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## Re: Quant Problems

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• Hi I came across a few quant probs. n wud like to noe the solutions to some. Q. A person goes between 4 and 5 and comes back between 5 and 6 and realises that
Message 1 of 11 , Jul 5, 2005
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Hi
I came across a few quant probs. n wud like to noe the solutions to some.

Q. A person goes between 4 and 5 and comes back between 5 and 6 and
realises that the clock hands (hour hand and minute hand) interchange
their respective positions. How much time did he remain out of his
respective house?

Q. 20 women can finish a job in 20 days. After each day, one woman
is replaced by a man or a boy alternatively starting with a man.Man is
twice efficient and boy is half efficient as a woman. On which day
does the job get completed?

Q.group of new students who total age is 221 years joins a class,
becoz if which the strength of the class goes up by 50% but the avg
age of the class comes down by 1 year. whats the new avg of the class?
if it is known to be a natural number after the new group of students
have joined, given that the original strength of the class was a digit
number greater than 30.

Please tell me the solutions to these probs.Expecting to hear from u soon.
• sumedh hastak wrote: 1. Two men are moving in opposite directions with speeds 18 kmph & ... 100 ... solution to this prob is....
Message 2 of 11 , Jul 15, 2005
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sumedh hastak <sumedhhastak@...> wrote:

1. Two men are moving in opposite directions with speeds 18 kmph &
> 36 kmph in a straight line. The faster man can see a person from
100
> m. After seeing the slower man he takes 2 s for reacting (applying
> brakes) and reduces his speed by 1 m/sec in each second. If the
> slower man is as efficient as the faster one in all respects, find
> out the minimum rate of reduction of speed to avoid collision.
> Assume that both stop at the same time.
solution to this prob is....

36kmph=10m\s    18kmph=5m\s

now the seperation betwn them when they apply brakes is 70m i.e.(100-(20+10))  if they avoid collision sum of their distance travelled shud be 70... and time is same... therefore

50+25\2a=70... solving for a we get 0.625 which is the ans

Well try this

V(total)=14

so v*v=196

and final v=0

so 196=2a(70)

ans 1.4

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• Hi Well the acceleration can never be greater than 1 as it has to be less for the slower person.But if we see both will stop at same time: so by v= u + at,
Message 3 of 11 , Jul 18, 2005
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Hi
Well the acceleration can never be greater than 1 as it has to be less
for the slower person.But if we see both will stop at same time:
so by v= u + at, time will 10 sec as we know dat for faster person v
=0; u=10 and a=1
so again aplying the equation for second person we get a = 0.5.Now can
anybody explain me the flaw in this as 0.625 is looking appt.

On 7/15/05, siddharth juneja <siddharth_2002005@...> wrote:
>
>
> sumedh hastak <sumedhhastak@...> wrote:
> 1. Two men are moving in opposite directions with speeds 18 kmph &
> > 36 kmph in a straight line. The faster man can see a person from
> 100
> > m. After seeing the slower man he takes 2 s for reacting (applying
> > brakes) and reduces his speed by 1 m/sec in each second. If the
> > slower man is as efficient as the faster one in all respects, find
> > out the minimum rate of reduction of speed to avoid collision.
> > Assume that both stop at the same time.
> solution to this prob is....
>
> 36kmph=10m\s 18kmph=5m\s
>
> now the seperation betwn them when they apply brakes is 70m
> i.e.(100-(20+10)) if they avoid collision sum of their distance travelled
> shud be 70... and time is same... therefore
>
> 50+25\2a=70... solving for a we get 0.625 which is the ans
>
> Well try this
>
> V(total)=14
>
> so v*v=196
>
> and final v=0
>
> so 196=2a(70)
>
> ans 1.4
>
>
>
>
>
>
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• ... for this question , is the answer...18th day???
Message 4 of 11 , Jul 24, 2005
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> Q. 20 women can finish a job in 20 days. After each day, one woman
> is replaced by a man or a boy alternatively starting with a man.Man is
> twice efficient and boy is half efficient as a woman. On which day
> does the job get completed?

for this question , is the answer...18th day???
• Hi vaibhav, i am writting the answers to each one here.. please let me know if u have any further query after this.. 1. when guy leaves home, lets say minutes
Message 5 of 11 , Jul 26, 2005
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Hi vaibhav,

i am writting the answers to each one here.. please let me know if u
have any further query after this..

1. when guy leaves home, lets say minutes hand are at x minute unit
frm 5 mark, and and when he comes back, they are at y distance from 4
mark.
so, we get two equations.. by comparing hour and minutes hands
position..they are ..

==> (25+x)/60 = y/5; ==> 25+x = 12y;
==> (20+y)/60 = x/5; ==> 20+y = 12x;

solution is x=1.69, y=2.23
This translates to 4:26:42 Hrs and 5:22:12 Hrs
and a time difference of 0:55:30 ==>55 minutes and 30 seconds.

