Math, asked by roshnimohan172, 10 hours ago

if PQ-QP =27then find the value of P and Q

Answers

Answered by corjasmin11042020

Answer:

P=7

Q=4

Step-by-step explanation:

PQ-QP=27

74-47=?

-47

____

cancelation:

7 in 74 will become 6 and 4 will become 14

then

6-4 =2

and 14-7=7

so ..

74-47=27

PQ-QP=27

Answered by joelpaulabraham

Answer:

P can be 4 and Q can be 1

41 - 14 = 27

P can be 5 and Q can be 2

52 - 25 = 27

P can be 6 and Q can be 3

63 - 36 = 27

P can be 7 and Q can be 4

74 - 47 = 27

P can be 8 and Q can be 5

85 - 58 = 27

P can be 9 and Q can be 6

96 - 69 = 27

Any of these values is possible.

Step-by-step explanation:

We have,

QP - PQ = 27

But we must know that,

PQ = 10P + Q

QP = 10Q + P

This is another way of representing two digit numbers.

So,

QP - PQ = 27

(10P + Q) - (10Q + P) = 27

10P + Q - 10Q - P = 27

10P - P + Q - 10Q = 27

9P - 9Q = 27

9(P - Q) = 27

P - Q = 27/9

P - Q = 3

Here, we know that,

PQ and QP are two digit numbers then, P or Q has to be single digit numbers because otherwise they will become 3 digit numbers.

For ex:-

If P = 10

Then,

PQ = 10(10) + Q = 100 + Q

So, it becomes a 3 digit number

So we can safely say that P < 10

Now,

Can P be 0?

Let's see,

IF P = 0

then,

PQ = 10(0) + Q = 0 + Q

That will make it a 1 digit number

So, 0 is also not possible

So,

0 < P < 10

Now,

Let's check Q,

P - Q = 3

Q = P - 3

Let's say P was less than 3 then Q will become negative which will not give us QP because then it will be negative.

Also IF P = 3

Then,

Q = 3 - 3 = 0

It is possible for PQ, there is no problem but when we take QP it becomes 03 which is again a 1 digit number. So we understand that,

3 < P < 10

P = {4, 5, 6, 7, 8, 9}

Then we can find Q by substituting the values of P by Trial and error method using Q = P - 3

We get that,

0 < Q < 7

So,

Q = {1, 2, 3, 4, 5, 6}

Hence,