1. To divide a given number into two having a given difference.
Given number 100, given difference 40.
Lesser number required x. Therefore
2x + 40 = 100,
x = 30.
The required numbers are 70, 30.
2. To divide a given number into two having a given ratio.
Given number 60, given ratio 3:1.
Two numbers x, 3x. Therefore
4x = 60,
x = 15.
The required numbers are 45, 15.
3. To divide a given number into two numbers such that one is a given ratio of the other plus a given difference.
Given number 80, ratio 3:1, difference 4.
Lesser number x. Therefore the larger is 3x+4, and
4x + 4 =80,
x = 19.
The required numbers are 61, 19.
4. To find two numbers in a given ratio and also with a given difference.
Given ratio 5:1, given difference 20.
Numbers are 5x, x. Therefore
4x = 20,
x =5.
The required numbers are 25, 5.
5. To divide a given number into two numbers such that given (different) fractions of each number when added together produce a given number.
Necessary condition. The latter given number must lie between the numbers arising when the given fractions respectively are taken of the first given number.
First given number 100, given fractions 1/3 and 1/5, given sum of fractions 30.
Second part 5x. Therefore first part = 3(30-x). Hence
90 + 2x = 100,
x = 5.
The required parts are 75, 25.
6. To divide a given number into two numbers such that a given fraction of the first exceeds a given fraction of the other by a given number.
Necessary condition. The latter number must be less than that which arises when that fraction of the first number is taken which exceeds the other fraction.
Given number 100, given fractions ¼ and 1/6 respectively, given excess 20.
Second part 6x. Therefore first part = 4(x+20). Hence
10x + 80 = 100,
x =2.
The required parts are 88, 12.
7. From the same (required) number to subtract two given numbers so as to make the remainders have to one another a given ratio.
Given numbers 100, 20, given ratio 3:1.
Required number x. Therefore
x - 20 = 3 (x - 100)
x = 140.
The required number is 140.
8. To two given numbers to add the same (required) number so as to make the resulting numbers have to one another a given ratio.
Necessary condition. The given ratio must be less than the ratio which the greater of the given numbers has to the lesser.
Given numbers 100, 20, given ratio 3:1.
Required number x. Therefore
3x + 60 = x + 100
x = 20.
The required number is 20.
9. From two given numbers to subtract the same (required) number so as to make the remainders have to one another a given ratio.
Necessary condition. The given ratio must be greater than the ratio which the greater of the given numbers has to the lesser.
Given numbers 20, 100, given ratio 6:1.
Required number x. Therefore
120 – 6x = 100 – x
x = 4.
The required number is 4.
10. Given two numbers, to add to the lesser and to subtract from the greater the same (required) number so as to make the sum in the first case have to the difference in the second case a given ratio.
Given numbers 20, 100, given ratio 4:1.
Required number x. Therefore
(20 + x) = 4 (100 – x)
x = 76.
The required number is 76.
11. Given two numbers, to add the first to, and subtract the second from, the same (required) number, so as to make the resulting numbers have to one another a given ratio.
Given numbers 20, 100, given ratio 3:1.
Required number x. Therefore
3x – 300 = x + 20
x = 160.
The required number is 160.
12. To divide a given number twice into two numbers such that the first of the first pair may have to the first of the second pair a given ratio, and also the second of the second pair to the second of the first pair another given ratio.
Given number 100, ratio of greater of first parts to lesser of second 2:1, and ratio of greater of second parts to lesser of first parts 3:1.
x lesser of second parts.
The parts then are (2x and 100-2x) and (300-6x and x). Therefore
300 – 5x = 100
x = 40.
The required numbers are (80, 20) and (60, 40).
13. To divide a given number thrice into two numbers such that one of the first pair has to one of the second pair a given ratio, the second of the second pair to one of the third pair another given ratio, and the second of the third pair to the second of the first pair another given ratio.
