问题1.小数的分子比分母小4。如果分子减2，分母增加1，则分母是分子的八倍。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x = y – 4,
x – y = − 4
and y + 1 = 8(x – 2)
y + 1 = 8x – 16
8x – y = 1 + 16
8x – y = 17
Therefore, we have two equations
x – y = -4 —————-(i)
8x – y = 17 ——————(ii)
Subtracting the second equation from the first equation, we get
(x – y) – (8x – y) = – 4 – 17
x − y − 8x + y = −21
−7x = −21
−7x = −21
x = 21/7 = 3
Substituting the value of x in the (i) eqn, we have
3 – y = – 4
y = 3 + 4 = 7
Hence the fraction is 3/7
问题：2如果分子和分母都加2，则小数变为9/11。如果分子和分母都加3，则变为5/6。找出分数
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x+2 / y+2 = 9/11
11(x+2) = 9(y+2)
11x + 22 = 9y + 18
11x – 9y = 18 – 22
11x – 9y + 4 = 0 —————-(i)
and x+3 / y+3 = 5/6
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x – 5y = 15 –18
6x – 5y + 3 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we have
x / (-9 x 3 -(-5) x 4) = -y / (11 x 3 – 6 x 4) = 1 / (11 x (-5) – 6 x (-9))
x / 7 = y / 9 = 1
x = 7 and y = 9
Hence, the fraction is 7/9.
问题3。如果分子和分母都减去1，则分数变成1/3。如果分子和分母都加1，则变为1/2。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y.
According to given condition’s,
x-1 / y-1 = 1/3
3(x – 1) = (y – 1)
3x – 3 = y – 1
3x – y – 2 = 0 —————–(i)
and x+1 / y+1 = 1/2
(2x + 1) = (y + 1) ⇒ 2x + 2 = y + 1
2x – y + 1 = 0 ——————(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / -1-2 = -y / 3+4 = 1 / -3+2
x / -3 = -y / 7 = 1 / -1
x = 3 and y = 7
Hence, the fraction is 3/7.
问题4。如果我们将1加到分子上并从分母中减去1，则小数变为1。如果只将1加到分母上，则分数也变成1/2。分数是多少?
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y.
According to given condition’s,
x+1 / y-1 = 1
(x + 1) = (y – 1)
x + 1– y + 1 = 0
x – y + 2 = 0 ————–(i)
and x / y+1 = 1/2
2x = (y + 1)
2x – y – 1 = 0 ————-(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / 1+2 = -y / -1-4 = 1 / -1+2
x/3 = y/5 = 1
x = 3 and y = 5
Hence, the fraction is 3/5.
问题5.小数的分子和分母之和为12。如果分母增加3，则小数变为1/2。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 12
x + y – 12 = 0 —————(i)
and x / y+3 = 1/2
2x = (y + 3)
2x – y – 3 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-3-12) = -y / (-3+24) = 1 / (-1-2)
x/15 = y/21 = 1/3
x = 5 and y = 7
Hence the fraction is 5/7.
问题6.当分母加3并从分子中减去2时，分数变为14。并且，当分母加6且分母乘以3时，分数变为23。找到分数。
解决方案：
Let’s assume that the numerator of a fraction be x and denominator be y,
According to given condition,
x-2 / y+3 = 1/4
4x – 8 = y + 3
4x – y = 11 ————–(i)
and x+6 / 3y = 2/3
3x + 18 = 6y
x – 2y = -6 ————–(ii)
x = 2y – 6 (from eqn. (ii))
substitute value of x in eqn. (i)
4(2y – 6) – y = 11
8y – 24 – y = 11
y = 5
x = 2 x 5 – 6 = 4
Hence, x / y = 4 / 5
问题7：分数的分子和分母的总和为18。如果分母增加2，则分数减小为1/3。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 18
x + y – 18 = 0 —————(i)
and x / y+2 =1/3
3x = (y + 2)
3x – y – 2 = 0
3x – y – 2 = 0 —————-(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / (-2-18) = -y / (-2+54) = 1 / (-1-3)
x/-20 = -y/52 = 1/-4
x = 5 and y = 13
Hence, the fraction is 5/13
问题8.如果将2加到分数的分子上，则减为1/2，如果从分母中减去1，则减为1/3。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
2(x + 2) = y
2x + 4 = y
2x – y + 4 = 0 ————-(i)
and x / y-1 = 1/3
3x = (y – 1)
3x – y + 1 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-1+4) = -y / (2-12) = 1 / (-2+3)
x / 3 = y / 10 = 1
x = 3 and y = 10
Hence, the fraction is 3/10.
问题9.小数的分子和分母的总和是分子的两倍的4多。如果分子和分母增加3，则它们的比率为2：3。确定分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 2x + 4
2x + 4 – x – y = 0
x – y + 4 = 0 —————-(i)
and x + 3 : y + 3 = 2 : 3
3(x + 3) = 2(y + 3)
3x + 9 = 2y + 6
3x – 2y + 3 = 0 ——–(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-120+60) = y / (200-75) = 1 / (-20+15)
x / 60 = y / 125 = 1 / 5
x = 5 and y = 9
Hence, the fraction is 5/9.
问题10.如果分数的分子乘以2，而分母减小5，则分数变为6/5。并且，如果分母加倍并且分子增加8，则分数变为2/5。找到分数。
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
2x / y-5 = 6/5
10x = 6(y – 5)
10x – 6y + 30 = 0
2(5x – 3y + 15) = 0
5x – 3y + 15 = 0 ————–(i)
and x+8 / 2y = 2/5
5(x + 8) = 4y
5x + 40 = 4y
5x – 4y + 40 = 0 ———–(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-120+60) = -y / (200-75) = 1 / (-20+15)
x / 60 = y / 125 = 1 / 5
x = 12 and y = 25
Hence, the fraction is 12/25.
问题11：分数的分子和分母之和比分母的两倍小3。如果分子和分母减小1，则分子变为分母的一半。确定分数
解决方案：
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x-1 = 1/2 x (y-1)
x-1 / y-1 = 1/2
x + y = 2y – 3
x + y – 2y + 3 = 0
x – y + 3 = 0 ————(i)
and 2(x – 1) = (y – 1)
2x – 2 = (y – 1)
2x – y – 1 = 0 —————(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (1+3) = -y / (-1-6) = 1 / (-1+2)
x / 4 = y / 7 = 11
x = 4 and y = 7
Hence, the fraction is 4/7.