NOTES
1. This tutorial in Taglish is based on Module: Equations 2 prepared and published by the Department of Education of the Philippines for the Alternative Learning System (ALS) program.
2. Sorry for any typographical and/or grammatical error that has been missed.
3. Please comment for any incorrect answers.
Lesson 5 – Applications of Linear Equations: Part 2
Sa nakaraang mga aralin ay nai-apply natin ang ating natutunang konsepto at pamamaraan sa paglutas ng mga word problems na kinasasangkutan ng linear equations na may kinalaman sa number, geometric, at work problems. Ang ating leksyon sa kasalukuyan ay nakapokus naman sa mga word problems na may kinalaman sa uniform motion, investment, at mixture problems.
Uniform Motion Problems
Ang mga problemang may kinalaman sa distansya, d, oras, t, at uniform rate, r, ay tinatawag na mga uniform motion problems. Ang uniform motion ay isang paggalaw sa isang tuwid na linya na may pare-parehong bilis o velocity.
Ang mga problemang ito ay madalas na nililinaw sa pamamagitan ng paggawa ng sketch ng mga relasyon na kasangkot.
Sa ganitong mga klase ng word problem, dapat nating alalahanin ang ugnayan ng speed, time, at rate sa isa’t isa. Ito ay sa pamamagitan ng pagkuha ng isa sa kanilang mga formula.
Karaniwan na nating naririnig ang salitang “80 km/hr” (80 kilometers per hour) na kumakatawan sa bilis (speed o rate (r) ng isang sasakyan o bagay na gumagalaw. Ang 80 km ay kinakatawan ng distance (d) o distansyang tinakbo, ang hour naman ay ang time (t) o oras. Samakatuwid, ang rate = distance/time o r = d/t.
Mula sa pormulang ito ay makukuha rin natin ang relationship ng distance (d) sa time (t) at rate (r), gayundin ang time (t) sa distance (d) at rate (r). Gawin natin.
r = d/t ==> t(r = d/t) ==> rt = d or d = rt, ibig sabihin, ang distance (d) ay rate (r) multiply by time (t).
Mula sa d = rt ==> d/r = rt/r ==> d/r = t or t = d/r.
Ibig sabihin nito, ang time (t) ay distance (d) divided by rate (r).
Isulat nating muli ang ating nakuhang pormula o relationship, kung saan ang
r = rate, d = distance, at t = time. Kailangan natin ang kaalamang ito sa pagsagot ng uniform motion problems.
r = d/t
d = rt
t = d/r
Hindi kailangang saulunin lahat ang mga pormula. Isang pormula ay sapat na dahil makukuha natin ang iba pang pormula mula rito tulad nang ginawa natin sa nakaraang pahina.
HALIMBAWA 1
A policeman on a motorcycle is pursuing a car the speed of which is 115 kilometers per hour. The policeman is 6 km behind the car and is moving at a speed of 130 kilometers per hour. How long will it take the policeman to overtake the car?
I-sketch natin ang ating problema.
Gumawa tayo ng table o talahanayan upang madaling malutas ang ating problema.
Upang makumpara ang distansya ng dalawang sasakyan, dapat ay iisa ang starting point ng mga ito. Ang pinili kong starting point ay ang car na kakatawanin ng x. Ang motorcycle ay naging x + 6, dahil ang pagitan nito sa car ay 6 km na.
Mamaya ay kukunin natin ang starting point ng motorcycle at ikumpara natin kung pareho ang ating magiging sagot.
Distance traveled by motorcycle:
x + 6 = 130t
Isolate x:
x = 130t – 6 (Transposition)
Distance traveled by car:
x = 115t
Equate the two equations:
Since they have the same distance traveled as x.
130t – 6 = 115t
130t – 115t – 6 + 6 = 115t – 115t + 6 (Transposition) 15t = 6
15t/15 = 6/15
t = 6/15
t = 2/5 (Simplification)
Thus, the motorcycle will catch up with the car after 2/5 of an hour or 24 minutes ==> ( 2/5 hr (60 min/hr) = 120/5 = 24 min.)
