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srcc25anu

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01 Apr 2013, 15:30

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Q1.

Distance between y axis and K = 1/3 ( distance between x axis and K)
Given co-ordinates of Point K = (-3, Y)
hence distance from Y axis = |-3| or 3 units
therefore distance from X-axis should be |-9| or 9 units

hence E

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01 Apr 2013, 18:20

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6. A team contributes total of $399 from its members. If each member contributed at least $10, and no one contributed $19, what is the greatest number of members the club could have?

Going by the options given : Suppose each member paid exactly 10 dollars.

A. 37 = 37*10 = 370, balance = 29 which we cant tell who paid what some one might have paid more then 19 alos
B. 38 = 38*10 = 380 , balance = 19 which No one paid 19 dollars so remove
C. 39 = 39*10 = 390 , balance 9 dollars which can be paid by any member
D. 40 = exceeds 399 dollar
E. 41 = exceeds 399 dollars

So , taking 39 members as the answer I dont know whether my approach is right or not.

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1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K. If the coordinates of K are (-3, y), what is the distance between point K and X-axis?

A. 1/2
B. 1
C. 3
D. 4.5
E. 9

Since (-3, y) are the coordinates of K, -3 is the distance from K to the Y-axis and y is the distance from the X-axis to K.

Given that the distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K, Thus : 3 = 1/3 * y --> y = 9

Answer : E

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01 Apr 2013, 22:16

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3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours re ,spectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x?

A. 1
B. 1.25
C. 2
D. 2.5
E. 4

A alone can do the work in 10 hours
B alone can do the work in 5 hours
C alone can do the work in x hours

Hence, in 1hour A alone will do W/10 and B alone W/5 and C alone W/x

From the figure below, W = W/5 + 3W/5 + W/40 + W/20 + W/4x -----> x = 2

Answer : C

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Rock750

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01 Apr 2013, 23:08

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4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool?

A. 9 liters
B. 18 liters
C. 27 liters
D. 36 liters
E. 45 liters

To empty X liters, 4 hours are needed

To empty X + (3*6) liters, 6 hours are needed

Hence, $X/4 = (X+18)/6$ --> $X=36$

agrchandan

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02 Apr 2013, 05:07

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3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x?

A. 1
B. 1.25
C. 2
D. 2.5
E. 4

A alone can complete the task in 10 hours, in 1 hour he can do 1/10 of the task
B alone can complete the task in 5 hours, in 1 hour he can do 1/5 of the task
C alone can complete the task in x hours, in 1 hour he can do 1/x of the task
A and B combined can do (1/10 + 1/5) = 3/10 of the task in 1 hour hour
A, B and C combined can do (1/10 + 1/5 + 1/x) = (3/10 + 1/x) of the task in one hour

As per question, A starts the work and works for 2 hours so he completes 1/10 * 2 of the task i.e. 1/5 of the task
Then A and B work for next two hours i.e they complete 3/10 *2 = 3/5 of the task
Remaining task = 1 - (1/5 + 3/5) = 1/5
This remaining task is done by A, B and C combined in 15 mins.
So 1/5 = (1/10 + 1/5 + 1/x)* 15/60, that gives x =2 (option c)

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04 Apr 2013, 09:37

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2. What is the area of a region enclosed by |x/3|+|y/9|=10?

A. 675
B. 1350
C. 2700
D. 5400
E. 10800

Area is enclosed by following four lines:
3x + y = 90 Passing through (0, 90) and (30, 0)
3x - y = 90 Passing through (0, -90) and (30, 0)
- 3x + y = 90 Passing through (0, 90) and (-30, 0)
- 3x - y = 90 Passing through (0, -90) and (-30, 0)
This makes rhombus area with diagonals of length 60 and 180.
Area of the region = d1*d2/2 = 60*180/2 = 5400

Correct answer is D.

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04 Apr 2013, 09:39

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4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool?

A. 9 liters
B. 18 liters
C. 27 liters
D. 36 liters
E. 45 iters

Rain inflow in 6 hrs = 18 lt.
==> Draining pipe took extra 2 hrs to drain out extra 18 lt. water
==> Draining pipe can clear 36 lt. in 4 hrs.
==> Capacity of the pool is 36 liters.

Correct answer is D.

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7. Mary spent 64 percent of her salary on food (including meat) and 16% of her salary on meat. What percent of the salary spent on food were not spent on meat?

A. 16%
B. 25%
C. 32%
D. 48%
E. 75%

Assuming salary as $100, $64 is spend on food and $16 spent on meat.
So, 16/64 = 1/4 = 25% of the food expenditure was spent on meat.
This means 75% of the salary spent on food was not spent on meat.

Correct answer is E.

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10. If n is a non-negative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true?

