Problem Solving

Three girls Reshma Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip and Mandip to Reshma. if the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip.

A: 5.0m
B: 6.0m
C: 7.2m
D: 8.4 m
E: 9.6 m

maihuna wrote:Three girls Reshma Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip and Mandip to Reshma. if the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip.

A: 5.0m
B: 6.0m
C: 7.2m
D: 8.4 m
E: 9.6 m

circumference = 2pi*5 = 31.4
distance between reshma and mandip = 31.4 - (6+6) = 18.4
is "1" missing in option D?

I believe here we have to find straight line distance between mandip and reshma, not the length of arc between mandip and reshma.

Refer the attachment..

IMO answer is 6.
Whats OA maihuna?

Thanks aroon, pradip,

no 1 is not missing. pradip has tried the best. Ibut 6 is not correct option. have the OA and the explanation that is not fully understood by me as well, so let us try again, i will post the oa soon.

IMO, the answer is E

Draw a perpendicular line CQ from the centre of the circle to the chord RM.

Since angle CQM is a right angle, CQ bisects RM

Also, since triangle RSM is an isosceles triangle, a line which bisects and is perpendicular to its base will also pass through the vertex at S, i.e SCQ is a straight line.

In triangle CQM, CQ^2 = 25 - x^2, and in triangle SMQ,
x^2 = 36 - (5 + CQ)^2

Solving the two equations, CQ = -1.4 (i.e. triangle RSM is actually inside the semi-circle) and x = sqrt(25-1.4^2) = 4.8 and therefore MR = 9.6

Yep thats correct.

Daimaomee, sqrt(25-1.4^2) = 4.8.......any easy way of calculating this precisely??

Daimaomee, sqrt(25-1.4^2) = 4.8.......any easy way of calculating this precisely??

Not that I know of. However, 1.4 is approx. sqrt(2), so -1.4^2 = ~2, which means x=sqrt(23). sqrt(23) is a bit less than sqrt(25), so you'll look for an option choice that is a bit less than 2*sqrt(25), i.e. <10. Ans E (9.6) is the best option.

The next closest is D (8.4), but as such, x would have to be much closer to sqrt(16) to get an answer that's roughly 2*4.

Such approximation wouldn't work if the answer choices were bunched more closely in the 9-10m range.

The GMAT is apt to ask for which answer is "closest" if it wants such an approach, so I'd still like to hear how best to crunch the numbers manually.

GID09 wrote:Daimaomee, sqrt(25-1.4^2) = 4.8.......any easy way of calculating this precisely??

Hi GID09, I have an "ancient" way to find the square root of a number:

(i) Separate the number in group of 2 digits (if it is a odd-digit number, leave the leftmost number as single-digit)

(ii) Starting from the leftmost group, find a number which square is nearest to the group (in this case, 4, as 4x4 = 16 is closest to 23) (Balloon 1)

(iii) Subtract the square of that number from the group and write down the difference (7 in this case)

(iv) Draw down the second group next to the difference (704)

(v) Double the number pointed by Balloon 1 and write it on the left hand side of the difference (4 x 2 = 8 in this case, Balloon 2)

(vi) Find a number and place it next to the "8" and place the same number next to the "4" on the top (Balloon 3) such that the product of the number formed (88) and the newly added number on the top (8) is closest to 704. In this case 88x8 = 704 exactly.

Even if the number is not a perfect square, this method can still find an approximate number whose square is closest to that number.

Hope this helps.

I dont have a way to prepare diagram, but will try to give verbal explanation: And no squaring is needed:

Represent Reshma, Salma, and Mandip by R S and M respectively. Let R and M are parallel below the centre, keep S below O connect R->S, S.?M amd O-?M(down):

Let Line OM intersect RM at K:

Let KR = xm,

Area of Triangle, ORS = (1/2)xX5

Also area of triiangle ORS = 1/2 RSXOL = (1/2)X6X4

so 5x = 24 or x = 24/5 = 4.8

We need 2KR (as the line from centre bisects any chord), =2*4.8 = 9.6

HUUHHHH no squaring...