**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,Â their sum 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……………….(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 â€” 5y) = 5Â Ã— 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) â€” (10y + x)= 45

=> 110x + y â€” 10y â€” x = 45

=> 9x â€” 9y = 45

=> 9(xÂ â€” y) = 45

=> xÂ â€” 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) + (xâ€”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 â€“ 9

=> y = 4

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

**Â **

**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 â€“ 10y â€“ x = 9

=> 9x â€“ 9y = 9

=> 9(x â€“ y) = 9

=> x â€“ 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 â€“ 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 10Â Ã— 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 â€“ 10y â€“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, 10Â Ã— 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 = Â±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 andÂ x +y = 6

x â€“ y = -2 andÂ 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 = 2Â …………..(3)

x + y = 6 …………..(4)

On adding the equations 3 and 4, we get;

(x â€” 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 10Â Ã— 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 10Ã—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;

=> x^{2} â€“y^{2} = 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 â€”y = Â±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 = 9Â ……………(2)

So, now we have two systems of simultaneous equations.

x-y= 3 andÂ x+y = 9

x-y = -3 andÂ 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 â€“ y)+(x + y)= 3 + 9

=> xâ€“y+x+y=12

=> 2x =12

=> x = 12/2

=> x =6

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

=> 6â€“y=3

=>y= 6-3

=> y=3

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

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

x â€“y = â€“3 ……….(5)

x+y=9Â …………..(6)

Adding the equations 5 and 6, we get;

=> (x â€“ y)+(x + y)= â€“3 +9

=> xâ€“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 10Ã—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 â€” 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 =2Â …………..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 – 2Â Ã— 2=0

=> x – 4=0

=> x = 4

Hence, the required number is 10Ã—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 â€” 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 Ã— 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 â€“ 2

=> x = 4

Hence, the required number is 10Â Ã— 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 = 2xyÂ …………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=2Ã—xÃ—2y

=>12y = 4y^{2}

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

=> 4y(y -3) = 0

=>y(y â€” 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 10Â Ã— 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:Â **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 +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

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.Â Hence, the number is 10Â Ã— 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-8Ã—1=0

=> x-8 = 0

=> x = 8

Hence, the number is 10Â Ã— 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 = Â±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 â€” 2y) = 0

=> x – 2y = 0 …………2

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

x – y = 3 andÂ 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 â€“ 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 â€“ 3

=> y=3

Hence, the number is 10Â Ã— 3 + 6 = 36

Now, l**et 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 â€“ 8 and 6xâ€“ 8

As per the given condition,

(5x â€“ 8)/ (6x â€“ 8) = 4: 5

On cross multiplication we get,

5(5x â€“ 8) = 4(6x â€“ 8)

=> 25x â€“ 40 = 24x â€“ 32

=> x = 8

Substituting the value of x, we get;

5x = 5Â Ã— 8 = 40

6x = 6Â Ã— 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) â€“ 5

10y + x = 8x + 8y â€“ 5

7x â€“ 2y = 5â€¦â€¦(1)

As per the given statement, 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)

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

21x â€“ 6y = 15

17x â€“ 6y = 3

4x = 12

x = 12/4 = 3

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

7Â Ã— 3 â€“ 2y = 5

2y = 21 â€“ 5

2y = 16

y = 16/2 = 8

Therefore, number at unit place is 3 and tenâ€™s digit is 8.

Hence, the required number is 83.