Official Quant thread for CAT 2013

MBA CAT 2024
9841
@chillfactor said:
Yup, 448 and n*2⁽ⁿ⁻¹⁾ is correct for last question.Sum of the areas of all the triangles having vertices as vertices of cube having dimension 1 x 1 x 1 is (m + √n + √p). Find (m, n, p).
12 + rt288 + rt27?

I got 3 sizes of triangle, 24 + 24 + oops not 6 8 of each

Edit: 12 + rt288 + rt48?

regards
scrabbler
9842
@chillfactor said:
Yup, 448 and n*2⁽ⁿ⁻¹⁾ is correct for last question.Sum of the areas of all the triangles having vertices as vertices of cube having dimension 1 x 1 x 1 is (m + √n + √p). Find (m, n, p).
(12,288,48) ? EDIT

area of triangles 2 sides on edges of the cube = 1/2 * 1 * 1
total = 1/2 * 6 * 4 = 12

area of triangles with one side as the diagonal of a face and othe side as the edge of the cube = 1/2 * rt(2) * 1
total = 1/rt(2) * 12 * 2 = 12 * rt(2)

area of triangles with all sides as diagonals of faces of the cube = rt(3)/4 * 2 = rt(3)/2
total = rt(3)/2 * 8 = 4 * rt(3)

sum = 12 + 12rt(2) + rt(48)

m = 12
n = 288
p = 48
9843
@chillfactor said:
Yup, 448 and n*2⁽ⁿ⁻¹⁾ is correct for last question.Sum of the areas of all the triangles having vertices as vertices of cube having dimension 1 x 1 x 1 is (m + √n + √p). Find (m, n, p).
16+8rt(2)+4rt(3)?
9844

Yup, 12 + 12√2 + 4√3 is correct for last one.


Basically there are 24 triangles having dimension 1, 1, √2; 24 triangles having dimension 1, √2, √3; and 8 triangles having dimension √2, √2, √2

If P is product of all numbers (y - x) {not necessarily distinct}, where x and y are integers such that 1 ≤ x
9845
@chillfactor said:
Yup, 12 + 12√2 + 4√3 is correct for last one.Basically there are 24 triangles having dimension 1, 1, √2; 24 triangles having dimension 1, √2, √3; and 8 triangles having dimension √2, √2, √2If P is product of all numbers (y - x) {not necessarily distinct}, where x and y are integers such that 1 ≤ x
640 ?

x = 1
y = 2,3,4,5,6,7,....,40
difference = 1,2,3,4,...,39

x = 2
difference = 1,2,3,4,...,38

.
.

x = 38
difference = 1,2

x = 39
difference = 1

Product = 1! * 2! * 3! *... * 39!

Max power of 2 in each factorial = 0,1,1,3,3,4,4,7,7,8,8,9,9,10,10,15,15,16,16,17,...,22,22, 31,31,32,32,33,33,34,34

Sum = 1*2 + (3+4)*2 + (7+8+9+10)*2 + (15+16+...+22)*2 + (31+32+33+34)*2
= 2*(1 + 7 + 34 + 148 + 130)
= 640
9846
@chillfactor

38+2*36+34+3*32+..................2
9847
@chillfactor said:
Yup, 12 + 12√2 + 4√3 is correct for last one.Basically there are 24 triangles having dimension 1, 1, √2; 24 triangles having dimension 1, √2, √3; and 8 triangles having dimension √2, √2, √2If P is product of all numbers (y - x) {not necessarily distinct}, where x and y are integers such that 1 ≤ x
680?
9848
@chillfactor said:
Yup, 12 + 12√2 + 4√3 is correct for last one.Basically there are 24 triangles having dimension 1, 1, √2; 24 triangles having dimension 1, √2, √3; and 8 triangles having dimension √2, √2, √2If P is product of all numbers (y - x) {not necessarily distinct}, where x and y are integers such that 1 ≤ x
(40 - 1)*(40 - 2)*.......*(40-39) = 39!
(39 - 1)*(39 - 2)*........(39 - 38) = 38!
.
.
(3 - 1)*(3 - 2) = 2!

