问题1：找出所有成对的连续奇数正整数，两者均小于10，以使它们的总和大于11。
解决方案：
Let x and x + 2 be the two consecutive odd positive integers.
Given that the integers are smaller than 10 and their sum is more than 11.
Therefore,
x + 2 < 10 and x + (x + 2) > 11
x < 10 – 2 and 2x + 2 > 11
x < 8 and 2x > 11 – 2
x < 8 and 2x > 9
x < 8 and x > 9/2
9/2 < x < 8
Therefore, the two consecutive odd positive integers are x = 5, 7
The pairs of consecutive odd integers are
Let x = 5, then (x + 2) = (5 + 2) = 7.
Let x = 7, then (x + 2) = (7 + 2) = 9.
Therefore, the required pairs of odd integers are (5, 7) and (7, 9).
问题2：找出所有连续的奇数自然数对，两者均大于10，以使它们的总和小于40。
解决方案：
Let x and x + 2 be the two consecutive odd positive integers.
Given that both the odd natural numbers are greater than 10 and their sum is less than 40.
Therefore,
x > 10 and x + x + 2 <40
x > 10 and 2x < 38
x > 10 and x < 38/2
x > 10 and x < 19
10 < x < 19
Therefore, the two consecutive odd positive integers are x = 11, 13, 15, 17
The pairs of consecutive odd integers are
Let x = 11, then (x + 2) = (11 + 2) = 13
Let x = 13, then (x + 2) = (13 + 2) = 15
Let x = 15, then (x + 2) = (15 + 2) = 17
Let x = 17, then (x + 2) = (17 + 2) = 19.
Therefore, the required pairs of odd natural numbers are (11, 13), (13, 15), (15, 17), and (17, 19)
问题3：找出所有成对的连续偶数均大于5的正整数，以使它们的总和小于23。
解决方案：
Let x and x + 2 be the two consecutive even positive integers.
Given that both the even integers are greater than 5 and their sum is less than 23.
Therefore,
x > 5 and x + x + 2 < 23
x > 5 and 2x < 21
x > 5 and x < 21/2
5 < x < 21/2
5 < x < 10.5
Therefore, the consecutive even positive integers are x = 6, 8, 10
The pairs of consecutive even integers are
Let x = 6, then (x + 2) = (6 + 2) = 8
Let x = 8, then (x + 2) = (8 + 2) = 10
Let x = 10, then (x + 2) = (10 + 2) = 12.
Therefore, the required pairs of even positive integer are (6, 8), (8, 10), and (10, 12)
问题4.罗希特在两次测验中得到的分数分别为65和70。找到他在第三次测验中应获得的最低分数，以使平均分数至少达到65。
解决方案：
Given: marks scored by Rohit in two tests are 65 and 70.
Let marks scored by Rohit in the third test be x.
The average marks in the three papers ≥ 65
Average = (marks in 1st two papers + marks in the third test)/3
(65 + 70 + x)/3 ≥ 65
(135 + x)/3 ≥ 65
(135 + x) ≥ 65 × 3
(135 + x) ≥ 195
x ≥ 195 – 135
x ≥ 60
Since, the minimum marks to get an average of 65 marks is 60.
Therefore, the minimum marks required in the third test is 60.
问题5：溶液应保持在86°至95°F之间。如果摄氏(C)/华氏(F)转换公式由F = 9 / 5C + 32给出，则摄氏温度范围是多少?
解决方案：
Let us consider F1 = 86° F and F2 = 95°
Given, F = 9/5C + 32
F1 = 9/5 C1 + 32
F1 – 32 = 9/5 C1
C1 = 5/9 (F1 – 32)
C1 = 5/9 (86 – 32)
C1 = 5/9 (54)
C1 = 5 × 6
C1 = 30° C
Now,
F2 = 9/5 C2 + 32
F2 – 32 = 9/5 C2
C2 = 5/9 (F2 – 32)
C2 = 5/9 (95 – 32)
C2 = 5/9 (63)
C2 = 5 × 7
C2 = 35° C
Therefore, the range of temperature of the solution is 30°C and 35°C.
问题6：溶液应保持在30°C至35°C之间。以华氏度为单位的温度范围是多少?
解决方案：
Let us consider C1 = 30°C and C2 = 35°C
We know, F = 9/5C + 32
F1 = 9/5 C1 + 32
= 9/5 × 30 + 32
= 9 × 6 + 32
= 54 + 32
= 86°F
Now,
F2 = 9/5 C2 + 32
= 9/5 × 35 + 32
= 9 × 7 + 32
= 63 + 32
= 95°F
Therefore, the range of temperature of the solution is 86°F and 95°F.
