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# 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+ (x + 1)= 85

= x2 + x2 + 2x + 1 = 85

= 2x2 + 2x + 1 - 85 = 0

= 2x2 + 2x + - 84 = 0

= 2(x2 + x + -42) = 0

Now applying factorization method, we get, = x2 + 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, = x2 + (29 - x)2 = 425

= x2 + x2 + 841 + -58x = 425

= 2x2 - 58x + 841 - 425 = 0

= 2x2 - 58x + 416 = 0

= x2 - 29x + 208 = 0

Now, applying the factorization method = x- 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 cm2

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

Area of the square = x(x + 4) cm2

Given that the sum of the areas is 656 cm2

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

= 2x(x + 4) = 656 = x+ 4x = 328

Now by applying factorization method, = x+ 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 - x2 = 432

= x2 - 48x + 432 = 0

= x2 - 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 + x2 = 90

= x2 + x - 90 = 0

= x2 + 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,

x2 - 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

= x2 + x - 20 = 0

= x2 + 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)= 394

4x+1 - 4x + 4x2 +1 + 4x = 394

Now cancelling out the equal and opposite terms,

8x2 + 2 = 394

8x2 = 392

x2 = 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 - x2)

= 120 = 64x - 8x2

= 8x2 - 64x + 120 = 0

= 8(x2 - 8x + 15) = 0

= x2 - 8x + 15 = 0

= x2 - 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 = y2,

we get y + y2 = 6/25

= 25y2 + 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 = y2 = 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.

= x2 + (x + 1)(x + 2) = 154

= x2 + x2 + 3x + 2 = 154

= 3x2 + 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

= x2 + x = 12

= x2 + x - 12 = 0

= x2 + 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: x2 + (2x - 3)2 = 233

= x2 + 4x2 + 9 - 12x = 233

= 5x2 - 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)

= x2 - 4x - 21 = 0

Applying factorization theorem, = x- 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 x2 + (x - 3)2 = 117

= x2 + x2 + 9 - 6x - 117 = 0

= 2x2 - 6x - 108 = 0

= x2 - 3x - 54 = 0

= x2 - 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 x2 + (x +1 )2 + (x + 2)2 = 149

x2 + x2 + x+ 1 + 2x + 4 + 4x = 149

3x2 + 6x - 144 = 0

x2 + 2x - 48 = 0

Now applying factorization method, x+ 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 - x2 = 48

= -16x + x2 + 48 = 0

= +x2 - 16x + 48 = 0

= +x2 - 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

= 3x2 + 3x = 90

= 3x2 + 3x - 90 = 0

= x2 + x - 30 = 0

= x2 + 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(x2+1) = 17x

= 4x2 + 4 - 17x = 0

= 4x2 + 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

= x2 - 2x - 8 = 0

= x2 - 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(x2 + 4x - 12) = 0

= (x2 + 4x - 12) = 0

= x2 + 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 = 10x2 + 16 - 160 + x2 = 54

= 9x2 - 54x - 144 = 0

= x2 - 6x - 16 = 0

= x2 - 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

= x2 - 3x - 504 = 0

= x2 - 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 = x2 - 4x - 192 = 0

= x2 - 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 = 4x2

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 = x2

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

Therefore,

= x2 + 18x = 208

= x2 + 18x - 208 = 0

Applying factorization theorem,

= x2 + 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,

= x2 - 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 = 3x2 - 45x + 150 = 0

= x2 - 15

x + 50 = 0

Applying factorization theorem, = x2- 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 - x2 = 18

= x2 - 9x + 18 = 0

= x2 - 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 X2 + (x + 1)(x + 2) = 46

= x2 + x2 + 3x + 2 = 46

= 2x2 + 3x + 2 = 46

= 2x2 + 3x + 2 - 46 = 0

= 2x2 - 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 - x2 = 88

= 4x2 + 25 - 20x - x2 = 88

= 3x2 - 20x - 63 = 0

= 3x2 - 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 X2 - 8x = 180

X2 - 8x - 180 = 0

= X2 + 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.