Spec Math

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SPECIAL MATH PROJECT 1. Translating Word Problems 2. Integer Problem 3. Age Problem 4. Digit Problem 5. Money Problem 6. Motion Problem Martinez, Irish Benette M. Tongo, Faith Bes Sabino, Hans Ferdinand Jimenez, Mel Jericho

Transcript of Spec Math

SPECIAL MATH PROJECT1. Translating Word Problems2. Integer Problem3. Age Problem4. Digit Problem5. Money Problem6. Motion Problem

Martinez, Irish Benette M.Tongo, Faith Bes

Sabino, Hans FerdinandJimenez, Mel Jericho

Translating Word Problems

Mr. Fredrickson asked Russle to buy himRed, Blue, and Green Balloons.

He asked him to buy blue balloons that is 6 more than the number of red balloons.

X + 6

and green balloons that is 4 times 3 less than the number of blue balloons.

4(x+6-3))

If the total number of the balloons is 60, how many red, blue, and green balloons are needed to make Mr. Frederick’s house float?

X+X+6

+4(x+6-3))=

60

X+x+6+4(x+6-3)=60X+x+6+4x+24-12=60

6x+30-12=606x-18=606x=60-18

6x=42=

X=7

Therefor there are: Red Balloon : x =7Blue BalloonX+6- 7+6=13Green Balloon4(x+6-3) – 4(7+6-3)=40

Thankyou for helping Mr.Frederick buy his balloons now he can travel to their wife’s drem place !

Age Problems

Gru was planning to have a surprise birthday party for dave

While Gru was making dave’s birthday cake he noticed that the birthdaycandles was missing

?

Gru decided to go back to his home to ask his daughter , Agnes about Dave’s Age

AgnesHey dad!

Do you know how old dave

is?

Dave is 7 years younger

than me.

2 years ago, Agnes is 5 years less than the half of Gru’s age and dave is 7 years younger than agnes, What is dave’s present age if their ages is equal to 33?

Name Age 2 years ago

Gru X X-2

Agnes - 5 -2

Dave -5-7 -5-7-2

X-2+ -2+-7-2 = 332[X-2+ -2+-7-2 = 33]2

2x-4+x-10-4+x-10-14-4=662x+x+x=

4+10+4+10+14+4+664x=112

=X=28

Name Age 2 years ago

Present age

Gru X-2 26+2 = 28

Agnes – 5-2 7+2 = 9

Dave – 5-7-2 0 +2 = 2

Therefor Dave’s age is 2.

Now Gru is ready to surprise Dave

for his second birthday

Happy 2nd Birthday Dave!!!

THANK YOU FRIENDS FOR HELPING GRUYOU SURPRISED ME!

Digit Word Problems

It’s Sunday and so Rico was bored.

He went to the living room and decided to watch something in the TV.

Today we’re going to learn how to solve Digit Word Problems!

Let’s try to solve this Digit Word Problem and analyze it.

The digit at the ten’s place of a two digit number is twice the digit at the unit’s place. If the sum of this number and the number formed by reversing the digits is 66. Find the number.

x2x

(2x) (x)

Solution : Let the unit place digit = x

Ten’s place digit = 2x 

TENS UNIT

10 1

Number formed = 10(2x ) + x = 20 x + x = 21x 

+

(2x) (x)

Solution : Let the unit place digit = x

Ten’s place digit = 2x 

TENS UNIT

10 1

Number formed = 10(2x ) + x = 20 x + x = 21x 

+

The digit at the ten’s place of a two digit number is twice the digit at the unit’s place.

If the sum of this number and the number formed by reversing the digits is 66. Find the number.

Reversing the digit : 

Unit digit = 2x 

Ten’s digit = x 

10

(X) + (2x)1

Reversed number formed = 10 (x) + 2x = 10x + 2x = 12x 

As the sum of the number is 66. The equation will be:

21x + 12x = 66

33x = 66 

X = 2

Original number formed = 10(2x ) + x = 20 x + x = 21x 

So unit digit = 2

x = 22x = 10[2(2)]

4 2The number and the answer is:

That’s it! You already know how to solve and analyze DigitWord Problems. Remember to make your solution neat andclear so you will not be confused.

THANK YOU FOR WATCHING!

“Oh, that’s how!” Rico exclaimed. He was happy becausehe learned to solve Digit Word Problems.

MONEY PROBLEM

Sandy wants to have a lunch with her two friends, Patrick and Squidward.

Sandy is buying 5 krabby patty that cost $ 1.50 each. She wants to share her burger with her two friends. She asks her friends to pay for her for their share. Including Patrick, How much does each person spend on burger?

5 x $1.50 = $7.50(5 x $1) = $5.00($0.50 x 5) = $2.50($5.00 + $2.50) = $7.50convert $7.50 into cents750 cents ÷ 3 = 250 cents250 cents= $2.50

MOTION PROBLEM

A car and another car, who live 14 miles apart, started at noon to drive towards each other at the rate of 3 mph and 4 mph respectively. In how

many hours will they meet?

WORKING EQUATION:3X + 4X = 14

7X = 147X/7 = 14/7

X=2, THEREFORE THEY WILL MEET WITHIN TWO HOURS

FIRST DO A TABLE CONSISTING. THE RATE OR SPEED OF THE CARS.

THEN MAKE YOUR OWN WORKING EQUATION.

AFTER THAT COMPUTE FOR THE FINAL ANSWER, THEN START CONCLUDING ABOUT YOUR

ANSWER.

RATE TIME DISTANCE

CAR 1 3 X 3X

CAR 2 4 X 4X