RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.7

RD Sharma Class 10 Solutions Chapter 3 Ex 3.7 PDF Free Download

Exercise 3.7

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

Solution: Let the two numbers be x and y.

Let us say, x is greater than or equal to y.

Now as per the given question;

The sum of the two numbers, x + y = 8………….(1)

Next statement is,  their sum is four times their difference. Thus, we have;

x + y = 4(x— y)

=>x + y = 4x-4y

=>4x — 4y—x—y = 0

=> 3x – 5y = 0……………….(2)

Now we have to solve equation 1 and 2 to find the value of x and y.

Multiply equation.(1) by 5 and then add with equation (2), we will get here;

5 (x + y) + (3x — 5y) = 5 × 8 + 0

=> 5x + 5y + 3x – 5y =40

=>8x = 40

=> x = 5

Putting the value of x in eq.(1), we get;

=> 5 + y = 8

=> y = 8 – 5

=> y = 3

Therefore, the two numbers are 5 and 3.

 

Q. 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: Assume, the digits at unit’s place is x and at ten’s place is y, respectively. Then as per the given question, the number is 10y + x.

Given, the sum of the digits of the number,

x + y = 13…………(1)

If we interchange the position of digits, the new number will be, 10x+y.

Again given here, the difference between the new number and the original number is equal to 45. Therefore, we can write it as;

(10x + y) — (10y + x)= 45

=> 110x + y — 10y — x = 45

=> 9x — 9y = 45

=> 9(x — y) = 45

=> x — y = 5………..(2)

Now we have to solve equation 1 and 2 to find the value of x and y.

Adding the eq. (1) and (2), we get;

=> (x+y) + (x—y) = 13 + 5

=> x + y + x – y = 18

=> 2x = 18

=> x = 9

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

=> 9 + y = 13

=> y = 13 – 9

=> y = 4

Therefore, the required number is,  10 × 4 + 9 = 49

 

Q.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: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number becomes 10y+x.

Now as per the given statement, the sum of the digits of the number is equal to 5.

Therefore, x + y = 5 …………..(1)

When we interchange the place of digits, the new number will be 10x+ y.

Again as per the given statement, the new number obtained is greater by 9 from the original number. Therefore, we can write;

=> 10x + y = 10y + x +9

=> 10x + y – 10y – x = 9

=> 9x – 9y = 9

=> 9(x – y) = 9

=> x – y = 1……………….(2)

Now, we have to solve equation 1 and 2 to find the value of x and y.

Adding the eq. 1 and 2, we get;

=> (x + y) + (x – y) = 5+1

=> x + y + x – y = 5+1

=> 2x = 6

=> x = 6/2

=> x = 3

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

3 + y = 5

=> y = 5-3

=> y = 2

Hence, the required number is 10 × 2 + 3 = 23

 

Q.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 unit’s place be x and ten’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 ……………(1)

Upon interchanging the digit’s place, the new number will be 10x + y.

Given, the new number obtained is exceeding from the original number by 9 . Therefore, we can write the given condition as;

=> 10x + y = 10y + x + 9

=> 10x + y – 10y –x = 9

=> 9x – 9y = 9

=> 9(x – y) = 9

=> x – y = 9/9

=> x – y = 1 …………………..(2)

Now we have to solve equation 1 and 2 to find the value of x and y.

Adding the equations 1 and 2, we get;

=> (x+ y) + (x – y) = 15 + 1

=> x + y + x – y = 16

=> 2x = 16

=> x = 16/2

=> x = 8

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

=> 8+ y = 5

=> y = 15 – 8

=> y = 7

Hence, the required number is, 10 × 7 + 8 = 78

 

Q.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: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number we need to find is 10y + x.

As per the given statement, the two digits of the number are differing by 2. Thus,

x – y = ±2…………..(1)

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

Again given, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, we can write the given condition as;

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

=> 10x+y+10y+x = 66

=> 11x +11y= 66

=> 11(x + y) = 66

=> x + y = 66/11

=> x + y = 6…………..(2)

Now, we have two systems of simultaneous equations

x – y = 2 and x +y = 6

x – y = -2 and x + y = 6

Now we have to solve both the systems, to find the value of x and y.