2.
lets consider unit equivalence for a woman=2, a man =4, one child=1
1st Day ==> 20 woman = 40 units;
(a, b, c) stands here for daily addition of deduction for units from
(woman, man, child);

2nd Day ==> sum(-2, 4, 0) = 2 ==> units = 42
3rd Day ==> sum(-2, 0, 1) = -1 ==> units = 41
4th day ==> sum(-2, 4, 0) = 2 ==> units = 43
5th Day ==> sum(-2, 0, 1) = -1 ==> units = 42

anyways, the sum of the units should just cross 20 * 20 *2 mark = 800

therefore,

==> 40 + sum [ (40+2) + ( 40+1) + (40+3) + (40+2) + ....] <= 800;
Either you can claculate this sum manually or go in mathematically ..
doing ..
==> sum [ (40+2) + ( 40+1) + (40+3) + (40+2) + ....] <= 760;
rearranging terms ..
==> sum[(40+2)+(40+3)+(40+4)+...] +sum[(40+1)+(40+2)+(40+3)+..] <=760;
==> 40*n + n + n*(n+1)/2 + 40*n + n*(n+1)/2 <=760
==> 82*n + n^2 <=760
==> n^2 + 82*n <=760;

solving, we get n = 8.43 this implies LHS for n =8 is 720, the work
will be completed one day later, as we found solution for 8 complete
turns in above summation as this adds to effort less than or equal to
800.

the work will be done on 1+ 2*8 + 1 th Day = 18th Day!!

3.
lets say initial avg is n; and initial strength is x; final strength
= 3/2 *x;
inital total age = n*x;
later total age = 3/2 *(n-1)*x;
we have ..
==> [n*x +221] = [3/2 *x * (n-1)];
==> (n-3) * x = 442;

again 442 = 2 * 13 * 17

it implies that x can only be 2, 13, 17, 36, 34, ....
But again, x and 3x/2 will be integer , implying x is even ..
so, x = 26, 34, 442

forget 442,
for x = 26 , n = 20;
for x = 34, n = 16;
But given x <30, we get solution as x = 26, and n = 16.

The final avg age is 15. This holds true as we cross-check the
answers.

Thanks,
Sachin
sachin.sk@...
• ... some. ... Solution The man remains out for 55 mins. Since he goes out between 4 & 5 the hour hand should be on 4 ,& also as he returns between 5 & 6 the
Message 6 of 11 , Jul 27, 2005
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--- In ascent4cat@yahoogroups.com, Vaibhav Gupta <vaibhavgupta007@g...>
wrote:
> Hi
> I came across a few quant probs. n wud like to noe the solutions to
some.
>
> Q. A person goes between 4 and 5 and comes back between 5 and 6 and
> realises that the clock hands (hour hand and minute hand) interchange
> their respective positions. How much time did he remain out of his
> respective house?
>
Solution

The man remains out for 55 mins.
Since he goes out between 4 & 5 the hour hand should be on 4 ,&
also as he returns between 5 & 6 the hour hand is on 5. As the hour &
minute hand have exchanged, he has left at 4.25 and returned at 5.20.
So,he remained out for 55 mins.
Is this the answer . Plz inform me if it is correct or not.
• well your solution is wrong.... Consider the acute angle between the minutes hand and the hours hand when he goes out and comes back. Since the hands have
Message 7 of 11 , Aug 13, 2005
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well your solution is wrong....
Consider the acute angle between the minutes hand and the hours hand when he goes out and comes back. Since the hands have exchanged, the angle will remain constant.... Let the angle be x.

For the minutes hand:
Angle moved by the hand when the person is outside his house is 360-x.
Minutes hand moves 6 degrees every minute.
Time for which the man stays outside = (360-x)/6

Applying similar analogy
For the hours hand:
Angle moved when the person is outside the house = x
Hours hand moves 1/2 degree in a minute.

Time for which the man stays outside = 2x minutes

Now since we could calculate the time taken in both the cases, they should be equal.

i.e. 2x = (360-x)/6 => x = 360/13

We could find out the angle between the minutes hand and hours hand.

The man stays out for 2x minutes i.e. 720/13 = 55 5/13 minutes.

This is the exact answer.

Hope you understood the solution.

Prakash

dear_rita2001 <dear_rita2001@...> wrote:
--- In ascent4cat@yahoogroups.com, Vaibhav Gupta <vaibhavgupta007@g...>
wrote:
> Hi
> I came across a few quant probs. n wud like to noe the solutions to
some.
>
> Q.  A person goes between 4 and 5 and comes back between 5 and 6 and
> realises that the clock hands (hour hand and minute hand)  interchange
> their respective positions. How much time did he remain  out of his
> respective house?
>
Solution

The man remains out for 55 mins.
Since he goes out between 4 & 5 the hour hand  should be on 4 ,&
also as he returns between 5 & 6 the hour hand is on 5. As the hour &
minute hand have exchanged, he has left at 4.25 and returned at 5.20.
So,he remained out for 55 mins.
Is this the answer . Plz inform me if it is correct or not.

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