Official Quant thread for CAT 2013

MBA CAT 2024
23187
@Faruq said:
There are “n” necklaces in a safe box (n > 1). Every necklace has the same number of diamonds. Each necklace has at least 2 diamonds. The total number of diamonds in these “n” necklaces is between 500 and 600. If this data is sufficient to find the value of n, then what is the value of “n”?a. 19 b. 23 c. 29 d. None of these
23 ?
23188
@Faruq said:
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
m getting 6/7
23189
@Faruq said:
There are “n” necklaces in a safe box (n > 1). Every necklace has the same number of diamonds. Each necklace has at least 2 diamonds. The total number of diamonds in these “n” necklaces is between 500 and 600. If this data is sufficient to find the value of n, then what is the value of “n”?a. 19 b. 23 c. 29 d. None of these
No. of diamonds per necklace = x

n*x lies between 500 and 600 , and n > 1, x >=2

now, if n and x are distinct then it would not be sufficient to find the value of n, as they can be interchanged

so, n and x are equal.

Then again, if n and x are not prime then we can have multiple possibilities of n.
So, Total diamonds = square of a prime number lying between 500 and 600

=> 529 => n = 23 ?
23190
@Faruq said:
There are “n” necklaces in a safe box (n > 1). Every necklace has the same number of diamonds. Each necklace has at least 2 diamonds. The total number of diamonds in these “n” necklaces is between 500 and 600. If this data is sufficient to find the value of n, then what is the value of “n”?a. 19 b. 23 c. 29 d. None of these
n*x must be a prime square to answer the ques...

23^2 satisfies...hence 23...
23191
@ScareCrow28 said:
@bodhi_vriksha Please tell us the solution P&C; question wherein 4couples were to be arranged.. I got 21 ways? What's the answer??
I guess you missed my earlier query regarding the question - "Let's make it easy.....First tell me the answer for only two pairs and then for three pairs...see if you can get up to something from this.."

By the way, answer for four pairs is = 60 :)

For those, who may have lost the question in this math forest here it is - 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired? Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.
23192

Bit easy one now - please don't use paper/pen, just use your brain :)

[Geo] In a 3-4-5 right triangle, semicircle is drawn whose diameter lie on side with length 4. Also the semicircle touches the other two sides as well.
Find the radius of the semicircle.

23193
@bodhi_vriksha said:
Bit easy one now - please don't use paper/pen, just use your brain [Geo] In a 3-4-5 right triangle, semicircle is drawn whose hypotenuse lie on side with length 4. Also the semicircle touches the other two sides as well.Find the radius of the semicircle.
1.5? Visualising similar triangles here...

regards
scrabbler
23194
@scrabbler said:
1.5? Visualising similar triangles here...regardsscrabbler
kaise kara....mujhe toh ques hi nahi samajh aya...
23195
@bodhi_vriksha said:
Bit easy one now - please don't use paper/pen, just use your brain [Geo] In a 3-4-5 right triangle, semicircle is drawn whose hypotenuse lie on side with length 4. Also the semicircle touches the other two sides as well.Find the radius of the semicircle.
1.5...

"diameter" rather than "hypotenuse" lies on 4cm length side..
23196
@Logrhythm said:
kaise kara....mujhe toh ques hi nahi samajh aya...
i guess.. hypotenuse lies on 4 cm length side me confuse ho gaya hoga... it must be diameter of semicircle which lies on 4 cm length side..
23197
@bodhi_vriksha said:
I guess you missed my earlier query regarding the question - "Let's make it easy.....First tell me the answer for only two pairs and then for three pairs...see if you can get up to something from this.."By the way, answer for four pairs is = 60 For those, who may have lost the question in this math forest here it is - 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired? Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.
Let the couples be AB, CD, EF, GH (doesn't matter who is the guy, who the girl).

Now A can be paired with any of the other 6 (leaving B). Suppose A is paired with C.

Now look at B and D. There are 2 cases:

1) B is paired with D. In this case, E can be paired with either G or H and F with the remaining person so 2 ways.

