Another application of linear equations in two variables is solving problems based on numbers. The RD Sharma Solutions Class 10 created by experts at BYJUâ€™S can serve as an important tool for reference and preparing for exams. Students wishing to get a good grip on this particular application of LPE can utilise the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.7 PDF given below.

## RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.7 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.7

**1. The sum of two numbers is 8. If their sum is four times their difference, find the numbers.**

**Solution:**

Letâ€™s assume the two numbers to be x and y.

Also letâ€™s consider that, x is greater than or equal to y.

Now, according to the question

The sum of the two numbers, x + y = 8â€¦â€¦â€¦â€¦. (i)

Also given that, their sum is four times their difference. So, we can write;

x + y = 4(x – y)

â‡’ x + y = 4x-4y

â‡’ 4x – 4y â€“ x – y = 0

â‡’ 3x â€“ 5y = 0â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Solving (i) and (ii), we can find x and y, so the required two numbers.

On multiplying equation (i) by 5 and then add with equation (ii), we get here;

5 (x + y) + (3x – 5y) = 5Â Ã— 8 + 0

â‡’ 5x + 5y + 3x â€“ 5y = 40

â‡’ 8x = 40

â‡’ x = 5

Putting the value of x in (i), we find y

5 + y = 8

â‡’ y = 8 â€“ 5

â‡’ y = 3

Therefore, the two numbers are 5 and 3.

**2. The sum of digits of a two-digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number?**

**Solution:**

Letâ€™s assume the digit at the unitâ€™s place as x and at tenâ€™s place as y. Then the required number is 10y + x.

Also itâ€™s given that, the sum of the digits of the number is 13,

So, x + y = 13â€¦â€¦â€¦â€¦ (i)

On interchanging the position of digits, the new number so formed will be 10x+y.

Again is itâ€™s given that, the difference between the new number so formed upon interchanging the digits and the original number is equal to 45. Therefore, this can be expressed as;

(10x + y) – (10y + x) = 45

â‡’ 110x + y – 10y – x = 45

â‡’ 9x – 9y = 45

â‡’ 9(xÂ – y) = 45

â‡’ xÂ – y = 5â€¦â€¦â€¦..(ii)

Solving (i) and (ii) we can find x and y,

Now, adding (i) and (ii), we get;

(x + y) + (x – y) = 13 + 5

â‡’ x + y + x â€“ y = 18

â‡’ 2x = 18

â‡’ x = 9

Putting the value of x in the equation (i), we find y;

9 + y = 13

â‡’ y = 13 â€“ 9

â‡’ y = 4

Hence, the required number is,Â 10 Ã— 4 + 9 = 49.

**3. A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.**

**Solution:**

Letâ€™s assume the digit at unitâ€™s place as x and tenâ€™s place as y. Thus, the number to be found is 10y + x.

From the question itâ€™s given as, the sum of the digits of the number is equal to 5.

Thus we can write, x + y = 5 â€¦â€¦â€¦â€¦.. (i)

On interchange the place of digits, the new number so formed will be 10x+ y.

Again from the question itâ€™s given as, the new number so obtained after interchanging the digits is greater by 9 from the original number. Therefore, this can be written as;

10x + y = 10y + x +9

â‡’ 10x + y â€“ 10y â€“ x = 9

â‡’ 9x â€“ 9y = 9

â‡’ 9(x â€“ y) = 9

â‡’ x â€“ y = 1â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Solving (i) and (ii), we can find x and y

Adding the eq. 1 and 2, we get;

(x + y) + (x â€“ y) = 5+1

â‡’ x + y + x â€“ y = 5+1

â‡’ 2x = 6

â‡’ x = 6/2

â‡’ x = 3

Putting the value of x in the equation 1, we get;

3 + y = 5

â‡’ y = 5-3

â‡’ y = 2

Hence, the required number is 10Â Ã— 2 + 3 = 23

**4. The sum of digits of a two-digit number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.**

**Solution:**

Let the digits at unitâ€™s place be x and tenâ€™s place be y, respectively. Thus, the number we need to find is 10y + x.

