This equation is an identity, and is therefore satisfied by any finite value of x.

If it be solved in the usual way, we obtain

That is, the conditions of the problem will be satisfied by any number of yards in the first piece.

Infinite Solutions.

17. Pr. A cistern has three pipes. Through the first it can be filled in 24 minutes; through the second in 36 minutes; through the third it can be emptied in a minutes. In what time will the cistern be filled if all the pipes be opened at the same time?

Let x stand for the number of minutes after which the cistern will be filled. In one minute of its capacity enters through the first pipe, and hence in a minutes of its capacity enters. For a similar reason, of its capacity enters through the second pipe in x minutes; and in the same 1

time x of its capacity is discharged through the third pipe.

a

Therefore, after x minutes there is in the cistern

of its capacity. But by the condition of the problem, that the cistern is then filled, we have

If we now let a approach, then a becomes infinite.

This result would mean that the cistern would never be filled. This is also evident from the data of the problem, since the third pipe in a given time would discharge from the cistern as much as would enter it through the other pipes.

The Problem of the Couriers.

18. Pr. Two couriers are travelling along a road in the direction from M to N; one courier at the rate of m, miles an hour, the other at the rate of m, miles an hour. The former

is seen at the station A at noon, and the other is seen h hours later at the station B, which is d miles from A in the direction in which the couriers are travelling. Where do the couriers meet?

Assume that they meet to the right of B at a point C1, and let x stand for the number of miles from B to the place of meeting C (Fig. 2).

The first courier, moving at the rate of m1 miles an hour, travels dx miles, from A to C1, in hours; the second courier, moving at the rate of m, miles an hour, travels x miles,

d + x mi

from B to C1, in hours. By the condition of the problem

m2

it is evident that, if the place of meeting is to the right of B, the number of hours it takes the first courier to travel from A to C exceeds by h the number of hours it takes the second courier to travel from B to C1. We therefore have

A

(i.) A Positive Result. The result will be positive either when hm,>d and m2>m1, or when hm,<d and m1⁄2<m. positive result means that the problem is possible with the assumption made; i.e., that the couriers meet at a point to the right of B.

(ii.) A Negative Result. The result will be negative either when hm, >d and m1⁄2 < m1, or when hm,<d and m, > m1. Such

a result shows that the assumption that the couriers meet to the right of B is untenable, since, as we have seen, in that case the result is positive.

That under the assumed conditions the couriers can meet only at some point to the left of B can also be inferred from the following considerations, which are independent of the negative result: If hm,>d, the first courier has passed B when the second courier is seen at that station; that is, the second courier is behind the first at that time. And since also m<m1 the first courier is travelling the faster, and must therefore have overtaken the second, and at some point to the left of B.

On the other hand, if hm, >d, the first courier has not yet reached B when the second is seen at that station; that is, the first courier is behind the second at that time. And since also m2 > m1, the second courier is travelling the faster, and must therefore have overtaken the first, at some point to the left of B. Similar reasoning could have been applied in (i.).

=

(iii.) A Zero Result. A zero result is obtained when hm, d, and m, is not equal to m1; that is, the meeting takes place at B. This is also evident from the assumed conditions. For the first courier reaches Bh hours after he was seen at A; and since the second courier is seen at Bh hours after the first was seen at A, the meeting must take place at B.

(iv.) Indeterminate Result. - An indeterminate result is obtained if hm1 =÷d, and m2 = m1. In this case every point of the road can be regarded as their place of meeting. For the first courier evidently reaches B at the time at which the second courier is seen at that station; and since they are travelling at the same rate, they must be together all the time. The problem under these conditions becomes indeterminate.

(v.) An Infinite Result. An infinite result is obtained when hm, ‡d, and m2 = m. In this case a meeting of the couriers is impossible, since both travel at the same rate, and when the second is seen at B the first either has not yet reached B or has already passed that station.

An infinite result also means that the more nearly equal m1 and m, are, the further removed is the place of meeting.

EXERCISES II.

Solve the following problems, and interpret the results:

1. In a number of two digits, the digit in the tens' place exceeds the digit in the units' place by 5. If the digits be interchanged, the resulting number will be less than the original number by 45. What is the number?

2. The sum of the first and third of three consecutive even numbers is equal to twice the second. What are the numbers?

3. A father is 26 years older than his son, and the sum of their ages is 26 years less than twice the father's age. How old is the son?

4. In a number of two digits, the digit in the units' place exceeds the digit in the tens' place by 4. If the sum of the digits be divided by 2, the quotient will be less than the first digit by 2. What is the number?

a

b

Discuss the solutions of the following general problems:

5. What number, added to the denominators of the fractions will make the resulting fractions equal?

ď

6. Having two kinds of wine worth a and b dollars a gallon, respectively, how many gallons of each kind must be taken to make a mixture of n gallons worth c dollars a gallon?

7. Two couriers, A and B, start at the same time from two stations, distant d miles from each other, and travel in the same direction. A travels n times as fast as B. Where will A overtake B?

is a decreasing geometrical progression, whose ratio is . It follows from Ch. XXI., Art. 26, that the sum of this series approaches a definite finite value as the number of terms is indefinitely increased.

By actual computation, we obtain

S1 = 1, S1, S21, S1 = 211, etc.

These sums approach 3 more and more nearly, as more and more terms are included.

This infinite series may therefore be regarded as having the finite sum 3.

But the sum of the series

1 + 2 + 4 +8 + ·

increases beyond any finite number, as the number of terms increases indefinitely.

2. The examples of the preceding article illustrate the following definitions: