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Thread: Quant Thread For CAT 2011

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    Default Doubts

    I have some doubts in questions based on time, speed, distance & time-work ! I am not really sure, whether, this is the right thread to post doubts. Nevertheless, this seems to be the most active at the moment. If there is any thread particularly dedicated to quant doubts, I would like to be informed about the same...... In any case, it would be gr8 to receive some good solutions to the problems mentioned below -

    Q1. Trains are traveling b/w stations A and B situated 120kms apart.
    There are 3 more stations b/w A and B. Each train halts at each station for 2 mins. What should be the minimum no.of trains, if the trains are traveling at 40kmph and a regular interval of 6mins is to be maintained b/w the departure of two trains from both the stations?
    (a) 31 (b) 30 (c) 62 (d) 63 (e) None of these

    Q2. A and B can each run at a uniform speed along a circular track. To cover the whole track, A need 5 seconds less than B. If they start simultaneously from the same place and run in the same direction, they meet 30 seconds later. At how many points will they meet if they run in the opposite directions?
    (a) 9 (b) 12 (c) 13 (d) 16 (e) None of these

    Q3. 545 crates were shifted by few men in 5 days . Every day after the 1st day , 6 more men are put onto the job and every day after the first day , each man shifted 5 fewer crates, than the day before .

    Q 1. what is the total number of crates shifted on 3rd day ?

    Q 2. what is the number of men on last day ?

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    guys plz solve ds..
    some olive sare purchased at the rate of 8 olives /Re and same number of olives at the rate of 6 olives/Re.now the whole quantity is sold at the rate of 7 olives/Re.What is the net percentage profit/loss?
    1) 0.6% profit
    2) 0.6%loss
    3) 1.2%loss
    4) no profit/loss

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    Quote Originally Posted by quixotic_me View Post
    guys plz solve ds..
    some olive sare purchased at the rate of 8 olives /Re and same number of olives at the rate of 6 olives/Re.now the whole quantity is sold at the rate of 7 olives/Re.What is the net percentage profit/loss?
    1) 0.6% profit
    2) 0.6%loss
    3) 1.2%loss
    4) no profit/loss
    As per my solution, answer is not in the option.

    LCM of 8,6 & 7 = 168. Therefore let the number of olive purchased be 168 each of 8 olive/Re type and 6 olive/Re type.

    Cost incurred due to purchase of 8 olive/ Re type = 21/-
    Cost incurred due to purchase of 6 olive/ Re type = 28/-
    Total sales realised by selling the olives at 7 olive/ Re = 24/-

    Net loss = (21+28)-(2 x 24) = 1/-

    Net loss % = 2.04% ...

    Another method of obtaining this is thru alligation rule .... In any case, i did not find the answer in the options. Kindly correct me, if I am wrong !

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    Quote Originally Posted by mohasan02 View Post
    As per my solution, answer is not in the option.

    LCM of 8,6 & 7 = 168. Therefore let the number of olive purchased be 168 each of 8 olive/Re type and 6 olive/Re type.

    Cost incurred due to purchase of 8 olive/ Re type = 21/-
    Cost incurred due to purchase of 6 olive/ Re type = 28/-
    Total sales realised by selling the olives at 7 olive/ Re = 24/-

    Net loss = (21+28)-(2 x 24) = 1/-

    Net loss % = 2.04% ...

    Another method of obtaining this is thru alligation rule .... In any case, i did not find the answer in the options. Kindly correct me, if I am wrong !
    yea! even m getting d same ans 2.04 %. options are nt correct i guess..

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    A shopkeeper marks up his goods by 20 % and then gives a discount of 20%.Besides he cheats both his supplier and customer by 100 grams i.e he takes 1100 gram from his supplier and sells only 900 grams to his customer.he also takes a discount of 20% from his supplier which he disregards while preparing his mark up.

    1) now he adulterates his goods in the ratio of 1:5 (adulterantriginal) Assuming the adulterant comes free of cost,what is the net profit percentage?
    Options
    1) 62.5 %
    2) 66.66%
    3)72.08%
    4) none of these
    2)If the cost of the adulterant is 20% of original substance ,what is the net profit percentage?
    Options
    1) 77.77%
    2) 81.08%
    3)86.75%
    4) none of these
    aman99 likes this.

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    Quote Originally Posted by quixotic_me View Post
    guys plz solve ds..
    some olive sare purchased at the rate of 8 olives /Re and same number of olives at the rate of 6 olives/Re.now the whole quantity is sold at the rate of 7 olives/Re.What is the net percentage profit/loss?
    1) 0.6% profit
    2) 0.6%loss
    3) 1.2%loss
    4) no profit/loss
    it is possible that the options are wrong... else that you wrote 8 olives /per Re instead of 8rupee per olive... cause that way answer would be no profit no loss...