Given number 100, ratio of greater of first parts to lesser of second 3:1, of greater of second to lesser of third 2:1, and of greater of third to lesser of first 4:1.
x lesser of third parts.
Therefore greater of second parts = 2x, lesser of second = 100-2x, greater of first = 300-6x.
Hence lesser of first = 6x-200, so that greater of third = 24x-800. Therefore
25x – 800 = 100
x = 36.
The required numbers are (84, 16), (72, 28) and (64, 36).
14. To find two numbers such that their product has to their sum a given ratio. [One is arbitrarily assumed]
Necessary condition. The assumed value of one of the two must be greater than the number representing the ratio.
Ratio 3:1, x one of the numbers, 12 the other (>3). Therefore
12x = 3x + 36
x = 4.
The required numbers are 4, 12.
15. To find two numbers such that each after receiving from the other a given number may bear to the remainder a given ratio.
Let the first receive 30 from the second, the ratio being then 2:1, and the second 50 from the first, the ratio being then 3:1; take x+30 for the second.
Therefore the first = 2x-30, and
(x + 80) = 3 (2x – 80)
x = 64.
The required numbers are 98, 94.
16. To find three numbers such that the sums of the pairs are given numbers.
Necessary condition. Half the sum of the three given numbers must be greater than any one of them singly.
Let (1) + (2) = 20, (2) + (3) = 30, (3) + (1) = 40.
x the sum of the three. Therefore the numbers are x-30, x-40, x-20. The sum
x = 3x – 90
x = 45.
The required numbers are 15, 5, 25.
17. To find four numbers such that the sums of all sets of three are given numbers.
Necessary condition. One-third of the sum of the four must be greater than any one singly.
Sums of threes 22, 24, 27, 20 respectively.
x the sum of all four. Therefore the numbers are x-22, x-24, x-27, x-20. Therefore
4x – 93 = x
x = 31.
The required numbers are 9, 7, 4, 11.
18. To find three numbers such that the sum of any pair exceeds the third by a given number.
Given excesses 20, 30, 40.
2x the sum of all three.
We have (1) + (2) = (3) + 20.
Adding (3) to each side, we have: twice(3) + 20 = 2x, so (3) = x-10.
Similarly the numbers (1) and (2) are x-15 and x-20 respectively. Therefore
3x – 45 = 2x
x = 45.
The required numbers are 30, 25, 35.
19.To find four numbers such that the sum of any three exceeds the fourth by a given number.
Necessary condition. Half the sum of the four given differences must be greater than any one of them.
Given differences 20, 30, 40, 50.
2x the sum of the required numbers. Therefore the numbers are x-15, x-20, x-25, x-10. Therefore
4x – 70 = 2x
x = 35.
The required numbers are 20, 15, 10, 25.
20. To divide a given number into three numbers such that the sum of each extreme and the mean has to the other extreme a given ratio.
Given number 100; and let (1) + (2) = 3.(3) and (2) + (3) = 4.(1).
x the third number. Thus the sum of the first and second = 3x, and the sum of the three = 4x = 100.
Hence x = 25, and the sum of the first two = 75.
Let y be the first. Therefore sum of second and third = 4y, 5y = 100 and y = 20.
The required parts are 20, 55, 25.
21. To find three numbers such that the greatest exceeds the middle number by a given fraction of the least, the middle exceeds the least by a given fraction of the greatest, but the least exceeds a given fraction of the middle number by a given number.
Necessary condition. The middle number must exceed the least by such a fraction of the greatest that, if its denominator be multiplied into the excess of the middle number over the least, the coefficient of x in the product is greater than the coefficient of x in the expression for the middle number resulting from the assumptions made.
Suppose greatest exceeds middle by 1/3 of least, middle exceeds least by 1/3 of greatest, and least exceeds 1/3 of middle by 10.
x+10 the least. Therefore middle = 3x, and greatest = 6x-30. Hence lastly
6x – 30 – 3x = (1/3)(x + 10)
x + 10 = 9x – 90
x = 25/2.
