For every positive, even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1 then p is, A) between 2 and 10 B) between 10 and 20, C) 20-30, D) 30-40, or E) greater than 40

ABCD is a parallelogram with angle B = 60.if the longer diagonal is of length 7 cm and the area of the parallelogram
ABCD is 15root(3)/2 sq.cm thne the perimeter of the parallelogram in cm is

A 15 B 15root(3) C 16 D 16root(3)

Is it 16.

lets adjacent sides be x and y
then area x.y.sin60=15root3/2
=>x.y=15-------------------1

and using cos formula in BDC
which gives x^2+y^2=34-------------2


solving 1 and 2

x=3 y=5

hence perimeter 16.

ABCD is a parallelogram with angle B = 60.if the longer diagonal is of length 7 cm and the area of the parallelogram
ABCD is 15root(3)/2 sq.cm thne the perimeter of the parallelogram in cm is

A 15 B 15root(3) C 16 D 16root(3)

coolmba001 Says

An article is sold for Rs 500 cash or for Rs 150 downpayment followed by 5 equal monthly installments. If rate of interest is 18% pa, compute the monthly installment.

My take is 73 (approximate value)

Rate of interset = 18% per year = (3/2)% per month

We have 150 downpayment.
So, difference to be paid = 350

Now, 350 paid now will become (350 + 5*(3/2)*(350/100)) after 5 months
Say, monthly installment is 'x'.

So, for first month we will get interst of 4 months. For second month's payment, we will get interest of 3 months and so on..
Total interest of (4+3+2+1+0) = 10 months on 'x'

So, we get (5x + 10*(3/2)*(x/100))
=> (350 + 5*(3/2)*(350/100)) = (5x + 10*(3/2)*(x/100))
=> 103x = 7525
=> x = 73 (approximately)

coolmba001 Says

An article is sold for Rs 500 cash or for Rs 150 downpayment followed by 5 equal monthly installments. If rate of interest is 18% pa, compute the monthly installment.

I am getting around Rs 55..not sure abt this one tho..

Akrosh007 Says

i think its 172.5...:-(..?

How Akrosh?

sumit99 Says

bro underline part smajh ni aaya

Bhai dekho..
_G_G_
There are 3 gaps here..call it a,b,c
now the reason why b>0 is because if b is zero then both the G's come together which violates the said condition..
Now if a is zero doesnt matter coz condition is still not violated..similarly for c..but b has to be greater than zero
=> a + b + c = 4
but b' = b + 1
=> a + b' + c = 3
=> (5-1)C3
=> 4C3 ways
Hope its clear now!

coolmba001 Says

An article is sold for Rs 500 cash or for Rs 150 downpayment followed by 5 equal monthly installments. If rate of interest is 18% pa, compute the monthly installment.

i think its 172.5...:-(..?

An article is sold for Rs 500 cash or for Rs 150 downpayment followed by 5 equal monthly installments. If rate of interest is 18% pa, compute the monthly installment.

you are right!!!

Please solve this one...

Q1. I bought a radio on EMI for 2 years. I paid Rs. 450 as down payment and emi of Rs. 680 and 590 for first and second year respectively. The rate of interest was 18%. What is the original price of radio?

its 1450

590 is 118% of amt remaining after first yr..
so, amt remaining after 1st yr is 500 after paying 680..

so 680+500=1180 is 118% of amt remaining after making down payment of 450. and that amt is 1000Rs.

so, total price is 1450..

Please solve this one...

Q1. I bought a radio on EMI for 2 years. I paid Rs. 450 as down payment and emi of Rs. 680 and 590 for first and second year respectively. The rate of interest was 18%. What is the original price of radio?

Bhai is it Rs. 1450?? I always have doubt in these types of questions :)

450*(1.18 )^2 + 680*(1.18 ) + 590 = P (1.18 )^2

P = 450 + 680/1.18 + 590/(1.18 )^2
= 1450

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

edited my previous post for that horse sum.. ans is 6!/2=360..

1)its 5C2*5C2=100

2)sum is 2=6320
(n-1)/2*(n+1)/2=6320
n^2-1=25280
n=159

:cheers::cheers:

Answer to the second ques is 159

:clap::clap::clap:

P.S. I have edited my post and also gave the answer to the second ques as 360...Is it correct??

:thumbsup::thumbsup:

17 nov D Day Says

how many 5 digit no. can be formed from 1,2,3,4,5 such that digit at tens place should be greater than digit at thousands place????

it is 60 if all the numbers has to distinct

Please solve this one...

Q1. I bought a radio on EMI for 2 years. I paid Rs. 450 as down payment and emi of Rs. 680 and 590 for first and second year respectively. The rate of interest was 18%. What is the original price of radio?

17 nov D Day Says

how many 5 digit no. can be formed from 1,2,3,4,5 such that digit at tens place should be greater than digit at thousands place????

When all the digits are distinct:

Select the two digits in C(5,2) ways and only one way to arrange them.
Rest of the digits can be arranged in 3! ways
So total numbers = 10*6 = 60 such numbers
When repetition is allowed:

Select the two digits in C(5,2) ways. Rest of the places have 5 options each
so 5^3 * 10 = 1250 numbers

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 ;)

bro underline part smajh ni aaya

check the calculations bro...



1.:clap::clap::clap::clap:

2.:nono::nono::nono::nono::nono:

1.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?


A. 100

B. 25

C. 50

D. 75

E. 3600

2.The sum of the even numbers between 1 and n is 79*80, where n is an odd number. n=?

edited my previous post for that horse sum.. ans is 6!/2=360..

1)its 5C2*5C2=100

2)sum is 2=6320
(n-1)/2*(n+1)/2=6320
n^2-1=25280
n=159

1.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?


A. 100

B. 25

C. 50

D. 75

E. 3600

2.The sum of the even numbers between 1 and n is 79*80, where n is an odd number. n=?

Answer to the second ques is 159
P.S. I have edited my post and also gave the answer to the second ques as 360...Is it correct??

how many 5 digit no. can be formed from 1,2,3,4,5 such that digit at tens place should be greater than digit at thousands place????

The number is of the form 20K+19
Sum = 10*n*(n+1) + 19n
= 10*50*51 + 19*50
= 24550

check the calculations bro...

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)5!=120 ways..

1.:clap::clap::clap::clap:

2.:nono::nono::nono::nono::nono:

1.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?


A. 100

B. 25

C. 50

D. 75

E. 3600

2.The sum of the even numbers between 1 and n is 79*80, where n is an odd number. n=?

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 ??