math answers, and they'll get their assignment done for a day. Teach someone how to find their answers, and they'll be able to do their assignments for a lifetime.

Good luck.

Answer 5

Any odd integer can be written as 2n+1 where n is some (odd or even) integer. So we have three consecutive odd integers and they would be written as
2n+1
2n+3
2n+5
The problem says
2n+3 + 2n+5 = 2n+1 + 43 so
4n+8 = 2n+44 and (subtracting 8 from both sides)
4n = 2n+36 (subtract 2n from both sides)
2n = 36 and (divide both sides by 2)
n = 18 so the integers are
2*18+1 = 37
2*18+3 = 39
2*13+5 = 41
check
39+41 = 80 = 37+43 Yup. All done.

Now..... **You** do the rest of them. Math is like any other sport, you won't get good at it if you just set around and watch somebody else play☺

Doug

Answer 6

let the *consecutive* *odd* integers be 2n+1,2n+3,2n+5
(2n+3)+(2n+5) = (2n+1)+43
2n = 43 +1 - 3 -5 = 36
so n = 18
and the numbers are 37,39,41


let brothers age be x
14 = 2(x-4)
22 = 2x
so x = 11


suppose the number of 50c, 25c, 10c are x,y,z
x+y+z = 64
x = 2y
z = y-8
x+y+z = 2y + y + y-8 = 64
y = 72 / 4 = 18
z = 18 - 8 = 10
x = 2*18 = 36
so he has 36*0.50 + 18*0.25 + 10*0.10 = $23.5

Ric wants to travel 180 km in 3 hours, so 60 km/h

Answer 7

1)Let the three consecutive odd integers be a-2,a and a+2(since consecutive odd integers differ by 2).
According to the problem,
a+(a+2)=43+a-2
2a+2=41+a
a=39
Required nos are 37,39 and 41
2)Let the age of sigs bro be x
4 yrs ago,
sigs bros age will be x-4
sigs age=14
Condition,
14=2(x-4)
SIMPLIFYING,
x=11

Answer 8

For question 1:

Since consecutive odd numbers are separated by 2, let the numbers be a, a+2 and a+4. Then we require
(a+2) + (a+4) = a + 43
2a + 6 = a + 43
a = 37

So the numbers are:
37,39,41

Answer 9

1. (2x+1) this equals the first odd number (if x is any number, then multiplied by 2 makes it even and added 1 makes it odd).
(2x+3) the second and (2x+5) the third. Now:

(2x+3) + (2x+5) = 43 + (2x+1)
x= 18

Number 1: 37
Number 2: 39
Number 3: 41

4. 180 km. in 4 hours. He wants to get there in 3. 180km/3h= 60km/h.

Answer 10

Glenn and I can help you out with 1 and 2:

1. 37, 39, 41

2. 11 yrs old

Have a good day :>)