As per the given statement, the sum of the digits of the number is 15. Thus, we have;

x+ y = 15 â€¦â€¦â€¦â€¦â€¦(i)

Upon interchanging the digitâ€™s place, the new number will so be 10x + y.

Also itâ€™s given from the question that, the new number obtained exceeds from the original number by 9. Therefore, we can write this as;

10x + y = 10y + x + 9

â‡’ 10x + y â€“ 10y â€“x = 9

â‡’ 9x â€“ 9y = 9

â‡’ 9(x â€“ y) = 9

â‡’ x â€“ y = 9/9

â‡’ x â€“ y = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

Solving (i) and (ii), we can find x and y

Now, adding the equations (i) and (ii), we get;

(x + y) + (x â€“ y) = 15 + 1

â‡’ x + y + x â€“ y = 16

â‡’ 2x = 16

â‡’ x = 16/2

â‡’ x = 8

Putting the value of x in the equation (i), to get y

8+ y = 5

â‡’ y = 15 â€“ 8

â‡’ y = 7

Hence, the required number is, 10Â Ã— 7 + 8 = 78

**5. The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?**

**Solution:**

Letâ€™s assume the digit at unitâ€™s place as x and tenâ€™s place as y. Thus from the question, the number needed to be found is 10y + x.

From the question itâ€™s told as, the two digits of the number are differing by 2. Thus, we can write

x â€“ y = Â±2â€¦â€¦â€¦â€¦.. (i)

Now after reversing the order of the digits, the number becomes 10x + y.

Again from the question itâ€™s given that, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as;

(10x+ y) + (10y+x) = 66

â‡’ 10x + y + 10y + x = 66

â‡’ 11x +11y = 66

â‡’ 11(x + y) = 66

â‡’ x + y = 66/11

â‡’ x + y = 6â€¦â€¦â€¦â€¦.. (ii)

Now, we have two sets of systems of simultaneous equations

x â€“ y = 2 andÂ x + y = 6

x â€“ y = -2 andÂ x + y = 6

Letâ€™s first solve the first set of system of equations;

x â€“ y = 2Â â€¦â€¦â€¦â€¦. (iii)

x + y = 6 â€¦â€¦â€¦â€¦.. (iv)

On adding the equations (iii) and (iv), we get;

(x – y) + (x + y) = 2+6

â‡’ x â€“ y + x + y = 8

â‡’ 2x =8

â‡’ x = 8/2

â‡’ x = 4

Putting the value of x in equation (iii), we get

4 â€“ y = 2

â‡’ y = 4 â€“ 2

â‡’ y = 2

Hence, the required number is 10Â Ã— 2 +4 = 24

Now, letâ€™s solve the second set of system of equations,

x â€“ y = -2 â€¦â€¦â€¦â€¦. (v)

x + y = 6 â€¦â€¦â€¦â€¦.. (vi)

On adding the equations (v) and (vi), we get

(x â€“ y)+(x + y )= -2 + 6

â‡’ x â€“ y + x + y = 4

â‡’ 2x = 4

â‡’ x = 4/2

â‡’ x = 2

Putting the value of x in equation 5, we get;

2 – y = -2

â‡’ y = 2+2

â‡’ y = 4

Hence, the required number is 10Ã—4+ 2 = 42

Therefore, there are two such possible numbers i.e, 24 and 42.

**6. The sum of two numbers is 1000 and the difference between their square is 256000. Find the numbers.**

**Solution:**

Letâ€™s assume the two numbers be x and y. And also assume that x is greater than or equal to y.

So as per the question, we can write the sum of the two numbers as

x + y = 1000 â€¦â€¦â€¦.. (i)

Again itâ€™s given that, the difference between the squares of the two numbers, thus writing it

x^{2}Â â€“ y^{2}Â = 256000

â‡’ (x + y) (x – y) = 256000

â‡’ 1000(x-y) = 256000

â‡’ x – y = 256000/1000

â‡’ x â€“ y = 256 â€¦â€¦â€¦â€¦.. (ii)

By solving (i) and (ii), we can find the two numbers

On adding the equations (i) and (ii), we get;

(x+ y) + (x- y) = 1000 + 256

â‡’ x + y + x – y =1256

â‡’ 2x = 1256

â‡’ x = 1256/ 2

â‡’ x = 628

Now, putting the value of x in equation (i), we get

628 + y =1000

â‡’ y = 1000 – 628

â‡’ y = 372

Hence, the two required numbers are 628 and 372.

