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. In 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 FIGURE
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<=r<=5 anc 1<=c<=5
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