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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Quadratic Equations |

Exercise | 8.7 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **8**** Quadratic Equations** Ex 8.7 Download PDF

**Quadratic Equations**Ex 8.7 Download PDF

**Chapter 8: Quadratic Equations Exercise – 8.7**

**Question: 1**

Find the consecutive numbers whose squares have the same sum of 85.

**Solution:**

Let the two consecutive two natural numbers be (x) and (x +1) respectively.

Given,

That the sum of their squares is 85.

Then, by hypothesis, we get, = x^{2 }+ (x + 1)^{2 }= 85

= x^{2 }+ x^{2 }+ 2x + 1 = 85

= 2x^{2 }+ 2x + 1 – 85 = 0

= 2x^{2 }+ 2x + – 84 = 0

= 2(x^{2 }+ x + -42) = 0

Now applying factorization method, we get, = x^{2 }+ 7x – 6x – 42 = 0

= x(x + 7) – 6(x + 7) = 0

= (x – 6)(x + 7) = 0

Either, x – 6 = 0 therefore, x = 6 x + 7 = 0 therefore x = -7

Hence the consecutive numbers whose sum of squares is 85 are 6 and -7 respectively.

**Question: 2**

Divide 29 into two parts so that the sum of the squares of the parts is 425.

**Solution:**

Let the two parts be (x) and (29 – x) respectively.

According to the question, the sum of the two parts is 425.

Then by hypothesis, = x^{2} + (29 – x)^{2} = 425

= x^{2 }+ x^{2 }+ 841 + -58x = 425

= 2x^{2 }– 58x + 841 – 425 = 0

= 2x^{2 }– 58x + 416 = 0

= x^{2 }– 29x + 208 = 0

Now, applying the factorization method = x^{2 }– 13x – 16x + 208 = 0

= x(x – 13) – 16(x – 13) = 0

= (x – 13)(x – 16) = 0

Either x – 13 = 0 therefore x = 13 Or, x – 16 = 0 therefore x = 16

The two parts whose sum of the squares is 425 are 13 and 16 respectively.

**Question: 3**

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2.find the sides of the squares.

**Solution:**

Given,

The sum of the sides of the squares are = x cm and (x + 4) cm respectively.

The sum of the areas = 656 cm^{2}

We know that, Area of the square = side * side

Area of the square = x(x + 4) cm^{2}

Given that the sum of the areas is 656 cm^{2}

Hence by hypothesis, = x(x + 4) + x(x + 4) = 656

= 2x(x + 4) = 656 = x^{2 }+ 4x = 328

Now by applying factorization method, = x^{2 }+ 20x – 16x – 328 = 0

= x(x + 20) – 16(x + 20) = 0

= (x + 20)(x –16) = 0

Either x + 20 = 0 therefore x = – 20 Or, x – 16 = 0 therefore x = 16

No negative value is considered as the value of the side of the square can never be negative.

Therefore, the side of the square is 16.

Therefore, x + 4 = 16 + 4 = 20 cm

Hence, the side of the square is 20 cm.

**Question: 4**

The sum of two numbers is 48 and their product is 432. Find the numbers.

**Solution:**

Given the sum of two numbers is 48.

Let the two numbers be x and 48 – x also the sum of their product is 432.

According to the question = x(48 – x) = 432

= 48x – x^{2 }= 432

= x^{2 }– 48x + 432 = 0

= x^{2 }– 36x – 12x + 432 = 0

= x(x – 36) – 12(x – 36) = 0

= (x – 36)(x – 12) = 0

Either x – 36 = 0 therefore x = 36 Or, x – 12 = 0 therefore x = 12

The two numbers are 12 and 36 respectively.

**Question: 5**

If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.

**Solution:**

Let the integer be x

Given that if an integer is added to its square, the sum is 90

= x + x^{2} = 90

= x^{2} + x – 90 = 0

= x^{2} + 10x – 9x – 90 = 0

= x(x + 10) – 9(x + 10) = 0

= (x + 10)(x – 9) = 0

Either x + 10 = 0

Therefore x = –10 Or, x – 9 = 0

Therefore x = 9

The values of the integer are 9 and -10 respectively.

**Question: 6**

Find the whole numbers which when decreased by 20 is equal to69 times the reciprocal of the numbers.

**Solution:**

Let the whole number be x cm

As it is decreased by

Now by applying factorization method,

x^{2} – 23x + 3x – 69 = 0

x(x – 23) + 3(x – 23) = 0

(x – 23)(x + 3) =0

Either, x = 23 Or, x = -3

As the whole numbers are always positive x = – 3 is not considered. The whole number is 23.

