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Mayor, Brian Allen L.
1.4 BSIT

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Shamalla Faith Callejas
1.4 bsit

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What is being asked: how old is peter now?
Given:
John=2x
Peter=x
Alice = x-5
Age in 5 years
John=2x+5
Peter=x+5
Alice = (x-5) +5
solution
2x + 5 = 3(x – 5 + 5)
2x + 5 = 3x
5 = 3x-2x
x = 5
Peter is now 5 years old.

What is being asked: How long would it take the faster roofer working alone?
Given:  speed of roofer1 = t
Speed of the other roofer = 3t
Speed of two roofer = 5 days

Solution
5/t + 5/3t = 1
(3t) 5/t + 5/3t = 1 (3t)
15 + 5 = 3t
20 =3t
20/3 = 3t/3
T = 20/3 hrs. or 6.67 days

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Elmer L Belandres
1.4BSIT
Assignment # 2

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Jhustin Rhoyce N. Juacalla
1.4 BSIT

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Stephen David Q. Condino
1.4 BSIT

Question # 1
John is twice as old as his friend Peter. Peter is 5 years older than alice.
In 5 years, John will be three times as old as Alice. How old is Peter now?

Asked: How old is Peter now?

Given: Age of John, Peter and Alice

	
Let X represent Peter
John is twice as old as his friend peter
John = 2x
Peter is 5 years older than alice
Alice = x-5

After 5 years
	2x + 5
	x + 5
	x - 5 + 5
	
	In 5 years, John will be three times as old as Alice

		= 2x + 5 = 3(x-5+5)		= Simplify the Phrase first then Distribute X to 3
		= 2x - 2x + 5 = 3x - 2x		= Subtractive Property
		= x = 5				= Final Answer
		Peter'age will be 5 years old

Question # 2
One roofer can put a new roof on a house three times faster than another.
Working together they can roof a house in 5 days. How long would it take the
faster roofer working alone?

Asked: Number of hours would it take the faster roofer working alone?

Given: 	Together they can roof a house in 5days = T
	Let x represent the faster roofer
	Slower roofer = 3x

	T/x + T/3x = 1

		= 5/x + 5/3x = 1		= Write the given
		= 3x(5/x +5/3x = 1)3x		= Cross multiply and divide both side at 3x
		= 15 + 5 = 3x		
		= 20 = 3x		 
		= 20/3 = 3x/3			= Simplify 3x to x by dividing 
		= 20/3 = x		
		= 6.66 = x			= Final Answer
		Faster roofer would take 6.66 days working alone
	

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Lance Jazzper I. Frondozo
BSIT 1.4

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Alexander Louie A. Broncano
1.4 BSIT

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Maurillo, Julie
1.4BSIT

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#1

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#2

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Alcantara, John Benedict G. bsit 1.4 assignment #2

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Alcantara, John Benedict G. bsit 1.4 assignment #2

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Alexander Louie A. Broncano
1.4 BSIT

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1.4 BSIT

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BSIT 1.4

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PACHECO, NOEL S. 1.4BSIT

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HICBAN, JOSHUA
1.4 BSIT

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Santos, Mica Ella G.

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Moses Christian A. Bantegui
1.4 BSIT

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Calalang, Keith Christian C.

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Arlos Razelle Brian L. 1.4BSIT Assignment #2

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Arlos Razelle Brian L. 1.4BSIT Assignment #2

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1.4 BSIT

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Garcia, Rodolfo Jr. M.                             1.4BSIT

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Garcia, Rodolfo Jr. M.                             1.4BSIT

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Assignment 2

How old is Peter now?

Given:
John =x2
Peter =x
Alice = (x-5) + 5

Solution
2x + 5 = ( x - 5 + 5)
2x + 5 = 3x
5 = 3x - 2x
X = 5

How long would it take the fast roofer working alone?
Given. Speed of roofer = t
Speed of other roofer =3t
Speed of two roofer= 5t

Solution
5/t + 5/3t = 1
(3t) 5/t + 5/3t = 1 (3t)
15 + 5 = 3t
20 = 3t
20/3 = 3t/3
T = 20/3 hrs
6.67 days

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Aristhotel R. Arroyo
BSIT 1.4

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Jimbo Areniego
BSIT 1.4

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Delos Angelea Edgardo Jr.BSIT 1.4

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ASSIGNMENT 2
CHRISTIAN KURL RUZOL
1.4 BSIT

What is being asked: how old is peter now?
Given:
John=2x
Peter=x
Alice=x-5
Age in 5 years
John=2x+5
Peter=x+5

Solution:
2x + 5 = 3(x- 5 + 5)
2x + 5 = 3x
5 = 3x- 2x
X=5

Peter is now 5 years old

What is being asked: How long would it take the faster roofrr working alone?
Given: Speed of roofer1 = t
Speed of the other roofer = 3t
Speed of two roofer = 5 days

Solution:
5/t + 5/3t = 1
(3t) 5/t + 5/3t = 1 (3t)
15 + 5 = 3t
20 = 3t
20/3 = 3t/3
T = 20/3 hrs or 6.67 days

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EUDES MARTIN M. MUNOZ 1.4 BSIT ASSIGNMENT 2

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EUDES MARTIN M. MUNOZ 1.4 BSIT ASSIGNMENT 2

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Maurillo, Julie
1.4BSIT

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Arca, Jaybee Juber M.
1.4BSIT