Break Up the Prompt: General principles

• Use variable names that represent the question's quantities. Suppose a machine performs a certain task you can jot down using the letter \(m =\) machine or \(t=\) task. At the end, when you set up a suitable equation, you immediately will be able to identify what the \(m\) or \(t\) stands for. If you use variables, such as x is too vague.
  
• Break down the problem into as many pieces as possible. Construct your equation one step at a time. Not try to write the entire situation in one shot, creating a supergiant equation. You will get lost on the way process, probably than not.
  
• When you have already set up the equation/s be sure to understand your final goal. Not only you must solve the equation per see but also you must solve it, understanding what the problem asks or means. Be sure to know EXACTLY what the question wants you to find out.
  
• if you have established the relationship between the two unknowns, label the smallest quantity of the two with the letter you pick.
  
• If the translation is a bit complex, paraphrase the sentences and personalize its words in a way you clearly understand what you are looking for.
  
• Organize step by step your translation of words into math. DO NOT mess up. Keep them clean with an understandable flowing. Do not rush: one of the students' most significant mistakes is running, pressured by the time. You have time. If you evaluate well up-front a problem, then you waste LESS time.
  
  
• I do not know where or how to start
  
  A passive student or someone who does not attack the question proactively waits for something happens to solve the question. On the other hand, those who aggressively attacks the question try always to find
  
  - a possible relationship.
  - try something, and if this does not lead anywhere, immediately switch in another direction without wasting time.
  - try multiple things simultaneously, seeing what best suits the question to solve in that precise moment. Sometimes, in solving a question, the students still process the previous question that they are unsure about. Focus 100% on the task you do have in front of you. An active thinker identifies unknowns and creates variables to represent them.
  
  
• I don’t know what they want me to do.
  
  If you wait until you know everything to start the price to solve a question, I.E. you wait like so many people do to be perfect, you will never begin. Begin small and figure out what you need for a piece by piece.
  
  Look at this problem and how we set up and solve the problem step by step.
  
  In 10 years, Bill will be twice as old as Jeffery. Today, Bill is 4 times as old as Jeffery. How old is Bill?
  
  Solution
  
  Today, Bill is 4 times as old as Jeffery
  Let J = Jeffery's PRESENT age
  So, 4J = Bill's PRESENT age
  
  In 10 years, Bill will be twice as old as Jeffery.
  Since we already know each person's PRESENT age, we know the following...
  J + 10 = Jeffery's age IN 10 YEARS
  4J +10 = Bill's age IN 10 YEARS
  
  If Bill is twice as old as Jeffrey in 10 years we can write: 4J + 10 = 2(J + 10)
  Expand to get: 4J + 10 = 2J + 20
  Solve to get: J = 5
  
  So, Jeffery's PRESENT age is 5
  Since 4J = Bill's PRESENT age, we know that Bill's PRESENT age = 4(5) = 20

Translating Rightly

Word Translation
Words	Math
A is half the size of B	\(A = \frac{1}{2} B\) NOT \(B =\frac{1}{2}A\)
A is 5 less than B	\(A=B-5\) NOT \(B=5-B\)
A is less than B	\(A < B\) NOT \(A > B\)
Carcass bought twice as many cars as bikes	\(C=2B\) NOT \(2C=B\)
P is X percent of Q	\(P=\frac{X}{100} Q \) NOT \(P=X\)%\(Q\)
Pay $30 per pair of boots, then $28 per additional pair	\(n=#\) the number of pair of boots purchased \(+\) \(T =\) the total amount paid for $ \(T=$30 + $28 \times (n-1)\)

Most common word problem translations






More Examples


1. Find two consecutive odd numbers the difference of whose squares is 296.

Sol. Let the numbers be 2X + 1 and 2X + 3 Then (2X + 3)² – (2X + 1)² = 296 ⇒ X = 36
Hence 2X+ 1 = 2 x 36 + 1 = 73 and 2X + 3 = 75 ⇒ The required numbers are 73 and 75. [Verification. (75)² – (73)²= 5625 – 5329 = 296]


2. A is 29 years older than B, B is 3 years older than C and D is 2 years younger than C. Two years hence A’s age will be twice the united ages of B, C and D. Find their present ages.

Sol. Let D’s age be = X years Then C’s age = (X + 2) years, B’s age = (X + 5) years and A’s age = (X + 34) years. Two yrs hence A’s, B’s, C’s and D’s ages will be X + 36, X + 7, X + 4 and X + 2 years respectively. ⇒ 2 (X + 2 + X + 4 + X + 7) = X + 36 ⇒ X = 2 ⇒ A’s age = 36 yrs; B’s age = 7 yrs, C’s age = 4 yrs; D’s age = 2 yrs.


