1. Five pieces of wood have an average (AM)length of 124 cm and a median length of 140 cm. What is the maximum possible length,in cm,of the shortest piece of wood?

a) 90 b) 100 c) 110 d) 130 e) 140

2. For any +ve integer n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75=3*5*5. How many two-digit +ve intergers have length 6?

a) None b)One c)Two d)three e) Four

Pls post ur answers with explanation...i will post the answers..tomorrow..

Hi...
Could someone help me in solving these problems. I have posted answers along with these. But haven't got a clue on how to arrive at these answers..

1) For every positive integer n, the function h(n) is defined to be the product of all event integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40
Ans: e

2) On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?
(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.
Ans (c)

3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)

4) Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussel sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussel sprouts. How many of teh students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans
Ans (D)

5) If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
(a) 2/11
(b) 1/3
(c) 41/99
(d) 2/3
(e) 23/37
Ans : e

1) For every positive integer n, the function h(n) is defined to be the product of all event integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40
Ans: e

h(n) will not include any prime no other than 2, but will include twice of prime nos. So all the nos less than 50 will be the factor of h(n). So, the largest prime factor of h(n) will be less than equal to 50. So the smallest prime factor of h(n) +1 will not less than 50 (will be greater than 50). So the best answer is E.

________________________________________________________________________________________

2) On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?
(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.
Ans (c)

From Question W/C=5/2 M=?
(1) C/M=5/11, W not known so not sufficient so BCE,
(2) W<30, No information for Men so not sufficient so CE,
(1+2) W/C * C/M = W/M=25/22 M=22W/25 As the no of Men is a whole no 22W/25 should be an integer, which means W should be divisible by 25. As W<30, the only possible no is 25.

So the answer is C
________________________________________________________________________________________
3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)


n=p^2q
or k.5= p^2q
or (k.5)^2= (p^2q)^2
or k^2 * 25=p^2q^2

SO the answer is D

________________________________________________________________________________________
4) Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussel sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussel sprouts. How many of the students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans
Ans (D)

From Question stem
Students = 2/3 of Total Students
Students =2/5 of
Students =2/5 of 2/3 of Total Students

From (1) Total Students = 120 Sufficient. (2/5*2/3*120)
From (2) As Students = 2/3 of Total Students
So Students = 1/3 of Total Students
So Total Students = 40*3 = 120 Sufficient (2/5*2/3*120)

So Answer is D
___________________________________________________________________________________
5) If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
(a) 2/11
(b) 1/3
(c) 41/99
(d) 2/3
(e) 23/37
Ans : e

This is based on general properties of numbers.
1/N Where N= 0 to 9
For every digit has its fixed recurring decimal nos.
1/1=1.000000
1/2=0.500000
1/3=0.333333
1/4=0.250000
1/5=0.200000
1/6=0.166666
1/7=0.142857
and so on

It is a clear fact that 7 has the most no recurring decimal places. So every no having 7 as its unit digit will follow the similar pattern.

So the answer is E.
_____________________________________________________________________________________

Need help with the following problem :

Q. How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
a. 216
b. 3152
c. 240
d. 600

Rgds
MSD

Here I go>>>.
Differenct combinations of 5 digits out of the six available are---
01234 total = 10
01235 total = 11
01245 total = 12
01345 total = 13
02345 total = 14
12345 total = 15
So in order to be divisible by 3 a combination of five digit must be either (01245) or (12345).
01245 no of arrangements = 4*4*3*2*1 (0 cannot be on the first place) = 96
12345 no of arrangements = 5*4*3*2*1 = 120
Total = 96+120 = 216

For a number to be divisible by 3, the sum of the digits must be divisible by 3.

Taking 0,1,2,3,4,5 - split this problem into 2 parts

1) take 1,2,3,4,5 alone => we can form 5! five-digit numbers from these digits = 120

2) now taking 0 along with the other 5 digits, we need to ensure two things A) 0 does not start the number and B) the sum is divisible by 3.

so with digits 0,1,2,4,5 (ignore the digit "3" as it does not give us a number divisible by 3) we can form 4x4! numbers = 4 x 24 = 96

summing them up = 120+96 = 216.. hope this helps!