Let us solve the first system of equations;

x – y = 2  …………..(3)

x + y = 6 …………..(4)

On adding the equations 3 and 4, we get;

(x — y)+(x + y) = 2+6

=>x-y+x+y=8

=>2x =8

=> x = 8/2

=> x = 4

Putting the value of x in equation 3, we get;

=> 4 – y = 2

=> y = 4 – 2

=> y = 2

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

Now, we have to solve the second system of equation,

x – y = -2 ………….(5)

x + y = 6 …………..(6)

On adding the equations 5 and 6, we get;

=> (x – y)+(x + y)= -2 + 6

=> x-y+x+y=4

=> 2x =4

=> x = 4/2

=> x=2

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

=>2-y = -2

=> y=2+2

=> y=4

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

Therefore, there are two such possible numbers.

 

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

Solution: Let the two numbers be x and y. And assume that x is greater than or equal to y.

As per the given statement, the sum of the two numbers,

x+y = 1000 ………..1

Again given, the difference between the squares of the two numbers;

=> x2 –y2 = 256000

=>(x+y)(x-y)= 256000

=>1000(x-y)= 256000

=> x-y = 256000/1000

=> x – y = 256 …………..2

Now we have to solve equation 1 and 2 to find the value of x and y.

On adding the equations 1 and 2, we get;

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

=>x+y+x-y =1256

=> 2x=1256

=> x= 1256/ 2

x=628

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

=> 628 + y =1000

=> y= 1000-628

=>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.

Soln: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number we need to find is 10y + x.

As per the given statement, the two digits of the number are differing by 3. Therefore,

x —y = ±3 …………..(1)

After reversing the digits, the number obtained is 10x + y.

Again given, the sum of the numbers obtained by reversing the digit’s places and the original number is 99. Thus, we can write the given situation as;

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

=> 10x+y+10y+x = 99

=>11x+11y = 99

=> 11(x + y)= 99

=>x+y = 99/11

=>x + y = 9  ……………(2)

So, now we have two systems of simultaneous equations.

x-y= 3 and x+y = 9

x-y = -3 and x+y=9

Now we have to solve both the systems, to find the value of x and y.

Let us solve the first system of equations;

x-y= 3 ………..(3)

x+y = 9 ……….(4)

Adding the equations 3 and 4, we get;

=> (x – y)+(x + y)= 3 + 9

=> x–y+x+y=12

=> 2x =12

=> x = 12/2

=> x =6

Putting the value of x in equation 3, we get;

=> 6–y=3

=>y= 6-3

=> y=3

Hence, the required number is 10×3+ 6 =36

Now, we have to solve the second system of equation,

x –y = –3 ……….(5)

x+y=9  …………..(6)

Adding the equations 5 and 6, we get;

=> (x – y)+(x + y)= –3 +9

=> x–y+x+y= 6

=> 2x = 6

=>x=3

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

=> 3-y = -3

=>y=3+3

=> y=6

Hence, the number is 10×6+3=63

Therefore, there are two such numbers.

 

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.

Soln: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number we need to find is 10y+x.

As per the given statement, the number is 4 times the sum of the two digits. Thus,

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

=>10y+x = 4x+ 4y

=> 4x +4y -10y -x =0

=> 3x-6y = 0

=> 3(x — 2y) = 0

=>x-2y=0 ………………1

After reversing the digits, the new number is 10x+y.

Again given here, if 18 is added to the original number, the digits are reversed. Thus, we get;

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

=>10x+y-10y-x=18

=> 9x-9y =18

=> 9(x -y)=18

=>x-y= 18/9

=>x-y =2  …………..2

Now we have to solve equation 1 and 2 to find the value of x and y.

On Subtracting the eq. 1 from eq.2, we get;

(x- y)- (x – 2y) = 2-0

=> x-y-x+ 2y =2

=> y=2

Putting the value of y in the eq.1, we get;

=>x – 2 × 2=0

=> x – 4=0

=> x = 4

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

 

9. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number we need to find, is 10y + x.

As per the given statement, the number is 3 more than 4 times the sum of the two digits. Thus,

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

=>10y+x=4x+4y+3

=> 4x+4y-10y-x=-3

=> 3x – 6y = -3

=> 3(x – 2y) = -3

=>x-2y= -3/3

=>x-2y= -1 …………….1

Alter reversing the digits, the new number is 10x+ y.

