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1

Qs. 1 There are these 8 numbers in a set 9.4,9.9,9.9,9.9,10.0,10.2,10.2,10.5 The mean and standard deviation of the 8 numbers are 10.0 and 0.3 respectively, what percent of the 8 numbers are within 1 standard deviation of the mean?
a) 90%
b) 85%
c) 80%
d) 75%
e) 70%

Answer: D
mean = 10
sd = 0.3
"one standard deviation" = 1*(sd) = 0.3.
1 stdev range would be mean plus/ minus std Dev
Lower range:10 - 0.3 = 9.7
Higher range:10 + 0.3 = 10.3
6 numbers out of 8 fall within the range
Then 6/8 or 3/4 are within range
3/4 = 75%
Hence answer is D

2

Qs. 2: The fourth grade at school X is made up of 300 students who have a total weight of 21,600 pounds.
If the weight of these four graders has a normal distribution and standard deviation equals 12 pounds, approximately what percentage of the fourth graders weighs more than 84 pounds?

a 12%
b 16%
c 36%
d 48%
e 60%

Answer: B
Mean = 21600/300 = 72
Mean + 1sigma (std dev) = 72 + 12 = 84
+/- 1 covers 68.2% (34.1% + 34.1%) of the normal distribution. Therefore, remain distribution of data 100% - 68.2% = 31.8%.
Since we care only about data points above 84 lbs => 31.8 / 2 = 15.9% = 16%

3

Qs. 3 A teacher prepares a test. She gives 5 objective type question out of which 4 have to be answered. Find the total ways in which they can be answered if the first 2 questions have 3 choice and last 3 have 4 choice.

Answer:816

Options available

1. Get first one wrong. X*3*4*4*4= 192
2. Get 2nd one wrong 3*X*4*4*4 = 192
3. Get 3rd one wrong 3*3*X*4*4= 144
4. Get 4th wrong 3*3*4*X*4=144
5. Get 5th wrong 3*3*4*4*X=144

Hence 144+144+144+192+192= 816

4

Qs. 4 The letters of the word PROMISE have to be arranged so that no two vowels come together. Find the number of arrangements.

word PROMISE has vowels O,I,E
if you chose two vowels then you will have following combinations

OI - it can also also be arranged as IO
IE - it can also be arranged as EI
OE - it can also be arranged as EO

Consider one combination OI
Now you have PRMSE and OI
Following are some of the combinations that can be obtained
ESMRPOI
ESMRPIO
and
SMRPEIO - > Here all the 3 vowels are together
So just accounting for two vowels at a time should cover everything.

Thus for each vowel pair we have invalid combinations as
2 * 5! -> for OI and IO
2 * 5! -> for IE and EI
2 * 5! -> for EO and OE
this is same as 2 * 3C2 * 5! = 720

Total combinations are 7! = 5040
so valid combinations are 7! - 720 = 4320

5

Qs. 5 Four questions repeated outcomes

Qs. 1 Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)

Qs. 2 Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)

Qs. 3 There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

Qs. 4 There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

Repeated Outcomes [#permalink] Wed Mar 09, 2005 9:29 am
2 This post received
KUDOS
Repeated Outcomes

If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

Explanation: For each object, there are k outcomes. So the total number of outcomes would be k*k*k ...*k, for n times.

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total number of fruits = 9
For each fruit there are 2 possible outcomes: included, not included
Total outcome = 2^9-1
(The minus one is to take out the one possible outcome where nothing is in the basket.)

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total outcome = (2^3-1)*(2^4-1)*(2^2-1)

Example:

There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

P = (good outcomes)/(total possible outcomes).

Let the 3 reports be A, B, and C.
Each report must be assigned to a secretary.

Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
3*3*3 = 27.

Good outcomes:
A good outcome occurs when each report is assigned to a DIFFERENT secretary.
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 2. (Either of the 2 remaining secretaries.)
Number of options for C = 1. (Only 1 secretary left.).
To combined these options, we multiply:
3*2*1 = 6.