0 + 1 + 1 + 3 + 3 + 4 + 4 ........... 34 + 34 = 640
9849
Yes, its 680.

Exponent of 2 in 1^39 * 2^38 * 3^37 * .... * 39^1

So, it will be:-
E = (2 + 4 + .... + 38) + (4 + 8 + ... + 36) + (8 + 16 + 24 + 16) + (8 + 24) + 8 = 680

A group of children participated in a competition. Winner got n points and the child in r-th position got (n + 2 - 2r) points. Total of all the points secured by all the children is 2009. Find the smallest possible value of n.
9850
@chillfactor said:
Yes, its 680.Exponent of 2 in 1^39 * 2^38 * 3^37 * .... * 39^1So, it will be:-E = (2 + 4 + .... + 38) + (4 + 8 + ... + 36) + (8 + 16 + 24 + 16) + (8 + 24) + 8 = 680A group of children participated in a competition. Winner got n points and the child in r-th position got (n + 2 - 2r) points. Total of all the points secured by all the children is 2009. Find the smallest possible value of n.
7 hoga shayad (of course 1 is technically the minimum but since it says children....)

regards
scrabbler
9851
@scrabbler said:
7 hoga shayad (of course 1 is technically the minimum but since it says children....)regardsscrabbler
No, n = 1 means winner got 1 point, first runner up got -1 points and so on...
1 + (-1) + (-3) + .... can never be 2009

n = 7, means winner got 7, 1st, 2nd 3rd runner up got 5, 3, 1 points and so on
7 + 5 + 3 + 1 + (-1) + ... can never be 2009
9852
@chillfactor said:
No, n = 1 means winner got 1 point, first runner up got -1 points and so on...1 + (-1) + (-3) + .... can never be 2009n = 7, means winner got 7, 1st, 2nd 3rd runner up got 5, 3, 1 points and so on7 + 5 + 3 + 1 + (-1) + ... can never be 2009

Oh sorry. I meant 7 children. n = 293.

regards
scrabbler
9853
@scrabbler said:
Oh sorry. I meant 7 children. n = 293
No, it can be further minimized
9854
@grkkrg
Price at market = 1/(x-2)

Price at Scrooge's shop = 1/(x)
ye kaise hua? please explain
9855

@chillfactor said:
Yes, its 680.Exponent of 2 in 1^39 * 2^38 * 3^37 * .... * 39^1So, it will be:-E = (2 + 4 + .... + 38) + (4 + 8 + ... + 36) + (8 + 16 + 24 + 16) + (8 + 24) + 8 = 680A group of children participated in a competition. Winner got n points and the child in r-th position got (n + 2 - 2r) points. Total of all the points secured by all the children is 2009. Find the smallest possible value of n.

Score are given in the seque. => n ,n-2,n-4,n-6,n-8 and so on.......

take only one term =n
Sum of first two terms =2n-2 =2(n-1)
Sum of the first three terms =3n-6 =3(n-2)
Sum of the first four terms =4n-12 =4 (n-3)
and so on.



Factors of 2009 =1,7,41,49,287,2009

take 1(n)=2009

take 7 = 7(n-6)=2009
n=293

take 41 =41(n-40)=2009
n=89

take 49 = 49(n-48) =2009
n=89

take 287 = 287(n-286) =2009
n=293

take 2009 = 2009(n-2008) =2009
n=2009

So minimum value of n is 89

9856
@IIM-A2013 said:
@grkkrgPrice at market = 1/(x-2)Price at Scrooge's shop = 1/(x)ye kaise hua? please explain
x chocolates at 1 rupee at Scrooge's shop
(x-2) chocolates at 1 rupee at market
9857

pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.

9 pm
7 pm
5 pm
1 pm

9858
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
7pm
9859
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
7 pm?

1/p = 1/8
1/q = -1/20
2/8 + x(1/8 - 1/20) = 1
x(6/80) = 3/4
x = 20/2 = 10
So 7 pm
9860
@19rsb said:
7pm
steps