A craftsperson made 126 ornaments and put them all into boxes.

Let x = No of boxes containing 6 ornaments and y = No of boxes containing 24 ornaments
6x+24y=126

Statement 1: Fewer than 4 of the boxes contained 6 ornaments
x<4
If x=3 or 2, then y wont be an integer.
Hence x=1 is the only solution.
Statement 1 is sufficient
Statement 2: More than 3 of the boxes contained 24 ornaments
y>3
If y=4 , x= 4
If y=5 , x=1
There are 2 solutions. So B is not possible
Hence A

Need KUDOS pls

Can someone please explain the reason ?

Solution

From the stem we are able to form an equation:
\(6x + 24y = 126\) ornaments we are asked to solve for y

We can simplify this equation to \(x + 4y = 21\) but remember that x and y must be integers as they relate to whole boxes.

(1) Fewer than 4 boxes contain 6 ornaments

let's test values
\(x = 3\\
3 + 4y = 21\\
4y = 18\)
18 is not divisible by 4 , x=3 is not a solution

\(x=2\\
2 + 4y = 21\\
4y = 19\)
again, not a solution

\(x=1\) (the last value we can test)
\(1 + 4y = 21\\
4y = 20 \\
y = 5\)
This is the only possible solution.

\((1)--> Sufficient\)

(2) More than 3 boxes contained 24 ornaments


This tells us \(y > 3\)

Let's test values for y using the original equation \(6x + 24y =126\)

\(y = 4\\
6x + 24(4) = 126 \\
6x = 54\\
x = 9\)
one possible solution

\(y = 5\\
6x + 24(5) = 126\\
6x = 6\\
x= 1\)
Another possible solution

2 possible answers for x--> Inconsistent therefore \(Insufficient\)

Let x = No of boxes containing 6 ornaments
Let y = No of boxes containing 24 ornaments

So, 6x+24y=126

=> x + 4y = 21

Possible solutions to this are:
x = 1, y = 5
x = 5, y = 4
x = 9, y = 3
x= 13, y = 2
x = 17, y= 1

Statement 1

Fewer than 4 boxes contain 6 ornaments

Only 1st case, where x = 1 is the only solution.

Sufficient.

Statement 2

More than 3 boxes contained 24 ornaments

There are 2 possible solutions for this.
Hence, insufficient

Do give a Kudos if you found the solution good and easy to understand.

We can let a = the number of boxes containing 6 ornaments each and b = the number of boxes containing 24 ornaments each. We can create the equation:

6a + 24b = 126

6a = 126 - 24b

Dividing both sides of the equation by 6, we have: a = 21 - 4b

We see that the value of b could be 0, 1, 2, 3, 4, or 5, and the respective values of a would then be 21, 17, 13, 9, 5, and 1.

Statement One Only:

Fewer than 4 of the boxes contained 6 ornaments.

In other words, a < 4. From the stem analysis, we see that if a < 4, then a must be 1, and hence b must be 5. So the number of boxes that contain 24 ornaments is 5. Statement one alone is sufficient.

Statement Two Only:

More than 3 of the boxes contained 24 ornaments.

In other words, b > 3. From the stem analysis, we see that if b > 3, then b must be 4 or 5, and hence, a must be 9 or 1, respectively. In the former case, the number of boxes that contain 24 ornaments is 4; however, in the latter case, the same number is 5. Statement two alone is not sufficient.

Answer: A
_________________

Given: A craftsperson made 126 ornaments and put them all into boxes. Each box contained either 6 ornaments or 24 ornaments.
In situations like this, where there aren't many possible cases, I like to invest a little time upfront to systematically list those cases.
case i: 0 24-ornament boxes and 21 6-ornament boxes
case ii: 1 24-ornament boxes and 17 6-ornament boxes
case iii: 2 24-ornament boxes and 13 6-ornament boxes
case iv: 3 24-ornament boxes and 9 6-ornament boxes
case v: 4 24-ornament boxes and 5 6-ornament boxes
case vi: 5 24-ornament boxes and 1 6-ornament boxes
Aside: I'd typically list the six possible outcomes in table form, which would take less than 15 seconds.

Target question: How many of the boxes contained 24 ornaments?

Statement 1: Fewer than 4 of the boxes contained 6 ornaments
This statement rules out cases i to v, leaving only case vi, which means 5 boxes contained 24 ornaments.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: More than 3 of the boxes contained 24 ornaments
This statement rules out cases i to iv, leaving cases v and vi, which means EITHER 4 boxes contained 24 ornaments OR 5 boxes contained 24 ornaments.
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Hi Brent. I agree with your approach and always list out test cases like this and use rues of divisibility. However, I find it time consuming. How does it take you 15 seconds to list out all of these cases?

Also, do you have other problem examples that I can practice similar to this one? Thanks.