MKJ Says

6^83+8^83/49=........(remainder???)

35

(7-1)^83mod49 = (7*83 - 1)mod 49 = -7-1

(7+1)^83mod49 = (7*83 + 1)mod49 = -7+1

So, (-14)mod49 = 35

2.find the number of common terms to the progressions 2,8,14,20....98 and 6,10,14,18,.....,102

8 ??

6m - 2 = 4n + 2

m=3___n=3
m=5___n=6
m=7___n=9
m=17__n=24

So 8 values possible.

6^83+8^83/49=........(remainder???)

1. A person has to select one of three capsules .One capsule makes his weight increase by 1 Kg . Another one ensures that his weight does not change.Third one makes him gain 2 Kg and also compels him to select one of the three capsules once again-all similar to first set of three capsules. the person has no way identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
1)3/4
2)5/4
3)3/2
4)5/3

3/2 ??

Let x be his gain in weight
x= 1/3(1) + 1/3(0) + 1/3(2 + x)
1/3 is the probability of selecting the tablet.

x = 1/3 + 2/3 + x/3
x = 3/2

1. A person has to select one of three capsules .One capsule makes his weight increase by 1 Kg . Another one ensures that his weight does not change.Third one makes him gain 2 Kg and also compels him to select one of the three capsules once again-all similar to first set of three capsules. the person has no way identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
1)3/4
2)5/4
3)3/2
4)5/3

Reqd. Prob. (RP) = 1/3*1 + 1/3*0 + 1/3

=> RP = 1/3 + 0 + 1/3(2+RP)

Thus, RP = 3/2

two cards are drawn from a well shuffled pack of cards one after another .wat is the probability that first frawn card is red and the second is an ace. when the cards that is drawn first is not replaced
1)29/442
2)3/104
3)6/169
4)17/442

prob (red)=26/52
1.prob (red ace)=2/52
2.prob (red without ace)=24/52

case 2:Second with 1
p(ace)=3/51

second with 2
p(ace)=4/51

so ans.d)1/26=17/442
total=

1.find the approx sum of the series 11/4,31/8,79/16..... upto 10 terms.

The given exp cab be written as

2+3/4+3+7/8+4+15/16+....

=65+3/4+7/8+.....
=65+10-(1/4+1/8+.....)
=75-(1/4+1/8....)
The sum in brackets is a gp series...compute sum and put the value to get ans....

what is the least possible value of exp
E=(x-1)(x-3)(x-4)(x-6)+10, for real values of x ?

is it 1?????

what is the least possible value of exp
E=(x-1)(x-3)(x-4)(x-6)+10, for real values of x ?

(x-1)(x-3)(x-4)(x-6)+10
=> (x^2-7x+6)(x*2-7x+12) + 10

take x^2-7x = y

(Y+6)(y+12) + 10

min will be at y = -(6+12)/2 = -9 (and y = x^2-7x = -9 satisfies for real x as (-7)^2 - 4*9 > 0)

so put y = -9

it is -3*3 + 10 = 1

what is the least possible value of exp
E=(x-1)(x-3)(x-4)(x-6)+10, for real values of x ?

:(.........:(

1.find the approx sum of the series 11/4,31/8,79/16..... upto 10 terms.

2.find the number of common terms to the progressions 2,8,14,20....98 and 6,10,14,18,.....,102

3.In all 3 digit number ,the hundred digit, tens digit and the units digit are in descending order and in AP.Each digit of the number is multiplied by the sum of other two digits. the sum if all such results is A. B is the product of tens digit an sum of all digits and A=(4/3)B. find the number of such numbers.

4.In an AP,first term is a natural number and the common difference is whle number.If the first term is dereased by 2 and the third term is increased by 10, the first three terms would be in GP. find the common difference

what is the least possible value of exp
E=(x-1)(x-3)(x-4)(x-6)+10, for real values of x ?

is the answer 10 ??

1. A person has to select one of three capsules .One capsule makes his weight increase by 1 Kg . Another one ensures that his weight does not change.Third one makes him gain 2 Kg and also compels him to select one of the three capsules once again-all similar to first set of three capsules. the person has no way identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
1)3/4
2)5/4
3)3/2
4)5/3

what is the least possible value of exp
E=(x-1)(x-3)(x-4)(x-6)+10, for real values of x ?

two cards are drawn from a well shuffled pack of cards one after another .wat is the probability that first frawn card is red and the second is an ace. when the cards that is drawn first is not replaced
1)29/442
2)3/104
3)6/169
4)17/442

Its d
there are 2 cases

1>The first one is red and ace,2nd one is ace
2>The first one is red and not ace and the end one is ace

hence 2*3/52*51 + 24*4/52*51

two cards are drawn from a well shuffled pack of cards one after another .wat is the probability that first frawn card is red and the second is an ace. when the cards that is drawn first is not replaced
1)29/442
2)3/104
3)6/169
4)17/442

(96 + 6)/52*51 = 102/52*51 = 1/26 = 17/442, which is option D

wovfactorAPS Says

.5 i think so

no SRY thats not the answer . options given are
1) 9/16
2) 1/2
3) 7/16
4)5/8
answer os option 3

Parnaam ________/\_______ Bhai log

Plz post The questions on Weight Measurement :)

i need Examples Plzz

is it 323/345 by any chance



??

no SRY thats not the answer . options given are
1) 9/16
2) 1/2
3) 7/16
4)5/8
answer os option 3

Total cases=52*51

fav. cases: First red and non ace and second ace=24*4
First red and ace and second ace=2*3

prob=(96+6)/52*51=1/26=17/442

yea thats correct thanks a lot

PrashantBS Says

Vishal and Anirudh decide to meet at a cafe between 5 pm to 6pm. they agree that the person who arrives first will wait for exactly 15 mins for the other. if each of them arrives at random betwen 5 pm to 6 pm what is the probability that the meeting takes place?

.5 i think so

no i dont understand ur method . plus the answer is potion D .. dont knw how it came though

PrashantBS Says

Vishal and Anirudh decide to meet at a cafe between 5 pm to 6pm. they agree that the person who arrives first will wait for exactly 15 mins for the other. if each of them arrives at random betwen 5 pm to 6 pm what is the probability that the meeting takes place?

is it 323/345 by any chance



??

two cards are drawn from a well shuffled pack of cards one after another .wat is the probability that first frawn card is red and the second is an ace. when the cards that is drawn first is not replaced
1)29/442
2)3/104
3)6/169
4)17/442

Total cases=52*51

fav. cases: First red and non ace and second ace=24*4
First red and ace and second ace=2*3

prob=(96+6)/52*51=1/26=17/442

vyomconfused Says

A&B; together complete a work in 6 min.But A works for 4 min and B works for 6 mins. After this only 1/5 work is remaining.how much time will the slowest person will take to do work all alone?

Pranam vyom bhai____/\____kafi din bad dekhe apko
pranam saurav bhai,bindita____/\____

suppose A does a units of work per min and B does b units
then total work=(a+b)*6
hence (a+b)*6=4a+6b+(a+b)*6/5
hence a/b=3/2
let them be 2x and 3x
hence total work=30x
hence req time=30x/2x=15

two cards are drawn from a well shuffled pack of cards one after another .wat is the probability that first frawn card is red and the second is an ace. when the cards that is drawn first is not replaced
1)29/442
2)3/104
3)6/169
4)17/442

15 min??

1/A + 1/B = 1/6

A&B; work for 4 min in which they can complete 2/3 of the work. After that B alone works for 2 min and complete (4/5 - 2/3) work

So 2/B = 4/5 - 2/3 = 2/15
B = 15 mins

1/A = 1/6 - 1/15
A = 10 mins

So B is slower

Damn such a silly mistake....i took it this way A working indiviually for 4 min and B working indiviually for 6 mins..:banghead:


Anyways thanx.

96 + y^2 = x^2..so 96= x^2-y^2 =(x+y)(x-y)
96=2^5*3..so it can be written as a product of two factors in 0.5*(5+1)*(1+1)= 6 ways..

how did you get 4.?.I guess some out of those 6 will not satisfy the criteria..

yes it can be written as a product of two factors in 6 ways
but both factors has to be even or odd to get integer values of x and y
so subtract 2 cases i.e 1 and 96 and 3 and 32

Vishal and Anirudh decide to meet at a cafe between 5 pm to 6pm. they agree that the person who arrives first will wait for exactly 15 mins for the other. if each of them arrives at random betwen 5 pm to 6 pm what is the probability that the meeting takes place?

When 96 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?
1. 4
2. 8
3. 6
4. 12
5. 8

a^2 + 96= b^2
b^2-a^2=96
(b+a) (b-a)=96
Total 4 values satisfy
(5,11),(23,25),(2,10),(10,14)

mkj Says

1!+2!...........+100!/720=...............(remainder??)

1! +.....+ 5!/720

=>153

:)

When 96 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?

1. 4
2. 8
3. 6
4. 12
5. 8

N=2 -> 2^2 +96 =10^2
N=5-> 5^2 +96 =11^2
N=10 -> 10^2 +96 =14^2
N=23-> 23^2 +96 -> 25^2

so 4 values satisfy

When 96 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?

1. 4
2. 8
3. 6
4. 12
5. 8

4 ??

96 + b^2 = a^2
(a-b)(a+b) = 96 = 2^5*3

(a,b) = (25, 23) (14,10) (10,2) (11,5)

is it opt 1 ... 4 ????

96 + y^2 = x^2..so 96= x^2-y^2 =(x+y)(x-y)
96=2^5*3..so it can be written as a product of two factors in 0.5*(5+1)*(1+1)= 6 ways..

how did you get 4.?.I guess some out of those 6 will not satisfy the criteria..

vyomconfused Says

A&B; together complete a work in 6 min.But A works for 4 min and B works for 6 mins. After this only 1/5 work is remaining.how much time will the slowest person will take to do work all alone?

15 mins


1/10 + 1/15 = 1/6

Let the set S = {1, 2, 3, 4, 5, 6, 7, 8}.
Find the number of ordered pairs (A, B), where A and B are subsets of S and A is a proper subset of B.
OPTIONS
1) 6305
2) 6561
3) 4096
4) 6433

Now no. of pairs
(A,B)=8c1(2c0)+(8C2*(2C1))+(8C3*(3C2+3C1))+(8C4(4C3+4c2+4c1)+.....+(8c8(8c7+......+8c1))
=8c2*(2^2-1)+8c3*(2^3-1)+8c4(2^4-1)+8c5(2^5-1)+.....+8c8(2^8-1)
=(8c2*2^2*1^6+..........+8c8*2^8*1^0)-2*(8c2+8c3+8c4+....+8c
=(3^8-8C1*2-1)-1*(2^8-8c1-1)+8c1
=6305

MKJ Says

1!+2!...........+100!/720=...............(remainder??)

6! onwards will be divisible by 720
Hence remainder wd be 1!+2!+3!+4!+5!=153

MKJ Says

1!+2!...........+100!/720=...............(remainder??)

720
2360
2|180
2|90
2|45
3|15
35
6!=6X5X4X3X2X1

Everything after 6!(including 6) is divisible by 720
1!+2!+3!+4!+5!=1+2+6+24+120=153 The ans...

MKJ Says

1!+2!...........+100!/720=...............(remainder??)

all nos from 6! will give remainder 0
so sum of nos till 5! = 153
so remainder 153

MKJ Says

1!+2!...........+100!/720=...............(remainder??)

6! = 720. So remainder 1! + 2! + 3! + 4! + 5! = 153

When 96 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?

1. 4
2. 8
3. 6
4. 12
5. 8

is it opt 1 ... 4 ????

vyomconfused Says

A&B; together complete a work in 6 min.But A works for 4 min and B works for 6 mins. After this only 1/5 work is remaining.how much time will the slowest person will take to do work all alone?

15 min??

1/A + 1/B = 1/6

A&B; work for 4 min in which they can complete 2/3 of the work. After that B alone works for 2 min and complete (4/5 - 2/3) work

So 2/B = 4/5 - 2/3 = 2/15
B = 15 mins

1/A = 1/6 - 1/15
A = 10 mins

So B is slower

1!+2!...........+100!/720=...............(remainder??)

vyomconfused Says

A&B; together complete a work in 6 min.But A works for 4 min and B works for 6 mins. After this only 1/5 work is remaining.how much time will the slowest person will take to do work all alone?

A+B=1/6
10A+10B=5/3
AX4+BX6=4/5
10A+15B=2
5B= 1/3
B=1/15
A=1/6-1/15
=5-2/30
A=1/10
B is slower he will take 15 minutes

When 96 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?