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    Quote Originally Posted by mohasan02 View Post
    I have some doubts in questions based on time, speed, distance & time-work ! I am not really sure, whether, this is the right thread to post doubts. Nevertheless, this seems to be the most active at the moment. If there is any thread particularly dedicated to quant doubts, I would like to be informed about the same...... In any case, it would be gr8 to receive some good solutions to the problems mentioned below -

    Q1. Trains are traveling b/w stations A and B situated 120kms apart.
    There are 3 more stations b/w A and B. Each train halts at each station for 2 mins. What should be the minimum no.of trains, if the trains are traveling at 40kmph and a regular interval of 6mins is to be maintained b/w the departure of two trains from both the stations?
    (a) 31 (b) 30 (c) 62 (d) 63 (e) None of these

    Q2. A and B can each run at a uniform speed along a circular track. To cover the whole track, A need 5 seconds less than B. If they start simultaneously from the same place and run in the same direction, they meet 30 seconds later. At how many points will they meet if they run in the opposite directions?
    (a) 9 (b) 12 (c) 13 (d) 16 (e) None of these

    Q3. 545 crates were shifted by few men in 5 days . Every day after the 1st day , 6 more men are put onto the job and every day after the first day , each man shifted 5 fewer crates, than the day before .

    Q 1. what is the total number of crates shifted on 3rd day ?

    Q 2. what is the number of men on last day ?
    ok well now Q1) i cant get the wording of the Question.. does one train on reaching a station turn around?? or what... as in what does this regular interval of 6 min mean??

    Q2) answer would be 5 points ...when they move in opposite direction.

    Q3) is this Question correct ?? cause if there are 545 crates... and it takes few men 5 days... it means each day they shift in all 109 crates... but 109 being a prime number... it means either there is only one man... or there are 109 men and each moves only one crate a day.... either way a problem arises according to the second condition in the question... i am assuming each man shifts same amount of crates on a particular day....?

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    Quote Originally Posted by quixotic_me View Post
    A shopkeeper marks up his goods by 20 % and then gives a discount of 20%.Besides he cheats both his supplier and customer by 100 grams i.e he takes 1100 gram from his supplier and sells only 900 grams to his customer.he also takes a discount of 20% from his supplier which he disregards while preparing his mark up.

    1) now he adulterates his goods in the ratio of 1:5 (adulterantriginal) Assuming the adulterant comes free of cost,what is the net profit percentage?
    Options
    1) 62.5 %
    2) 66.66%
    3)72.08%
    4) none of these
    2)If the cost of the adulterant is 20% of original substance ,what is the net profit percentage?
    Options
    1) 77.77%
    2) 81.08%
    3)86.75%
    4) none of these

    Here is my approach !

    a.) Let the original price of rice be 1000/- for 1 kg.

    But the bluffmaster got it for 800/- for 1kg.

    HOwever he marks up the price based on the original price of the rice. Thus, Markd up price, MP = 1.2 x 1000 = 1200/-

    SP = 960/- for 1 kg

    Since the bluffmaster fiddles around with the weight !!

    It means 800/- is used to purchase 1100gm of rice, whereas he sells only 900gm for 960/-.
    Further, he mixes impurities in it in the ratio 1:5. Thus he sells an effective weight of 750gm.


    800/- is used to purchase 1100gm, which means 1000/- may be used to purchase 1375gm rice.
    Similarly,
    750gm rice is sold for 960/-, which means 1000/- may be used to sell 781gm of rice (kindly check calculation of my divisions).
    THus net profit percentage;
    (1375-781)/781 = 0.7605 i.e. 76.05%

    option .4 - none of these

    b.) From the above steps we find -
    1100gm is purchased for 800/-
    So, 900gm may be purchased for 655/- (unitary method). However this is adulterated with impurities whose cost is 20% of the original rice i.e. 200/-.

    This gives us an effective cost of 580/- for 900gm
    Whereas, SP for 900gm rice is 960/-

    Thus profit % = 65.51%

    Option 4. None of these ( Comparing the two cases, we see that a decrease in profit percentage in the latter is due to the cost price of the adulterate, which was earlier taken to be zero)

    Kindly rectify me, incase I am wrong somewhere.

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    Quote Originally Posted by grimreaper View Post
    ok well now Q1) i cant get the wording of the Question.. does one train on reaching a station turn around?? or what... as in what does this regular interval of 6 min mean??

    Q2) answer would be 5 points ...when they move in opposite direction.

