d Question 11 & 12 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Question 11 & 12

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




We know

1 m = 100 mm


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?



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;


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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