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How to Solve a Problem: Basic Tips

How to Solve a Problem: Basic Tips. Read the problem thoroughly until you understand what is the given (data) and what you are asked to find (unknown).

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How to Solve a Problem: Basic Tips

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  1. How to Solve a Problem: Basic Tips • Read the problem thoroughly until you understand what is the given (data) and what you are asked to find (unknown). • Whenever possible draw a picture or make a table representing the problem, where the data and the unknowns could be represented and it shows the relation between them. • For some problems you should have previous knowledge of analyzing the current situation and apply the necessary formulas. • Use letters to represent the unknowns. Basic and simple problems usually can be solved using one unknown. • Try to translate the problem to one or more equations. • To be able to solve a problem you need to find as many equations as unknowns. • Solve the equation (or system of equations) for the unknown(s). • Check the solution found in step 7 into the problem and see whether the answer is correct or not. Then write out the answer at the end.

  2. The following table contains 6 examples illustrated with different type of problems. If you want to see them all, simply just mouse click till the end, otherwise you can mouse click the specific example in the table that you prefer to see.

  3. Ex: 1. Twice a number Is 9 more than 204 minus the number. Find the number. • “Twice a number is 9 more than 204 minus the number. • The unknown is the number, let’s call it x. • Replacing the number by x the problem will look like • Twice x is 9 more than 204 minus x. • So, the equation associated to this problem is 2x = (204 – x) + 9 • Removing the parenthesis … 2x = 204 – x +9 • Isolating the x in the left side … 3x = 213 • Dividing by 3 both sides… x = 71 • Checking 71 as solution (in the original problem). • Twice the number is 2(71) =142 • 204 – the number is 204 – 71 = 133 • And 142 = 133 + 9 TRUE ! • Answer:The number is 71 Return to Table

  4. Ex 2: The sum of three consecutive even numbers is 1014. Find the numbers. • From problem (1st number)+(2nd number)+(3rd. number) = 1014 • We need to learn how to write three consecutive even numbers. • For instance if the first number is 22, then the second is 22+2 and the third is 22+4. So, given an even number its consecutive is the given number plus 2. •  Calling x the first number.  First number x (even) • Second number x+2 2 more than 1st • Third number x+4 2 more than 2nd • So the associated equation is x+(x+2)+(x+4) = 1014 • Removing parenthesis … x+x+2+x+4 = 1014 • Collecting like terms … 3x+6 = 1014 • Isolating the x … 3x = 1014 - 6 • Dividing by 3 … x = 1008/3 =336  • Check. First number 336 (even) • Second number 336+ 2 = 338 • Third number 336+ 4 = 340 • 336 +3438 +340 =1014. OK!  • Answer: The numbers are 336, 338, and 340. Return to Table

  5. Draw right triangle where Leg 1 = 10 and length hypotenuse = x So Leg 1=10 , Leg 2 = x-2, Hyp = x Pythagorean Theorem: In a right triangle: (Hyp)2 = (leg1)2 + (leg2)2 So, x2 = 102 + (x-2)2 Expanding x2 = 100 + x2 –4x+4 Isolating … 4x = 104 Dividing by 4 … x = 104/4 =26 So Hyp = 26 and Leg 2 = 26-2 = 24 Checking … 242 +102 = 262. OK! Answer: The hypotenuse length is 26 in. Ex 3: The length of one of the legs of a right triangle is 10 inches. The other leg is 2 less than the hypotenuse. Find the hypotenuse length. x x - 2 10 Return to Table

  6. Ex 4:The sale price of a computer is $1054 after 15% discount. Find the price before discount. • Let call x the price before discount. • So price before discount = x • discount = 15% of x = (0.15)x • sale price = 1054 • We know that price before discount – discount = sale price • Replacing … x – 0.15x = 1054 • Collecting … 0.85x = 1054 • Dividing by 0.85 both sides x = 1054/0.85 = 1240 • Check! … 15% of 1240 = 0.15(1240) = 186 • And Retail price = 1240 – 186 =1054. OK! • Answer: $ 1054 is the price before the discount. Return to Table

  7. Read the problem and realize that: Jon is 8 years older than Sue So J > S and J = S+ 8 Sue is 3 years younger than Frank. So   S < F and F = S+ 8 The sum of their ages is 98. J+S+F =98 Let’s x be Sue’s age x+8 Jon’s age x+3 Frank’s age So x+(x+8)+(x+3)=98 Removing parenthesis… x+x+8+x+3 = 98 Collecting … 3x +11= 98 Isolating … 3x = 87 Dividing by 3 … x = 29 So Sue has 29 Jon has 37 & Frank has 32 Check! … Jon is 8 years older than Sue. 37-29 = 8. OK! Sue is 3 years younger than Frank. 32-29 = 3. OK! The sum of their ages is 98. 27+29+32=98. OK! Answer: John has 27, Sue 29 and Frank 32. Ex 5:Jon is 8 years older than Sue and Sue is 3 years younger than Frank. The sum of their ages is 98. Find the ages of each one Return to Table

  8. Distance = Rate x Time Replacing … 400 = 10 t Dividing … 40 = t Answer: 40 sec Ex 6: The moving sidewalk in some airport is 400 ft long and moves at a speed of 4 ft/sec. If Tom can walk at a speed of 6 ft/sec, how long does it take him to walk the 400 feet using the moving sidewalk? 400 ft End Start 6ft/s 6ft/s 4ft/s Moving Sidewalk Not moving Distance to be traveled 400 ft Speed of sidewalk 4 ft/sec + Speed of Tom on sidewalk 6 ft/sec Total speed of Tom 10 ft/sec Time required t sec Return to Table

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