Quant by Arun Sharma

MBA CAT 2022
630
(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.


i got it now...thanks..even i was following the same process but was not able to follow what was written..even i got d answer as -1,3..maybe d answer given in d book is wrong..
631
the game!! Says
i got it now...thanks..even i was following the same process but was not able to follow what was written..even i got d answer as -1,3..maybe d answer given in d book is wrong..



dont know but i found many mistakes ingiven ans in AS.i have done 2 chepters complete and i think i found more thn 6 ans worng 😐
632

Hi everyone,
I am stuck at Q 29 of LOD1 of P & C.
The question statement is
Q1) How many ways are there to choose 4 cards different suit and different values from a deck of 52 cards
a) 13.12.11.10 b) 52.36.22.10

I chose option b but the answer is a. If a is the correct answer can anyone please explain it to me?

-Rahul

633


Hi everyone,
I am stuck at Q 29 of LOD1 of P & C.
The question statement is
Q1) How many ways are there to choose 4 cards different suit and different values from a deck of 52 cards
a) 13.12.11.10 b) 52.36.22.10

I chose option b but the answer is a. If a is the correct answer can anyone please explain it to me?



ill try

see out of the 4 different suits

Card 1 can be selected in only 13 C 1 = 13 ways
Now we have to leave THIS particular suite and from the remaining three suits we have to pick up one suit but now that we have already chosen a card from suit 1 and we know that same card cannot be picked up from suit 2 , we are left with only 12 cards in suit two

so Card 2 can be picked up in 12 ways , ditto fr next two suits

Thus 13.12.11.10

hope im clear
634
ill try

see out of the 4 different suits

Card 1 can be selected in only 13 C 1 = 13 ways
Now we have to leave THIS particular suite and from the remaining three suits we have to pick up one suit but now that we have already chosen a card from suit 1 and we know that same card cannot be picked up from suit 2 , we are left with only 12 cards in suit two

so Card 2 can be picked up in 12 ways , ditto fr next two suits

Thus 13.12.11.10

hope im clear


Hi Fuzon, How cone first card be selected in 13 C1 ways? there are 52 cards and hence 1st card can be picked in 52 ways, isn't it?
635

I am finding difficulty in questions related to installments in CI or SI
'Question 20 of LOD2 od sinple interest and CI chapter

Question 20 A sum of Rs 8000 is borrowed at 5% P.a. compound interest and paid back in 3 equal annual installmemts.What is the amount of each statement.

Question 26 An amount of Rs 12820 due 3 yrs hence,is fully repaid in 3 annual installments starting after 1 year .The first installment is 1/2 the second instllment and the second installment is 2/3 of the third installment.If the rate of interst is 10% per annum ,find the first installment.

636

Is the answer (a) -> 2

Applying the Remainder Theorm, 3^20 would leave a remainder of 1 when divided by 50.
3^32 would leave the remainder of -9 ,i.e, 41 when divided by 51.

41/50 = 0.82, thus the value of x is 2.

Cheers..

this problem is from the number system chapter, LOD2, question no. 20

gives remainder and {.} denotes the fractional part of that. the fraction part is of the form (0.bx). the value of x could be

(a) 2, (b) 4, (c) 6, (d) 8, (e) none of these.

May be its an easy one, but i couldn't understand the question.:
plz answer and explain it.
637

Hi Puys
I saw dis problem in arun sharma can someone help me out in solving dis
find the number of zeros in 1^1*2^2*3^3*4^4*.............*98^98*99^99*100^100
the answer is 1300 .

