Solutions of Word Problems Involving Equations

In the solution of problems, by means of equations, two things are necessary: First, to translate the statement of the question from common to algebraic language, in such a manner as to form an equation: Secondly, to reduce this equation to a state in which the unknown quantity will stand by itself, and its value be given in known terms, on the opposite side. The manner in which the latter is effected, has already been considered.

It is one of the principal peculiarities of an algebraic solution, that the quantity sought is itself introduced into the operation. This enables us to make a statement of the con ditions in the same form, as though the problem were already solved. Nothing then remains to be done, but to reduce the equation, and to find the aggregate value of the known quantities. (Art. 52.) As these are equal to the unknown quantity on the other side of the equation, the value of that also is determined, and therefore the problem is solved.

Problem 1. A man being asked how much he gave for his watch, replied; If you multiply the price by 4, and to the product add 70, and from this sum subtract 50, the remainder will be equal to 220 dollars.

To solve this, we must first translate the conditions of the problem, into such algebraic expressions as will form an equation.

Let the price of the watch be represented by   x
This price is to be multiplied by 4, which makes   4x
To the product, 70 is to be added, making   4x + 70
From this, 50 is to be subtracted, making   4x + 70 - 50

Here we have a number of the conditions, expressed in algebraic terms; but have as yet no equation. We must observe then, that by the last condition of the problem, the preceding terms are said to be equal to 220.

We have, therefore, this equation   4x + 70 - 50 = 220
Which reduced gives     x = 50.

Here the value of x is found to be 50 dollars, which is the price of the watch.

To prove whether we have obtained the true value of the letter wnich represents the unknown quantity, we have - only to substitute this value, for the letter itself, in the equation which contains the first statement of the conditions of the problem; and to see whether the sides are equal, after the substitution is made. For if the answer thus satisfies the conditions proposed, it is the quantity sought. Thus, in the preceding example,
The original equation is      4x + 70 - 50 = 220
Substituting 50 for x, it becomes   4.50 + 70 - 50 = 220
That is,        220 = 220.

Problem 2. What number is that, to which, if its half be added, and from the sum 20 be subtracted, the remainder will be a fourth of the number itself?

In stating questions of this kind, where fractions are concerned, it should be recollected, that (1/3)x is the same as x/3; that (2/5)x = 2x/5,&c. (Art. 158.)

In this problem, let x be put for the number required.
Then by the conditions proposed,   x + x/2 - 20 = x/4
And reducing the equation    x = 16.
         Proof,    16 + 16/2 - 20 = 16/4.

Problem 3. A father divides his estate among his three sons, in such a manner, that:
The first has $1000 less than half of the whole;
The second has 800 less than one third of the whole;
The third has 600 less than one fourth of the whole;
What is the value of the estate?
If the whole estate be represented by x, then the several shares will be x/2 - 1000, and x/3 - 800, and x/4 - 600.

And as these constitute the whole estate, they are together equal to x.
We have then this equation x/2 - 1000 + x/3 - 800 + x/4 - 600 = x.
Which reduced gives         x = 28800
Proof 28800/2 - 1000 + 28800/3 - 800 + 28800/4 - 600 = 28800.

To avoid an unnecessary introduction of unknown quantities into an equation, it may be well to observe, in this place, that when the sum or difference of two quantities is given, both of them may be expressed by means of the same letter. Foi if one of the two quantities be subtracted from their sum, it is evident the remainder will be equal to the other. And if the difference of two quantities be subtracted from the greater, the remainder will be the less.

Thus if the sum of two numbers be      20
And if one of them be represented by     x
The other will be equal to      20 - x.

Problem 4. Divide 48 into two such parts, that if the less be divided by 4, and the greater by 6, the sum of the quotients will be 9.

Here, if x be put for the smaller part, the greater will be 48 - x.

By the conditions of the problem x/4 + (48 - x)/6 = 9.
Therefore     x = 12, the less.
And      48 - x = 36, the greater.

Letters may be employed to express the known quantities in an equation, as well as the unknown. A particular value is assigned to the numbers; when they are introduced into the calculation: and at the close, the numbers are restored.

Problem 5. If to a certain number, 720 be added, and the sum be divided by 125 ; the quotient will be equal to 7392 divided by 462. What is that number?

