Pencils and Jars

Let there be $x$ pencils and $y$ jars. If 4 pencils are put into each jar there will be one jar left over meaning that 4 pencils into (y-1) jars would divide evenly: $\frac{x}{4} = y-1$

And if 3 pencils go into each jar one pencil is left over so (x-1) pencils divided by 3 goes into y jars: $\frac{x-1}{3} = y$

These can then be arranged into 2 simultaneous equations by multiplying them by 4 and 3 respectively giving $x=4y-4$ and $x-1=3y$ We use substitution to find that $y=\boxed{5},\ x=\boxed{16}$ 16 pencils and 5 jars! :P

there is 16 pencils in the 5 jars. 4X4=16 one jar is remaining 3x3=15 one pencil is remaining simple answer

Because 4 options have been given, we can easily eliminate. If we devide 4 pencils in a number of jars and one jar is leftover, the number of pencils must be a multiplier of 4. Therefore, there is only one possible choice, which is 16 pencils and 5 jars.

In this case, there is no need to solve equations.

There are two distinct statements. When there are four pencils in each jar there will be one jar left over. We can say that the number of jars is equal to the number of pencils divided by four plus another one. This can be expressed using algebra: j = (p/4) + 1 (where p = pencils and j = jars) When there are 3 pencils in each jar there will be one pencil left. We now need to calculate how many pencils there are. We can say that the number of pencils will be equal to the number of jars times three plus another one. Using algebra again we have: p = 3j + 1

We have ended up with two linear equations. First rearrange the formula to make j the subject: j = (p/4) + 1 (equation 1) j= (p − 1) ÷ 3 (equation 2). with elimination and subtitution then you will get result p = 16 and j = 5

you can write two equations to solve : ........ y=4(x-1).........&........y=3x+1......... where y is the number of pens and x is the number of jars ........ that reveals :......... x=5........ y=16 .......

Well here is how i solved it:- Since if you put 4 pencils in each jar , 1 jar is left behind so the no. of jars is odd. Hence, (a) is out. Then since if you put 3 pencils in jar then 1 pencil is left behind and so the no. of pencils is more than a multiple of three and thus (c) is out. Now (b) has 16 pencils and (d) has 19 pencils. However since when you put 4 pencils in each jar then a jar is left so it has to be a multiple of 4 and so (b) fits. Thus the answer is 6 pencils and 5 jars.

16 pencils & 5 jars!! 16/4 then 1 jar is left now 16/3 then 1 pencil is left . so simple....

This problem could just as easily be drawn out, if you would be willing to take the time.

While there are many explanations using equations, consider this logical approach:

Assume that there are 4 pencils in every jar except 1 jar, as described.

Now, the question asks us to have 3 pencils in each jar, which is equivalent to removing 1 pencil from each jar with pencils.

With this we can put 3 pencils in the empty jar and still have 1 pencil left. This implies that we removed 4 pencils in total and hence the number of jars must be 1 more than that, that is 5 jars.

Now, there are 3 pencils in each jar and then 1 more, that is 16 pencils.

The first statement implies that the number of pencils is a multiple of 4 and only one option fulfils that condition and when double checked for the second statement that indeed is the answer. In the absence of multiple choice answers one can easily do it either by simultaneous equations or by hit and trial.

Let x be the number of pencils and y be the number of jars.

//If I put 4 pencils into each jar, I will have one jar left over.// If we rephrase this statement to mathematical equation we can state:

x /4 = y-1

then, //If I put 3 pencils into each jar, I will have one pencil left over.// become:

(x-1)/3= y

then we can multiply both equation by two different integers to remove one variable.

using simple algebra, Equation 1 become: x=4(4y-1) x=4y-4

And Equation two become: x=3y+1

So, we use 3 and 4 to remove the variable y.

3(x=4y-4) 4(x=3y+1)

if two equations added, it'll become: -x=-16 then x=16

we can now use substitution..

y= (x-1)/3 y=((16)-1)/3 y= 5

if among 5 jar each jar contains 4 pencils then 16 pencil and one jar left ... and among 5 jars if you distribute 3 in each then 15 pencil get its place but one is left among 16... so 16 pencils and 5 jars

This is how I approached the problem:

1) If I put 3 pencils into each jar, I will have one pencil left over.

  So, the number of pencil is some multiple of 3 plus 1. e.g 4,7,10,...

2) If I put 4 pencils into each jar, I will have one jar left over.

  So, the number of pencil % 4 = 0 bcs, no pencil left over

Finding the LCM of (x3 + 1 and x4) = 16 pencils

Going back to :

2) I will have one jar left over

Thus the number of jars = $\frac{16}{4}$ +1 = 5 jars

16 is the only option divisable by 4

I wrote out some multiples of 4 and 3:

4 8 12 16 20

3 6 9 12 15

When 3 pencils are placed in each jar, there is one left over, so I found two multiples which had a difference of 1. These were 16 and 15. I knew this was the answer because the multiples are 1 apart (4 * 4 and 5 * 3) so they were one 'jar' apart.

So if you guys don't want complex equations to solve this question use the "trial and error " method just as I did it. You'll naturally find the fourth option viz 5 jars and 16 pencils!! 😊

That means, I have X jar.

When 1 jar left over, Then I have =4(x-1) pencils.