3. A number consists of three consecutive digits, that in the unit’s place being the greatest of them three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digit. Find the number.

Sol. Let the hundred’s digit be X. Then the ten’s digit = X + 1 and the unit’s digit = X + 2 ⇒ The number = 100 x X + 10(X + 1) + X + 2 = 111X + 12. The number formed by reversing the digits = 100(X + 2) + 10(X + 1) + X = 111X + 210 ⇒ 111X + 210 – 111X – 12 = 22 (X + 2 + X + 1 + X) ⇒ X = 2. Hence the required number = 234.


4. The crew of a boat can row at the rate of 5 miles an hour in still water. If to row 24 miles, they take 4 times as long as to row the same distance down the river, find the speed at which the river flows.

Sol. Let X miles per hour be the speed of the river. Hence, on equating the times, we get: \(\frac{24}{(5 – X)} = 4 \times \frac{24}{(X + 5)} ⇒ X = 3\) Thus, the river flows at the rate of 3 miles an hour.


5. The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length decreases by 7 m and breadth increases by 5 m. Find the dimensions of the rectangle.

Sol. Let the length of the rectangle be X m and breadth of the rectangle = Y m. Area = XY sq. m.

I Case : Length = (X + 7) m and breadth = (Y – 3) m ⇒ Area = (X + 7) (Y – 3) sq. m. ⇒ (X + 7) (Y – 3) = XY or – 3X + 7Y – 21 = 0 ....

II Case : Length = (X – 7) m and breadth = (Y + 5) m ⇒ (X – 7) (Y + 5) = XY or 5X – 7Y – 35 = 0 .... (2) ⇒ Y = 15 and X = 28 Hence length = 28 m and breadth = 15 m Answer.

Note : Assume L = 28 and B = 15 as an option and try checking the conditions given in the problem. You will see that working backward is exceptionally fast in such cases.


6. The ratio of incomes of two persons is 9 : 7, and the ratio of their expenditures is 4 : 3. If they save Rs. 200 per month, find their monthly incomes.

Sol. Let the monthly income of the first person be Rs 9X and the monthly income of the second person be Rs 7X. Let the expenditure of the first person be 4Y and the expenditure of the second person be 3Y. ⇒ Saving of the first person = Rs (9X – 4Y) and solving of second person = Rs (7X – 3Y). Using the given informations, we have : 9X – 4Y = 200 .... (1) and 7X – 3Y = 200 .... X = 200. Hence, the monthly income of first-person = Rs. 9 x 200 = Rs. 1800 and the monthly income of second person = Rs. 7 x 200 = Rs. 1400


7. Find two consecutive even numbers such that 1/6th of the greater exceeds 1/10th of the smaller by 29.

Sol. Let the numbers be 2X and 2X + 2 Then (2X + 2)/6 – 2X/10 = 29. ⇒ X = 21 Hence 2X = 430 and 2X + 2 = 432. ⇒ The required numbers are 430 and 432. [Verification. \(\frac{432}{6} – \frac{430}{10} = 72 – 43 = 29\)]


8. A number consists of two digits whose sum is 12. The ten’s digit is three times the unit’s digit. What is the number?

Sol. Let the unit’s digit be X, Then the ten’s digit is 12 – X. ⇒ 3X = 12 – X ⇒ X = 3 Hence the number is 93. [Verification. 9 = 3 x 3; and 9 = 3 = 12]


9. A train traveled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the journey would have occupied 4 hours less; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance.

Sol. Let us suppose that X miles per hour is the speed of the train and Y hours is the time taken for the journey. ⇒ Distance travelled = XY = (X + 6) (Y – 4) = (X – 6) (Y + 6) This gives two simultaneous equations. Solving, we get : X = 30, Y = 24 ⇒ Distance = XY = 720 miles


10. A sum of money was divided equally among a certain number of persons; had there been six more, each would have received a rupee less, and had there been four fewer, each would have received a rupee more than he did; find the sum of money and the number of men.

Sol. Let X be the number of persons and Rs Y be the share of each. Then by the conditions of the problem, we have (X + 6) (Y – 1) = XY .......(1) (X – 4) (Y + 1) = XY .......(2). Thus the number of person is X = 24 and the share of each is Y = Rs 5. The sum of money = 5 x Rs 24 = Rs 120.