Yes. The answer is A. can you please provide explanation?

Rgds
MSD

Yes. The answer is A. can you please provide explanation?

Rgds
MSD

By the property of divisibility by 3 ie "a no: is divible by 3, if the sum of the digits is divisible by 3"(eg= 12-->1+2=3)

so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5(sum=12) & 1,2,3,4,5(sum=15)

from first set(0,1,2,4,5) no:s formed are 96 ie first digit can be formed from any 4 no: except 0, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

from second set(1,2,3,4,5) no:s formed are 120 ie first digit can be formed from any 5 digits, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

so otal 120+96=216

My pick is A.

If the ans is correct i'll give reasoning..........

Yes. The answer is A. can you please provide explanation?

Rgds
MSD

Need help with the following problem :

Q. How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
a. 216
b. 3152
c. 240
d. 600

Rgds
MSD

My pick is A.

If the ans is correct i'll give reasoning..........

Is the answer (A) - 216?

Need help with the following problem :

Q. How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
a. 216
b. 3152
c. 240
d. 600

Rgds
MSD

Need help with the following problem :

Q. How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
a. 216
b. 3152
c. 240
d. 600

Rgds
MSD

Originally Posted by msd_2008
Hello,
Someone please help me with the following question:
Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?
A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5
Thanks
MSD



i guess the correct answer should be D.msd_2008 ,can you please verify

The correct answer is C.

Ya buddy got it........Some calculation mistake .........
thanks a lot for explaining

Originally Posted by Inder.14
Originally Posted by msd_2008 (GMAT Problem Solving Discussions)"]
Hello,

Someone please help me with the following question:

Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?

A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5

Thanks
MSD



i guess the correct answer should be D.msd_2008 ,can you please verify

Distance travelled by Johanna = 3t miles = Circumfrence of lake = 2//r
so 3t = 22/7*2
or t = 2.095
So correct answer is choice C

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@ashishjha100 -
I hope this will help -

Avg (Arithmetic mean) cost for x calculators = 300/x
Selling price for 1 calculator = (300/x) + 5 (5 more than avg price of calculator)
No. of calculators sold by merchant = x-2 (2 less than no. of calculators bought)
Total revenue from the sale = 300 + 120 = 420

we know that - Total revenue = Selling price for 1 calculator * No. of calculators sold
=> * = 420
After simplifying quadratic equation, you'll get 30 as the answer. Also, you can try putting 30 directly in the above equation.

P.S - I read somewhere that when you need to chose and eliminate the choices, start from E to A.

Thanks Dopa,
I was in a confusion to take 300 as cargo charge or total price of x calculators......

A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
(A) 24
(B) 25
(C) 26
(D) 28
(E) 30

Pl help.....

avg for x calculators is 300/x.
he sold (x-2) calculators for (300/x+5).
so revenue (x-2)(300/x+5) = 300+20.

it is easy to solve it by substitution rather than solving the equation. one can save some time by substituting lowest and highest (here 24 and 30) and then moving on to next options. after doing with highest and lowest you kind of get the idea which value it could be. this just saves some seconds of your time.

A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
(A) 24
(B) 25
(C) 26
(D) 28
(E) 30

Pl help.....

@ashishjha100 -
I hope this will help -

Avg (Arithmetic mean) cost for x calculators = 300/x
Selling price for 1 calculator = (300/x) + 5 (5 more than avg price of calculator)
No. of calculators sold by merchant = x-2 (2 less than no. of calculators bought)
Total revenue from the sale = 300 + 120 = 420

we know that - Total revenue = Selling price for 1 calculator * No. of calculators sold
=> * = 420
After simplifying quadratic equation, you'll get 30 as the answer. Also, you can try putting 30 directly in the above equation.

P.S - I read somewhere that when you need to chose and eliminate the choices, start from E to A.

A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
(A) 24
(B) 25
(C) 26
(D) 28
(E) 30

Pl help.....

