QNR 3. For her parent’s 51st wedding anniversary celebration Malina decides to buy 200 balloons. Her supplier only sells balloons in bunches of 12 for Rs 250 and 7 for Rs 175. Malina buys exactly 100 balloons.
What is the total price she pays?
1) 4750
2) 4850
3) 4400
4) Cannot be determined.
Ans: This question is same as the QNR 2 except that on 51st wedding anniversary of her parents, Malini has decided to buy 200 balloons (four times the previous number). :-).
Let us assume Malini buys ‘x’ packets of the first kind and ‘y’ of the second. The equation becomes:-
12x + 7y = 200 ————- (1)
Re-writing the equation to make it meaningful:
(7x + 5x) + 7y = 7*28 + 4
=> (7x + 7y)= 7*7 + ( 4 -5x) ——— (2)
Repeating the logic:- 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 (4-5x).
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 (4-5x) should also be.
The question now is – how can we make (4-5x) 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 4 -> Result not a multiple of 7.
If x is 1, result is -1 -> Result not a multiple of 7.
If x is 2, result is -6 -> Result not a multiple of 7.
If x is 3, result is -11 -> Result not a multiple of 7.
If x is 4, result is -16 -> Result not a multiple of 7.
If x is 5, result is -21 -> 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’ =20. Now you know Malini buys ‘5‘ packets of the first kind and ‘20‘ of the second. So, total price is 5* 250 + 20* 175 = 1250 + 3000 = 4750. Option 1.
And that is Wrong! 🙂
If you put x=12 and y=8, we get 12x + 7y = 200; so very well Malini can get ‘12‘ packets of the first kind and ‘8‘ of the second. So, total price is 12* 250 + 8* 175 = 3000 + 1400 = 4400. Option 3.
So which one should be the answer; both the prices are correct and since there is no enough data available we have to go for option 4.
Let us try to understand why two answers:- if you look at the equation 12x + 7y = 200, you could very well view it geometrically as an equation of line. [Remember ax+by=c].
So, there could be multiple values of (x,y) that will satisfy the equation; it is just that ‘x’ and ‘y’ here cannot be negative- Malini cannot buy negative packets!
How many values of (x,y) are possible for 12x + 7y = 200, if there is NO restriction. The answer should be-> INFINITE as there would infinite points on this line. Given the restriction on this case let us look at the two solution sets:
(x,y) = { (5, 20) , (12, 8) }
You should note this pattern that x gets increased by 7 and y gets decreased by 12. And 7 and 12 are nothing but
co-efficients of y and x respectively. The bottomline is :-
The values of ‘x’ form an arithmetic progress with common difference equal to the co-efficient of ‘y’. Similarly ‘y’ values form an arithmetic progress with common difference equal to the co-efficient of ‘x’. Below figure should put the gist here:
Out of all the available sets, only two solutions are possible (x,y) = { (5, 20) , (12, 8) }. Option 4.
