two digit number

Question

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)

harsh.champ · Answer

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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ajith · Answer

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]

jeffedwards · Answer

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

jeffedwards · Answer

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.

ajith · Answer

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

jeffedwards · Answer

ajith, yeah I left the computer and it hit me. sorry about that silly mistake. thanks for the explanation

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