there are 16 multiples of 6 less than 100. now on adding2 to all we get 16 numbers which are leaving a remainder 2 when divided by 6 but there you have left 2 also which is less than 100 and leaves a remainder 2 when divided by 6.
so 17 is the answer.
======================================================== Can somebody please explain me what is this method of solving this question. I am not able to comprehend the statement
"100/6 = there are 16 multiples of 6 less than 100." It is true that there are 16 multiples of 6 less than hundred, but HOW DOES IT HELP IN SOLVING THIS QUESTION?
Suppose I Change the question a bit and say How many natural numbers below 100 will remainder 5 when divided by 6?
The answer should be count of following numbers {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} i.e. 16
But if you solve by the method described by various people i.e.
i) "100/6 = there are 16 multiples of 6 less than 100." i.e.16 ii) "Add 5, as 5 divided by 6 leaves a remainder 5" i.e. 1
So i) + ii) = 16 + 1 = 17 in total, which is wrong.
In case of original question with remainder "2", the answer just coincided with the correct answer. ====================================================== May be I am missing something. Please help.
======================================================== Can somebody please explain me what is this method of solving this question. I am not able to comprehend the statement
"100/6 = there are 16 multiples of 6 less than 100." It is true that there are 16 multiples of 6 less than hundred, but HOW DOES IT HELP IN SOLVING THIS QUESTION?
Suppose I Change the question a bit and say How many natural numbers below 100 will remainder 5 when divided by 6?
The answer should be count of following numbers {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} i.e. 16
But if you solve by the method described by various people i.e.
i) "100/6 = there are 16 multiples of 6 less than 100." i.e.16 ii) "Add 5, as 5 divided by 6 leaves a remainder 5" i.e. 1 So i) + ii) = 16 + 1 = 17 in total, which is wrong.
In case of original question with remainder "2", the answer just coincided with the correct answer. ====================================================== May be I am missing something. Please help.
H Rajat,
The method given by techo_nishi is applicable here also
Let me explain where you got it wrong here
The method used by you
i) "100/6 = there are 16 multiples of 6 less than 100." > its not 16 its 15 here {:snipersm:5,11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95}
In the step 2 you include five also like rahul said. {5,11,17....6n+5}
ii) "Add 5, as 5 divided by 6 leaves a remainder 5" i.e. 1 Dont think of it like an algorithm...get the funda which is All factors of 6 with remainder 5 are represented by {5,11,17....6n+5}
from the above set pick the factors which are less than 100 ie 16
So the second step is not needed at all if you included 5 in the first step.(You included 5 in the first step and again counted it in the second step) Hope this helps ;)
======================================================== Can somebody please explain me what is this method of solving this question. I am not able to comprehend the statement
"100/6 = there are 16 multiples of 6 less than 100." It is true that there are 16 multiples of 6 less than hundred, but HOW DOES IT HELP IN SOLVING THIS QUESTION?
Suppose I Change the question a bit and say How many natural numbers below 100 will remainder 5 when divided by 6?
The answer should be count of following numbers {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} i.e. 16
But if you solve by the method described by various people i.e.
i) "100/6 = there are 16 multiples of 6 less than 100." i.e.16 ii) "Add 5, as 5 divided by 6 leaves a remainder 5" i.e. 1
So i) + ii) = 16 + 1 = 17 in total, which is wrong.
In case of original question with remainder "2", the answer just coincided with the correct answer. ====================================================== May be I am missing something. Please help.
the concept here is we have to find all the numbers which are of the form 6n+5 where n=0,1,2,3,..... but here the condition applied is less than 100 and so while solving the question this thing should be kept in mind.
now 100/6 gives 16 multiples but check the 16th multiple you will get 101 which is more than 100 so we will leave the last multiple and hence we are left with 15 multiples plus 5 also as it also leaves a remainder 5 on dividing by 6 and is also less than 100. so, we get the answer as 16 and not 17.
the concept here is find out the total multiples less than that given in the question at the same time also keep in mind to check the last multiple whether it fulfils the condition that is mentioned in the question or not.
hope this makes it clear if not ask me again i will elaborate it more.
DS 9. The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees? 1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively. 2). c
DS 9. The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees? 1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively. 2). c
================= I go for (A), reasoning as follows
Using 1) we can find out the sides of the triangle, assuming pie = 3.14 e.g. (3.14 * a^2)/2 = 4 so a = (8/3.14)^(1/2) b = (10/3.14)^(1/2) c = (12/3.14)^(1/2)
If we know all three sides of a triangle we can draw the triangle (take a geometry compass and try to draw) and hence we can find (measure) all the 3 angles.
So (1) alone is sufficient to answer the question.
Using (2) alone we cannot find any of the sides or the angles. Hence (2) alone is not sufficient.
DS 9. The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees? 1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively. 2). c
I am making a correction to my earlier post
The triangle may be acute or obtuse from 1 but not right angled.
DS 9. The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees? 1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively. 2). c
statement 1:the 3 sides are a^2=28/11 b^2=35/11 c^2=42/11
clearly,c is the largest side. now,a^2+b^2>c^2 so,it is acute angled.
Hi Shashank, Thanks for the reply. I have understood the triangle property, as explained above. But my question is: How do you arrive at a^2+b^2>c^2 using (2) alone. Can you please explain me this. I am dying to know this.
Q. Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible? A. 10 B. 21 C. 210 D. 420 E. 1,260
Could someone please explain the solution to this?
my pick will be 7P3 ie option C. Pick three people out of 7 and arrange in 3! ways. alternatively, the first prize can go in 7 different ways, the second in 6 possible ways and third in 5 possible ways, creating a unique arrangement and total number would be a product of three.
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