Originally Posted by msd_2008
Hello,
Someone please help me with the following question:
Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?
A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5
Thanks
MSD



i guess the correct answer should be D.msd_2008 ,can you please verify

Distance travelled by Johanna = 3t miles = Circumfrence of lake = 2//r
so 3t = 22/7*2
or t = 2.095
So correct answer is choice C

Hi Bhavin,

I guess i made some mistake in approach towards the answer...Rather i took it as Square than Rhombus.....

Explanation goes like this....

ABCD is a Rhombus....

AB=BC=CD ...also =AD....

so AB=BC=CD=AD=r

Take centre of BD as O....

and by Rhombus properties ...AO=OC = AC/2 = r/2 ..
and AOB=COB ...is right angle

so by Pythagoras theorem ...AB^2 = AO^2 + OB^2....
==> r^2 = (r/2)^2 + OB^2...

so OB = root(3)/3....

so tan (x) = AO/OB....(r/2)/(root(3)*r)/2 = 1/root(3)
==> x = tan Inverse (1/root(3)) = 30 degrees....

Correct me if i am wrong...

Originally Posted by msd_2008
Hello,

Someone please help me with the following question:

Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?

A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5

Thanks
MSD

ashishjha100 Says

i think ans is C........if it is correct i'll give reasoning.....

i guess the correct answer should be D.msd_2008 ,can you please verify

Hello,

Someone please help me with the following question:

Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?

A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5

Thanks
MSD

i think ans is C........if it is correct i'll give reasoning.....

msd_2008 Says

In the point (1) you have mentioned 1/3 as being P<1/2 and just after that you've mentioned 1/3 as being P>1/2. If we indeed get P<1/2 in all cases....then the answer should actually be A and not E.

it was a typing mistake

8C2/10C2 = 28/45 =.633 so p>1/2

Hello,

Someone please help me with the following question:

Q.1 A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track at the average speed of 3 miles per hour. If t represents the number of hours it took Johanna to walk completely around the lake, which of the following is a correct statement?

A. 0.5< t < 0.75
B. 1.75< t < 2.0
C. 2.0 < t < 2.5
D. 2.5 < t < 3.0
E. 3 < t < 3.5

Thanks
MSD

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

I think ans E.
Total selection is 10C2
P= p is the probability that both representatives selected will be women
Is p>1/2
(1)More than 1/2 of the 10 employees are women.
Ie no: of women may be n=6,7,8,8,9 at extreme 10
So required probability, p=nC2/10C2
Consider each value of n.
When n=6 , p=6C2/10C2=1/3 p<1/2
When n=7 , p=7C2/10C2=7/15 p<1/2
When n=8 , p=8C2/10C2=1/3 p>1/2
So A alone not sufficient
(2)The probability that both representatives selected will be men is less than 1/10
Let a be the no: of men
So aC2/10C2 <1/10
Solving a<3.5
Ie a=1,2,3

So B alone not sufficient

Taking together when a=3 , p<1/2
when a=2 , p>1/2

So both not sufficient..
Anyone to validate my reasoning

In the point (1) you have mentioned 1/3 as being P<1/2 and just after that you've mentioned 1/3 as being P>1/2. If we indeed get P<1/2 in all cases....then the answer should actually be A and not E.

exactly two is = (AB)+(BC)+(AC) but not ABC..
to calculate exactly two classes, we hv to deduct 3 times ABC from (AB)+(BC)+(AC)..bcoz each one contains one time ABC.

more than one is = (AB)+(BC)+(AC) including one time ABC
to calculate more than one, we hv to deduct 2 times ABC from (AB)+(BC)+(AC)

this is better explained by venn diagrams.

Thanks a lot for the explanation dude. Will follow what you said.

ashishjha100 Says

Pls help with reasoning...........

