Official Quant thread for CAT 2013

MBA CAT 2024
12843
@naga25french said:
till 150 , we will have 15 blocks(1,6,1,6,5,6,1,6,1,0) .. but catch is 2^3! ends in 4 where as other 2 series will end in 6so till 150 the number ends in 5 - 2 = 3now add rest of three numbers3 + 1+ 6 + 1 = 1
😃 😃 for 1 to 10 sum 31 ==> nxt bocks of 10 ll hav sum of 33

so 😃 31 + 33*14+ 1+6+ 1 = 1+ 2+1+6 + 1= 11 😃 units digit =1
12844
@ravi.theja said:
ABC is a three digit number such that ABC = 4 [AB + BC + CA], where AB, BC and CA are all two digit numbers. Find the total number of possible values for the number ABC.a) 2b) 1c) 0d) 3e) more than 3
b) 1???
12845

The largest number which exactly divides 210 ,315,147 and 168?

A. 3

B. 7

C. 21

D. 4410

12846

The largest number which exactly divides 210,315,147 and 168?

A.3

B. 7

C. 21

D. 4410


Any faster method to solve above problem directly from options or some other interesting approach?

12847
@cadmium said:
The largest number which exactly divides 210,315,147 and 168? A.3B. 7C. 21D. 4410Any faster method to solve above problem directly from options or some other interesting approach?
hcf ( 168-147 , 315-210) = hcf(21, 105) = 21
12848
@cadmium 21 :)
12849

HCF of differences is the standard and fastest method for these kind of problmes

@cadmium said:
The largest number which exactly divides 210,315,147 and 168? A.3B. 7C. 21D. 4410Any faster method to solve above problem directly from options or some other interesting approach?
12850
@insane.vodka said:
b) 1???
yes 1 it is 😃 method?? gt the equation A= 34/ 56 *B + 43 / 56 * C
from this how did u get the number
12851
@ravi.theja said:
HCF of differences is the standard and fastest method for these kind of problmes
can you please elaborate that differences method? thanks.
12852
@rkshtsurana said:
hcf ( 168-147 , 315-210) = hcf(21, 105) = 21
@cadmium
12853
@rkshtsurana said:
hcf ( 168-147 , 315-210) = hcf(21, 105) = 21
What is the logic behind this? Can u plz explain
12854

Greatest number which can divide 1354,1866 and 2762 leaving the same remainder 10 in each case, is:

A.64

B.124

C.156

D.260


After taking differences , what we'll have to do?

Logical approach please.

12855
@bs0409 said:
LOL..........really good one..........From a circular sheet of paper with a radius of 20 cm, four circles of radius 5cm each are cut out. What is the ratio of the uncut to the cut portion? (1) 5 (2) 3 (3) 6 (4) 4 (5) none of these
4??
12856
@cadmium
it is A. 64.

Think of it this way.

greatest number which divides all these leaving remainder 10 is the highest common factor of all these numbers less 10.
12857
@ravi.theja said:
yes 1 it is method?? gt the equation A= 34/ 56 *B + 43 / 56 * C from this how did u get the number
see
100A + 10B + c = 44(A+B+C)
ABC/44 = (A+B+C)
so ABC must be divisible by 44 as well as A+B+C also it needs to be an even number divisible by 4
We need to check for only even multiples of 44 till it reaches 1000 ie 22
Only 792 gives A+B+C = 18, also 44 x 18 = 792
Long method but sure ans
PS this is only way???
12858
@cadmium said:
Greatest number which can divide 1354,1866 and 2762 leaving the same remainder 10 in each case, is: A.64B.124C.156D.260After taking differences , what we'll have to do?Logical approach please.
A.64
remainder is 10 so subtract 10 from all numbers and find divisors...
1354 - 10 = 1344 = 2^6 * 21
so 64 is greatest no.
12859
@insane.vodka said:
1344 = 2^6 * 21

Have you learned-by-heart all the 2^n powers ? Or did you implement some other logic which I am unable to figure it out?
How come you concluded that 1344=2^6 * 21?


12860
@insane.vodka said:
1344 = 2^6 * 21

Have you learned-by-heart all the 2^n powers ? Or did you implement some other logic which I am unable to figure it out?
How come you concluded that 1344=2^6 * 21?


12861
@insane.vodka @ravi.theja

it still eludes me. 3792 is the only number but the process is a bit too cumbersome.
Hope there's a good solution around the corner.
12862
@krum Deleted brother