FORUM  
 
 
 
 
     Recent Posts  |   Recent Topics  |   FAQ  |   Search  |   Forum Actions  |   Quick Links Prizes
 
Notices
 
Page 3 of 146 FirstFirst 1 2 3 4 5 6 7 8 13 23 53 103 ... LastLast
Results 41 to 60 of 2914
Like Tree2Likes

Thread: TestFunda Answers - Higher Maths

  1. #41
    Aficionado
    Join Date
    Sep 2008
    Locn
    Dewas India
    Posts
    50
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    In a class of 10 students there are 3 girls A,B,C.In how many different ways can they be arranged in a row such that no 2 of the 3 girls are consecutive?

  2. #42
    Aficionado
    Join Date
    Sep 2008
    Locn
    Dewas India
    Posts
    50
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    In how many ways can a lawn tennis mixed doubles be made up from 7 married couples if no husband & wife play in the same set?
    Additional Notes: plz give detailed expl.

  3. #43
    Enthusiast
    Join Date
    Sep 2010
    Locn
    Bareilly India
    Posts
    23
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

  4. #44
    Virtuoso
    Join Date
    Jul 2009
    Locn
    Kolkata India
    Posts
    489
    Thanks
    1
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths


  5. #45
    Virtuoso
    Join Date
    Jul 2009
    Locn
    Kolkata India
    Posts
    489
    Thanks
    1
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    i think  the answer is 8P3. Since there are 3 girls therefore there are 7 boys. Now there are 8 places where these girls can be arranged such that no two of them are consecutive . hence the permutations of arranging these 3 girls in 8 places is 8P3.

  6. #46
    Virtuoso
    Join Date
    Jul 2009
    Posts
    804
    Thanks
    51
    Thanked 25 Times
    in 18 Posts

    Post TestFunda Answers - Higher Maths

    Answer is: 8P3*7!*3!
    For each permutation, 7 boys can be arranged in 7! ways and 3 girls can be arranged as 3! ways.

  7. #47
    Virtuoso
    Join Date
    Jul 2009
    Posts
    804
    Thanks
    51
    Thanked 25 Times
    in 18 Posts

    Post TestFunda Answers - Higher Maths

    answer is 42.
    For all male player, 6 possible female partners are possible. So, 7*6=42 

  8. #48
    Virtuoso
    Join Date
    Jul 2009
    Posts
    804
    Thanks
    51
    Thanked 25 Times
    in 18 Posts

    Post TestFunda Answers - Higher Maths

    Oops, answer is: 8C3*3!*7! (not 8P3 * 3! * 7!)
    For each combination, 7!*3! arrangement possible. 

  9. #49
    Intern
    Join Date
    Apr 2010
    Posts
    49
    Thanks
    0
    Thanked 2 Times
    in 2 Posts

    Post TestFunda Answers - Higher Maths

    Pick male 1=m1 , now he can be paired with w2,w3,w4,w5 , w6 or w7 total 6 ways.

    Now for each of these pair say m1w2 , we have to make a pair not having w1 and m2 , so 5 men to choose from but 4 women ( because if you choose m3 then you cannot choose w3 ) in total 20 ways

    so final answer = 6*20=120 

  10. #50
    Stranger
    Join Date
    Sep 2010
    Locn
    Jaipur India
    Posts
    2
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    SET P= PRIME NOS<100. Q1,Q2.....Qn= SUBSET OF P & EACH HAS FIVE NOS IN AP.  Q IS SUPERSET OF Q1,Q2....Qn.AVG SET OF ELEMENTS=
    Options
    1) 
    13
    2) 
    26
    3) 
    45
    4) 
    17

  11. #51
    Enthusiast
    Join Date
    Oct 2010
    Locn
    Delhi India
    Posts
    15
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    see...lets get the boys seated first..when the boys' seats are taken the girls are left with the option of taking up 8 places:
    ...now the girls take up 3 places out of the available 8...and the boys arrange among themselves...and the girls arrange amongst themselves...
    so,  8C3*7!*3!...hope it helps

  12. #52
    Virtuoso dk78's Avatar
    Join Date
    Jun 2009
    Posts
    409
    Thanks
    5
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths


    Two cities A and B , at a distance of 50km , are connected by two separate roads.The speed of any vehicle travelling between the two cities on road 1 is 50 km per hr, while the speed on road 2 is (80/n) km per hr, where n is the number of vehicles (including the concerned vehicle).If you travel in a vehicle from A and B on road 1 and come back from B to A on road 2 (where there are already three vehicles plying), your approximate average speed is:


    Options
    1) 
    26 km per hr
    2) 
    29 km per hr
    3) 
    32 km per hr
    4) 
    35 km per hr

  13. #53
    Virtuoso
    Join Date
    Jul 2009
    Locn
    Kolkata India
    Posts
    489
    Thanks
    1
    Thanked 0 Times
    in 0 Posts

  14. #54
    Virtuoso
    Join Date
    Jul 2009
    Locn
    Kolkata India
    Posts
    489
    Thanks
    1
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    the ans is 28.57 which is approximately equal to 29.therefore option 2.

