Let p and q each represent a digit from 0 through 9 and let pq represe

given
10q+p- 10p-q = 27
q-p = 3
possible number such that pq is prime ; 47
sum is 11
option D

can any one explain to me why A is wrong ?
41 is a prime no and when we interchange the digits then we got 14 and according to question
(41-14=27) then the some of p+q is 5 how it is wrong?

Check question for range where pq can fall in..
shivanks

20<pq<99


Posted from my mobile device

I actually think it's just easy to list out the primes starting with 23 and flipping them to see your options. Doesn't take too long because they're aren't actually that many.

After listing out a few you're kinda just left with 41 -> 14 and 47 -> 74.

If you read the question, the lather number needs to be larger so 47 -> 74 is the only answer and adding the sums = 11.

Really only takes like 30 seconds this question.

why is 63-36=27 is wrong? it satisfies the equation. some questions just make me sick!

We are told that the smaller number, pq, is prime, whereas 36 in your example is not. You should read questions more carefully.

We are given that qp - pq = 27

\((10q + p) - (10p + q) = 27\)
\(9(q-p) = 27\)
\((q-p) = 3\)

Hence various values of p and q are possible. Neither p nor q can be 0 because both pq and qp are two digit numbers.
p = 1, q = 4, Then pq = 14 (Not prime)
p = 2, q = 5, Then pq = 25 (Not prime)
p = 3, q = 6, Then pq = 36 (Not prime)
p = 4, q = 7, Then pq = 47 (Prime!)

We needed pq to be prime (but not qp necessarily) and when p = 4, q = 7, pq is prime. Here p + q = 11.
Other values of p and q are also possible but we will not evaluate them because this is a PS question and here there will be only one correct answer.

Hence answer (D)

Given pq = 23, 29, 31, 39, 41, 43, 47, 53, 57, 59, 61, 67....97

10p + q = 27 + 10q + p
P + q=?

10p + q + 27 = 10q + p
27 = 9q – 9p
27/9 = q-p = 3

Therefore q can be 5,7,9 (Q cannot be even, and q must be greater than 3)
And p can be 2, 4, 6

So possible combinations are, 25, 47, 69
25 and 69 aren’t prime, so 47 is the only option

Lets check qp = 27 + pq, which is 27 + 47 = 74

Therefore, q+p = 7 + 4 =11

Why did you add the coefficient "10" to the equation?

2-digit integer qp can be written as 10q + p because q is 10's digit and p is units digit. Example, if qp = 74, here q=7 and p=4
10q + p = 10*(7) + 4 = 70 + 4 = 74.

Danou

Look there are 2 ways of talking about a two digit number. I can say a two digit number is x. I can say when we reverse its digits, the number becomes y. If the difference between them is 9, I can write it as x - y = 9. But how do I solve it now? I have lost some important information - that digits in y are reverse of digits in x.

So in this case I can say that say the first two digit number is pq (where p is tens digit and q is units digit). Then the second number is qp (where q is tens digit and p is unit digit). Now this captures the info of reversing of the digits. But when I want to now calculate the difference between them, I must write it as
10p + q because the value of pq = 10p+q.
Essentially, when p = 3 and q = 2, pq = 32 = 3*10 + 2
I must take the place value of the digit into account too.

A three digit number per = 100p + 10q + r where p, q and r are the digits of the number (say 8, 5 and 1).
851 = 100*8 + 10*5 + 1

Let p and q each represent a digit from 0 through 9 and let pq represent a 2-digit positive prime number such that \(20 < pq < 99\). For example, if \(p = 2\) and \(q = 3\), then the 2-digit prime number \(pq\) is 23.

If the 2-digit integer \(qp\) is 27 greater than \(pq\), what is the sum of the digits p and q?

20 < 10p + q < 99

(10q + p) - (10p + q) = 27
9(q-p) = 27
q - p = 3
q = p + 3 < 10
p < 7

20 < 10p + p + 3 < 99
20 < 11p + 3 < 99
17 < 11p < 96

p = {2,3,4,5,6}

p = 2 ; q = p+3 = 5; 25 is NOT a prime number
p = 3 ; q = p+3 = 6; 36 is NOT a prime number
p = 4 ; q = p+3 = 7; 47 is a prime number
p = 5 ; q = p+3 = 8; 58 is NOT a prime number
p = 6 ; q = p+3 = 9; 69 is NOT a prime number

pq = 47
Sum of digits of pq = 4+ 7 = 11

IMO D

Classic digit swap scenario, set up as usual:

they specified it is a prime number, I did the same mistake.