There are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two??pls explain with steps:-(
kittu12 SaysThere are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two??pls explain with steps:-(
my take is 3000...use venn diagram....
A=star tv network viewer
B=zee v
n(A)=9000
n(B)=12000
n(A union B)=4000
solve this you will get answer....as 3000..
p.s.-for me it is hard to explain this without diagram...and i don't know how to attach image:-(:-(
kittu12 SaysThere are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two??pls explain with steps:-(
It should be 3000.
Both Networks = 4000
Only Star TV = 9000 - 4000 = 5000
Only Zee TV = 12000 - 4000 = 8000
Total = 4000+5000+8000 = 17000
=> None = 20000 - 17000 = 3000
kittu12 SaysThere are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two??pls explain with steps:-(
itz 3000...20000-(5000+8000+4000)
kittu12 SaysThere are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two??pls explain with steps:-(
Kittu Ji you can refer to the venn diagram below for better understanding.
Star Tv only = 9000 - 4000 = 5000
Zee Tv only = 12000 - 4000 = 8000
Total subscribers = Star Tv only + Zee Tv only + both = 17000
Remaining = 20000 - 17000 = 3000
I) Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off.9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cutoff in Section 1 and Section 3.the number of people who cleared each section alone was equal and was 21 for each section.
1)how many cleared all the 3 sections? ans:3
2)how many cleared only one of the three sections?ans:63
3)the ratio of the number of students clearing the cutoff in one or more of the sections t the number of the sudents clearing the cutoff in section 1 alone is?78/21
II)in the indian athletic squad sent to the sydney olympics,21 athletes were in triathlon team,26 were in pentathlon team and 29 were there in the marathon team.14 athletes can take part in triathlon and pentathlon,12 can take part in marathon and triathlon 15 can take part in pentathlon and marathon and 8 can take part in all the three games.
how many players are there in all?
ans:43
pls explain with steps
I) Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off.9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cutoff in Section 1 and Section 3.the number of people who cleared each section alone was equal and was 21 for each section.
1)how many cleared all the 3 sections? ans:3
2)how many cleared only one of the three sections?ans:63
3)the ratio of the number of students clearing the cutoff in one or more of the sections t the number of the sudents clearing the cutoff in section 1 alone is?78/21
II)in the indian athletic squad sent to the sydney olympics,21 athletes were in triathlon team,26 were in pentathlon team and 29 were there in the marathon team.14 athletes can take part in triathlon and pentathlon,12 can take part in marathon and triathlon 15 can take part in pentathlon and marathon and 8 can take part in all the three games.
how many players are there in all?
ans:43
pls explain with steps
Ans)
1) There seems to be an error in the data...with the line...
"Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off.9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cutoff in Section 1 and Section 3.the number of people who cleared each section alone was equal and was 21 for each section."
Repost it please..
2) Consider a venn diagram for 3 paramateres ie Triathlon, Pentathlon and Marathon.
Now let
a = triathlon only
b=pentathlon only
c=marathon only
d=triathlon and pentathlon only
e=triathlon and marathon only
f=marathon and pentathlon only
g=all the 3.
Now as per the conditions
a+d+g+e = 21 ---> (1)
b+d+g+f = 26 ---->(2)
c+e+g+f = 29----->(3)
d+g=14 ---->(4)
e+g=12---->(5)
g+f=15----(6)
and g = 8 --->(7)
so automatically
d=6
e=4
f=7
Now on adding (1) (2) and (3) we get,
a+b+c+2(d+e+f)+3g = 76 ---->(
On substituting the above values we get
a+b+c = 18 ---->(9)
so total no of people = a+b+c+d+e+f+g = 18+6+4+7+8 = 43
How many 4 digit nos each consisting of 4 different digits can be formed with the digits 0,1,2,3?
a)18 b)20 c)21 d)24 e)14
How many ways can you arrange AKSHAY such that vowels do not start the words?'
a)2x5! b)5! c)180 d)7!/2! e)6!/2!
How many 4 digit nos each consisting of 4 different digits can be formed with the digits 0,1,2,3?
a)18 b)20 c)21 d)24 e)14
my take is option (A) what is the correct answer????
How many 4 digit nos each consisting of 4 different digits can be formed with the digits 0,1,2,3?
a)18 b)20 c)21 d)24 e)14
for first digit you have option of 3 nos, as 0 can't be used for first digit
for second digit you have option of 3 nos, as one no is already chosen for first digit
for third digit you have option of 2 nos, as two nos are already chosen for first two digits
for fourth you have only option hence1
no of combination=3*3*2*1=18
How many ways can you arrange AKSHAY such that vowels do not start the words?'
a)2x5! b)5! c)180 d)7!/2! e)6!/2!
Ans is a
Total way to form word"akshay" = 6!/2=360
less the ways where a i.e vovel place at the starting 5! ways
360-120=240 ans
How many 4 digit nos each consisting of 4 different digits can be formed with the digits 0,1,2,3?
a)18 b)20 c)21 d)24 e)14
How many ways can you arrange AKSHAY such that vowels do not start the words?'
a)2x5! b)5! c)180 d)7!/2! e)6!/2!
Ans:
1) One cannot start a number with zero. So digits left are 1,2,3. Taking 1 as the first digit, rest of the digits can be arranged in 3! ways. So total no of ways = 3x3! ways = 18 numbers .
2) The vowel here is A. COndition is that one cannot start a word with it. Total no of letters = 6. Number of consonants = 4, vowels =1 (repeated twice). Taking say K as the 1st letter, the rest 5 letters can be arranged in 5!/2 ways. So total number of arrangements = 4x5!/2 ways = 2x5! ways
how many 4 digit nos each consisting of 4 different digits can be formed with the digits 0,1,2,3?
