Quant Question of the Day Chat

Total distance=3x
Time B to A= Time A to B+ (1/3)
(2x/30)+(x/10)=(3x/20)+(1/3)
x/60= (1/3)
x= 20 Km
3x= 20*3= 60 Km

even i used the answer option it came 60

Total distance is distance from A to B + distance from B to A = 2d

Which is 2*60 = 120

Yes. Total distance is 120 KM

DS Question 1 - June 17

If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digits X and Y, is 3(2.00X) > 2(3.00Y) ?

(1) 3X < 2Y
(2) X < Y − 3

Source: Official Guide | Difficulty: Hard

PS Question 1 - June 17

The difference 942 — 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc — cba is a positive multiple of 7 ?

A. 142

B. 71

C. 99

D. 20

E. 18

Source: Official Guide | Difficulty: Hard

E, 100a+10b+c-(100c+10b+a)= 99(a-c)==> b can take any value from 1,2,..9 & 99(a-c) = 7*K, where (a-c) has to be the multiple of 7. It is possible only in 9_2 or in 8_1 case and we can put any of the digits from 9 available ones. Hence, 18 no. of combinations are possible.

Great Question!
We can rewrite the stem as:
3(2+x/1000)>2(3+y/1000)
3x>2y?

1)
The inverse of our choice.Suff.

2) x<y-3
This is where it gets slightly tricky.
First, you have to realize that x and y are digits. That means they can take on a value between 0 and 9 inclusive.
As x cannot be negative, y must be greater than 3 - otherwise the inequality would force x to take a negative value, which is not possible
so y=4,5,6,7,8,9
We can test cases:
if y=4
x<1, x must be 0 - clearly 2y>3x
if y=5
x<2, x must be 0 or 1, in either case 2y>3x
if y=6
x<3, x must be 0, 1, 3 - in any case 2y>3x

and so on
In all cases, 2y>3x.
So B is sufficient as well.

D

If n = s^a*t^b, where a, b, s and t are integers, is n an integer?

(1) a+b is an even number
(2) a is an even number

1) a+b is even, in this case say a is 3 and b is -1 then n may not be an integer but if a=3 and b=1 then n is integer --> not conclusive
2) a is even and say b is negative number then n is not integer but if b is positive then n is integer --> again not conclusive.
Same in case of 1)&2) both taken simultaneously.
Therefore both statement not sufficient

Realize this:
n=(s^a)(t^b)
the only way possible for n to not be an integer, is if a or b were negative - otherwise, n will always be some integer as S and T are integers.

1) a+b=even
A or B can be both negative, positive, or a mix. NS.

2) a is even
Again, this does not tell us what the sign of A is. A can be negative or positive

3) all we know is that a is some even integer, and b is some even integer. They can be positive or negative. NS.

E

[m] is defined as the greatest integer less than or equal to m, what is the value of [m]?

(1) 1 < m < 2
(2) |m| <1

can you explain how... a is right

statement 1 : 1<m<2

if

m=1.5 then also it is 1

taking any value between 1 aqnd 2 the [m] will always be 1

but in statement 2 : -1<m<1 if m= -0.2 [m]=-1 and if m=0.2 [m] =0

hence A is sufficient

1) Tells us that m can be some value between 1 and 2.
We can take two cases:
m=1.001
in this case, [m]=1
m=1.9999
in this case, [m]=1

[m] will always be 1 in this range, suff

2) abs(m)<1
-1<m<1
Here we have a few different cases
Say m=-0.5
[m]=-1
But if m=0.5
[m]=0

INS.
A

great guys, thank you

DS Question 1 - June 19

Is |1 - 4k| > k?

(1) k > 4x^3
(2) k < 2x – x^2 - 2

Source: GMATPrepNow | Difficulty: Hard

PS Question 1 - June 19

A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed?

(A) 18 days
(B) 27 days
(C) 26.67 days
(D) 16 days
(E) 12 days

Source: GMAT Paper Tests | Difficulty: Hard

They work for a total of t days
Person A will therefore contribute 1/20(d-10)
and person B will therefore contribute (1/30)(d)

which means:

1/20(d-10)+1/30(d)=1
d/20-1/2+d/30=1
d/20+d/30=3/2
3d+2d=90
5d=90
d=18
A


d days, not t days


Great Question!
First thing to notice is that the above inequality will always hold if k is some negative value (since abs has minimum value of 0, which is always greater than any negative number).
You can solve the equation to get more specific ranges:
k<1/5 or k>1/3

1)
Suppose x=2
then we have k>4*8>1/3 (always yes)
Suppose x=-2
k>-32 (maybe yes, maybe no - if k<1/5, yes if 1/5<k<1/3, no)
ins

2) k<-x^2+2x-2
This is a tricky quadratic to factorize. So lets plugin some values:
if x=0,
k<-2, always true
if x=1,
k<-1+2-2
k<-1, target statement always true
if x=2
k<-4+4-2
k<-2 (always true).
if x=-2
k<-4+2(-2)-2
In any case, k is some negative value - which means the target equation must always hold true, as the abs(anything) is 0 or greater.
B should be correct


2nd statement is probably best analyzed graphically - but that might be overkill for most gmat questions haha