Wizard's Quiz

Algebra Level 3

Dorothy while walking on the Yellow Brick Road meets an old wizard. The wizard has 6 6 identical coins all made up of 6 different 100% pure metals (coins are not alloys). He tells Dorothy that the lighter coin worth less than the heavier coin. The metal coins are made up of Copper, Zinc, Silver, Platinum, Gold and Iron . The value of the coins is in AP . The total value of all 6 6 coins is 240 240 and if the coins are arranged in increasing order of their weight then the ratio of the product of the values of 1 s t 1^{st} and 6 t h 6^{th} coin is to 2 n d 2^{nd} and 5 t h 5^{th} coin is to 3 r d 3^{rd} and 4 t h 4^{th} coin is 375 : 1159 : 1551 375:1159:1551 Help Dorothy to find the value of the Gold coin .

61 61 47 47 19 19 75 75

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1 solution

Mohd. Hamza
Feb 16, 2019

Since the coins are identical, so their weight depends on their densities. Arranging the metals in increasing order of their densities: Z n < F e < C u < A g < A u < P t Zn<Fe<Cu<Ag<Au<Pt According to wizard Zinc should have the least value while Platinum the most.

The values are in AP .

Let the AP be: a 5 d , a 3 d , a d , a + d , a + 3 d , a + 5 d a-5d,a-3d,a-d,a+d,a+3d,a+5d & S X 6 = 240 = a 5 d + a 3 d + . . . + a + 5 d \ce{S6}=240=a-5d+a-3d+...+a+5d 240 = 6 a \Rightarrow 240 =6a a = 40 \Rightarrow \boxed{a=40} It is given that: a X 1 × a X 6 a X 2 × a X 5 = 375 1159 \frac{\ce{a1}\times\ce{a6}}{\ce{a2} \times\ce{a5}} = \frac{375}{1159} ( 40 5 d ) ( 40 + 5 d ) ( 40 3 d ) ( 40 + 3 d ) = 375 1159 \Rightarrow \frac{(40-5d)(40+5d)}{(40-3d)(40+3d)}=\frac{375}{1159} 1600 25 d 2 1600 9 d 2 = 375 1159 \Rightarrow \frac{1600-25d^2}{1600-9d^2}=\frac{375}{1159} d = ± 7 \Rightarrow d=\pm7 In order to make Zinc with least value d d is to be positive.

So d = 7 \boxed{d=7}

& The AP is 5 , 19 , 33 , 47 , 61 , 75 5,19,33,47,61,75 Gold coin's value is at a X 5 \ce{a5}

So the value of Gold coin is 61 \boxed{61} .

Can you please elaborate more on what d represents?

Martin Karroum - 2 years, 3 months ago

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Yes, d is the common difference between each term of an AP

Mohd. Hamza - 2 years, 3 months ago

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