10类RD Sharma解–第3章两个变量的线性方程对–练习3.7

问题1.两个数字的和为8。如果两个数字的和是它们的差的四倍，则找到这些数字。

解决方案：

Let us assume the first number and second number to be x and y respectively.

Now, according to the given conditions, we have,

x + y = 8 ….(i)

and x + y = 4 (x – y)

⇒ 4 (x – y) = 8

⇒ x – y = 2 ….(ii)

On adding eq(i) and (ii), we get 

2x = 10 ⇒ x = 5

On subtracting eq(ii) from (i), we get

2y = 6 ⇒ y = 3

Therefore, the required numbers are 5 and 3 respectively. 

问题2。两位数字的位数总和是13。如果从通过互换数字获得的一位中减去该数字，则结果为45。什么是数字?

解决方案：

Let us assume the unit’s digit to be x and ten’s digit to be y respectively.

Therefore, the number = x + 10y

Now, as per the specified constraints, 

x + y = 13 ….(i)

Also, 

The same number after interchanging their digits = y + 10x

Now, we have,

y + 10x – x – 10y = 45

On solving, we get, 

9x – 9y = 45

⇒ x – y = 5

x – y = 5 ….(ii)

On adding eq(i) and (ii), we get

2x = 18 ⇒ x = 9

2y = 8 ⇒ y = 4

Substituting the value, we get,

Number = x + 10y = 9 + 4 x 10 = 9 + 40 = 49

问题3.一个数字由两个数字组成，其总和为5。当数字反转时，数字将增加9。查找号码。

解决方案：

Let us assume the unit’s digit to be x and ten’s digit to be y respectively.and ten’s digit = y

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + y = 5 ….(i)

and y + 10x = x + 10y + 9

On solving, we get,

⇒ y + 10x – x – 10y = 9

⇒ 9x – 9y = 9

On dividing by 9, we get,

⇒ x – y = 1 ….(ii)

On adding eq(i) and (ii), we get

2x = 6 ⇒ x = 3 

On subtracting eq(i) and (ii), we get

2y = 4 ⇒ y = 2 

Therefore, the number = x + 10y 

= 3 + 10 x 2 

= 3 + 20 = 23

问题4。两位数字的位数总和是15。通过反转给定数字的位数顺序获得的数字比给定数字超出9。找到给定数字。

解决方案：

Let us assume the unit’s digit to be x and ten’s digit to be y respectively.and ten’s digit = y

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + y = 15 ….(i)

y + 10x = x + 10y + 9

⇒ y + 10x – x – 10y = 9

⇒ 9x – 9y = 9

On dividing by 9, we get,

⇒ x – y = 1 ……..(ii)

Now adding eq(i) and (ii) equations together, we get,

2x = 16

=> x = 8

On subtracting eq(i) and (ii), 

2y = 14 ⇒ y = 7

Therefore, the number = x + 10y 

= 8 + 10 x 7

= 8 + 70 = 78

问题5.两位数的总和与数字顺序颠倒形成的数字的总和为66。如果两位数相差2，则找到该数字。有多少个这样的数字?

解决方案：

According to the given constraints, sum of two-digit number and 

number formed by reversing its digits = 66

Let us assume the units digit to be x and the tens digit to be x + 2

Therefore, the number = x + 10 (x + 2)

= x + 10x + 20 = 11x + 20

The number obtained by reversing its digits, is, 

assuming the units digit to be x + 2 and the tens digit to be x.

Now, number = x + 2 + 10x = 11x + 2

11x + 20 + 11x + 2 = 66

⇒ 22x + 22 = 66

⇒ 22x = 66 – 22 = 44

Solving, we get,

⇒ x = 2

The required number = 11x + 20 = 11 x 2 + 20 = 22 + 20 = 42

Also, the number by reversing its digits will be 11x + 2 = 11 x 2 + 2 = 22 + 2 = 24

Therefore, the numbers are 42 and 24 respectively.

问题6.两个数字的总和为1000，两个平方之间的差为256000。找到数字。

解决方案：

Let us assume the first number and second number to be x and y respectively.

