## Online Education for RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

These Solutions are part of **Online Education** RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

**Question 1.**

Find two consecutive numbers whose squares have the sum 85. **(C.B.S.E. 2000)**

**Solution:**

Let first number = x

Then second number = x + 1

According to the condition

x² + (x + 1)^{2} = 85

=> x² + x² + 2x + 1 = 85

=> 2x² + 2x + 1 – 85 = 0

=> 2x² + 2x – 84 = 0

=> x² + x – 42 = 0

=> x² + 7x – 6x – 42 = 0

=> x (x + 7) – 6 (x + 7) = 0

=> (x + 7) (x – 6) = 0

Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6

(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6

(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7

Hence numbers are -7, -6 or 6, 7

**Question 2.**

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

**Solution:**

Total = 29

Let first part = x

Then second part = 29 – x

According to the condition

x² + (29 – x)^{2} = 425

=> x² + 841 + x² – 58x = 425

=> 2x² – 58x + 841 – 425 = 0

=> 2x² – 58x + 416 = 0

=> x² – 29x + 208 = 0 (Dividing by 2)

=> x² – 13x – 16x + 208 = 0

=> x(x – 13) – 16 (x – 13) = 0

=> (x – 13) (x – 16) = 0

Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16

(i) If x = 13, then First part =13 and second part = 29 – 13 = 16

(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13

Parts are 13, 16

**Question 3.**

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm^{2}. Find the sides of the squares. **(C.B.S.E. 1997)**

**Solution:**

Side of the first square = x cm

Its area = (side)^{2} = x² cm^{2}

Side of the second square = (x + 4) cm

Its area = (x + 4)^{2} cm^{2}

According to the condition,

x² + (x + 4)^{2} = 656

=> x² + x² + 8x + 16 = 656

=> 2x² + 8x + 16 – 656 = 0

=> 2x² + 8x – 640 = 0

=> x² + 4x – 320 = 0 (Dividing by 2)

=> x² + 20x – 16x – 320 = 0

=> x (x + 20) – 16 (x + 20) = 0

=> (x + 20) (x – 16) = 0

Either x + 20 = 0, then x = -20 Which is not possible being negative

or x – 16 = 0, then x = 16

Side of the first square = 16 cm

and side of the second square = 16 + 4 = 20 cm

**Question 4.**

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

**Solution:**

Sum of two numbers = 48

Let first number = x

The second number = 48 – x

According to the condition,

x (48 – x) = 432

=> 48x – x² = 432

=> – x² + 48x – 432 = 0

=> x² – 48x + 432 = 0

=> x² – 12x – 36x + 432 = 0

=> x (x – 12) – 36 (x – 12) = 0

=> (x – 12) (x – 36) = 0

Either x – 12 = 0, then x = 12 or x – 36 = 0, then x = 36

(i) If x = 12, then First number = 12 and second number = 48 – 12 = 36

(ii) If x = 36, then First number = 36 and second number = 48 – 36 = 12

Numbers are 12, 36

**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 given integer be = x

According to the condition

x² + x = 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, then x = -10 or x – 9 = 0, then x = 9.

The integer will be -10 or 9

**Question 6.**

Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

**Solution:**

Let the given whole number = x

Then its reciprocal = \(\frac { 1 }{ x }\)

According to the condition,

x – 20 = 69 x \(\frac { 1 }{ x }\)

=> x – 20 = \(\frac { 69 }{ x }\)

=> x² – 20x = 69

=> x² – 20x – 69 = 0

=> x² – 23x + 3x – 69 = 0

=> x (x – 23) + 3 (x – 23) = 0

=> (x – 23) (x + 3) = 0

Either x – 23 = 0, then x = 23

or x + 3 = 0, then x = -3, but it is not a whole number

Required whole number = 23

**Question 7.**

Find two consecutive natural numbers whose product is 20.

**Solution:**

Let first natural number = x

Then second number = x + 1

According to the condition,

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, then x = -5 which is not a natural number

or x – 4 = 0, then x = 4

First natural number = 4 and second number = 4 + 1=5

**Question 8.**

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

**Solution:**

Let first odd number = 2x + 1

Then second odd number = 2x + 3

According to the condition

(2x + 1)^{2} + (2x + 3)^{2} = 394

=> 4x² + 4x + 1 + 4x² + 12x + 9 = 394

=> 8x² + 16x + 10 = 394

=> 8x² + 16x + 10 – 394 = 0

=> 8x² + 16x – 384 = 0

=> x² + 2x – 48 = 0 (Dividingby8)

=> x² + 8x – 6x – 48 = 0

=> x(x + 8) – 6(x + 8) = 0

=> (x + 8) (x – 6) = 0

Either x + 8 = 0, then x = 8 but it is not possible as it is negative

or x – 6 = 0, then x = 6

First odd number = 2x + 1 = 2 x 6 + 1 = 13

and second odd number = 13 + 2 = 15

**Question 9.**

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

**Solution:**

Sum of two numbers = 8

Let first number = x

Then second number = 8 – x

According to the condition,

(ii) If x = 5, then First number = 5 and second number = 8 – 5 = 3

Numbers are 3, 5

**Question 10.**

The sum of a number and its positive square root is \(\frac { 6 }{ 25 }\). Find the number.