For 2nd Question
--------------------------
AB=BC=CD=DA= r ..This becomes a rhombus...
Take centre of BD as O....
AC^2 = AD^2 + CD^2 = r^2 + r^2 = 2* (r^2)
From this...... AC=root(2) * r...AO becomes (root(2)*r)/2 = r/root(2)...
Similarly..BO = r/root(2)....
Tan(x) = AO/BO = 1..x = Tan Inverse of(1) = 45degrees

Hey jpmadhav...ABCD is a rhombus and not a square...
statement in bold is incorrect...angle ADC is not 90 degrees and pythogaras rule is not applicable..

we get 2 eq triangles in form of ABC and ADC .
angle ABD = 1/2 angle ABC
= 1/2 * 60
x = 30

hope this helps..

ashishjha100 Says

But guys OA is E

Thats what I marked in my previous post :

5.The figure above represents a rectangular parking lot that is 30 meters by 40 meters and an attached semicircular driveway that has an outer radius of 20 meters and an inner radius of 10 meters. If the shaded region is not included, what is the area, in square meters, of the lot and driveway?
(A) 1,350?
(B) 1,200 + 400?
(C) 1,200 + 300?
(D) 1,200 + 200?
(E) 1,200 + 150?

30*40 + (pi*(20)(20) - pi*(10)(10))/2 => 1200 + 150pi

Thanks for the explanation. But I am a bit confused.

The first ques asks - how many students are registered for exactly two classes ? (asking two classes, so how do i figure out how much to deduct)? what if it asked 3 classes or one.

In the other one it was more than one of the three products. This means two or three. Isnt the above one two. I can't figure out the difference.

exactly two is = (AB)+(BC)+(AC) but not ABC..
to calculate exactly two classes, we hv to deduct 3 times ABC from (AB)+(BC)+(AC)..bcoz each one contains one time ABC.

more than one is = (AB)+(BC)+(AC) including one time ABC
to calculate more than one, we hv to deduct 2 times ABC from (AB)+(BC)+(AC)

this is better explained by venn diagrams.

a+b+c=19..thats correct. but this contains three times ABC. So u need to deduct three times ABC to get the answer which is ONLY 2 langs.

in the second question, question asked for MORE THAN ONE = TWO + THREE. in which case, you will deduct two times ABC to get the answer.

Thanks for the explanation. But I am a bit confused.

The first ques asks - how many students are registered for exactly two classes ? (asking two classes, so how do i figure out how much to deduct)? what if it asked 3 classes or one.

In the other one it was more than one of the three products. This means two or three. Isnt the above one two. I can't figure out the difference.

Help needed with SETS question -

I came across this question..

Q-1 In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 19
B. 13
C. 10
D. 8
E. 7

OA is 10...My ans was 13.

I am using the following method -

AUBUC = A + B + C -AB - BC - CA + ABC + none.

ok, here A = history, b=math, c=english
As per question -

68 = 25 + 25 + 34 - a - b - c + 3
68 - 50 - 34 - 3 = -a -b -c
68 - 87 = - a - b - c

thus a + b + c = 19...

Now since this 19 has 3 included three times we need to elimate 3 thats included twice. So we do 19 - 3*2 => 19 - 6 or 13
but the answer is 10. Why ??? In the ques (another sets ques) below the above method did work.

a+b+c=19..thats correct. but this contains three times ABC. So u need to deduct three times ABC to get the answer which is ONLY 2 langs.

in the second question, question asked for MORE THAN ONE = TWO + THREE. in which case, you will deduct two times ABC to get the answer.

Help needed with SETS question -

I came across this question..

Q-1 In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 19
B. 13
C. 10
D. 8
E. 7

OA is 10...My ans was 13.

I am using the following method -

AUBUC = A + B + C -AB - BC - CA + ABC + none.

ok, here A = history, b=math, c=english
As per question -

68 = 25 + 25 + 34 - a - b - c + 3
68 - 50 - 34 - 3 = -a -b -c
68 - 87 = - a - b - c

thus a + b + c = 19...

Now since this 19 has 3 included three times we need to elimate 3 thats included twice. So we do 19 - 3*2 => 19 - 6 or 13
but the answer is 10. Why ??? In the ques (another sets ques) below the above method did work.

dude !!!

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi => This will give area for complete circle...whereas we need for a semi-circle....this should be 300pi/2 = 150pi

Total Area = 1200+300pi....

