PS Problem Set 02

[gdk800]

Q1) Jim invested $1000 @ 10% compounded annually. Laura invested $2000 @ 5% compounded annually. Total interest of Jim is how much more than total interest of Laura after 2 years?
a. 5
b. 15
c. 50
d. 100
e. 105

[spoiler]OA is: A
[/spoiler]

Q2) A set of 15 different integers has median of 25 and a range of 25. What is greatest possible integer that could be in this set?
a. 32
b. 37
c. 40
d. 43
e. 50

[spoiler]OA is D.

[/spoiler]

Q3) If n is multiple of 5, and n = p2q where p and q are prime, which of the following must be a multiple of 25?
a. p2
b. q2
c. pq
d. p2q2
e. p3q


OA is D.

My question is why can't it be B? Because if q2 is needed for n to be a multiple of 25 than option B also has it.

[beat_gmat_09]

Q1) Jim invested $1000 @ 10% compounded annually. Laura invested $2000 @ 5% compounded annually. Total interest of Jim is how much more than total interest of Laura after 2 years?
a. 5
b. 15
c. 50
d. 100
e. 105

Jim's 1st interest = 1000*0.1 = 100, compounding: new principle =1000+100 = 1100, after 2nd yr interest = 1100*0.1=110
Total interest earned on $1000 = $100+$110 = $210

Laura's 1st interest = 2000*0.05 = $100, compounding: new principle = 2000+100 = 2100, after 2nd year interest = 2100*0.05 = 105
Total interest earned = 100+105 = $205
Difference between interest earned = 210-205 = $5
Jim earns $5 more than Laura in interest money.

[beat_gmat_09]

Q3) If n is multiple of 5, and n = p2q where p and q are prime, which of the following must be a multiple of 25?
a. p2
b. q2
c. pq
d. p2q2
e. p3q

I'm assuming its p^2 not 2*p that makes sense.
This is must be true question, so option answer will always be multiple of 25.
n = (p^2)*q
n is multiple of 5. p and q are prime numbers.
In what cases can n be multiple of 5 - when q =5 or when p=5
when q=5 p cannot be =5 as p is prime number, reverse is the case when p=5 q cannot be=5
1st case q=5 : n=5*p^2
From option answers (option B) q^2 is only one which can be multiple of 5. Second case (p=5) is not satisfied in this.
2nd case p=5: when p=5, p^2 will be multiple of 25, option A is true.
Option D = (p^2)*(q^2) will always be true regardless of p=5 or q=5
if p=5 p^2*q^2 will be multiple of 25, but option B cannot be multiple of 25 - hence option B is not always true but option D always has to be true.
if q = 5 p^2*q^2 will be multiple of 25, but option A cannot be multiple of 25 - hence option A is not always true but option D is always true.

[goyalsau]

Q2) A set of 15 different integers has median of 25 and a range of 25. What is greatest possible integer that could be in this set?
a. 32
b. 37
c. 40
d. 43
e. 50

[spoiler]OA is D.

Excellent Question.

25 is the middle term - Then it should be 8th term of the series.

There must be 7 terms before 25, and all should be different 25-7 = 18,

18 must be the first term and the range is 25 so .. 18+25 = 43.

[Rahul@gurome]

Q1) Jim invested $1000 @ 10% compounded annually. Laura invested $2000 @ 5% compounded annually. Total interest of Jim is how much more than total interest of Laura after 2 years?
a. 5
b. 15
c. 50
d. 100
e. 105

Interest of Jim after 1st year = $(1000)*(10/100) = $100
Interest of Jim after 2nd year = $(1000 + 100)*(10/100) = $110
Total interest of Jim = $(100 + 110) = $210

Interest of Laura after 1st year = $(2000)*(5/100) = $100
Interest of Laura after 2nd year = $(2000 + 100)*(5/100) = $105
Total interest of Jim = $(100 + 105) = $205

After 2 years,

• Interest of Jim - Interest of Laura = $(210 - 205) = $5

The correct answer is A.

[Rahul@gurome]

Q2) A set of 15 different integers has median of 25 and a range of 25. What is greatest possible integer that could be in this set?
a. 32
b. 37
c. 40
d. 43
e. 50

Say, the largest integer is x and the lowest integer is y.
Now, (x - y) = 25 => x = (y + 25). Thus to maximization of x means maximization of y too. (There difference is constant)

What could maximum value the lowest integer?
25 is the median of 15 different integers. As 15 is odd, 25 must be the middle term i.e. the 8th term of the set when arranged in increasing order. Thus there are 7 different integers before 25. The lowest integer will be maximum when each of these 7 integers differ by 1 (As 1 could be the minimum possible difference between any two consecutive integers)

Thus maximum possible value of y = (25 - 7) = 18
Thus maximum possible value of x = (18 + 25) = 43

The correct answer is D.

[Rahul@gurome]

Q3) If n is multiple of 5, and n = p²q where p and q are prime, which of the following must be a multiple of 25?
a. p²
b. q²
c. pq
d. p²q²
e. p³q

If n is multiple of 5, and n = p²q where p and q are prime, then either p or q or both of them must be equal to 5. Let's analyze each of the cases. (Note that only one of the following can happen at a time)

• 1. p = 5, p² is multiple of 25, q² not
  2. q = 5, q² is multiple of 25, p² not
  3. p = q = 5, p² = q² = multiple of 25

We have to find a generalized expression containing p and q such that it becomes multiple of 25. From above analysis we know p² or q² is not that expression as they may or may not be a multiple of 25. But in p²q² both of them are present and simultaneously all the three cases are merged into one! For any of the above cases p²q² will be always a multiple of 25.

The correct answer is D.