New Set of Mixed Questions!!!

8. Usually Holly leaves home to school at 9:00, however today she left home 20 minutes later. In order to be at school on time she increased her usual speed by 20% and still was at school 15 minutes later than usual. What is her usual time from home to school?
A. 15 minutes
B. 20 minutes
C. 25 minutes
D. 30 minutes
E. 210 minutes

Let the usual speed be \(s\) and usual time \(t\) minutes, then as the distance covered is the same we will have: \(st=1.2s*(t-20+15)\) --> \(t=30\) minutes.

Answer: D.

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: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.

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

After 2 hours \(2*\frac{1}{10}=\frac{1}{5}\) of the taks will be done (as only A works);

After 4 hours \(\frac{1}{5}+2*(\frac{1}{10}+\frac{1}{5})=\frac{4}{5}\) of the task will be done and \(\frac{1}{5}\) will be left to be done;

We are told that \(\frac{1}{5}\)th of the task is done in 15 minutes (1/4th of an hour) by all three workers: \(\frac{1}{4}*(\frac{1}{10}+\frac{1}{5}+\frac{1}{x})=\frac{1}{5}\). From which we can find that \(x=2\) hours.


Answer: C

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%

64% of her salary on food;
16% of her salary on meat;

64%-16%=48% on food but not on meat --> 48/64=3/4=75% of the salary spent on food were not spent on meat.

Answer: E.

5. For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)

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

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:

\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:

\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:

\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D.

SOLUTIONS:

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

Point K has the coordinates (-3, y) means that it's somewhere on the line x=-3. Hence the distance from this point to the Y-axis is 3 units.

Since the distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K, then the distance from K to the X-axis is 9 units.

Answer: E.

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

3^0=1 --> the remainder when 1 is divided by 4 is 1;
3^1=3 --> the remainder when 3 is divided by 4 is 3;
3^2=9 --> the remainder when 9 is divided by 4 is 1;
3^3=27 --> the remainder when 27 is divided by 4 is 3;
...

We can see that in order the condition to hold true n must be odd.

I. n^2 divided by 4 yields the reminder of 1 --> odd^2 divided by 4 always yields the reminder of 1. So, this statement must be true.

II. (-2)^n is less than 0 --> (-2)^odd<0. So, this statement must be true.

III. n is a prime number. Not necessarily true.

Answer: D.

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?
A. 37
B. 38
C. 39
D. 40
E. 41

Obviously, the team could not have 40 or more members, since $10*40 = $400 > $399. What about 39? If 37 members contribute $10 each ($10*37 = $370) and the remaining two members contributed, for example, $11 and $18, respectively, then the total number of members the team would have is 37 + 1 + 1 = 39.

Answer: C.

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

Note that the question asks which of the following statements must be true, rather than could be true.

Let's evaluate the options:

A. Both \(x\) and \(y\) are negative. This option is not always true, consider \(x = -20\) and \(y = 8\).

B. \(xy > 0\). This option is not always true either, consider \(x = -20\) and \(y = 8\).

C. If \(y < 0\), then \(x > 0\). This option is not always true, consider \(y = -2\) and \(x = -10\).

D. If \(y > 0\), then \(x < 0\). If \(y\) is positive, then \(x\) must be negative in order for the sum of \(x\) and \(y\) to be negative.

E. \(x - y > 0\). This option is not always true, consider \(x = -20\) and \(y = 8\).

Therefore, only option D is ALWAYS true: if \(y\) is positive, then \(x\) must be negative in order for the sum of \(x\) and \(y\) to be negative.


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.

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?

To empty X liters ==> 4 h.
To empty X + 3*6 liters ==> 6 h.

\(\frac{x}{4}=\frac{x+18}{6}\)

\(x=36\)

IMO D

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

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.

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?

rates : A=1/10 B=1/5 c=1/x

A*2h+(A+B)*2h+(A+B+C)*15min = 1 work.

(1/10)*2 +(1/10+1/5)*2+(1/10+1/5+1/x)*1/4=1

x=2 IMO C

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

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

Work done by :
A in 2 hrs = 2/10.-------(1)
A and B in 2 hrs = 2(1/10 + 1/5)--------(2)
A,B and C in 15 mins = 1/4(1/10 + 1/5 + 1/x)---------(3)
(1)+(2)+(3) =1

on simplification
4/5 + 1/4(3/10 + 1/x) = 1

or 1/x= 4/5 -3/10 = 5/10 = 1/2

X= 2 hrs

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.