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Cpt maths question

Nabeel (CA) (3288 Points)

28 April 2012

Hey Frnds, plz solve this CPT maths question...............

Q. The sum of the digits of the two digit numbers are 10. If 18 is subtracted from it, the digits in the resulting number will be equal. The number is ???

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9 Replies

chandu (student) (1051 Points)
Replied 28 April 2012

ans is 73

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chandu (student) (1051 Points)
Replied 28 April 2012

Let x = 10's digit of a two digit number

Let y = units digit

Let z = the resulting one

let u assume an equton

"the sum of a two digit number is 10."

x + y = 10----eq1

y = (10-x)

18 be subtracted from it,

then eq-2 is:

10x + y - 18 = 10z + z

10x + y = 11z + 18---eq2

Substitute (10-x) for y in eq2

10x + (10-x) = 11z + 18

10x - x = 11z + 18 - 10

9x = 11z + 8

The single digit integer values for x and z are limited to x=7 and z=5 in this equation:

x = 7, y = 3, z = 5

so d 2digit num is:73

3 Like

intermediate(ipc)course (no) (1460 Points)
Replied 28 April 2012

ans.73....

1 Like

Nabeel (CA) (3288 Points)
Replied 28 April 2012

Frnds, i knew the answer........ I want the process of solving.......

Nabeel (CA) (3288 Points)
Replied 28 April 2012

Originally posted by : chandu
	Let x = 10's digit of a two digit number Let y = units digit Let z = the resulting one let u assume an equton "the sum of a two digit number is 10." x + y = 10----eq1 or y = (10-x) 18 be subtracted from it, then eq-2 is: 10x + y - 18 = 10z + z 10x + y = 11z + 18---eq2 Substitute (10-x) for y in eq2 10x + (10-x) = 11z + 18 10x - x = 11z + 18 - 10 9x = 11z + 8 The single digit integer values for x and z are limited to x=7 and z=5 in this equation: x = 7, y = 3, z = 5 so d 2digit num is:73

Thnks a lot, but i wont be able to understand the last step....... Plz explain me...........

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Tejaswi Kasturi (student-cpt) (427 Points)
Replied 28 April 2012

the equation is 9x-11z =8 where x,z are integers

The equation is a diophantine equation. It can be solved in the following way.

greatest common divisor of 9 and 11 is 1

11 = 9+2

9 = 4*2 + 1

implies

1 = 9-4*2

1 = 9 - 4(11-9)

1 = 9*5 -(11*4)

now multiply both sides with 8

8 = 9*40-11*32

we can add and subract 99k where k is any integer

8 = 9*(40-11k)-11*(32-9k) (as both the extra terms cancel) for k belonging to integers

so (40,32),(29,23),(18,14),(7,5) ,(-4,-4)......all form the roots of the above equation

in them we need theroots which are greater than 0 and less than 10 hence &(7,5)

usually we don't do all this stuff in exam. we merely try to substitute the answers in multiple choice question in the equation and see whether it is satisfied or not.check out the following link for more information

en.wikipedia.org/wiki/Diophantine_equation

Tejaswi Kasturi (student-cpt) (427 Points)
Replied 28 April 2012

For solution to general diophantine equation ax+by=c where x,y are integers check out the following linkhttps://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

Harsh Modi (1) (27 Points)
Replied 01 May 2012

However it is better to solve such questions by Method of Options.

Out of four given options in exam only 1 will be correct.

So take the first option and check whether sum of its digit is 10 or not. If it is 10, then subtract 18 from it and see whether resulting number satisfies given condition.

Repeat the process for other options.

This is the fastest method of solving.