    Q3) is this Question correct ?? cause if there are 545 crates... and it takes few men 5 days... it means each day they shift in all 109 crates... but 109 being a prime number... it means either there is only one man... or there are 109 men and each moves only one crate a day.... either way a problem arises according to the second condition in the question... i am assuming each man shifts same amount of crates on a particular day....?
    Q1. What i understood was - trains leave alternately from stations A and B in intervals of 6 mins, towards each other. !!!... I could not get the hang of the questions, could be possible that the language is faulty !

    Q2. Even I got the answer to be 5 times. Can you plz elaborate the solution, so that I may compare my approach with yours !!.... According to my approach :

    number of distinct points = (time taken by A&B to reach starting point for the 1st time) / (time taken by anyone of them to cover the circumference at a relative speed of x+y)

    x = speed of A ; y = speed of B

    Q3. In my thinking - sme data is missing regarding the rate of work done by each man. OR may just in case 545 becomes 595 !!... then 109 becomes 119 = 17x 7 !!!....
    In any case, thnxz for ur responses !!

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    Quote Originally Posted by mohasan02 View Post
    Q1. What i understood was - trains leave alternately from stations A and B in intervals of 6 mins, towards each other. !!!... I could not get the hang of the questions, could be possible that the language is faulty !

    Q2. Even I got the answer to be 5 times. Can you plz elaborate the solution, so that I may compare my approach with yours !!.... According to my approach :

    number of distinct points = (time taken by A&B to reach starting point for the 1st time) / (time taken by anyone of them to cover the circumference at a relative speed of x+y)

    x = speed of A ; y = speed of B

    Q3. In my thinking - sme data is missing regarding the rate of work done by each man. OR may just in case 545 becomes 595 !!... then 109 becomes 119 = 17x 7 !!!....
    In any case, thnxz for ur responses !!
    yes for Q3) thats what most likely i guess.

    as for Q2) post your solution.... elaborate it...
    cause how do you knw what is the time taken by a and b to reach the starting point for first time ? meeting the first time and the starting point need not be the same always....

    anyways good set of questions nonetheless ....
    just try numbering them... that way others can find it easy to follow.... as in.. number according to the forum numbering... like the last Q to be posted was Q 4... so makes yours as Q 5 Q6 Q7 so on....

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    Quote Originally Posted by mohasan02 View Post
    I have some doubts in questions based on time, speed, distance & time-work ! I am not really sure, whether, this is the right thread to post doubts. Nevertheless, this seems to be the most active at the moment. If there is any thread particularly dedicated to quant doubts, I would like to be informed about the same...... In any case, it would be gr8 to receive some good solutions to the problems mentioned below -

    Q5. Trains are traveling b/w stations A and B situated 120kms apart.
    There are 3 more stations b/w A and B. Each train halts at each station for 2 mins. What should be the minimum no.of trains, if the trains are traveling at 40kmph and a regular interval of 6mins is to be maintained b/w the departure of two trains from both the stations?
    (a) 31 (b) 30 (c) 62 (d) 63 (e) None of these

    Q6. A and B can each run at a uniform speed along a circular track. To cover the whole track, A need 5 seconds less than B. If they start simultaneously from the same place and run in the same direction, they meet 30 seconds later. At how many points will they meet if they run in the opposite directions?
    (a) 9 (b) 12 (c) 13 (d) 16 (e) None of these

    Q7. 545 crates were shifted by few men in 5 days . Every day after the 1st day , 6 more men are put onto the job and every day after the first day , each man shifted 5 fewer crates, than the day before .

    a. what is the total number of crates shifted on 3rd day ?

    b. what is the number of men on last day ?
    Solutions --

    Q5. As per what i understood - the condition states that "a train halts at each of th three stations between A & B for 2 mins each". Thus total time taken for one train to travel A to B at 40km/hr = 186 mins

    So lets say, if train T1 leaves A at 6.00am, it wud reach B at 9.06am. Therefore we need to find the number of trains that can depart alternately from A & B in intervals of 6 mins in this 186 mins interval !! i.e.
    T1 leaves A at 6.00am
    T2 leaves B at 6.06 am
    T3 leaves A at 6.12 am ..... & so on. For the number of trains to be minimum, the last train would be Tn-1 departing A at 9.00pm because in the next 6 mins, T1 reaches B ( so we gta avoid the case in which T1 & Tn would arrive & depart B respectivly, at the same time -- for the minimum case).

