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# 12 Easy Pieces (or not?)

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12 Easy Pieces (or not?) [#permalink]

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21 Jan 2012, 06:10
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After posting some 700+ questions, I've decided to post the problems which are not that hard. Though each question below has a trap or trick so be careful when solving. I'll post OA's with detailed solutions after some discussion. Good luck.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?
A. 1/5
B. 2/5
C. 3/4
D. 4/5
E. 1

Solution: 12-easy-pieces-or-not-126366.html#p1033919

2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?
A. 5
B. 6
C. 7
D. 18
E. 20

Solution: 12-easy-pieces-or-not-126366.html#p1033921

3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Solution: 12-easy-pieces-or-not-126366.html#p1033924

4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

Solution: 12-easy-pieces-or-not-126366.html#p1033925

5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?
I. c>a+b
II. c^2>a^2+b^2
III. c/a/b=10/6/2

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

Solution: 12-easy-pieces-or-not-126366.html#p1033930

6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Solution: 12-easy-pieces-or-not-126366.html#p1033932

7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1?
A. 11/14
B. 13/18
C. 4/7
D. 3/7
E. 3/14

Solution: 12-easy-pieces-or-not-126366.html#p1033933

8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?
A. 3
B. 5
C. 6
D. 16
E. 19

Solution: 12-easy-pieces-or-not-126366.html#p1033935

9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M?
A. 22
B. 30
C. 38
D. 46
E. 54

Solution: 12-easy-pieces-or-not-126366.html#p1033936

10. If $$n$$ is an integer and $$\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}$$, then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Solution: 12-easy-pieces-or-not-126366.html#p1033938

11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

Solution: 12-easy-pieces-or-not-126366-20.html#p1033939

12. If $${-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}$$ and $${-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}$$, what is the least value of $$x^2*y$$ possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

Solution: 12-easy-pieces-or-not-126366-20.html#p1033949

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Re: 12 Easy Pieces (or not?) [#permalink]

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29 Mar 2017, 13:59
Bunuel wrote:
12. If $${-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}$$ and $${-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}$$, what is the least value of $$x^2*y$$ possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of $$x^2*y$$, which obviously will be negative, try to maximize absolute value of $$x^2*y$$, as more is the absolute value of a negative number "more" negative it is (the smallest it is).

To maximize $$|x^2*y|$$ pick largest absolute values possible for $$x$$ and $$y$$: $$(-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}$$. Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).

Is it not the other way around where -1/100 is the least number and -1/18 is the max negative number. I would have picked A as the answer.

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Re: 12 Easy Pieces (or not?) [#permalink]

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29 Mar 2017, 21:38
vs224 wrote:
Bunuel wrote:
12. If $${-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}$$ and $${-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}$$, what is the least value of $$x^2*y$$ possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of $$x^2*y$$, which obviously will be negative, try to maximize absolute value of $$x^2*y$$, as more is the absolute value of a negative number "more" negative it is (the smallest it is).

To maximize $$|x^2*y|$$ pick largest absolute values possible for $$x$$ and $$y$$: $$(-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}$$. Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).

Is it not the other way around where -1/100 is the least number and -1/18 is the max negative number. I would have picked A as the answer.

No.

-1/18 = ~-0.06 and -1/100 = -0.01

-0.06 < -0.01

As you can see -1/100 is to the right of -1/18.

[Reveal] Spoiler:
Attachment:

MSP3701ihg50f15748di7300006532e8b04fda4c6c.gif [ 1.41 KiB | Viewed 475 times ]

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12 Easy Pieces (or not?) [#permalink]

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30 Mar 2017, 08:24
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Bunuel wrote:
SOLUTIONS:

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?
A. 1/5
B. 2/5
C. 3/4
D. 4/5
E. 1

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

Hello @Benuel,

I have a few queries related to the solution provided by you. Please throw some light on my following doubts:

1. We are give PAIRS of socks and the question asks us to pick up A SOCK and not a PAIR of socks.
2. Most importantly, nowhere in the question it's stated that each of the type of sock has to be picked. So i can have a case in which i pick up all 4 white socks.

Please guide me if my understanding of the question is wrong.

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Re: 12 Easy Pieces (or not?) [#permalink]

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28 May 2017, 22:08
Bunuel wrote:
11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211=prime.

Was it implicit that we could only use available digits only once ? I wasted time using same digits multiple times to form different no . Though I ended up with ( 77,73,61 ) which too gave 211 but I arrived at it by brute force. Was there something in the language that I should be careful to identify such scenarios so I don't make this mistake again ?

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Re: 12 Easy Pieces (or not?) [#permalink]

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28 May 2017, 23:34
booksknight wrote:
Bunuel wrote:
11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211=prime.

Was it implicit that we could only use available digits only once ? I wasted time using same digits multiple times to form different no . Though I ended up with ( 77,73,61 ) which too gave 211 but I arrived at it by brute force. Was there something in the language that I should be careful to identify such scenarios so I don't make this mistake again ?

We are given a data set which has three 7's in it. If we could use each number in the set multiple times, then why 7's were written three times?
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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 04:44
Bunuel wrote:
2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?
A. 5
B. 6
C. 7
D. 18
E. 20

Also tricky. Notice that $$x$$ can take positive, as well as negative values to satisfy $$9<x^2<99$$, hence $$x$$ can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of $$x_{max}-x_{min}$$, ans since $$x_{max}=9$$ and $$x_{min}=-9$$ then $$x_{max}-x_{min}=9-(-9)=18$$.

