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pacifist85
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350

Let Jelly beans in jar \(X = x\) and \(Y = y\)

\(x + y = 1600\) ----- (\(i\))

Given jar \(X\) has \(100\) fewer jelly beans than three times the number of beans in jar \(Y\).

Therefore; \(x + 100 = 3y\)

\(x = 3y - 100\)

Substituting value of \(x\) in equation (\(i\)), we get;

\(3y - 100 + y = 1600\)

\(4y = 1600 + 100 = 1700\)

\(y = \frac{1700}{4} = 425\)

Substituting value of \(y\) in equation (\(i\)), we get;

\(x + 425 = 1600\)

\(x = 1600\) \(-\) \(425 = 1175\)

Answer D
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Hi,
If X has 100 fewer jelly beans than three times the number of beans in jar Y.
Then why did we write X+100 ?
Shouldnt it be X-100?
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x+y=1600 (I)

And

3*y = x-100

Reorganizing it

3*y-100 = x (II)

Subs. (II) in (I)

3*y-100+y=1600
4*y=1700
y=425

Then

x=1175

Ans (D)
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3*y = x-100

Reorganizing it

3*y-100 = x (II)

But if -100 go to the other side of equation, shouldnt it become +100?
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hsn81960
3*y = x-100

Reorganizing it

3*y-100 = x (II)

But if -100 go to the other side of equation, shouldnt it become +100?

Sorry, it's 3*y = x+100 instead
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pacifist85
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350

We can let y = the number of jelly beans in jar Y; thus, there are 3y - 100 jelly beans in jar X. We can now create the equation:

3y - 100 + y = 1600

4y = 1700

y = 425

Therefore, there are 3(425) - 100 = 1275 - 100 = 1175 jelly beans in jar X.

Answer: D
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pacifist85
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350


x+y=1600 , X=1600-Y
x+100=3y, 1700=4Y
Y=425

x=1600-y=1600-425=1175
D
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Given, x + y = 1600
Also, x + 100 = 3*y

After subtracting the two equations,
y - 100 = 1600 - 3y
4y = 1700
y = 425

Thus, x = 1600 - 425 = 1175

Thus, the correct option is D.
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There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?


Let the no. of jelly beans in Jar Y = n
Therefore, the no. of jelly beans in Jar X = 3n - 100

=> n + 3n - 100 = 1600
=> 4n = 1700
=> n = 425

Therefore, the no. of jelly beans in Jar X = 3n - 100 = 3x425 - 100 = 1275 - 100 = 1175

Hence D
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