gre quant questions practice (daily min 1 question will be posted)

[cicerone]

Guys, i am a faculty member for GRE and CAT in one of the leading institutes in India. I will be posting daily in this thread GRE quant questions for ur practice. Hoping that this move will be encouraged here comes our first query

 

1. http://img242.imageshack.us/img242/5838/rghttrg7sr.gifIn the above figure ABC is a right angled isosceles triangle right angled at vertex B. Six lines which are parallel to BC are drawn such that the distance between any two consecutive lines is same. Find the sum of the lengths of those six lines if AB= 8.4 cms

[manish8109]

Ans is 25.2

[cicerone]

manish8109, try to explain ur answers man, merely giving answers would not help our memebers, i hope u got the point.

 

manish8109, try to explain ur answers man, merely giving answers would not help our memebers, i hope u got the point.

[rainth]

A

 

 

B------C

 

 

Right triangled and isoscaled

AB=BC=8.4

AB is divided into 7 part 1.2 each

The most upper small triangle is also isoscaled and right angular

Hence Upper most parrallel line is of length 1.2

And it is increased by 1.2 at each step

 

Hence the sum = 1.2*6*(6+1)/2 = 21*1.2 =25.2

[dipen01]

Well what i did was same as rainth...except final calculation..

 

I did it longer way... 1.2+2.4+3.6+4.8+6.0+7.2 = 25.2

was little time consuming compared to rainth's..

 

rainth can u elaborate ur final step..

 

Regards,

Dipen

[rainth]

take 1.2 common

then sum = 1.2(1+2+3+4+5+6) = 1.2 * 6 * 7 / 2

[dipen01]

Too silly..:P

 

Too silly..:P

 

Too silly..:P

[cicerone]

gUYS, ALL OF U R RIGHT, BUT I HAVE A SMALL AND EFFECTIVE WAY OF ANSWERING THIS. LOOK AT THE BELOW FIGUREhttp://img216.imageshack.us/img216/1563/rgtsol8ls.gif

nOW LENGHT OF EACH LINE IS SAME AS BC WHICH IS EQUAL TO 8.4 CMS

sO THE ANSWER IS SIMPLY 6X8.4/2 =25.2

i HOPE ALL OF U GOT THIS.

 

gUYS, LITTLE BIT BUSY IN THE LAST TWO DAYS,

Any way here comes our next set of queries

 

1. Find the total no. of different mixed doubles tennis games that can be conducted among 7 married couples, if no husband and wife play in the same game.

 

2. Find the total no. of ways in which a black square and a whit square can be selected from a chess board such that neither of them lie in the same row nor in the same column.

 

3. Find the total of only rectangles on a 5X5 square board.

 

4. There are 18 stations in between two stations A and B. Find the total no. of different second class tickets to be printed so that a passenger can travel from one place to any other place.

 

5. In an infinite GEOMETRIC PROGRESSION each term is equal to sum of the all its following terms. Find the common ratio.

 

6. A father has 6 children. He takes every 3 out of them to a park everyday with no 3 being repeated.

a. How many times does father go to the park? (in one complete cycle)

b. How many times does each child go to the park?(in one complete cycle)

 

Guys, I'll hold myself here for today. U people be on this job.

[pavelbuet]

1.

choose husbands in 7C2 ways

then, choose wives 5C2 ways (wives of the previously choosen 2 husbands are left out of choice)

selected 2 husbands and 2 wives can make team in 2 ways {(H1,W1),(H2,W2)},{(H1,W2),(H2,W1)}

total game = 7C2*5C2*2 = 420

 

2. considering order of choosing black and white is not important.

first choice can be met in 32 ways

second choice can be met in (32-8)=24 ways (8 of opposite color in same column or row)

so , total =32*24 =768

 

3.

 

(1X1) rectangle total 5*5

(1X2) ..........total 5*4

(1X3) ..........total 5*3

(2X2) ..........total 3*3

....one pattern is clear

 

so in general total rectangle of form (rXc) is (5-r+1)*(5-c+1)=(6-r)*(6-c)

here 1

say, r=1 then total rectangle is= (6-1)*(6-1)+(6-1)*(6-2)+(6-1)*(6-3)+....+(6-1)*(6-5)

=5*(5+4+3+2+1)

=5*(5*6/2)

=5*15

if r=2 then=4*15

if r=3 then=3*15

 

goes this way, so total=15*(5+4+3+2+1)=15*15=225

 

[becomes lengthy maybe there is some quick reasoning]

 

4.

if i have understood the problem correctly then its simple 20*19=380

 

5. a=ar+ar^2+ar^3+....................

=>a=ar(1+r+r^2+...........)

