Quant Question of the Day Chat

PS Question 1 - May 29

A farmer decides to plant a row of trees along one side of a road. He decides to plant one tree every 15 meters. If the road is 460 meters long what is the maximum number of trees the farmer could plant?

A. 28
B. 29
C. 30
D. 31
E. 32

Source: GMATNinja | Difficulty: Medium

DS Question 1 - May 29

If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A), is the area of the triangle greater than 15 ?

(1) A < 3
(2) The triangle is right

Source: GMAT Club Test | Difficulty: Hard

starting from the "0" meter mark: 1
adding increments of 15 until the last set of 15:
should be around 30
15*3=30+15=450
so a total of 31 trees. - 1 at the 0 meter mark, and then 30 for each 15 afterr that, untill the 450 meter mark in which no more trees can be planted

Not clear, does anyone have another approach?

Sorry to hear that the explanation was not clear. Let me take another stab at it.

I am taking smaller values of the same question for this explanation. The concept can be extended to the question that you’ve posted.

Julian (J), Derek (D) and George (G) together have 4 identical
sugarcane stalks to share. If each of them can be
given any whole number of stalks from O to 4, in how
many different ways can the stalks be distributed?


Note: The question doesn’t mention any constraint, which means J D and G can get 0 stalks or can get all 4.

The sugarcane stalks are represented as X.

Let’s assume that the four stalks are represented as -

X X X X

We will partition (divide & distribute) the stalks among the three individuals using |. To do so we need two partitions as shown below -

X | X X | X

In this representation ⇒ One stalk is distributed to one person, two stalks are distributed to the second person and the last stalk is distributed to the third person.

One of the other possible arrangements could be

X X X | X |

In this representation ⇒ Three stalks are distributed to one person, one stalk is distributed to the second person and zero (no stalk) is distributed to the third person

Similarly we can have other arrangments as shown below

| X | X X X
X | X | X X
..
..
..
..

So essentially the number ways the stalks can be distributed to J, D and G equals various arrangments of the four Xs & two |s (similar finding number of possible arrangements the alphabets in the word MISSISSIPPI)

We have 4 stalks (Xs) & 2 partitions (|s). Hence, the number of ways of arranging the six items is

6! / (2!*4!) = 15.

You can now extend this concept to the problem shared earlier. Hope this helps !

Pls. excuse typos.

Sorry to hear that the explanation was not clear. Let me take another stab at it.

I am taking smaller values of the same question for this explanation. The concept can be extended to the question that you’ve posted.

Julian (J), Derek (D), and George (G) together have 4 identical
sugarcane stalks to share. If each of them can be
given any whole number of stalks from O to 4, in how
many different ways can the stalks be distributed?


Note: The question doesn’t mention any constraint, which means J D and G can get 0 stalks or can get all 4.

The sugarcane stalks are represented as X.

Let’s assume that the four stalks are represented as -

X X X X

We will partition (divide & distribute) the stalks among the three individuals using |. To do so we need two partitions as shown below -

X | X X | X

In this representation ⇒ One stalk is distributed to one person, two stalks are distributed to the second person, and the last stalk is distributed to the third person.

One of the other possible arrangements could be

X X X | X |

In this representation ⇒ Three stalks are distributed to one person, one stalk is distributed to the second person, and zero (no stalk) is distributed to the third person

Similarly, we can have other arrangments as shown below

| X | X X X
X | X | X X
..
..
..
..

So essentially the number of ways the stalks can be distributed to J, D, and G equals various arrangments of the four Xs & two |s (similar to finding the number of possible arrangements of the alphabet in the word MISSISSIPPI)

We have 4 stalks (Xs) & 2 partitions (|s). Hence, the number of ways of arranging the six items is

6! / (2!*4!) = 15.

You can now extend this concept to the problem shared earlier. Hope this helps!

Pls. excuse typos.

Given: Julin + Derek + George = 7

let 7 sugarcane be 7 sticks,
| | | | | | |

And we have to asigen ’+’ symbol in between these sticks,
we are doing this because we can have multiple ways
example - ( | | | + | | + | | )
this is one way of doing it, where J got 3, D got 2, G got 2.

So, Directly we can do it as we have 6 gaps between these 7 sticks and we have to put-
- two ’+’ symbol to satisfy above equation in 6 gaps so we can do it in,
6C2 ways which is:
6!/2!.4! = 15

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?

(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040

Let’s bring some twist to this question.

The number of donuts (all identical) is 10 and each person (L, M, and D) must get to get at least two donuts. In how many different ways can the donuts be distributed?

Let’s bring some twist to this question.

The number of donuts (all identical) is 10 and each person (L, M, and D) must get at least two donuts. In how many different ways can the donuts be distributed?

If 2^x+2^y=x^2+y^2, where x and y are nonnegative integers, what is the greatest possible value of |x−y|?
a.01234
Manhattan GMAT

Q. 2^x+2^y=x^2+y^2, where x and y are nonnegative integers, what is the greatest possible value of |x−y|?
a. 0 b. 1 c. 2 d. 3 e. 4
Source: Manhattan GMAT: 700 level

No. How are you solving this?

gmatophobia, Can you teach concepts of circles,tangents,parabola teted in gmat

A

Because SD always positive or zero. Never negative.

as sd cant be negative

But SD is zero only when all values are identical right?

How to scroll up in this chat. Whenever I try to go up, it reloads the chat

2^3+2^0=3^2+0^2=9
Therefore 3-0=3

DS Question 1 - May 31

Is p > q ?

(1) p^2 > q

(2) p^3 < q

Source: Expert’s Global | Difficulty: Hard

PS Question 1 - May 31

Let x be a two-digit number. If the sum of the digits of x is 9, then the sum of the digits of the number (x + 10) is

(A) 1

(B) 8

(C) 10

(D) either 8 or 10

(E) either 1 or 10

Source: Others | Difficulty: Hard

Option D
1. When x is negative: sum will be 8
2. When x is positive: sum will be 10

But intial statement states that sum of digits is 9.. how can negative number have sum of digits as 9?