Thx sausi for the explanation...I seriously doubt if this is a GMAT Q?
-Deepak.
This cannot be a GMAT question, simply beacause the wording was so ambiguous that most puys had to request for a decryption
This cannot be a GMAT question, simply beacause the wording was so ambiguous that most puys had to request for a decryption
a, b,c are all used ofr counting the no of 7's
3a (counts the no of 7's used by 777)
2a (counts the no of 7's used by 77)
a (counts the no of 7's used by 7)
So
777a+77b+7c=7000 (@Suez, please try to detail the steps for others to understand)
Thus, 111a+11b+c=1000
Now we need to find N, which will be = 3a+2b+c
Thus, substracting the 2 eqns,
we get, 108a+9b=1000-N
or, 9(12a+b)=1000-N
now taking 12a+b = x
9x = 1000-N, where N varies from 1 to 1000
This is where i will stop, i dont think i will check all the values which satisfy this eqn,
@Suez: how can anyone be sure that after 37, there is no value which does not satisfy this eqn?
You neednt check all the values that satisfy the equation. This is where is becomes a boundary value problem. For N = 37
12a+b = 108 which gives a = 8 b = 11 BUT 3a +2b >37 ( 3a+2b increases with decreasing r) which makes p
For N = 46 ... 1000 , 12a+b decreases from 108, 107...0 Hence the b
Thx sausi for the explanation...I seriously doubt if this is a GMAT Q?
-Deepak.
sausi007 SaysThis cannot be a GMAT question, simply beacause the wording was so ambiguous that most puys had to request for a decryption
I have honestly encountered questions which were equally difficult if not more in MGMAT and Kaplan tests and few other forums like gmatclub.
Anyways its good practice 😃
Hi all,
I came across the following quant question and looking for a solution/procedure to solve it..
A test has 80 questions. There is 1 mark for a correct answer, while there is a negative penalty of -1/2 for a wrong answer and -1/4 for an unattempted question. If a sutdent has scored a net total of 34.5 marks, what is the number of questions answered correctly
a) 45 b) 48
c) 54 d) Cannot be determined
Hoping for some help... 
Thanks,
Insiya
Hi all,
I came across the following quant question and looking for a solution/procedure to solve it..
A test has 80 questions. There is 1 mark for a correct answer, while there is a negative penalty of -1/2 for a wrong answer and -1/4 for an unattempted question. If a sutdent has scored a net total of 34.5 marks, what is the number of questions answered correctly
a) 45 b) 48
c) 54 d) Cannot be determined
Hoping for some help...
Thanks,
Insiya
I would go with D, although 45,48 are possible values which can get the student to 34.5, we cannot point any number specifically.
45 : 45C+7W+28NA => 45-(7/2)-(28/4) =45-3.5-7 =34.5
48 : 48C+22W+10NA=> 48-(22/2)-(10/4) =48-11-2.5 =34.5
Thats why D, cannot be determined.
I would go with D, although 45,48 are possible values which can get the student to 34.5, we cannot point any number specifically.
45 : 45C+7W+28NA => 45-(7/2)-(28/4) =45-3.5-7 =34.5
48 : 48C+22W+10NA=> 48-(22/2)-(10/4) =48-11-2.5 =34.5
Thats why D, cannot be determined.
Thats right I think. In fact the following question was if the number of unattempted questions are 10, then what is the number of correct answers. So in that case b) 48 would be right.
I was not able to deduce the equation. Got it now. Thanks!
Hi all,
I came across the following quant question and looking for a solution/procedure to solve it..
A test has 80 questions. There is 1 mark for a correct answer, while there is a negative penalty of -1/2 for a wrong answer and -1/4 for an unattempted question. If a sutdent has scored a net total of 34.5 marks, what is the number of questions answered correctly
a) 45 b) 48
c) 54 d) Cannot be determined
Hoping for some help...
Thanks,
Insiya
We essentially have two equations
a+b+c = 80
ab/2 c/4 = 34.5
Solving we get 6b+5c =182 => 5c = 182 6b => c is even
Therefore b can take values 2, 7, 12, 17, 22, 27 (for 32, c And (a,c )can take corresponding values (44,34), (45, 28 ), (46, 22), (47, 16),(48, 10), (49, 4)
We have 6 possible sets so its option D
We essentially have two equations
a+b+c = 80
ab/2 c/4 = 34.5
Solving we get 6b+5c =182 => 5c = 182 6b => c is even
Therefore b can take values 2, 7, 12, 17, 22, 27 (for 32, c And (a,c )can take corresponding values (44,34), (45, 28 ), (46, 22), (47, 16),(48, 10), (49, 4)
We have 6 possible sets so its option D
Actually i solved through options and found it much shorter...
So taking option a)45, we are left with 2 equations :
X/2 + Y/4 = 10.5
X + Y = 35
which gives a solution of X=7 and Y=28
and so does option b) give a solution of X=22 and Y=10
Hence answer is d)
1. If and
, and
does not equal zero, then
a)
b)
c)
d)
e)
2. If ,
, and
are positive integers, with
, are
,
, and
consecutive integers?
1. Ifand
, and
does not equal zero, then
a)
b)
c)
d)
e)
Its quick to use values and eliminate options.
Let a = 2, b = 1
a^2 - b^2 = 3
x = 2, y =2
Using in options, we can eliminate option c,d,e. Left with a,b
Let a = 2,b = 2
a^2 - b^2 = 0
x=4, y=1
Using in options, we can eliminate option b
So option a) is correct
2. If,
, and
are positive integers, with
, are
Suppose 3 integers are: k-1, k, k+1
Statement 1:
1/(k-1) - 1/k = 1/(k+1)
1/k(k-1) = 1/(k+1)
k^2 - 2k -1 =0
This equation doesnt have integral roots
So statement 1 sufficient.
