I will give minimalistic answers....the way I think one should think in the exam....hope it helps.

@vivaciousgal said: hey puys....can anyone help me in solving these questions...1. A number is mistakenly divided by 5 instead of being multiplied by 5. Find the percentage change in the result due to mistake.?(a) 96% (b) 95% (c) 2400% (d) 200% (e) none

2. In a mixture of 80 litres of milk and water, 25% of the mixture is milk.How much water should be added to the mixture so that milk becomes 20% of the mixture.?(a) 20 litres (b) 15 litres (c) 25 litres (d) 24 litres (e) none

3. At an election, the candidate who got 56% of the votes cast won by 144 votes. Find the total number of the voters on the voting list if 80% people cast their vote and there were no invalid votes.?(a) 360 (b) 720 (c) 1800 (d) 1500 (e) 1600

4. Ram spends 20% of his monthly income on his house hold expenditure, 15% of the rest on books, 30% of the rest on clothes and saves the rest. On counting, he comes to know that he has finally saved Rs. 9520.Find his monthly income.?(a) 10000 (b) 15000 (c) 20000 (d) 12000 (e) none

5. In an examination, Mohit obtained 20% more than Sushant but 10% less than Rajesh. If the marks obtained by Sushant is 1080,find the percentage marks obtained by Rajesh if the full marks is 2000.? (a) 86.66% (b) 72% (c) 78.33% (d) 77.77% (e) none

profit earned by selling an article for rs 1060 is 20% more than the loss incurred by selling the article for rs 950 at what price should the article be sold to earn 20%%profit

what was the percentage discount given ?
a) 23.5% profit was earned by selling an almirah for Rs.12350.
b)if there were no discount , the earned profit would have been 30%
c)the cost price of almirah is 10000 rs
option
1)only a and b is required for answering the question
2)only b and c
3) only a and c
4) any two of three
5) none of these

what was the percentage discount given ?
a) 23.5% profit was earned by selling an almirah for Rs.12350.
b)if there were no discount , the earned profit would have been 30%
c)the cost price of almirah is 10000 rs
option
1)only a and b is required for answering the question
2)only b and c
3) only a and c
4) any two of three
5) none of these

WHAT WAS THE COST PRICE OF WATCH ?
1) THE SHOPKEEPER LABELLED THE PRICE OF THE WATCH 20% ABOVE THE COST PRICE
2) AFTER ALLOWING A DISCOUNT OF 15 % ON THE LABELLED PRICE , THE SHOPKEEPER CHARSES RUPEES 408 FOR THE WATCH
3)HAD THERE BEEN NO DISCOUNT , THE SHOPKEEPER WOULD HAVE EARNED 20% PROFIT .
OPTINS :-
A) 1 AND EITHER 2 OR 3 IS REQUIRED FOR ANSWER
B) 2 AND EITHER 1OR 3
C) 3 AND EITHER 1 OR 2
D) 1 AND 2 ONLY
E) ANY TWO OF THE THREE


question from R S AGGRAWAL and pls provide solution in detail

@ScatterBrain16 said: A train encounters 38 intermediate stations between two cities. In how many ways can the train stop at exactly 3 intermediate stations such that no two of them are consecutive?1) 65452) 59843) 71404) 8436

@techsurge said: A person borrowed a certain sum at the rate of 20% p.a compound interest. He paid a total of Rs 4000 in two annual equal installments. what is the interest paid ?a - 955b - 944c - 933d - 922could anyone tell me what's wrong my approachlet loan be P and compounded annually for 2 years on 20 % meansA= P(1.2)^2 =1.44 PNow let installment be equal to xx=4000interest on 1st installment =0.2x => amount =1.2x= 4800no interest on 2nd installment so amount =4000total =88008800 =1.44 PP = 6111.11interest =8000-6111.11 = 1888.888where am i wrong

hey puys....

No matter hw hard I study I get only 50-60%ile in mocks... Pls help wat to do?

The answer says its 36C3.But that is in the case of no consecutive stations as

3 stations can be considered as 1
38-3=35

35+1(3 taken together) = 36

36C3

How is it the answer in case of no consecutive stations??

A train encounters 38 intermediate stations between two cities. In how many ways can the train stop at exactly 3 intermediate stations such that no two of them are consecutive?

