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Let x represent A's income, and y B's income.
+y=570; and 3x+5y=2350.
These Equations will give a $250, and y=$320.
19. How old are we?" said a person to his father : ago,” replied the latter, "I was a third more than 3 times as old as you were; and in three years, if I multiply your age by 2, it will then be equal to mine." What were their ages?
Let x represent the father's age, and y the son's age; then
x+3 will be the father's age 3 years hence; and
x-6=3(y-6)+ ; and
from which we shall find x=36, and y=15 years.
20. Find a number such that if we subtract it from 4980, divide the remainder by 8, and subtract 123 from the quotient, we shall find a remainder equal to the number itself.
Let x represent the number; then
4980- represents the remainder after subtracting the number from 4980; and
represents the quotient after dividing the remainder
Then, subtracting 123 from this quotient, we shall have the Equation 4980-x -123=x;
which will give x=444.
21. A laborer engaged for 40 days upon these conditions; that for every day he worked he should receive 80 cents, but for every day he was idle he should forfeit 32 cents. At the end of the time he was entitled to $15.20. How many days did he work, and how many was he idle?
Let x represent the number of days he worked, and y the number he was idle; then
80x represents the number of cents he earned; and
These Equations will give x=25, and y=15 days.
22. A cistern containing 820 gallons is filled in 20 minutes by 3 pipes, the first of which conveys 10 gallons more, and the second 5 gallons less than the third, per minute. How much flows through each pipe in a minute?
Let x, y, and z respectively represent the number of gallons that flow through each pipe per minute.
Then, since 20x, 20y, and 20z respectively represent the number of gallons that flow through each pipe in 20 minutes, we shall have the Equation
20x+20y+20z=820. And the other Equations will evidently be
From these three Equations we shall find x=22, y=7, and z=12 gallons.
23. A trader maintained himself for 3 years at an expense of £50 a year, and each year augmented that part of his stock which was not thus expended by thereof. At the end of the third year his original stock was doubled. What was that stock?
Let x represent his stock; then
x-50 is what was left after the 1st expenditure;
ing his stock the 1st time;
is what he had after augment-.
is what was left after the 2d ex
+3 of 4x-350),
And since A lost
(x + 40 — 2 +40),
had after augmenting his stock the 3d time. Then, by the problem, the Equation will be
Let x represent the sum; then
x+40 is what A had at the end of the 1st year;
which will give x=£740.
24. A and B began to trade with equal sums of money. The first year A gained $40; and B lost $40; the second year A lost of what he had at the end of the first, and B gained $40 less than twice what A lost; when it appeared that B had twice as much money as A. What sum did each begin with?
is what was left after the
is what he had
is what A had at the end of the
and B gained $40 less than twice that
end of the second year.
Then, by the problem, the Equation will be
From this Equation we shall find x= = $320.
is what he
-40 represents what B gained the 2d year; and x+40
is what B had at the
25. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes; but the denominator being doubled, and the numerator increased by 2, the value becomes ?
Let-represent the fraction; then the Equations will be
From these Equations we shall find x=4, and y=5. Hence the fraction is
26. A and B together can perform a piece of work in 8 days, A and C in 9 days, and B and C in 10 days. How many days would it take each person to perform the work alone?
Let x, y, and z respectively represent the number of days in which A, B, and C could do it; then
represents the part A could do in 1 day;
represents the part B could do in 1 day;
represents the part C could do in 1 day.
And, since A and B together, in 1 day, would do and C together of it, and B and C together will be
Subtracting the 2d Equation from the 1st, we shall have
Adding this to the 3d Equation, we shall have
of the work, A of it, the Equations
which will give y=1723. The values of x and z will now be found equal to 1434, and 23, respectively.
27. From two places which are 154 miles apart, two persons set out at the same time to meet each other, one traveling at the rate of 3 miles in 2 hours, and the other at the rate of 5 miles in hours; in how many hours will they meet?
Let x represent the number of hours; then, since
is the number of miles the first goes per hour; and
is the number of miles the second goes per hour;
is the number of miles the first goes in x hours; and
which will give x=
is the number of miles the second goes in x hours.
Then, since both together traveled 154 miles, the Equation will be
28. In a naval engagement, the number of ships captured was 7 more, and the number burned was 2 less, than the number sunk. Fifteen escaped, and the fleet consisted of 8 times the number sunk. Of how many ships did the fleet consist?
Let x represent the number of ships; then represents the number sunk;
which will give x=32.
+7 represents the number captured; and
-2 represents the number burned.