**Solution:**

**Question 11.**

The sum of a number and its square is \(\frac { 63 }{ 4 }\) , find the numbers.

**Solution:**

**Question 12.**

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

**Solution:**

Let first integer = x

Then second integer = x + 1

and third integer = x + 2

According to the condition,

x² + (x + 1) (x + 2) = 154

=> x² + x² + 3x + 2 = 154

=> 2x² + 3x + 2 – 154 = 0

=> 2x² – 16x + 19x – 152 = 0

=> 2x(x – 8) + 19 (x – 8) = 0

=> (x – 8) (2x + 19) = 0

Either x – 8 = 0, then x = 8

or 2x + 19 = 0, then 2x = -19 => x = \(\frac { -19 }{ 2 }\) But it is not an integer

First number = 8

Second number = 8 + 1=9

and third number = 8 + 2 = 10

**Question 13.**

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

**Solution:**

Let first multiplie of 5 = 5x

Then second multiple = 5x + 5

According to the condition,

5x (5x + 5) = 300

=> 25 x² + 25x – 300 = 0

=> x² + x – 12 = 0 (Dividing by 25)

=> x² + 4x – 3x – 12 = 0

=> x (x + 4) – 3 (x + 4) = 0

=> (x – 4) (x – 3) = 0

Either x + 4 = 0, then x = -4

or x – 3 = 0, then x = 3

(i) When x = -4, then

Required multiples of 5 will be

5 (-4) = -20, -20 + 5 = -15

or when x = 3, then

Required multiples will be

5 x 3 = 15, 15 + 5 = 20

Required number are -20, -15 or 15, 20

**Question 14.**

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

**Solution:**

Let first number = x

Then second number = 2x – 3

According to the condition,

x² + (2x – 3)^{2} = 233

=> x² + 4x² – 12x + 9 = 233

=> 5x² – 12x + 9 – 233 = 0

=> 5x² – 12x – 224 = 0

=> 5x² – 40x + 28x – 224 = 0

=> 5x (x – 8) + 28 (x – 8) = 0

=> (x – 8) (5x + 28) = 0

Either x – 8 = 0, then x = 8

or 5x + 28 = 0, then 5x = -28 => x = \(\frac { -28 }{ 5 }\) But it is not possible

x = 8

First number = 8

Second number = 2x – 3 = 2 x 8 – 3 = 16 – 3 = 13

Number are 8, 13

**Question 15.**

Find two consecutive even integers whose squares have the sum 340.

**Solution:**

Let first even integer = x

The second even integer = x + 2

According to the condition,

x² + (x + 2)^{2} = 340

x² + x² + 4x + 4 = 340

=> 2x² + 4x + 4 – 340 = 0

=> 2x² + 4x – 336 = 0

=> x² + 2x – 168 = 0

=> x² + 14x – 12x – 168 = 0

=> x (x + 14) – 12 (x + 14) = 0

=> (x + 14) (x – 12) = 0

Either x + 14 = 0, then x = -14

or x – 12 = 0, the x = 12

(i) If x = -14, then

First number = -14

and second number = -14 + 2 = -12

(ii) If x = 12, then

First number =12

and second number =12 + 2 = 14

Hence even numbers are 12, 14 or -14, -12

**Question 16.**

The difference of two numbers is 4. If the difference of their reciprocals is \(\frac { 4 }{ 21 }\), find the numbers. **(C.B.S.E. 2008)**

**Solution:**

Let first number = x

Then second number = x – 4

According to the condition,

Either x – 7 = 0, then x = 7

or x + 3 = 0, then x = -3

(i) If x = 7, then

First number = 7

and second number = 7 – 4 = 3

(ii) If x = -3, then

First number = -3

and second number = -3 – 4 = -7

Number are 7, 3 or -3, -7

**Question 17.**

Find two natural numbers which differ by 3 and whose squared have the sum 117.