The numbers are 45, 75/2, 45/2.
22. To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal.
Let first give 1/3 of itself to second, second ¼ of itself to third, third 1/5 of itself to first.
Assume first to be a number of x’s divisible by three, say 3x, and second to be a number of units divisible by 4, say 4.
Therefore second after giving and taking becomes x+3.
Hence the first also after giving and taking must become x+3; it must therefore have taken x+3-2x, or 3-x; 3-x must therefore be 1/5 of third, or third = 15-5x.
Lastly,
15 – 5x – (3-x) + 1 = x + 3
13 – 4x = x + 3
x = 2.
The numbers are 6, 4, 5.
23. To find four numbers such that, if each give to the next following a given fraction of itself, the results may all be equal.
Let first give 1/3 of itself to second, second ¼ of itself to third, third 1/5 of itself to fourth, and fourth 1/6 of itself to first.
Assume first to be number of x’s divisible by 3, say 3x, and second to be number of units divisible by 4, say 4.
The second, after giving and taking, becomes x+3.
Therefore first after giving x to second and receiving 1/6 of fourth = x+3; therefore fourth = 6(x+3-2x) = 18-6x.
But fourth after giving 3-x to first and receiving 1/5 of third = x+3; therefore third = 30x-60.
Lastly, third after giving 6x-12 to fourth and receiving 1 from second = x+3. That is
24x – 47 = x + 3
x = 50/23.
The numbers are therefore 150/23, 4, 120/23, 114/23; or, after multiplying by the common denominator, 150, 92, 120, 114.
24. To find three numbers such that, if each receives a given fraction of the sum of the other two, the results are all equal.
Let first receive 1/3 of (second + third), second ¼ of (third + first), and third 1/5 of (first + second).
Assume first = x, and for convenience’s sake take for sum of second and third a number of units divisible by 3, say 3.
Then sum of the three = x + 3
and first + 1/3 (second + third) = x + 1.
Therefore second + ¼ (third + first) = x + 1;
hence 3 times second + sum of all = 4x + 4,
and, therefore second = x + 1/3.
Lastly, third + 1/5 (first + second) = x + 1,
or 4 times third + sum of all = 5x + 5,
and third = x + ½.
Therefore
x + (x + 1/3) + (x + ½) = x + 3,
x = 13/12.
The numbers, after multiplying by the common denominator, are 13, 17, 19.
25. To find four numbers such that, if each receives a given fraction of the sum of the remaining three, the four results are equal.
Let first receive 1/3 of the rest, second ¼ of the rest, third 1/5 of the rest, and fourth 1/6 of the rest.
Assume first to be x and sum of rest a number of units divisible by 3, say 3.
Then sum of all = x + 3.
Now first + 1/3 (second + third + fourth) = x + 1.
Therefore second + ¼ (third + fourth + first) = x + 1,
whence 3 times second + sum of all = 4x + 4,
and therefore second = x + 1/3.
Similarly third = x + ½,
and fourth = x + 3/5.
Adding, we have
4x + 43/30 = x + 3,
x = 47/90.
The numbers, after multiplying by a common denominator, are 47, 77, 92, 101.
26. Given two numbers, to find a third number which, when multiplied into the given numbers respectively, makes one product a square and the other the side of that square.
Given numbers 200, 5; required number x.
Therefore
200x = (5x)^2,
x = 8.
The required number is 8.
27. To find two numbers such that their sum and product are given numbers.
Necessary condition. The square of half the sum must exceed the product by a square number.
Given sum 20, given product 96.
2x the difference of the required numbers.
Therefore the numbers are 10+x, 10–x.
Hence
100 – x^2 = 96,
x = 2.
The required numbers are 12, 8.
28. To find two numbers such that their sum and the sum of their squares are given numbers.
Necessary condition. Double the sum of their squares must exceed the square of their sum by a square.
Given sum 20, given sum of squares 208.
Difference 2x.