=>1=r/(1-r)

=>1-r=r

=>r=1/2

 

6.

 

a) 6C3=20

b) 5C2=10

 

tired giving explanation(right/wrong) of previous ones...so left the last one.

 

if there are flaws in the above answers pls correct me.

[pankaj.chawla]

ans to question no 3 according to me is like this-

 

1x5 rectangles = 10(5 vertical +5 horizontal)

1x4 rectangles = 20

similarly

1x3 = 30

1x2 = 40

 

then next series-

2x5 = 8

2x4 = 16

2x3 = 24

 

3x5 = 6

3x4 = 12

 

4x5 = 4

 

total = 170

[pavelbuet]

where is 1X1, 2X2, 3X3 ,4X4, 5X5......they are not only squares but also rectangles

[pankaj.chawla]

sorry pavelbuet.

i failed to realize d simple fact that square is also a rectangle.

 

u r correct.

[rainth]

1. Find the total no. of different mixed doubles tennis games that can be conducted among 7 married couples, if no husband and wife play in the same game.

Women's double tennis tournament.

Men's double tennis tournament.

I reckon the answer should be...........

 

AP1 AP2

------------------

 

BP1 BP2

 

14*12*10*8=13440

 

the answer comes out by filling the gaps.

 

Yah people got attraction to other's wives.

Besides we have to consider abt the people who may not be straight.

hey dudes dont take me otherwise.

 

hey . correct me if I am wrong.

 

2. Find the total no. of ways in which a black square and a whit square can be selected from a chess board such that neither of them lie in the same row nor in the same column.

 

Pavel boss thik ase bondho.

 

32*24 = 768 is the answer

 

3. Find the total of only rectangles on a 5X5 square board.

 

pavel boss tor ta thiki ase.

ami ektu matbori korlam.

equation dekhlei matha ghure.

 

1-2-3-4-5

X-X-X-X-X

X-X-X-X-X

X-X-X-X-X

X-X-X-X-X

 

never include previous number element.Becaues there will be repetation in tat case.

 

starting from row 1 . only element at row 1 are included here.

 

here we can make up 5 rectangle including 1 [necessary condition to build rect]

here we can make up 4 rectangle including 2 and excluding 1[necessary condition to build rect]

here we can make up 3 rectangle including 3 and excluding 1,2[necessary condition to build rect]

here we can make up 2 rectangle including 4 and excluding 1,2,3[necessary condition to build rect]

here we can make up 1 rectangle including 5 and excluding 1,2,3,4[necessary condition to build rect]

 

hence 15 rectangle down here

 

 

apply this strategy to column

 

now we will include and exclude a total row.Never go backwords.

Becaues there will be repetation in tat case.

 

1-2-3-4-5

X-X-X-X-X

X-X-X-X-X

X-X-X-X-X

X-X-X-X-X

 

including row one and taking next rows we will get 5 case with 15 rectangle in each case

 

never go backwords

hence now discard row 1

take row 2 we taking next rows will get 4 case with 15 rect each

 

row 3 ...............3 case 15 rect each

row 4 ...............2 case 15 rect each

row 5 ...............1 case 15 rect each15*15=

 

hence result is 15*(1+2+3+4+5) = 225

 

 

long explanation.

sorry dudes.

4. There are 18 stations in between two stations A and B. Find the total no. of different second class tickets to be printed so that a passenger can travel from one place to any other place.

 

380 is the answer. Pavel correct, boss.

 

 

5

pavel correct.

r = 1/2

 

6

pavel correct.

 

ans----A

[pavelbuet]

Q1.

as i know mixed double tennis means one team made of one girl +1 boy. if this information is correct then rainth no way other than being straight.

[learn_dec]

guys,

i am wondering are these type of questions will ask in GRE????:yuck:

I cant answer these?:(

need help,so that i can learn these kinda tactics.:blush:

 

thanks in advance.

[rainth]

Hey cicerone....watz the ansr of the 1st question.

So long dude.. Come up with new Q and the sollution for the 1st one.

Tat was a promise 1 Q per day or Promise r made 2 b broken ?

[cicerone]

Hey rainth,

 

I am waiting for the simple reason of giving some time to the questions posted. Anyhow here r the answers along with explanations.

 

1.Find the total no. of different mixed doubles tennis games that can be conducted among 7 married couples, if no husband and wife play in the same game.