Let x = the number required.
a = 720          d = 7392
b = 125          h = 462
Then by the conditions of the problem      (x + a)/b = d/h
Therefore          x = (bd - ah)/h
Restoring the numbers,x = [(125.7392) - (720.462)]/462 = 1280.

When the resolution of an equation brings out a negative answer, it shows that the value of the unknown quantity is contrary to the quantities which, in the statement of the question,"are considered positive.

Problem 6. A merchant gains or loses, in a bargain, a certain sum. In a second bargain, he gains 350 dollars, and, in a third, loses 60. In the end he finds he has gained 200 dollars, by the three together. How much did he gain or lose bv the first ?

In this example, as the profit and loss are opposite in their nature, they must be distinguished by contrary signs. If the profit is marked +, the loss must be -.
Let x = the sum required.
Then according to the statement      x + 350 - 60 = 200
And          x = -90.

The negative sign prefixed to the answer, shows that there was a loss in the first bargain ; and therefore that the proper sign of x is negative also. But this being determined by the answer, the omission of it in the course of the calculation can lead to no mistake.

Problem 7. A ship sails 4 degrees north, then 13 S. then 17 N. then 19 S. and has finally 11 degrees of south latitude. What was her latitude at starting ?
Let x = the latitude sought.
Then marking the northings +, and the southings -;
By the statement      x + 4 - 13 + 17 - 19 = -11
And        x = 0.

The answer here shows that the place from which the ship started was on the equator, where the latitude is nothing.

Problem 8. If a certain number is divided by 12, the quotient, dividend, and divisor, added together, will amount to 64. What is the number?

Let x = the number sought.
Then          x/12 + x + 12 = 64.
And          x - 624/13 = 48.

Problem 9. An estate is divided among four children, in such a manner that,
The first has 200 dollars more than 1/4 of the whole,
The second has 340 dollars more than 1/5 of the whole,
The third has 300 dollars more than 1/6 of the whole,
The fourth has 400 dollars more than 1/8 of the whole,
What is the value of the estate?      Answer 4800 dollars.

Problem 10. What is that number which is as much less than 500, as a fifth part of it is greater than 40?      Answer 450.

Problem 11. There are two numbers whose difference is 40, and which are to each other as 6 to 5. What are the numbers?      Answer 240 and 200.

Problem 12. What number is that, which is to 12 increased by three times the number, as 2 to 9 ?      Answer 8.

Problem 13. A ship and a boat are,descending a river at the same time. The ship passes a certain fort, when the boat is 13 miles below. The ship descends five miles, while the boat descends three. At what distance below the fort will they be together?      Answer 32,5 miles.

Problem 14. What number is that, a sixth part of which exceeds an eighth part of it by 20?      Answer 480.

Problem 15. Divide a prize of 2000 dollars into two such parts, that one of them shall be to the other, as 9:7.
         Answer The parts are 1125, and 875.

Problem 16. What sum of money is that, whose third part, fourth part, and fifth part, added together, amount to 94 dollars?      Answer 120 dollars.

Problem 17. A man spent one third of his life in England, one fourth of it in Scotland, and the remainder of it, which was 20 years, in the United States. To what age did he live?      Answer to the age of 48.

Problem 18. What number is that 1/4 of which is greater than 1/5 of it by 96?

Problem 19. A post is in the earth, 3/7 in the water and 13 feet above the water. What is the length of the post?
         Answer 35 feet.

Problem 20. What number is that, to which 10 being added, 3/5 of the sum will be 66?

Problem 21. Of the trees in an orchard, 3/4 are apple trees, 1/10 pear trees, and the remainder peach trees, which are 20 more than 1/8 of the whole. What is the whole number in the orchard?      Answer 800.

Problem 22. A gentleman bought several gallons of wine for 94 dollars; and after using 7 gallons himself, sold 1/4 of the remainder for 20 dollars. How many gallons had he at first?
         Answer 47.

Problem 23. What number is that, of which if 1/3, 1/4, and 2/7 be added together the sum will be 73?      Answer 84.

