Mathematical problems — Problems 31 to 40
The ten problems are presented first, followed by the solutions to the ten problems.
Problem 31:
Find two consecutive odd integers such that when 2 is subtracted from the square root of the smaller integer and that result is multiplied by 9, the product equals the larger integer.
Problem 32:
Almonds that cost $4.50 per pound are mixed with walnuts that cost $2.50 per pound. How many pounds of each were used to make a 100-pound bag of mixed nuts that costs $3.24 per pound?
Problem 33:
If you were to take the square root of Bob’s age 10 years ago, it would be one-tenth of Jim’s age 10 years ago. If you were to double Bob’s current age, it would equal Jim’s age 10 years from now. How old are Bob and Jim currently?
Problem 34:
Ten years ago Jim was 4-times as old as Bob. In 10 years Jim will be twice as old as Bob. How old are the two men now?
Problem 35:
At what exact time between 8:00 and 9:00 do the minute hand and the hour hand of a clock form a straight line for the first time? For the second time?
Problem 36:
In a certain family, each daughter has the same number of brothers as she has sisters, and each son has twice as many sisters as he has brothers. How many sons and daughters are in this family?
Problem 37:
What is the probability that a three-child family has at least one girl?
Problem 38:
If it was 2 hours later that it is right now, it would be 1/3 as long to noon as it would be if it were 2 hours earlier than it is right now. What time is it currently?
Problem 39:
If 2 typists can type 2 pages in 2 minutes, how many typists can type 18 pages in 6 minutes?
Problem 40:
A nut vendor obtained peanuts at $1.00 per pound, almonds at $2.00 per pound, and cashews at $3.00 per pound. He combined the nuts and made a 105-pound bag with the mixture costing $1.57 per pound. In the bag there are twice as many pounds of peanuts as almonds, and twice as many pounds of almonds as cashews. How many pounds of each kind of nut comprise the mixture?
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Solution to Problem 31:
- x = smaller integer, so x + 2 = larger integer
- 9 (√x – 2) = x + 2
- 9√x – 18 = x + 2
- 9√x = x = 20
- (9√x)2 = (x + 20)2
- 81x = x2 + 40x + 400
- x2 – 41x + 400 + 0
- (x – 16)(x – 25) = 0
- So, x = 16 or 25
- The two desired integers are 25 and 27
- The integers 16 and 18 work also, but they are even numbers
Solution to Problem 32:
- a = almonds and w = walnuts
- a + w = 100 pounds
- a($.50 per pound) + w($2.50 per pound) = 100($3.24 per pound)
- 4.5a + 2.5 w = 324
- Since w = 100 – a, we can substitute and obtain 4.5a + 2.5(100 – a) = 324
- 4.5a + 250 – 2.5a = 324
- 2a = 74
- a = 37 pounds of almonds, so walnuts = 100 – 37 = 63 pounds
Solution to Problem 33:
- B = Bob’s current age, J = Jim’s current age
- 10 (√(B – 10)) = J – 10 , so squaring both sides of the equation yields 100 (B – 10) = J2 – 20J +100
- 2B = J + 10, so B = 1/2 J + 5
- Substituting yields: 100((1/2 J + 5) – 10) = J2 – 20J + 100
- Thus, 50J – 500 = J2 – 20J + 100
- J2 – 70J + 600 = 0
- So, (J – 60)(J – 10) = 0
- J (Jim’s current age) = 60, so Bob’s age equals 35
Solution to Problem 34:
- B = Bob’s current age and J = Jim’s current age
- 2(B + 10) = J +10, so 2B + 20 = J + 10, and thus 2B + 10 = J
- 4(B – 10) = J – 10, so 4B – 40 = J – 10, and thus 4B – 30 = J
- Thus, 2B + 10 = 4B – 30
- So, 2B = 40
- B = 20 = Bob’s current age. Jim’s age must be 50.
Solution to Problem 35:
- A straight line forms an angle of 180°.
- At 8:00, the hands of a clock form an angle of 120°.
- As the hands move, the minute hand enlarges the angle between the two hands at a rate of 6° per minute (360°/ 60 min).
- As the hands move, the hour hand diminishes the angle between the two hands at a rate of 0.5° per minute (30°/ 60 min).
- Thus, 180° = 120° + (6°/ min)(t) – (0.5°/ min)(t)
- 180 = 120 + 5.5 t, so t = 10.9091 min = 10 minutes 54.5 seconds
- The exact time at which the hands form a straight line is 8:10:54.5.
- For the second part of the problem, the larger angle between the two hands at 8:00 is 240°.
- This time the minute hand is diminishing the angle and the hour hand is enlarging the angle.
- Thus, 0° = 240° – 6(t) + 5.5(t), so 5.5(t) = 240 and t = 43.6363 minutes
- The time for the second straight line is 8:43:37.8 .
Solution to Problem 36:
- b = boys and g = girls
- From the statements in the problem, we know that b = g – 1 and 2 (b – 1) = g
- Thus, 2b – 2 = b + 1
- Hence, b = 3 and g = 4
Solution to Problem 37:
- 23 = 8, so there are 8 combinations of boys and girls in the family
- There is only one combination that does not fit the stipulation, and that is boy-boy-boy
- 7 ÷ 8 = 87.5%
Solution to Problem 38:
- Solution method number one:
- x = current time
- 12 – (x – 2) = 3 (12 – (x + 2))
- 12 – x + 2 = 36 – 3x – 6
- 14 – x = 30 – 3x
- 2x = 16
- x = 8 (o’clock)
- Solution method number two:
- 12 – (x + 2) = 1/3 (12 – (x – 2))
- 10 – x = 1/3 (14 – x)
- 30 – 3x = 14 – x
- 2x = 16
- x = 8 (o’clock)
Solution to Problem 39:
- p = pages, m = minutes, t = typists
- (2p / 2m) / 2t = (18p / 6m) / xt
- 1/2 = 3/x
- x = 6 typists
Solution to Problem 40:
- Pounds of peanuts = p, pounds of almonds = a, and pounds of cashews = c.
- ($1.57 / pound) x 105 pounds = (($1.00 / pound) x p) + (($2.00 / pound) x a) + ((#3.00 / pound) x c)
- 2c = a and 2a = p, so 4c = p .
- Substituting these values in the above equation yields ($1.57 / pound) x 105 pounds = (($1.00 / pound) x 4c) + (($2.00 / pound) x 2c) + ((#3.00 / pound) x c)
- So, 1.57 x 105 = 4c + 4c + 3c
- 165 = 11c, so c = 15.
- Thus, pounds of cashews equals 15, pounds of almonds = 30, and pounds of peanuts = 60.
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