Official Quant thread for CAT 2013

MBA CAT 2024
30009
@albiesriram said:
From a point P, the tangents PQ and PT are drawn to a circle with centre O and radius 2 units. From the centre O, OA and OB are drawn parallel to PQ and PT respectively. The length of the chord TQ is 2 units. Find the measure of the ∠AOB.(a) 30° (b) 90° (c) 120° (d) 45°
120
30010
@ScareCrow28 said:
Distance between them = 300 = (a+b) where a and b are perpendicular distances to each other.Also, Shortest distance between them = _/(a^2 + b^2) = _/((a+b)^2 - 2ab) = _/(300^2 - 2ab)Minimise the shortest distance...Maximise aba+b/2 >= _/abMaximum ab = 300^2/4Shortest distance = _/(300^2/2) = 150_/2 ??
OA : 200
30011
@heylady said:
can this be looked the other way as: number of solutions of the equation: 2x+3y+5z=10?
No, we need x+y+z = 10 and we want to find number of possible values of 2x+3y+5z.

Problem is we need distinct values, so enumeration is safest given that small number of cases...

regards
scrabbler
30012
@DivineSeeker said:
A person can buy 20 sparrows for a rupee, a pigeon for a rupee, and a peacock for Rs. 5. Find the number of birds of each type he needs to buy if he wants to buy total 100 birds for Rs. 100 so that he buys at least one bird of each type.ans is 19 peacock=95rs+80 sparrows=4 rs+ 1 pigeon=1 rs.is there any mathematical approach for these kindaa questions if we pile up the data???
S + Pig + Pea = 100
S/20 + Pig + 5*Pea = 100

Subtracting, 19S/20 = 4*Pea
S = 80/19 * Pea ...Now Pea has to be multiple of 19 and can only be 19
So S = 80
Pea = 19
And Pig = 1

30013
@DivineSeeker said:
A yearly payment to a servant is Rs. 90 plus one turban. the servant leaves the job after 9 months and receives Rs. 65 and a turban. find the price of the turban..a little confusion...plz post with approach...thanks
Orally karna ho to...

12 months = 90 + turban
9 months = 65 + turban

Check the difference between the two cases => 3 months salary = 90 - 65 = 25

So 12 months salary should be 100 which is 90 + turban => turban = 10.

regards
scrabbler
30014
@audiq7 said:
120??
u r right bro...me also getting 120.
30015
@amresh_maverick said:
OA : 200
But, 150_/2 to 200 se bhi kam hai sir.. Dekhlo ap khud satisfy karra hai ye Q ko.. length = breadth = 150
Distance between them = 150_/2

P.S. Sorry sir 😛 200 se zada hai... Slotion hai??
30016
There are five consecutive integers a, b, c, d and e such that a a2 + b2 + c2 = d2 + e2 . What is/are the possible value(s) of b?

(a) 0 (b) 11 (c) 0 and €“11 (d) €“1 and 11.
30017
@ScareCrow28 said:
But, 150_/2 to 200 se bhi kam hai sir.. Dekhlo ap khud satisfy karra hai ye Q ko.. length = breadth = 150Distance between them = 150_/2
150_/2= 212.1
30018
@albiesriram said:
There are five consecutive integers a, b, c, d and e such that a a2 + b2 + c2 = d2 + e2 . What is/are the possible value(s) of b?(a) 0 (b) 11 (c) 0 and €“11 (d) €“1 and 11.
(a-2), (a-1), a, (a+1), (a+2)

3a^2 -6a + 5 = 2a^2 + 6a + 5
=> a^2 -12a = 0
a can be 0 or 12

b = -1 or 11 .. ??
Option (d)
30019
@ScareCrow28 said:
But, 150_/2 to 200 se bhi kam hai sir.. Dekhlo ap khud satisfy karra hai ye Q ko.. length = breadth = 150Distance between them = 150_/2P.S. Sorry sir 200 se zada hai... Slotion hai??
PFA


30020
@amresh_maverick said:
PFA
Sir inhone case lia hai..where A'B = B'C = 50m
How to prove that this is possible?
30021
@ScareCrow28 said:
(a-2), (a-1), a, (a+1), (a+2)3a^2 -5a + 5 = 2a^2 + 6a + 5=> a^2 -11a = 0a can be 0 or 11b = -1 or 10
logic is sound
but calculation mistake in line 2... 3a^2 - 5a 6a+ 5 = 2a^2 + 6a + 5 :P
30022
@albiesriram said:
There are five consecutive integers a, b, c, d and e such that a a2 + b2 + c2 = d2 + e2 . What is/are the possible value(s) of b?(a) 0 (b) 11 (c) 0 and €“11 (d) €“1 and 11.
d. 11 and -1
a,a+1,a+2,a+3,a+4 for a,c,d,e respectively
we get after substituting in a2 + b2 + c2 = d2 + e2
a^2-8a-20=0
a=10,-2
so b=11,-1 ?

30023
@iLoveTorres said:
bro yeh kaise aaya? mujhe toh kafi time laga to project and calculate the radius of the circle
30024
@cynara said:
logic is soundbut calculation mistake in line 2... 3a^2 - 5a 6a+ 5 = 2a^2 + 6a + 5with a=0,12so answer will come out to be -1,11 OA= D @albiesriram
Ye sab galtiyaan @scrabbler ki wajah se hori h Orally karne ke chakkar me sab galat hora hai.. 😛
30025
Walking 7/11 of his usual speed, a man is 16 minutes late. The usual time taken by him to cover that distance is:

1. 1 hour 2. 28 min. 3. 12 min. 4. 8 min.20 sec.
30026
@albiesriram said:
Walking 7/11 of his usual speed, a man is 16 minutes late. The usual time taken by him to cover that distance is:1. 1 hour 2. 28 min. 3. 12 min. 4. 8 min.20 sec.
2. 28 min
30027
@albiesriram said:
Walking 7/11 of his usual speed, a man is 16 minutes late. The usual time taken by him to cover that distance is:1. 1 hour 2. 28 min. 3. 12 min. 4. 8 min.20 sec.
16+t = 11/7 t
t = 28
30028
Raju has 128 boxes with him. He has to put atleast 120 oranges in one box and 144 at the most. Find the least number of boxes which will have the same number of oranges.

1. 5 2. 103 3. 6 4. Cannot be determined