**Solution:**

Let first number = x

Then second number = x – 3

According to the condition,

x² + (x – 3)^{2} = 117

=> x² + x² – 6x + 9 = 117

=> 2x² – 6x + 9 – 117 = 0

=> 2x² – 6x – 108 = 0

=> x² – 3x – 54 = 0 (Dividing by 2)

=> x² – 9x + 6x – 54 = 0

=> x (x – 9) + 6 (x – 9) = 0

=> (x- 9) (x + 6) = 0

Either x – 9 = 0, then x = 9

or x + 6 = 0, then x = -6 which is not a natural number

First natural number = 9

and second number = 9 – 3 = 6

**Question 18.**

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

**Solution:**

Let first number = x

Then second number = x + 1

and third number = x + 2

According to the condition,

x² + (x + 1)^{2} + (x + 2)^{2} = 149

=> x² + x² + 2x + 1 + x2 + 4x + 4 = 149

=> 3x² + 6x + 5 – 149 = 0

=> 3x² + 6x – 144 = 0

=> x² + 2x – 48 = 0 (Dividing by 3)

=> x² + 8x – 6x – 48 = 0

=> x (x + 8) – 6 (x + 8) = 0

=> (x + 8) (x – 6) = 0 .

Either x + 8 = 0, then x = -8, But it is not a natural number

or x – 6 = 0, then x = 6

Numbers are 6, 6 + 1 = 7, 6 + 2 = 8 or 6, 7, 8

**Question 19.**

The sum of two numbers is 16. The sum of their reciprocals is \(\frac { 1 }{ 3 }\). Find the numbers. **(C.B.S.E. 2005)**

**Solution:**

Sum of two numbers = 16

Let first number = x

Then second number = 16 – x

According to the condition,

Either x – 12 = 0, then x = 12

or x – 4 = 0, then x = 4

(i) If x = 12, then

First number = 12

and second number = 16 – 12 = 4

(ii) If x = 4, then First number = 4

and second number = 16 – 4 = 12

Hence numbers are 4, 12

**Question 20.**

Determine two consecutive multiples of 3 whose product is 270.

**Solution:**

Let first multiple of 3 = 3x

Then second multiple of 3 = 3x + 3

According to the condition,

3x (3x + 3) = 270

=> 9x² + 9x – 270 = 0

=> x² + x – 30 = 0 (Dividing by 9)

=> x² + 6x – 5x – 30 = 0

=> x (x + 6) – 5 (x + 6) = 0

=> (x + 6) (x – 5) = 0

Either x + 6 = 0, then x = -6

or x – 5 = 0, then x = 5

(i) When x = -6, then

First number = 3x = 3 x (-6) = -18 and second number = -18 + 3 = -15

(ii) If x = 5, then

First number = 3x = 3 x 5 = 15 and second number =15 + 3 = 18

Hence numbers are 15, 18 or -18, -15

**Question 21.**

The sum of a number and its reciprocal is \(\frac { 17 }{ 4 }\) , Find the number.

**Solution:**

**Question 22.**

A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number.

**Solution:**

Product of two digits = 8

Let units digit = x

Then tens digit = \(\frac { 8 }{ x }\)

Number = x + 10 x \(\frac { 8 }{ x }\) = x + \(\frac { 80 }{ x }\)

**Question 23.**

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. Determine the number.

**Solution:**

**Question 24.**

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:**

**Question 25.**

Two numbers differ by 3 and their product is 504. Find the numbers. **(C.B.S.E. 2002C)**

**Solution:**

Difference of two numbers = 3

Let first number = x

Then second number = x – 3

According to the condition,

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, then x = 24

or x + 21 = 0, then x =-21

(i) If x = 24, then

First number = 24

and second number = 24 – 3 = 21

(ii) If x =-21, then

First number = -21

and second number = -21 – 3 = -24

Hence numbers are 24, 21 or -21, -24

**Question 26.**

Two numbers differ by 4 and their product is 192. Find the numbers. **(C.B.S.E. 2000C)**

**Solution:**

Let first number = x

Then second number = x – 4

According to the condition,

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, then x = 16

or x + 12 = 0, then x = -12

(i) If x = 16, then

First number = 16

and second number = 16 – 4 = 12

(ii) If x = -12, then

First number = -12

and second number = -12 – 4 = -16

Hence numbers are 16, 12 or -12, -16

**Question 27.**

A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number. **(C.B.S.E. 1999C)**

**Solution:**

Let units digit of the number = x

and tens digit = y

Number = x + 10y

According to the given conditions,

x + 10y = 4 (x + y)

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

=> 10y – 4y = 4x – x

=> 3x = 6y

=>x = 2y …(i)

and x + 10y = 2xy ….(ii)

