Quant by Arun Sharma

MBA CAT 2024
4101
testdjjaydev Says
can someone plz tell me d link for download of arun sharma solutions,,stuck wid some questions,dint get d solutions as well..plz help !!!!


I dont think any link is avail. please post here.
We will togetherly try to find out the solutions
Thanks
Ashish
4102

wats d remainder wen 50^51^52/11???
ne shortcut or good approach???

4103
wats d remainder wen 50^51^52/11???
ne shortcut or good approach???


If it is 50^(51^52) then remainder is 5.
If it is (50^51)^52 hen remainder is 3.

Euler's number for 11 is 10.
So, 50^10k will leave remainder by 11 as 1.
And 50 leaves remainder by 11 as 6.

So, 50^(51^52) = 50^(10k+1) => Remainder => 6^1 => 6
And (50^51)^52 = 50^(10k+2) => Remainder = 6^2 => 3
4104

wats d remainder wen 50^51^52/11???
ne shortcut or good approach???

If it is 50^(51^52) then remainder is 5.
If it is (50^51)^52 hen remainder is 3.

Euler's number for 11 is 10.
So, 50^10k will leave remainder by 11 as 1.
And 50 leaves remainder by 11 as 5.

So, 50^(51^52) = 50^(10k+1) => Remainder => 5^1 => 5
And (50^51)^52 = 50^(10k+2) => Remainder = 5^2 => 3

but howcome d ans given is 6,,,check wid anodr one...32^33^34%11=???
and d answer given is 10!!!!!how???

4105

ya...i got it,,u did a calc mistake ...50%11 is 6..so 6^1%11=6..n so is 32^33^34%11=10.... :)

BUT AGAIN!!!!!
30^72^87%11=??...n d ans is given 5???bhai koi help karo!!!!

4106
ya...i got it,,u did a calc mistake ...50%11 is 6..so 6^1%11=6..n so is 32^33^34%11=10.... :)

BUT AGAIN!!!!!
30^72^87%11=??...n d ans is given 5???bhai koi help karo!!!!



30^72^87%11 = 30^72^7%11... now, 72^7%10(euler num. of 11) = 8 => 30^8%11 = (-3)^8%11 = 6561%11 = 5.
4107
In four throws with a pair of dices, what is the chance of throwing a double twice?
a) 11/216
b) 25/216
c) 35/126
d) 45/216

ru sure its pair of dice i m getting some huge answer , but if we take a dice thn 25/216 is the answer
4108

the sample space is ... 6X6X6X6

and the chances of getting a double ie 11,22,33,44,55,66 is 6

and the other two dices will have anything other than the double so it is 5X5

so P=6X5X5/6X6X6X6

ie 25/216

4109
the sample space is ... 6X6X6X6

and the chances of getting a double ie 11,22,33,44,55,66 is 6

and the other two dices will have anything other than the double so it is 5X5

so P=6X5X5/6X6X6X6

ie 25/216

for a pair of dice sample space will be 6^8 rite ? here u have considred only 1 dice
4110

try this...

getting same values on two dices and differnt values on the others for one throw gives us

1/6*1/6*5/6/*5/6

for this to happen 2 times out of for we multiply it by 4C2.

this gives us 25/216

4111
the master Says
What if the question was (30^72^87) % 11?? Plz solve


30^72^87%11 = 30^72^7%11... now, 72^7%10(euler num. of 11) = 8 => 30^8%11 = (-3)^8%11 = 6561%11 = 5.
4112

Hi puys,
Can anyone solve this one from Profit and Loss LOD 3 Question 34 :

A newspaper vendor sells three kinds of periodicals-dailies,weeklies and monthlies.
The weeklies sell for Rs 12 at a profit of 20% , the monthlies sell at Rs 50 at a profit of 25% , while the dailies sell at Rs 3 at a profit of 50% . If there is a government restriction on the total number of peroidicals that one particular news vendor can sell , and Kalu a newspaper vendor has sufficient demand for all the three types of periodicals what should he do to maximise profits ?

a) Sell maximum weeklies
b) Sell maximum monthlies
c) Sell maximum dailies
d) Cannot be determined

Answer : b)

Without the restriction mentioned in the problem above,what should the newspaper vendor do to maximise his profits if his capital is limited ?

a) Sell maximum weeklies
b) Sell maximum monthlies
c) Sell maximum dailies
d) Cannot be determined

Answer : c)

4113

Originally Posted by mohdhuzaifa2010 View Post
LOD 1 ques 46 chapter 5
in an exam 80% students passed in physics. 70% in chemistry while 15% failled in both the subjects. if 325 students passed in boh the subjects. find the total number of students who appeared in the examination.

