2 DS questions

A request: Try to post one question per post. That makes it easy to follow.

Ques 1: 4+7n is divided by 3

we can write it as, (3+1)+(6n+n) = 3 + 6n + n + 1
3 and 6n will be divisible by 3, remainder will be zero
we have to concentrate on (n+1)

st1: n+1 is divisible by 3, what we are looking for: Sufficient

st2: n>20: n=22, remainder of 23/3 = 2
n=23, remainder of 24/3 = 0: Insufficient

ANs: A

Ques 2 has been posted recently
pls search the forum

hi mental,

how can "4+7n is divided by 3" be rewritten as "(3+1)+(6n+n)"? (How do you come from the first to the secon term?)

ocean

P.S.: sorry for the bad post!

how can "4+7n is divided by 3" be rewritten as "(3+1)+(6n+n)"? (How do you come from the first to the secon term?)

7n = 6n+n
4 = 3+1

6n+n+3+1 can be rearranged as

(n+1) + (6n+3)

Stmt I

n+1 is divisible by 3. 6n is divisible by 3 and 3 is divisible by 3

Whenever u add terms that are each divisible by a number then the sum is also divisible by that number.

So the expression (n+1) + (6n+3) is always divisible by 3 with a remainder 0

One defnite value for remainder hence SUFF


Stmt II

n> 20

n=21

7n+4 = 151 gives a remainder of 1 when divided by 3

n=23

7n+4 = 165 when divided by 3 gives a remainder 0

2 different answers(more different answers possible based on n value) INSUFFICIENT

Choose A)

Q2:

Refer here:

https://www.beatthegmat.com/multiple-que ... 23076.html