# CLASS-10QUADRATIC EQUATIONS - PROBLEMS ON NUMBERS

Problem on Numbers

Example.1) The product of two numbers in 192 and the sum of these numbers is 28. Find the numbers.

Ans.) As per the given condition sum of the two numbers is 28. Let one number is x and another number would be (28 – x), when x + (28 – x) = 28

As per the given condition, product of two numbers is 192.

So, x (28 – x) = 192

=> 28x - x²- 192 = 0

=> x²- 28x + 192 = 0

=> x²- (16 + 12)x + 192 = 0

=> x²- 16x – 12x + 192 = 0

=> x(x – 16) – 12(x – 16) = 0

=> (x – 16) (x – 12) = 0

=> (x – 16) = 0 or (x – 12) = 0

=> x = 16 or x = 12

Hence the required numbers are 16 & 12.     (Ans.)

Example.2) 50 is divided into two parts such that the sum of their

1

reciprocals is ------. Find the two parts.

12

Ans.) Let the required parts be x & (50 – x).

1               1

and their reciprocals are ------- and --------- Then,

x            (50 – x)

as per given condition, we have -

1              1               1

------- + ----------- = ---------

x         (50 – x)            12

(50 – x)+ x          1

=> -------------- = -------

x (50 – x)           12

=>   (50 X 12) = 50x - x²

=>  x²- 50x + 600 = 0

=>  x²- (30 + 20)x + 600 = 0

=>  x²- 30x – 20x + 600 = 0

=>  x (x – 30) – 20(x – 30) = 0

=>  (x – 30)(x – 20) = 0

=> (x – 30) = 0 or (x – 20) = 0

=> x = 30 or x = 20

Hence the required two parts are 30 & 20 (Ans.)

Example.3) A positive number is divided into two parts such that the sum of the squares of the two parts is 208. The square of the larger part is 18 times the smaller part. Taking x as smaller of the two parts, find the number.

Ans.) Let the smaller part is ‘x’, and the larger part be ‘y’.

So, the required number be (x + y)

As per the given condition, x² + y² = 208

Or, y² = (208 - x²) …………………..(i)

And, y² = 18 x ………………..(ii)

Now we will substitute the value of (ii) in (i) and we get,

18 x = 208 - x²

=> x² + 18 x – 208 = 0

=> x² + (26 – 8)x – 208 = 0

=> x² + 26 x – 8 x – 208 = 0

=> x(x + 26) – 8(x + 26) = 0

=> (x + 26) (x – 8) = 0

=> (x + 26) = 0 or (x – 8) = 0

=> x = -26 or x = 8

=> x = 8   [where each part is positive]

Putting x = 8 in (ii), we get –

y²= (18 X 8) = 144

or, y² = 12²

or, y = 12       [neglecting y = - 12]

so, required number = (8 + 12) = 20    (Ans.)

Example.4) Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 96. Assume the middle number to be ‘x’ and form a quadratic equation satisfying the above statement. Solve the equation and hence find the three numbers.

Ans.) Let the middle natural number be ‘x

Then, the other two natural number are (x – 1) and (x + 1)

By, given condition we have –

x²- {(x + 1)² - (x – 1)²} = 96

=> x²- {(x² + 2x + 1) – (x² - 2x + 1)} = 96

=> x²- (x² + 2x + 1 - x² + 2x – 1) = 96

=> x²- 4x = 96

=> x²- 4x – 96 = 0

=> x²- (12 – 8)x – 96 = 0

=> x²- 12x + 8x – 96 = 0

=> x (x – 12) + 8(x – 12) = 0

=> (x – 12) (x + 8) = 0

=> (x – 12) = 0 or (x + 8) = 0

=> x = 12 or x = -8

=> x = 12      [when – 8 ∉ N]

So, the three numbers are (12 – 1), 12, and (12 + 1), i.e., 11, 12, 13 (Ans.)

Example.5) The sum of the numerator and denominator of a fraction is 8. If 1 is added to both the numerator and denominator, the fraction

1

is increased by -------. Find the fraction.

15

Ans.) Let, the denominator of the required fraction be x. Then, its numerator = (8 – x)

(8 – x)

So, the fraction is ---------.

x

As per the given condition –

(8 – x) + 1         (8 – x)          1

------------- = ---------- + -------

x + 1                x             15

9 – x          8 – x           1

or, --------- - --------- = -------

x + 1           x             15

{x (9 – x)} – {(8 – x) (x + 1)}            1

or, ----------------------------- = --------

x (x + 1)                       15

9x - x² - (8x - x² + 8 – x)             1

or, ------------------------------ = --------

x² + x                           15

9x - x²- 7x + x² - 8            1

or, ------------------------ = --------

x² + x                      15

=> 15 (2x – 8) = x² + x

=> 30 x – 120 = x² + x

=> x²- 30 x + x + 120 = 0

=> x²- 29 x + 120 = 0

=> x²- (24 + 5) x + 120 = 0

=> x²- 24 x – 5 x + 120 = 0

=> x (x – 24) – 5 (x – 24) = 0

=> (x – 24) (x – 5) = 0

=> (x – 24) = 0 or (x – 5) = 0

=> x = 24 or x = 5

Now, if x = 24, then (8 – x) = (8 – 24) = - 16 which is not possible where

x        - 16

------- = ------- is not a fraction

24          24

Now, we will take x = 5, and we get (8 – x) = (8 – 5) = 3 which is possible

x           3

because ------- = -------     (Ans.)

5           5