    Therefore, the minimum no. of trains will be = (186/6) - 1= 30
    ( Maximum number of trains would be 31)

    option. b

    Q.6) Let circumference = C
    speed of A = x ( Assuming x>y)
    speed of B = y

    Movng simultaneously in the same direction they catch up in 30 sec which is equivalent to A meeting up B at the starting point only after travelling with a relative speed of (x-y).

    *Note: In absolute sense, A may or may not be catching up with B at the starting point, but since we are using relativity, thus we are asuming B to be fixed & A to be moving at (x-y) speed. So A can catch up B only after it has completed a relative distance of C with (x-y) speed !!

    Thus,

    C = 30(x-y) ......... (i)
    & from other conditions -

    [C/x] = [C/y] - 5 ......... (ii)

    Solving i & ii, we get -

    (x-y)2/(xy) = 1/6

    x & y can assume a range of values
    x = {3,6,12,...} ; y = {2,4,8,.....} *However as we would need to measure ratio, therefore any ordered set (x,y) would do be fine. For calculation ease, I choose (3,2)

    Thus, C = 30(3-2)=30
    Time taken by B to complete one round = 15 sec
    Time taken by A to complete one round = 10 sec.

    As we need to find the number of times they meet up travelling in opposite direction. Let us choose an unit time interval within which these number of points is to be counted !!!... That interval will be the first time A & B would meet at the starting point (in absolute sense) travelling in opposite direction = LCM of 10 & 15 = 30 sec

    Time taken by A&B to meet each time= time taken by A to cover a distance C to catch up with B at (x+y) speed = [C/x+y] = 6 sec

    Therefore number of distinct meeting points = 30sec/6 sec = 5 !!

    Q7. Values given are doubtful !! Data missing


    Folks kindly post your opinions for the above solutions, so that incase of any conceptual error, it may be rightly pointed out !..Thanxz

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    Quote Originally Posted by mohasan02 View Post

    Q3. 545 crates were shifted by few men in 5 days . Every day after the 1st day , 6 more men are put onto the job and every day after the first day , each man shifted 5 fewer crates, than the day before .

    Q 1. what is the total number of crates shifted on 3rd day ?

    Q 2. what is the number of men on last day ?

    Ans1: 169
    Ans2: 25
    CAT2011 : 99.97%ile
    XAT2012 : 99.07%ile
    FMS 2012-14
    Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR

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    Quote Originally Posted by shaunak_87 View Post

    Ans1: 169
    Ans2: 25
    Thanxz for ur resposne. Can u plz elaborate the method of obtaining the number of men = 1 & crates moved = 23 for 1st day.... For the subsequent days, its pretty obvious.... But how did u arrive at the day 1`s combination ?!!

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    Quote Originally Posted by shaunak_87 View Post

    Ans1: 169
    Ans2: 25

    i see your point... but then how did u arrive at this... i mean... u take the condition of each day 6 men more to the initial problem itself right...

    but even then... i can only think of trial and error to get at this table....
    even then i wnt be sure till my total comes upto 545....

    i mean yes i would have started with only one man on the first day... but nonetheless... i still wld have to calculate all the way till day 5 to be sure....
    but yes the table is right

    @ mohsan - post your method for that circular Question..... lets see what you did , and try the previous Qs posted in this forum...

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    Default Q5)

    Pls follow the question numbering....

    Q5) How many integral values of x are possible such that 2^70+2^1039+2^x is a perfect square of a whole number

    a) two
    b) three
    c)infinite
    d) one
    e) four

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    Quote Originally Posted by grimreaper View Post
    i see your point... but then how did u arrive at this... i mean... u take the condition of each day 6 men more to the initial problem itself right...

    but even then... i can only think of trial and error to get at this table....
    even then i wnt be sure till my total comes upto 545....

    i mean yes i would have started with only one man on the first day... but nonetheless... i still wld have to calculate all the way till day 5 to be sure....
    but yes the table is right

    @ mohsan - post your method for that circular Question..... lets see what you did , and try the previous Qs posted in this forum...
    Let circumference = C
    speed of A = x ( Assuming x>y)
    speed of B = y

    Movng simultaneously in the same direction they catch up in 30 sec which is equivalent to A meeting up B at the starting point only after travelling with a relative speed of (x-y).

    *Note: In absolute sense, A may or may not be catching up with B at the starting point, but since we are using relativity, thus we are asuming B to be fixed & A to be moving at (x-y) speed. So A can catch up B only after it has completed a relative distance of C with (x-y) speed !!