Thus:
(good outcomes)/(total possible outcomes) = 6/27 = 2/9.

Example:

There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

P = (good outcomes)/(total possible outcomes).

Let the 4 reports be A, B, C and D.
Each report must be assigned to a secretary.

Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
Number of options of D = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
3*3*3*3 = 81.

Good outcomes:
For each secretary to be assigned at least 1 report, exactly 1 secretary must receive a PAIR of reports, while the other 2 secretaries receive 1 report each.
Number of pairs that can be formed from the 4 reports = 4C2 = (4*3)/(2*1) = 6.
Number of ways to assign this pair = 3. (Any of the 3 secretaries.)
Number of ways to assign the next report = 2. (Either of the 2 remaining secretaries.)
Number of ways to assign the last report = 1. (Only 1 secretary left.)
To combine these options, we multiply:
6*3*2*1 = 36.

Thus:
(good outcomes)/(total possible outcomes) = 36/81= 4/9.

6

Qs. 6 Three questions on probability of repeated experiments.

Qs. 1 When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?

Qs. 2 The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days?

Qs. 3 The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

Binomial Distribution

In each individual test, the probability of A happening is p and not happening is 1-p. What is the probability of A happening exactly k times in n repeated tests?

Formula:
C(n,k) * p^k * (1-p) ^ n-k

Example:
When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?
C(4, 2) * (1/2) ^2 * (1/2) ^2 = 6/16 = 3/8

Example:
The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days?
C(7, 3)*0.3^3*0.7^4

Example:
The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

Probability that at least two babies are boys
= Probability that two babies are boys + probability that three babies are boys + ... + Probability that ten babies are boys
(also) = 1- Probability that non are boys - probability that only one is a boy
Choose the easier route
P=1-C(10,0)*0.5^0*0.5^10-C(10,1)*0.5^1*0.5^9
=1-0.5^10-10*0.5^10
=1-11*0.5^10

7

Qs.8 Of the 10 employees at a certain company, 5 had annual salaries of $20,000, 4 had annual salaries of $25,000, and 1 had an annual salary of $30,000. If a bonus equal to 10 percent of annual salary was given to each employee, what was the total amount of he bonuses?
(A) $230,000
(B) $75,000
(C) $30,000
(D) $23,000
(E) $7,500

Answer: D

10% of 20k = 2k
For 5 employees = 5*2k = 10k

10% of 25k = 2.5k
For 4 employees = 4*2.5k = 10k

10% of 30k = 3k
For 1 employee = 3k

Add all = 10 + 10 + 3 = 23k

8

Qs. 9 A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16

Answer : A

Prob (not blue) * Prob ( 2nd not blue) = 6/8*5/7 = 15/28

9

Qs. 10 In a group of 30 students, 25 are taking mathematics, 22 English, and 19 history. Every
student is taking at least one of the courses. The greatest number of students who COULD be
taking all three courses is x. The least number of students who COULD be taking all three
courses is y. What is the value of x + y?
a. 17
b. 19
c. 22
d. 23
e. 24

E

T=M+E+H-(ME+MH+EH)-2MEH
30=25+22+19-(ME+MH+EH)-2MEH
(ME+MH+EH) -2MEH = 36

To maximize MEH, the (ME+MH+EH) must be minimum. Hence (ME+MH+EH) must be zero
Putting it in the above equation MEH = 18

Similarly for Minimum value (ME+MH+EH) must be maximum
Sine M=25 the maximum value of EH would be 30-25= 5
Similarly MH would be 8
And ME would be 11

Therefore MEH = 6

Thus x+y would be 18+6= 24

10

Qs. 11 a company has 15 disrtibution centres and uses color coding to identify each center.either a single color or a pair of two different colors is chose to represent each center uniquely.what is the mnimum number of colors needed for the coding and the order of the colors doesnot matter?

