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Word problems , translating English into equations

The most testable application of this chapter is the word problem: an English description hiding two linear equations. Once you have the equations, any of the methods (substitution, elimination, even graph) will solve them. The art is the translation.

A four-step plan

  1. Read carefully, identifying the two unknowns. Give them names like xx and yy.
  2. Set up two equations from the conditions in the problem. Most well-posed word problems give exactly two pieces of independent information.
  3. Solve by substitution or elimination.
  4. Interpret the answer: convert back to the units / meanings the problem uses, and check for consistency (no negative ages, no fractional people, etc.).

Common templates

Ages. "Five years ago, AA was four times as old as BB" \Rightarrow A5=4(B5)A - 5 = 4(B - 5). "Ten years hence, AA will be twice BB" \Rightarrow A+10=2(B+10)A + 10 = 2(B + 10).

Two-digit numbers. Let the tens digit be xx and units digit be yy. The number is 10x+y10 x + y. The reversed number is 10y+x10 y + x.

Fractions. "When 11 is added to the numerator and 11 subtracted from the denominator, the fraction becomes 1/21/2" , let the fraction be x/yx/y.

Speed–time–distance. Distance == speed ×\times time. With a stream/current of speed uu: upstream speed =vu= v - u, downstream =v+u= v + u.

Work. If AA alone takes aa days and BB alone bb days, together they take aba+b\dfrac{ab}{a + b} days. Word problems usually give the combined work time and a relation between a,ba, b.

Mixtures. "33 kg of tea at ₹100100/kg and 22 kg of tea at ₹xx/kg gives a mix at ₹yy/kg" , track total cost and total weight separately.

A general template

If two variables xx and yy are linked by two independent conditions, set them up as a pair of linear equations and solve. The hardest part is correctly writing the equations; we'll practise on the standard examples below.

Worked examples

Example 1 (ages). The present ages of a father and his son are in the ratio 7:37 : 3. After 1010 years, the ratio becomes 2:12 : 1. Find their present ages.

Let father =7x= 7 x, son =3x= 3 x (years). 7x+103x+10=217x+10=2(3x+10)=6x+20x=10.\frac{7 x + 10}{3 x + 10} = \frac{2}{1} \Rightarrow 7 x + 10 = 2(3 x + 10) = 6 x + 20 \Rightarrow x = 10. Father =70= 70 years, son =30= 30 years.

Example 2 (two-digit number). A two-digit number is such that the sum of its digits is 99. When 2727 is added to the number, the digits are reversed. Find the number.

Let tens digit =x= x, units digit =y= y. Number =10x+y= 10 x + y.

Conditions: x+y=9x + y = 9 and 10x+y+27=10y+x10 x + y + 27 = 10 y + x, i.e., 9x9y=27xy=39 x - 9 y = -27 \Rightarrow x - y = -3.

Adding to x+y=9x + y = 9: 2x=6x=32 x = 6 \Rightarrow x = 3, y=6y = 6. Number =36= 36.

Check: 3+6=93 + 6 = 9 ✓; 36+27=6336 + 27 = 63 which is digit-reversed. ✓

Example 3 (boats and current). A boat goes 3030 km upstream and 4444 km downstream in 1010 hours. In 1313 hours it can go 4040 km upstream and 5555 km downstream. Find the speed of the boat in still water and the speed of the stream.

Let boat speed =u= u km/h, stream speed =v= v km/h. Upstream =uv= u - v, downstream =u+v= u + v.

30uv+44u+v=10,40uv+55u+v=13.\frac{30}{u - v} + \frac{44}{u + v} = 10, \qquad \frac{40}{u - v} + \frac{55}{u + v} = 13. Let p=1/(uv),q=1/(u+v)p = 1/(u - v), q = 1/(u + v). System: 30p+44q=1030 p + 44 q = 10, 40p+55q=1340 p + 55 q = 13.

Multiply first by 44 and second by 33: 120p+176q=40120 p + 176 q = 40, 120p+165q=39120 p + 165 q = 39. Subtract: 11q=1q=1/1111 q = 1 \Rightarrow q = 1/11, so u+v=11u + v = 11. From first: 30p+4=10p=1/530 p + 4 = 10 \Rightarrow p = 1/5, so uv=5u - v = 5.

Hence u=8,v=3u = 8, v = 3. Boat speed =8= 8 km/h, stream =3= 3 km/h.

Example 4 (fraction). A fraction becomes 1/31/3 when 11 is subtracted from the numerator and it becomes 1/41/4 when 88 is added to its denominator. Find the fraction.