**Question: 7**

Find the consecutive natural numbers whose product is 20

**Solution:**

Let the two consecutive natural number be x and x + 1 respectively.

Given that the product of natural numbers is 20 = x(x + 1) = 20

= x^{2 }+ x – 20 = 0

= x^{2 }+ 5x – 4x – 20 = 0

= x(x + 5) – 4(x + 5) = 0

= (x + 5)(x – 4) = 0

Either x + 5 = 0

Therefore x = – 5

Considering the positive value of x. Or, x – 4 = 0

Therefore x = 4

The two consecutive natural numbers are 4 and 5 respectively.

**Question: 8**

The sum of the squares of two consecutive odd positive integers is 394. Find the two numbers?

**Solution:**

Let the consecutive odd positive integer are 2x – 1 and 2x + 1 respectively.

Given, that the sum of the squares is 394.

According to the question,

(2x – 1)^{2 }+ (2x + 1)^{2 }= 394

4x^{2 }+1 – 4x + 4x^{2 }+1 + 4x = 394

Now cancelling out the equal and opposite terms,

8x^{2 }+ 2 = 394

8x^{2} = 392

x^{2} = 49

x = 7 and – 7

Since the value of the edge of the square cannot be negative so considering only the positive value.

That is 7 Now,

2x – 1 = 14 -1 = 13

2x + 1 = 14 + 1 = 15

The consecutive odd positive numbers are 13 and 15 respectively.

**Question: 9**

The sum of two numbers is 8 and 15 times the sum of the reciprocal is also 8. Find the numbers.

**Solution:**

Let the numbers be x and 8 – x respectively.

Given that the sum of the numbers is 8 and 15 times the sum of their reciprocals.

According to the question,

= 120 = 8(8x – x^{2})

= 120 = 64x – 8x^{2}

= 8x^{2 }– 64x + 120 = 0

= 8(x^{2 }– 8x + 15) = 0

= x^{2 }– 8x + 15 = 0

= x^{2 }– 5x – 3x + 15 = 0

= x(x – 5) – 3(x – 5) = 0

= (x – 5)(x – 3) = 0

Either x – 5 = 0 therefore x = 5 Or, x – 3 = 0 therefore x = 3

The two numbers are 5 and 3 respectively.

**Question: 10**

The sum of a number and its positive square root is 6/25. Find the numbers.

**Solution:**

Let the number be x

By the hypothesis, we have

Let us assume that x = y^{2},

we get y + y^{2} = 6/25

= 25y^{2 }+ 25y – 6 = 0

The value of y can be determined by:

Where a = 25, b = 25, c = – 6

y = 1/5 and y = −11/10 = x = y^{2 }= 15/2 =1/25

The number x is 1/25

**Question: 11**

There are three consecutive integers such that the square of the first increased by the product of the other two integers gives 154. What are the integers?

**Solution:**

Let the three consecutive numbers be x, x + 1, x + 2 respectively.

= x^{2 }+ (x + 1)(x + 2) = 154

= x^{2 }+ x^{2 }+ 3x + 2 = 154

= 3x^{2 }+ 3x – 152 = 0

The value of x can be obtained by the formula

Here a = 3, b = 3, c = 152

x = 8 and x = – 19/2

Considering the value of x If x = 8 x + 1 = 9 x + 2 = 10

The three consecutive numbers are 8, 9, 10 respectively.

**Question: 12**

The product of two successive integral multiples of 5 is 300. Determine the multiples.

**Solution:**

Given that the product of two successive integral multiples of 5 is 300

Let the integers be 5x and 5(x+1)

According to the question,

5x[5(x + 1)] = 300

= 25x(x + 1) = 300

= x^{2 }+ x = 12

= x^{2 }+ x – 12 = 0

= x^{2 }+ 4x – 3x – 12 = 0

= x(x + 4) – 3(x + 4) = 0

= (x + 4)(x – 3) = 0

Either x + 4 = 0

Therefore x = -4 Or, x – 3 = 0

Therefore x = 3 x = – 4

5x = – 20

5(x + 1) = -15

x = 3

5x = 15

5(x + 1) = 20

The two successive integral multiples are 15, 20 and -15 and -20 respectively.

**Question: 13**

The sum of the squares of two numbers is 233 and one of the numbers is 3 less than the other number. Find the numbers.

**Solution:**

Let the number is x Then the other number is 2x – 3

According to the question: x^{2 }+ (2x – 3)^{2} = 233

= x^{2 }+ 4x^{2 }+ 9 – 12x = 233

= 5x^{2 }– 12x – 224 = 0

The value of x can be obtained by

Here a = 5, b = – 12, c = – 224

x = 8 and x = −28/5

Considering the value of x = 8 2x – 3 = 15 The two numbers are 8 and 15 respectively.

