@MSD, here are my explantaions for 1, 2 and 5th :-
1. The number has 2 digits and is greater than 20 so it can be any number between 21 - 99. Option -A: The tens digit of n is a factor of the units digit of n, that means if tens digit is 2, tens digit could only be 2,4,6,8 making the number as 22,24,26,28 (similar for an odd number at tens place). All of these number are composite. Sufficient.
Option -B: The tens digit of n is 2 that makes n a even number and all even numbers are composite except 2 and n >20. So, n is composite. Sufficient.
Answer is D.
2. We have to find value of n, in other terms we have to find no. of terms in the sequence.
Option -A: Sum of all the terms is 3124 but we dont know as what are the actual values of terms so there are various possible terms which can make sum 3124. Not sufficient.
Option -B: Average of n terms is 4. This information is of no good as we dont know the sum (don't get in the trap and consider sum as 3124, consider B option alone). Not sufficient.
Option A and B simultaneously: Sum = 3124, Average = 4. As we know (Sum of n terms/n = Average) => n = 3124/4 = 781. Sufficient.
Answer is C.
5. does line K intersect quandrant II?
Option -A: All the lines having a negative slope will pass through IInd quadrant as slope = tan (angle made by line K at x axis) and tan is negative in II quadrant. Sufficient. This can also be solved just by drawing a line having negative slope on rectangular coordinate (please refer figure A in the attached document)
Option -B: Y intercept of line K is negative. A line having negative intercept can or can not pass through II quadrant. Not sufficient please refer figure B in the attached document for a line having negative intercept on Y but its not passing through II quadrant.
Answer is A.
Regards, Ashutosh
With reference to Q.1, how can we say that the answer is D? I agree with your explanation as far as choice A is concerned. However, in choice B it is given that the tens place digit is 2. This means that the units digit can be even or odd....we dont know. Hence, we cannot be certain whether such a 2 digit number will be composite or not....for eg:- 29 which has 2 in tens place and is still not a composite number.
y=mx+c From statement 1: y=3cx+c-----not sufficient From statement 2: 0=-m/3+c m=3c...........same as statement 1 hence not sufficient Neither are both taken together sufficient since both imply the same. Hence answer is E
y=mx+c From statement 1: y=3cx+c-----not sufficient From statement 2: 0=-m/3+c m=3c...........same as statement 1 hence not sufficient Neither are both taken together sufficient since both imply the same. Hence answer is E
With reference to Q.1, how can we say that the answer is D? I agree with your explanation as far as choice A is concerned. However, in choice B it is given that the tens place digit is 2. This means that the units digit can be even or odd....we dont know. Hence, we cannot be certain whether such a 2 digit number will be composite or not....for eg:- 29 which has 2 in tens place and is still not a composite number.
Regards MSD
oops...i am sorry dost, my bad...somehow read 'tens' as 'ones'. So, the answer for that will be A and that's the OA.
another one guyss.. How many prime numbers between 1 and 100 are factors of 7150? a) 1 b) 2 c) 3 d) 4 e) 5 I chose 'e" 'cuz factors are - 2,5,5,11,13. It's asking us for primer numbers NOT "DIFFERENT prime numbers". Then answer would be D. But OA gives D as the answer. Is it wrong? thanks!
another one guyss.. How many prime numbers between 1 and 100 are factors of 7150? a) 1 b) 2 c) 3 d) 4 e) 5 I chose 'e" 'cuz factors are - 2,5,5,11,13. It's asking us for primer numbers NOT "DIFFERENT prime numbers". Then answer would be D. But OA gives D as the answer. Is it wrong? thanks!
Whereas for clarity, it should be mentioned in the question that its asking for "different prime numbers" but if its not then I think we should assume as its asking for different numbers only and as its asking for factors it also strengthens the assumption. Btw, from where you got this question ? Hope we dont get any ambiguity in the questions on the G-Day!
Hey - Didnot get this - May be I am missing something elementary. Now my logic is - Sq A side measures "a". The diagonal square therefore is 2a^2 (Pythagoras thrm a^2+a^2 = 2a^2) Sq B side measures "b". The diagonal square therefore if 2b^2
Given that = 2a^2 = 1/2 (2b^2) Ratios of area = b^2 : a^2 = 2:1
:idea: - Pls help point out my error
sorry for catching up late...was away...but i beliv u r well anserd...;)
Among 400 students, 56% study sociology, 44% study mathematics, and 40% study biology. If 30% of the students study both sociology and mathematics, what is the largest possible number of students who study biology but neither of the other two subjects?
(A) 30 (B) 90 (C) 120 (D) 172 (E) 188
I arrived at the right answer.. but i'm a bit worried if my reasoning is right! Can someone plz help? thanks a lot!
Among 400 students, 56% study sociology, 44% study mathematics, and 40% study biology. If 30% of the students study both sociology and mathematics, what is the largest possible number of students who study biology but neither of the other two subjects?
(A) 30 (B) 90 (C) 120 (D) 172 (E) 188
I arrived at the right answer.. but i'm a bit worried if my reasoning is right! Can someone plz help? thanks a lot!
In a game of chess, the moves of white and black pieces alternate, with white pieces having the first move. During a certain chess tournament, the white pieces have made 2319 moves altogether, while the black pieces have made a total of 2315 moves. In every game, either the white side wins, black side wins, or there is a draw. If it is impossible for the losing side to make the last move in a game, which of the following could be true about the tournament?
1) the black sides lost 5 games 2) the black sides won more games than the white sides. 3) all games ended in a draw.
A) 3 only B) 1 & 2 only C) 1 & 3 only D) 2 & 3 only E) 1, 2, & 3.
I arrived at 120 too.. can u tell me how u got to the answer? I need to know if i'm right.. or whether it was just luck all the way
s-sociology m-maths b-biology
n(s)=224 n(m)=176 n(b)=160 n(s and m)=120 n(s U b U c)=400
n(s U b U c)=n(s)+n(m)+n(b)-n(s and m)-n(s and b)-n(b and m)+n(s and m and b)
sustituting values, n(s and b)+n(b and m)-n(s and m and b)=120---(1)
now, the value we need is, x=.
the motive is to minimise y=n(s and m and b) i.e. it must be 0 or else, if it is non-zero, n(s and b)+n(b and m) has to increase to satisfy (1) and hence, the overall {n(s and b)+n(b and m)+n(s and m and b)} value will increase resulting in lower x.....which is not desired... hence, when y is 0...we get the max value of x
In a game of chess, the moves of white and black pieces alternate, with white pieces having the first move. During a certain chess tournament, the white pieces have made 2319 moves altogether, while the black pieces have made a total of 2315 moves. In every game, either the white side wins, black side wins, or there is a draw. If it is impossible for the losing side to make the last move in a game, which of the following could be true about the tournament?
1) the black sides lost 5 games 2) the black sides won more games than the white sides. 3) all games ended in a draw.
A) 3 only B) 1 & 2 only C) 1 & 3 only D) 2 & 3 only E) 1, 2, & 3.
ans should be option A...let me kno if dat's correct