Quant Question of the Day Chat

What all are them?


Yeah even I got the same.

Did I do this right? A

x^3-3x^2+2x
x(x-2)(x-1)=0
x=0,1,2.
X cannot be 0 because positive. X can also not be 1, otherwise Z would be 0 (z is pos). X must therefore be 2. /2 has R0 or 1. because Z cannot be 0 it must be 1.

Absolutely correct...

PS Question 1 - Dec 28

The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36

B. 30 < x < 37

C. 31 < x < 37

D. 31 < x < 38

E. 32 < x < 38

Source: GMAT Club Tests | Difficulty: Hard
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DS Question 1 - Dec 28

If p, q and r are integers, and pq + r is an odd integer, is p an even integer?

(1) pq + pr is an even integer.
(2) p + qr is an odd integer.

Source: Expert’s Global | Difficulty: Hard
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DS Question 2 - Dec 28

Is A > B?

(1) A/B > 1
(2) A > 2B

Source: EMPOWERgmat | Difficulty: Hard
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1) We can have the case of integers, in which case A>B. We can also have the case of fractions: (1/2/(1/3)) for example. Insuff.
2) If A>2B
Iwe can have A=-4, B=-3. Which is no, or A=2 and B=-3 which is yes.
3) Putting the two, we only know that they share the same sign.
We can still have the case of A=-4 and -3 or A=-2 and B=-3.
Clearly insuff.

E.

Tricky sign based question!


For these problems its handy to list out the cases possible in the stem.
If pq+r=O, then we can have
pq=e, r=o or pq=o, r=e.
This means that we can have:
p=e, q=e, r=o
p=o, q=e, r=o
P=e, q=o r=o
P=o, q=o, r=e

now statement 1)
p(q+r)=e.
Lets check the cases where p=odd
if p=odd, q+r must be even. that means q and r are either both odd or both even.
Notice that from the working in the stem, if p is odd, q and e must be opposing odd/even. So this is not possible. A is suff.

2)
p+qr=odd
we can have p=e, and qr=odd, which would be the 3rd case from the stem
or
p=o, qr=even, which is case 2 from the stem. INS.

A.

PS Question 2 - Dec 28

When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3
B. 1/3
C. 1/4
D. 1/6
E. 1/9

Source: GMAT Club Tests | Difficulty: Hard
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PS Question 1 - Dec 29

x is a positive integer and has 8 factors while (x+1) has 4 factors. If x<100 then how many values of x are possible?

A) 0
B) 1
C) 2
D) 3
E) 4

Source: GMAT Whiz | Difficulty: Hard

DS Question 1 - Dec 29

If x and y are nonzero integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2y, the remainder is 4.

(2) When x + y is divided by y, the remainder is 4.

Source: Kaplan | Difficulty: Hard
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B. Surprised to find this in the Hard category to be honest

[quote="gmatophobia"]PS Question 1 - Dec 29 x is a positive integer and has 8 factors while (x+1) has 4 factors. If x100, so the cases stop here.

Case 2: x=a^3*b
note that a can only be 2. for 3, its already at 81 and any other factor will bring x over 100.
x=8*3=24, 25 is not prime but it does not work (25=5^2 which has 3 factors, not 4).
x=8*5=40, 41 is prime.
x=8*7=56, 57 is prime.
x=8*11=88, 89 is prime.
x=8*13 goes over 100, so we can stop here.

I hope im right, I tend to forget cases with long qs like this

aaand i just realized that i said 3^3 was 81, and not 27

The question is marked as 75% hard

DS Question 2 - Dec 29

If BC = BD, what is the length of BC?

(1) x = 30
(2) AD = 1

Source: Magoosh | Difficulty: Hard
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1) Using x = whatever degree value , one cannot find length of any particular side

In triangle BDC , if BC=BD then angle D=C=x, Also angle ABD = 2x

In triangle ABD, angle A=B=2x, consequently side AD=BD=1

Taking all the details into account, AD=BD=BC=1, Hence Statement 2 is sufficient.
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haha, and 57 is not prime, 57 = 3*19. I found 2 values 54 & 56. My answer is C

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DS Question 1 - Dec 30

Is x^2 + y^2 > 100?

(1) 2xy 200

Source: Vertias Prep | Difficulty: Hard

PS Question 1 - Dec 30

List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E - S ?

I. -16
II. 6
III. 10

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Source: Official Guide | Difficulty: Hard
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PS Question 2 - Dec 30

If F < 0, which of the following represents the largest value?

A. (2F - 5)/2

B. (4F - 1)/4

C. (6F - 2)/6

D. (7F - 5)/7

E. F - 1

Source: Kaplan | Difficulty: Medium
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neat question. Can be solved with some logic
A and E can be instantly eliminated - they take away more than -1.
D is F-(5/7), B is F-(1/4) and C is F-(1/3). Clearly, because 1/4 is the smallest fraction, B will give the F closest to 0, hence the biggest value.

Was reviewing some Magoosh quant content…. Is the calculation on the photo right?

Excuse me I got it now