问题1.数字的五分之四大于数字的四分之三乘以4。找到数字。
解决方案：
Let us consider the number as ‘x’
Three-fourth of the number = 3x/4
Fourth-fifth of the number = 4x/5
4x/5 – 3x/4 = 4
Now we will take LCM of 5 and 4 is 20
(16x – 15x)/20 = 4
Now by doing cross-multiplying we get,
16x – 15x = 4(20)
x = 80
The number is 80.
问题2。两个连续数字的平方之间的差是31。找到数字。
解决方案：
Let us assume that two consecutive numbers be x and (x – 1)
According to question
x2 – (x-1)2 = 31
As we know the formula (a-b)2 = a2 + b2 – 2ab
x2 – (x2 – 2x + 1) = 31
x2 – x2 + 2x – 1 = 31
2x – 1 = 31
2x = 31+1
2x = 32
x = 32/2 = 16
Two consecutive numbers are x and (x-1) : 16 and (16-1) = 15
The two consecutive numbers are 16 and 15.
问题3.查找一个数字，该数字的两倍数比其一半大45。
解决方案：
Let us assume that the number is ‘x’
2x – x/2 = 45
(4x-x)/2 = 45
Now we will do cross-multiplying,
3x = 90
x = 90/3 = 30
The number will be 30.
问题4.找到一个数字，使得当从该数字的5倍中减去5时，结果是该数字的两倍多4。
解决方案：
Let us assume that number is ‘x’
Then, five times the number will be 5x (According to the question)
two times the number will be 2x
5x – 5 = 2x + 4
5x – 2x = 5 + 4
3x = 9
x = 9/3
x = 3
问题5.第五部分增加5的数字等于第四部分减少5的数字。找到该数字。
解决方案：
Let us assume that number is ‘x’
x/5 + 5 = x/4 – 5
x/5 – x/4 = -5 – 5
Now take LCM for 5 and 4 which is 20
(4x-5x)/20 = -10
Now we will do cross-multiplying,
4x – 5x = -10(20)
-x = -200
x = 200
The number is 200.
问题6.一个数字由两个数字组成，其总和为9。如果从该数字中减去27，则这些数字将颠倒。查找号码。
解决方案：
Let us assume that one of the digit be ‘x’
The other digit is 9-x
The two digit number is 10(9-x) + x
The number obtained after interchanging the digits is 10x + (9-x) [According to question]
10(9-x) + x – 27 = 10x + (9-x)
By doing simplification,
90 – 10x + x – 27 = 10x + 9 – x
-10x + x – 10x + x = 9 – 90 + 27
-18x = -54
x = 54/18
= 9/3
= 3
The two digit number is 10(9-x) + x
By substituting the value of x we get,
10(9-x) + x
10(9 – 3) + 3
10(6) + 3
60+3 = 63
The number is 63.
问题7.将184分为两部分，以使一部分的三分之一可以超过另一部分的七分之八。
解决方案：
Let assume that one of the number be ‘x’
And the other number as 184 – x
According to question, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM of 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
By doing cross-multiplying,
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
10x = 720
x = 720/10 = 72
One of the number is 72
other number is 184 – x
= 184 – 72 = 112.
问题8.小数的分子比分母小6。如果将3加到分子上，则分数等于2/3。原始分数等于多少?
解决方案：
Let us assume that denominator as x and numerator as (x-6)
As we know that formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
Now By doing cross-multiplying
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
x = 9
The denominator is x = 9,then numerator is (x-6) = (9-6) = 3
fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3
问题9：总和为800卢比的面额为10卢比和20卢比。如果纸币的总数为50，请查找每种类型的纸币的数量。
解决方案：
Let us assume that number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount of 10Rs notes = 10 × x = 10x
Amount of 20Rs notes = 20 × (50 – x) = 1000 – 20x
The total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10 = 20
The number of 10Rs notes = 20
Number of 20Rs notes are 50 – 20 = 30
问题10。Seeta Devi的硬币为9卢比，五十法郎和二十五法郎。她的二十五分钱硬币是其五十分钱硬币的两倍。她有几种硬币?
解决方案：
Let assume that number of fifty paise coins be x
Number of twenty-five paise coins will be 2x
Amount of fifty paise coins = (50×x)/100 = 0.50x
Amount of twenty-five paise coins = (25×2x)/100 = 0.50x
The total amount is Rs 9
0.50x + 0.50x = 9
1x = 9
x = 9
The number of fifty paise coins is x = 9
Number of twenty-five paise coins 2x = 2×9 = 18
问题11. Sunita的年龄是Ashima的两倍。如果从Ashima的年龄中减去6岁，再加上Sunita的年龄4年，那么Sunita的年龄将是Ashima的4倍。他们两年前几岁?
解决方案：
Let assume that present age of Ashima be ‘x’ years
Let the present age of Sunita is 2x years
So , Ashima’s new age = (x – 6) years
Sunita’s new age = (2x + 4) years
(2x + 4) = 4 (x – 6)
2x + 4 = 4x – 24
2x – 4x = -24 – 4
-2x = -28
x = -28/-2 = 14
Age of Ashima is x years = 14 years
Age of Sunita is 2x years = 2(14) = 28 years
Now ,
Two years ago, age of Ashima is 14 – 2 = 12 years
Age of Sunita = 28 – 2 = 26 years.
问题12. Sonu和Monu的年龄比例为7：5十年，因此，他们的年龄比例将为9：7。
解决方案：
Let us assume that present age of Sonu be 7x years
Present age of Monu will be 5x years
Sonu’s age after 10 years will be = (7x + 10) years
Monu’s age after 10 years will be = (5x + 10) years
(7x + 10) / (5x + 10) = 9/7
By doing cross-multiplication,
7(7x + 10) = 9(5x + 10)
49x + 70 = 45x + 90
49x – 45x = 90 – 70
4x = 20
x = 20/4 = 5
Present age of Sonu wil be 7x = 7(5) = 35years
Present age of Monu will be 5x = 5(5) = 25years
问题13。五年前，一个男人的年龄是儿子的七倍。因此，五年后，父亲的年龄将是儿子的三倍。找到他们现在的年龄。
解决方案：
Let us assume that age of son five years ago be x years
The age of man five years ago will be 7x years
After five years, son’s age will be x + 5 years
After five years father’s age will be 7x + 5 years
So, five years the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
4x = 20
x = 5
Present father’s age will be 7x + 5 = 7(5) + 5 = 35 + 5 = 40 years
Present son’s age will be x + 5 = 5 + 5 = 10 years