Therefore the numbers are 10+x, 10-x.
Thus
200 + 2x^2 = 208,
x = 2.
The required numbers are 12, 8.
29. To find two numbers such that their sum and the difference of their squares are given numbers.
Given sum 20, given difference of squares 80.
Difference 2x.
The numbers are therefore 10+x, 10-x.
Hence
(10 + x)^2 – (10 – x)^2 = 80,
40x = 80,
x = 2.
The required numbers are 12, 8.
30. To find two numbers such that their difference and product are given numbers.
Necessary condition. Four times the product together with the square of the difference must give a square.
Given difference 4, given product 96.
2x the sum of the required numbers.
Therefore the numbers are x + 2, x – 2; accordingly
x^2 – 4 = 96,
x = 10.
The required numbers are 12, 8.
31. To find two numbers in a given ratio and such that the sum of their squares also has to their sum a given ratio.
Given ratios 3:1 and 5:1 respectively.
Lesser number x. Therefore
10x^2 = 5.4x
x =2.
The required numbers are 2, 6.
32. To find two numbers in a given ratio and such that the sum of their squares also has to their difference a given ratio.
Given ratios 3:1 and 10:1.
Lesser number x. Therefore
10x^2 = 10.2x
x = 2.
The required numbers are 2, 6.
33. To find two numbers in a given ratio and such that the difference of their squares also has to their sum a given ratio.
Given ratios 3:1 and 6:1.
Lesser number x. Therefore
8x^2 = 6.4x
x = 3.
The required numbers are 3, 9.
34. To find two numbers in a given ratio and such that the difference of their squares also has to their difference a given ratio.
Given ratios 3:1 and 12:1.
Lesser number x. Therefore
8x^2 = 12.2x
x = 3.
The required numbers are 3, 9.
Similarly by the same method can be found two numbers in a given ratio and (1) such that their product is to their sum in a given ratio, or (2) such that their product is to their difference in a given ratio.
35. To find two numbers in a given ratio and such that the square of the lesser also has to the greater a given ratio.
Given ratios 3:1 and 6:1 respectively.
Lesser number x. Therefore
x^2 = 6.3x
x = 18.
The two numbers are 18, 54.
36. To find two numbers in a given ratio and such that the square of the lesser also has to the lesser itself a given ratio.
Given ratios 3:1 and 6:1.
Lesser number x. Therefore
x^2 = 6.x
x = 6.
The two numbers are 6, 18.
37. To find two numbers in a given ratio and such that the square of the lesser also has to the sum of both a given ratio.
Given ratios 3:1 and 2:1.
Lesser number x. Therefore
x^2 = 2.4x
x = 8.
The two numbers are 8, 24.
38. To find two numbers in a given ratio and such that the square of the lesser also has to the difference between them a given ratio.
Given ratios 3:1 and 6:1.
Lesser number x. Therefore
x^2 = 6.2x
x =12.
The two numbers are 12, 36.
Similarly can be found two numbers in a given ratio and
(1) such that the square of the greater also has to the lesser a given ratio, or
(2) such that the square of the greater also has to the greater itself a given ratio, or
(3) such that the square of the greater also has to the sum or difference of the two a given ratio.
39. Given two numbers, to find a third such that the sums of the several pairs multiplied by the corresponding third number give three numbers in arithmetical progression.
Given numbers 3, 5.
Required number x.
The three products are therefore 3x+15, 5x+15, 8x.
Now 3x+15 must be either the middle or the least of the three, and 5x+15 either the greatest or the middle.
(1) 5x+15 greatest, 3x+15 least. Therefore
5x + 15 + 3x + 15 = 2.8x
x = 15/4.
(2) 5x+15 greatest, 3x+15 middle. Therefore
5x + 15 – (3x + 15) = 3x + 15 – 8x
x = 15/7.
(3) 8x greatest, 3x+15 least. Therefore
8x + 3x + 15 = 2(5x + 15)
x = 15.