Question

A one-digit number, which is the ten’s digit of a
two digit number X, is subtracted from X to give Y
which is the quotient of the division of 999 by the
cube of a number. Find the sum of the digits of X.
(a) 5
(b) 7
(c) 6
(d) 8​

Answer 1

Answer:

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Let the unit's place digit be x, the tenth place digit be y and  the hundredth place digit be z of the three digit number  

So the number will be 100z+10y+x

Now given, sum of digits =x+y+z=12......(1)

And y=2x,z=3x

Now using (1)  

⇒x+2x+3x=12

⇒6x=12⇒x=2

So y=2x=4 and z=3x=6

Hence the number is =100z+10y+x=642

So the reversed number is 246

Answered By AMEYA

Step-by-step explanation:

Answer 2

Given : A one digit number which is the ten's digit of a two digit number X,  is substracted from X to give Y which is the quotient of the division of 999 by the cube of a number.

To  find :   the sum of the digits of X

(a) 5

(b) 7

(c) 6

(d) 8​

Solution:

two digit number X,  = AB

Tens digit = A

10A + B  -  A =  9A  + B

Y = 9A + B

Y    =  999 / z³

  999 = 3 * 3 * 3 * 37  = 3³ x 37

or  1 * 1 * 1 * 999  =  1³ x 999

Y can not be  999

Hence  z = 3

=> Y    =  999 / 3³  = 37

Y = 9A + B  =  37

A = 4  and  B =  1

X  = 10A + B = 41

sum of digits of X  = 4 + 1 = 5

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