sud21 wrote:When divided by 5, R, S, and T have the same remainder. T=?
1). R+S=T
2). 20<=T<=24

1.R + S = T and all have same remainder when divided by 5

if the reminder != 0 for R and T then reminder of T = reminder of R + reminder of S and hence all three remainders won't be same. So only multiples of 5 will work out here.

R 10, s 20 t 30
r 40 s 50 t 90. T takes more than one value. INSUFF

2.20<=T<=24
T again can take any value between 20 and 24. INSUFF

Combining 1 and 2

From 1 - R, S and T can be Multiples of 5.
From 2 - only 20 fits the criteria.

IMO C.

@Sam..Though your reasoning are 99% Correct I just have ONE point about your statement 1 explanation.

If I take Remainder = 0 any of multiples of 5 satisfies the Question and statement 1

How multiples of 10 are going to have a Remainder !=0 if divided by 5?

ArunangsuSahu wrote:@Sam..Though your reasoning are 99% Correct I just have ONE point about your statement 1 explanation.

If I take Remainder = 0 any of multiples of 5 satisfies the Question and statement 1

How multiples of 10 are going to have a Remainder !=0 if divided by 5?

Edited the explanation to understand better. Hope that's clear

@sam..The following line is INCORRECT

"if R S and T are multiples of 5 then R + S will yield T as a multiple of 10."

Example:

R=5, S=10, T=15

T is not a multiple of 10

ArunangsuSahu wrote:@sam..The following line is INCORRECT

"if R S and T are multiples of 5 then R + S will yield T as a multiple of 10."

Example:

R=5, S=10, T=15

T is not a multiple of 10

if you are confused, lets see another approach without plugging in numbers!!

it is given that the R , S and T have the same remainder when divided by 5.

R = 5K1 + X
S = 5K2 + X
T = 5K3 + X

We have to find tha value of T.

1. R+S=T : 5 ( K1 + k2 ) + 2X = 5 ( K3 ) + X
5 ( K1 + K2) + X = 5 k3

We dont have any value at hand : All we can know is that X = 5n ( where n is any whole number)

2. 20<=T<=24 : it provide a wide range of values : T = 20, 21, 22,23 or 24.

Now lets combine these two.

T = 5K3 + X :

for 20 -- X = 0
for 21 -- X = 1
for 22 -- X = 2
for 23 -- X = 3
for 24 -- X = 4

from here X can be only 0 as proved earlier. so T = 20. hence both are sufficient.

Hope this helps

ArunangsuSahu wrote:@sam..The following line is INCORRECT

"if R S and T are multiples of 5 then R + S will yield T as a multiple of 10."

Example:

R=5, S=10, T=15

T is not a multiple of 10

Right again. I know how it works but not able to explain it in clearly. Will try to change it.