Worded Problems on age, clock, mixture.pptx

Worded problems

1. Worded Problems

2. Coverage: Age Problems Coin Problems Clock Problems Digit Problems Lever Problems Mixture Problems Motion / Rate Problems Work Problems

3. Age Problems Case 1: For a single person How old is Gio now if in 10 years he will be twice his age 4 years ago? Solution: G = age now G – 4 = age 4 years ago G + 10 = age in 10 years to come Subject Past Present Future Giovanni G - years passed G G + years to come G + 10 = 2 ( G – 4) G = 18 years old

4. Age Problem Case 2: Comparing ages of two persons: Two years ago, a father was six times as old as his son. In 18 years, he will be twice as old as his son. How many years older is the father than his son? Equation 1: F – 2 = 6 ( S – 2) Equation 2 : 2(F + 18) = S + 18 Subjects Past Present Future Person1 X – years passed X X + years to come Person2 Y – years passed Y Y + years to come Subjects Past Present Future Father F – 2 F F + 18 Son S – 2 S F + 18 Solving the equation: F = 32, S = 7. Difference = 32 – 7 = 25 years

5. Nino is 4 years younger than his sister Kyla. In 10 year, Nino will be twice the age of Kyla now. What are their ages now? Solution: let x = Nino x + 10 = 2 (x + 4) x = 2 Subjects Present Future Nino X X + 10 Kyla x + 4 X + 14

6. Example: Eight years ago, the sum of the ages of Ana and Bill was 26. In five years, Ana will be 35 less than twice the age of Bill. How old is Bill? Solution: A = age of Ana B = age of Bill (A - 8) + (B – 8) = 26 A + B = 42  equation 1 (A + 5) = 2(B + 5) – 35 A – 2B = -30  equation 2 B = 24 years old

7. Coin Problems Coin Denomination Coin Value Total Value Penny, p 1 cent p Nickel, n 5 cents 5n Dime, d 10 cents 10d Quarter 25 cents 25q Half, h 50 cents 50h

8. Solution: Let q = no. of quarters 2q = no. of dimes 2(2q) – 2 = no. of nickels 25(q) + 10(2q) + 5(4q-2) = 185 25q + 20q + 20q -10 = 185 65q = 195 q = 3  no. of quarters Geoffrey has nickels, dimes, and quarters amounting to $1.85. If he has twice as many dimes as quarters, and the number of nickels is two less than twice the number of dimes, how many quarters does he have?

9. Harry bought a camera and received a change worth $13. He received 10 more dimes than nickels and 22 more quarters than dimes. How many dimes did he receive? Solution: 5(d – 10) + 10d + 25(d + 22) = 1300 5d + 10d + 25d = 1300 + 50 - 550 40d = 800 D = 20 Quantity Value Total Nickels d – 10 5 cents 5(d – 10) Dimes D 10 cents 10d Quarters D + 22 25 cents 25(d + 22) together 1300 cents

10. Clock Problems Basis of Analysis: x = no. of minute spaces covered by the minute hand = no. of minute paces traveled by the hour hand. This means that the minute hand travels 12 times faster than the hour hand.   Hands of the clock: Separation in degrees and minutes Together: 0 ° or 0 minutes apart Perpendicular: 90° or 15 minutes Opposite: 180° or 30 minutes Positions of the Hands of the Clock and Angle between them:

11. Shortcut: To determine the time (in minutes) elapsed from a given start-off hourly position after which the hands of the clock move to the desired position anywhere of being together, opposite each other, or that in which they form a right angle is given by the formula: Where: x = number of minutes elapsed from a given start-off hourly position n = the number on the face of the clock behind the minute hnd when it has moved to the desired position.  

