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A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/3 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/4 of the distance. By what factor should hare increase its speed so as the win the race?
(a) 4
(b) 3
(c) 12
(d) 5
Read Solution (Total 3)
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- The circle circumference is given by the formula 2*pi*r or(pi x diameter) = 100pi yards.Therefore this is the length of the race.
The tortoise has a 100pi/ 3 = 33.33 pi head start
The hare sets off leisurely and gets 1/4 of the way round ie 1/4 x 100 pi= 25 pi yards when he meets the tortoise
therefore by d time hare move 25 pi ,the tortoise move (100-33.33-25)=41.66pi yards
Therefore just to go at the same speed as the tortoise, the hare must increase his speed by a factor of 41.66/25 = 1.664
now the tortoise only has 25pi yards to the finish, while the hare has 75pi yards to go. hence hare must increase his speed by a factor of 75/25 = 3 to allow for larger distance to go
hence to tie d race he mst increase his speed by a factor of
1.664*3 = 4.992 approx (5) - 13 years agoHelpfull: Yes(12) No(3)
- 1/3,1/4
3*4=12
12-3=9
9-4=5
9*5/3*3=45/9=5.00
ans is d. - 13 years agoHelpfull: Yes(8) No(6)
- 100pie.......1/8(100)::12.5..100-12.5...87.5,....87.5-20:::67.5.......compare the speed.....ans is 5
- 10 years agoHelpfull: Yes(0) No(0)
TCS Other Question
Alok is attending a workshop How to do more with less and today's theme is working with fever digits. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fever digits. The problem posed at
the end of the workshop is, How many 10 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4? Can you help Alok to find the answer?
(a) 1953125
(b) 781250
(c) 2441407
(d) 2441406
Subha Patel is an olfactory scientist working for International Flavors and Fragrances.
She specializes in finding new scents recorded and reconstituted from nature thanks to Living
Flower Technology. She has extracted fragrance ingredients from different flowering plants
into bottles labeled herbal, sweet, honey, anisic and rose. She has learned that a formula for a
perfume is acceptable if and only if it does not violate any of the rules listed: If the perfume
contains herbal, it must also contain honey and there must be twice as much honey as herbal.
If the perfume contains sweet, it must also contain anisic, and the amount of anisic must
equal the amount of sweet. honey cannot be used in combination with anisic. anisic cannot be
used in combination with rose. If the perfume contains rose, the amount of rose must be
greater than the total amount of the other essence or essences used. Which of the following
could be added to an unacceptable perfume consisting of two parts honey and one part rose to
make it acceptable?
(a) Two parts rose
(b) One part herbal
(c) Two parts honey
(d) One part sweet