问题7：要获得一门课程的A级成绩，必须在5篇论文中平均获得90分或以上，每100分。如果Shikha在前四篇论文中分别获得87、95、92和94分，那么她在最后一篇论文中必须获得的最低分才能在课程中获得“ A”级。
解决方案：
Given: marks scored by Shikha in the first four papers are 87, 95, 92, and 94.
Let marks scored by Shikha in the fifth test be x.
The average marks in the five papers ≥ 90
Average = (marks in 1st four papers + marks in the fifth test)/5
(87 + 95 + 92 + 94 + x)/5 ≥ 90
182 + 186 + x ≥ 90 × 5
368 + x ≥ 450
x ≥ 450 – 368
x ≥ 82
Since, the minimum marks to get an average of 0 marks is 82.
Therefore, the minimum marks required in the fifth paper is 82.
问题8：某公司生产卡式盒，其一周的成本和收益函数分别为C = 300 +(3/2)x和R = 2x，其中x是生产的卡式盒数量并在一个焊道中为实心。公司要实现利润必须售出多少盒磁带?
解决方案：
Given: Cost = 300 + (3/2)x and Revenue = 2x
As profit = Revenue – Cost
Therefore, to earn a profit, the revenue should be greater than the cost.
Revenue > Cost
2x > 300 + (3/2)x
2x – (3/2)x > 300
(4x – 3x) > 600
x > 600
Therefore, the manufacture must sell more than 600 cassettes to gain profit.
问题9：如果三角形的周长至少为61cm，则三角形的最长边是最短边的三倍，而第三边要比最长边短2厘米。找到最短边的最小长度。
解决方案：
Let the length of the shortest side be x.
Given, the longest side of a triangle is three times the shortest side = 3x
and the third side is 2 cm less than the longest side = 3x – 2
Also given that the perimeter of the triangle ≥ 61
Therefore,
x + 3x – 2 + 3x ≥ 61
7x ≥ 61 + 2
7x ≥ 63
x ≥ 63/7
x ≥ 9
Therefore, the minimum length of the shortest side is 9cm.
问题10：必须向1125升的45％的酸溶液中添加几升水，以便所得混合物中酸含量超过25％但不到30％?
解决方案：
Let the quantity of water to be added in liters be x
Therefore,
25% of (1125 + x) < 45% of 1125
25/100(1125 + x) < 45/100 1125
1125 + x < (45 × 1125)/25
1125 + x < 45 × 45
1125 + x < 2025
x < 2025 – 1125
x < 900 ……. (1)
Also given that 45% of 1125 < 30% of (1125 + x)
Therefore,
45/100 × 1125 < 30/100 (1125 + x)
45/30 × 1125 < 1125 + x
3/2 × 1125 < 1125 + x
1687.5 < 1125 + x
1687.5 – 1125 < x
562.5 < x ……. (2)
Now, by using equation 1 and 2, we get
562.5 < x < 900
Therefore, the quantity of water to be added will be between 562.5 liters and 900 liters.
问题11：通过向其中添加2％的硼酸溶液来稀释8％的硼酸溶液。所得混合物应大于4％但小于6％的硼酸。如果有640升的8％溶液，则必须添加多少升的2％溶液?
解决方案：
Let the 2% solution be added to 640 liters of the 8% solution be x.
Therefore, the total quantity of mixture = (640 + x)
The total acid in (640 + x) liters of mixture is
(2/100)x + (8/100)640
Given that the resulting mixture should be more than 4% and less than 6%.
4/100(640 + x) < (2x/100 + (8 ×640)/100 < 6/100(640 + x)
4(640 + x) < (2x + 8640) < 6(640 + x)
2560 + 4x < 2x + 8640 and 2x + 8640 < 3840 + 6x
2560 – 8640 < 2x – 4x and 2x – 6x < 3840 – 8640
x < 1280 and x > 320
Therefore, more than 320 liters but less than 1280 liters of 2% to be added.
问题12：当三个每日测量的平均pH读数在7.2至7.8之间时，池中的水酸度被认为是正常的。如果前两个pH读数分别为7.48和7.85，请找到第三个读数的pH值范围，这将导致酸度水平正常。
解决方案：
Let x be the pH value of third reading.
Therefore,
7.2 < (7.48 + 7.85 + x)/3 < 7.8
21.6 < 7.48 + 7.85 + x < 23.4
21.6 < 15.33 + x < 23.4
21.6 – 15.33 < x and x < 23.4 – 15.33
6.27 < x and x < 8.07
Therefore, the range of pH values for the third reading lies between 6.27 and 8.07.