    Thus,

    C = 30(x-y) ......... (i)
    & from other conditions

    [C/x] = [C/y] - 5 ......... (ii)

    Solving i & ii, we get -

    (x-y)2/(xy) = 1/6

    x & y can assume a range of values
    x = {3,6,12,...} ; y = {2,4,8,.....} *However as we would need to measure ratio, therefore any ordered set (x,y) would do be fine. For calculation ease, I choose (3,2)

    Thus, C = 30(3-2)=30
    Time taken by B to complete one round = 15 sec
    Time taken by A to complete one round = 10 sec.

    As we need to find the number of times they meet up travelling in opposite direction. Let us choose an unit time interval within which these number of points is to be counted !!!... That interval will be the first time A & B would meet at the starting point (in absolute sense) travelling in opposite direction = LCM of 10 & 15 = 30 sec

    Time taken by A&B to meet each time= time taken by A to cover a distance C to catch up with B at (x+y) speed = [C/x+y] = 6 sec

    Therefore number of distinct meeting points = 30sec/6 sec = 5 !!

    I have posted my method for both the questions - circular track one & the train - station problem in earlier posts. Yet, I am posting my method for the circular track question again.

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    Quote Originally Posted by mohasan02 View Post
    Let circumference = C
    speed of A = x ( Assuming x>y)
    speed of B = y

    Movng simultaneously in the same direction they catch up in 30 sec which is equivalent to A meeting up B at the starting point only after travelling with a relative speed of (x-y).

    *Note: In absolute sense, A may or may not be catching up with B at the starting point, but since we are using relativity, thus we are asuming B to be fixed & A to be moving at (x-y) speed. So A can catch up B only after it has completed a relative distance of C with (x-y) speed !!

    Thus,

    C = 30(x-y) ......... (i)
    & from other conditions

    [C/x] = [C/y] - 5 ......... (ii)

    Solving i & ii, we get -

    (x-y)2/(xy) = 1/6

    x & y can assume a range of values
    x = {3,6,12,...} ; y = {2,4,8,.....} *However as we would need to measure ratio, therefore any ordered set (x,y) would do be fine. For calculation ease, I choose (3,2)

    Thus, C = 30(3-2)=30
    Time taken by B to complete one round = 15 sec
    Time taken by A to complete one round = 10 sec.

    As we need to find the number of times they meet up travelling in opposite direction. Let us choose an unit time interval within which these number of points is to be counted !!!... That interval will be the first time A & B would meet at the starting point (in absolute sense) travelling in opposite direction = LCM of 10 & 15 = 30 sec

    Time taken by A&B to meet each time= time taken by A to cover a distance C to catch up with B at (x+y) speed = [C/x+y] = 6 sec

    Therefore number of distinct meeting points = 30sec/6 sec = 5 !!

    I have posted my method for both the questions - circular track one & the train - station problem in earlier posts. Yet, I am posting my method for the circular track question again.
    sorry i cldnt see tht post till now...
    hmmmm seems alright.... but once you find out the ratio.... why are you going further??

    remember always the number of points two ppl meet wld be the sum of speeds ratio when in opp direction... and the difference of speed ratio when in same direction...

    important - the ratio shld be in its simplest form... that is 6/4 doesnt mean they meet at ten points.... cause the simple ratio wld be 3:2 hence they meet at 5 points....

    as for the rest of ur solution its proper....

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    Default Q5. solution

    Quote Originally Posted by grimreaper View Post
    Pls follow the question numbering....

    Q5) How many integral values of x are possible such that 2^70+2^1039+2^x is a perfect square of a whole number

    a) two
    b) three
    c)infinite
    d) one
    e) four
    option (d)

    The given expression has three terms and therefore it should be of the form (a+b)2 for it to be perfect square.

    270 = 22(35)
    2x = 22(y)
    21039 = 2. 235.2y
    Therefore x = 1003

    Thus x has a unique value

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    Quote Originally Posted by grimreaper View Post
    sorry i cldnt see tht post till now...
    hmmmm seems alright.... but once you find out the ratio.... why are you going further??

    remember always the number of points two ppl meet wld be the sum of speeds ratio when in opp direction... and the difference of speed ratio when in same direction...

    important - the ratio shld be in its simplest form... that is 6/4 doesnt mean they meet at ten points.... cause the simple ratio wld be 3:2 hence they meet at 5 points....

    as for the rest of ur solution its proper....
    Thanxz man !!... ur insight saved me lot of unnecessary wrk !

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    Quote Originally Posted by mohasan02 View Post
    option (d)

    The given expression has three terms and therefore it should be of the form (a+b)2 for it to be perfect square.

    270 = 22(35)
    2x = 22(y)
    21039 = 2. 235.2y
    Therefore x = 1003

    Thus x has a unique value
    yup.... an easy prob ... ehmmm bt x=2006 :P not 1003....

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