Latest TCS Aptitude Question SOLUTION: A long thin strip of width 7cm is kept on a flat surface. Another identical strip is kept on it in an overlapping manner such that the combined width of the two is 11 cm. What is t

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(#M40002106) TCS Numerical Ability question Keep an EYE

A long thin strip of width 7cm is kept on a flat surface.
Another identical strip is kept on it in an overlapping
manner such that the combined width of the two is 11 cm.
What is the width of 55 such strips kept like this?

Asked In TCS gowri (12 years ago)

Solved SUBHAJIT CHATTOPADHYAY Read Solution (14)

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Read Solution (14) :

Aditya Krishna Awasthi  (12 years ago)

11-7=4 cm

4*54=216

216+7=223 cm

ans= 223

Is this solution Helpfull? Yes (19) | No (3)  

SUBHAJIT CHATTOPADHYAY  (12 years ago)

I believe..
first strip width is 7cm
And after keeping another strip the width become 11cm
so now both are different strip.
now 55 such strip are kept.means 27*11+7=309(as 54/2=27 and the width of last one is 7)

Is this solution Helpfull? Yes (16) | No (9)  

ananth  (12 years ago)

first strip width is 7cm,
second strip width is 4cm
according to this, the combined 54 strips width is 594cm,55th strip is 7cm
so that 55 strips width is 601cm

Is this solution Helpfull? Yes (2) | No (5)  

Harsh Gupta  (12 years ago)

Their is 55 strips..width of single strip is 7cm and pair having 11 cm width.So
Given 55 strip and their 27 pair and one single strip.
so that 27*11+11=304

Is this solution Helpfull? Yes (2) | No (1)  

UTHRAKUMAR  (12 years ago)

11*55=605cm

Is this solution Helpfull? Yes (1) | No (6)  

damodar reddy  (12 years ago)

304..
two strips=>11cms
27*2+1=54+1=55
so 27*11+7=304cms

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

Mayank kulshrestha  (12 years ago)

11-7=4
4*54=216
216+7=223

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

priya jaisinghani  (12 years ago)

(11/2)*55 i have done this

Is this solution Helpfull? Yes | No (4)  

rohit negi  (12 years ago)

after putting 2nd strip extra width is 11-7 = 4cm
we are putting 54 such strip over by over on 55th strip and so on.....
then total width will be = 7cm + 54* 4 = 223 cm
so the ANS is 223 cm.

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Mithu  (12 years ago)

Width of two strips(overlapping)=11cm
width of 54 such strips(taking 2 at a time i.e.54/2=27)=27*11=297cm
adding one more strip(width of each is 7cm)=297+7=304cm
ANS=304cm

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nayyera  (12 years ago)

605...as we multiply 55*11=605

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Ravinder Yadav  (12 years ago)

if their are two such strips are combined then 3 cm is common in between two , & 4cm and 4cm are uncovered in the two
if their are 5 such strips then 4cm are uncommon in between them .
and their are 3 overlapped strips means n-2 where n is no of strips oflength 3cm
and their are 3 unoverlapped strips means n-2 of length 1 cm
so total length is 4+(3*3)+(3*1)+4=20cm
so if their are 55 strips then length is =
4+(3*(55-2))+(1*(55-2))+4=220cm

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simran gade  (12 years ago)

According to the question,
the overlapping part must be 3 cm or in other words each time a new strip is added , the total width increases by 4cm.
As there are total 55 strips,
54 more strips were added to the initial 7cm strip.
This gives us that there was total 54 times 4cm width was added to the 7cm strip width.
therefore, width of such 55 strips= [7 + 54(4)]cm
= 7+216 cm
= 223 cm

Is this solution Helpfull? Yes | No (1)  

Vinay Mahobia  (7 years ago)

in general form, we can write the condition as
length of n strips = 7n-3(n-1)
{Checking :- put n=2 (i.e., when 2 strips kept together), we get length as 11cm.}
Now,
putting n=55, we will get our answer = 223.

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