**Question: 14**

The difference of two number is 4. If the difference of the reciprocal is 421. Find the numbers.

**Solution:**

Let the two numbers be x and x – 4 respectively.

Given, that the difference of two numbers is 4.

By the given hypothesis we have,

= 84 = 4x(x – 4)

= x^{2 }– 4x – 21 = 0

Applying factorization theorem, = x^{2 }– 7x + 3x – 21 = 0

= (x – 7)(x + 3) = 0

Either x – 7 = 0 therefore x = 7 Or, x + 3 = 0 therefore x = -3 Hence the required numbers are – 3 and 7 respectively.

**Question: 15**

Let us find two natural numbers which differ by 3 and whose squares have the sum 117.

**Solution:**

Let the numbers be x and x – 3

According to the question x^{2 }+ (x – 3)^{2 }= 117

= x^{2 }+ x^{2 }+ 9 – 6x – 117 = 0

= 2x^{2 }– 6x – 108 = 0

= x^{2 }– 3x – 54 = 0

= x^{2 }– 9x + 6x – 54 = 0

= x(x – 9) + 6(x – 9) = 0

= (x – 9)(x + 6) = 0

Either x – 9 = 0 therefore x = 9 Or, x + 6 = 0 therefore x = – 6

Considering the positive value of x that is 9 x = 9 x – 3 = 6 The two numbers are 6 and 9 respectively.

**Question: 16**

The sum of the squares of these consecutive natural numbers is 149. Find the numbers.

**Solution:**

Let the numbers be x, x + 1, and x + 2 respectively.

According to given hypothesis x^{2 }+ (x +1 )^{2 }+ (x + 2)^{2} = 149

x^{2 }+ x^{2} + x^{2 }+ 1 + 2x + 4 + 4x = 149

3x^{2} + 6x – 144 = 0

x^{2 }+ 2x – 48 = 0

Now applying factorization method, x^{2 }+ 8x – 6x – 48 = 0

x(x + 8) – 6(x + 8) = 0

(x + 8)(x – 6) = 0

Either x + 8 = 0 therefore x = – 8 Or, x – 6 = 0 therefore x = 6

Considering only the positive value of x that is 6 and discarding the negative value. x = 6 x + 1 = 7 x + 2 = 8

The three consecutive numbers are 6, 7, and 8 respectively.

**Question: 17**

Sum of two numbers is 16. The sum of their reciprocal is 1/3.find the numbers.

**Solution:**

Given that the sum of the two natural numbers is 16 Let the two natural numbers be x and 16-x respectively

According to the question

= 16x – x^{2} = 48

= -16x + x^{2 }+ 48 = 0

= +x^{2 }– 16x + 48 = 0

= +x^{2 }– 12x – 4x + 48 = 0

= x(x – 12) – 4(x – 12) = 0

= (x – 12)(x – 4) = 0

Either x – 12 = 0 therefore x = 12 Or, x – 4 = 0 therefore x = 4

The two numbers are 4 and 12 respectively.

**Question: 18**

Determine the two consecutive multiples of 3 whose product is 270.

**Solution:**

Let the consecutive multiples of 3 are 3x and 3x + 3

According to the question 3x(3x + 3) = 270

= x(3x + 3) = 90

= 3x^{2 }+ 3x = 90

= 3x^{2 }+ 3x – 90 = 0

= x^{2 }+ x – 30 = 0

= x^{2 }+ 6x – 5x – 30 = 0

= x(x + 6) – 5(x + 6) = 0

= (x + 6)(x – 5) = 0

Either x + 6 = 0 therefore x = – 6 Or, x – 5 = 0 therefore x = 5

Considering the positive value of x

x = 5 , 3x = 15, 3x + 3 = 18

The two consecutive multiples of 3 are 15 and 18 respectively.

**Question: 19**

The sum of a number and its reciprocal is 17/4. Find the numbers.

**Solution:**

Let the number be x According to the question

= 4(x^{2}+1) = 17x

= 4x^{2 }+ 4 – 17x = 0

= 4x^{2 }+ 4 – 16x – x = 0

= 4x(x – 4) – 1(x – 4) = 0

= (4x – 1)(x – 4) = 0

Either x – 4 = 0 therefore x = 4 Or, 4x – 1 = 0

therefore x = 1/4 The value of x is 4

**Question: 20**

A two digit is such that the products of its digits is 8when 18 is subtracted from the number, the digits interchange their places. Find the number?