1) 6545
2) 5984
3) 7140
4) 8436

SI / CI :

1) what annual installment will discharge a debt of Rs 1815 due in 3 years at 10% simple interest ?

a - 500
b - 520
c - 550
d - 580


4) What annual installment will discharge a debt Rs 717.60 due in 4 years at 20% p.a simple interest, if installments are paid at the end of each year?

a - 128
b - 134
c - 138
d - 144

Kindly do post the explanations. thanks puys

soln....

1. there will be three installments.....

on 1st installment he will get 2 years interest
on 2nd installment he will get 1 year interest
on 3rd installment he will get no interest as it would be final installment

let each installment be x
interest on 1st installment for the vendor = 0.2x=> amount =1.2 x
interest on 2nd installment for the vendor = 0.1x=> amount =1.1 x
3rd installment = x

so total money the vendor will make from 3 installments = 3.3 x =1815
x=550

4. there will be four installments.....

on 1st installment vendor will get 3 years interest
on 2nd installment vendor will get 2 year interest
on 3rd installment vendor will get 1 year interest
on 4th installment vendor will get no interest as it would be final installment

let each installment be x
interest on 1st installment for the vendor = 0.6x=> amount =1.6 x
interest on 2nd installment for the vendor = 0.4x=> amount =1.4 x
interest on 2nd installment for the vendor = 0.2x=> amount =1.2 x
4th installment = x

so total money the vendor will make from 3 installments = 5.2 x =717.6
x=138


3) A person borrowed a certain sum at the rate of 20% p.a compound interest. He paid a total of Rs 4000 in two annual equal installments. what is the interest paid ?

a - 955
b - 944
c - 933
d - 922

could anyone tell me what's wrong my approach
let loan be P and compounded annually for 2 years on 20 % means

A= P(1.2)^2 =1.44 P
Now let installment be equal to x
x=4000
interest on 1st installment =0.2x => amount =1.2x= 4800
no interest on 2nd installment so amount =4000
total =8800

8800 =1.44 P
P = 6111.11
interest =8000-6111.11 = 1888.888

where am i wrong

SI / CI :

1) what annual installment will discharge a debt of Rs 1815 due in 3 years at 10% simple interest ?

a - 500
b - 520
c - 550
d - 580



2) A person invested Rs 20,000 in a bank which is offering 10% per annum simple interest. After two years, he withdrew the money from the bank and deposited the total amount in another bank which gives an interest rate of r% p.a compounded annually. After 2 years, he received an amount of Rs 2460 more than what he had invested in that bank. Find r ?

a - 5
b - 10
c - 15
d - 12


3) A person borrowed a certain sum at the rate of 20% p.a compound interest. He paid a total of Rs 4000 in two annual equal installments. what is the interest paid ?

a - 955
b - 944
c - 933
d - 922


4) What annual installment will discharge a debt Rs 717.60 due in 4 years at 20% p.a simple interest, if installments are paid at the end of each year?

a - 128
b - 134
c - 138
d - 144

Kindly do post the explanations. thanks puys

Hi;
We get 2 Equations. as per the question solving which we get the req. ans.
(s-c)/c = 20/100 ............................(1)
(s-100)-(c-100)=(c-100)(104/100).....(2)

Solve simultaneous eq. to get the ans..


'Q2 a person sells his table at a profit of 25/2% and the chair at a loss of 25/3% but on the whole he gains rs 25 . on the other hand if he sells the table at a loss of 25/3% and the chair at a profit of 25/2% then he neither gain nor loss
find the cost price .'

Let CP be x,y respectively::again 2 simultaneous equations:-
x(1+25/200) + y(1-25/300) - (x+y) =25......(1)
x(1-25/300) + y(1+25/200) = (x+y)............(2)



Regards,
Never Back Down

Having Doubt...
Plz explain
In this
"
(s-c)/c = 20/100 ............................(1)
(s-100)-(c-100)=(c-100)(104/100).....(2)
"
Why its not 120/100 in eq (1) ???

And
How these two eq made ??
"
x(1+25/200) + y(1-25/300) - (x+y) =25......(1)
x(1-25/300) + y(1+25/200) = (x+y)............(2)
"
TIA

good work puys!!!!!!!!!

here are some fundas/concepts which could be useful..

PS: I found them in old QUANT threads in PG... hope it is useful..

Ton of thanks....

3) Let us assume that each gram costs 1 Rs .
So The price of 1 kg = 1000.
He says that the price is Increased by 20%.
Now the new price = 1200
But the Actual weight is 800gm. Its cost = 800
Profit = 1200 - 800 = 400
Profit % = 400/800 * 100 = 50%

4) Let quantity of milk be 100 Liters.
Let the cost of each liter be 1 Rs. So the total cost = 100.
He wants 25% gain which means 100 + 25% of 100 = 125
Now he must add 25 liters of water to the milk.
Percentage of water in the mixture is 25/125 * 100 = 20%

5) Profit % = S.P - C.P / C.P * 100
22 = (S.P - 200 / 200) * 100
44 = S.P - 200
S.P = 244

I had done "5" with "S.P. = (100 + profit%) C.P 100 "
and my answer is 266.
Plz correct me ??

freakonomist Says

Thanks for the reply,what would have been your approach if it was 3 people in 10 chairs with no adjacent sitting..

An alternate approach (seems logical, though I can't quite prove it correct):

Let me represent a person by a "1" and a chair by a "0". So first I will put in the sequence "1 0 1 0 1" which satisfies the condition that the people should be separated. Now the remaining 5 "0"s can be placed in any of the positions - before the first person, between any of the people, or beyond the last; in other words we can distribute the remaining 5 identical "0"s into 4 groups which can be done in (5+4-1)C(4-1) or 8C3 or 56 ways. In each case the 3 people can be arranged in 3! ways ie 56 x 6 = 336 ways.

If I try it with the earlier question i.e. the 5 people 10 chairs case, I start with "1 0 1 0 1 0 1 0 1" and say that the remaining chair can be put into any of the 6 groups formed by the 5 people in (1+6-1)C(6-1) or 6 ways (this could also have been counted manually, in this case!) and hence we get 6 x 5! = 720 ways. Of course, for this question, MadaboutIIMs method is far superior as it is much quicker and more elegant and easier to understand! :)

~scrabbler

Again there is no option but to count the cases individually.
Select 3 chairs out of 10 such that no two are consecutive. This is done by subtracting cases when 2 consecutive chairs are selected from total no. of ways of selecting 3 chairs out of 10, which is 10 C 3 = 120.

To count ways where atleast 2 are consecutive =
Nos. in brackets are serial nos. of chairs selected, and no. in front is the no. of ways of selecting those 2 chairs.
(1,2) = 8 ways, (1,2,3); (1,2,4) etc.
(2,3) = 7
(3,4) = 6
(4,5) = 5
(5,6) = 4
(6,7) = 3
(7, 8 ) = 2
(8,9) = 1

Total = 36

So, required ways = 120-36 = 84
And total ways of seating 3 people = 84X3! = 504

Hi crab365,

I am afraid this won't work. When you have counted the 36 cases, you have missed cases where the 2nd and 3rd chairs are consecutive. This adds another 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 cases such as (1, 3, 4), (1, 4, 5) (2, 4, 5)....till (7, 9, 10)

So we must do 120 - 36 - 28 = 56 cases x 3! = 336 ways.

I also have another way of trying this, but will describe that in a separate post as this one is already confused...that also gives 336 as an answer (and will work for the earlier problem too)

~scrabbler

--------------------------

-----------------------

sameer verma Says

what is the remainder when 18^323/325?

=>18^322*18
=>(324)^161*18 (taking 18^2=324)
=>(325-1)^161*18
=>(-1)^161*18
=>-1*18=-18

Remanider = 325-18=307

What is the answer?

what is the remainder when 18^323/325?

Somebody please solve this, Thanks.

How many even integers n, where 100<=n<=200, are divisible neither by seven nor by nine?

is it 39

total even numbers 51
numbers divisible by 7 -7
numbers divisible by 9- 6
and one number is common in both so total even numbers will be
51-(7+6-1)
39

For Qa please participate in this thread also to get your fundas clear...

http://www.pagalguy.com/forum/prep-resources/82497-how-crack-section-i-cat.html

ps- Number system week

Somebody please solve this, Thanks.

How many even integers n, where 100<=n<=200, are divisible neither by seven nor by nine?

_x_x_x_x_x_

out of 6 dash we need to select 5....so 6C5*5!

Junta please note that the solution quoted above is true, and mine was wrong.

I had missed out on 4 additional cases:
1,3,5,7,10
1,3,5,8,10
1,3,6,8,10
1,4,6,8,10
So total no. of ways = 6, and no. of seating arrangements = 6X5! = 720.

The above quoted text is obviously a better way to go about this problem.

Thanks to MadaboutIIMs for pointing out my bad

That will not work....the last assumption you made is itself wrong "any number ending with unit digit 5 divided by 4 will have remainder 1" - for example consider 15 divided by 4, leaves remainder 3! Units digit alone is not sufficient - this is an easy trap to fall into (speaking from experience here )

~scrabbler

thanks buddy ,that was really a silly mistake

freakonomist Says

Thanks for the reply,what would have been your approach if it was 3 people in 10 chairs with no adjacent sitting..

Again there is no option but to count the cases individually.
Select 3 chairs out of 10 such that no two are consecutive. This is done by subtracting cases when 2 consecutive chairs are selected from total no. of ways of selecting 3 chairs out of 10, which is 10 C 3 = 120.

To count ways where atleast 2 are consecutive =
Nos. in brackets are serial nos. of chairs selected, and no. in front is the no. of ways of selecting those 2 chairs.
(1,2) = 8 ways, (1,2,3); (1,2,4) etc.
(2,3) = 7
(3,4) = 6
(4,5) = 5
(5,6) = 4
(6,7) = 3
(7, = 2
(8,9) = 1

Total = 36

So, required ways = 120-36 = 84
And total ways of seating 3 people = 84X3! = 504

Hi Freakonomist
n+r-1 C r-1 won't work because that formula is used to distribute n identical things in r groups.

Thanks for the reply,what would have been your approach if it was 3 people in 10 chairs with no adjacent sitting..

Can any one post me the approach for this type of question:
How to make 5 people sit in 10 chairs so that no 2 people sit adjacent.Here n+r-1Cr-1 doesnt work.How to go about this? Please illustrate the approach.

_x_x_x_x_x_

out of 6 dash we need to select 5....so 6C5*5!

Hi guys,

Can anyone tell me about the weightage of quant and DI in paper one i.e how many questions out of 30 are of quant and how many out of 30 are of DI? Also, tell the distribution for paper II i.e how many questions out of 30 are of VA and how many out of 30 are of LR?

Let me know the approximate cut-off for both the sections as well.

Thanks - Sameer

Can any one post me the approach for this type of question:
How to make 5 people sit in 10 chairs so that no 2 people sit adjacent.Here n+r-1Cr-1 doesnt work.How to go about this? Please illustrate the approach.

Hi Freakonomist :)
n+r-1 C r-1 won't work because that formula is used to distribute n identical things in r groups.

To solve your question, number the chairs 1 through 10.
If the chairs are to be occupied in such a way that only alternate ones are occupied, there are two possibilities that the 5 people can be distributed in the 10 chairs:
1,3,5,7,9 are occupied OR
2,4,6,8,10 are occupied.
There is no other way of seating the people in different chairs such that no 2 people sit together.
In either case, no. of ways of seating 5 people = 5! = 120
So, total no. of ways = 240

Can any one post me the approach for this type of question:
How to make 5 people sit in 10 chairs so that no 2 people sit adjacent.Here n+r-1Cr-1 doesnt work.How to go about this? Please illustrate the approach.

profit and loss :

1) i purchased a dining table for rs 8000 two years ago. present price of a similar dining table is rs 15000. if i sell my table, i would get rs 10000 for it. if instead i polish it, i would have to spend rs 3000 on it but can sell it at the price of a new one. at what discount should i sell my dining table now if i expect the same percentage profit as i would have in the event of my selling it without polishing ?

a - 25/3 %
b - 10 %
c - 23/2 %
d - 77/5 %
2) i purchased items worth rs 1500. i sold some items and found out that i had made a profit of 20% which incidentally was equal to 5% of my total purchase. what is the woth of the items that i had sold ?

a- 225
b- 375
c- 675
d- 1125


please explain the approach too. thanks

1.Profit % in 1st case :
10000-8000=2000
2000/8000=25%

.Profit % in 2nd case :
15000-11000=4000
4000/11000=?
to make profit %=25
11000+25*11000=13750,But S.P=15000.so he has to give a discount of 1250 on 15000 to make the sp =13750
so disount%=1250/15000=8.33%

2. Let the CP of his purchase be worth x
1/5*x=75 Given
x=375

help pls..
Q1. Which numbers have exactly 3 factors?


thanks..

all perfect squares of primes have exactly 3 factors

remember all natural numbers have even number of factors except for perfect squares which have odd number of factors.

help pls..
Q1. Which numbers have exactly 3 factors?

Q2. x, a and b are natural numbers. If x is divisible by a and beach then find a condition for which x is definitely divisible by a*b?

Q3.If LCM of a, b, c is x and that of d, e is y then, would the LCM of x,y be the LCM of a, b, c, d, e.
Similarly for HCF what is the answer?

thanks..

profit and loss :

1) i purchased a dining table for rs 8000 two years ago. present price of a similar dining table is rs 15000. if i sell my table, i would get rs 10000 for it. if instead i polish it, i would have to spend rs 3000 on it but can sell it at the price of a new one. at what discount should i sell my dining table now if i expect the same percentage profit as i would have in the event of my selling it without polishing ?

a - 25/3 %
b - 10 %
c - 23/2 %
d - 77/5 %



2) i purchased items worth rs 1500. i sold some items and found out that i had made a profit of 20% which incidentally was equal to 5% of my total purchase. what is the woth of the items that i had sold ?

a- 225
b- 375
c- 675
d- 1125


please explain the approach too. thanks

4) What is the sum of the product of all possible pairs of distinct numbers of the first ten natural numbers.
a) 1320
b) 2640
c) 1705
d) 1920
1705 my take

Should be 1320

~scrabbler

i think it can also be done like this :

7^5^7^5=>
taking unit digit of powers from up to down=> consider 7^5 its unit digit will be 7 as cyclicity of 7 is 4.....

now unit digit of 5^(the number whose unit digit is 7) is 5...as we multiply 5 as many times we get unit digit 5

now unit digit of 7^(the number whose unit digit is 5)=> cyclicity of 7 is 4...and any number ending with unit digit 5 divided by 4 will have remainder 1...so unit digit of 7^(5^(7^5)) will be 7

That will not work....the last assumption you made is itself wrong "any number ending with unit digit 5 divided by 4 will have remainder 1" - for example consider 15 divided by 4, leaves remainder 3! Units digit alone is not sufficient - this is an easy trap to fall into (speaking from experience here )

~scrabbler

4) What is the sum of the product of all possible pairs of distinct numbers of the first ten natural numbers.
a) 1320
b) 2640
c) 1705
d) 1920
1705 my take

No it is not correct

4) What is the sum of the product of all possible pairs of distinct numbers of the first ten natural numbers.
a) 1320
b) 2640
c) 1705
d) 1920
1705 my take

It will be 7.

If we look at powers of 7 i.e. 7^1, 7^2, 7^3 etc, we find that the last digits follow the pattern 7, 9, 3, 1, 7, 9, 3, 1...repeating after every 4 steps - i.e. the cyclicity of 7 is 4 steps.

So we need to take the given power (in this case (5^(7^5))) and see what happens when this is divided by 4. Now if we observe carefully, this is of the form 5^x which is nothing but (4+1)^x, which on binomial expansion will give all terms divisible by 4 except the last term 1. In other words we can say that (5^(7^5)) is of the form 4k+1.

Hence 7^(4k+1) can be written as (7^4k) * (7^1) = (7^4)^k * (7^1) using laws of indices. But (7^4) ends in 1 and hence (7^4)^k will also end in 1, while (7^1)ends in 7 so that when they are multiplied the last digit will be 1 * 7 = 7.

Now comes the fun part, learning how to do all this orally!

~scrabbler

i think it can also be done like this :

7^5^7^5=>
taking unit digit of powers from up to down=> consider 7^5 its unit digit will be 7 as cyclicity of 7 is 4.....

now unit digit of 5^(the number whose unit digit is 7) is 5...as we multiply 5 as many times we get unit digit 5

now unit digit of 7^(the number whose unit digit is 5)=> cyclicity of 7 is 4...and any number ending with unit digit 5 divided by 4 will have remainder 1...so unit digit of 7^(5^(7^5)) will be 7

Progressions SET-6 :-

1) If a+b, b+c and c+a are in harmonic progression, then
a) a^2, b^2 and c^2 will be in AP
b) a^2, bc and c^2 will be in HP
c) 2ab, a^2 and b^2 will be in AP
d) b^2, a^2 and c^2 will be in AP

2) If 1/(1+x) , 1/(1-x), 4/(1-x^2) are in Arithmetic progression, then
a) x lies between 0 and 2
b) x lies between -2 and 0
c) x lies between 2 and 4
d) there is no possible values of x

3) If the sides of a right-angled triangle are in AP, then the minimum possible difference between the magnitude of the area and perimeter of such a triangle is..
a) 0
b) 6
c) 12
d) 4.5

4) What is the sum of the product of all possible pairs of distinct numbers of the first ten natural numbers.
a) 1320
b) 2640
c) 1705
d) 1920

5) If log 0.5 , log(2^x -1) and log(2^x+5) are in arithmetic progression, then x=
a) 3/2
b) log 3 base 2
c) log 2 base 3
d) - log 2 base 3

thanks.. dude

hey.. can anyone pls tell me what is the answer and solving method for the question:
unit's digit of 7^(5^(7^5))

thanks..

It will be 7.

If we look at powers of 7 i.e. 7^1, 7^2, 7^3 etc, we find that the last digits follow the pattern 7, 9, 3, 1, 7, 9, 3, 1...repeating after every 4 steps - i.e. the cyclicity of 7 is 4 steps.

So we need to take the given power (in this case (5^(7^5))) and see what happens when this is divided by 4. Now if we observe carefully, this is of the form 5^x which is nothing but (4+1)^x, which on binomial expansion will give all terms divisible by 4 except the last term 1. In other words we can say that (5^(7^5)) is of the form 4k+1.

Hence 7^(4k+1) can be written as (7^4k) * (7^1) = (7^4)^k * (7^1) using laws of indices. But (7^4) ends in 1 and hence (7^4)^k will also end in 1, while (7^1)ends in 7 so that when they are multiplied the last digit will be 1 * 7 = 7.

Now comes the fun part, learning how to do all this orally!

~scrabbler

hey.. can anyone pls tell me what is the answer and solving method for the question:
unit's digit of 7^(5^(7^5))

thanks..

The answer should be 7

hey.. can anyone pls tell me what is the answer and solving method for the question:
unit's digit of 7^(5^(7^5))

thanks..

My take 7...

hey.. can anyone pls tell me what is the answer and solving method for the question:
unit's digit of 7^(5^(7^5))

thanks..

hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100:shocked::shocked:

The best way to find the unit's digit is to find the remainder of a number when it's divided by 10. So, for any x!, x>=10, this will always be zero!!!

hi guys,
please let me know which topics you are discussing ..

Regards
kapil

hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100:shocked::shocked:

Units digit will be Zero..!! Because it is getting multiplied by numbers which have their units digit as zero...!!!

Progressions SET-5 :-

1) M and N are natural numbers such that by M = (5N 4) (5N + 1).
If 1N200,what is the harmonic mean of all the possible values of M?
a) 200^2/1001
b) 1001
c) 1001/200^2
d) 1/1001

2) A sequence 192, 360, 576. is formed by multiplying the corresponding terms of two different Arithmetic Progressions. What is the eighth term of the sequence?
(a) 2376 (b) 1040 (c) 2116 (d) None of these

3) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term?
(a) 16 (b) 22 (c) 25 (d) 26

1) M=(5N-4)(5N+1) , 1<= N <= 200
When N=1, M = 1*6
When N=2, M = 6*11
When N=200, M = 996*1001
Harmonic Mean = 200/(1/5+1/5+....+1/5)
= 200/(1/5) = 1001

2) The nth term of an AP can be written in the form of a linear equation an+b.
So the product of 2 A.P's will have terms of the form pn^2+qn+r where n=0,1,2....
Putting n=0,1,2 we get r=192, p+q+r = 360 and 4p+2q+r = 576
Solving the above equations we get p=24 and q=144
Putting n=7 and values of p,q,r we get 8th term as 2376.

3) Led d be the common difference and a be the first term of AP.
The fifth term of the series will be a+4d. If a+4d is divisible by d then a should also be divisible by d. Hence the cases are
d=1 , a=1,2,3....16
d=2, a=2,4,6....16
d=3, a=3,6
d=4, a=4
so answer is 16+6+2+1 = 25

hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100:shocked::shocked:

here the answer is 16 i think.
as the three numbers are 14,15,16 with common difference 1

No answer is 17. Just see the solution clearly...!!! Common difference will not be 1. It is 2.

vikranth504 Says

hey can u pls explain this model of problem in detail

What didn't you understand in this problem...???

3) 9+180+2700
=2889

25494 /4
remainder is 2

number till 28381 term

6373+999=7372

so next 2 digits from left 73

so the 28383rd number is 3

hey can u pls explain this model of problem in detail

1) Sum of three numbers in AP = 45
(a-d)+a+(a+d) = 45
3a = 45
a = 15
Sum of squares of numbers in AP = (a-d)^2+a^2+(a+d)^2 = 683
3a^2+2d^2 = 683
2d^2 = 683 - 675
d^2 = 4
d = +/- 2
Largest number among them = a+d = 15+12 = 17

here the answer is 16 i think.
as the three numbers are 14,15,16 with common difference 1

aimcats are going to start from 25 th may ...

Progressions SET-5 :-

1) M and N are natural numbers such that by M = (5N 4) (5N + 1).
If 1N200,what is the harmonic mean of all the possible values of M?
a) 200^2/1001
b) 1001
c) 1001/200^2
d) 1/1001

2) A sequence 192, 360, 576. is formed by multiplying the corresponding terms of two different Arithmetic Progressions. What is the eighth term of the sequence?
(a) 2376 (b) 1040 (c) 2116 (d) None of these

3) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term?
(a) 16 (b) 22 (c) 25 (d) 26

Progressions SET-4 :-

1) In an arithmetic progression, if the ratio of (r+2)th term and the (r+5)th term is (r+3):(r+6) and the sum of the first 4r terms and sum of the first 7r terms are in the ratio 1:3. Find value of r.
a) 15 b) 18 c) 27 d) 29

2) Consider 20 infinite geometric progressions, whose first terms are 2,3,4,....21 respectively and the common ratios are 1/3,1/4,1/5....1/22 respectively. If s1,s2,s3....s20 denote the sums of these 20 geometric progressions. Find s1+s2+s3+...+s20.
a) 80 b) 320 c) 160 d) 250

3) Find 28383rd term in series 1234567891011..
a) 6 b) 7 c) 3 d) 4

1) [a+(r+2-1)d]/[a+(r+5-1)d] = (r+3)/(r+6)
By solving this we get
a=2d ----> I
Given that
(4r/2 * [ 2a+(4r-1)d] ) / (7r/2* [2a+(7r-1)d] = 1/3
Substitute 2d in place of a
4/7 * [(4rd+3d)/(7rd+3d)] = 1/3
48r+36 = 49r+21
r = 36-21 = 15

2) Sum of infinite GP = a/(1-r)
s1 = 2/(1-1/3) = 3
s2 = 3/(1-1/4) = 4
s3 = 4/(1-1/5) = 5
.
.
s20 = 21/(1-1/22) = 22
s1+s2+...+s20 = 20*(3+22)/2 = 10*25 = 250

3) 9+180+2700
=2889

25494 /4
remainder is 2

number till 28381 term

6373+999=7372

so next 2 digits from left 73

so the 28383rd number is 3

Progressions SET-4 :-

1) In an arithmetic progression, if the ratio of (r+2)th term and the (r+5)th term is (r+3):(r+6) and the sum of the first 4r terms and sum of the first 7r terms are in the ratio 1:3. Find value of r.
a) 15 b) 18 c) 27 d) 29

2) Consider 20 infinite geometric progressions, whose first terms are 2,3,4,....21 respectively and the common ratios are 1/3,1/4,1/5....1/22 respectively. If s1,s2,s3....s20 denote the sums of these 20 geometric progressions. Find s1+s2+s3+...+s20.
a) 80 b) 320 c) 160 d) 250

3) Find 28383rd term in series 1234567891011..
a) 6 b) 7 c) 3 d) 4