Kung ang distance ng motorcycle ang ating magiging starting point, ganito na ang kalalabasan:
Distance traveled by motorcycle:
x = 130t
Distance traveled by car:
x – 6 = 115t
Kaya x – 6 dahil mas malayo na ng 6 km ang car mula sa motorcycle.
Isolate x:
x – 6 = 115t
x = 115t + 6
Equate the two equations:
130t = 115t + 6
130t – 115t = 115t – 115t + 6 (Transposition)
15t = 6
15t/15 = 6/15
t = 6/15 or 2/5 or 24 minutes
Makikita na pareho lang ang ating nakuhang sagot.
HALIMBAWA 2
A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.
Illustrate the problem:
Ilagay natin sa table ang ating mga given data:
Distance traveled by car:
d + 20 = (2r – 30)2 (Equation 1)
Distance traveled by bus:
d = 2r (Equation 2)
We can substitute Equation 2 into Equation 1 to find r.
d + 20 = (2r – 30)2
2r + 20 = (2r – 30)2
2r + 20 = 4r – 60
2r + 20 – 20 – 4r = 4r – 60 – 4r – 20
-2r = -80
-2r/-2 = -80/-2
r = 40 mph ==> rate of bus
2r – 30 = 2(40) – 30
= 80 - 30
= 50 mph ==> rate of car
Thus, the rate of the car is 50 mph.
Let us check our answer by substituting our values into our original equations.
For Equation 1:
d + 20 = (2r – 30)2
80 + 20 ≟ [2(40) – 30]2
80 + 20 ≟ (80 – 30)2
80 + 20 ≟ 100
100 = 100
For Equation 2:
d = 2r
80 ≟ 2(40)
80 = 80
Since our left and right terms for our original equations are balanced, our answer is correct.
Investment Problems
Ang ilang mga problema sa pamumuhunan ay gumagamit ng formula ng interes na I = PRT. Sa formula na ito, kinakatawan ng I ang interes, P, ang principal o halagang pinuhunan, R (%), ang rate ng interes at T, ang oras. Para sa mga layunin ng ating talakayan, dapat nating ipagpalagay na R ang taunang rate ng interes at ang T ay ibibigay sa mga taon (years).
Karamihan sa mga problema sa pamumuhunan ay nagsasangkot ng mga equation na may mga decimal coefficient. Ang nasabing equation ay maaaring malutas sa pamamagitan ng pagpaparami ng magkabilang panig ng equation sa pamamagitan ng power of 10 upang makakuha ng katumbas na equation na may mga whole number coefficients.
HALIMBAWA 3
An investment counselor invested 75% of a client’s money into a 9% annual simple interest money market fund. The remainder was invested in 6% annual simple interest government securities. Find the amount invested in each if the total annual interest earned is ₱3300.
Let x = total amount invested
0.75x = amount invested in the market fund
0.25x = amount invested in government securities
Gumawa tayo ng table para maging malinaw ang lahat.
Set up our equation:
0.75x(0.09) + 0.25x(0.06) = 3300
.0675x + .015x = 3300
1000(.0825x = 3300)
825x/825 = 3300000/825
x = ₱40000 ==> total amount invested
0.75x = 0.75(40000) = ₱30000 ==> invested in money market
0.25x = 0.25(40000) = ₱10000 ==> invested in government securities
Substitute 40,000 as the value of x into the original equations to check if our answer is correct.
0.75x(0.09) + 0.25x(0.06) = 3300
0.75(40,000)(0.09) + 0.25(40,000)(0.6) ≟ 3300
2,700 + 600 ≟ 3300
3,300 = 3300
Since we got equality, our answer is correct.
HALIMBAWA 4
A man invests x pesos in bonds and ₱8000 more than this in stocks. How much did he invest in stocks if his total investment is ₱15000?