Substituting the value of x in (i)

2y + 10y = 2 x 2y x y

=> 12y = 4y2

=> 4y2 – 12y = 0

=> y2 – 3y = 0

=> y (y – 3) = o

Either y = 0, but it is not possible because y is tens digit number

or y – 3 = 0, then y = 3

x = 2y = 2 x 3 = 6

and number = x + 10y = 6 + 10 x 3 = 6 + 30 = 36

**Question 28.**

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers. **[CBSE 2014]**

**Solution:**

Let first large number = x

and smaller number = y

According to the condition,

x^{2} – y^{2 }= 180 …(i)

and y^{2} = 8x

From (i) and (ii),

x^{2} – 8x – 180 = 0

=> x^{2} – 18x + 10x – 180 = 0

=> x (x – 18)+ 10 (x – 18) = 0

=> (x – 18) (x + 10) = 0

Either x – 18 = 0, then x = 18

or x + 10 = 0, then x = -10 But it is not possible being negative

x = 18

First number =18

Then second number y^{2} = 8x

y^{2} = 8 x 18 = 144 = (12)^{2}

=> y = 12

Numbers are 18, 12

**Question 29.**

The sum of two numbers is 18. The sum of their reciprocals is \(\frac { 1 }{ 4 }\). Find the numbers. **(C.B.S.E. 2005)**

**Solution:**

Sum of two numbers = 18

Let one number = x

Then second number = 18 – x

According to the condition,

=> x^{2} – 12x – 6x + 72 = 0

=> x (x – 12) – 6 (x – 12) = 0

=> (x – 12) (x – 6) = 0

Either x – 12 = 0, then x = 12

or x – 6 = 0, then x = 6

(i) If x = 12, then

First number = 12

Second number =18 – 12 = 6

(ii) If x = 6, then

First number = 6

Then second number = 18 – 6 = 12

Numbers are 6, 12

**Question 30.**

The sum of two numbers a and b is 15, and the sum of their reciprocals \(\frac { 1 }{ a }\) and \(\frac { 1 }{ b }\) is \(\frac { 3 }{ 10 }\). Find the numbers a and b. **(C.B.S.E. 2005)**

**Solution:**

or b – 5 = 0, then b = 5

(i) a = 15 – 10 = 5

(ii) or a = 15 – 5 = 10

Numbers are 5, 10 or 10, 5

**Question 31.**

The sum of two numbers is 9. The sum of their reciprocals is \(\frac { 1 }{ 2 }\). Find the numbers. **[CBSE 2012]**

**Solution:**

Sum of two numbers = 9

Let first number = x

Then second number = 9 – x

According to the condition,

By cross multiplication

18 = 9x – x^{2}

=> 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, then x = 6

or x – 3 = 0, then x = 3

Numbers are 6 and (9 – 6) = 3, or 3 and (9 – 3) = 6

Numbers are 3, 6

**Question 32.**

Three consecutive positive integers are such that the sum of the square of the’ first and the product of other two is 46, find the integers. **[CBSE 2010]**

**Solution:**

Let first number = x

Then second number = x + 1

and third number = x + 2

According co the condition,

(x)^{2 }+ (x+ 1) (x + 2) = 46

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

=> 2x^{2} + 3x + 2 – 46 = 0

=> 2x^{2} + 3x – 44 = 0

=> 2x^{2} + 11x – 8x – 44 = 0

=> x (2x + 11) – 4 (2x + 11) = 0

=> (2x + 11) (x – 4) = 0

Either 2x + 11 = 0, then x = \(\frac { -11 }{ 2 }\) which is not possible being fraction

or x – 4 = 0, then x = 4

Numbers are 4, 5, 6

**Question 33.**

The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.** [CBSE 2010]**

**Solution:**

Let smaller number = x

Then larger number = 2x – 5

According to the condition,

(2x – 5)^{2} – x^{2} = 88

=> 4x^{2} – 20x + 25 – x^{2} – 88 = 0

=> 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, then x = 9

or 3x + 7 = 0, then x = \(\frac { -7 }{ 3 }\) which is not possible

Smaller number = 9

and greater number = 2x – 5 = 2 x 9 – 5 = 18 – 5 = 13

Hence numbers are 13, 9

**Question 34.**

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find two numbers.** [NCERT]**

**Solution:**

**Question 35.**

Find two consecutive odd positive integers, sum of whose squares is 970.

