Official Quant thread for CAT 2013

MBA CAT 2024
7321
@maneeshp said:
square terms -2
is 1 - 2 = 0 😲 ?
7322
@rachit_28 it's 23
7323
@chillfactor said:
a1, a2, a3, ...., an are positive integers such that a1 + a2 + .... + an = 2005. Find the maximum possible value of a1*a2*a3*.....*an
3^668 * 1 :mg:
7324
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11
a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
7325

4, 5, 9, 18, 34, ?


43
49
50
59
7326
@Cat.Aspirant123 said:
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
A's speed = 3x
B's speed = 2x

14*(3x) + 9*(2x) = 120 => 60x = 120 => x = 2
A's speed = 3x = 6 km/hr ?
7327
@Cat.Aspirant123 said:
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
A's speed --->6kmph
B's speed---->4 kmph
7328
@Cat.Aspirant123 said:
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
a = 3x
b = 2x

(120 - 15x)/5x = 9

120 - 15x = 45x
=> 2

so , speed of A = 3*2 = 6 km/hr
7329
@rachit_28 said:
4, 5, 9, 18, 34, ?43495059
59

7330
@19rsb said:
125369872*98658314631236878027AB9882736,also A is 4 times B,then find the values of A and B.
Digital sum
7*8=A+B+87
2=A+B+87
B=1,A=4
B=2 A=8 no solution
A=4 ,B=1
7331
@rachit_28 said:
4, 5, 9, 18, 34, ?43495059
Difference is in squares of 1, 2,3,4 etc
4 + 1^2 = 5
5 + 2^2 = 9

=> 34 + 5^2 = 59 ?
7332
@rachit_28 said:
4, 5, 9, 18, 34, ?43495059
59??
5 = 4+1^2
9 = 5+2^2
18 = 9+3^2...and so on
7333
@rachit_28 said:
4, 5, 9, 18, 34, ?43495059
59
7334
@jain4444 said:
3^668 * 1

@chillfactor said:a1, a2, a3, ...., an are positive integers such that a1 + a2 + .... + an = 2005. Find the maximum possible value of a1*a2*a3*.....*an

Answer is 4*3^667

Always express the number as 2a + 3b and maximize b
So, 2a + 3b = 2005
Maximum value of b is 667 (when a = 2)

So, maximum product = 4*3^667

Why 2a + 3b??
We know that al the terms should be equal to get the maximum product. Suppose the no of terms is n.

So, max product = (A/n)^n
We know that x^(1/x) is maximum for x = e
But in our case numbers are integers. Integers that are closest to e are 2 and 3, so the numbers have to be 2 or 3
Now, 2*2*2

Sum of digits of n consecutive integers is not divisible by 9. What is the maximum possible value of n.
7335
@maneeshp said:
wat about this?0*1 = 00+1*2 = 21+2*3 = 72+3*4 =143+4*5 = 234+5*6 = 34
Tried this bro..still can't explain the logic for the 1st term
7336
@maneeshp said:
wat about this?0*1 = 00+1*2 = 21+2*3 = 72+3*4 =143+4*5 = 234+5*6 = 34
first term should be -1 + 0*1 as per ur logic..
7337
@Cat.Aspirant123 said:
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
A has traveled for 5 hours => distance = 5x
B's speed = x/1.5
(120 - 5x)/(2.5x/1.5) = 9
3(120 - 5x) = 45x
360 - 15x = 45x
60x = 360
x = 6kmph
A's speed = 6kmph
B's speed = 4kmph
7338
@chillfactor said:
@chillfactor said:a1, a2, a3, ...., an are positive integers such that a1 + a2 + .... + an = 2005. Find the maximum possible value of a1*a2*a3*.....*anAnswer is 4*3^667Always express the number as 2a + 3b and maximize bSo, 2a + 3b = 2005Maximum value of b is 667 (when a = 2)So, maximum product = 4*3^667Why 2a + 3b??We know that al the terms should be equal to get the maximum product. Suppose the no of terms is n.So, max product = (A/n)^nWe know that x^(1/x) is maximum for x = eBut in our case numbers are integers. Integers that are closest to e are 2 and 3, so the numbers have to be 2 or 3Now, 2*2*2
Isnt it distinct integers 😞 ?
7339
@chillfactor said:
@chillfactor said:a1, a2, a3, ...., an are positive integers such that a1 + a2 + .... + an = 2005. Find the maximum possible value of a1*a2*a3*.....*anAnswer is 4*3^667Always express the number as 2a + 3b and maximize bSo, 2a + 3b = 2005Maximum value of b is 667 (when a = 2)So, maximum product = 4*3^667Why 2a + 3b??We know that al the terms should be equal to get the maximum product. Suppose the no of terms is n.So, max product = (A/n)^nWe know that x^(1/x) is maximum for x = eBut in our case numbers are integers. Integers that are closest to e are 2 and 3, so the numbers have to be 2 or 3Now, 2*2*2
yeah I wasted 1 term by taking it as "1"

a ship is moving at a speed of 30 km/hr. To know the depth of the ocean beneath it, it sends a radiowave which travels at a speed 200m/s. The ship receives the signal after it has moved 500 m. The depth of the ocean is.
(A) (square root 143)/2 km (B) 12 km (C) square root6 (D) 8 km
7340
@maneeshp said:
wat about this?0*1 = 00+1*2 = 21+2*3 = 72+3*4 =143+4*5 = 234+5*6 = 34
got some weird logic..

1^3-1=0 2^3-1=7 3^3-1=26 (odd terms)

even terms
2^2-2=2

4^2-2=14

6^2-2=34

tried to fit smtng.. :P