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two digit number

by shashank.ism » Mon Feb 08, 2010 10:55 am
A two-digit number N is 'q' times the sum of its digits. A two-digit number formed by reversing the digits of N is 'p' times the sum of the digits. Which of the following is equal to 'p'?

1. (9 - q)
2. (q + 1)
3. (11 - q)
4. (q - 1)
5. (10 - q)
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by harsh.champ » Mon Feb 08, 2010 10:59 am
shashank.ism wrote:A two-digit number N is 'q' times the sum of its digits. A two-digit number formed by reversing the digits of N is 'p' times the sum of the digits. Which of the following is equal to 'p'?

1. (9 - q)
2. (q + 1)
3. (11 - q)
4. (q - 1)
5. (10 - q)
Let the number be xy.

∴ 10x + y = q(x + y)
∴ (10 - q)x = y(q - 1)
or
x/y = (q-1)/(10-q)...(i)
If number xy is reversed, let it be equal to 'p' times the sum of
digits.
Then 10y + x = p(x + y)
⇒ x(p - 1) = y(10 - p)
or
x/y = (10-p)/(p-1)
− ...(ii)
Comparing
(10-p)/(p-1)=(q-1)/(10-q)
[spoiler]Solving, we get p = 11 - q i.e. C.[/spoiler]
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by ajith » Mon Feb 08, 2010 11:46 am
shashank.ism wrote:A two-digit number N is 'q' times the sum of its digits. A two-digit number formed by reversing the digits of N is 'p' times the sum of the digits. Which of the following is equal to 'p'?

1. (9 - q)
2. (q + 1)
3. (11 - q)
4. (q - 1)
5. (10 - q)
say 10a+b is the number

10a+b = q(a+b)

10b+a = p (a+b)

p = (10b+a)/(a+b) = 9a/(a+b) +1
q = 9b/(a+b) +1

p+q = 9+2

p = 11-q
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by jeffedwards » Mon Feb 08, 2010 12:21 pm
Just pick numbers for the variables. That's what I did.

N = 24
Sum = 6
q times = N/Sum = 24/6 = 4
N reverse = 42
Sum = still 6
p times = Nr/Sum = 42/6 = 7
plug in the answer for q (4) to find the answer for p (7).
11-q is the only one that works

I did the same for another number 63 and (11-q) was the equation to work
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by jeffedwards » Mon Feb 08, 2010 12:41 pm
ajith wrote:
shashank.ism wrote:A two-digit number N is 'q' times the sum of its digits. A two-digit number formed by reversing the digits of N is 'p' times the sum of the digits. Which of the following is equal to 'p'?

1. (9 - q)
2. (q + 1)
3. (11 - q)
4. (q - 1)
5. (10 - q)
say 10a+b is the number

10a+b = q(a+b)

10b+a = p (a+b)

p = (10b+a)/(a+b) = 9a/(a+b) +1
q = 9b/(a+b) +1

p+q = 9+2

p = 11-q
ajith, where are you getting 9+2 for a p & q respectively?

I see your logic to get p+q= 11

(10a+b)/(a+b) + (10b+a)/(a+b) =
(11a + 11b)/(a+b)=
(11(a+b))/(a+b)=
11

But where are you getting 9 and 2? When I plugged in (in the above response), the first time I got 7 & 4. The second time I got 4 & 7. And if you plug in for 29 you get approx. 8.36 and 2.63.
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by ajith » Mon Feb 08, 2010 12:45 pm
jeffedwards wrote:
ajith wrote: p = (10b+a)/(a+b) = 9a/(a+b) +1
q = 9b/(a+b) +1

p+q = 9+2
well , 2 is 1+1 and 9 is 9a/(a+b) + 9b/(a+b) [ (9a+9b)/(a+b) = 9]

sorry for not explaining the steps
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by jeffedwards » Mon Feb 08, 2010 1:12 pm
ajith, yeah I left the computer and it hit me. sorry about that silly mistake. thanks for the explanation
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