  15. #55
    Enthusiast
    Join Date
    Sep 2010
    Locn
    Bareilly India
    Posts
    23
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

  16. #56
    Master jayavignesh's Avatar
    Join Date
    Apr 2009
    Posts
    110
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    Hans and Bhaskar have salaries that jointly amount to Rs.10,000 per month.They spend the same amount monthly and then it is found that the ratio of the savings is 6:1.What is Hans' salary?
    Options
    1) 
    Rs.4000
    2) 
    Rs.5000
    3) 
    Rs.6000
    4) 
    Rs.3000
    Additional Notes: this question is from arun sharma book .
    but there is no detailed explanation for the answer
    Last edited by jayavignesh; 24-Oct-10 at 5:00 PM.

  17. #57
    Master jayavignesh's Avatar
    Join Date
    Apr 2009
    Posts
    110
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    Hans and Bhaskar have salaries that jointly amount to Rs.10,000 per month.They spend the same amount monthly and then it is found that the ratio of the savings is 6:1.Which of the following can be Hans's  salary?
    Options
    1) 
    Rs.6000
    2) 
    Rs.5000
    3) 
    Rs.4000
    4) 
    Rs.3000
    Additional Notes: this question is from arun sharma book .
    but there is no detailed explanation for the answer

  18. #58
    Intern
    Join Date
    Aug 2010
    Locn
    Palakkad India
    Posts
    29
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    consider option 1..hans salary =6000,then baskrs salary =4000
    now according to the question (6000-x)/(4000-x) = 6/1
    solve for x to obtain x = 3600
    this satisfies the condition,
    this is the normal procedure..but there is a simple shortcut..
    if amount spent is equal then if the ratio of their savings is to be greater than 1(in this case 6/1=6)
    then hans salary must be greater than baskrs salary..only option 1 satisfies this
    Last edited by ayyada; 24-Oct-10 at 4:18 PM.

  19. #59
    Intern
    Join Date
    Aug 2010
    Locn
    Palakkad India
    Posts
    29
    Thanks
    0
    Thanked 0 Times
    in 0 Posts

    Post TestFunda Answers - Higher Maths

    speed on road 1 = 50
    speed on road 2 = 80/4 =20
    thus avg speed = 2*50*20/(50+20)=28.56 =29

  20. #60
    Virtuoso
    Join Date
    Feb 2009
    Posts
    6,678
    Thanks
    0
    Thanked 17 Times
    in 12 Posts

    Post TestFunda Answers - Higher Maths


    Let h and b be Hans' and Bhaskar's salary

    So,

    h+b = 10,000

    Let x be the money spend by each of them

    (h-x)/(b-x) = 6/1

    Trying out option 1,

    h=6000 and b=4000

    So, x=3600

    Equations are satisfied

    If amount spent is equal then if the ratio of their savings is to be greater than 1(in this case 6/1=6)

    Then Hans' salary must be greater than Baskrs salary

    Only option 1 satisfies this

     

    Let h and b be Hans' and Bhaskar's salary
    So,h+b = 10,000
    Let x be the money spend by each of them
    (h-x)/(b-x) = 6/1
    Trying out option 1,
    h=6000 and b=4000
    So, x=3600
    Equations are satisfiedIf amount spent is equal then if the ratio of their savings is to be greater than 1(in this case 6/1=6)
    Then Hans' salary must be greater than Baskrs salary
    Only option 1 satisfies this

Similar Threads

  1. TestFunda Answers - Modern Maths
    By TestFunda in forum Quantitative Ability
    Replies: 5569
    Last Post: 19-Sep-17, 6:02 PM
  2. TestFunda Answers - Geometry
    By TestFunda in forum Quantitative Ability
    Replies: 1317
    Last Post: 14-Sep-17, 7:15 PM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
 
 
 
 
 

 
 

 
©2008-2017   Enabilon Learning Private Limited. All rights reserved