1 Colour codes: 4
2 Colour codes: we need to choose 2 colors from 4. This can be accomplished in 4c2 ways
4c2= 4x3x2x1/(2x(4-2))= 6
Similarly 3 Colour codes: 4c3 =4
For 4 Colour codes: 4c4=1
Hence 4+6+4+1=15

Therefore 4 is the correct answer.

11

Qs. 12 At a certain financial institution, 30% of the clients use both credit card and a cheque book, but 40% of the clients who use the credit card do not use the cheque book. What percentage of the members of the institution use the credit card?

A. 35
B. 40
C. 50
D. 65
E. 75


Pleas help me and give me a nice explanation, of course an explanation which takes less than 2 min. to do.


Good post? |
IMO C.

Let the total clients = 100
No. of clients using both credit card and cheque book = 30.

Let total no of clients using credit card = x
So, no. of clients using credit card but no cheque book = 0.4x

According to question,
0.4x+30 = x
x= 50 (which is the total number of clients using credit card)
Since we had assumed total no. of client to be 100, this value of x is the percentage answer.

You can aslo do it by making table as:

CC no CC Total
CB 30

no CB 0.4x

Total x 100

It took little less than a min to solve this. I hope this was helpful.

12

Qs. 13 Could the answer be an integer if x is an integer greater than 1?
a) x^(10) + x^(–10)
b) x^(1/6) + x^(1/2)

Sufficient or not sufficient?

Answer is A only

13


Good post? |
PLZ solve this question ........


# One variable is equal to 40% more than X. The same variable is equal to 20% less than Y. By how much percentage is Y greater than X?


Good post? |
Originally Posted by sanaulcse
# One variable is equal to 40% more than X. The same variable is equal to 20% less than Y. By how much percentage is Y greater than X?
Let x be 10

So the other variable ( Z ) be 14

Now it is given 14 is 20% less than Y

So the other variable Z is 80% of Y

Hence Y = 17.5

Now we have all the 3 variables , X , Y and Z

X = 10

Y = 17.5

Z = 14


So let's find out how much is Y greater than X ,

Y - X is 7.5

So percentage is Y greater than X is -

( 7.5 / 10 ) * 100 => 75%


Hope this helps...

14

If the sequence X 1 (X one), X 2 (X two), X 3 (X three), .....Xn is such that X 1= 3 and X n+1 = 2 (Xn)-1 for n=1 , then X 20 - X 19 =

A) 2 ^ 19
B) 2 ^ 20
C) 2 ^ 21
D) (2 ^ 20) -1
E) (2 ^ 21) -1


Can some one please explain how this problem can be solved without making the lengthy calculations.


Good post? |
Xn+1 = (2Xn) - 1 or Xn = (Xn-1) - 1

X1 = 3 = 2^1 + 1
X2 = 2*(3) - 1 = 5 = 2^2 + 1
X3 = 2*5 - 1 = 9 = 2^3 + 1

So, it follows that the nth term of the sequence is obviously Xn = 2^n + 1

Therefore, X^20 - X^19 = 2^20 + 1 - 2^19 - 1 = 2^20 - 2^19 = 2^19 * (2-1) = 2^19.
(Choice A is the answer)

15

If x < 0 , SQRT ( -x *|x|)
1) -x
2) -1
3) 1
4) x
5) sqrt(x)

SPOILER: Official Answer A. I guess its D. Am I missing something here
If x < 0 , SQRT ( -x *|x|)
A) -x
B) -1
C) 1
D) x
E) sqrt(x)
A quick way to solve this question is to plug in a value for x.
Since xs say that x= -5

So, the sqrt[-x (|x|)] = sqrt[5 (|-5|)] = sqrt[5(5)] = sqrt[25] = 5

Answer choice A is -x, so if x= -5, then -x = 5. Since our output above is 5, answer choice A is correct.