Now, 

x + y = 1000 ……..(i)

and  x²  – y ² = 256000

Dividing by eq(i), we obtain, 

$\frac{x^2-y^2}{x+y} = \frac{256000}{1000}$

We get, x – y = 265 … (ii) 

On solving eq(i) and (ii), we get, 

2x = 1256

x = 628 

Now substitute the value of x in eq(ii), we get the value of y

y = 372.

问题7.一个两位数的数字与通过颠倒其数字顺序而获得的数字的总和为99。如果数字相差3，则找到该数字。

解决方案：

Let the unit’s digit of the number and ten’s digit be x and y respectively.

Therefore, the number = x + 10y

Upon reversing the digits, the number = y + 10x

Given,

x + 10y + y + 10x = 99

⇒ 11x + 11y = 99

⇒ x + y = 9 ….(i)

and x – y = 3 ….(ii)

On adding eq(i) and (ii), we get,

2x = 12

x = 6

On subtracting eq(i) and (ii), we get,

2y = 6

y = 3

Therefore,

The required number = x + 10y = 6 + 10 x 3 

= 6 + 30 = 36

问题8.两位数字是其数字总和的4倍。如果将18加上数字，数字将被反转。查找号码。

解决方案：

Let the unit digit and tens digit of the number be x and y respectively.

 Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + 10y = 4 (x + y)

⇒ x + 10y = 4x + 4y

⇒ 4x + 4y – x – 10y = 0

⇒ 3x – 6y = 0

⇒ x – 2y = 0

⇒ x = 2y …. (ii)

and x + 10y + 18 = y + 10x

⇒ x + 10y – y – 10x = -18

⇒ – 9x + 9y = -18

⇒ x – y = 2 ….(ii)

 From eq(i) put the value of x in eq(ii), we get

⇒ 2y – y = 2 

⇒ y = 2

x = 2y = 2 x 2 = 4

Therefore,

The required number = x + 10y = 4 + 10 x 2 

= 4 + 20 = 24

问题9.两位数是3的四位数之和。如果将18加上数字，数字将被反转。查找号码。

解决方案：

Let us considered the unit digit and tens digit of the number be x and y.

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + 10y = 4 (x + y) + 3

⇒ x + 10y = 4x + 4y + 3

⇒ x + 10y – 4x – 4y = 3

⇒ -3x + 6y = 3

⇒ x – 2y = -1 ….(i)

and x + 10y + 18 = y + 10x

⇒ x + 10y – y – 10x = -18

⇒ -9x + 9y = -18

⇒x – y = 2 ….(ii)

On subtracting eq(i) from (ii), we get,

y = 3

x – 3 = 2

⇒ x = 2 + 3 = 5  [From eq(ii)]

Therefore, 

The required number = x + 10y = 5 + 10 x 3 

= 5 + 30 = 35

问题10：两位数是4的总和，是其位数总和的6倍。如果从数字中减去18，数字将反转。查找号码。

解决方案：

Let us considered the unit digit and tens digit of the number be x and y.

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + 10y = 6 (x + y) + 4

⇒ x + 10y = 6x + 6y + 4

⇒ x + 10y – 6x – 6y = 4

⇒ -5x + 4y = 4 ….(i)

and x + 10y – 18 = y + 10x

⇒ x + 10y – y – 10x = 18

⇒ -9x + 9y = 18

⇒ x – y = -2 ….(ii)

⇒ x = y – 2

On substituting the value in eq(i),

-5 (y – 2) + 4y = 4

-5y + 10 + 4y = 4

-y = 4 – 10 = – 6

On solving, we get,

y = 6

The required number = x + 10y = 4 + 10 x 6 

= 4 + 60 = 64

问题11：两位数是其数字总和的4倍，是数字乘积的两倍。查找号码。

解决方案：

Let the unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x + 10y = 4(x + y) 

⇒ x + 10y = 4x + 4y

⇒ x + 10y – 4x – 4y = 0

⇒ -3x + 6y = 0

⇒ x = 2y …(i)

Also,

x + 10y = 2xy …(ii)

Substituting the value of x, in eq(ii)

2y + 10y  = 2 * 2y * y ⇒ 12y = 4y²

⇒ 3y = y²

⇒ y(y – 3) = 0

y = 0 is not possible. Then y = 3.

