Official Quant thread for CAT 2013

@amresh_maverick said:
Q>78 identical cubes each with 2 cm edge are joined together to form a cuboid. If the perimeter of the base of the cuboid is 64 cm, then the number of cubes along the height of the cuboid is
height = 4cm..
2 cubes along the height ??

@ScareCrow28 said:
P.S. Either my brain is faster than I think OR My keyboard needs hospitalization ( Silly mistakes)..
Latter ! 😛 :mg:
sry for spam :P
@amresh_maverick said:
options are : a 1381 b 1281 c 1481 d None of theseQ>78 identical cubes each with 2 cm edge are joined together to form a cuboid. If the perimeter of the base of the cuboid is 64 cm, then the number of cubes along the height of the cuboid isPS: @ChirpiBird , kaun si movie dekhi aaj ?
OA : 2

Q>
Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
@amresh_maverick said:
OA : 2Q>Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
2 / 285?

8 / 20C3...


Edit: missed something, recalculating...

11 / 20C3 ? i.e. 11/1140...

regards
scrabbler

@amresh_maverick said:
OA : 2Q>Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
10/20C3 ?
@rnishant231 said:
10/20C3 ?
need to go to basics of counting.... 8/20C3
@amresh_maverick said:
OA : 2Q>Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
8/20C3?.. Highly Doubtful
@amresh_maverick said:
OA : 2Q>Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
OA : 11/1140 some how could do it myself 😃

hint : take r as 3/2 and 4/3 and u will get 3 more series
@scrabbler said:
2 / 285?8 / 20C3...Edit: missed something, recalculating...11 / 20C3 ? i.e. 11/1140...regardsscrabbler
11/20C3 .........

haye!!!
mine too 11cases... second time same answer as @scrabbler ...
#dear diary moment again!!

@amresh_maverick : watched kung fu panda
#love_Po!

Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x – 1), (x – 2), (x – 3),...and (x – 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders.

Find the value of f(52) + f(0).
@amresh_maverick said:
Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x – 1), (x – 2), (x – 3),...and (x – 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders.Find the value of f(52) + f(0).
F(x) = (x-1)*....*(x-51) + x

Put values, F(52) + F(0) = 52 ??
@amresh_maverick said:
OA : 2Q>Three distinct numbers are randomly selected from the first 20 natural numbers. Find the probability that the selected numbers are in a geometric progression having common ratio greater than 1.
total = 20C3
fav ways:
1 2 4
1 3 9
1 4 16
2 4 8
2 6 18
3 6 12
4 8 16
5 10 20
above case hav ratio as natural no..but ratio cud be fraction too..
r=3/2
8 12 18

9cases i guess.
prob= 9/20C3 ? else i missd few cases....

..................................
Ramesh and Suresh are running on a circular track of length 1200m in opposite direction. everytime they meet, they exchange their speed as well as directions. what is the shortest distance btw their 13th and 23rd meeting points, if the ratio of their speeds is 1:5?
a) 200m
b)400m
c) 800m
d) 1000m

Plz quote me
@amresh_maverick said:
Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x – 1), (x – 2), (x – 3),...and (x – 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders.Find the value of f(52) + f(0).
52
@amresh_maverick said:
Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x – 1), (x – 2), (x – 3),...and (x – 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders.Find the value of f(52) + f(0).
52? when you divide by x-1 it means if you substitute x=1 in f(x) you get 1 i.e f(1)=1;f(2)=2....f(51)=51
@amresh_maverick said:
Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x – 1), (x – 2), (x – 3),...and (x – 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders.Find the value of f(52) + f(0).
OA: 52

In three distinct regular polygons, it is known that the measure of the internal angle of one regular polygon exceeds the measure of the internal angle of the other two regular polygons by 15° and 27° respectively. Furthermore, the sum of the measures of the external angles of all the three regular polygons is 177°. What is the sum of the number of sides of all these three regular polygons ?
@aditi88 said:
Ramesh and Suresh are running on a circular track of length 1200m in opposite direction. everytime they meet, they exchange their speed as well as directions. what is the shortest distance btw their 13th and 23rd meeting points, if the ratio of their speeds is 1:5?a) 200mb)400mc) 800md) 1000mPlz quote me
look at 13th and 23rd meet the speed doesnt change.. but the catch here is the direction..
Now see the pattern for the distance R will travel
1st meet 1/6*1200=200
2nd meet 5/6*1200=1000(remember this is in opposite direction so it is 800 in the opposite direction from the starting point or 400 in the normal direction from the starting point)
3rd meet 1/6*1200=200. Now he is at 600 from the starting point
4th meet 5/6*1200=1000. Now he is 400 in the opposite direction from the starting point or 800 in the normal direction.
5th meet 1/6*1200=200. Now he is 1000 from the starting point.
6th meet 5/6*1200=1000. Here he will be at the starting point.

therefore at every 6th meeting R will be at the starting point. So for 13th meet he will be 200 from the starting point and for 23rd meet he will be 1000 from the starting point. but the shortest distance will be 1000-1200 and 0-200 so 400.
@amresh_maverick said:
OA: 52In three distinct regular polygons, it is known that the measure of the internal angle of one regular polygon exceeds the measure of the internal angle of the other two regular polygons by 15째 and 27째 respectively. Furthermore, the sum of the measures of the external angles of all the three regular polygons is 177째. What is the sum of the number of sides of all these three regular polygons ?
5 + 6 + 8 = 19?

regards
scrabbler

@aditi88 said:
Ramesh and Suresh are running on a circular track of length 1200m in opposite direction. everytime they meet, they exchange their speed as well as directions. what is the shortest distance btw their 13th and 23rd meeting points, if the ratio of their speeds is 1:5?a) 200mb)400mc) 800md) 1000mPlz quote me
It is clear that they will meet at 6 points = sum of ratios of speeds
l/6, 2l/6,3l/6,4l/6,5l/6 and l - those meeting points

first meeting at l/6 and start counting
13th meeting will be at the same place i.e l/6

similarly 23 meeting will be at l/6 but in opposite to the 13th one

distance b/w = 2l/6 = 400
@amresh_maverick said:
OA: 52In three distinct regular polygons, it is known that the measure of the internal angle of one regular polygon exceeds the measure of the internal angle of the other two regular polygons by 15째 and 27째 respectively. Furthermore, the sum of the measures of the external angles of all the three regular polygons is 177째. What is the sum of the number of sides of all these three regular polygons ?
19?