**Exercise 3.7**

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

**Soln: **Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 8. Thus, we have x+y = 8

The sum of the two numbers is four times their difference. Thus, we have

x+y = 4(xâ€” y)

=>x+y= 4x-4y

=>4x â€” 4yâ€”xâ€”y = 0

=> 3x-5y = 0

So, we have two equations

x+y=8

3x-5y = 0

Here x and y are unknowns.

We have to solve the above equations for x and y.

Multiplying the first equation by 5 and then adding with the second equation, we have

5(x + y)+(3x â€” 5y) = 5×8+0

=>5x+5y+3x-5y =40

=>8x = 40

=> x = 5

x = Substituting the value of x in the first equation, we have

5+y=8

=>y=8-5

=>y=3

Hence, the 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?**

**Soln:** Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The sum of the digits of the number is 13. Thus, we have x+y =13

After interchanging the digits, the number becomes10x+y.

The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have

(10x + y) â€” (10y + x)= 45

=> 110x+y-10yâ€”x= 45

=> 9x-9y=45

=> 9(x – y) = 45

=>.x-y=5

So, we have two equations

x+y=13

x- y =5

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

(x+y)+(xâ€”y)=13+5

=>x+y+x-y=18

=> 2x=18

=> x= 9

Substituting the value of x in the first equation, we have

9+y = 13

=> y = 13 â€“ 9

=> y = 4

Hence, the number is 10 x 4 + 9 = 49

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**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.**

**Soln:** Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The sum of the digits of the number is 5. Thus, we have x+ y = 5

After interchanging the digits, the number becomes 10x+ y.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

10x+ y =10y + x +9

=> 10x + y â€“ 10y â€“ x = 9

=> 9x â€“ 9y = 9

=> 9(x â€“ y) = 9

=> x â€“ y = 1

So, we have two equations

x +y=5

x-y=1

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

(x + y)+(x – y) = 5+1

=> x + y + x â€“ y = 5+1

=> 2x = 6

=> x = 6/2

=> x = 3

Substituting the value of x in the first equation, we have

3 + y = 5

=> y = 5-3

=> y = 2

Hence, the number is 10 x 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.**

**Soln: **Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The sum of the digits of the number is 15. Thus, we have x+ y = 15

After interchanging the digits, the number becomes 10x + y.

The number obtained by interchanging the digits is exceeding by 9 from the original number. Thus, we have

10x+y=10y+x+ 9

=> 10x + y â€“ 10y â€“x = 9

=>9x- 9y = 9

=>9(x -y) = 9

=> x-y = 9/9

=>x-y = I

So, we have two equations

x+y =15

x-y = I

Here x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have

(x+ y)+(x – y)=15+1

=>x+y+x-y=16

=>2x = 16

=> x = 16/2

=> x = 8

Substituting the value of x in the first equation, we have

8+ y = 5

=> y = 15-8

=> y = 7

Hence, the number is 10 x 7 + 8 = 78

**Q.5: The sum of 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?**

**Soln: **Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The two digits of the number are differing by 2. Thus, we have x-y=Â±2

After interchanging the digits, the number becomes 10x + y.

The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have

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

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

=> 11x +11y= 66

=> 11(x + y) = 66

=> x + y = 66/11

=> x + y = 6

So, we have two systems of simultaneous equations

x-y = 2,

x +y = 6

x â€“ y = -2,

x + y = 6

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

**(i)** First, we solve the system

x-y= 2,

x+y=6

Adding the two equations, we have

(xâ€” y)+(x + y) = 2+6

=>x-y+x+y=8

=>2x =8

=> x = 8/2

=> x = 4

Substituting the value of x in the first equation, we have

4-y=2

=>y=4-2

=> y=2

Hence, the number is 10 x 2 +4 = 24

**(ii)** Now, we solve the system

x-y= -2,

x+y=6

Adding the two equations, we have

(x – y)+(x + y)= -2 + 6

=>x-y+x+y=4

=> 2x =4

=> x = 4/2

=> x=2

Substituting the value of x in the first equation, we have

2-y = -2

=> y=2+2

=> y=4

Hence, the number is 10×4+ 2 = 42

There are two such numbers.

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

**Soln: **Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 1000. Thus, we have x+y = 1000

The difference between the squares of the two numbers is 256000. Thus, we have

x^{2} â€“y^{2} = 256000

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

=>1000(x-y)= 256000

=> x-y = 256000/1000

=>x-y =256

So, we have two equations

x+y =1000

x- y= 256

Here x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have

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

=>x+y+x-y =1256

=> 2x=1256

=> x= 1256/ 2

x=628

Substituting the value of x in the first equation.we have

628+y =1000

=> y= 1000-628

=>y = 372

Hence, the 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 units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The two digits of the number are differing by 3. Thus, we have x â€”y = Â±3

After interchanging the digits, the number becomes 10x + y.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

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

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

=>11x+11y = 99

=> 11(x + y)= 99

=>x+y = 99/11

=>x = 9

So, we have two systems of simultaneous equations

x-y= 3,

x+y = 9

x-y = -3,

x+y=9

Adding the two equations, we have

(x â€“ y)+(x + y)= 3 + 9

=>xâ€“y+x+y=12

=> 2x =12

=> x= 12/2

=>x =6

Substituting the value of x in the first equation, we have

6â€“y=3

=>y= 6-3

=> y=3

Hence, the number is 10×3+ 6 =36

(ii) Now, we solve the system

x â€“y = â€“3,

x+y=9

Adding the two equations we have

(x â€“ y)+(x + y)= â€“3 +9

=> xâ€“y+x+y= 6

=> 2x = 6

=>x=3

Substituting the value of x in the first equation, we have

3-y = -3

=>y=3+3

=>y=6

Hence, the number is 10×6+3=63

Note that 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 units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The number is 4 times the sum of the two digits. Thus, we have

10y+x = 4(x + y)

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

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

=> 3x-6y = 0

=> 3(x â€” 2y) = 0

=>x-2y=0

After interchanging the digits, the number becomes10x+y.

If 18 is added to the number, the digits are reversed. Thus, we have

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

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

=> 9x-9y =18

=> 9(x -y)=18

=>x-y= 18/9

=>x-y =2

So, we have the systems of equations

x- 2y =0,

x-y=2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Subtracting the first equation from the second, we have

(x- y)- (x – 2y) = 2-0

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

=> y=2

Substituting the value of y in the first equation, we have

x-2×2=0

=> x-4=0

=> x = 4

Hence, the number is 10x 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.**

**Soln: **Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

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

Alter interchanging the digits, the number becomes10x+ y.

If 18 is added to the number, the digits are reversed. Thus, we have

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

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

=>9x-9y =18

=> 9(x â€” y) =18

=> x-y=18/9

=> x-y= 2

So, we have the systems of equations x-2y=-1,

x-y= 2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Subtracting the first equation from the second, we have

(x- y)-(x-2y) = 2 – (-1)

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

=> y=3

Substituting the value of y in the first equation, we have

x-2×3=-1

=>x-6=-1

=> x=-1+6

=> x =5

Hence the number is 10 x3 + 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.**

**Soln:** Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The number is 4 more than 6 times the sum of the two digits. Thus, we have

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

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

=> 6x +6y -10y – x = -4

=> 5x-4y = -4

After interchanging the digits, the number becomes 10x + y.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

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

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

=> 9x-9y=-18

=> 9(x -y)= -18

=>x-y = -18/9

=>x-y = -2

So, we have the systems of equations

5x – 4y = -4,

x – y = -2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Multiplying the second equation by 5 and then subtracting from the first, we have

(5x – 4y)- (5x -5y) = -4 – (-2 x 5)

=> 5x – 4y – 5x + 5y= -4 +10

=> y = 6

Substituting the value of y in the second equation, we have

x-6 = -2

=> x = 6 â€“ 2

=> x = 4

Hence, the number is 10 x 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.**

**Soln:** Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 4 times the sum of the two digits. Thus, we have

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

After interchanging the digits, the number becomes 10x + y.

The number is twice the product of the digits. Thus, we have I0y+x = 2xy

So, we have the systems of equations

x = 2y,

10y + x = 2xy

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Substituting x = 2y in the second equation, we get

10y+2y=2x2yxy

=>12y = 4y^{2}

=> 4y^{2} -12y = 0

=> 4y(y -3) = 0

=>y(y â€” 3) = 0

=> y=0 or y=3

Substituting the value of y in the first equation, we have

y | 0 | 3 |

x | 0 | 6 |

Hence, the number is 10 x 3 +6 = 36

Note that 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.**

**Soln:**Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y+ x.

The product of the two digits of the number is 20. Thus, we have xy = 20

After interchanging the digits, the number becomes 10x+ y

If 9 is added to the number, the digits interchange their places. Thus, we have

(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

So, we have the systems of equations

xy = 20,

x-y=1

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting x = I + y from the second equation to the first equation, we get

(1+ y) y = 20

=> y +y^{2} = 20

=> y^{2}+y-20=0

=> y^{2} +5y-4y -20 = 0

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

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

=> y=-5 or y=4

Substituting the value of y in the second equation, we have

y | -5 | 4 |

x | -4 | 5 |

Hence, the number is 10 x 4 +5 = 45

Note that in the first pair of solution the values of x and y are both negative. But the digits of the number canâ€™t be negative. So, we must remove this pair.

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

**Soln:** Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The difference between the two numbers is 26. Thus, we have x -y = 26

One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have x = 3y

So, we have two equations

x-y = 26

x = 3y

Here x and y are unknowns. We have to solve the above equations for x and y.

Substituting x = 3y from the second equation in the first equation, we get

3y – y = 26

=> 2y = 26

=> y = 13

Substituting the value of y in the first equation, we have

x-13 =26

=> x=13+ 26

=> x = 39

Hence.the 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.**

**Soln:** Let the digits at units and tens place of the given number be x and y respectively Thus, the number is 10y+x.

The sum of the two digits of the number is 9.

Thus, we have x + y = 9

After interchanging the digits, the number becomes 10x + y.

Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have

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

So, we have the systems of equations

x+ y = 9,

x – 8y = 0

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Substituting x = 8y from the second equation to the first equation, we get

8y + y = 9

=>9y = 9

=> y =9/9

=> y = 1

Substituting the value of y in the second equation, we have

x-8×1=0

=> x-8 = 0

=> x = 8

Hence, the number is 10 x 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 units and tens place of the given number be x and y respectively. Thus, the number is 10y+x.

The difference between the two digits of the number is 3. Thus, we have x-y = Â±3

After interchanging the digits, the number becomes 10x + y.

Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have

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

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

=> 40x + 4y – 70y-7x = 0

=> 33x – 66y = 0

=>33(x â€” 2y) = 0

=> x – 2y = 0

So, we have two systems of simultaneous equations

x – y = 3,

x-2y=0

x – y = -3,

x – 2y = 0

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

x – y = 3,

x – 2y = 0

Multiplying the first equation by 2 and then subtracting from the second equation, we have

(x -2y)- 2(x – y) = 0 â€“ 2 x 3

=> x – 2y – 2x + 2y = -6

=> -x = -6

=> x =6

Substituting the value of x in the first equation, we have

6-y=3

=>y=6 â€“ 3

=> y=3

Hence, the number is 10 x3 + 6 = 36

(ii) Now, we solve the system

x – y = -3,

x-2y=0

Multiplying the first equation by 2 and then subtracting from the second equation we have

(x -2y)- 2(x – y) = 0 – (-3 x 2)

=> x- 2y – 2x + 2y =6

=> -x = 6

=> x = -6

Substituting the value of x in the first equation, we have

-6 – y = -3

=> y = -6 + 3

=> y = -3

But, the digits of the number can’t be negative. Hence, the second case must be removed.

**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 numbers be 5x and 6x

Now subtracting 8 we get the numbers as

5x â€“ 8 and 6xâ€“ 8

Thus, (5x â€“ 8)/ (6x â€“ 8) = 4: 5

By cross multiplying we get,

5(5x â€“ 8) = 4(6x â€“ 8)

=> 25x â€“ 40 = 24x â€“ 32

=> x = 8

Hence, the numbers are

5x = 5 x 8 = 40

6x = 6 x 8 = 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 unit digit and tenâ€™s digit of the number be x and y respectively.

Therefore the number = 10y + x

Sum of digits = x + y

10y +x = 8(x + y) â€“ 5

10y + x = 8x + 8y â€“ 5

7x â€“ 2y = 5â€¦â€¦(1)

Difference of the digits = y â€“ x [if x < y]

10y + x = 16(y â€“ x) + 3

10y + x = 16y â€“ 16x +3

17x â€“ 6y = 3â€¦â€¦(2)

Multiply equation (1) and (2) and subtracting equation (2)

21x â€“ 6y = 15

17x â€“ 6y = 3

4x = 12

x = 12/4 = 3

Putting the value of x = 3 in equation (1)

7 x 3 â€“ 2y = 5

2y = 21 â€“ 5

2y = 16

y = 16/2 = 8

Thus the unit digit of the number is 3 and tenâ€™s digit is 8.

Therefore the number is 83.