**Solution:**

Let the digits be x and x – 2 respectively.

The product of the digits is 8 According to the question x(x – 2) = 8

= x^{2 }– 2x – 8 = 0

= x^{2 }– 4x + 2x – 8 = 0

= x(x – 4) + 2(x – 4) = 0

Either x – 4 = 0 therefore x = 4 Or, x + 2 = 0 therefore x = -2

Considering the positive value of x = 4 x – 2 = 2

The two digit number is 42.

**Question: 21**

A two digit number is such that the product of the digits is 12, when 36 is added to the number, the digits interchange their places .find the number.

**Solution:**

Let the tens digit be x Then, the unit digit = 12/x

Therefore the number = 10x + 12/x

And, the number obtained by interchanging the digits

= 9(x^{2 }+ 4x – 12) = 0

= (x^{2 }+ 4x – 12) = 0

= x^{2 }+ 6x – 2x – 12 = 0

= x(x + 6) – 2(x + 6) = 0

= (x – 2)(x + 6) = 0

Either x – 2 = 0 therefore x = 2 Or, x + 6 = 0 therefore x = – 6

Since a digit can never be negative. So x = 2

The number is 26.

**Question: 22**

A two digit number is such that the product of the digits is 16 when 54 is subtracted from the number, the digits are interchanged. Find the number.

**Solution:**

Let the two digits be: Tens digit be x Units digit be 16/x

Numbers = 10x + 16/x …. (i)

Number obtained by interchanging

= 54 = 10x^{2 }+ 16 – 160 + x^{2} = 54

= 9x^{2 }– 54x – 144 = 0

= x^{2 }– 6x – 16 = 0

= x^{2 }– 8x + 2x – 16 = 0

= x(x – 8) + 2(x – 8) = 0

=(x – 8)(x + 2) = 0

Either x – 8 = 0 therefore x = 8 Or, x + 2 = 0 therefore x = – 2

A digit can never be negative so x = 8

Hence by putting the value of x in the above equation (i) the number is 82.

**Question: 23**

Two numbers differ by 3 and their product is 504. Find the numbers.

**Solution:**

Let the numbers be x and x – 3 respectively.

According to the question = x(x – 3) = 504

= x^{2 }– 3x – 504 = 0

= x^{2 }– 24x + 21x – 504 = 0

= x(x – 24) + 21 (x – 24) = 0

= (x – 24)(x + 21) = 0

Either x – 24 = 0 therefore x = 24 Or, x + 21 = 0, therefore x = -21 x = 24 and x = -21 x – 3 = 21 and x – 3 = – 24

The two numbers are 21 and 24 and – 21 and – 24 respectively.

**Question: 24**

Two numbers differ by 4 and their product is 192. Find the numbers.

**Solution:**

Let the two numbers be x and x-4 respectively

Given that the product of the numbers is 192

According to the question = x(x – 4) = 192 = x^{2 }– 4x – 192 = 0

= x^{2 }– 16x + 12x – 192 = 0

= x(x – 16) + 12(x – 16) = 0

= (x – 16) (x + 12) = 0

Either x – 16 = 0 therefore x = 16 Or, x + 12 = 0 therefore x = -12

Considering only the positive value of x

x = 16

x – 4 = 12

The two numbers are 12 and 16 respectively.

**Question: 25**

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

**Solution:**

Let the digit in the tens and the units place be x and y respectively.

Then it is represented by 10x + y

According to the question,

10x + y = 4(sum of the digits) and 2xy

10x + y = 4(x + y) and 10x + y = 2xy

10x + y = 4x + 4y and 10x + y = 2xy

6x – 3y = 0 and 10x + y – 2xy = 0

y = 2x and 10x + 2x – 2x(2x) = 0

12x = 4x^{2}

4x(x- 3) = 0

Either 4x = 0 therefore x = 0 Or, x –3 = 0 therefore x = 3 We have y = 2x When x = 3, y= 6

**Question: 26**

The sum of the squares of two positive integers is 208. If the square of the large number is 18 times the smaller. Find the numbers.

**Solution:**

Let the smaller number be x Then, square of the large number be = 18x

Also, square of the smaller number be = x^{2}

It is given that the sum of the square of the integer is 208.

Therefore,

= x^{2} + 18x = 208

= x^{2 }+ 18x – 208 = 0

Applying factorization theorem,

= x^{2} + 26x – 8x – 208 = 0

= x(x + 26) – 8(x + 26) = 0

= (x + 26)(x – 8) = 0

Either x + 26 = 0 therefore x = -26 Or, x – 8 = 0 therefore x = 8

Considering the positive number, therefore x = 8.

