If a and b are different positive integers and a+b=a(a+b)

Zero is neither positive not negative integer.

I pick numbers for this problem

a+b = a(a+b). to make the statement true a=1,b=2
1+2=1(1+2)
3 =3.

from this we know ,a=1 & a<b. answer should be E

Number picking might not be the best way to solve MUST BE TRUE questions.

The question asks which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

So the set you chose just proves that II is not always true and hence it's not a part of a correct choice. As for I and III: they might be true for this particular set of numbers but not true for another set, so you can not say that I and III are always true just based on one set of numbers (it just happens to be that I and III are always true).

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

Hope it helps.

tricky question, i went with A only to realize that both a and b are positive integers, so b must be greater than a

Tricky one indeed. I marked E but was wrong in solutioning. I thought a=-b means a is smaller than b. Didn't realise that this equation is not possible since both a and b are positive

a+b=a(a+b)
1(a+b)-a(a+b)=0
(1-a)*(a+b)=0 from this either a-1=0 or a+b=0, but since a and b are positive their could not be equal to 0.
Thus, 1-a=0 or a=1.
1. holds true
2.not true because a and b are different positive integers and we know that a=1
3. true- As a, b are "different positive integers" when 1 is the smallest , thus a<b.

hence E.

1 is the smallest positive integer.
0 is a signless integer. These two are required.

from the equation we get a=1 as (a+b) cannot be zero.
so
I. a = 1 always true
II. b = 1 not always true
III. a < b always true


so the ans is E

\(a+b = a (a+b)\)
\(\frac{a+b}{a+b}=a\)
\(a=1\)

I. a = 1 always TRUE
II. b = 1 b must not be the same as a. Thus, b is not equal to 1. FALSE
III. a < b b is a positive integer but not equal to 1 then it must be 2 onwards. always TRUE

Answer: E

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we can rewrite the equation as:
(a+b)/a = a+b
a/a=1
a+b/a = a+b
we can subtract a from both sides
b/a=b
this is true only when A is 1.
since we are told that the numbers are different, and both positive, we know for sure that b MUST be >1, or b>a.

E

a + b = a(a + b)
=> (a+b)(a-1) = 0
=> a=-b , a =1

a and b are both positive. So, a = -b is not possible.

a=1,
Also a and b are both different positive integers. So, b >1.

and hence a<b

So, I & II both are correct.

Answer E

Since a and b are positive integers, a + b ≠ 0. So we can divide the given equation by a + b and obtain:

1 = a

Therefore, Roman numeral I is true. Furthermore, since a and b are different positive integers, b > a. So Roman numeral III is also true.

Answer: E

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