Question 11 & 12

Past Exam Questions 2018 0

Question 11: 1 millimetre thick paper if folded 50 times then find the thickness of the fold.

FPSC AD IB 2018

Solution:

We know

$1 m = 100 mm$

and

$1 km = 1000 m$

$1 km = 1000 \times 100 mm$

$\Rightarrow 1 km = 100000mm$

So, we can other way say that;

$\Rightarrow 100000mm=1 km$

$\Rightarrow 1mm=\frac{1}{100000}km$

$\Rightarrow 1mm=0.000001km$

so, 1 millimeter is equal to 0.000001 kilometer

or we can say that single paper is 0.000001 km thick

If we fold this paper one time then there will be two layers of papers;

One time folding

$One Time Folding = 2 Times Thickness=2\times0.000001= 2^{1}$ times of 0.000001

Second time folding, (there will be four layers of papers)

$Second Time Folding =4 times thickness= 2^{2} Times Of 0.000001$

Third time folding ( there will be 8 layers of papers)

$Third Time Folding = 8 Times Thickness= 2^{3} Times of 0.000001$

Fourth time folding (there will be 16 layers of paper)

$Fourth Time Folding = 16 Times Thickness= 2^{4} Times of 0.000001$

,,,,,,,,,,,,,,,,,,,,,, so on ,,,,,,,,,,,,,,,,,,>,,,,,,,,,,,,,,,,,so on,,,,,,,,,,,,,,,,,,,>,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,>,,,,,,,,, so on ,,,,,,,,,,,,,,,,,,,,,,,,>,,,,,,,,,,,,,,,,,,,,,,,,,,,, till

Fiftieth time folding

$Fiftieth Time Folding = 2^{50} Times Of 0.000001$

Now let’s evaluate the term $2^{50} Times Of 0.000001$ = $2^{50}\times0.000001=1125899906.842624 km$

So, if a paper of 1 mm thickness is folded 50 times then it’s thickness will be equal to 11258999906.842624 kilometers

Or 112 billion kms approximately

Question 12: Total age of 3 men if added is 80 years. What was the sum three years ago?

FPSC AD IB 2018

Solution:

Let there be three persons A, B and C and their ages at present be x, y and z, Then according to the given condition;

$x+y+z=80 ---(1)$

Three years ago the ages of A, B and C was x-3, y-3 and z-3 respectively. Now the sum of ages three years ago will be;

$(x-3)+(y-3)+(z-3)$

Now let us simplify this term;

Sum of ages three years ago = $(x-3)+(y-3)+(z-3) = x+y+z-3-3-3=x+y+z-9 -----(2)$

Putting value of x+y+z from eq (1)

$\Rightarrow Sum Of Ages Three Years Ago = 80-9 = 71$

So sum of ages three years ago is 71

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Question 11 & 12

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Question 11 & 12

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