Vikram,
Here is the solution:
Tr.y to do it by set theory.
let x = no of student in math and german,
y = no of student in math and eng
z = no of student in eng and german
so total student in only in math = 40 - (x +y + 15 )
similarly only in eng = 35- (z +y + 15 ) and
only in german = 30 - (x +z + 15 ).
now, 40 - (x +y + 15 ) + 35- (z +y + 15 ) + 30 - (x +z + 15 ) + (x +y+z + 15 ) = 70
which gives x + y + z = 5 = no of student in exactly two of the cources.
Sujoy
Vikram,
Here is the solution:
Tr.y to do it by set theory.
let x = no of student in math and german,
y = no of student in math and eng
z = no of student in eng and german
so total student in only in math = 40 - (x +y + 15 )
similarly only in eng = 35- (z +y + 15 ) and
only in german = 30 - (x +z + 15 ).
now, 40 - (x +y + 15 ) + 35- (z +y + 15 ) + 30 - (x +z + 15 ) + (x +y+z + 15 ) = 70
which gives x + y + z = 5 = no of student in exactly two of the cources.
Sujoy
Vikas,
try to draw 3 intersectional circle as we do for set theories(each circle for separate language) and then put the values..
x + y + z + 15 = student in math and german + student in math and eng + student in german and enf + student for all three language..
Varun: nice to see a extra option buddy .. you rock:-)
Hey.. I got this very different answer in here..
I followed simple probability rule;
P(A U B U C) = P(A) + P(B) + P(C) - P(A^B) - P(B^C) - P(C^A) + P(A^B^C)
Thus,
P(A U B U C) = 70
P(A) = 40
P(B) = 35
P(C) = 30
P(A^B^C) = 15
P(A U B U C) - [ P(A) + P(B) + P(C) + P(A^B^C)] = - [P(A^B) + P(B^C) + P(C^A)]
70 - [40 + 35 + 30 + 15] = 70 - (120) = -50
i.e. [P(A^B) + P(B^C) + P(C^A)] = 50
Maybe some calculation mistake or there is some mistake in the question, since I am getting 50 as my answer..
Answer is option F: 50
Vikram,
Here is the solution:
Tr.y to do it by set theory.
let x = no of student in math and german,
y = no of student in math and eng
z = no of student in eng and german
so total student in only in math = 40 - (x +y + 15 )
similarly only in eng = 35- (z +y + 15 ) and
only in german = 30 - (x +z + 15 ).
now, 40 - (x +y + 15 ) + 35- (z +y + 15 ) + 30 - (x +z + 15 ) + (x +y+z + 15 ) = 70
which gives x + y + z = 5 = no of student in exactly two of the cources.
Sujoy
Tom, Jerry, and Donald and other three people sit in a line. From left to right, if Tom cannot sit on the first seat, Jerry cannot sit on the second seat, and Donald cannot sit on the fourth seat, how many different arrangements are possible?
(A) 720
(B) 426
(C) 432
(D) 438
(E) 444
Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?
a) 1/3
b) 2/5
c) 5/12
d) 7/16
e) 1/2
A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?
* 5 minutes
* 6 minutes
* 8 minutes
* 9 minutes
* 10 minutes
Thanks everyone !
Just like vikas I missed the x+y+z+15 part and then got boggled. I liked the P(AUBUC) formula...when I start to think in those terms, things get easier.
needless to say the correct answer is 5 :)
70 students are enrolled in Math, English or German. 40 students are in Math, 35 in English and 30 in German. 15 students are enrolled in all three of the courses. How many students are enrolled in exactly two of the courses: Math, English and German?
A. 10 B. 5 C. 15 D. 25 E. Cannot be determined
I have been having problems with these count problems. Please provide a complete explanation on how you arrived at the answer.
Thanks.
Vikram,
Here is the solution:
Tr.y to do it by set theory.
let x = no of student in math and german,
y = no of student in math and eng
z = no of student in eng and german
so total student in only in math = 40 - (x +y + 15 )
similarly only in eng = 35- (z +y + 15 ) and
only in german = 30 - (x +z + 15 ).
now, 40 - (x +y + 15 ) + 35- (z +y + 15 ) + 30 - (x +z + 15 ) + (x +y+z + 15 ) = 70
which gives x + y + z = 5 = no of student in exactly two of the cources.
Sujoy
vikas130678 SaysCan you xplain the bold part?
Hey.. I got this very different answer in here..
I followed simple probability rule;
P(A U B U C) = P(A) + P(B) + P(C) - P(A^B) - P(B^C) - P(C^A) + P(A^B^C)
Thus,
P(A U B U C) = 70
P(A) = 40
P(B) = 35
P(C) = 30
P(A^B^C) = 15
P(A U B U C) - [ P(A) + P(B) + P(C) + P(A^B^C)] = - [P(A^B) + P(B^C) + P(C^A)]
70 - [40 + 35 + 30 + 15] = 70 - (120) = -50
i.e. [P(A^B) + P(B^C) + P(C^A)] = 50
Maybe some calculation mistake or there is some mistake in the question, since I am getting 50 as my answer..
Answer is option F: 50
Vikas,
try to draw 3 intersectional circle as we do for set theories(each circle for separate language) and then put the values..
x + y + z + 15 = student in math and german + student in math and eng + student in german and enf + student for all three language..