12. Example: In how many minutes after 7 o’clock will the hands of the clock Be together for the first time? Perpendicular for the first time? Opposite for the first time? Solution: a) x = no. of minutes traveled by the minute hand x = 35 + x/12 11x/12 = 35 x = 38.18 minutes

13. b) perpendicular: X = no. of minutes traveled by the minute hand (x/12 + 35) – x = 15 35 – 15 = x – x/12 20 = 11x / 12 X = 21.82 minutes

14. c) Opposite: X = no. of minutes traveled by the minute hand (x/12 + 35) – x = 30 35 – 30 = x – x/12 5 = 11x/12 X = 5.4545 minutes

15. What time after 7 o’clock will the line joining the center of the clock and the 7 o’clock mark bisect the angle between the hands of the clock? Solution: From the figure, equate the angle: x/12 = 35 – x X = 32.31 minutes

16. Digit Problems: Basis of Analysis: For 2 digit number Let u = units digit t = tens digit t + u = sum of digits Valuation: 10t + u = original number 10u + t = reversed number For 3 digit number: Let u = units digit t = tens digit h = hundreds digit h + t + u = sum of digits Valuation: 100h + 10t + u = original number 100u + 10t + h = reversed number

17. The sum of the digits of a certain two digit number is 10. If the digits are reversed, a new number is formed which is one less than twice the original number. Find the original number. Solution: Let: t = tens digit u = units digit 10t + u = original number 10u + t = reversed number t + u = 10  equation 1 10u + t = 2(10t + u) -1 10u – 2u + t – 20t = -1 8u – 19t = -1  equation 2 t = 10 – u 8u – 19(10 – u) = -1 8u + 19u = -1 + 190 27u = 189 u = 7 t = 10 – 7 = 3 Original number = 37

18. Lever Problems Basic Principle: The basic principle of the lever is that the weight times the length of the arm on the left side is equal to the weight times the length on the right side of the lever. Basic formula : w 1 (x 1 ) = w 2 (x 2 ) Where w 1 and w 2 are weights X 1 and x 2 are lengths/distance from the fulcrum.

19. An iron bar 8 meters long has a 50 lb boy on one end and a 150 plb man at the opposite end. How far from the man should the fulcrum be located to balance the bar? Solution: X = distance of the man from the fulcrum 50(8 - x) = 150x 400 – 50x = 150x 200x = 400 X = 2 m

20. On a 16-foot seesaw Gio, weighing 80 lb , stands on one end. Next to Gio stands Christiansen, weighing 84 lbs. Christiansen is 4 feet from the fulcrum. On the other side at the end stands, Kristine, weighing 95b lbs. Where should Nino, weighing 75 lbs stand in order to balance the seesaw. Solution: 80(8) + 84(4) = 75x + 95(8) X = 2.88 ft

21. Mixture Problems Basic Principle: Analyze the mixture based on the pure stuff The amount of mixture times the percent of pure stuff equals the amount of pure stuff in a container.

22. If x mL of a mixture of coffee and cream containing 20% cream is added to a 50 mL mixture containing 45% cream to produce a mixture that is 30% cream, find x. Solution: X mL + 50 mL = x + 50 20% cream 45% cream = 30% cream 0.20x + 0.45(50) = 0.3(x+50) X = 75 mL

23. How many gallons of water must be added to 80 gallons of a 40% salt solution to produce a solution that is 25% salt. Solution: Using salt analysis: 80(0.4) + x (0) = (80 + x) (0.25) 32 + 0 = 20 + 0.25x 12 = 0.25x X = 48 Using water analysis: 80(0.6) + x(1)= (80+x)(0.75) 48 + x =60+0.75x 0.25x = 12 X = 48

24. Motion / Rate Problems Motion in a Straight Path: Case 1: Meeting Situation D = d 1 + d 2 Case 2: Departing Situation Case 3: Overtaking Situation D = d 1 + d 2 t 1 = t 2  

25. Motion / Rate Problems Motion in a Circular Path Working Equations: Moving in opposite direction and meet each other every time, t: Moving in the same direction and pass each other every time, t: Where: x = rate of the faster runner y = rate of the slower runner T = time D = total distance (equal to the circumference of the circular path)  

26. Two friends move at different but constant speeds along a circle of circumference 500 m. Starting at the same instant and from the same place, when they move in opposite directions they passed each other every 30 seconds and when they move in the same direction, they pass each other every 55 seconds. Determine their rates. Solution: X and y = speeds of faster and slower runners, respectively Opposite direction: The same direction: Solving for x and y we get: x = 12.89 m/s y = 3.78 m/s  

