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

These Solutions are part of 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 =

According to the condition,

x – 20 = 69 x

=> x – 20 =

=> 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 . Find the number.

**Solution:**

**Question 11.**

The sum of a number and its square is , 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 = 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 = 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 , 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 . 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 , 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 =

Number = x + 10 x = 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 . 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 and is . 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 . 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 = 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 = 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 . 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 . 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 =

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Reet wadhwa says

March 5, 2019 at 8:38 pmThank you so much for the answers . I have no words seriously . These have helped me so much . God bless you all