Let the fraction be x/yx/y. Then (x1)/y=1/3(x - 1)/y = 1/3 and x/(y+8)=1/4x/(y + 8) = 1/4.

From the first: 3(x1)=yy=3x33(x - 1) = y \Rightarrow y = 3 x - 3. From the second: 4x=y+8y=4x84 x = y + 8 \Rightarrow y = 4 x - 8.

Equate: 3x3=4x8x=53 x - 3 = 4 x - 8 \Rightarrow x = 5. So y=12y = 12. Fraction =5/12= 5/12.

Example 5 (work). 22 men and 77 boys can do a piece of work in 44 days, while 44 men and 44 boys do the same work in 33 days. How long would 11 man alone take, and 11 boy alone?

Let one man do mm units of work per day, one boy do bb units. Total work =1= 1.

(2m+7b)4=1(2 m + 7 b) \cdot 4 = 1 and (4m+4b)3=1(4 m + 4 b) \cdot 3 = 1, i.e., 8m+28b=18 m + 28 b = 1 and 12m+12b=112 m + 12 b = 1.

Subtract suitable multiples. Multiply first by 33 and second by 22: 24m+84b=324 m + 84 b = 3 and 24m+24b=224 m + 24 b = 2. Subtract: 60b=1b=1/6060 b = 1 \Rightarrow b = 1/60, so a boy alone does 1/601/60 of the work per day , 6060 days for the boy. From 12m+12/60=112 m + 12/60 = 1: 12m=4/5m=1/1512 m = 4/5 \Rightarrow m = 1/15, so a man takes 1515 days alone.

Try it yourself

  1. The sum of the digits of a two-digit number is 1212. The number obtained by interchanging the digits exceeds the original by 1818. Find the number.
  2. A man rows downstream 2020 km in 22 hours and upstream 44 km in 22 hours. Find his rowing speed in still water and the stream speed.
  3. The cost of 22 kg apples and 11 kg grapes is ₹160160. After a month the cost is ₹300300 for 44 kg apples and 22 kg grapes. (Same prices.) Find the cost per kg.
  4. Five years ago Nuri was thrice as old as Sonu. Ten years later Nuri will be twice as old as Sonu. How old are they now?
  5. A railway half-ticket costs half the full fare and the reservation charge is the same on both. Two reservation tickets from station AA to BB cost ₹25302530 (one full and one half). Find the full first-class fare from AA to BB if reservation charge is ₹1010.
  6. The denominator of a fraction is 11 more than twice the numerator. If the sum of the fraction and its reciprocal is 216212 \tfrac{16}{21}, find the fraction.
  7. A father is 33 times as old as his son. In 1212 years' time he will be twice his son's age. Find their present ages.
  8. In a rectangle, if length is increased by 22 m and breadth decreased by 22 m, the area decreases by 2828 m². If length is decreased by 11 m and breadth increased by 22 m, area increases by 3333 m². Find the dimensions.
  9. The sum of the two-digit number 10x+y10x + y and the number formed by reversing its digits is 9999. Find x+yx + y. (One equation; comment on why this doesn't pin down the number.)
  10. Two pipes can fill a tank in 2020 min and 3030 min respectively. If both are open, how long to fill the tank?

Pitfalls / Insight

  • Define variables clearly with units. "Let xx be the speed" is not enough , "Let xx km/h be the speed of the boat in still water" is.
  • Check that the answer makes sense. Negative ages, fractional people, or speeds exceeding light should make you re-check.
  • In digit problems, the digits are integers between 00 and 99; the tens digit must be at least 11.

Insight. Almost every word problem of this chapter follows the pattern two unknowns, two conditions. If you can spot the unknowns and write the conditions, the algebra is routine.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Word problems
6 questions · pick the best answer
Q1

Father:son age ratio is 7:37:3; after 1010 years it becomes 2:12:1. Son's present age is:

Q2

Sum of digits is 99; reversing increases the number by 2727. The number is:

Q3

Boat goes 3030 km up and 4444 km down in 1010 h; 4040 up, 5555 down in 1313 h. Boat speed in still water is:

Q4

Fraction becomes 1/31/3 if 11 subtracted from numerator; 1/41/4 if 88 added to denominator. Fraction is:

Q5

Two pipes fill a tank in 2020 and 3030 minutes. Both together fill it in:

Q6

Length increased by 22, breadth decreased by 22: area down by 2828 m². Length decreased by 11, breadth increased by 22: area up by 3333 m². The original length is: