问题1.如果从一个数字中减去1/2，然后将结果乘以1/2，则得出1/8是多少?
解决方案：
Let the number be ‘a’.
According to the question,
(a – 1/2) × 1/2 = 1/8
a/2 – 1/4 = 1/8
a/2 = 1/8 + 1/4
a/2 = 1/8 + 2/8
a/2 = (1 + 2)/8
a/2 = 3/8
a = (3/8) × 2
So,
a = 3/4
问题2.矩形游泳池的周长为154 m。它的长度比其宽度的两倍大2 m。泳池的长度和宽度是多少?
解决方案：
Given that,
Perimeter of rectangular swimming pool = 154 m
Let the breadth of rectangle be ‘a’
Length of the rectangle = 2a + 2 We know that,
Perimeter = 2 × (length + breadth)
So,
2(2a + 2 + a) = 154
2(3a + 2) = 154
3a + 2 = 154/2
3a = 77 – 2
3a = 75
a = 75/3
a = 25
Therefore, Breadth = 25 m
Length = 2a + 2
= (2 × 25) + 2
= 50 + 2
Length = 52 m
问题3.等腰三角形的底边是4/3厘米。三角形的周长为62/15厘米。其余均等边的长度是多少?
解决方案：
Base of isosceles triangle = 4/3 cm
Perimeter of triangle = 62/15
Let the length of equal sides of triangle be ‘a’.
So,
2a = (62/15 – 4/3)
2a = (62 – 20)/15
2a = 42/15
a = (42/30) × (1/2)
a = 42/30
a = 7/5
So, length of either of the remaining equal sides are 7/5 cm each.
问题4.两个数字的总和为95。如果一个数字超过另一个数字15，则找到数字。
解决方案：
Let one of the numbers be ‘a’.
Then, the other number becomes (a + 15) Given in the question,
Also given that,
a + (a + 15) = 95
2a + 15 = 95
2a = 95 – 15
2a = 80
a = 80/2
a = 40
So, First number = 40
And, other number is = (a + 15) = 40 + 15 = 55
问题5.两个数字之比为5：3。如果它们相差18，那么数字是多少?
解决方案：
Let the two numbers be ‘5a’ and ‘3a’. So, according to the question,
5a – 3a = 18
2a = 18
a = 18/2
a = 19
Thus,
The first numbers is (5a) = 5 × 9 = 45
And another number (3a) = 3 × 9 = 27.
问题6.三个连续的整数加起来等于51。这些整数是什么?
解决方案：
Let the three consecutive integers be ‘a’, ‘a + 1’ and ‘a + 2’. So, according to the question,
a + (a + 1) + (a + 2) = 51
3a + 3 = 51
3a = 51 – 3
3a = 48
a = 48/3
a = 16
So, the integers are
First integer will be (a) = 16
Second integer will be (a + 1) = 17
& third integer will be (a + 2) = 18
问题7. 8的三个连续倍数之和是888。找到倍数。
解决方案：
Let the three consecutive multiples of 8 be ‘8a’, ‘8(a+1)’ and ‘8(a+2)’. According to the question,
Given,
8a + 8(a + 1) + 8(a + 2) = 888
8 (a + a + 1 + a + 2) = 888 (Taking 8 as common)
8 (3a + 3) = 888
3a + 3 = 888/8
3a + 3 = 111
3a = 111 – 3
3a = 108
a = 108/3
a = 36
Thus, the three consecutive multiples of 8 are:
First no. = 8a = 8 × 36 = 288
Second no. = 8(a + 1) = 8 × (36 + 1) = 8 × 37 = 296
Third No. = 8(a + 2) = 8 × (36 + 2) = 8 × 38 = 304
问题8.三个连续的整数是这样的：当按递增顺序将它们分别乘以2、3和4时，它们的总和为74。找到这些数字。
解决方案：
Let the three consecutive integers are ‘a’, ‘a+1’ and ‘a+2’. According to the question,
Given,
2a + 3(a + 1) + 4(a + 2) = 74
2a + 3a +3 + 4a + 8 = 74
9a + 11 = 74
9a = 74 – 11
9a = 63
a = 63/9
a = 7
Thus, the numbers are:
First integer. = a = 7
Second integer = a + 1 = 8
and Third integer = a + 2 = 9
问题9.拉胡尔和哈隆的年龄之比为5：7。四年后，他们的年龄总和将为56岁。他们现在的年龄是几岁?
解决方案：
Let the ages of Rahul and Haroon be ‘5a’ and ‘7a’.
Four years later,
The ages of Rahul and Haroon will be (5a + 4) and (7a + 4) respectively. According to the question,
Given, (5a + 4) + (7a + 4) = 56
5a + 4 + 7a + 4 = 56
12a + 8 = 56
12a = 56 – 8
12a = 48
a = 48/12
a = 4
Therefore, Present age of Rahul = 5a = 5 × 4 = 20
And, present age of Haroon = 7a = 7 × 4 = 28
问题10.班上男孩和女孩的比例为7：5。男孩的数量比女孩的数量多8。总的班级人数是多少?
解决方案：
Let the number of boys be ‘7a’ and girls be ‘5a’.
According to the question,
Given, 7a = 5a + 8
7a – 5a = 8
2a = 8
a = 8/2
a = 4
Therefore, Number of boys = 7 × 4 = 28
And, Number of girls = 5 × 4 = 20
Total number of students = 20 + 28 = 48
问题11：Baichung的父亲比Baichung的祖父小26岁，比Baichung大29岁。这三个年龄的总和是135年。他们每个人的年龄是多少?
解决方案：
Let age of Baichung’s father be ‘a’.
Then, age of Baichung’s grandfather = (a + 26)
and, Age of Baichung = (a – 29) According to the question,
Given, a + (a + 26) + (a – 29) = 135
3a + 26 – 29 = 135
3a – 3 = 135
3a = 135 + 3
3a = 138
a = 138/3
a = 46
Age of Baichung’s father = a = 46
Age of Baichung’s grandfather = (a + 26) = 46 + 26 = 72
Age of Baichung = (a – 29) = 46 – 29 = 17
问题12.从现在起十五岁将是拉维的年龄的四倍。拉维现在的年龄是几岁?
解决方案：
Let the present age of Ravi be ‘a’.
Fifteen years later, Ravi age will be (a+15) years. According to the question,
Given, a + 15 = 4a
4a – a = 15
3a = 15
a = 15/3
a = 5
Therefore, Present age of Ravi = 5 years.
问题13：一个有理数是这样的：当您将其乘以5/2并乘积2/3时，您将得到-7/12。号码多少?
解决方案：
Let the rational be ‘a’.
According to the question,
Given, a × (5/2) + 2/3 = -7/12
5(a/2) + 2/3 = -7/12
5(a/2) = -7/12 – 2/3
5(a/2) = (-7- 8)/12
5(a/2) = -15/12
5a/2 = -5/4
a = (-5/4) × (2/5)
a = – 10/20
a = -1/2
Therefore, the rational number will be -1/2.
问题14.拉克希米是一家银行的出纳员。她的纸币面额分别为₹100，₹50和₹10。这些音符的数量比例为2：3：5。 Lakshmi的现金总额为₹4,00,000。她每种面额有多少张钞票?
解决方案：
Let the numbers of notes of ₹100, ₹50 and ₹10 be ‘2a’ , ‘3a’ and ‘5a’ respectively.
Value of ₹100 = 2a × 100 = 200a
Value of ₹50 = 3a × 50 = 150a
Value of ₹10 = 5a × 10 = 50a According to the question,
Given, 200a + 150a + 50a = 400000
400a = 400000
a = 400000/400
a = 1000
Numbers of ₹100 notes = 2a = 2000
Numbers of ₹50 notes = 3a = 3000
Numbers of ₹10 notes = 5a = 5000
问题15：我总共有₹300的面额分别为₹1，₹2和₹5的硬币。 ₹2硬币的数量是₹5硬币的数量的3倍。硬币的总数为160。每个面额有多少枚硬币?
解决方案：
Let number of ₹5 coins be ‘a’.
Then,
Number ₹2 coins = 3a
and, number of ₹1 coins = (160 – 4a) Now,
Value of ₹5 coins= a × 5 = 5a
Value of ₹2 coins = 3a × 2 = 6a
Value of ₹1 coins = (160 – 4a) × 1 = (160 – 4a)
According to the question,
Given, 5a + 6a + (160 – 4a) = 300
11a + 160 – 4a = 300
7a = 140
a = 140/7
a = 20
Number of ₹5 coins = a = 20
Number of ₹2 coins = 3a = 60
Number of ₹1 coins = (160 – 4a) = 160 – 80 = 80
问题16.论文竞赛的组织者决定，竞赛的优胜者将获得₹100的奖励，未赢得竞赛的参与者将获得₹25的奖励。分配的总奖金为₹3,000。如果参与者总数为63，则找到获胜者的数量。
解决方案：
Let the numbers of winner be ‘a’
Then, the number of participant who didn’t win will be (63 – a)
Total money given to the winner = a × 100 = 100a
Total money given to participant who didn’t win = 25 × (63 – a)
According to the question,
Given, 100a + 25 × (63 – a) = 3000
100a + 1575 – 25a = 3000
75a = 3000 – 1575
75a = 1425
a = 1425/75
a = 19
So, the number of winners are 19.