**Mathematical problems — Problems 21 to 30**

The ten problems are presented first, followed by the solutions to the ten problems.

**Problem 21:**

Mary travels 30 miles to work. She found that if she increases her normal speed by 10 mph, she arrives 6 minutes earlier to work. What is her normal speed?

*Problem 22:*

Three positive integers exist such that c^{2} – a^{2} – b^{2} = 101 and a • b = 72. What is the value of a + b + c ?

*Problem 23:*

If 86 in base 10 is represented by 321 in base x, what does 123 in base x equal in base 10?

*Problem 24:*

The square of a given number is 6 more than 5 times the number. What is the number?

*Problem 25:*

The ‘ones’ digit of a 2 digit number is 4 less than the ‘tens’ digit. Dividing the number by the sum of the digits yields 7. What is the number?

*Problem 26:*

Divide 30 by half and add 10. What is the answer?

*Problem 27:*

If you take 12 apples from a pile of 20 apples, how many apples do you have?

*Problem 28:*

Twelve years from now, Ralph will be 4 times as old as he was 15 years ago. How old is Ralph now?

*Problem 29:*

If Tom is twice as old as Dick was 10 years ago, and if the sum of their ages 5 years ago was 90, how old is Tom now?

*Problem 30:*

Three times a certain positive integer is one less than the square of another positive integer. The difference of the square of these two integers equals 9. What are the integers?

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**Solution to Problem 21:**

- r = rate or speed, t = time
- r
_{1}x t_{1}= 30 miles, r_{2}x t_{2}= 30 miles, r_{1}x t_{1}= r_{2}x t_{2} - r
_{2}= r_{1}+ 10, t_{2}= t_{1}– 6 min = t_{1}– 1/10 hour - So, r
_{1}x t_{1}= (r_{1}+ 10) ( t_{1}– 1/10 ) - rt = rt – r/10 +10t – 1
- 10t – r/10 = 1, or 100t – r = 10
- Since rt = 30, then t = 30/r
- Thus, 100 (30/r) – r = 10
- (3000/r) – r = 10
- 3000 – r
^{2}= 10r - r
^{2}+ 10r – 3000 = 0 - (r + 60) (r – 50) = 0
- r = 50 mph

**Solution to Problem 22:**

- With three unknowns and only 2 equations, this problem cannot be solved mathematically, but it can be solved intuitively.
- With a and b as integers, there are only 6 combinations where a • b = 72 ( 1 • 72, 2 • 36, 3 • 24, 4 • 18, 6 • 12, and 8 • 9 ).
- With c
^{2}= a^{2}+ b^{2}+ 101, we only need to find the a – b pair which forms an integer square root with 101. - Only the pairing of 4 and 18 works ( 212 = 42 + 182 + 101), or 441 = 441 .
- Thus, a + b + c = 4 + 18 + 21 =
**43**

*Solution to Problem 23:*

- 1 (x
^{0}) + 2 (x^{1}) + 3 (x^{2}) = 86 - 3x
^{2}+ 2x + 1 = 0 - (3x + 17)(x – 5) = 0
- x = 5
- 3(5
^{0}) + 2(5^{1}) + 1(5^{2}) = y - 3 + 10 + 25 = 38, so 123 in base 5 equals 38 in base 10

*Solution to Problem 24:*

- Thus, x
^{2}– 6 = 5x - This yields a quadratic equation: x2 – 5x – 6 + 0
- Factoring yields (x – 6)(x + 1) = 0
- Hence, the number is 6. [
*Note*: -1 is also a correct response]

*Solution to Problem 25:*

- xy equals the number, with x = the ‘tens’ digit and y = the ‘ones’ digit
- x = y + 4 and (10x + y) / (x + y) = 7
- Simplifying the second equation yields: 10x + y = 7x +7y
- Thus, 3x = 6y
- Substituting y + 4 for x yields 3(y + 4) = 6y
- So, 3y + 12 = 6y
- So, y = 4
- Hence, the original number must be 84

**Solution to Problem 26:**

- Did I trick you? Notice that the problem says ‘by half’, not ‘in half’. Dividing 30 by half yields 60, not 15. The correct answer is 70.

**Solution to Problem 27:**

- Did I ‘get’ you again? If YOU take 12 apples from a pile, no matter how many apples are in the pile, YOU have 12 apples! Did you say 8?

*Solution to Problem 28:*

- R = Ralph’s current age
- R + 12 = 4 (R-15)
- R + 12 = 4R – 60
- 3R = 72
- R = 24, so Ralph is presently 24 years old.

*Solution to Problem 29:*

- T = Tom’s age now; D = Dick’s age now
- T = 2(D – 10) and (T – 5) + (D – 5) = 90
- Substituting yields: ((2(D – 10)) – 5) + (D – 5) = 90
- So, 2D – 20 – 5 + D – 5 = 90
- 3D – 30 = 90, so 3D = 120, and D = 40
- So, T = 2(40 – 10), or T = 60
- Tom is 60 and Dick is 40

*Solution to Problem 30:*

- From the first statement: 3x + 1 = y
^{2} - From the second statement: x
^{2}– y^{2}= 9 - Substituting from the 1st statement into the 2nd statement yields: x
^{2}– (3x + 1) = 9 - Thus, x
^{2}– 3x – 1 = 9 - So, x
^{2}– 3x – 10 = 0 - Factoring yields (x – 5)(x + 2) = 0
- So x = 5
- If x = 5, then y = 4

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