That is tricky. I took 3<x<10
So minimum value of x= 4
Maximum value of x= 9

Bunuel
When a question involves ^2 or any even power, then we have to consider the negative value?

Will that be the case in all situations?

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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 04:50
Shiv2016 wrote:
Bunuel wrote:
2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?
A. 5
B. 6
C. 7
D. 18
E. 20

Also tricky. Notice that $$x$$ can take positive, as well as negative values to satisfy $$9<x^2<99$$, hence $$x$$ can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of $$x_{max}-x_{min}$$, ans since $$x_{max}=9$$ and $$x_{min}=-9$$ then $$x_{max}-x_{min}=9-(-9)=18$$.

That is tricky. I took 3<x<10
So minimum value of x= 4
Maximum value of x= 9

Bunuel
When a question involves ^2 or any even power, then we have to consider the negative value?

Will that be the case in all situations?

Yes, x^2 = (positive integer) has two solutions for x, positive and negative.
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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 04:54
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

12 easy pieces ? I don't think so.
Bunuel what is the level of these questions? 700 in disguise.

I have a doubt. Why did you take max value of y as 9 when its not included in the range?

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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 04:58
Shiv2016 wrote:
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

12 easy pieces ? I don't think so.
Bunuel what is the level of these questions? 700 in disguise.

I have a doubt. Why did you take max value of y as 9 when its not included in the range?

Check here: http://gmatclub.com/forum/12-easy-piece ... l#p1268538
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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 05:03
Bunuel wrote:
10. If $$n$$ is an integer and $$\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}$$, then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Bunuel . First easy question I got correct.

But I solved it differently:

1/10^(n+1) <737/10^5 <1/10^n

= 10^5/10^(n+1) <737< 10^5/10^n

= 10^(4-n)<737<10^(5-n)

This is possible only when n= 2

Then
100<737<1000

Does this makes sense?

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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 05:10
Shiv2016 wrote:
Bunuel wrote:
10. If $$n$$ is an integer and $$\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}$$, then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Bunuel . First easy question I got correct.

But I solved it differently:

1/10^(n+1) <737/10^5 <1/10^n

= 10^5/10^(n+1) <737< 10^5/10^n

= 10^(4-n)<737<10^(5-n)

This is possible only when n= 2

Then
100<737<1000

Does this makes sense?

Yes, that's correct. +1.
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Re: 12 Easy Pieces (or not?) [#permalink]

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11 Jul 2017, 05:19
Bunuel wrote:
mamacta wrote:
Bunuel Can you please explain ques no. 12 ? how is the least value -1/18 ? why is it not -1/100 ? I am not understanding it.

Because -1/18<-1/100.

Exactly. Now I get it.

On a number line, -1/18 is on the left of -1/100.

or simply: -0.05 <-0.01

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Re: 12 Easy Pieces (or not?) [#permalink]

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12 Jul 2017, 17:28
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

Hi Bunuel,

Is this option possible: *R*R_R*R*R*? Or "in a row" means we can not have any space in between?
Thank you

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Re: 12 Easy Pieces (or not?) [#permalink]

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12 Jul 2017, 21:26
1
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Soul777 wrote:
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

Hi Bunuel,

Is this option possible: *R*R_R*R*R*? Or "in a row" means we can not have any space in between?
Thank you

In R*R*R*R*R* each * is a place for 2 blue, 2 green and 1 yellow marbles. So, * cannot accommodate say 2 marbles.
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Re: 12 Easy Pieces (or not?) [#permalink]

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17 Aug 2017, 12:35
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

I completely understand this way,but can you please help me with an another approach of total-always together?

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Re: 12 Easy Pieces (or not?) [#permalink]

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17 Aug 2017, 21:30
himanshukamra2711 wrote:
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

I completely understand this way,but can you please help me with an another approach of total-always together?

It will be very time consuming as there will be many cases to consider.
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Re: 12 Easy Pieces (or not?) [#permalink]

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18 Aug 2017, 14:21
Bunuel wrote:
3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

I couldn't understand (25+65)*1.5=135 miles apart i above solution. Why are we multiple combined rate with 1.5.

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Re: 12 Easy Pieces (or not?) [#permalink]

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19 Aug 2017, 00:06
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LalaB wrote:

4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
(y-x )min=Y min -Xmax =-7-5=-12
(y-x )max=Y max -Xmin =9-(-3)=12

Hi All,

Just a query, here why are we considering Ymin as -7 when the inequality states y lies between -7 and 9 but the greater than equal to sign is missing.

Pls help in case I am failing to understand a very basic concept .

Thanks :)

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Re: 12 Easy Pieces (or not?) [#permalink]

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19 Aug 2017, 03:39
ammuseeru wrote:
Bunuel wrote:
3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

I couldn't understand (25+65)*1.5=135 miles apart i above solution. Why are we multiple combined rate with 1.5.

Because 1.5 hours before they meet, the distance left to cover would be (combined rate)*(time) = (25+65)*1.5 = 135 miles.
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Re: 12 Easy Pieces (or not?) [#permalink]

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19 Aug 2017, 03:40
narayani20 wrote:
LalaB wrote:

4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
(y-x )min=Y min -Xmax =-7-5=-12
(y-x )max=Y max -Xmin =9-(-3)=12

Hi All,

Just a query, here why are we considering Ymin as -7 when the inequality states y lies between -7 and 9 but the greater than equal to sign is missing.

Pls help in case I am failing to understand a very basic concept .

Thanks :)

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Kudos [?]: 124124 [0], given: 12072

Re: 12 Easy Pieces (or not?)   [#permalink] 19 Aug 2017, 03:40

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