Problem 24. A person after spending 100 dollars more than 1/3 of his income, had remaining 35 dollars more than 1/2 of it. Required his income

Problem 25. In the composition of a quantity of gunpowder
The nitre was 10 lbs. more than 2/3 of the whole,
The sulphur 4,5 lbs. less than 1/5 of the whole,
The charcoal 2 lbs. less than 1/7 of the nitre.
What was the amount of gunpowder? Answer 69 lbs.

Problem 26. A cask which held 146 gallons, was filled with a mixture of brandy, wine, and water. There were 15 gallons of wine more than of brandy, and as much water as the brandy and wine together. What quantity was there of each?

Problem 27. Four persons purchased a farm in company for 4755 dollars; of which B paid three times as much as A; C paid as much as A and B; and D paid as much as C and B. What did each pay?      Answer 317, 951, 1268, 2219.

Problem 28. A father divided a small sum among four sons.
The third had 9 shillings more than the fourth;
The second had 12 shillings more than the third;
The first had 18 shillings more than the second;
And the whole sum was 6 shillings more than 7 times the sum which the youngest received.
What was the sum divided?      Answer 153.

Problem 29. A farmer had two flocks of sheep, each containing the same number. Having sold from one of these 39, and from the other 93, he finds twice as many remaining in the one as in the other. How many did each flock originally contain?

Problem 30. An express, travelling at the rate of 60 miles a day, had been dispatched 5 days, when a second was sent after him, travelling 75 miles a day. In what time will the one overtake the other?      Answer 20 days.

Problem 31. The age of A is double that of B, the age of B triple that of C, and the sum of all their ages 140. What is the age of each ?

Problem 32. Two pieces of cloth, of the same price by the yard, but of different lengths, were bought, the one for five pounds, the other for 6,5. If 10 be added to the length of each, the sums will be as 5 to 6. Required the length of each piece.

Problem 33. What number is that, which being severally added to 36 and 52, will make the former sum to the latter, as 3 to 4?

Problem 34. A gentleman bought a chaise, horse, and harness, for 360 dollars. The horse cost twice as much as the harness ; and the chaise cost twice as much as the harness and horse together. What was the price of each?

Problem 35. Out of a cask of wine, from which had leaked 1/3 part, 21 gallons were afterwards drawn; when the cask was found to be half full. How much did it hold?

Problem 36. A man has 6 sons, each of whom is 4 years older than his next younger brother; and the eldest is three times as old as the youngest. What is the age of each?

Problem 37. Divide the number 49 into two such parts, that the greater increased by 6, shall be to the less diminished by 11, as 9 to 2.

Problem 38. What two numbers are as 2 to 3; to each of which, if 4 be added, the sums will be as 5 to 7?

Problem 39. A person bought two casks of porter, one of which held just 3 times as much as the other ; from each of these he drew 4 gallons, and then found that there were 4 times as many gallons remaining in the larger, as in the other. How many gallons were there in each?

Problem 40. Divide the number 68 into two such parts, that the difference between the greater and 84, shall be equal to 3 times the difference between the less and 40.

Problem 41. Four places are situated in the order of the letters A. B. C. D. The distance from A to D is 34 miles. The distance from A to B is to the distance from C to D as 2 to 3. And 1/4 of the distance from A to B, added to half the distance from C to D, is three times the distance from 2? to C. What are the respective distances?
     Answer From A to B = 12; from B to C = 4; from C to D = 18.

Problem 42. Divide the number 36 into 3 such parts, that 1/2 of the first, 1/3 of the second, and 1/4 of the third, shall be equal to each other.

Problem 43. A merchant supported himself 3 years, for 50 pounds a year, and at the end of each year, added to that part ot his stock whicn whs iiot thus expended, a sum equal to one third of this part. At the end of the third year, his original stock was doubled. What was that stock?
         Answer 740 pounds.

Problem 44. A general having lost a battle, found that he had oniy half of his army +3600 men left fit for action ; 1/8 of the army +600 men being wounded ; and the rest, who were 1/5 of the whole, either slain, taken prisoners, or missing. Of how many men did his army consist?      Answer 24000.

For the solution of many algebraic problems, an acquaintance with the calculations of powers and radical quantities is required. It will therefore be necessary to attend to these before finishing the subject of equations.

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