16


Good post? |
DS- Multiplication by zero
Official Guide Quant Review - 2nd Edition - DS-Question 46


What is the value of x^2 - y^2 ? (x square minus y square)


(1) x - y = y + 2
(2) x - y = 1 / (x+y)


Statement 2 tells us that x-y = 1/(x+y). From this, we can conclude that x+y does not equal zero. If x+y did equal zero, then 1/(x+y) would be undefined in which case it couldn't equal some other value. So, knowing that 1/(x+y) equals some other value, it's safe to conclude that x-y does not equal zero.

17


Good post? |
GMAT Prep Set theory Problem
Of the 200 members of a certain association, each member who speaks german also speaks english, and 70 of the members speak only spanish. if no member speaks all three languages, how many of the members speak two of the three languages?

(1) 60 of the members speak only english.
(2) 20 of the members do not speak any of the three languages.

more specifically:

there are eight subsets:
none
E only
G only
S only
ES
EG
GS
EGS

let's fill in the list with the information that we already have from the problem:

none = 20 (from statement 2)
E only = 60 (from statement 1)
G only = 0 (because they all speak english too)
S only = 70 (given)
ES = _______
EG = _______
GS = _______
EGS = 0 (given)

the only blanks combine to give the desired quantity. we can't find the values of the individual blanks, but we don't care; all that matters is their sum, which is easily found by subtracting 20, 60, and 70 (as well as the two 0's, if you want) from the total of 200. there's no need to perform this calculation, because it's data sufficiency and we know there's going to be a unique numerical answer.

ans = c

18


Good post? |
Easy data sufficiency question that's confusing. OG 12th ed
Is X an integer?
1. X/2 is an integer
2.2x is an integer.

The answer is A.

19

In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture?

(1) The difference between Albs captured and Berks captured is 7.

(2) The number of Albs captured is divisible by 5.

.Actually, this problem is testing knowledge of modular arithmetic - that is arithmetic with remainders.

Consider how many ticks you get for each Alb captured

Alb Tick
0 0
1 0
2 0
3 1
4 1
5 1
6 2

The point here is that you do not get fractional ticks (T) for each Alb captured - rather you get one for every third Alb (A) and one click (C) for every fourth Berk (B).

So from the question stem, we know the T + C = 77

Now consider 1). We are told that A - B = 7. It is tempting to see this as sufficient. Turn Alb and Berk into ticks and clicks and you might get A/3 + B/4 = 77. You have two equations in two variables. If you solve, you get A = 135 and B = 128. This gives you 45 ticks and 32 clicks. Now when you look at 2) and see that Alb is divisible by four, there appears to be a conflict between the two statements.

The problem with the above is that the equation you should have sent up above is Q (A/3) + Q (B/4) =77 where Q(x/y) is the integer quotient, excluding the remainder. So eg Q(10/5) =2 and Q(13/5) =2. Now, (1) does not yield a unique solution. A =135, B=128 and A =136, B=129 and A=137 and B=130 all yield T =45 C = 32. Looking just at the Alb for this example, see 135/3 = 45 r0, 136/3 = 45 r1, 137/3 = 45 r2.

Now consider 2. It's clearly insufficient by itself. Consider 1 and 2 together. Of the three possible solutions we found in 1), only A=136, B=129 has A divisible by 4. So 1) and 2) are sufficient together and the answer is C.


I hope you add this problem back into the database - after a couple of problems like this, the real exam will seem easy. ;)

20


Good post? |
800 level Difficult Exponent question.
How can i solve this Data Sufficiency problem, anyone pull me out from the dark........
If m is a positive integer, is the value p+q at least twice the value of 3^m+4^m ?
1) p=3^(m+1) and q=2^(2m+1)
2) m=4

First, let's rewrite the target question as "Is p+q > 2(3^m + 4^m) ?"