**7. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.**

**Solution:**

Letâ€™s assume the digit at unitâ€™s place is x and tenâ€™s place is y. Thus from the question, the number we need to find is 10y + x.

From the question since the two digits of the number are differing by 3. Therefore,

x – y = Â±3 â€¦â€¦â€¦â€¦. (i)

And, after reversing the digits, the number so obtained is 10x + y.

Again itâ€™s given from the question that, the sum of the numbers obtained by reversing the digitâ€™s places and the original number is 99. Thus, this can be written as;

(10x + y) + (I0y + x) = 99

â‡’ 10x + y + 10y + x = 99

â‡’ 11x + 11y = 99

â‡’ 11(x + y) = 99

â‡’ x + y = 99/11

â‡’ x + y = 9Â â€¦â€¦â€¦â€¦â€¦ (ii)

So, finally we have two sets of systems of equations to solve. Those are,

x â€“ y = 3 andÂ x + y = 9

x â€“ y = -3 andÂ x + y = 9

Now, letâ€™s solve the first set of system of equations;

x â€“ y = 3 â€¦â€¦â€¦.. (iii)

x + y = 9 â€¦â€¦â€¦. (iv)

Adding the equations (iii) and (iv), we get;

(x â€“ y) + (x + y) = 3 + 9

â‡’ x â€“ y + x + y =12

â‡’ 2x = 12

â‡’ x = 12/2

â‡’ x = 6

Putting the value of x in equation (iii), we find y

6 â€“ y = 3

â‡’ y = 6 – 3

â‡’ y = 3

Hence, when considering this set the required number should be 10Ã—3 + 6 =36

Now, when solving the second set of system of equations,

x â€“ y = â€“3 â€¦â€¦â€¦.(v)

x + y = 9Â â€¦â€¦â€¦â€¦.. (vi)

Adding the equations (v) and (vi), we get;

(x â€“ y) + (x + y) = â€“3 + 9

x â€“ y + x + y = 6

2x = 6

x = 3

Putting the value of x in equation 5, we get;

3 – y = -3

â‡’ y = 3 + 3

â‡’ y = 6

Hence, when considering this set the required number should be 10Ã—6+3=63

Therefore, there are two such numbers for the given question.

**8. A two- digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.**

**Solution:**

Letâ€™s assume the digit at unitâ€™s place is x and at tenâ€™s place is y. Thus from the question, the number we need to find is 10y + x.

From the question since the number is 4 times the sum of the two digits. We can write,

10y + x = 4(x + y)

â‡’ 10y + x = 4x+ 4y

â‡’ 4x + 4y – 10y -x = 0

â‡’ 3x – 6y = 0

â‡’ 3(x – 2y) = 0

â‡’ x – 2y = 0 â€¦â€¦â€¦â€¦â€¦â€¦ (i)

Secondly, after reversing the digits, the new number formed is 10x + y.

Again itâ€™s given from the question that if 18 is added to the original number, the digits are reversed. Thus, we have

(10y+x) + 18 = 10x+y

â‡’ 10x + y- 10y – x = 18

â‡’ 9x – 9y = 18

â‡’ 9(x -y) = 18

â‡’ x â€“ y = 18/9

â‡’ x-y =2Â â€¦â€¦â€¦â€¦. (ii)

Now by solving equation (i) and (ii) we can find the value of x and y and thus the number.

On subtracting the equation (i) from equation (ii), we get;

(x- y) – (x â€“ 2y) = 2-0

â‡’ x â€“ y â€“ x + 2y = 2

â‡’ y=2

Putting the value of y in the equation (i) to find x, we get

x â€“ 2Â Ã— 2=0

â‡’ x â€“ 4=0

â‡’ x = 4

Hence, the required number is 10Ã—2 +4 = 24