OR

2) B is paired with any of E, F, G or H (4 cases) in each case D has 2 options (reason: if B is paired with E, D cannot be with F (as that would leave G and H paired), so DG or DH will happen. Similarly for any other pairing of B) So this gives us 4 x 2 = 8 ways.

So A (6 ways) and B (2 or 8 = 10 ways) gives 6 x 10 = 60 ways total.

regards
scrabbler
23198
@bodhi_vriksha said:
Bit easy one now - please don't use paper/pen, just use your brain [Geo] In a 3-4-5 right triangle, semicircle is drawn whose hypotenuse lie on side with length 4. Also the semicircle touches the other two sides as well.Find the radius of the semicircle.
1.5 ?
23199
@bodhi_vriksha said:
Bit easy one now - please don't use paper/pen, just use your brain [Geo] In a 3-4-5 right triangle, semicircle is drawn whose hypotenuse lie on side with length 4. Also the semicircle touches the other two sides as well.Find the radius of the semicircle.
It will be the in-radius of the the triangle with sides 6,5,5 ?

r = Area/s = ((1/2)*(6*4))/(1/2)*(6 + 5 + 5) = 24/16 = 3/2 = 1.5 ?
23200
@Logrhythm said:
kaise kara....mujhe toh ques hi nahi samajh aya...

Note that the side of 4 will be divided into 2 parts of r and (4-r)... also note that the radius dropped to the hypotenuse is perpendicular to it and so will form a right triangle similar to original big one. So r / 4-r should be 3 : 5 and hence we divide 4 in the ratio 3 : 5 to get 1.5 : 2.5

Could also do by realising that the 3 and 5 (hyp) sides are both tgts to the semicircle, so the point where the circle meets the hyp divides it in 3 and 2...now the radius and the "2" part are in the ratio 3 : 4 using similar triangles hence 1.5....

regards
scrabbler
23201
@YouMadFellow said:
It will be the in-radius of the the triangle with sides 6,5,5 ?r = Area/s = ((1/2)*(6*4))/(1/2)*(6 + 5 + 5) = 24/16 = 3/2 = 1.5 ?
Yeah I saw that and thought the calculation would be too hard to do mentally so left it and went for another way.... stupid of me.

regards
scrabbler
23202
@Logrhythm said:
kaise kara....mujhe toh ques hi nahi samajh aya...
can do like this..also..
23203
@scrabbler said:
Let the couples be AB, CD, EF, GH (doesn't matter who is the guy, who the girl).

Now A can be paired with any of the other 6 (leaving B). Suppose A is paired with C.

Now look at B and D. There are 2 cases:

1) B is paired with D. In this case, E can be paired with either G or H and F with the remaining person so 2 ways.

OR

2) B is paired with any of E, F, G or H (4 cases) in each case D has 2 options (reason: if B is paired with E, D cannot be with F (as that would leave G and H paired), so DG or DH will happen. Similarly for any other pairing of B) So this gives us 4 x 2 = 8 ways.

So A (6 ways) and B (2 or 8 = 10 ways) gives 6 x 10 = 60 ways total.
regardsscrabbler

but wat is wrong wid 8C2 -4 .......
y m i getting only 24 while ur nswer sounds more sensible now....
23204
@Dexian said:
but wat is wrong wid 8C2 -4 .......y m i getting only 24 while ur nswer sounds more sensible now....
8C2 is choosing just 1 couple (and leaving the other 6 to do group dance). You need to pair up the other 6 people as well :)

regards
scrabbler
23205
@scrabbler said:
8C2 is choosing just 1 couple (and leaving the other 6 to do group dance). You need to pair up the other 6 people as well
regardsscrabbler

ohhhkkkkkk.....
u r right.........
23206
The guards of two goods trains are in the respective last carriages of their trains. The two trains A and B of length 200 m and 300 m respectively, begin to enter a 500-metre long tunnel from opposite ends. The guard of one of the trains wants to pick up a flag from the other guard. How long after the instant when the two trains begin to enter the tunnel, does this exchange occur? (Given that speed of train A = 2 m/s, speed of train B = 3 m/s)