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One Equation with two variables- How to solve contextual questions!

QNR 2. For her parent’s 50th wedding anniversary celebration Malina decides to buy 50 balloons. Her supplier only sells balloons in bunches of 12 for Rs 250 and 7 for Rs 175. Malina buys exactly 50 balloons.

What is the total price she pays?

1) 1042

2)1100

3)1175

4)1225

Ans:  Malini has to buy exactly 50 balloons and she has to buy them in bunches of 12 and 7. Consider, bunch to be a ‘packet’- two kinds of packets available, one containing 12 and the other variety containing 7. Let us assume Malini buys ‘x’ packets of the first kind and ‘y’ of the second. The equation becomes:-

12x + 7y = 50  ————- (1)

[Let us not think about prices for now as we don’t know how many packets each did she buy.]

We can put values of ‘x’ and ‘y’ and solve this but that will take time- think about the case if the R.H.S would have been a bigger value say 500 in place of 50. Instead, let us make it more meaningful by re-writing the equation as:

(7x + 5x) + 7y = 7*7 + 1

=> (7x + 7y)= 7*7 + ( 1 -5x) ——— (2)

In the above two lines I have tried to put every term of the original equation to be shown as a ‘multiple of 7’ and whatever remains is put together separately; here the separate part is (1-5x).

Now from the equation (2), you can tell that L.H.S is a multiple of 7. R.H.S should also be. First term of the R.H.S (7*7) is a multiple of 7 and thus (1-5x) should also be.

The question now is – how can we make (1-5x) a multiple of 7. Let us now start putting in sequential values of x to see what values do we get.

If x is 0, result is 1 -> Result not a multiple of 7.

If x is 1, result is -4 -> Result not a multiple of 7.

If x is 2, result is 1 -> Result not a multiple of 7.

If x is 3, result is -14 -> Result is a multiple of 7. [Don’t worry about negative; does not matter].

Now substitute the value of x in the equation (1), you get ‘y’ =2. Now you know Malini buys ‘3‘ packets of the first kind and ‘2‘ of the second. So, total price is 3* 250 + 2* 175 = 1100. Option 2.

Although this takes many lines of text to explain but it is very fast if you start practising this way. And the advantage is if there were a big “number” on the right hand side, this would have been a better and faster approach. I will shortly post a similar question that has a big R.H.S. and explain you how this question then becomes a Arithmetic Progression question. 🙂 . Stay tuned and keep practising!!!

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