Statement 1: If we replace p and q with their respective values, we can reword the target question as:
Is 3^(m+1)+2^(2m+1) > 2(3^m + 4^m) ? (do we have sufficient information to answer this question? let's find out)

To simplify the right-hand-side, first recognize that 4^m = (2^2)^m = 2^2m
So, we can reword the target question as: Is 3^(m+1)+2^(2m+1) > 2(3^m + 2^2m) ?
If we expand the right-hand-side, we get: Is 3^(m+1)+2^(2m+1) > (2)3^m + 2^(2m+1)?
At this point, we can subtract 2^(2m+1) from both sides to get: Is 3^(m+1) > (2)3^m?
Now, if we divide both sides by 3^m, we get: Is 3 > 2?
Yes, 3 is greater than 2.
Since we can answer the reworded target question with certainty, statement 1 is sufficient.

Statement 2: Since we have no information about p and q, we cannot answer the target questions. So, statement 2 is not sufficient and the answer is
SPOILER: A
.

Cheers,
Brent

21


Good post? |
Triangle Problem
ABC and PQR are both triangles. Is the area of triangle ABC greater than the area of triangle PQR?



1.Sides AB=PQ, BC=QR, and AC= PR

2. Angles A=P, B=Q, and C=R

A. Together the statements are sufficient, but alone neither statement is sufficient
B. Together the statements are not sufficient
C. Both statements are sufficient alone
D. Statement 1 alone is sufficient, but statement 2 alone is not sufficient
E. Statement 2 alone is sufficient, but statement 1 alone is not sufficient

Lolz, this is a funny one

The answer should be (A)

Using 1
If the 3 sides of a triangle are equal, then by SSS (side,side,side) the 2 triangles are congruent
we do not need to know much about the angles

Using 2
If the 3 angles have the same angle, then by AAA (angle, angle, angle) the 2 triangles are similar, but the area may be different

22

Of the students in the class, 55% of the females and 35% of the males passed the exam. Did more than half the students pass the exam?

1) More than half the students in the class are females
2) The number of female students is 20 more than the the number of male students.

SPOILER: Official Answer - B

Let M - number of males, F - number of females

Need to compare (M+F)/2 and 0,55F+0,35M given that F = M+20

(M + M + 20)/2 compare with 0,55(M +20) + 0,35M

(2M+20)/2 compare with 0,55M+0,35M+11

M+10

23

Average
A student finds the average of 10 positive integers. Each integers contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this the average becomes 1.8 less than the previous one.What was the difference of the two digits b and a?

A. 8
B. 6
C. 2
D. 4
E. None of the above


Originally Posted by shanssv
A student finds the average of 10 positive integers. Each integers contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this the average becomes 1.8 less than the previous one.What was the difference of the two digits ba and ab?

Answer: C. 2

Let Sc be the sum of 10 current numbers then the average (Current) is Avc = Sc / 10

Similalrly let Sp be the sum of 10 previous numbers then the average (previous) is Avp = Sp / 10

But as per the questions
Avc = Avp - 1.8
then Sc/10 = Sp/10 - 1.8
therefore Sc/10 = (Sp - 18)/10
therefore Sc = Sp -18
or Sc -Sp = 18

Now Let A be the sum of all other numbers except ba then Sc = A + (b=10a)
Similaly Sp = A + (a + 10b)

Put in above equation
A + (b + 10a) - A - (a + 10b) = 18
hence 9(b-a) = 18
therefore b - a = 2
Hence C is teh correct answer

24

on Average
Three math classes: X, Y, and Z, take an [COLOR=blue ! important][COLOR=blue ! important]algebra[/COLOR][/COLOR] test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81. What is the average for all the three classes?
81
81.5
82
84.5
Ans : B



Can someone help me on this one?