1. 4
2. 8
3. 6
4. 12
5. 8

What is the approximate sum of the following series:

1+3/11+7/11^2+13/11^3+21/11^4.........

1.4

1.30

1.34

1.5

any other shortcut apart from usual multiplication and subtraction methodbec it takes time to solve

Let the set S = {1, 2, 3, 4, 5, 6, 7, 8}.

Find the number of ordered pairs (A, B), where A and B are subsets of S and A is a proper subset of B.
OPTIONS

1)6305
2)6561
3)4096
4)6433

6305.

2 page piche Aizen bhai ne mast samjha diya sum

In how many ways can the natural no's from 1 to 9 be divided into 3 groups of 3 numbers each,such that the sum of the numbers in each group is divisible by 3?
a.9
b.18
c.37
d.38

37..i think so

A&B; together complete a work in 6 min.But A works for 4 min and B works for 6 mins. After this only 1/5 work is remaining.how much time will the slowest person will take to do work all alone?

let the set s = {1, 2, 3, 4, 5, 6, 7, 8}.

Find the number of ordered pairs (a, b), where a and b are subsets of s and a is a proper subset of b.
Options

1)6305
2)6561
3)4096
4)6433

c(8, 1)*(2 - 1) + c(8, 2)*(4 - 1) + c(8, 3)*(2^3 - 1) + ... + c(8, 8 )*(2^8 - 1)
= 3^8 - 1 - (2^8 - 1)
= 6305

Let a,b and c be the positive real numbers. Determine the minimum value of (a+3c)/(a+2b+c) + 4b/(a+b+2c) -8c/(a+b+3c)

Options:
A. 122-17
B. 62-17
C. 122+17
D. 62+17

keep a=b=c=1
1+1-1.6=0.4
so this means c and d cannot be ans.
6*1.4-17=-8.xx
12*1.4=16.8-17=-0.2
let a=1,b=1,c=2
7/5=1.4
4/6=0.666
16/8=2
so=0.0666

So ans. possibly A?? as a , b,c when close are near option A

Let the set S = {1, 2, 3, 4, 5, 6, 7, 8}.

Find the number of ordered pairs (A, B), where A and B are subsets of S and A is a proper subset of B.
OPTIONS

1)6305
2)6561
3)4096
4)6433

10000! = (100!)K P, where P and K are integers. What can be the maximum value of K?



p.s - Warm up question :)

Exponent of 2 in 100! = 97
Exponent of 2 in 10000! = 9995
= 103

Exponent of 97 in 100! = 1
Exponent of 97 on 10000! = 104

So, lower of 103 and 104, i.e, 103 will be the answer

Namaste sabko _____/\______

Hello puys !..
Been long!... I ws out for 3 days !

A set P consists of the first 25 natural numbers. How many minimum numbers should one pick from the set P so that there is always atleast one pair (x,y) among them satisfies 3x=2y?

A: 13 B: 16 C:17 D:18 E:20

is 18 is the answer...???

koi mere bhi ques kardo nahi to muje spam karna padega aise.

10000! = (100!)^K P, where P and K are integers. What can be the maximum value of K?



p.s - Warm up question :)

Shouldn't this question have 100^K*P. Then the answer we can get as 104 as 100! has 24 zeroes and 10000! has 2499 zeroes. Otherwise K can be highest as 101*102*....10000

Hello puys !..
Been long!... I ws out for 3 days !

A set P consists of the first 25 natural numbers. How many minimum numbers should one pick from the set P so that there is always atleast one pair (x,y) among them satisfies 3x=2y?

A: 13 B: 16 C:17 D:18 E:20

Swati04 Says

OA is if C is any number other than 3.

Right if C=3, then (1) - (2) and (2) - (3) becomes identical lines in X-Z plane. Then the point of intersection of all three lines will come out to be a single point.

Whereas if C

Swati04 Says

AIMCAT1317Q:For what values of C does the following system of equations have no solution? i. 2x + y + 3z = 4 ii. 3x + y- z = 5 iii. x + y + 7z = C

iX2 gives 4x+2y+6z=8
substracting 2 -3x-y+z=-5
x+y+7z=3
Therfore c = 3