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Ankur
@Ankurb
1,473

If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3

solve for any one variable..

1/(4-p) + 3p/(7p-3) = 1

9 - 12p + 4p^2 = 0

p = 3/2

r = 5/3

q = 2/5

- 1 Like

Ankur
@Ankurb
1,473

hey puys,it seems u have used euler's theore to get the remainder 116.

but isnt it true that it can be used only when the two numbers are coprime

here 8 and 132 are not coprime!!

what u r saying is true but you can simplify it further, to apply Euler:-

2^1929 mod 4*33 =

2^1927 mod 33 = now apply euler

E(33) = 20

2^1927 mod 33 = 2^7 mod 33 = 29

remainder = 29*4 = 116

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Ankur
@Ankurb
1,473

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y) and y - 2| = 1. .. What is the area of the triangle?my doubts :given |y - 2 = 1=> y=1 or y=3 ,

in both cases this is distant by 1 unit from given y co-ordinate(-2, 2), (3, 2)in the graph

but then what ? can we still find the area of the triangle ? my main doubt is i cant form the triangle . i cant visualize the triangle . where is the triangle formed ?

take point (x,y) anywhere on line y = 3 or on y = 1,

area will always be = 0.5 * base * height = 0.5 * 5*1 =

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Ankur
@Ankurb
1,473

janvats SaysWhat is the remainder when 8^643 is divided by 132?

8^643 mod 4 = 0

8^643 mod 3 = -1

8^643 mod 11 = 6

11p + 6 = 3q-1

q = 11p + 7)/3,==>p = 1

17 + 33r = 4t,==>r = 3

- 1 Like

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Ankur
@Ankurb
1,473

Q)f(x+y) =f(x)f(y) for all "x" and "y".

f(x) = 1+x*g(x) where limit g(x) = T (x-->0) where T is a positive integer.

If fn(x) = kf(x) then k is equal to:

A) T

B) T^n

C)logT

D)(logT)^n

fn refers to nth derivative of the function or thats what I think it refers to. Its a XAT 2005 problem. U can check it if you have the same.

Cauchy Equation:-

f(x) = e^ax

f1(x) = ae^ax =af(x)

.

.

fn(x) = a^nf(x), ==> k = a^n

given condition:-

g(x) = (f(x) - 1)/x = (e^ax - 1)/x

as limit(x-->0)g(x) = T(a constant), ==>

T = a

==>

- 1 Like

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Ankur
@Ankurb
1,473

A number is known asmultiplicatively perfectif it is equal to the product of its positive divisors ( except 1 and the number itself).For example, 15 is such a number because 15 = 3 5.How many two digit multiplicatively perfect numbers are there?(a) 28(b) 30(c) 32(d) None of these

N = a^p b^q c^r

condition:-

(N^(p+1)(q+1)(r+1)/2)/N = N

==>

(p+1)(q+1)(r+1) = 4

N will either be of the form a^3 or a.b

a^3--- 1 number

a.b-- 29 numbers

- 2 Likes

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Ankur
@Ankurb
1,473

:w00t:....Edited

There are 10 bags each bag contain each bag contain 10 ball of 100 gms . but one of these bag contain ball of 90 gms. Each bag contain same color ball and every bag contain diffrent color ball then another bag's ball. You are allowed to wieght some / one / all balls once using elctronic wieghing machine and you has to find which bag has 90gms ball.

select 1 ball from first bag, 2 balls from 2nd bag and so on.

if balls wer identical, reading wud be = 5500 gms

if 1st bag have lighter balls , reading = 5500 - 10 = 5490 gms

if 2nd bag have lighter balls , reading = 5500 - 20 = 5480 gms

.

.

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Ankur
@Ankurb
1,473

Dats right but m asking how do we calculate for powers in decimals 1.05^1.2

How to go about it

binomial:-

(1+0.05)^1.2 = 1+0.05*1.2 = 1.06(approx)

- 6 Likes

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Ankur
@Ankurb
1,473

There are 10 students out of which three are boys and seven are girls. In how many different ways can

the students be paired such that no pair consists of two boys?

(1) 630 (2) 1260 (3) 105 (4) 210 (5) None of these

did it this way...

with no condition:-

9*7*5*3

with a pair consisting two boys:-

3*7*5*3

9*7*5*3 - 3*7*5*3 = 6*7*5*3 =

- 4 Likes

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Ankur
@Ankurb
1,473

3.If a,b,c,d are each +ve, a+b+c+d=8,a^2+b^2+c^2+d^2=25 and c=d ..Then what is the greatest value of c..?

(A)1/2 (B) 3/2 (C) 5/2 (D) 7/2

how to solve such kind of problems??

a+b = 8 - 2c

2c^2 = 25 -(a^2+b^2)

using above two:-

2c^2 = 25 -((8-2c)^2 - 2ab)

2c^2 = 25 -64 - 4c^2 + 32c + 2ab

6c^2 = 2ab + 32c - 39

also:-

ab

for max:-

6c^2 = 2(4 - c)^2 + 32c - 39

6c^2 - 24c + 21/2

c = 1/2 or 7/2

- 1 Like

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