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[administration]

The Tracks are set, the festive season for MBA Aspirants is on – CAT, XAT … Lets enjoy this festive season with an Official Q1 Race.

Rules:

1. No spamming on the thread. No Flirting, No Chit Chat other than related to question. You have your profile and Wire for the same 
2. Small Chit CHat allowed but within the domain of question.
3. Everyone is free to put the questions just mind few thing

• Solve the set of questions which is already there in the race. The guy/gal who solves the last question of the set will put next question set. A set of questions can contain atmost 4 questions at a time. But only after the solution of previous question is already there. 
• Numbering of the questions will continue.(i.e. if the last set of questions end at 14 . You will give next set of
• Nobody should put the next set of questions until the previous one has been fully discussed.
• Nobody should give just the answer without solution, i.e;”Is the answer x”..or” I think the answer is y”…whatever you think x or y just put it with the solution.
• You can post solution of 1 question at a time…Not a boundation to solve all at a time. 🙂
• If anybody want to share any valuable theory, go ahead..!!

1. To add pictures if any . Just load it at http://postimage.org and enter link in tab with green tree.
2. You need to login with your profile to post in the Q1 Race. Use Facebook login for quick (1 second) wild card entry 🙂

The official Race starts Now 

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[administration]

So let me give a green signal to the Race with these 4 questions 🙂

 

Q1. A candidate must secure 40% of the total aggregate to pass an examination in which each question has equal marks. He attempts only 8 questions out of every 10 questions. On 75% of his answers, he secures only 26% of the marks and on the remaining of the answers, 100% of the marks. Thus he gets 22 marks, too few to pass. The maximum number of marks is
a) 425  b) 500  c) 380  d) 620

 

Q2.   In how many ways we can form committee of 7 members out of 6 boys nd 4 girls if at least one of the president and the secretary is female?

a) 768   b) 812 c) 1680 d) 3360  e) None

 

Q3. One plans to prepare schedule for 7 day period in which he shall play exactly one sport out of A,B,C,D on each day,number of ways in which a schedule can be prepared such that each sport is played on atleast one day in the 7 day period is ?
a) 4200  b) 6300   c) 8400   d) 53760   e) None

 

Q4.  All 4 digit no. using digits 1,2,3,4,5 once are taken and their product is found How many zeroz at the end will such a product have ?
a)18  b) 24  c) 30 d)34

[md shadab hussain]

Q1)  let no of questions = Maximum marks be N;
Passing marks = 40% of N = 2N/5

he attempted 4/5 N questions;

out of these in 3/4 * (4/5N) questions he secured 26% marks;

= 0.26 * 3/4 * 4N/5 = 0.78N/5

of the remaining 1/4 * (4N/5) questions he secured full marks = N/5

He fell short of the passing marks by 22marks;

2N/5 – [N/5 + 0.78N/5] = 22

0.22N/5 = 22 or N=500…..ANS….(b)

[administration]

ok right answer @magnumopus22605 shadab :)…. carry on guys/gals Q2, Q3, Q4 … go , go , go …

[md shadab hussain]

Q4)Numbers of form XXX5…will give 1 ’0′ = 4C3 * 3! = 24….[1]
Numbers of form XX25…will give 2 ’0′ at the end of which 1 is already counted in the previous case = 3C2*2! = 6…[2]
Numbers of form X125…will give 3 ’0′ at the end of which 2 are already counted = 2…..[3]

NOTE : the number 3125 = 5^5 will give 5 ’0′s of which 3 are already counted = 2 more zeros….[4]

Total number of zeros = 24+6+2+2 = 34…ANS

[administration]

Yup @magnumopus22605 shadab… this one is also correct … 🙂 Ans is 34… 2 more to go 🙂

[md shadab hussain]

Q5. and Q6. above 🙂

[cat]

http://mba.webmaggu.com/groups/quantitative-ability/forum/topic/q1-the-official-quant-race/#post-208

1) -> 8Q in decimal system will be 105. From options 2Q^2+5Q+3 = 105 => Options 2
2) -> 7Q = 94 => Option 4

 

[cat]

http://mba.webmaggu.com/groups/quantitative-ability/forum/topic/q1-the-official-quant-race/#post-200

3) The 4 sports can be played in 4111, 3211, 2211 ways
=> 4c1*7!/4! + 4c1*3c1*7!/3!*2! + 4c2*7!/2!*2!
=> 840 +  5040 + 15120 => 21000 ways I am getting :/

[administration]

The correct answers are … Q1)  b i.e. 500 Q2) d i.e. 3360 Q3) c i.e. 8400 Q4) d i.e. 34 🙂 @cattack

[Wonder Boy]

Q3) ->

4,1,1,1: C(4,1)*C(7,4)*C(3,1)*C(2,1)=840
3,2,1,1:C(4,1)*C(7,3)*C(3,1)*C(4,2)*C(2,1)=5040
2,2,2,1:C(4,3)*C(7,2)*C(5,2)*C(3,2)=2520
Thus,Total:840+5040+2520=8400 Ans (c)

 

Q4) ->

Numbers ending with 5 = 24
Numbers ending with 25 = 6
Numbers ending with 125 = 2
Numbers ending in 3125 = 1 (bit here we will get 2 extra)

So, 24 + 6 + 2 + 2 = 34 Ans (d)

[Wonder Boy]

Ok a few questions from my side :). The race must go on 😉

Q7. The pendulum of a clock takes 7 sec to strike 4 O’clock. How much time will it take to strike 11 O’clock?

a. 18 sec  b. 20 sec   c. 19.25 sec  d. 23.33 sec

 

Q8. Suppose the function f is defined so that f(m)= 3m if m (m-1)² if m>1. If n is negative then f(1-n)= __ ?

a. (2-n)² b. n² c. 3-3n d.3n e.(n-1)²

 

Q9. Find the remainder when 1×2+2×3+3×4+….+99×100 is divided by 101 ?
a. 2 b. 100. c. 0 d. 7

Q10. If p and q are roots of quadratic equation ax² + bx + c = 0 such that unit digit of p and q are same. It is also given that D/a² is not divisible by 10. If n is the remainder when (p + q)/2 is divided by 10, then how many values n can take:- (D is discriminant)
a) 0 b) 1 c) 2 d) 3 e) More than 3

[Nayyar Abdul Rasheed]

Q8. I did it graphically, rotating the graph around m = 0.5. Ans is n²

[Wonder Boy]

I didn’t got your solution @Nayyar. Though your answer is correct. 🙂 . Can you attach the graph ?

[Nayyar Abdul Rasheed]

if u plot the graph of f(m), it’s the curve (m-1)² for m>1 and the straight line 3m for m<1 with discontinuity at m=1. now, when in a function y=f(x), x is replaced by a-x, the new function’s graph is just the mirror image of the graph of y=f(x) around x = a/2. in this case the new graph is just the mirror image of the previous graph around m=1/2. Sorry, unable to post the graph. 🙁

[Author]

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