Quant by Arun Sharma

MBA CAT 2022
610
no dude .........
ans is 13..................
611
Quest from Ratio, Proportion and variation

LOD - III : Quest :24

There are two quantities of milk - Amul and Sudha having diff prices per litre, there volumes being 130 litres and 180 litres respectively. After equal amounts of milk was removed from both, the milk removed from amul was added to sudha and vice-versa. The resulting two types of milk now have the same price. find the amount of milk drawn out from each type of milk ?

Options : (a) 58.66 (b) 75.48 (c) 81.23 (d) None

Answer given : 75.48
612

Nice Thread Can Anyone Share E Book Of Arun Sharma Cat

Thanks A Ton

613
Quest from Ratio, Proportion and variation

LOD - III : Quest :24

There are two quantities of milk - Amul and Sudha having diff prices per litre, there volumes being 130 litres and 180 litres respectively. After equal amounts of milk was removed from both, the milk removed from amul was added to sudha and vice-versa. The resulting two types of milk now have the same price. find the amount of milk drawn out from each type of milk ?

Options : (a) 58.66 (b) 75.48 (c) 81.23 (d) None

Answer given : 75.48


hmmm....
i think some data is seriously missing....
614

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615
mani_nandwani Says
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refrain from spamming in the threads..
616
Can someone explain how to solve this problem from LOD II, Problem 32, Pg. 72, Progressions - Arun Sharma 2nd edition:

Q: The Sum of the series

1/(2+1) + 1/(2+3)++1/(120+121) is

Options given: a. 10 b. 11 c. 12 d. None of these

The answer is supposed to be a. 10. How do you solve this problem?

Thanks,
BlackSabbath
617
Can someone explain how to solve this problem from LOD II, Problem 32, Pg. 72, Progressions - Arun Sharma 2nd edition:
Q: The Sum of the series
1/(2+1) + 1/(2+3)++1/(120+121) is
Options given: a. 10 b. 11 c. 12 d. None of these
The answer is supposed to be a. 10. How do you solve this problem?

Thanks,
BlackSabbath


Rationalise the denominator for each term
Eg.
1/(2+1) = (2-1)/=(2-1)

Hence the question reduces to
2-1+3-2+4-3+.......+120-119+121-120
= 121-1
=11 -1
=10
618

heyya puys i am late to this thread

now cannot go through all the 60 odd pages in this thread
if anybody has the archives of the fundas and problems posted in this thread please upload them

they will be useful to the late starters like me ........

619

hey guys pls solve these questions.
1> Find the number of divisors of 1080 excluding the throghout divisors,which are perfect squares.

2> FInd the number of divisors of 544 excluding 1 and 544

3> Find the number of divisors of 544 which are greater than 3

4> find the sum of divisors of 544 excluding 1 and 544

5> find the sum of odd divisors of 544

6> fine sum of even divisors of 4096

620
hey guys pls solve these questions.
1> Find the number of divisors of 1080 excluding the throghout divisors,which are perfect squares.

2> FInd the number of divisors of 544 excluding 1 and 544

3> Find the number of divisors of 544 which are greater than 3

4> find the sum of divisors of 544 excluding 1 and 544

5> find the sum of odd divisors of 544

6> fine sum of even divisors of 4096


2) 544=2^5*17^1.....num of divisors will be (5+1)*(1+1)=12-2=10
5+1=6(the power of 2 is 5 so according to the formula of getting number of divosrs or factors we have to add 1 and same we will do in case of power of 17 ) now we have subtract 2 factos(1 and 544)thn we get totoal divisors 12-2=10 for 544

3) again 12-2(1 and 2 and less thn 3)=10

4)sum of divosrs...2^5*17^1.....(2^0+2^1+2^2+2^3+2^4+2^5)*(17^0+17^1)=1134-545=589

5) sum of odd devisors= 2^0*=18

6) sum of even divsors of 4096=2^12
(2^1+2^2...............2^12)=8190...we wont add 2^0 as we are taking sum of even factors :)


plz someone ans q1 😐
621

Hi,
I am solving Block 1 number system LOD II and got stuck in Question No.-64.
Could anyone help me please to find out the logic to solve these type of Question?
Well pasting Question here--
find the 28383rd term in the series 123456789101112.....

622
Hi,
I am solving Block 1 number system LOD II and got stuck in Question No.-64.
Could anyone help me please to find out the logic to solve these type of Question?
Well pasting Question here--
find the 28383rd term in the series 123456789101112.....


i just came to post same question well i got 3 😞 but given ans is 9 😐
623

well i got the same ans for this question.....

i am sure answer is 3

624

hi
i am solving quadratic equation from arun sharma..
got stuck with sum in lod 2..
question no.13..
please help out me with this
the sum is
value of the expression (x^2-x+1)/(x-1) cannot lie between
a. 1,3
b. -1,-3
c. 1,-3
d. -1,2
e -1,3

Thanks in advance!!

625
hi
i am solving quadratic equation from arun sharma..
got stuck with sum in lod 2..
question no.13..
please help out me with this
the sum is
value of the expression (x^2-x+1)/(x-1) cannot lie between
a. 1,3
b. -1,-3
c. 1,-3
d. -1,2
e -1,3

Thanks in advance!!

Well I m not sure if Qeqn funda alone can solve this problem. But graph certainly can. Here I m giving my approach.

Given exp can be written as x+1/(x-1) = (x-1)+1/(x-1) +1.
Now for x>0, (x+1/x)>=2 and for x
626
Well I m not sure if Qeqn funda alone can solve this problem. But certainly can. Here I m giving my approach.

Given exp can be written as x+1/(x-1) = (x-1)+1/(x-1) +1.
Now for x>0, (x+1/x)>=2 and for x


thanks 4 d answer..but they are sayin -1,2 is the answer..
i was not able to follow ur first step 'x+1/(x-1) = (x-1)+1/(x-1) +1'..
can u plz explain this part..

regards
the game!!
627
hi
i am solving quadratic equation from arun sharma..
got stuck with sum in lod 2..
question no.13..
please help out me with this
the sum is
value of the expression (x^2-x+1)/(x-1) cannot lie between
a. 1,3
b. -1,-3
c. 1,-3
d. -1,2
e -1,3

Thanks in advance!!


Ans is (e).

We can consider x^2-x+1/(x-1) = k
Hence ,x^2-(k+1)x + (k+1) = 0
For this eqn to have real roots,
(k+1)^2 -4(k+1) >=0
or k^2 -2k - 3 >=0
or(k-3)(k+1) >=0
or k cannot lie in range -1 to 3
628
Ans is (e).

We can consider x^2-x+1/(x-1) = k
Hence ,x^2-(k+1)x + (k+1) = 0
For this eqn to have real roots,
(k+1)^2 -4(k+1) >=0
or k^2 -2k - 3 >=0
or(k-3)(k+1) >=0
or k cannot lie in range -3 to 1


Hey every thing is perfect ..except the last line.
K will not fall in -1 to 3, but u have written it as -3 to 1. So that vindicates my answer too.
629
thanks 4 d answer..but they are sayin -1,2 is the answer..
i was not able to follow ur first step 'x+1/(x-1) = (x-1)+1/(x-1) +1'..
can u plz explain this part..

regards
the game!!

(x^2 -x +1)/(x-1) = /(x-1)=x+1/(x-1)=(x-1)+1/(x-1)+1.

So Mr Arun says that the exp can take values in 2 to 3 as well. let's c.

(x^2 -x +1)/(x-1)=5/2 => 2x^2-7x+7 = 0, for which D is -7, so no real value of x is possible.
I think our answer is correct.