A)18 b)20 c)21 d)24 e)14
_ _ _ _ = 3*3*2*1 = 18
Hi All,
I have just started my preparations and I am surprised that I have gone so bad in my quant:nono:
But still I believe that I will pick up gradually. 
I have started with number systems and with the starting practice exercise of LCM AND HCF I got stuck with two questions.
Sorry for asking so obvious questions but still I have no other option.
Kindly bear with me
First question
The HCF of 2472,1284 and a third number N is 12. If their LCM is 2*2*2*3*3*5*103*107, then the number N is
a) 2*2*3*3*7
b)2*2*3*3*3*103
c)2*2*3*3*5
d)none
The correct answer is b. But how I am not able to get it.
I was trying to do it by
lcm(n1,n2,n3)*hcf(n1,n2,n3) = n1*n2*n3
and that gives 15 as answer but I dont know whats wrong I am doing here.
Second Question
Find the greatest number of four digits which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively.
a)9907
b)9903
c)9893
d)none
I know that I have to proceed with finding the LCM. The LCM is 9900
but what to do next.
Kindly Help
All the best to everyone
Cheers
:)
Hi All,
I have just started my preparations and I am surprised that I have gone so bad in my quant:nono:
But still I believe that I will pick up gradually.
I have started with number systems and with the starting practice exercise of LCM AND HCF I got stuck with two questions.
Sorry for asking so obvious questions but still I have no other option.
Kindly bear with me
First question
The HCF of 2472,1284 and a third number N is 12. If their LCM is 2*2*2*3*3*5*103*107, then the number N is
a) 2*2*3*3*7
b)2*2*3*3*3*103
c)2*2*3*3*5
d)none
The correct answer is b. But how I am not able to get it.
I was trying to do it by
lcm(n1,n2,n3)*hcf(n1,n2,n3) = n1*n2*n3
and that gives 15 as answer but I dont know whats wrong I am doing here.
Second Question
Find the greatest number of four digits which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively.
a)9907
b)9903
c)9893
d)none
I know that I have to proceed with finding the LCM. The LCM is 9900
but what to do next.
Kindly Help
All the best to everyone
Cheers
:)
Ans)
1) I agree with your approach there. It has to be 15. I guess the answer is (d).
2)Difference between the divisors and remainders in each case is 7. Now LCM of 10,11,15 and 22 = 330. SO largest 4 digit number of this order = 330x30 = 9900. So the required number = 9900 - 7 = 9893. Option (c)
First question
The HCF of 2472,1284 and a third number N is 12. If their LCM is 2*2*2*3*3*5*103*107, then the number N is
a) 2*2*3*3*7
b)2*2*3*3*3*103
c)2*2*3*3*5
d)none
The correct answer is b. But how I am not able to get it.
I was trying to do it by
lcm(n1,n2,n3)*hcf(n1,n2,n3) = n1*n2*n3
and that gives 15 as answer but I dont know whats wrong I am doing here.
Second Question
Find the greatest number of four digits which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively.
a)9907
b)9903
c)9893
d)none
I know that I have to proceed with finding the LCM. The LCM is 9900
but what to do next.
Kindly Help
All the best to everyone
Cheers
:)
1. HCF * LCM = Product of the numbers
=> 12 * 2*2*2*3*3*5*103*107 = 2472 * 1284 * n
=> n = 15 is correct
2. By observation we see that remainder of 7 is present in all the divisors..
10 - 3 = 11 - 4 = 7 and so on
LCM = 9900
Now subtract 7 from 9900 to get 9983 as ur answer
Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off in section2 and 32 in section 3.10 cleared in section 1 and section 2.9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cutoff in Section 1 and Section 3.the number of people who cleared each section alone was equal and was 21 for each section.
1)how many cleared all the 3 sections? ans:3
2)how many cleared only one of the three sections?ans:63
3)the ratio of the number of students clearing the cutoff in one or more of the sections t the number of the sudents clearing the cutoff in section 1 alone is?78/21
Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off in section2 and 32 in section 3.10 cleared in section 1 and section 2.9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cutoff in Section 1 and Section 3.the number of people who cleared each section alone was equal and was 21 for each section.
1)how many cleared all the 3 sections? ans:3
2)how many cleared only one of the three sections?ans:63
3)the ratio of the number of students clearing the cutoff in one or more of the sections t the number of the sudents clearing the cutoff in section 1 alone is?78/21
kittu ji yeh question toh direct ho rha h....simple case of venn diagram
a+b+e+f=33(passed in section1)
b+e+d+c=34(passed in section 2)
e+f+d+g=32(passed in section 3)
now ,b+e=10
e+d=9
e+f=8 and it is also given that g=c=a=21
i hope now it is clear...but if still there is any mistake or problem then kindly let me know
1) a person on an excercise regime decides to loose 500 cal. in a day. For every 30 minute of excercise he loses 80 cal. , but gains 25 cal. per cake that he eats. In the morning he excercise for a certain time and he eats a certain no of cakes. in the afternoon he excercise for 1.5 times the times he spends in the morning session ans eats 4 times the no of cakes he eats in the morning. Had he not eaten any cake, he would have exceeded his targate by 220 cal. if he actually falls by 180 cal. ,then, what is the no of cakes he has eaten in the evening?
2)A shopkeeper sold a certain no of toys. the no of toys as well as the price of each toy was a two digit no. By mistake, he reversed the digits of both the no of toys he sold and the price of each toy. As a result he found that his stock account at the end of the day showed 81 items more than it was actally was
a)find the actual no of toys sold
b) if the faulty calculations show a total sale of Rs 882, find the actual selling price of each toy.
c)using the info in the previous Question find his actual sales revenue in Rs