I've made a simple question complicated

[MathsFormula]

The question is:
The total mass of a jar one-quarter full of jam is 250g. The total mass of the same jar three-quarter full of jam is 350g. What is the mass of the empty jar? The answer is 200g
I think I’ve gone about the question the wrong way. I think it’s meant to be a simple arithmetic sum but I’ve made it into basic algebra.

J = JAM X = WEIGHT OF JAR

So (3/4)J +X = 350 .......... (1/4)J +X = 250

X = 350-(3/4)J ................. X = 250-(1/4)J



Using simultaneous equations J = 200

Now putting J = 200 into the formula X = 250-(1/4)J gives X = 200g (the weight of the empty jar).

Someone did actually show me a basic arithmetic way of doing the sum and it took them 2 minutes from first glance at the question. I just nodded my head in agreement but I didn’t know what they were talking about. Please advise. Thanks.

[Can someone show me how to post proper formula symbols on here like the ones they post on the algebra threads. Thanks]

 

[Deleted member 4993]

The question is:
The total mass of a jar one-quarter full of jam is 250g. The total mass of the same jar three-quarter full of jam is 350g. What is the mass of the empty jar? The answer is 200g
I think I’ve gone about the question the wrong way. I think it’s meant to be a simple arithmetic sum but I’ve made it into basic algebra.

J = JAM X = WEIGHT OF JAR

So (3/4)J +X = 350 .......... (1/4)J +X = 250

X = 350-(3/4)J ................. X = 250-(1/4)J



Using simultaneous equations J = 200

Now putting J = 200 into the formula X = 250-(1/4)J gives X = 200g (the weight of the empty jar).

Someone did actually show me a basic arithmetic way of doing the sum and it took them 2 minutes from first glance at the question. I just nodded my head in agreement but I didn’t know what they were talking about. Please advise. Thanks.

[Can someone show me how to post proper formula symbols on here like the ones they post on the algebra threads. Thanks]

You are probably talking about a process like shown below:

3*[1/4*J + X] - [3/4*J + X] = 3*250-350 = 750 - 350 = 400

3X - X = 400 → 2X = 400 → X = 200

You did just fine with simultaneous equations. That method will work "always". The short-cut shown above can only work sometime. So stick with simultaneous equations - better be a little late but correct (than very quick but wrong).

 

[HallsofIvy]

A slightly different but equivalent method:
(3/4)J+ X= 350
(1/4)J+ X= 250

Subtracting as they are gives (3/4- 1/4)J+ (1- 1)X= 100. (1/2)J= 100, J= 200. X= 250- (1/4)(200)= 250- 50= 200.

Yet another (perhaps more like what you were trying to do): From (3/4)J+ X= 350, X= 350- (3/4)J. From (1/4)J+ X= 250, X= 250- (1/4)J. X= X of course so 350- (3/4)J= 250- (1/4)J. Subtract 250 from both sides and add (3/4)J to both sides: 350- 250= (3/4)J- (1/4)J. 100= (1/2)J again so J= 200.