gmatophobia
DS Question 2 - Dec 5The top surface area of a square tabletop was changed so that one of the dimensions was reduced by 1 inch and the other dimension was increased by 2 inches. What was the surface area before these changes were made?
(1) After the changes were made, the surface area was 70 square inches.
(2) There was a 25 percent increase in one of the dimensions.
Source:
Official Guide | Difficulty: Hard
You can create equation:
(x-1)(x+2)=area2 and x^2=area2
1) we have (x-1)(x+2)=70
x^2+2x-x-2=70
x^2+x-72=0
(x+9)(x-8)=0
x is -9 or 8. Remeber, lengths cannot be neg so x=8. Suff
2) Only 1 dimension had increase and that was by 2 inches. Therefore 2=x(1+1/4)
Can clearly solve for x, sufficient.
D.
gmatophobia
DS Question 1 - Dec 5
Are the numbers k/4, z/3, and r/2 in increasing order?
(1) 3 < z < 4
(2) r < z < k
Source: Others | Difficulty: Medium
1 is clearly insuff
2) If k is pos and the rest are negative, no
If k is negative and the rest are negative, no
if all are pos maybe
3) z is between 3 and 4 and r
If we take extreme value of k, k/4 is much larger than z/3
But if K is a smaller value, say 6, then it is increasing order since z/3 will be larger.
E.
Rohstar750
gmatophobia
gmatophobia
DS Question 1 - Dec 04
Set T is a finite set of positive consecutive multiples of 14. How many of these integers are also multiples of 21?
(1) Set T consists of 30 integers.
(2) The smallest integer in set T is a multiple of 21.
Source: Veritas Prep | Difficulty: Hard
Explanation please.
LCM of 14 and 21=
7*3*2=42 Therefore any m42 in the set will be a m of 21.
14,28,42 - every third multiple of 14 will be a m42 which is also a m21.
The problem tells us that they are consectuive multiples - so all integers will be increasing and will be unique in the set.
1) tells us that there are 30 integers in the set.
We can test a bit:
If the set starts with m14, then:
m14,m14,m42,m14,m14,m42 - every second value of the set is m42 so 9 integers in the set will be m42.
If rhe set starts with m42 the pattern is the same:
m42, m14,m14,m42,m14,m14
if the second number is m42 the same:
m14,m42,m14,m14,m42
So we will always have 9 m42s in the set, so 9 m21s.
B) is not sufficient for us to determine
The smallest integer can be m21, if there is only 1 member of the set, there is only 1 m21 but if there are 9 members, it will be more than 1. NS.
A