Official Quant thread for CAT 2013

MBA CAT 2024
12682
@pavimai said:
the sum of the first 30 terms of an AP is4500.the ratio of the first 20 terms and the last ten terms is 4:5.Find the first term???????????
6.5
7.5
6
5
S30=4500
S20/S30-S20 =4/5
9S20=4s30
S20=2000
15(2a+29d)=4500
2a+29d=300
10(2a+19d)=2000
2a+19d=200
d=10
2a=10
a=5
12683
@ravi.theja OAis 5

12684

@pavimai said:
A and B start running simultaneously on a circular track from point O in the same direction .if the ratios of the speed is 6:1 respectively ,then how many times A is ahead of B by a quarter of the length of the track before they meet at O for the first time?4578
as A is 6 times faster, he will overtake B 5 times before they meet at O for the first time

so 5
12685
@pavimai a=5 😃 15 [ 2*a + 29d ] = 4500

10 [ 20a+ 19d ]/ 5 [ 2 (a+20d) + 9d ] = 4/5 ==> 6a=3d substituting u get a=5

p.s : sum to n terms = n/2 [ 2*a +(n-1)d ] ...a-frst term d-common diff
12686
@ravi.theja said:
@pavimai a=5 15 [ 2*a + 29d ] = 450010 [ 20a+ 19d ]/ 5 [ 2 (a+20d) + 9d ] = 4/5 ==> 6a=3d substituting u get a=5p.s : sum to n terms = n/2 [ 2*a +(n-1)d ] ...a-frst term d-common diff
12687
@pavimai said:
@ravi.theja OAis 5
probably they hav considerd as L +L/6 😃 then ans ll b 5
12688
@krum said:
as A is 6 times faster, he will overtake B 5 times before they meet at O for the first timeso 5
yes dont know y do i sit for calculating simple problem
12689
@pavimai said:
the sum of the first 30 terms of an AP is4500.the ratio of the first 20 terms and the last ten terms is 4:5.Find the first term???????????6.57.565

Though puys have solved it before I write a different method using which it can be done almost verbally without use of any pen and paper. (Observe that calculations are really simple in this method)

sum of last 10 terms = 5/(4+5)*4500 = 2500
mean of AP = 4500/30 = 150

since excess of last 10 terms from average = deficit of first 10 terms from the average

now excess of last 10 terms from average = 2500- 10*150 = 1000

=>sum of first 10 terms = 10*150 - 1000 = 500

ATDH.
12690
@manoj_msr said:
shouldnt it be 599 days..?Assuming total work to be of 300 units.. A works at 6 units/ day B at 5U/day.. so in 2 days 1 unit.. so for 300 units 600 days.. but the wall will be complete one day before so 599..?Edit : it should be 593 days.. See ATDH sir's reply below..
ya right ... i do not know what i was doing
12691
@anytomdickandhary said:
say wall is 300 bricks ( LCM of 50 and 30)A places (300/50) = 6 bricks per daysB demolishes (300/60) = 5 bricks per daySo in a block of 2 days net bricks laid = (6-5) = 1after 294 bricks are laid..... next day A will put 6 bricks and wall will be complete..so 2*294 + 1 = 593 daysATDH.
sir, it should have been 589. :-|
12692
@Zedai said:
sir, it should have been 589. :-|
yes sir.....

please refer to my signature...

ATDH.
12693
@anytomdickandhary said:
yes sir.....please refer to my signature...ATDH.


So, apparently i have witnessed GOD committing mistake! :D
12694
@pavimai said:
A and B start running simultaneously on a circular track from point O in the same direction .if the ratios of the speed is 6:1 respectively ,then how many times A is ahead of B by a quarter of the length of the track before they meet at O for the first time?4578
5?
12695
In a convex polygon,interior angle is in AP with samllest angle 120 and common difernce 5.find numbr of sides of polygon?
15
17
18
9
12696

Define a number k such that it is the sum of the squares of the first M natural nos(i.e k=1^2+2^2+...+m^2), where ma) 10 b) 11
c) 12 d) 13

12697
@ravi.theja 18..?
12698
@sails said:
@ravi.theja 18..?
OA nahi pata..post ur solution bro
12699
@ravi.theja n/2(240+(n-1)5) must be a whole number....so n is a multiple of 2
12700
@ravi.theja said:
In a convex polygon,interior angle is in AP with samllest angle 120 and common difernce 5.find numbr of sides of polygon?1517189
Smallest Interior angle= 120...Thus, Largest Exterior angle=60

Now, 2nd smallest interior angle=125..thus, 2nd largest exterior angle=55

Thus, Ext angle forms an AP with a=60, and CD=-5

Let, n be the Number of sides..

Thus, 360=n/2*[2*60 + (n-1)*(-5)]

=>720 = n(125 - 5n)

=>n*(25-n) = 144

Hence, n= 9..

@ravi.theja said:
Define a number k such that it is the sum of the squares of the first M natural nos(i.e k=1^2+2^2+...+m^2), where ma) 10 b) 11c) 12 d) 13
k = M(M + 1)(2M + 1)/6

For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.

For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47; (M + 1) will be divisible by 8

So, for total 12 values of M 55, k will be divisible by 4..

Courtesy: @chillfactor Sir 😃 [4 the 2nd ques]
12701
@pavimai 5..?