Latest TCS Aptitude Question SOLUTION: a uniform circular coin required too much metal relative to its value.. it is decided that the diameter of the coin can be reduced by 10% by how much will weight of the coin reduce

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a uniform circular coin required too much metal relative to its value.. it is decided that the diameter of the coin can be reduced by 10% by how much will weight of the coin reduce if everything else about the coin remains unchanged.

Asked In TCS ratna (12 years ago)

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Read Solution (7) :

Garima  (12 years ago)

It will be in same ratio as change in volume of coin.
considering thickness remains unchanged,
change in volume = change in area of one side.
Earlier area = pi*r^2
new area = pi*(0.9*r)^2 =pi*0.81*r^2 as radius is also reduced by 10%

% reduction in area = 100*0.19/1 = 19%

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Sankhadeep  (12 years ago)

how could u tell area=weight of the material?

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mahalakshmi  (12 years ago)

% of weight reduced=19%

Is this solution Helpfull? Yes (1) | No (3)  

Vinay Sipani   (12 years ago)

%reduce in weight = %reduce in volume
This is because density remains constant
i.e mass/volume = constant
=> mass ∝ volume ∝ weight.

Also volume = area x thickness,
thickness is constant
=> volume ∝ area

Therefore
Weight ∝ area

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chandra sehkhar gupta  (11 years ago)

m=d/v
here d means denisity remain constant
v means volume
m1m2=(v2)^3/(v1)3
m1=.729m2
so the reduced weight is 1-.729=.271

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Ravi Bhole  (11 years ago)

19% weight will be reduced diameter is reduced by 10% means diameter 2r will be 1.8 r. Therefore new radius = 0.9 r Volume formula Pir r^2h/Pie (0.9r^2) h.It implies it is 0.81 means 81% Weight reduced is 100-81 19%

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saurabh joshi  (9 years ago)

There is short cut method for this types i.e net-equivalent
for reduce -
increase +
10% reduce so
-10-10+(-10*-10/100)
-20+100/100
-20+1=-19
ie19%

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