Again given here, if 18 is added to the original number, the digits are reversed. Thus, we get;

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

=>10x+y-10y-x =18

=>9x-9y =18

=> 9(x — y) =18

=> x-y=18/9

=> x-y= 2 ……………2

Now we have to solve equation 1 and 2 to find the value of x and y.

On subtracting the eq.1 from eq.2, we get;

=> (x- y)-(x-2y) = 2 – (-1)

=> x- y-x+2y =3

=> y=3

Putting the value of y in the eq. 1, we get;

=>x-2×3=-1

=>x-6=-1

=> x=-1+6

=> x =5

Hence, the required number is 10 × 3 + 5 = 35

 

10. A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.

Solution: Let the digits at unit’s place is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y+x.

As per the given statement, the number is 4 more than 6 times the sum of the two digits. Thus,

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

=> 10y+x=6x+6y+4

=> 6x +6y -10y – x = -4

=> 5x-4y = -4 …………….1

After reversing the digits, the new number is 10x + y.

Again given here, if 18 is added to the original number, the digits are reversed. Thus, we get;

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

=> 10x+y-10y-x=-18

=> 9x-9y=-18

=> 9(x -y)= -18

=>x-y = -18/9

=>x-y = -2 ………….2

Now we have to solve equation 1 and 2 to find the value of x and y.

On multiplying the eq.2 by 5 and then subtracting from the eq.1, we get;

=> (5x – 4y)- (5x -5y) = -4 – (-2 x 5)

=> 5x – 4y – 5x + 5y= -4 +10

=> y = 6

Putting the value of y in the eq.2, we get;

=> x-6 = -2

=> x = 6 – 2

=> x = 4

Hence, the required number is 10 × 6 + 4 = 64

 

11. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Solution: Let the digits at unit’s is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y + x.

As per the given statement, the number is 4 times the sum of the two digits. Thus,

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

=> 10y + x = 4x + 4y

=> 4x +4y -10y – x = 0

=> 3x – 6y = 0

3(x – 2y) = 0

=> x – 2y = 0

=> x = 2y ………..1

After reversing the digits, the new number we get is 10x + y.

Given, the number is twice the product of the digits. Thus, we can write,

10y+x = 2xy  …………2

Now we have to solve equation 1 and 2 to find the value of x and y.

Putting x = 2y in the eq.2, we get

=> 10y+2y=2×x×2y

=>12y = 4y2

=> 4y2 -12y = 0

=> 4y(y -3) = 0

=>y(y — 3) = 0

=> y=0 or y=3

Putting the value of y in eq.1, we get;

y 0 3
x 0 6

Hence, the number is 10 × 3 +6 = 36

Therefore, the first pair of solution does not give a two-digit number.

 

12. A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Solution: Let the digits at unit’s is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y + x.

As per the given statement, the product of the two digits of the number,

xy = 20…………(1)

After reversing the digits, the new number is 10x + y.

Again given here, if 9 is added to the original number, the digits are reversed. Thus, we get;

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

=>10y+x+9=10x+y

=> 10x+y-10y-x = 9

=> 9x- 9y = 9

=> 9(x – y) = 9

=> x-y = 9/9

=> x-y=1………….(2)

Now we have to solve equation 1 and 2 to find the value of x and y.

Putting x = 1 + y from the eq.2 to the eq.1, we get;

=> (1+ y) y = 20

=> y +y2 = 20

=> y2+y-20=0

=> y2 +5y-4y -20 = 0

=>y(y + 5)- 4(y + 5) = 0

=>(y+5) (y-4)= 0

=> y=-5 or y=4

Putting the value of y in the eq.2, we get;

y -5 4
x -4 5

We can see, in the first pair of solution, the values of both x and y are negative. But, we cannot consider negative digits. Hence, the number is 10 × 4 +5 = 45

 

13. The difference between two numbers is 26 and one number is three times the other. Find them.

Soln: Let the two numbers be x and y. Assume that x is greater than or equal to y, to match the given condition.

As per the given statement, the difference between the two numbers,

=> x -y = 26 …………1

Also, given, x is three times y. Thus, we get;

=> x = 3y ……………2

Now we have to solve equation 1 and 2 to find the value of x and y.