**Solution:**

Let two consecutive positive integers be x and x + 2

A.T.Q.,

(x)^{2} + (x + 2)^{2} = 970

=> x^{2} + x^{2} + 4x + 4 – 970 = 0

=> 2x^{2} + 4x – 966 = 0

=> x^{2} + 2x – 483 = 0 (Dividing by 2)

=> x^{2} + 23x – 21x – 483 = 0

=> x (x + 23) – 21 (x + 23) = 0

=> (x – 21) (x + 23) = 0

Either x – 21 = 0 or x + 23 = 0

x = 21 or x = – 23 (rejected being -ve)

As integers should be +Ve

x = 21 and x + 2 = 21 + 2 = 23

Hence integers are 21, 23

**Question 36.**

The difference of two natural numbers is 3 and the difference of their reciprocals is \(\frac { 3 }{ 28 }\). Find the numbers.

**Solution:**

y(y + 7) – 4(y + 7) = 0

(y – 4) (y + 7) = 0

y – 4 = 0 or y + 7 = 0

y = 4 or y = -7 (rejected being natural no.)

When y = 4, x = 3 + 4 = 7 [From (ii)]

Number are 7, 4

**Question 37.**

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

**Solution:**

Let two consecutive positive integers be x and x + 2

A.T.Q.,

(x)^{2} + (x + 2)^{2} = 394

x^{2} + x^{2} + 4x + 4 – 394 = 0

2x^{2} + 4x – 390 = 0

x^{2} + 2x – 195 = 0 (Dividing by 2)

x^{2} + 15x – 13x – 195 = 0

x (x + 15) – 13 (x + 15) = 0

(x – 13) (x + 15) = 0

Either x – 13 = 0 or x + 15 = 0

x = 13 or x = -15 (rejected)

Number should be x = 13 and x = 13 + 2 = 15

or x = -15 and x = -15 + 2 = -13

Hence odd numbers are 13, 15 or -15, -13

**Question 38.**

The sum of the squares of two consecutive multiple of 7 is 637. Find the multiples.** [ICSE 2014]**

**Solution:**

Let first multiple of 7 = 7x

Then second = 7x + 7

(7x)^{2} + (7x + 7) = 637

49x^{2} + 49x^{2} + 98x + 49 = 637

98x^{2} + 98x + 49 – 637 = 0

98x^{2} + 98x – 588 = 0

x^{2} + x – 6 = 0 (dividing by 98)

x^{2} + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

Either x + 3 = 0, then x = -3, but not possible being negative

or x – 2 = 0, then x = 2

Numbers will be 14, 21

**Question 39.**

The sum of the squares of two consecutive even numbers is 340. Find the numbers. **[CBSE 2014]**

**Solution:**

Let first even number = 2x

Then second number = 2x + 2

(2x)^{2} + (2x + 2)^{2} = 340

4x^{2} + 4x^{2} + 8x + 4 – 340 = 0

8x^{2} + 8x – 336 = 0

x^{2} + x – 42 = 0 (Dividing by 8)

x^{2} + 7x – 6x – 42 = 0

x (x + 7) – 6 (x + 7) = 0

=> (x + 7) (x – 6) = 0

Either x + 7 = 0, then x = -7 but not possible being negative

or x – 6 = 0, then x = 6

Numbers are 12, 14

**Question 40.**

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \(\frac { 29 }{ 20 }\). Find the original fraction.

**Solution:**

**Question 41.**

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. **[NCERT Exemplar]**

**Solution:**

Let n be a required natural number.

Square of a natural number diminished by 84 = n^{2} – 84

and thrice of 8 more than the natural number = 3 (n + 8)

Now, by given condition,

n^{2} – 84 = 3 (n + 8)

=> n^{2} – 84 = 3n + 24

=> n^{2} – 3n – 108 = 0

=> n^{2} – 12n + 9n – 108 = 0 [by splitting the middle term]

=> n (n – 12) + 9 (n – 12) = 0

=> (n – 12) (n + 9) = 0

=> n = 12 [n ≠ – 9 because n is a natural number]

Hence, the required natural number is 12.

**Question 42.**

A natural number when increased by 12 equals 160 times its reciprocal. Find the number. **[NCERT Exemplar]**

**Solution:**

Let the natural number be x.

According to the question,

x + 12 = \(\frac { 160 }{ x }\)

On multiplying by x on both sides, we get

=> x^{2} + 12x – 160 = 0

=> x^{2} + (20x – 8x) – 160 = 0

=> x^{2} + 20x – 8x – 160 = 0 [by factorisation method]

=> x (x + 20) – 8 (x + 20) = 0

=> (x + 20) (x – 8) = 0

Now, x + 20 = 0 => x = -20 which is not possible because natural number is always greater than zero

and x – 8 = 0 => x = 8.

Hence, the required natural number is 8.

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 are helpful to complete your math homework.

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