Substituting y = 3, we obtain, x = 6

Therefore, 

The required number = x + 10y = 36

问题12.一个两位数的数字等于其数字的乘积为20。如果将9加到该数字上，则这些数字将互换其位置。查找号码。

解决方案：

Let the unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

xy = 20

x + 10y + 9 = y + 10x

=> -9x + 9y = -9

x- y +1 

On substituting in eq (i), we get

(1 + y)y = 20

y² + y – 20 = 0

y² + 5y – 4y – 20 = 0 

y(y + 5) – 4(y + 5) = 0

(y – 4)(y + 5) = 0

y = -5 is not possible. Therefore, y = 4

Therefore, 

x = 1 + y = 1 + 4 = 5

Therefore, the required number is x + 10y = 5 + 40 = 45

问题13：两个数字之间的差是26，一个数字是另一个的三倍。找到他们。

解决方案：

Let us assume the first number and second number to be x and y respectively.

Now, according to the given constraints, 

x – y = 26 ….(i)

x = 3y ….(ii)

On substituting the value of x in eq.(i), we get,

3y – y = 26

⇒ 2y = 26

⇒ y = 13

On substituting the value of y, in eq(ii) we get,

x = 3y = 3 x 13 = 39

Therefore, the numbers are 39 and 13 respectively.

问题14.两位数字的位数之和为9。此外，此数字的9倍是通过反转数字的顺序获得的数字的两倍。查找号码。

解决方案：

Let the unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x + y = 9 …..(i)

9 (x + 10y) = 2 (y + 10x)

⇒ 9x + 90y = 2y + 20x

⇒ 9x + 90y – 2y – 20x = 0

⇒ -11x + 88y = 0

On dividing the equation by -11, we get,

⇒ x – 8y = 0 

⇒ x = 8y

On substituting the value of x in eq(i)

8y + y = 9

⇒ 9y = 9

⇒ y= 1

x = 8y = 1 x 8 = 8

Therefore,

The required number = x + 10y = 8 + 10 x 1 

= 8 + 10 = 18

问题15：两位数的七倍等于四位数字反转得到的数的四倍。如果数字之间的差为3。找到数字。

解决方案：

Let the unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x – y = 3 ….(i)

Also,

7 (x + 10y) = 4 (y + 10x)

⇒ 7x + 70y = 4y + 40x

⇒ 7x + 70y – 4y – 40x = 0

⇒ -33x + 66y = 0

⇒ x – 2y = 0 (Dividing by -33)

⇒ x = 2y

On substituting the value of x in eq(i),

2y – y = 3 ⇒ y = 3

x = 2y = 2 x 3 = 6

Therefore,

The required number = x + 10y = 6 + 10 x 3 

= 6 + 30 = 36

问题16：两个数字的比例为5：6。如果从每个数字中减去8，则该比例为4：5。找到数字。

解决方案：

Let the two numbers be x and y respectively.

Ratio of these two numbers = 5 : 6

that is, x : y = 5 : 6

⇒ x/y = 5/6

⇒ y = 6x/5

If 8 is subtracted from both the numbers, the ratio becomes 4:5

That is, x – 8 /(y – 8) = 4/5 

Now, 

⇒ 5x – 40 = 4y – 32 

⇒ 5x – 4y = 8

Now, substituting the value, 

5x – 4(6x/5) = 8

x = 40 

Substituting x = 40, we get,

y = 48.

Therefore, the numbers are 40 and 48 respectively.

问题17.通过将数字的总和乘以8，然后减去5，或通过将数字的差乘以16，然后再加上3，来获得两位数。找到数字。

解决方案：

Let us assume the two-digit number to be 10x + y 

Case I : Multiplying the sum of the digits by 8 and then subtract 5 

⇒ 8 x (x + y) – 5 = 10x + y

⇒ 8x + 8y – 5 = 10x + y

⇒ 2x – 7y = – 5

Case II : Multiplying the difference of the digits by 16 and then add 3

16 * (x – y) + 3 = 10x + y

⇒ 6x – 17y = -3 …(i)

Multiplying eq(i) by 3, we obtain, 

6x – 21y = -15 … (ii)

On subtracting eq(ii) from (i), 

4y = 12

y = 3

On substituting the value of y, in eq(i) we get

x = 8

Hence, 

The required number = 10x + y

⇒ 10 * 8 + 3

⇒ 83