Putting x = 3y from the eq.2 in the eq.1, we get;

=> 3y – y = 26

=> 2y = 26

=> y = 13

Putting the value of y in the eq.1, we get;

=>x-13 =26

=> x=13+ 26

=> x = 39

Hence.the two numbers are 39 and 13.

 

14. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution: Let the digits at unit’s is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y + x.

As per the given statement, the sum of the two digits of the number,

=> x + y = 9 …………..1

After reversing the digits, the new number is 10x + y.

Also given, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we can write the given situation;

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

=> 90y+ 9x = 20x + 2y

=> 20x + 2y – 90y – 9x = 0

=>11x -88y = 0

=>11(x -8y) = 0

=>x-8y=0 …………..2

Now we have to solve equation 1 and 2 to find the value of x and y.

Putting x = 8y from the eq.2 to the eq.1, we get;

=> 8y + y = 9

=>9y = 9

=> y =9/9

=> y = 1

Putting the value of y in the eq.2, we get;

=>x-8×1=0

=> x-8 = 0

=> x = 8

Hence, the number is 10 × 1 +8 = 18

 

15. Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.

Soln: Let the digits at unit’s is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y + x.

As per the given statement, the difference between the two digits of the number,

=> x – y = ±3 …………1

After reversing the digits, the new number is 10x + y.

Also given, seven times the number is equal to four times the number produced by reversing the order of the digits. Thus, we get;

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

=> 70y + 7x = 40x + 4y

=> 40x + 4y – 70y-7x = 0

=> 33x – 66y = 0

=>33(x — 2y) = 0

=> x – 2y = 0 …………2

So, now we have two systems of simultaneous equations to find x and y value;

x – y = 3 and x-2y=0

x – y = -3 and x – 2y = 0

Let us solve the first two systems of equations, i.e.;

x – y = 3 ……….3

x – 2y = 0 ………….4

On multiplying eq.3 by 2 and then subtracting from the eq.4, we get;

=> (x -2y)- 2(x – y) = 0 – 2 x 3

=> x – 2y – 2x + 2y = -6

=> -x = -6

=> x =6

Putting the value of x in the eq.3, we get;

=> 6-y=3

=>y=6 – 3

=> y=3

Hence, the number is 10 × 3 + 6 = 36

Now, let us solve the next two systems of equations, i.e.;

x – y = -3 ……….5

x-2y = 0 ………..6

On multiplying the eq.5 by 2 and then subtracting from eq.6, we get;

=>(x -2y)- 2(x – y) = 0 – (-3 x 2)

=> x- 2y – 2x + 2y =6

=> -x = 6

=> x = -6

Putting the value of x in eq.5, we get;

=> -6 – y = -3

=> y = -6 + 3

=> y = -3

But, as we know the digits of the number cannot be negative. Thus, only first case satisfies the condition.

 

16. Two numbers are in the ratio 5: 6. If 8 is subtracted from each of the numbers the ratio becomes 4: 5. Find the numbers.

Soln: Let the two numbers be 5x and 6x, which we need to find.

Now, on subtracting 8 from both the numbers, we get;

5x – 8 and 6x– 8

As per the given condition,

(5x – 8)/ (6x – 8) = 4: 5

On cross multiplication we get,

5(5x – 8) = 4(6x – 8)

=> 25x – 40 = 24x – 32

=> x = 8

Substituting the value of x, we get;

5x = 5 × 8 = 40

6x = 6 × 8 = 48

Hence, the two numbers are 40 and 48.

 

17. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Soln: Let the digits at unit’s is x and ten’s place is y, respectively. Thus, the number, we need to find is 10y + x.

As per the given statement,Sum of digits = x + y

10y +x = 8(x + y) – 5

10y + x = 8x + 8y – 5

7x – 2y = 5……(1)

As per the given statement, difference of the digits = y – x [if x < y]

10y + x = 16(y – x) + 3

10y + x = 16y – 16x +3

17x – 6y = 3……(2)

On Multiplying eq. (1) by 3 and subtracting from eq.(2), we get;

21x – 6y = 15

17x – 6y = 3

4x = 12

x = 12/4 = 3

Putting the value of x = 3 in equation (1), we get;

7 × 3 – 2y = 5

2y = 21 – 5

2y = 16

y = 16/2 = 8

Therefore, number at unit place is 3 and ten’s digit is 8.

Hence, the required number is 83.

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