Question

32 timesof a 2 digit number is 23 times the number obtained by reversing its digit .The sum of the digit is 15 find the number

Answer 1

Answer:

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Let the units digit be x and tens digit=15-x

according to the question,

32[10*(15-x)+x]=23[10x+(15-x)]

32[150-9x]=23[9x+15]

4800-288x=207x+345

4800-345=207x+288x

4455=495x

x=4455/495=9

therefore no.=10*(15-9)+9=69

Answer 2

Answer:  The required number is 69.

Step-by-step explanation:  Given that 32 times of a two digit number is 23 times the number obtained by reversing its digits and the sum of the digits is 15.

We are to find the number.

Let x and y represents the digits in the ten's place and unit's place of the number.

Then, the number will be

$n=10\times x+y=10x+y.$

According to the given information, we have

$32(10x+y)=23(10y+x)\\\\\Rightarrow 320x+32y=230y+23x\\\\\Rightarrow 320x-23x=230y-32y\\\\\Rightarrow 297x=198y\\\\\Rightarrow x=\dfrac{198}{297}y\\\\\Rightarrow x=\dfrac{2}{3}y~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$x+y=15\\\\\Rightarrow \dfrac{2}{3}y+y=15~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{from equation (i)}]\\\\\Rightarrow \dfrac{5}{3}y=15\\\\\Rightarrow y=3\times3\\\\\Rightarrow y=9.$

So, from equation (i), we get

$x=\dfrac{2}{3}\times9=2\times3=6.$

Therefore, the required number is

$n=10\times6+9=60+9=69.$

Thus, the required number is 69.