Hi!

a. IMO C
(1) d = 2x
A = xy
statement (1) ALONE is INSUFFICIENT
We should eliminate answers A and D.
(2) d = x + 0.1
A = (d - 0.1)*y
statement (2) ALONE is INSUFFICIENT
We should eliminate answer B.
(1) & (2)
d = 2x & d = x + 0.1
x = 0.1
d = 0.2
We know from Pythagorean theorem that
d^2 = x^2 + y^2
y^2 = d^2 - x^2
y^2 = 0.03
y = √0.03
A = 0.1*√0.03
Statements (1) & (2) both are TOGETHER SUFFICIENT
We should eliminate answer E and choose answer C.

c. IMO C
(1) P = 2,3, 5, 7, ...
Let us plug in numbers:
P = 2
P/2 = 1 and the remainder is 0.
P = 3
P/2 = 1 and the remainder is 1.
statement (1) ALONE is INSUFFICIENT
We should eliminate answers A and D.
(2) P > 10
statement (2) ALONE is INSUFFICIENT
We should eliminate answer B.
(1) & (2) P = 11, 13 , 17, 19, 23, ...
P is odd, so P/2 will have the remainder 1
Statements (1) & (2) both are TOGETHER SUFFICIENT
We should eliminate answer E and choose answer C.

Hope it helps!
Best,
Maciek

b. Is â""šx + 1â""š < 2 ?

1) (x-1)2(square) < 1
2) x2(square) - 2 < 0

Maciek wrote:Hi!

a. IMO C
(1) d = 2x
A = xy
statement (1) ALONE is INSUFFICIENT
We should eliminate answers A and D.

(2) d = x + 0.1 How can one be sure of that this side is x OR the same side as mentioned in the above statement.

Maciek wrote: A = (d - 0.1)*y
statement (2) ALONE is INSUFFICIENT
We should eliminate answer B

From (1) and (2)
d = 2x & d = x + 0.1

IF we consider this as y + 0.1 , Then i don't think we can solve this problem.

Maciek wrote: x = 0.1
d = 0.2
We know from Pythagorean theorem that
d^2 = x^2 + y^2
y^2 = d^2 - x^2
y^2 = 0.03
y = √0.03
A = 0.1*√0.03
Statements (1) & (2) both are TOGETHER SUFFICIENT
We should eliminate answer E and choose answer C.

a.What is the area of rectangle ?

1. Diagonal is twice as long as side
2. Diagonal is 0.1 meters longer than side

let us call one side A, the other side B, diagonal C; Area of rectangle is A*B

(1) C=2A? or C=2B? Values A and B? Insufficient
(2) C=0.1+A? C=0.1m+B? Insufficient

Combining (1)&(2): C=2B and C=0.1m+B [e.g. B=10cm, C=20cm, C=10cm+B]-need A;C=2A and C=0.1m+A [e.g. A=10cm, C=20cm, C=10cm+A]-need A; C=2B and C=0.1m+A [e.g. A=10cm, C=20cm, C=10cm+A, B=C/2 or B=10]
three possibilities. Insufficient

Answer E.

b. Is |x+1|<2 ?

1) (x-1)^2 < 1
2) x^2 - 2 < 0

critical points on the number line:
-x-1<2 => x+1>-2 => x>-3
x+1<2 => x<1

the value set {-3<x<1}

(1) (x-1)^2 < 1 => x^2 - 2x + 1 <1 => x^2 - 2x <0 => x(x-2)<0 => x<0 and x-2<0 or x<2 => x<0 Not sufficient
(2) x^2 - 2 < 0 => x^2 < 2 => |x|<2 => -2<x<2 Not sufficient

Combining (1)&(2): -2<x<2 and x<0 new value set for x {-2<x<0} Sufficient.

Answer C.


c. What is the remainder of p/2 ?

1. p is a prime integer.
2. p > 10.

p/2=i+r/2 => p=2i+r

(1) p-prime number => if p=|2|, then r=0 otherwise r=|1| Not sufficient.
(2) p>10 ~ without (1) not sufficient

Combining (1)&(2) p-prime number and p>2 Satisfies.

Answer C.