Varun: nice to see a extra option buddy .. you rock:-)
no calculation error buddy ....just a interpretation error ...
P(A^B) is interpreted as atleast 2 grps (i.e A and B)
exactly 2 grps = p(A^B) - p(A^B^C)
similarly, for p(A^C) and p(B^C)
hence, p(exactly 2 grps) = [P(A^B) + P(B^C) + P(C^A)] - 3(A^B^C)
= 50 - 3*15
= 5 ....same Ans...
Actually, for 3 way venn diagram
atleast 1 grp = Grp1+Grp2+Grp3-(Grp12+Grp13+Grp23)+Grp123
where,
Grp1 = only grp1 + only Grp12+ only Grp13 + Grp123
similarly for Grp2 and Grp3
Grp12= Only Grp12 + Grp123
similarly fr Grp13 and Grp23
Hope, this helps ...things become very clear by drawing a venn diagram ..
sujog has already explained an alternate way ...
Exactly ... the best approach..
Please see below the ven diagram used for this approach...
Tom, Jerry, and Donald and other three people sit in a line. From left to right, if Tom cannot sit on the first seat, Jerry cannot sit on the second seat, and Donald cannot sit on the fourth seat, how many different arrangements are possible?
(A) 720
(B) 426
(C) 432
(D) 438
(E) 444
Tom, Jerry, and Donald and other three people sit in a line. From left to right, if Tom cannot sit on the first seat, Jerry cannot sit on the second seat, and Donald cannot sit on the fourth seat, how many different arrangements are possible?
(A) 720
(B) 426
(C) 432
(D) 438
(E) 444
-----------------
Ans: 426
Explanation:
Let our candidates be: J T D x y z
1st position can have: J D x y or z
When J is in 1st position: others can be arranged in 5! ways. But this would include 4! ways in which D is in 4th position. So 5!-4! = 96 ways they can be seated.
When D is in first position: and J cannot be in second position. So 96 ways.
when x is in first position: the arrangements for 1st two positions can be xD, xT, xy or xz. When xD the rest can be in 4! ways, but with other choices, the arrangement needs to be (4! - 3!) as D cannot be in 4th seat.
So 4! + 3x(4!-3!) = 78ways
Same with y and z in first positions.
-----------
Sum it up: 96 + 96 + (3 x 78 ) = 426
-------------------------------------
Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?
a) 1/3
b) 2/5
c) 5/12
d) 7/16
e) 1/2
Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?
a) 1/3
b) 2/5
c) 5/12
d) 7/16
e) 1/2
good question ....vikram has done it in d best poss way...
just to add an alternate approach ...
required arrangements = total arrangements - arrangements in which atleast 1 person occupies incorrect position
total arrangements of 6 ppl = 6! = 720
atleast 1 person is in incorrect position = 3*5! - 3*4! + 3! (this is like 3 way intersection venn diagram) = 360-72+6
= 294
Hence, required arrangements = 720 - 294
= 426
Cheers !!
I could be wrong on this one...tried it anyway.
Ans: A (1/3) (?? doubltful)
I will use the approach P(x) = 1 - p(-x)
p(-x) is for two adjacent seats to be vacant. If candidate 1 is the chosen one, the either candidate 10 or candidate 2 will need to be selected for two consecutive positions to become vacant.
So it is: 3C1 x 2/9
so p(-x) = 6/9 or 2/3
so p(x) = 1/3
Is the ans 1/2 ....pls clarify ...nt sure abt d ans ...will post d approach in any case ...
p(adjacents seats r not empty) = 1 - p (adjacents seats r empty)
= 1- p(atleast a pair of nominees is seated besides each other)
10 persons can seated in a round table in 9! ways .
poss such that atleast a pair of nominees is seated besides each other
= 3*8!*2! - 2*7!*3!
= 7! (48 - 12)
= 7! * 36
Hence, p(adjacents seats r not empty) = 1 - 7!*36/9!
= 1 - 36/72
= 1/2 ...Ans ...
looking forward to a better approach and corrections, if any in above approach ..
The answer is neither 1/2 nor 1/3 😉
its 5/12 ... here is the explanation..
Total number of ways = 9! (Circular arrangement)
for fav ways lets take out 3 from 10 people. people left = 7
number of ways there 7 people can sit = 6!
now we need to fit these 3 people (X,Y,Z)
X can go at any of the 7 places availabe = 7 ways
Y can go at 6 places availabe = 6 ways (cant go left or right of X)
Z can go at 5 places available = 5 ways (same logic as above)
probability = 6! * 7 * 6 * 5 / 9! = 30/72 = 5/12 ... answer
Guys..I found a question in one of the simulated tests. It follows.
Two intersecting lines form four angles. Are the lines perpendicular?
(1) Each of the angles is equal to exactly one of the other three angles.
(2) The sum of three angles does NOT equal to 270 degrees.
A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D) Either statement BY ITSELF is sufficient to answer the question.
E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
Now I think the answer is B because (1) is nothing but a restatement of the property of intersection of two straight lines. It is always true. But the answer in the test says that it should be D.
What should be the correct answer?
A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?
* 5 minutes
* 6 minutes
* 8 minutes
* 9 minutes
* 10 minutes
star_highway SaysAny takers ?
shud be E....both are insufficient and in combination also they tell u nothing....
ab=cd=5 bc=10 cdbc yes
no specific answer.