638

I saw dis problem in arun sharma can someone help me out in solving dis
find the number of zeros in 1^1*2^2*3^3*4^4*.............*98^98*99^99*100^100
the answer is 1300 .


okay one more try from my side

see the only people responsible for a zero are 5 and 2

now we have plenty of 2's in such a expression as compared to 5's for obvious reasons

so just check how many 5's are there in the expression , and that shall suffice your need

now the first 5 you will encounter is at 5^5 , next at 10^10 , 15^15 etc etc

thus this addition of 5s will be (5+10+15+......100)

but are we missing something ? ... yes .... the fact that 25 will have Two 5's and so will 50 , 75 and 100

25 = 5*5
50 = 5*5*2
75=5*5*3
100= 5*5*4

thus we have to add these extra fives also which are (25+50+75+100)

thus the total expression is addition of both the fives which is
(5+10+15+20+.....+100) +(25+50+75+100) = 1300

hope i am lucidly clear 😃
639

Hai Puys
I found this problem in LOD 3 of arun sharma......... chapter- Numbers.
What is the remainder of 32^32^32 divided by 9 ......
answer provided is 4 .

another problem stated the remiander of 32^32^32 divided by 7 ..in this case also the answer is 4.

i didnt understand the solution that they provided .can someone help me out???:idea:

640
Hai Puys
I found this problem in LOD 3 of arun sharma......... chapter- Numbers.
What is the remainder of 32^32^32 divided by 9 ......
answer provided is 4 .

another problem stated the remiander of 32^32^32 divided by 7 ..in this case also the answer is 4.

i didnt understand the solution that they provided .can someone help me out???:idea:


see this
this should help you

1)remainder when (32^32^32)/7
= {32}^32^32 =
start from top exponent....

32 mod 7 = 4 mod 7 = 4
32^2 mod 7 = 4^2 mod 7 = 16 mod 7 = 2
32^4 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
32^8 mod 7 = 4^2 mod 7 = 16 mod 7 = 2
32^16 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
32^32 mod 7 = 4^2 mod 7 = 16 mod 7 = 2

so , 2
(32^32)^2 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
(32^32)^4 mod 7 = 4^2 mod 7 =16 mod 7 = 2
(32^32)^8 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
(32^32)^16 mod 7 = 4^2 mod 7 =16 mod 7 = 2
(32^32)^32 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

so ans is : 4


2)remainder when (32^32^32)/9
= {32}^32^32 =
start from top exponent....

32 mod 9 = 5 mod 9 = 5
32^2 mod 9 = 5^2 mod 9 = 25 mod 9 = 7
32^4 mod 9 = 7^2 mod 9 = 49 mod 9 = 4
32^8 mod 9 = 4^2 mod 9 = 16 mod 9 = 7
32^16 mod 9 = 7^2 mod 9 =49 mod 9 = 4
32^32 mod 9 = 4^2 mod 9 = 16 mod 9 = 7

so , 7
(32^32)^2 mod 9 = 7^2 mod 9 = 49 mod 9 = 4
(32^32)^4 mod 9 = 4^2 mod 9 = 16 mod 9 = 7
(32^32)^8 mod 9 = 7^2 mod 9 = 49 mod 9 = 4
(32^32)^16 mod 9 = 4^2 mod 9 = 16 mod 9 = 7
(32^32)^32 mod 9 = 7^2 mod 9 = 49 mod 9 = 4

so ans is : 4
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this question is from ARUN SHARMA, NUMBER SYSTEM chapter, LOD-2

a set 'S' is formed by including some of the first One thousand natural numbers. 'S' contains the maximum number of numbers such that they satisfy the following conditions:
1. no numbers in the set 'S' is prime
2. when the numbers of the set 'S' are selected two at a time, we always see co prime numbers
what is the number of elements in the set 'S'?

. 11, . 12, . 13, . 7


plz explain the process.


fst u have to select those number which r nt prime....
say 1,4,8,9,10......
now see if we select 1 and 4 at a time..these r co-prime to each other ..but if we take 4 and 8 at a time...these r not co-prime...now take 4 and 9 at a time..both are co-prime to each other...again 8 and 10 or 4 and 10 are not co-prime....but if u take 4 and 25 both are co prime...25 and 49 are co-prime..and so on.
so we can say tht here co-prime nos are squares of prime numbers.....1,4,9,25,49,121 169 289 361 529 841 961 which are not prime numbers but if we take any 2 together...are co-prime to each other..hence ans is (2)12
642
here is another problem from arun sharma, number system

arun, bikas and chetakar have a total of 80 coins among them. arun triples the number of coins with the others by giving them some coins from his own collection. next, bikas repeats the same process. after this bikas has 20 coins. find the number of coins he had at the beginning:

. 11, . 10, . 9, . 12


c suppose arun is having A coins,bikas B and chetakar C
so A+B+C=80.......(1)
now after fst transaction
arun would have A-2B-2c
bikas 3B
chetkar 3C
now after second transaction
arun 3(A-2B-2C)
bikas 3B-2(A-2B-2C)-6C
chetkar 9c

now it is given in qs tht after 2nd transaction bikas has 20 coins
so 3B-2(A-2B-2C)-6C=20
=>3B-2A+4B+4C-6C=20
=>7B-2A-2C=20........(2)

multiply eq(1) by 2
we will get
2A+2B+2C=160
2A+2C=160-2B------(3)

by eq 2 and 3 we will get
7B-160+2B=20
9B=180
B=20

i think given ans is wrong 😐
643

here is another problem from arun sharma, number system

arun, bikas and chetakar have a total of 80 coins among them. arun triples the number of coins with the others by giving them some coins from his own collection. next, bikas repeats the same process. after this bikas has 20 coins. find the number of coins he had at the beginning:

. 11, . 10, . 9, . 12


a , b and c be the inititals of the guys above

Now initial case : a+b+c = 80

After a gives some x coins the status is

(a-x) + 3b + 3c = 80

And After b does the same thing , viz : gives back y coins

3(a-x) + (3b -y) + 9c = 80

3(a-x) + 20 + 9c = 80 (Given)

3(a-x) + 9c = 60 => (a-x) + 3c = 20

put this value in

(a-x) + 3b + 3c = 80 => 3b = 60 => b= 20 !!

Hey is the answer wrong or have i goofed up again ?
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fuzon seems ans is worng

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Hi Puys
This is a problem from Arun Sharma -Averages.can someone help me out with this?
A team of miners planned to mine 1800 tons of ore during a certain number of days,Dut to technical difficulties in one third of the planned no of days the team was able to make 20 tons of ore less then the planned output.to make up for this they over achieved for ther rest of the days by 20 tons.The end result was that the team completed the task one day ahead of time .How many tons of ore did the team initially plan to ore per day ?
1)50 2)100 3)150 4)200
The answer provided is 100.

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can anyone have ebook of quant by arun sharma

647
Hi Puys
This is a problem from Arun Sharma -Averages.can someone help me out with this?
A team of miners planned to mine 1800 tons of ore during a certain number of days,Dut to technical difficulties in one third of the planned no of days the team was able to make 20 tons of ore less then the planned output.to make up for this they over achieved for ther rest of the days by 20 tons.The end result was that the team completed the task one day ahead of time .How many tons of ore did the team initially plan to ore per day ?
1)50 2)100 3)150 4)200
The answer provided is 100.


yeah i too have same prb..can anyone post ans in detail?
648
nidhisoni24 Says
yeah i too have same prb..can anyone post ans in detail?
thanks a lot
649

Assume that they plan to complete the work in n days.

So the amount of ore that they plan to mine on a single day is 1800/n tonnes
but on n/3 number of days they mine only ( 1800/n - 20) tonnes
So the amount of ore mined in n/3 days = ( 1800/n - 20) * n/3 ---(1)

By completing it one day before the scheduled time they complete the remaining it in (2n/3 -1) days.
i.e. for 2n/3 -1 days they mine ( 1800/n + 20 ) tonnes each day
So the amount of ore mined in 2n/3 -1 days = ( 1800/n + 20 ) * (2n/3 -1) ------(2)

We know the total amount of ore mined is 1800
Adding (1) and (2)
So ( 1800/n - 20) * n/3 + ( 1800/n + 20 ) *(2n/3 -1) = 1800

600 + 20n /3 + 1200 - 1800/n + 40n/3 -20 = 1800

20n/3 - 1800/n -20 = 0

n^2 - 270 - 3n = 0

this quadratic equation has two roots 18 and -15. So number of days is 18 therefore amount of ore mined will be 100 tonnes