Question

The difference between the squares of two number is 72 .Eight times the numerically smaller number 1 more than 5 times the other number.Find the numerically greatest number.​

Answer 1

Let the numbers be X & Y and X > Y.

X^2 - Y^2 = 72 ——> 1

8*Y = 5*X + 1 ——> 2

Y = (5*X + 1)/8

Substituting Y in above equation:

X^2 - (5*X + 1)/8)^2 = 72

X^2 - ((5*X + 1)^2/64)= 72

64*X^2 - (5*X + 1)^2 = 64*72

64*X^2 - (25*X^2 + 10*X + 1) = 64*72

64*X^2 - 25*X^2 - 10*X - 1 = 64*72

39*X^2 - 10*X - 1 = 64*72

39*X^2 - 10*X - 1 = 4608

39*X^2 - 10*X - 4609 = 0

Using factorisation method,

39*X^2 - 429*X + 419*X - 4609 = 0

39*X(X - 11) + 419(X - 11) = 0

(39*X + 419)(X - 11) = 0

X - 11 = 0, 39*X + 419 = 0

X = 11, X = -419/31

Discarding X = - 419/31 since it’s a decimal number.

Therefore the greater number which is X = 11 ——> Answer.

For information:

11^2 - y^2 = 72

Y^2 = 121 - 72

Y^2 = 49

Y = +/- 7

Y = -7 has to be discarded, and hence can have only one value Y = 7.

Answer 2

Let smaller number be x and larger number be y

According to the question,

$⇒ {y}^{2} − {x}^{2} =88$

------ ( 1 )

$⇒ y=2x−5$

----- ( 2 )

Substituting value of y from equation (2) in ( 1 )$$,

(

${(2x−5)}^{2} − {x}^{2} =88$

$⇒ {4x}^{2} −20x+25− \: {x}^{2} −88=0$

$⇒ {3x}^{2} −20x−63=0$

$⇒ {3x}^{2} −27x+7x−63=0$

$⇒ 3x(x−9)+7(x−9)=0$

$⇒ (x−9)(3x+7)=0$

$⇒ x−9=0 \: and \: 3x+7=0$

$⇒ x=9 and x=− \frac{3}{7} </p><p>$

Number cannot be negative.

∴ Smaller number =9

⇒ Larger number =2x−5=2×9−5=18−5=13

#SP J2