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two DS problems

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two DS problems

by kishokbabu » Wed Dec 28, 2011 8:36 am
problem -1.: A collection of 36 cards consists of 4 sets of 9 cards each. The 9 cards in each set are numbered 1 through 9. If one card has been removed from the collection, what is the number on that card?

a) The units digit of the sum of the numbers on the remaining 35 cards is 6.
b) The sum of the numbers on the remaining 35 cards is 176

problem -2.:

In the X-Y co-ordinate plan, point (r,s) lies on a circle with center at the origin. What is the value of r^2+s^2?

a) The circle has radius 2.
b) The point ( √2, √-2) lies on the circle.

Kindly help with some methods for solving the above 2 problems
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by sam2304 » Wed Dec 28, 2011 9:57 am
kishokbabu wrote:problem -1.: A collection of 36 cards consists of 4 sets of 9 cards each. The 9 cards in each set are numbered 1 through 9. If one card has been removed from the collection, what is the number on that card?

a) The units digit of the sum of the numbers on the remaining 35 cards is 6.
b) The sum of the numbers on the remaining 35 cards is 176
We have 36 cards with 4 sets of numbers from 1 - 9. So total face value of all the cards will be 180.
Sum of all cards in one set = 45 (1 + 2 + ... +9).
so for 4 sets its 180.

What is the value of the card that is removed ?

I. Total value is 180, it cannot be less than 171 as only one card is removed and the maximum value can be 9. So when the units digit of sum of 35 cards is given as 6 then the number can only be 176 as anything less that such as 166, 156 needs removal of more than one card.

Hence SUFF

II.Directly given as 176. So the removed card will have the face value of 4.

Hence SUFF

IMO D.
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by sam2304 » Wed Dec 28, 2011 10:05 am
kishokbabu wrote: problem -2.:

In the X-Y co-ordinate plan, point (r,s) lies on a circle with center at the origin. What is the value of r^2+s^2?

a) The circle has radius 2.
b) The point ( √2, √-2) lies on the circle.

Kindly help with some methods for solving the above 2 problems
Equation of a circle is (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center and r is the radius

since (a,b) is the origin the above eqn will be x^2 + y^2 = r^2

I. given r = 2. Substituting in the above eqn we get x^2 + y^2 = 4
II. given (x,y) = ( √2, √-2), x^2 + y^2 = 4.

IMO D.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
https://gmatandbeyond.blogspot.in/
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