Again, When 1 pencils left over, Than I have = 3x+1 pencils.

So That, 4(x-1)=3x+1 => 4x-4 =3x+1 => x = 5

So, I have 5 jars & I have (3*5+1)=15+1 =16 pencils.

Simple, the amount of pencils had to be a number of 4 so the only option was 16 pencils

Let there be x pencils & y jars. 1st case: x/4 + 1= y 2nd case: (x-1)/3=y Solving, we get, x=4 and y=5

Let's No. of jar is n And No. of pencil is m At 4 m=4(n-1), At 3 m=3n +1, 3n+1=4n-4, then n=5 jar & m=16 pencil

There could, indeed, be an infinite number of pencils and an infinite number of jars. But, given the four options as possible solutions, the simplest approach is to eliminate the options that do not conform to the logistics of the stated problem. In that case, the only correct option is 16 pencils and 5 jars. If no options were stated for solutions, then multiples of the correct solution would be equally valid, causing an infinite number of pencils and jars.

Suppose there are 5 jars . in each jar i put 4 pencils . 1 jar is left .. Therefore 4+4+4+4 = 16. And there is a option of 16 pencils . and thats the answer .

jars minus 1 can divide for 4 pencils minus 1 can divide for 3 -> look at the answer we got 16 and 5

I first observed that if four pencils were put into four jars that means one jar is left over. That means the amount of pencils is a multiple of four.

The only answer which has the amount of pencils being a multiple of four is : There are 16 pencils and 5 Jars.

4×4=16 pencils left one jar and 3×5=15 pencils left one pencil

Pensils = 3 * Jars + 1Pensils ===> by putting jars=5 , so pensils=16

Looking at the answers, we see that the only answer where the pencils are divisible by 4 is 16 pencils, thus the answer.

Let there by n jars. So, 4(n - 1) = 3n + 1

4n - 4 = 3n + 1

n = 5

No. of jars = 5

Total pencils = 3n + 1 = 16

X=4Y-4;X=3Y+1 simply solve two equations ..X=16 &Y=5

When I read the question without looking at the answers my first way of working it out was right 😋

Idiots, there are infinite pencils and infinite jars

If 4 pencils are put into jars nothing is left ,so no.of pencils is a multiple of 4. And if 3 pencils are put into each jar one is left ,so no.of pencils one more than the multiple of 3. Now, looking at the options 16 is the only number which is multiple of 4 and one greater than multiple of 3

It's easy to guess if the choices are shown. Otherwise, Katie's solution is the right approach.

Well instead of calculating eqs. U shud just look at options carefully

My idea was:

1(mod 4) for jars 1(mod 3) for pencils

When 16 is divided by 3, the remainder is 1 (3 x 5 = 15, 1 remainder). When 5 is divided by 4, you get 1 remainder (4 x 1 = 4, 1 remainder).

Thus, I got 16 pencils and 5 jars.

See all option for such problems It solve........ simple!

4*4=16 is 4 jars and each jar 4 pencils. Remains 1 jar empty

from the multiple answers, minus one jar and then remaining jars multiply with 4 pencils. then we got the total pencils..then we get the correct ans as 16 pencils and 5 jars..

In all of the given options, check for which number of pencils P is
$P \equiv 0 (mod 4)$
The only answer satisfying it is 16 pencils and 4 Jars. Which is the correct answer :D

get the lowest factor of 4 that if you divided it to 3 you'll get remainder 1, which is 16 the number of pencils and divide it to 3 and you'll get 5 the number of jars

Let x equal the total number of jars.

If you put 4 pencils into each jar and have one jar leftover, the expression for the total number of pencils would be: 4(x-1) , where (x-1 ) equals the total # of jars minus the one remaining

If you put 3 pencils into each jar with one pencil leftover, the expression for the total number of pencils would be: 3x + 1

Set both to each other:

4(x-1) = 3x + 1 x =5 jars

Plug in x = 5 jars to either side of the equation, and you'll get 16 pencils!

When I take the pencils in groups of 4 of them I know that the number of pencils is a four multiply, and if I take the pencils in groups of 3 the number of pencils is a three multiply plus one, then I have: $4x=3y+1$ Now by the first sentence I know that the number of jars is: $\frac{4x}{4}+1=x+1$ Then by both of sentences I know that the number of jars I will use when I take the pencils in groups of four is one unity larger than when I take them in goups of 3, because in the first case I left one jar over and blah blah blah, so: $y=x+1$ Now i have: $4x=3(x+1)+1$ $4x=3x+4$ $x=4$ Then the number of pencils is: $4x=16$ And the number of jars is: $x+1=5$ Tadaaaaaa!

it will be easier to start with any of the four options and proceed!!!!

For making the solution easy make it in the terms of pencils let the pencil be x and the jars be y

so form first statement 4(y-1)=x ............. eq(1)

here 1 is subtracted because there is only one jar left so total no. of jars i.e. (y-1)

from next statement

here we have one pencil left so our equation is (x-1)/3=y

(x-1)=3y

x=3y+1 ...............eq(2) subtracting both the equations

4y-4=x

-( 3y+1=x)

now

y-5=0 so y=5

As per first constraint,when there are 4 pencils in each jar and one jar is left when all jars are filled. So it's obviously 16 pencils and 5 jars.