Following are my answers:

Q1. In the figure shown above,the measure of angle PRS is how many degrees greater than the measure of angle PQR?

(1) The measure of angle QPR is 30
(2) The sum of the measures of angles PQR and PRQ is 150

We have to find the difference between <PRS and <PQR.



1.
<PQR + <QPR = <PRS ( Exterior angle theorem : Sum of two interior angles = opposite exterior angle)
<PQR + 30 = <PRS
Hence, <PRS - <PQR = 30 degrees.

So, 1 is sufficient.

2. <PQR + <PRQ = 150

<PQR + <PRQ +<QPR = 180 Hence <QPR = 30;
If we have <QPR, we have already proved in 1 that we can find <PRS - <PQR.

Hence Sufficient.

As each statement alone is sufficient, the answer is D.

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Q2. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A and inversely proportional to the concentration of chemical B present.If the concentration of chemical B is increased by 100%,which one of the following is the closest to the percentage change in the concentration of chemical A required to keep the reaction rate unchanged?

a)100% decrease
b)50% decrease
c)40% decrease
d)40% increase
e)50% increase

Let the rate of the reaction be R
Let concentration of chemical A be A
Let concentration of chemical B be B

Then R is proportional to A²
R is also proportional to 1/B

Hence, R is proportional to A²/B

If C is a constant, R=C*(A²/B)

If the concentration of B is increased 100% B becomes 2B ( B+(100/100)*B = 2B)

Let A2 be the new concentration of chemical A for the rate to be constant

Then R=C*(A²/B) = C*(A2²/(2*B))
Hence, A² = A2²/(2) So A² = A2/√2

A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

So, there is a 41% increase in A.

Answer is D

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Q3.For every integer k from 1 to 10,inclusive,the kth term of a certain sequence is given by (-1)^k+1 * (1/2^k).If T is the sum of the 1st 10 terms in the sequence then T is,

a)Greater than 2
b)between 1 and 2
c)between ½ and 1
d)between ¼ and ½
e)less than ¼

Kth term of a sequence is =
Rk = (-1)^k+1 * (1/2^k)

R1 = (-1)^2 * (1/2)^1 = 1/2
R2 = (-1)^3 * (1/2)^2 = -1/4
similarly, R3= 1/8, R4 = -1/16 etc

So sum of 1st 10 terms in the sequence =

S = 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 +1/128 etc upto the 10th term ( there will be total of 11 terms as S is inclusive of 1st and 10 term)

S = 1/2 +(- 1/4 + 1/8) + (- 1/16 + 1/32) etc ( there will be a total of 5 pairs like this as the sequence has 11 terms. )
S = 1/2 -1/8 -1/32 etc

If we sum up the negative terms
1/8 + 1/32 +.... we can see that sum is greater than 1/8 but less than 1/4 ( as 1/32+1/128 etc <1/8)

So the sum is greater than 1/2-1/4 =1/4 but less than 1/2 as we are subtracting 1/8 + 1/32 from 1/2
Hence the sum is between 1/2 and 1/4

Answer is D

Thanks a lot!

magnolia wrote: Let A2 be the new concentration of chemical A for the rate to be constant

Then R=C*(A²/B) = C*(A2²/(2*B))
Hence, A² = A2²/(2) So A² = A2/√2

A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

Magnolia,

Can you help me clarify the quoted step? I dont understand how you arrived at A2.

The rate =R =C*(A²/B)

Let A2 be the new concentration of chemical A for the rate to be constant.

The new concentration of chemical B = B+ 100% B ( B is increased 100%) = B+(100/100)*B= B+B=2B

So, the new rate is propotional to A2²/(2B) ( square of A2 divided by twice B)

As C is our constant new rate = C *{A2²/(2B)}

As new rate = old rate

C*(A²/B) = C*(A2²/(2B))
canceling common terms C and B from both sides we get
A² = A2²/2

If we take the square root of both sides, we get
A = A2/√2 (sorry, there was a typo here in my earlier explanation)


A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

Thanks magnolia! This helps