27. Motion / Rate Problems Water Motion against the current (upstream): Let R = rate / speed of the plane / object D = distance traveled C = rate of the current Motion with the current (downstream)   Wind Motion against the wind (headwind): Let R = rate / speed of the plane / object D = distance covered W = rate of the wind Motion with the wind (tailwind)  

28. The speed of a plane is 120 mph in calm air. With the wind it can cover a certain distance in 4 hours, but against the wind it can only cover 3/5 of that distance in the same time. Find the velocity of the wind. Solution: D against the wind = 3/5 ( D with the wind ) (R – W)t = 3/5(R + W)t (120 – W)t = 3/5(120 + W)t 600 – 5W = 360 + 3W 240 = 8W W = 30 mph

29. A boat propelled to move at 25 mph in still water, travels 4.2 miles upstream in the same time that it can travel 5.8 miles downstream. Find the speed of the stream. Solution: C = speed of current  

30. Number Problems A number 142 is divided into two parts such that when the greater part is divided by the smaller, the quotient is 3 and the remainder is 14. Find the larger part. Solution: x = smaller part 142 – x = larger part  

31. The denominator of a certain fraction is three more than twice the numerator. If 7 is added to both terms of the fraction, the resulting fraction is 3/5. Find the original fraction. Solution: x = numerator y = denominator Adding 7 to both numerator and denominator we get, The fraction is: 5/13  

32. Consecutive Numbers Integer Consecutive Integer Consecutive Even Integer Consecutive Odd Integers 1 st x x x 2 nd x + 1 x + 2 x + 2 3 rd x + 2 x + 4 x + 4 4 th x + 3 x + 6 x + 6 5 th x + 4 x + 8 x + 8 - - - - - -

33. Find three consecutive integers so that the sum of the first two is 35 more than the largest. Solution: x = first integer X + 1 = 2 nd consecutive number X + 2 = 3 rd number x + x + 1 = x + 2 + 35 2x + 1 = x + 37 x = 36 The numbers are:36, 37 and 38

34. Find three consecutive even integer if twice the smaller integer is 18 less than thrice the largest integer. Ans: 6,8,10

35. Find the sum of three consecutive odd integers if three times the largest is seven times the smallest. Ans: 3,5,7 = 15

36. Work Problems Basic Analysis Work rate of an individual (Rate analysis) Example: If a person can finish a certain job in 8 days, then his rate of doing work is Rate = 1/8 If two people A and B worked together and finished the job in T days, then the sum of their individual rate is equal to their combined rate Where T = total time to finish the job if all working together  

37. Work Problems Work done by an Individual For two or more individual working together: Complete work:  

38. Case 1: Determining Total Time to finish a job A father can do a job in 5 days, and his son can do the same job in 9 days. How long will it take them to finish the job if they work together? Solution: T = number of days it take for them working together 1/5 = rate of the father 1/9 = rate of the son  

39. Case 2: Determining additional time to finish a job started by another A mechanic can do a repair job in 6 hours. His helper can do the same job in 10 hours. On a given day the mechanic begins to work and after 2 hours he is joined by his helper. In how many hours will they complete the job? Solution: T = number of hours they can finish the job 1/6,1/10 = rate of mechanic and helper respectively By simplifying, we get:  

40. Case 3: Determining additional time to finish a job left by another who undoes the job An input pipe can fill a swimming pool in 7 hours. While a drain pipe can empty the pool in 11 hours. If for the first 4 hours, the drain was left open while the pipe was filling it, how much longer would it have taken the input pipe to fill the pool full if the pool was emptied at the start? Solution: X = additional time the inputpipe needs to fill the swimming pool alone  

41. Case 4: Determining Individual time to finish the job given rate relation An experienced carpenter works three times as fast as his helper. Working together, it would take them 6 hours to finish a certain job. How long would it take the helper to do the job alone? Solution: X = number of hours the helper can do the job alone 1/x = rate of the helper 3(1/x) = rate of carpenter