In this problem (and problems similar to these) I get confused in the language of the problem ....like in the above question when they say that "15% students failed in both the subjects" does it mean the total students failed or do they mean the students in common from the two subjects who failed ?? ..similarly for the statement "325 students passed in both the subjects"
While getting the solution , the approach puys have taken is to take 325 as the common students passed in both the subjects and 15% as the total students who failed in both the subjects ...even though they have been introduced in the same way in the problem ...can anyone please solve my doubt

Regards,

u r really confusing..................it means that those students which are not passed in any subjects,are 15%.
so.................
n(AUB)=n(A)+n(B)-n(A@B)
suppose,total nnumber of students are x,then...........
n(AUB)=0.8x+0.7x-325-------------------(1)
while n(AUB)--->those students who are passed in either physics or chemistry

now,
it is given that,15% students are failed
x-n(AUB)=0.15x------------------(2)
now calculate and u will get 500 is answer correct ??I think it should be correct 100% surity.............



Sorry for replying late , but I am still not clear ....
If 80% of the students pass in Physics doesn't it mean that the remaining 20% students fail in Physics...similarly if 70% students pass in chemistry...i guess it means that the remaining 30% students have failed in chemistry ...by consdering this and also the fact that 15% failed in both the subjects ...we get a total of 35% students failing ..if you find my LOGIC sane enough please try to solve it .
4114
Originally Posted by mohdhuzaifa2010 View Post
LOD 1 ques 46 chapter 5
in an exam 80% students passed in physics. 70% in chemistry while 15% failled in both the subjects. if 325 students passed in boh the subjects. find the total number of students who appeared in the examination.






Sorry for replying late , but I am still not clear ....
If 80% of the students pass in Physics doesn't it mean that the remaining 20% students fail in Physics...similarly if 70% students pass in chemistry...i guess it means that the remaining 30% students have failed in chemistry ...by consdering this and also the fact that 15% failed in both the subjects ...we get a total of 35% students failing ..if you find my LOGIC sane enough please try to solve it .


Yup you are correct 35% people have failed in atleast one of the subject. So, the number of people who passed in both the subjects = 65% of total = 325

Solving gives the total no. of students as 500.

Hope this helps.
4115

Please help me with this one :
A newspaper costs Rs 11 to print on a daily basis . Its sale price (printed) is Rs 3 . The newspaper gives a sales incentive of 40% on the printed price , to the newspaper vendors . The newspaper makes up for the loss through advertisements , which are charged on the basis of per column centimetre rates . The advertisement rates of the newspaper are Rs 300 per cc (column centimetre) . It has to give an incentive of 15% on the advertisement bill to the advertising agency . If the newspaper has a circulation of 12000 copies, what is the approximate minimum advertising booking required if the newspaper has a break-even on a particular day (Assume there is no wastage)

a) 300 cc b) 350 cc c) 435cc d)450 cc

For the above question , if it is known that the newspaper house is unable to recover 20% of it's dues , what would be the approximate advertising booking target on a particular day in order to ensure the break even point?
a) 375 cc b) 438 cc c) 544 cc d) 562.5 cc

4116

Hi puys,

Can someone please provide me a link to ArunSharma's Quant/DI ebook..??

4117
Please help me with this one :
A newspaper costs Rs 11 to print on a daily basis . Its sale price (printed) is Rs 3 . The newspaper gives a sales incentive of 40% on the printed price , to the newspaper vendors . The newspaper makes up for the loss through advertisements , which are charged on the basis of per column centimetre rates . The advertisement rates of the newspaper are Rs 300 per cc (column centimetre) . It has to give an incentive of 15% on the advertisement bill to the advertising agency . If the newspaper has a circulation of 12000 copies, what is the approximate minimum advertising booking required if the newspaper has a break-even on a particular day (Assume there is no wastage)

a) 300 cc b) 350 cc c) 435cc d)450 cc

For the above question , if it is known that the newspaper house is unable to recover 20% of it's dues , what would be the approximate advertising booking target on a particular day in order to ensure the break even point?
a) 375 cc b) 438 cc c) 544 cc d) 562.5 cc


Ans1) 435
Ans2) 375

What's ur doubt bro...!!
4118
Ganu02 Says
it's 495...12C4...

by askin 2 select 4 socks i think he means 2 pairs of socks=>each pair shud have socks of same color
4119

Originally Posted by 2ndApr2011 View Post
Please help me with this one :
A newspaper costs Rs 11 to print on a daily basis . Its sale price (printed) is Rs 3 . The newspaper gives a sales incentive of 40% on the printed price , to the newspaper vendors . The newspaper makes up for the loss through advertisements , which are charged on the basis of per column centimetre rates . The advertisement rates of the newspaper are Rs 300 per cc (column centimetre) . It has to give an incentive of 15% on the advertisement bill to the advertising agency . If the newspaper has a circulation of 12000 copies, what is the approximate minimum advertising booking required if the newspaper has a break-even on a particular day (Assume there is no wastage)
a) 300 cc b) 350 cc c) 435cc d)450 cc
For the above question , if it is known that the newspaper house is unable to recover 20% of it's dues , what would be the approximate advertising booking target on a particular day in order to ensure the break even point?
a) 375 cc b) 438 cc c) 544 cc d) 562.5 cc
Ans1) 435
Ans2) 375
What's ur doubt bro...!!
Yes you have got the first one right but not the second one ...then answer of the secnd one is option c however can you tell me how you got the first one ...
i mean the detailed steps...here the CP is Rs 11 the price of the paper
and the SP is Rs 3 ,,and the fixed cost is (30% of 3) 1.2 * 12000 ..i cannot proceed after this ...

4120
Originally Posted by 2ndApr2011 View Post
Please help me with this one :
A newspaper costs Rs 11 to print on a daily basis . Its sale price (printed) is Rs 3 . The newspaper gives a sales incentive of 40% on the printed price , to the newspaper vendors . The newspaper makes up for the loss through advertisements , which are charged on the basis of per column centimetre rates . The advertisement rates of the newspaper are Rs 300 per cc (column centimetre) . It has to give an incentive of 15% on the advertisement bill to the advertising agency . If the newspaper has a circulation of 12000 copies, what is the approximate minimum advertising booking required if the newspaper has a break-even on a particular day (Assume there is no wastage)
a) 300 cc b) 350 cc c) 435cc d)450 cc
For the above question , if it is known that the newspaper house is unable to recover 20% of it's dues(means it recovered 80% of the cost) , what would be the approximate advertising booking target on a particular day in order to ensure the break even point?
a) 375 cc b) 438 cc c) 544 cc d) 562.5 cc
Ans1) 435
Ans2) 375
What's ur doubt bro...!!
Yes you have got the first one right but not the second one ...then answer of the secnd one is option c however can you tell me how you got the first one ...
i mean the detailed steps...here the CP is Rs 11 the price of the paper
and the SP is Rs 3 ,,and the fixed cost is (30% of 3) 1.2 * 12000 ..i cannot proceed after this ...


Ans1) CP= (11-3+1.2)*12000= 9.2*12000=110400
earning from adv'= 255 per adv'
min no. of adv' needed to break even= 110400/255= this value should be equal to nearest integer value >= quotient.

Ans 2) total earning= 110400*(4/5)=88320
as the newspaper house recovered only 80% of the cost.
min bookings needed= 88320/255= ans.

short trick: this value should be less than Ans1 as we need to achieve only 80% booking target. only 1 option.

Now I have two questions.First, how can 80% target be greater than 100%..?? either u missed to mention something abt question or the answer is wrong. The value would actually be 348 & I marked the nearest answer.
Second, Can u help me get an ebook of Arun Sharma Quant/DI..!!!