Quant by Arun Sharma

MBA CAT 2024
2298
There is a 0 in 30 and a 5 and an even number.So,last 2 digits have to be 00

option (1)

Some literature on Euler Theorem here:

Euler's theorem - Wikipedia, the free encyclopedia

Basically if we have to find the remainder of

a^n mod b

where a and b are coprime,and b's Euler number is say x.

Now,if n is a multiple of x then the remainder is 1.

Difficult?Lets see an example

Find the remainder of 29^72 mod 11

Now,Euler number of any integer n=A^x*B^y*C^z where A,B,C are the prime factors of n is given by:

n(1-1/A)(1-1/B)(1-1/C)

For prime numbers,Euler number is equal to (n-1)

So,Euler number of 11 is 10.

So,as 29 and 11 are co prime.If the numerator is of the format 29^10x,then the remainder will be 1.

So, 29^70 mod 11 is 1

So,it effectively becomes 29^2 mod 11

841 mod 11

the answer is 5.

Now,there are some variations to these kind of sums,you can find them on the net.

EDIT: Coprime numbers are those which do not share any common factors except 1.



Hi...can you please tell me about the fermat,chinese remainder theorems?
2299

how many chapters have all of you covered???

2300
Hi guys I am not being able to solve the following questions. These are from LOD2.
17. According to the 2001 census, populations growth rate of lucknow is going to be an increasing AP with 1st year's rate as 5% and common difference as 5%, but simultaneously the migration rate is an increasing GP with the first term as 1% and common ratio of 2. If the population on 31st December 2000 is 1million, then find which year will lucknow witness its first dall in population.


20. A sum of Rs. 8000 is borrowed at 5% p.a. compound interest and paid back in three equal installments. What is the amount of each installment?


22. A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly.

Please guys I know you all must be solving these very easily. But I am not requesting for an answer, that I have in the book itself.I am really wanting to know how to solve this. So please explain.
2301
Hi guys I am not being able to solve the following questions. These are from LOD2.
17. According to the 2001 census, populations growth rate of lucknow is going to be an increasing AP with 1st year's rate as 5% and common difference as 5%, but simultaneously the migration rate is an increasing GP with the first term as 1% and common ratio of 2. If the population on 31st December 2000 is 1million, then find which year will lucknow witness its first dall in population.


20. A sum of Rs. 8000 is borrowed at 5% p.a. compound interest and paid back in three equal installments. What is the amount of each installment?


22. A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly.

Please guys I know you all must be solving these very easily. But I am not requesting for an answer, that I have in the book itself.I am really wanting to know how to solve this. So please explain.

hey

1. u can have the AP as 5,10,15,20,25,30,35 etc
second gp is 1,2,4,8,16,32,64

so much is data available. next u need to know how to spot the method of solving. they have given beginning DEC 2000.

next u use the data:
so from the series u know that in 2001 the popln increases by 5% and decreases by 1%. this implies a net increase. so check like that for every year till the net becomes a decrease.

2. The concept is that each annual equal installment has a present value that is different. and the sum of their present values has to be 8000.
check out the basics expln in percentages

3. this one confused me too initially but then finally i decided that it was too much data 😃 jus take out the first part of the data. SI at 6% for a year calculated every six months. end value 4704.
that simplifies it. if anyone got a better soln please let me know too 😃
2302

Hi Friends,

I am facing problems in few questions in Quadratic equations and Inequality. I am mentioning questions below:

1.If both the roots of the quadratic equation ax^2 +bx +c =0 lies int the interval(0,3) then a lies in
Options:
a (1,3)
b(-1,-3)
c(-sqrt(121)/91,-sqrt()
d(-1,3)
e None of these

2.For what values of p would the equation x^2 +2(p-1)x +p+5 =0 posses at least one positive root?
options
a 10
b 7
c 6
d 12
e none of these

3 If one root of the equation (I-M)x^2 +Ix +1 =0 is double of the other and is real,find the greatest value of m
options:
a 9/8
b 8/7
c 8/6
d 7/5
e 5/7

Please suggest me approach of solving these type of questions

Thanks in advance,

2303

i m havng dbts from chapter 1 thats is number system
lod 1
quest no 11 ,13,55,68,70,71,82,and in quest no 94 i m getng answer b but the ans is given c.
pls guys guv me solutions of ths questions .

2304
i m havng dbts from chapter 1 thats is number system
lod 1
quest no 11 ,13,55,68,70,71,82,and in quest no 94 i m getng answer b but the ans is given c.
pls guys guv me solutions of ths questions .

Please post the questions here too so that puys who don't have the book could attempt at solving these.
2305

Hi all,
The following is a question that I came across in the LOD-2 block of Number systems.
Find the 28383rd term in the number 12345678910111213...
By the method i used, i found the answer to be 3. However, the answer that was provided in the key was 9. Can anyone of you please explain how this could be 9?
Here is the method I had used to arrive to the solution as 3.
Since the numbers start from 1 and continue, i tried finding the n digit number, which would have the 28383rd term. so 9*1 + 90*2+900*3 = 2889 terms would be occupied by 1,2 and 3 digit numbers. since 9000*4 = 36000 which i greater than 28383, i councluded that the term will be in a 4 digit number. I subtracted 2889 from 28383 to get the number of terms that will be occupied by 4 digit number, ie. 25494. each 4 digit number starting from 1000 will occupy 4 terms in the series, Hence 25494/4 => Q=6373 R = 2 => 2nd term in the 6374th 4-digit number will be the 28383rd term in the series.
Thus 999+6374 = 7373, that leaves the solution of 3.
Please let me know if my arrival to the solution is the correct way, if not kindly clarify on the same.

2306
Hi all,
The following is a question that I came across in the LOD-2 block of Number systems.
Find the 28383rd term in the number 12345678910111213...
By the method i used, i found the answer to be 3. However, the answer that was provided in the key was 9. Can anyone of you please explain how this could be 9?

The answer is indeed 3. There are many wrong answers in Arun Sharma.
2307

hey pls explain the above question one more time.
my dbts are
1.if 2a. 6 b. 7 c. 8 d.4

2.amitesh buys a pen ,a pencil and an eraser for rs 41.if the lest cost of any of the three items is 12 and it is knwn that a pen cost less than a pencil and an eraser cost more than a pencil .if it is knwn that the eraser's cost i snot divisible by 4 the cost of pencil would b
a. 12 b. 13 c. 14 d.15:)

2308
hey pls explain the above question one more time.
my dbts are
1.if 2a. 6 b. 7 c. 8 d.4

2.amitesh buys a pen ,a pencil and an eraser for rs 41.if the lest cost of any of the three items is 12 and it is knwn that a pen cost less than a pencil and an eraser cost more than a pencil .if it is knwn that the eraser's cost i snot divisible by 4 the cost of pencil would b
a. 12 b. 13 c. 14 d.15:)


better if u post the questions in official quant thread...that thread is really active
2309

for the 2nd one is it 14 ?
if its correct i'll post solution

2310

summation of n^2=

2311
hey pls explain the above question one more time.
my dbts are
1.if 2a. 6 b. 7 c. 8 d.4

2.amitesh buys a pen ,a pencil and an eraser for rs 41.if the lest cost of any of the three items is 12 and it is knwn that a pen cost less than a pencil and an eraser cost more than a pencil .if it is knwn that the eraser's cost i snot divisible by 4 the cost of pencil would b
a. 12 b. 13 c. 14 d.15:)

1. 2
Upper limit of x+y = 3.99 + 2.99 = 6.98 (almost 7)
Lower limit of x-y = 3.99 - 2.99 = 1

Thus, ratio is 7, option (b)

2. Pen (P) + Pencil (L) + Eraser (E) = 41

P
2312

hi all,
This is my first post in this thread.. somebody solve this...


In a bag there are exactly 6 one rupee coin and at least one five rupee coins and one ten rupee coins, if the number of ten rupee coins were changed to the number of five rupee coins, the numbers five rupee coins were changed to the number of one rupee coin, and the number of one rupee coin were changed to the number of ten rupee coins, the sum would remain unchanged. Find the largest possible amount of money the bags contains.

a) Rs.163 b) Rs.156 c) Rs.183 d) Rs.83 e) cant be determined

2313
hi all,
This is my first post in this thread.. somebody solve this...


In a bag there are exactly 6 one rupee coin and at least one five rupee coins and one ten rupee coins, if the number of ten rupee coins were changed to the number of five rupee coins, the numbers five rupee coins were changed to the number of one rupee coin, and the number of one rupee coin were changed to the number of ten rupee coins, the sum would remain unchanged. Find the largest possible amount of money the bags contains.

a) Rs.163 b) Rs.156 c) Rs.183 d) Rs.83 e) cant be determined


Let there be x 5 Rs coins and y 10 Rs coins

10x+6*5+y=6+5x+10y

9y-5x=24

(y,x)=(6,6)(11,15)(16,24)....

As you can see,there is no definite answer for this.So,option 5
2314

Thanks shashank...:)

2315

From Arun Sharma Quant's:
Refer Attached Fig:

Somebody help me to get solution for this problem?
From the Fig: if AB=BC=CD=DE=EF=FG=GA. Find the angle DAE.

Answer Choice are :

1)15
2) 25
3) 30
4) 20.

2316
Answer should be a) 3.

We need to count the no. of digits:
1-9 = 9
10-99 = 90*2 = 180
100-999 = 900*3 = 2700
1000-9999 = 9000*4 = 36000

Thus, the number would be a 4 digit no. No. of digits till 999 is 2889. We need the 28383rd digit, i.e. we need to find the 25494th digit starting from 1000 (28383-2889 = 25494).

Since, we know it'll be a 4 digit number, we can find the last number by dividing 25492/4 (not 25494, as we're trying to find the number now. Once we find the number, we can find the next two digits to get to the 25494th digit).

25492/4 = 6373rd number starting from 1000. It means the number is 1000+6373-1 = 7372. Till 7372, we have got 25492+2889 = 28381 digits. Next number will be 7373, we're interested in only the 2nd number as then we'll reach the 28383rd digit, and this number is 3.

Hope you understood :)


cud u please explain why 1 has been subtracted from the sum of 1000 and 6373???
2317
guneet.makkar1 Says
cud u please explain why 1 has been subtracted from the sum of 1000 and 6373???

This is because we need to find the 6373rd number from 1000.

The 1st number from 1000 is 1000 itself, which is 1000+1-1
50th number = 1000+50-1 = 1049

Thus, 1000+6373-1 = 7372

Hope you've understood now.