===================

But guys OA is E

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

I think ans E.
Total selection is 10C2
P= p is the probability that both representatives selected will be women
Is p>1/2

(1)More than 1/2 of the 10 employees are women.
Ie no: of women may be n=6,7,8,8,9 at extreme 10
So required probability, p=nC2/10C2
Consider each value of n.
When n=6 , p=6C2/10C2=1/3 p<1/2
When n=7 , p=7C2/10C2=7/15 p<1/2
When n=8 , p=8C2/10C2=1/3 p>1/2

So A alone not sufficient

(2)The probability that both representatives selected will be men is less than 1/10
Let a be the no: of men
So aC2/10C2 <1/10
Solving a<3.5
Ie a=1,2,3

So B alone not sufficient

Taking together when a=3 , p<1/2
when a=2 , p>1/2

So both not sufficient..
Anyone to validate my reasoning

Hi Bunty,

I dont think you need to use Sector formula here...its a simple Semi-circle...so area of circle is enough....Anyways i took it as a circle....see the below post of Vicky...he has corrected my mistake above...

Yes Vicky....I missed it..its a semi-circle...so area will be half..thanks Dude...

Hi MSD,

Let hourly wage of John or Mary be 'J' dollars/hour

John worked on the job for 10 hrs ..So Amount John got is=10J
Mary worked for 2 hours less than John....So Amount Mary got is = 8J

Initial amount given to John and Mary each is = x dollars

Mary gave john y dollars of her payment so that they receive the same hourly wage

so John's amount ==> 10J = x + y(Amount from Mary)
Mary's Amount ==> 8J = x - y(Since Mary gave y dollars to John)


Since both recieve the same hourly wage....

(x+y)/10 = (x-y)/8
solving both
we get x = 9y

Hope you got the answer

For 2nd Question
--------------------------
AB=BC=CD=DA= r ..This becomes a rhombus...
Take centre of BD as O....

AC^2 = AD^2 + CD^2 = r^2 + r^2 = 2* (r^2)

From this...... AC=root(2) * r...AO becomes (root(2)*r)/2 = r/root(2)...

Similarly..BO = r/root(2)....

Tan(x) = AO/BO = 1..x = Tan Inverse of(1) = 45degrees

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi

Total Area = 1200+300pi....

dude !!!

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi => This will give area for complete circle...whereas we need for a semi-circle....this should be 300pi/2 = 150pi

Total Area = 1200+300pi....

===================

Hi madhav...I am getting different answer..my approach is as follows

i am using area of sector formula...i,e (theta/360) * pi * r^2
theta is the angle of the sector...

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> (180/360)*pi *30^2 - (180/360)*pi*10^2
==>Total Area = 1200+400pi....

pls correct me if I am wrong..

For 2nd Question
--------------------------
AB=BC=CD=DA= r ..This becomes a rhombus...
Take centre of BD as O....
AC^2 = AD^2 + CD^2 = r^2 + r^2 = 2* (r^2)
From this...... AC=root(2) * r...AO becomes (root(2)*r)/2 = r/root(2)...
Similarly..BO = r/root(2)....
Tan(x) = AO/BO = 1..x = Tan Inverse of(1) = 45degrees
For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200
Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi
Total Area = 1200+300pi....

1. A grocer purchased a quantity of bananas at 3 pounds for $0.50 and sold the entire quantity at 4 pounds for $1.00. How many pounds did the grocer purchase if the profit from selling the bananas was $10.00?
(A) 40
(B) 60
(C) 90
(D) 120
(E) 240

2. If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75

3. In each production lot for a certain toy, 25 percent of the toys are red and 75 percent of the toys are blue. Half the toys are size A and half are size B. If 10 out of a lot of 100 toys are red and size A, how many of the toys are blue and size B?
(A) 15 (B) 25 (C) 30
(D) 35 (E) 40

4. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140 (D) 150 (E) 160

5.The figure above represents a rectangular parking lot that is 30 meters by 40 meters and an attached semicircular driveway that has an outer radius of 20 meters and an inner radius of 10 meters. If the shaded region is not included, what is the area, in square meters, of the lot and driveway?
(A) 1,350?
(B) 1,200 + 400?
(C) 1,200 + 300?
(D) 1,200 + 200?
(E) 1,200 + 150?