Statement 2:
(k-1) + (k+1) = k^2 -1
k^2 - 2k - 1 = 0
Same equation as previous case.
So statement 2 also sufficient.
Both statements alone are sufficient
Hi ....
Ystdy was doin thes sums ... Found it very tricky ... It goes like this ....
In a class of 27 what is the probability that at least 3 students share the same month of birth ?
Another one -
The Probability of a particular event (Rains on Monday and does not rain on Tuesday ) is 6/10 , similarly the prob of another event ( Rains on Tuesday and does not rain on Monday ) is 4/10 ... What is the probability that it will Rain both on Monday as well as Tuesday ....
Will see who solves it. 😃
Hi...can anyone solve this 4me..wit steps pl...
3 partners A B C invest 5000, 8000, 12k, in business.if c is a workin partner and gets fourth of the overall profit as remuneration for his effort and the balance gets divided in ratio of their investments then c gets 61k in a yr.how much did a get in a tat yr?
We have to make 4 Teams out of the available 8 people.....
8C2*6C2*4C2*2C2 should give us the answer = 2520
Thus, B
But I doubt if this is correct, because 2520 is an insanely high number for just 8 teams, what is it that I am missing? Can someone please figure out.
i think 2520 is correct even i gt d same answer
Hi ....
Ystdy was doin thes sums ... Found it very tricky ... It goes like this ....
In a class of 27 what is the probability that at least 3 students share the same month of birth ?
100% probability.
12 months in a year.
if max 2 ppl have birthdays in each month, total count up to 12*2 =24.
We're still left with 3 ppl.
Whichever months these 3 ppl have their birthdays in, atleast 3 students would definitely end up sharing the same month of birth.
Another one -
The Probability of a particular event (Rains on Monday and does not rain on Tuesday ) is 6/10 , similarly the prob of another event ( Rains on Tuesday and does not rain on Monday ) is 4/10 ... What is the probability that it will Rain both on Monday as well as Tuesday ....
Will see who solves it. :)
I'd say its ZERO.
Both events are mutually exclusive since they add up to 1.
Its like saying, probability of obtaining a head with a coin toss is 'P', that of obtaining a tail is '1-P', what is the probability that both heads and tails occur.
Hi...can anyone solve this 4me..wit steps pl...
3 partners A B C invest 5000, 8000, 12k, in business.if c is a workin partner and gets fourth of the overall profit as remuneration for his effort and the balance gets divided in ratio of their investments then c gets 61k in a yr.how much did a get in a tat yr?
Let profit = 100P
C's share = 100P/4 + 12/25*(3/4*100P) = 61000
25P + 36P = 61000
61P = 61000
P = 1000
A's share = 5/25 * (3/4 * 100P) =15P = 15000
Hi ....
Ystdy was doin thes sums ... Found it very tricky ... It goes like this ....
In a class of 27 what is the probability that at least 3 students share the same month of birth ?
Another one -
The Probability of a particular event (Rains on Monday and does not rain on Tuesday ) is 6/10 , similarly the prob of another event ( Rains on Tuesday and does not rain on Monday ) is 4/10 ... What is the probability that it will Rain both on Monday as well as Tuesday ....
Will see who solves it. :)
Q1. Ans=1, class has 27 people.
Say 1 student shares his month, not possible since there are only 12 months and 27 students
Say 2 students share their month of birth, again, not possible, 2*12 = 24, still 3 students are left
Thus, we can safely say that at least 3 students WILL share the same month of birth, thus Probability = 1
Q2.
P(M) = 6/10 = 0.6
P(T) = 4/10 = 0.4
For mutually exclusive events P(M) + P(T) = 1,
thus, P(M) and P(T) are mutually exclusive,
Ans = 0
Hi...can anyone solve this 4me..wit steps pl...
3 partners A B C invest 5000, 8000, 12k, in business.if c is a workin partner and gets fourth of the overall profit as remuneration for his effort and the balance gets divided in ratio of their investments then c gets 61k in a yr.how much did a get in a tat yr?
Let Total Profit be x,
C's share = (12/25)(3/4)x + (1/4)x = 61000
x = 100000
A's Share = (5/25 )(3/4)x = 15000
Yups grt ,
Sausi and Angadbir .... Questions of Zambori 😃
Q1. Ans=1, class has 27 people.
Say 1 student shares his month, not possible since there are only 12 months and 27 students
Say 2 students share their month of birth, again, not possible, 2*12 = 24, still 3 students are left
Thus, we can safely say that at least 3 students WILL share the same month of birth, thus Probability = 1
Q2.
P(M) = 6/10 = 0.6
P(T) = 4/10 = 0.4
For mutually exclusive events P(M) + P(T) = 1,
thus, P(M) and P(T) are mutually exclusive,
Ans = 0
also in Question No.1 .... if I wanna know whose probability will be more , between , at least 3 students amongst 27 sharing the same birth month or at least 4 students amongst 36 students sharing the same birth month ?? what will be the answer
coolrahine Saysalso in Question No.1 .... if I wanna know whose probability will be more , between , at least 3 students amongst 27 sharing the same birth month or at least 4 students amongst 36 students sharing the same birth month ?? what will be the answer
I did not really understand what you are trying to ask, but i will give it a shot, let me know if you meant to know something else.
For case 1: P(Case1) = 1(as we saw earlier)
For Case 2: 4 among 36, will be always less than 1 since, 12*3 = 36, there is a possibility that without 4 people sharing the same birth month, there can be a group of 36 students. So P(Case2)
Correct answer Sausi .... Now lemme know have yu already gven GMAT and if so then what is yur score ... My prep tests are workin somwat fine with avgs varying between 700 to 750 ... 😃