This is a problem on weighted ave.

x*83+y*76/(x+y)= 79

y*76+z*85/(y+z) = 81

We need to find
x*83+y*76+z*85/(x+y+z)

25


Good post? |
PS Questions (GMAT Guru's Erin and 800Bob Please Help) Part 2
9. 1 – [2 –(3 – [4 – 5] + 6) + 7] =
(A) –2
(B) 0
(C) 1
(D) 2
(E) 16

SPOILER: D


11. If 0.497 mark has the value of one dollar, what is the value to the nearest dollar of 350 marks?
(A) $174
(B) $176
(C) $524
(D) $696
(E) $704

SPOILER: D


12. A right cylindrical container with radius 2 meters and height 1 meter is filled to capacity with oil. How many empty right cylindrical cans, each with radius 1/2 meter and height 4 meters, can be filled to capacity with the oil in this container?
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16

SPOILER: C



16. A family made a down payment of $75 and borrowed the balance on a set of encyclopedias that cost $400. The balance with interest was paid in 23 monthly payments of $16 each and a final payment of $9. The amount of interest paid was what percent of the amount borrowed?
(A) 6%
(B) 12%
(C) 14%
(D) 16%
(E) 20%

SPOILER: D


12. An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrect answers from the number of correct answers. If the student answered all of the questions and received a score of 38, how many questions did that student answer correctly?
(A) 19
(B) 38
(C) 41
(D) 44
(E) 46

SPOILER: D

1) D....Apply PEMDAS or BODMAS...ans:2

2) E....350/0.497 ~704

3) C....2^2*1=n*(1/2)^2*4...n=4

4) D ..amnt borrowed=400-75=325
Total amnt paid=75+23*16+9=452
amont paid as interest=452-400=52
->52/325=16%

5) I think there is something missing in the question..it is not mentioned the marks awarded for the correct answer.
If we assume it to be 1 than ans is 38+(12/2)=44...D

26

Average Question
A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A.74
B.75
C.76
D.77
E.78


Good post? |
b (65*4)+(80*6)+(77*5)/15

27

average score
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A. 2x - 8
B. x - 4
C. 8 - 2x
D. 16 - x
E. 8 - x/2

IMO A.

If the score for 5 students was 8 --> the total score for these 5 students is 40

(40 + 5y) / 10 = x

solve for Y and it is answer A. 2x - 8

28

an easy one
An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrect answers from the number of correct answers. If the student answered all of the questions and received a score of 38, how many questions did that student answer correctly?
(A) 19
(B) 38
(C) 41
(D) 44
(E) 46

for me the answer is e
correct=x
incorrect=50-x
x-2(x-50)=38
solve for x=46 correct

29


Good post? |
mean-median
Class A and B took a same test. For class A, median score is 80, average score is 82; for class B, median score is 78, average score is 74. Combining A and B, is the average greater than the median?
1). A has 37 students and B has 40 students.
2). A and B have 77 students.


Good post? |
Stmt 1:
A = 37, B = 40
=> median is the 38th element
A has 18 students below 80(median)
B has 20 students below 78 (median = (20 th element +21st element/2)
=> median of the combined set can be 80 or atleast > 78 as 38 students are below 80

Avg = 37*82 + 40*74/77 => avg < 78 (only if equal number of students are there, the avg will be 78)

Median > avg

Sufficient

Stmt 2:
No info on how many students are in A & B individually.
Insufficient

Ans A.

30


Good post? |
Averages
A student was able to score 600 in 12 tests.He scored less than or equal to 80% of his average score per test in the four of these tests.If he did not score more than 60 in any of the test,wht is the minimum number of tests in which he should have scored more than 50

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


Good post? |
i think i would agree with mastermind.

as per what karmaholic says:
__________________________________________________ ____________
Marks obtained in four tests = 160 (4*40) [40 = 80% of 50 and 50 is the average score]
Total Marks = 600. So the student needs another 440 marks from remaining 8 tests.
Now the max he can get in these tests is 60.
__________________________________________________ _____________

now we need the min number of tests in which he has to score max ie 60
if
no of tests with max score =2 then 2x60 + 6x50 = 420 .. option D incorrect
no of tests with max score =3 then 3x60 + 5x50 = 430 .. option C incorrect
no of tests with max score =5 then 4x60 + 4x50 = 440 .. correct option B