6.In ?PQS above, if PQ =3 and PS = 4, then
(A)
(B) 12/5
(C)
(D)
(E)

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

For third question


Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

by using statement one and two independently we can not answer this question

lets try with bot the statemnets
Student with brown hair = 30 (from statement 1)
male student = 20(from statement 2)

probability of male student with brown hair = p1(probability of stdent is male)*p2(probability of stdent is having brown hair)
=20/60*30/60=1/3*1/2=1/6

hence it can be answered by using both the statement i.e C

correct me if i am wrong

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

for the first question

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

n is the number of defective bulbs and n<5 ( as mentioned fewer than half are defective)
lets take first statement
1) P=total defective bulbs/Total number of bulbs
p=nC2/10C2=1/15(given in first statement)
n!*8!/(n-2)!*10! =1/15
n*(n-1)/10*9=1/15
n(n-1)=6
and on solving we will get n=3 and n=-2
hence it can be answered by first statement

Now lets take second statement
2) nC1 *10-nC1/10C2=7/15
on solving this we will get
n(10-n)=21
=>n=7,3 and as n<5 => n=3
it can be answered by second statemnet as well

so answere will be D i.e it can be answered 1 as well as second statement indipendently

Some please correct m eif i am wrong

Ashish ..for 3rd q..with table u can solve quickly...

Red Blue | %
_________________
Size A 10 | 40 50
--------------------
Size B | ? 50
--------------------
% 25 |75 100

so, Blue & Size B = 75-40 = 35

|ashishjha100 Says

Pls help with reasoning...........

Hello,

Please help me with the following questions (with step by step explanation).

1. John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hrs and Mary worked for 2 hours less than him. If Mary gave John y dollars of her payment so that they receive the same hourly wage, what was the dollar amount, in terms of y, that John was paid in advance?

a. 4y b.5y c.6y d.8y e.9y

2. A certain junior class has 1000students and a senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior, 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

a. 3/40,000 b.1/3,600 c.9/2,000 d.1/60 e.1/15

Thanks
MSD

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

reddy.harikisho Says

I think we can solve this very simply. if n is multiple of 5 then n^2 (n square) would be multiple of 25.therefore, n^2= (p^2q)^2, which is answer D.

dude..option D given is p^2 * q^2 but not (p^2q)^2 ..

Neo

For 2nd Question
--------------------------
AB=BC=CD=DA= r ..This becomes a rhombus...
Take centre of BD as O....

AC^2 = AD^2 + CD^2 = r^2 + r^2 = 2* (r^2)

From this...... AC=root(2) * r...AO becomes (root(2)*r)/2 = r/root(2)...

Similarly..BO = r/root(2)....

Tan(x) = AO/BO = 1..x = Tan Inverse of(1) = 45degrees

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi

Total Area = 1200+300pi....

1. Cost he purchased = (0.5x)/3...Where x is the number of pounds....
Cost he sold = x/4
Profit => x/6 - x/4 = 10
.....................x=120

3. Consider there are 100 toys in total
So we have 25 Red toys and 75 Blue Toys
.... 50 of which are Size A and 50 Size B

If 10 out of 100 are Red and Size A ...
remaining 40 (50-10) of Size A are Blue....
from this we have (75-40) 35 blue toys remaining which are of Size B....
So the answer is 35...

4. From the first sentence
(S+5)(T+1) = ST+70
from (2) we have
(S+10)(T+2) = ST+P ...where P is the more miles would he have covered

From 1 & 2 we get P = 150..

6. PR^2 + RS^2 = 16
PR^2 + QR^2 = 9
QR+QS = 5 ....So QR = 5-QS

Solving the above equations we get QR = 9/5...
By using QR...we get PR = 12/5

Hi Madhav,

Its crystal clear ! Thank u very much...i was making mistake at this step
" if we equal both cases.....(CAF^2)/200 = (100*100)/100"

Cheers
Bunty..