 

Sol. For a mixed doubles tennis game we need 2 males and 2 females. And for every set of 2 males and 2 females we can conduct two different games i.e if we select two males say M1,M2 and two females say F1,F2 we can conduct two games i.e

game 1:M1F1 as one team and M2F2 as the second team

game 2:M1F2 as one team and M2F1 as the second team

 

So since in the given question there are 7 males and 7 females we can select the required two males in 7C2 ways. Now we have to select the required two females from 5 females (because no husband and wife plays in the same game) and this can be done in 5C2.

 

So, total no. of different games that can be conducted will be 2*7C2*5C2 i.e 420 different games

 

2.Find the total no. of ways in which a black square and a whit square can be selected from a chess board such that neither of them lie in the same row nor in the same column.

 

Sol. Since there are 32 black and 32 white squares on a chess board we can select the required black square in 32 ways. Now in the row from which we have selected the black square there will be 4 white squares and in the column there will be 4 white squares. So the required white square can be selected excluding these 8 squares i.e in 24 different ways.

 

Hence the answer has to be 32*24=768

 

3.Find the total of only rectangles on a 5X5 square board.

 

Sol. Guys, this is really tricky one.

 

Firstly only rectangles = total rectangles- total squares.

 

Now total rectangles:

For a rectangle i need two horizontal lines and two vertical lines.

Since the given board is a 5X5 board it will have 6 horizontal and 6 vertical lines. Now if we select any two horizontal lines and any two vertical lines we will definitely get a rectangle(Remember that every square is a rectangle)

 

So total no. of rectangles will be 6C2*6C2 = 225 (remember there r some squares in these 225)

 

Now total no. of squares will be 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55.

 

Hence total no. of only rectangles will be 225-55 =170

 

4.There are 18 stations in between two stations A and B. Find the total no. of different second class tickets to be printed so that a passenger can travel from one place to any other place.

 

Sol: Totally there r 20 stations including A & B . For a ticket a passenger can select any two stations which is 20C2. The same no. of tickets must be printed while he is coming back from B-A.

So total no. of tickets will be 2*20C2 = 380

Some other guys have calculated this in other way which is also correct.

 

5. In an infinite GEOMETRIC PROGRESSION each term is equal to sum of the all its following terms. Find the common ratio.

 

Sol: Let the series be a, ar, ar^2, ar^3,...............

Given that a = ar + ar^2 + ar^3 + ar^4 +.................

i.e a = ar/1-r

i.e 1-r= r

Hence r = 1/2

 

6. A father has 6 children. He takes every 3 out of them to a park everyday with no 3 being repeated.

a. How many times does father go to the park? (in one complete cycle)

b. How many times does each child go to the park?(in one complete cycle)

 

Sol: Father has to go for 6C3 = 20 times

Now each child goes with the other two of his brothers selected from

the remaining 5 i.e 5C2 = 10 times

 

 

I hope this is clear to everyone.

 

Here comes our next set of questions( I'll continue the numbering of the queries)

 

7. A solution contains 18% alcohol. 8 litres of the solutions is taken out and replaced with water. The resultant solution contains 15% alcohol. Find the total volume of the solution.

 

8. Fresh grapes contain 90% water and Dry grapes contain 10% water. Find the amount of Dry grapes that can be obtained from 20 Kg of Fresh grapes

 

9. A piece of paper which is in the shape of a right angled triangle is cut along a line parallel to the hypotenuse such that the length of the hypotenuse is reduced by 35%. Find the area of the resulting smaller right angled triangle if the area of the original triangle is 30 square units.

 

10. A basketball team has played 25 matches and won 40% of them. Find the minimum no. of matches that the team should still play inorder to attain an overall success rate of 75%.

 

11. A ball is dropped from a height of 50 mts. It touches the ground and bounces back 4/5 of the height from which it was dropped, again it reaches the ground and bounches back 4/5 of the previous height from which it reached the ground . In this way the ball finally came to rest. Find the total distance travelled by the ball by the time it came to rest.

 

12. I have 6 coins of different denominations in my pocket. If i take them out of my pocket one after another, what is the probability of taking out the 6 coins in descending order of their denominations.

 

Waiting for the answers

[rainth]

Hey cicerone and all,

 

sry I missed the mixed double in the question.

My math sucks indeed. Gotta postpone GRE 4 abt 4 months more.

 

Hey does the only rectangle really mean a rect ab witha!=b

I m not getting such def in mathworld.

They says a square is a degenerated rect

 

 

12. I have 6 coins of different denominations in my pocket. If i take them out of my pocket one after another, what is the probability of taking out the 6 coins in descending order of their denominations.

 

is it 1/6!