I. n^2 divided by 4 yields the reminder of 1
II. (-2)^n is less than 0
III. n is a prime number

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

When 3^n is divided by 4, remainder is a multiple of 3 ==> n is odd.
I. For all odd n’s, remainder will be 1 if n^2 is divided by 4. Correct.
II. As n is odd, (-2)^n is –ve. Correct.
III. n can be 9 or 15 or some other composite odd number. Incorrect.

Correct answer is D.

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11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Total number of marbles is 8.
For any given position, probability for red marble is 5/8 and blue marble is 3/8.
So, the probability that the seventh marble drawn is NOT blue is 5/8

Correct answer is D.

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Bunuel

2. What is the area of a region enclosed by |x/3|+|y/9|=10?
A. 675
B. 1350
C. 2700
D. 5400
E. 10800

Find the x and y intercepts.

When y=0, then x=30 or x=-30.
When x=0, then y=90 or x=-90.

So, we have 4 points: (30, 0), (-30, 0) (0, 90), (0, -90). When joining these points we get the rhombus:

Attachment:

2.png

The area of a rhombus is $\frac{d_1*d_2}{2}$ (where $d_1$ and $d_2$ are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400.

Answer: D.

Bunuel what about the following points that also meets the given conditions of |x/3| + | y/9| = 10 but come with a different area.

(x,y) = (-3,81)
(x,y) = (-3,-81)
(x,y) = (3,-81)
(x,y) = (3,81)

In this case Area of the rectangle will be = 972

or you can have

(x,y) = (6,72)
(x,y) = (6,-72)
(x,y) = (-6,-72)
(x,y) = (-6,72)

In this case the area of the rectangle will be = 1728.

Can you please clarify

Bunuel

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20 Apr 2013, 05:15

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pikachu

Bunuel

Attachment:

2.png

The area of a rhombus is $\frac{d_1*d_2}{2}$ (where $d_1$ and $d_2$ are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400.

Answer: D.

Below is the graph of |x/3|+|y/9|=10:

The points you mention are on it and not making some other figure.

Nilabh_s

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Bunuel

11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Basically we need to find the probability that the seventh marble drawn is red (so not blue).

Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.

The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).

Answer: D.

Hi Bunuel,
Why is the probability not changing? If we keep withdrawing marbles withoout replacing then following things are changing:-
1. total no. of marble (hence the denominator in probability)
2. No. of Red marbles left.
3. No. of blue marbles left.

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Nilabh_s

Bunuel

11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Basically we need to find the probability that the seventh marble drawn is red (so not blue).

Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.

The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).

Answer: D.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope this helps.

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01 Jan 2014, 19:21

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Bunuel

Nilabh_s

Bunuel

.Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Hope this helps.

For the statement
" NOW that you pick a spade out of 5 remaining cards"

If you throw 3 cards then how are we sure than 5 spades are still remaining. Following three cards may have been thrown:-
1. All 3 spades.
2. All three hearts.
3. 2 spade & 1 heart & vice-versa.

Hence how can we be sure that 5 spades are till remaining?

Bunuel

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02 Jan 2014, 05:51

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Nilabh_s

Bunuel

Yes, but we don't know which cards are thrown and this won't affect the probability. Please follow the links in my previous post for better understanding of such problems.

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17 Jan 2014, 05:57

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Bunuel

Nice question Bunuel!!!! +1 for it.

Here is my take.

We have 5 red marbles and 3 red marbles. We just want that in the 7th pick blue marble should not appear. That indirectly means Red marble should appear.
We can visualize it as arrangement of 8 marbles (of which 5 are red and 3 are blue) in such a way that 7th marble will be red.
We can fix 1 red marble at 7th place and can arrange remaining marbles (Total 7 = 4 Red + 3 Blue) as $\frac{7!}{4!3!}$ = This is our desired outcome

$Probability = \frac{Desired Outcomes}{Total Outcomes}$

Desired Outcomes = $\frac{7!}{4!3!}$ = 5

Total outcomes = Arrangement of all 8 marbles ( 8 = 5 Red + 3 Blue) at 8 places = $\frac{8!}{5!3!}$ = 8

Probability = $\frac{5}{8}$ = D

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Bunuel

9. If x and y are integers and x + y = -12, which of the following must be true?
A. Both x and y are negative
B. xy > 0
C. If y < 0, then x > 0
D. If y > 0, then x < 0
E. x - y > 0

Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.

Answer: D.

I think C & D are right. If x is +ve i.e. 12 and y is -ve i.e. -24 then sum of x and y will be negative. Please explain
-Neha

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01 Jun 2014, 02:33

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nehamodak

Bunuel

I think C & D are right. If x is +ve i.e. 12 and y is -ve i.e. -24 then sum of x and y will be negative. Please explain
-Neha

C is not always true: if y is negative it's not necessary x to be positive. Consider y=-1 and x=-11.

But D must be true: if y is positive, then x must be negative in order the sum of x and y to be negative.

Hope it's clear.