Hi Bunty,

For the first Question
---------------------------
Let CA be concentration of chemical A and CB be concentration of chemical B .....initially
Hence rate of a certain chemical reaction => x proportional to (CA^2)/CB

Consider CA and CB as 100 and 100 before start....

so x proportional to (100*100)/100

After the concentration change in B .suppose CBF....present change in the chemical A required to keep the reation rate unchanged be CAF....

now.... x proportional to (CAF^2)/(CBF)
Since the concentration change in B increased is 100 percent...
CBF = CB+100....

so x will be proportional to (CAF^2)/(CB+100)

i.e ..(CAF^2)/200

if we equal both cases.....(CAF^2)/200 = (100*100)/100
...................................(CAF^2) = 200
...................................CAF = square root of 200 = 141... = CA+41 percent.

So change in the chemical A required to keep the reation rate unchanged is approx 40% increase.......

Hope you got the clear explanation....

Pls help with reasoning...........

Pls help in solving these probs...highlighted ones are OG's answers..

1. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent.Which of the following is closest to the present change in the chemical A required to keep the reation rate unchanged?

a 100% decrease
b 50% decrease
c 40% decrease
d 40% increase
e 50% increase

2. The number 75 can be written as the sum of the squares of 3 different positive intergers.What is the sum of these 3 integers.

a 17
b 16
c 15
d 14
e 13

3. Three grades of milk are 1 percent ,2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of 2 percent grade,and z gallons of the 3 percent grade are mixed to give x+Y+z gallons of a 1.5 percent grade.What is x in terms of y and z?

a. y+3z
b (y+z)/4
c 2y+3z
d 3y+z
e 3y+4.5z

1. R=kA^2/B

If B increases by 100% we get R=kA^2/2*B

To keep this same are before we need A=Root(2)*A => A 41% increase for A

2. Hit and trial : The three numbers are 1,7,5

1 + 49 + 25 = 75 => Ans=1+7+5=13

3. 0.01x + 0.02y + 0.03z = 0.015(x+y+z)

Solving the above eqn. we get x=y+3z

Hi Bunty,

For the first Question
---------------------------
Let CA be concentration of chemical A and CB be concentration of chemical B .....initially
Hence rate of a certain chemical reaction => x proportional to (CA^2)/CB

Consider CA and CB as 100 and 100 before start....

so x proportional to (100*100)/100

After the concentration change in B .suppose CBF....present change in the chemical A required to keep the reation rate unchanged be CAF....

now.... x proportional to (CAF^2)/(CBF)
Since the concentration change in B increased is 100 percent...
CBF = CB+100....

so x will be proportional to (CAF^2)/(CB+100)

i.e ..(CAF^2)/200

if we equal both cases.....(CAF^2)/200 = (100*100)/100
...................................(CAF^2) = 200
...................................CAF = square root of 200 = 141... = CA+41 percent.

So change in the chemical A required to keep the reation rate unchanged is approx 40% increase.......

Hope you got the clear explanation....

so here's your soln..

p,q are prime integers and n is a multiple of 5=> atleast one of p or q must be 5 . mathematically,

p^2 * q = 5k (k is integer) => k = (p^2 * q)/5 = (p^2/5)*q = p^2*(q/5)

for k to be a integer you can see that either p^2 must be a multiple of 5 or q must be a multiple of 5 (since p,q are primes)

If p^2 is multiple of 5 => p is a multiple of 5 (since p is prime).

so you can eliminate options A, B, C clearly. coz we are not sure whether p or q is a multiple of 5.

Now p^2 * q^2 = (p^2 * q) * q

so if both cases where p or q can be a multiple of 5, the above will always be a multiple of 25. Got it?? You can just give some values to p,q and check..

Neo

I think we can solve this very simply. if n is multiple of 5 then n^2 (n square) would be multiple of 25.therefore, n^2= (p^2q)^2, which is answer D.

. There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

p=9h

36000+h+9h=441000
h=40500
so p=364500

. There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

Pls help in solving these probs...highlighted ones are OG's answers..

1. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent.Which of the following is closest to the present change in the chemical A required to keep the reation rate unchanged?

a 100% decrease
b 50% decrease
c 40% decrease
d 40% increase
e 50% increase

2. The number 75 can be written as the sum of the squares of 3 different positive intergers.What is the sum of these 3 integers.

a 17
b 16
c 15
d 14
e 13

3. Three grades of milk are 1 percent ,2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of 2 percent grade,and z gallons of the 3 percent grade are mixed to give x+Y+z gallons of a 1.5 percent grade.What is x in terms of y and z?

a. y+3z
b (y+z)/4
c 2y+3z
d 3y+z
e 3y+4.5z

Hi...
Could someone help me in solving these problems. I have posted answers along with these. But haven't got a clue on how to arrive at these answers..

3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)

so here's your soln..

p,q are prime integers and n is a multiple of 5=> atleast one of p or q must be 5 . mathematically,

p^2 * q = 5k (k is integer) => k = (p^2 * q)/5 = (p^2/5)*q = p^2*(q/5)

for k to be a integer you can see that either p^2 must be a multiple of 5 or q must be a multiple of 5 (since p,q are primes)

If p^2 is multiple of 5 => p is a multiple of 5 (since p is prime).

so you can eliminate options A, B, C clearly. coz we are not sure whether p or q is a multiple of 5.

Now p^2 * q^2 = (p^2 * q) * q

so if both cases where p or q can be a multiple of 5, the above will always be a multiple of 25. Got it?? You can just give some values to p,q and check..

Neo

can someone help me to solve this problem?

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

1. 
   
   1. A. 20
      
   
   

1. 
   
   1. B. 40
      
   2. C. 50
      
   3. D. 80
      
   4. E. 120
      
   
   

another soln..

first select 3 ppl from 10 members = 10C3

remove the possibility of forming a couple in those selection of 3
= 5C1 * 8 (selecting one couple and the third member as any remaining one)

so 10C3- 5*8 = 120-40 = 80

Neo

Hi...
Could someone help me in solving these problems. I have posted answers along with these. But haven't got a clue on how to arrive at these answers..
3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)

I have adoubt in the above question:
Is it n = (p^2)*q or is it n = p^(2*q), I mean q is in the index or is it just a multiplicant with p^2?

I guess 3rd questions is tricky...need someone to solve this out....

2nd Question
---------------
women:children = 5:2

From(1)
children:men = 5:11

From(2)
women<30

Let children be 10....then women will be 25 ..this might hold true since women < 30..
and men will be (11/5) * 10 = 22

If only (1) is given..we may get multiples of the above results...
If only (2) is provided, with out men ratio...it is difficult to calculate men

So the answer will be D

Will go from bottom to top...
5th Question
Simply using division
2/11=0.222 (repeating decimal is 1 digit)
1/3=0.333 (repeating decimal is 1 digit)
41/99 = 0.414141 (repeating decimal is 2 digits)
2/3=0.6666 (repeating decimal is 1 digit)
23/37 =0.621621(repeating decimal is 3 digits)

so the number with longest sequence is option e

4th question
----------------
2/3 dislike lima beans ....
From (1)..of total 120 ..its 80 students (2/3 * 120)
out of 80 ...3/5 dislike brussel sprouts...so its 48 students ..
so students like brussels sprouts but dislike lima beans 80-48 = 32

From (2) ...40 students like lima beans
If 2/3 dislike lima beans of all students...1/3 like lima beans
so (1/3 * total students) = 40 ...so total students =120
and the from above we can achieve students like brussels sprouts but dislike lima beans....

So the answer is D..

Hi...
Could someone help me in solving these problems. I have posted answers along with these. But haven't got a clue on how to arrive at these answers..

1) For every positive integer n, the function h(n) is defined to be the product of all event integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40
Ans: e

2) On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?
(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.
Ans (c)

3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)

4) Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussel sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussel sprouts. How many of teh students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans
Ans (D)

5) If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
(a) 2/11
(b) 1/3
(c) 41/99
(d) 2/3
(e) 23/37
Ans : e