[manish8109]

7.

alcohol=18v/100

Amount of solution taken out=8

amount of alcohol present in it=8*18/100

18v/100 - 8*18/100 =15v/100

=> 3v/100=8*18/100

=>V=48

 

8. In 1 kg of fresh grapes (100/9 + 100) gm of dry grapes will be produced

 

so,

for 20 kg of fresh grapes ,

amount of dry grapes= 20(100/9 + 100) = 2+2/9 = 2.22 kg(ANS)

 

9. area of triangle= 1/2 a* b=30 => ab=60----(1)

 

hypotenuse=sqrt(a^2+b^2)

new hypotenuse= sqrt(a^2+b^2)*65/100

= sqrt[(a^2 +b^2)* 65^2/100^2]

 

new sides be x & y,so,

sqrt(x^2+y^2)= sqrt(a^2*65^2/100^2 + b^2*65^2/100^2)

 

x^2+y^2 = a^2 *65^2/100^2 +b^2*65^2/100^2

 

so, x^2= a^2*65^2/100^2 & y^2= b^2*65^2/100^2

x= 65a/100=0.65a & y=65b/100=0.65b

 

new area = 1/2 * (0.65)^2 *ab = 1/2 * (0.65)^2 *60 =12.8(ANS)

 

10. matches played=25

won 40% i.e= 25 *40/100 =10 mathches

so,

minimum matches be x

(10+x)/(25+x) =75/100=3/4

=> 40+4x= 75+3x

=> x=30(ANS)

 

11.

50 + 2*50(4/5 + 16/25 +64/125+------)

=> 50 +100(4/5/(1-4/5) )

=> 50 + 400

=>450 mts(ANS)

[krishnakanthc]

12. I have 6 coins of different denominations in my pocket. If i take them out of my pocket one after another, what is the probability of taking out the 6 coins in descending order of their denominations.

 

Ans=> there is only one way to take out the coins in dscending order .. or imagine taing out is like placing the coins in descending order on a table .. there are 6! ways to do this

 

and there is only way in which we can place the coins in that order ... 1/6!

 

= 1/ 720

must be the answer

[krishnakanthc]

12. I have 6 coins of different denominations in my pocket. If i take them out of my pocket one after another, what is the probability of taking out the 6 coins in descending order of their denominations.

 

Ans=> there is only one way to take out the coins in dscending order .. or imagine taing out is like placing the coins in descending order on a table .. there are 6! ways to do this

 

and there is only way in which we can place the coins in that order ... 1/6!

 

= 1/ 720

must be the answer

[pavelbuet]

q3.

cicerone, ur sol contradicts with the definition of rectangle

 

q9.

resulting new triangle is congruent with the old one. so, all corresponding sides will be of same ratio, a1/a2=b1/b2=c1/c2. in this case the ratio is .65 ,assuming old triangle with sides a2,b2,c2 where c2 is hypotenuse.

 

so, new triangle area= 1/2 * a2 * b2

=1/2 * (.65*a1 * .65*b1)

= (.65)^2 *(1/2*a1*b1)

= 12.8

[cicerone]

Hi pavelbullet, i am 100% true as far as my answers r concerned. Please find out the flaw with my logic

[Gurleen Kaur]

Even i think the reply for ques.3 is 170.

i took the results row-wise For row1, no. of 1*2,1*3,1*4,1*5 rectangles are 4,3,2,1 Total=10total no. of single rows r 5. so we have in all 10*5=50.

For 2-row combinations, we have 2*1,2*3,2*4,2*5 possibilities each with 5,3,2,1 with total 11. No.of two row combi's are 4.

Total =11*4=44.

similarly for 3-row,4-row and 5-row combis. we have score as 12*3,13*2,14*1

in all we have=50+44+36+26+14=170 possible rectangles.

Someone tell the correct reply.....

"Me+God=Majority"

 

Even i think the reply for ques.3 is 170.

i took the results row-wise For row1, no. of 1*2,1*3,1*4,1*5 rectangles are 4,3,2,1 Total=10total no. of single rows r 5. so we have in all 10*5=50.

For 2-row combinations, we have 2*1,2*3,2*4,2*5 possibilities each with 5,3,2,1 with total 11. No.of two row combi's are 4.

Total =11*4=44.

similarly for 3-row,4-row and 5-row combis. we have score as 12*3,13*2,14*1

in all we have=50+44+36+26+14=170 possible rectangles.

Someone tell the correct reply.....

"Me+God=Majority"

[pavelbuet]

by definition a square is always a rectangle but the opposite may not be true. so we can not drop 1X1, 2X2, 3X3, 4X4 and 5X5 squares from the result. you want to say that merely writing 'only rectangles' is enough to preclude squares. i don't think so..it just tell us not to consider other shapes...with more than 4 sides, thats all..no more no less.

 

cicerone, throw new questions man..tired with old ones.