Let x = investment in bonds
x + 8000 = investment in stocks
Our equation:
x + x + 8000 = 15000
2x + 8000 = 15000
2x + 8000 – 8000 = 15000 – 8000
2x = 7000
2x/2 = 7000/2
x = ₱3500 ==> invested in bonds
x + 8000 = 3500 + 8000 = 11500
Thus, ₱11500 is invested in stocks.
Substitute 3500 as the value of x into our original equation to check if our answer is correct.
x + x + 8000 = 15000
3500 + 3500 + 8000 ≟ 15000
15000 = 15000
Since we got equality, our answer is correct.
Mixture Problems
Ang isang paraan ng pagsusuri at paglutas ng mga mixture problems ay inilalarawan sa sumusunod na halimbawa.
HALIMBAWA 5
Tom and Jerry blended a chocolate mix that sells for ₱5 by mixing two types of chocolate. If they used 80 mL of chocolate that costs ₱6, how much of another chocolate costing ₱3 did they mix with the first?
Let x = amount of ₱3 chocolate mixed with ₱6 chocolate
Kung hindi pa bihasa sa ganitong klaseng problema, makatutulong nang malaki ang paggawa ng table upang ilagay ang ating mga datos.
Set up our equation:
3x + 6(80) = 5(80 + x)
3x + 480 = 400 + 5x
3x + 480 – 5x – 480 = 400 + 5x – 5x – 480
-2x = - 80
-2x/-2 = -80/2
x = 40
Thus 40 mL of chocolate @ ₱3 is needed for the mix.
Ihalili natin ang 40 bilang value ng x sa ating orihinal na equation upang matiyak na tumpak ang ating sagot.
3x + 6(80) = 5(80 + x)
3(40) + 6(80) ≟ 5(80 + 40)
120 + 480 ≟ 5(120)
600 = 600
Dahil ang ating nakuha ay balanse, tumpak ang ating sagot.
HALIMBAWA 6
A chemist has 10 mL of a solution that contains 30% acid. How many mL of pure acid must be added to it in order to increase its acid content to 50%?
Let x = amount of pure acid to be added
Set up our equation:
(10 + x) (0.5) = 3 + x
5 + 0.5x = 3 + x
10(5 + 0.5x = 3 + x)
50 + 5x = 30 + 10x
50 + 5x – 50 – 10x = 30 + 10x – 10x – 50
-5x = -20
-5x/-5 = -20/-5
x = 4
Thus, 4 ml of pure acid must be added to the solution to make its acid content 50%.
Substitute 4 for x into our original equation to check if our answer is correct.
(10 + x) (0.5) = 3 + x
(10 + 4) (0.5) ≟ 3 + 4
14(0.5) ≟ 7
7 = 7
Since we got equality, our answer is correct.
Pagsasanay
Solve the following word problems and check your answers:
1. Kiko and Kikay start from the same point and jog in opposite directions. Kiko jogs 4 km/h faster than Kikay. After 3 hours, they are 30 kilometers apart. How fast did each jog?
2. Bongbong leaves his house riding a bike at 20 km/h. Leni leaves 4 hours later on a motorcycle to catch up with him traveling at 60 km/h. How long will it take her to catch up with him?
3. Manny invested ₱10,000, part at 6%, and the rest at 5%, in interest-bearing accounts. At the end of the year, his total investment income was ₱566. How much did Manny invest at each rate?
4. An investment of ₱6,000 is made at an annual simple interest rate of 6%. How much additional money must be invested at an annual simple interest rate of 10% so that the total annual interest earned is 9% of the total investment?
5. White sugar worth ₱70 per kilo is mixed with brown sugar worth ₱50 per kilo to obtain 50 kilos of a mixture worth ₱65 per kilo. How many kilos of each type is used?
6. Solution A is 60% hydrochloric acid, while solution B is 85% hydrochloric acid. How many liters of each solution should be used to make 90 liters of a solution which is 75% hydrochloric acid?
ANSWERS:
