The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.
25450
24550
50900
49100

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720

(B) 360

(C) 120

(D) 24

(E) 21

1)
terms common are 19,39,59,..419

a=19 and d=20
sum of 50 terms= 19*50+*20=50(19+490)=50*509=25450

2)its 6!/2=360

The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.
25450
24550
50900
49100

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720

(B) 360

(C) 120

(D) 24

(E) 21

19+39+59............


25*(38+980)

1018*25=20360+5090=25450..sorry bhai no pen paper so yahi calculation kar di..

The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.
25450
24550
50900
49100

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720

(B) 360

(C) 120

(D) 24

(E) 21

1.) The number is of the form 20K+19
Sum = 10*(n-1)*(n) + 19n ====> (n-1)(n) because n varies from 0 to 1
= 10*49*50 + 19*50
= 25450

2.) Is it 360 ??

The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.
25450
24550
50900
49100

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720

(B) 360

(C) 120

(D) 24

(E) 21

yup answer is 70 but i didnt get u pls explain me again. :-(

69*68*67*66*65*64*61!mod71=1
-2*-3*-4*-5*-6*-7*69!*mod71=1
-1*69!mod71=1

actually i didnt get the bold ones earlies but ab agaye samaj
thanxx to u n @varun bhai :)

sry yaar typo tha..i mean isse ganda typo nahi ho sakta kabhi:splat:

17 nov D Day Says

find remainder when 63! divided by 71

70 ???
pranaam sabhiko !!!

69!mod71=1
69*68*67*66*65*64*61!mod71=1
-2*-3*-4*-5*-6*-7*69!*mod71=1
-1*69!mod71=1
69!mod71=-1=70..hope no mistakes

yup answer is 70 but i didnt get u pls explain me again. :-(

69*68*67*66*65*64*61!mod71=1
-2*-3*-4*-5*-6*-7*69!*mod71=1
-1*69!mod71=1

actually i didnt get the bold ones earlier but after reading varun bhai post samaj a gaya
thanxx to u n @varun bhai

17 nov D Day Says

find remainder when 63! divided by 71

Wilson bhaia ne kaha hai ki (p-2)!/p gives remainder 1 where p is a prime number

so 69! = 1 Mod(71)
69*68*67*66*65*64*63! = 1 Mod(71)
-2*-3*-4*-5*-6*-7 * 63! = 1 Mod(71)
120*42*63! = 1 Mod(71)
49*42*63! = 1 Mod(71)
70*63! = 1 Mod(71)
-1*63! = 1 Mod(71)
so 63! = -1 Mod(71) = 70 Mod(71)

@coolmba001: Please check post no 8146 on the previous page
@PK bhai: Pranaam bhai!! Aapki he kami thi yahan pe....ab aap apne methods se sabka man moh he lijiye bas :)

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways

Varun bhai would u pls explain this in detail as i dint get .

EXPLAIN PLS
3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?

For these questions.. Follow this methodology

Given Equations are :
3M + 4G + 5W = 750
6M + 9G + 10W = 1580

Multiply first and second by A and B respectively
We'll get:
M(3A-6B) + G(4A-9B) + W(5A-10B) = 750*A - 1580*B

Required to find:
6M + 4G + 10W

Compare the co-efficients
3A-6B=6 ----->1
4A-9B=4 ----->2
5A-10B=10 ----->3

Take any two equations and find the value of A and B
I took first and second and got A=10, B=4. Now these values should satisfy equation 3, which they are :D

So A=10 and B=4
750*10 - 1580*4 = 7500 - 6320 = 1180

17 nov D Day Says

find remainder when 63! divided by 71

69!mod71=1
69*68*67*66*65*64*63!mod71=1
-2*-3*-4*-5*-6*-7*63!*mod71=1
-1*63!mod71=1
63!mod71=-1=70..hope no mistakes

hi5blast Says

ya it is 7

Only for positive real values and so 7 it is..correcto.

find remainder when 63! divided by 71

h T c Says

Its 1180. check again bro

3m+4g+5w=750
6m+9g+10w=1580

Solving these guava=80

6m+10w=860

so 6m+4g+10w=1180

sorry for spam..par mantri ji papaer kaisa hua??

varun bhai kya akele akele bethe ho???

theek hua yaar..ab bas sab bhagwaan bharose hai...

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways

sumit99 Says

varun bhai plz ise explain kar do samajh ni aaya ye

Bhai dekho:

Case 1: When no green ball is there
That is possible in 1 way
Case 2: 1 Green ball
It can occupy any 6 places so 6 ways
Case 3: 2 Green balls
_G_G_
Three gaps a,b,c
a+b+c = 4, where b>0
so a+b+c = 3 => C(5,2) = 10 ways
Case 4: 3 Green balls

Green balls can occupy 1,3,5 or 2,4,6 or 1,3,6 or 1,4,6 number boxes=> 4 ways

Total = 1+6+10+4 = 21 ways

@Kalra bhai: Bhai sahi mein akela pad gaya tha....Aap aaye bahar aayi ;)

hi5blast Says

ya it is 7

sorry for spam..par mantri ji papaer kaisa hua??

varun bhai kya akele akele bethe ho???

it is 7 koi condition hi ni hai jo option mein minimum hai wo ans,..
15x+10y=380

y=8
x=20

ho gya satisfy...

ya it is 7

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways
bhai ye explain kar do i always have problem in permutation/combination

Hi Puys,

Q- 3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?

geetting 1180..2*ii-i se aa jayega..

15x + 10y + 7z = 429

What is the minimum value z can have?
a)7
b)12
c)17
d)22

it is 7 koi condition hi ni hai jo option mein minimum hai wo ans,..
15x+10y=380

y=8
x=20

ho gya satisfy...

Ganu02 Says

is it 1080rs..with 1 guava of cost 80...

Its 1180. check again bro

15x + 10y + 7z = 429

What is the minimum value z can have?
a)7
b)12
c)17
d)22

what about x,y,z..are they real no's or integers?

hey all tell me
what is the remainder when( 3^2011)/7

Reminder is 3..

15x + 10y + 7z = 429

What is the minimum value z can have?
a)7
b)12
c)17
d)22

Substitute the values...when z=7,15(20)+10(+7(7)=429

The cost of 6 mangoes, 4 Guavas and 10 Water melons = 750*10 - 1580*4 = 1180


EXPLAIN PLS

15x + 10y + 7z = 429

What is the minimum value z can have?
a)7
b)12
c)17
d)22

z=7 , x=20, y=8

hey all tell me
what is the remainder when( 3^2011)/7

remainder = 3

2011 = 6K + 1 ;(E (7)= 6)

hey all tell me
what is the remainder when( 3^2011)/7

3 it should be.

hey all tell me
what is the remainder when( 3^2011)/7

Euler (7) = 6

3^6K = 1 Mod(7)
3^2010 = 1 Mod(7)

The remainder is 3

15x + 10y + 7z = 429

What is the minimum value z can have?
a)7
b)12
c)17
d)22

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways

varun bhai plz ise explain kar do samajh ni aaya ye

Hi Puys,

Q- 3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?

is it 1080rs..with 1 guava of cost 80...

Hi Puys,

Q- 3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?

The cost of 6 mangoes, 4 Guavas and 10 Water melons = 750*10 - 1580*4 = 1180

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

Hi Puys,

Q- 3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?

pq, pr, qp and pp are two-digit numbers, pqr is a three-digit number and prrp is a four-digit number (p, q, r are all numerals). If pq pr = pqr, then the remainder when (qp prrp) is divided by pp is
(a) 0
(b) 1
(c) 7
(d) None of these

If we take p=1, r=0, q can be anything from 1 to 9.
say q=6

=> pq = 16, pr = 10, qp = 61, pp = 11
pqr = pq*pr = 16*10 = 160
prrp = 1001

(qp prrp)/pp = 61*1001/11 = 0 remainder

bhaiyo
i use to have have problems in identical ball and box concept in combinatories/permutation
how to tackle those problems
the discussed question related to filling of boxes from 1 to 6 so that no two adjacent boxes contain green coloured box for case iii)when there are two balls and three balls vo case clear nahi hua
plz help
thx

If the Pth term of an AP is 1/q and the qth term is 1/P, the sum to pq terms is:

1. (pq-1)/4 2. (pq+1)/4 3. (pq+1)/2 4. (pq-1)/2 ?

Post the soln......

Taking a sample AP 1/2,1...where q is 2 and p is 1...
so sum to 2 terms is 3/2 ...
now using options.... (pq+1)/2 satisfies

If the Pth term of an AP is 1/q and the qth term is 1/P, the sum to pq terms is:

1. (pq-1)/4 2. (pq+1)/4 3. (pq+1)/2 4. (pq-1)/2 ?

Post the soln......

Option 3 it would be.

vishal99999 Says

thx bhai par ye 14c2 kaise aaya

Integral solutions of the equation:
a+b+c+d+.....upto r terms = n
is given by C(n+r-1,r-1)...In this case n = 12 and r = 3
C(14,2) = 91

@Koshti bhai: Welcome back bhai!!!! How was your CAT ??

bhaiyo one question
There is an unlimited supply of identical red, blue and green coloured balls . In how many ways can 12 balls be selected from the supply?
364
91
55
65

We are looking for the

Coeff. of x^12 in (1+x+x+x+..+x^12)

=> Coeff. of x^12 in (1-x^13) * (1-x)^(-3)

=> C(14,12)

=> 91

Or

r+b+g =12

where 'r','b' &'g' represents the number of each type of colord ball chosen.

Hence total ways = C(12+2 ,2) = 91 .


PS: Posting after a long tym !! Hope i haven't became too rusty !!

If the Pth term of an AP is 1/q and the qth term is 1/P, the sum to pq terms is:

1. (pq-1)/4 2. (pq+1)/4 3. (pq+1)/2 4. (pq-1)/2 ?

Post the soln......

Option 3.) (pq+1)/2

a+(p-1)d = 1/q
a+(q-1)d = 1/p

From the two equations we get a=d=1/pq
Sum = pq/2*
= 1/2 *

vishal99999 Says

thx bhai par ye 14c2 kaise aaya

U have to find the integral solns of the eqtn.Nothing else.

bhaiyo one question
There is an unlimited supply of identical red, blue and green coloured balls . In how many ways can 12 balls be selected from the supply?
364
91
55
65

91 kya..did it mechanically.

bhaiyo one question
There is an unlimited supply of identical red, blue and green coloured balls . In how many ways can 12 balls be selected from the supply?
364
91
55
65

91..

R + B + G = 12

14c2 = 91

Red + Blue + Green = 12

The number of ways is C(14,2) = 7*13 = 91 ways

thx bhai par ye 14c2 kaise aaya

If the Pth term of an AP is 1/q and the qth term is 1/P, the sum to pq terms is:

1. (pq-1)/4 2. (pq+1)/4 3. (pq+1)/2 4. (pq-1)/2 ?

Post the soln......

bhaiyo one question
There is an unlimited supply of identical red, blue and green coloured balls . In how many ways can 12 balls be selected from the supply?
364
91
55
65

Red + Blue + Green = 12

The number of ways is C(14,2) = 7*13 = 91 ways

17 nov D Day Says

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

subscribing ...
No started with 1-10 cases
with 2-6 cases
with 3-3 cases
with 4- 1 case

total=216 cases

20/216=5/54

bhaiyo one question
There is an unlimited supply of identical red, blue and green coloured balls . In how many ways can 12 balls be selected from the supply?
364
91
55
65

17 nov D Day Says

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

So this is just like the question in how many ways can you arrange 3 digits in increasing order.

so the ans for the dice situation is just 6c3/6^3.

17 nov D Day Says

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

Sample space = 6^3 = 216

Case 1: When one is the first number
Cases = 4+3+2+1 = 10

Case 2: When 2 is the first number
Cases = 3+2+1 = 6

Case 3: When 3 is the first number
Case = 2+1 = 3

Case 4: When 4 is the first number
Cases = 1
Total cases = 10+6+3+1 = 20

Probability = 20/216 = 5/54

17 nov D Day Says

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

Probability of getting all distinct numbers: 6*5*4/6*6*6=5/9

Out of these, in one sixth of the cases, we'll have the required order.

So, it's 5/54

17 nov D Day Says

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

is dans 5/18

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

ans is 21..i suppose

die is rolled three times. the probability of getting a larger no. than the previous no. each time is....

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways

Total = 21 ways

six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?

1. 15
2. 16
3. 20
4. 21

pq, pr, qp and pp are two-digit numbers, pqr is a three-digit number and prrp is a four-digit number (p, q, r are all numerals). If pq pr = pqr, then the remainder when (qp prrp) is divided by pp is
(a) 0
(b) 1
(c) 7
(d) None of these

a) 0
p=q=1,r=0

pq, pr, qp and pp are two-digit numbers, pqr is a three-digit number and prrp is a four-digit number (p, q, r are all numerals). If pq pr = pqr, then the remainder when (qp prrp) is divided by pp is
(a) 0
(b) 1
(c) 7
(d) None of these

pq=11
pr=10

So, qp=11, prrp=1001
pp=11

Remainder of (qp*prrp)%pp=0