Question

1) The sum of the squares of the two natural no is 52 ,If the first no is 8 less than twice the second no ,find the nos.

2)Two water taps together can fill a tank in 
$9 \frac{3}{8}$

hrs .The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank respectively .Find the time in which each tap can separately fill a tank.

Kindly give both answers

Answer 1

Hey user

Here is your answer :-

Q.1

Let the one number be ' x ' .

So other number will be ' 2x - 8 '

By the condition given in the question

${x}^{2} + {(2x - 8)}^{2} = 52 \\ \\ {x }^{2} + {(2x)}^{2} - 2(2x)(8) + {8}^{2} = 52 \\ \\ {x}^{2} + 4 {x}^{2} - 32x + 64 = 52 \\ \\ 5 {x}^{2} - 32x + 64 - 52 = 0 \\ \\ 5 {x}^{2} - 32x + 12 = 0 \\ \\ 5 {x}^{2} - 30x - 2x + 12 = 0 \\ \\ 5x(x - 6) - 2(x - 6) = 0 \\ \\ (x - 6)(5x - 2) = 0 \\ \\ (x - 6)= 0 \: \: \: \: \: \: or \: \: \: \: \: \: 5x - 2 = 0 \\ \\ x = 6 \: \: \: \: \: \: or \: \: \: \: \: \: x = \frac{5}{2} \\ \\ x \: is \: a \: natural\: number \: \\ \\ hence \: x = 6 \\ \\ other \: number \: = 2x - 8 \\ \\ = 2(6) - 8 \\ \\ = 12 - 8 \\ \\ = 4 \\ \\ so \: the \: numbers \: are \: 6 \: and \: 4$
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Q.2

Let the time taken by tap of larger diameter to fill the tank be " x hours " .

The tap can fill 1/x part of tank in a hour

So the time taken by tap of smaller diameter is " x + 10 hours " .

The tap can fill 1/x+ 10 part of rank in a hour .

$both \: the \: taps \: can \: fill \: the \: tank \: in \: 9 \times \frac{3}{8} = \frac{75}{8} \\ \\ so \: they \: can \: fill \: \frac{8}{75} \: part \: of \: tank \: \\ \\ so \: we \: get \: equation \: \\ \\ \frac{1}{x} + \frac{1}{x + 10} = \frac{8}{75} \\ \\ \frac{(x + 10) + x}{x(x + 10)} = \frac{8}{75 } \\ \\ \frac{2x + 10}{ {x}^{2} + 10x} = \frac{8}{75} \\ \\ 8( {x}^{2} + 10x) = 75(2x + 10) \\ \\ 8 {x}^{2} + 80x = 150x + 750 \\ \\ 8 {x}^{2} + 80x - 150x - 750 = 0 \\ \\ 8 {x}^{2} - 70x - 750 = 0 \\ \\ comparing \: the \: equation \: with \: a {x}^{2} + bx + c = 0 \\ \\ here \: a \: = 8 \\ b = - 70 \\ c = - 750 \\ \\ {b}^{2} - 4ac = {( - 70)}^{2} - 4(8)( - 750) \\ \\ = 4900 + 24000 \\ \\ = 28900 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a } \\ \\ = \frac{ - ( - 70) + - \sqrt{28900} }{2 \times 8} \\ \\ = \frac{70 + - 170}{16} \\ \\ x = \frac{70 + 170}{16} \: \: \: \: or \: \: \: \: x = \frac{70 - 170}{16} \\ \\ x = \frac{240}{16} \: \: \: \: or \: \: \: \: x = \frac{ - 100 }{16} \\ \\ x = 15 \: \: \: \: or \: \: \: \: x = \frac{ - 25}{4} \\ \\ \: time \: cannot \: be \: negative \: \\ \\ x = 15 \\ \\ x + 10 = 15 + 10 = 25 \\ \\ so \: the \: taps \: with \: large \: and \: small \:diameter \: can \: fill \: the \: tank \: in \: 15 \: and \: 25 \: hours \: respectively \: indivitually$

Answer 2

Hi there

Ans 1 :-

Number of perfect squares between 0 to 52 are :-

4 , 9 , 16 , 25 , 36 , 49 .

36 + 16 = 52

6^2 + 4^2 = 52

But 36 - 32 ( twice of 16 ) = 4 which is contradicting the question that first number is less than twice the second number and also the first number becomes bigger than the second number .

If you take square of 36 and 34 then also :-

8 ( twice of 4 ) - 6 = 2 which is contradicting the question that first number is less than twice the second number .


So , It is not possible because the first number is 8 less than twice the second number, that would imply that the second number is at least 8. The square of 8 is 64 which is more than 52.


Ans 2 :- Refer to attachment .

Note :- Larger diameter of the tap can fill in 15 hours and smaller diameter of tge tank fill in 25 hours.



Thank You