Square of the largest number = 18x = 18 × 8 = 144

Largest Number = √144 = 12

Hence the numbers are 8 and 12 respectively.

**Question: 27**

The sum of two numbers is 18. The sum of their reciprocal is 1/4 .find the numbers.

**Solution:**

Let the numbers be x and (18 – x) respectively.

According to the given hypothesis,

Applying factorization theorem, we get,

= x^{2} – 6x – 12x + 72 = 0

= x(x – 6) – 12(x – 6) = 0

= (x – 6)(x – 12) = 0

Either, x = 6 Or, x = 12

The two numbers are 6 and 12 respectively.

**Question: 28**

The sum of two numbers a and b is 15 and the sum of their reciprocals 1/a and 1/b is 3/10. Find the numbers a and b.

**Solution:**

Let us assume a number x such that

= 3x^{2 }– 45x + 150 = 0

= x^{2} – 15

x + 50 = 0

Applying factorization theorem, = x^{2}– 10x – 5x + 50 = 0

= x(x – 10) – 5(x – 10) = 0

= (x – 10)(x – 5) = 0

Either, x – 10 = 0 therefore x = 10 Or, x – 5 = 0 therefore x = 5

Case (i) If x = a, a = 5 and b = 15 – x , b= 10

Case (ii) If x = 15 – a = 15 – 10 = 5, x = a = 10, b = 15 – 10 = 5

Hence when a = 5, b = 10 a = 10, b = 5

**Question: 29**

The sum of two numbers is 9. The sum of their reciprocal is 1/2. Find the numbers.

**Solution:**

Given that the sum of the two numbers is 9

Let the two number be x and 9 – x respectively

According to the question

= 9x – x^{2 }= 18

= x^{2 }– 9x + 18 = 0

= x^{2 }– 6x – 3x + 18 = 0

= x(x – 6) – 3(x – 6) = 0

= (x – 6)(x – 3) = 0

Either x – 6 = 0 therefore x = 6 Or x – 3 = 0 therefore x = 3

The two numbers are 3 and 6 respectively.

**Question: 30**

Three consecutive positive integers are such that the sum of the squares of the first and the product of the other two is 46. Find the integers.

**Solution**

Let the consecutive positive integers be x, x + 1, x + 2 respectively

According to the question X^{2 }+ (x + 1)(x + 2) = 46

= x^{2 }+ x^{2 }+ 3x + 2 = 46

= 2x^{2 }+ 3x + 2 = 46

= 2x^{2 }+ 3x + 2 – 46 = 0

= 2x^{2 }– 8x + 11x + – 44 = 0

= 2x(x – 4) + 11(x – 4) = 0

= (x – 4)(2x + 11) = 0

Either x – 4 = 0 therefore x = 4 Or, 2x + 11 = 0 therefore x = −11/2

Considering the positive value of x that is x = 4

The three consecutive numbers are 4, 5 and 6 respectively

**Question: 31**

The difference of squares of two numbers is 88. If the large number is 5 less than the twice of the smaller, then find the two numbers.

**Solution:**

Let the smaller number be x and larger number is 2x – 5

It is given that the difference of the squares of the number is 88

According to the question (2x – 5)^{2 }– x^{2 }= 88

= 4x^{2 }+ 25 – 20x – x^{2} = 88

= 3x^{2 }– 20x – 63 = 0

= 3x^{2 }– 27x + 7x – 63 = 0

= 3x(x – 9) + 7(x – 9) = 0

= (x – 9)(3x + 7) = 0

Either x – 9 = 0 therefore x = 9 Or, 3x + 7 = 0 therefore x = −7/3

Since a digit can never be negative so x = 9

Hence the number is 2x – 5 = 13

The required numbers are 9 and 13 respectively

**Question: 32**

The difference of squares of two numbers is180. The square of the smaller number is 8 times the larger number. Find the two numbers

**Solution:**

Let the number be x

According to the question X^{2 }– 8x = 180

X^{2 }– 8x – 180 = 0

= X^{2 }+ 10x – 18x – 180 = 0

= x(x + 10) – 18(x – 10) = 0

= (x – 18)(x + 10) = 0

Either x – 18 = 0 therefore x = 18 Or, x + 10 = 0 therefore x = -10 Case (i) X = 18 8x = 144

Larger number = √144 = 12

Case (ii) X = -10

Square of the larger number 8x= -80

Here in this case no perfect square exist

Hence the numbers are 18 and 12 respectively.

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