Latest TCS Aptitude Question SOLUTION: In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Arithmetic

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40153450) TCS Numerical Ability question numbers Keep an EYE

In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Asked In TCS swathi choda (9 years ago)

Unsolved Read Solution (14)

Is this Puzzle helpful?   (66)   (24) Submit Your Solution Arithmetic

Read Solution (14) :

subham mandal  (9 years ago)

sorry for the prev wrong ans..it will be..
48+56+64+........upto 20th term
so,
s=n/2[2a+(n-1)d]=2480

Is this solution Helpfull? Yes (20) | No (4)  

Hashavardhini  (9 years ago)

Explanation:
Given, total number of potatos = 20.
First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
48,56,64...........20 terms.
a = 48, d= 8, n = 20.
Sum of n terms in A.P = Sn=n2[2a+(n−1)d]
S20=202[2×48+(20−1)8]
S20=202[96+152]
S20=10×248=2480
∴ 2480 meters he run in bringing back all the potatoes.

Is this solution Helpfull? Yes (16) | No (1)  

ankit kumar verma  (9 years ago)

For every potato to bring it to iys starting point contestant has to run twice of the same distance and total distance as
24*2+28*2+......100*2=2*4[(1+2+3+4+5+6+7+8+...25)-(1+2+3+4+5)]
=8*[(25*(25+1)/2)-(5*(5+1)/2)]
=8*[(25*13)-(5*3)]
=8*310
=2480 meters

Is this solution Helpfull? Yes (6) | No (2)  

Pankaj Chaudhary  (9 years ago)

Given, total number of potatos = 20.
First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
48,56,64...........20 terms.
a = 48, d= 8, n = 20.
Sum of n terms in A.P = Sn=n2[2a+(n−1)d]
S20=202[2×48+(20−1)8]
S20=202[96+152]
S20=10×248=2480
∴ 2480 meters he run in bringing back all the potatoes.

Is this solution Helpfull? Yes (4) | No   

Harsh Chamedia  (9 years ago)

Ans:- 2090. Asto bring back 1st potato he required 24+24 i.e. 48m, for 2nd 56m, 3rd 64m -------------- 20th 200m. Hence addition of 48, 56,64,72,----------192,200 is 2090.

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

Anik Bhattachaya  (9 years ago)

Why Do every one see something with so much difficulty ?

1st potato =24 meters
then all the potatoes are 4 meters appart = 19x4= 76meters
therefore to collect all the potatoes, 24+76= 100 meters
and to come back, = 100 more meeters

thus total = 100 + 100 = 200 meters!

Is this solution Helpfull? Yes (1) | No (9)  

subham mandal  (9 years ago)

240..cz..48+56+64+72=240

Is this solution Helpfull? Yes | No (4)  

Aritra Chakraborty  (9 years ago)

24*2=48, 48*20=960, 8+16+24+...19terms = 8*(1+2+3+..+19)= 8*19*20/2=1520, 960+1520=2480 meters

Is this solution Helpfull? Yes | No (1)  

Niharika Gupta  (9 years ago)

24+(4*20)=104
2*104=208

Is this solution Helpfull? Yes | No (4)  

Joy cindrella  (9 years ago)

@ ankit kumar verma.. could you please explain how did you get 2*4[(1+2+3+4+5+6+7+8+...25)-(1+2+3+4+5)] this term? I am little confused!!! :(

Is this solution Helpfull? Yes | No (1)  

aditya raj  (8 years ago)

we first apply the fourmula to find the last term ie. L=a+(n-1)d
then sum of series we apply ie. sum=n/2(a+L)
ans is 2480..

Is this solution Helpfull? Yes | No (3)  

rajeswari  (8 years ago)

please anyone can explain clearly

Is this solution Helpfull? Yes | No   

sujana patwal  (6 years ago)

Given, total number of potatos = 20.
First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
48,56,64...........20 terms.
a = 48, d= 8, n = 20.
Sum of n terms in A.P =
Sn=n/2*[2a+(n−1)d]
S20=20/2*[2×48+(20−1)8]
S20=20/2*[96+152]
S20=10×248=2480
∴ 2480 meters he run in bringing back all the potatoes.

Is this solution Helpfull? Yes | No (1)  

Sabiyana Begum  (6 years ago)

I think the answer is unsolved here, coz while running for second potato, he will go diagonally, we need to find the diagonal distance to reach to second potato by using phytogerous theorem.

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.