Math, asked by VijayaLaxmiMehra1, 1 year ago

I want proper answer☆☆☆

Only Genius can solve this.....

3. 'a' and 'b' are two given constant positive integers. A student is asked to find two positive integers 'q' and 'r' are such that

a = bq + r , 0 < r < b. The findings of the student are q = 3 and 5 ; r = 2 and 4.

Is the student correct in his finding ? Justify your answer.

Standard:- 10

Content Quality Solution Required

No Spamming otherwise reported on the spot

Note : If you don't know then no need to answer it.

Answers

Answered by Anonymous

Your answer is --

No, student is not correct in his finding

Because , it is given that 'a' and 'b' are the two CONSTANT number .

It is possible only when q = 1 and r = 0.

as ,

➡ 6 = 6×1 + 0

➡ 4 = 4×1 + 0

➡ 10 = 10×1 + 0

➡ 3 = 3×1 + 0

➡ 5 = 5×1 + 0

In the above example a and b are constant .
So, q = 1 and r = 0

According to question
student take q = 3 and 5 ; r = 2 and 4.

✔✔ hence, his finding is not correct .

-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-

【 Hope it helps you 】

QGP: Similarly a=5q+4 is also similarly valid.

QGP: But now, the question asks whether the two relations: a=3q+2 and a=5q+4 can be simultaneously valid?

QGP: In other words, if a number gives a remainder of 2 when divided by 3, will the number give a remainder of 4 when divided by 5? We have to find whether any case exists for which the two previous statements are valid simultaneously

QGP: Try the question now.

Anonymous: i think u dont understand question

Anonymous: a and b are constant

QGP: Yes "a" and "b" are constant. But we don't know their values. From what I understand of the question, We have to check if there can be any constant values of a and b which can satisfy both a=3b+2 and a=5b+4 simultaneously

Anonymous: sir , give me example

QGP: Well, we have to simply solve the two equations a=3b+2 and a=5b+4, and check whether the values of a and b are positive or not. If not, then there cannot be any possible values of a and b under the given conditions, and the student has found wrong value of q and r

Anonymous: dhnyawad

Answered by Anonymous

Here is your solution :

When, q = 3 and r = 2.

=> a = bq + r

Substitute the value of q and r,

•°• a = 3b + 2 ---------- ( 1 )

When, q = 5 and r = 4.

=> a = bq + r

Substitute the value of q and r,

=> a = 5b + 4 ---------- ( 2 )

From , ( 1 ) and ( 2 ) we get ,

=> 3b + 2 = 5b + 4

Subtracting 3b to both sides,

=> 3b + 2 - 3b = 5b + 4 - 3b

=> 2 = 2b + 4

Subtracting 2 to both sides,

=> 2 - 2 = 2b + 4 - 2

=> 0 = 2b + 2

Subtracting 2 to both sides,

=> 0 - 2 = 2b + 2 - 2

=> -2 = 2b

Dividing both sides by 2,

=> ( -2 ) / 2 = 2b / 2

=> -1 = b

•°• b = -1

Substitute the value of b in ( 1 ),

=> a = 3b + 2

=> a = 3( -1 ) + 2

=> a = -3 + 2

=> a = -1.

We have the values of a and b as negative numbers but in the question it is given that a and b are positive numbers.

Hence, the student is wrong.

Hope it helps !!

Anonymous: :-)

QGP: Good Answer!

Anonymous: Thanks !!

VijayaLaxmiMehra1: its so long

Anonymous: I didn't get what do you want to say ?

Previous Question

Next Question