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Pratik Sahoo
@pratiksahoo33
77

Posted 01 Aug '11

> If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:
a>50
b>60
c>70
d>80
Please justify .............
Is it (b)60?
take the min pythagorean triplets 3*4*5 = 60

If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:

a>50

b>60

c>70

d>80

Please justify .............

Is it (b)60?

take the min pythagorean triplets 3*4*5 = 60

- 1 Like

Pratik Sahoo
@pratiksahoo33
77

Posted 01 Aug '11

> If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:
a>50
b>60
c>70
d>80
Please justify .............
Is it (b)60?
take the min pythagorean triplets 3*4*5 = 60

If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:

a>50

b>60

c>70

d>80

Please justify .............

Is it (b)60?

take the min pythagorean triplets 3*4*5 = 60

- 1 Like

Pratik Sahoo
@pratiksahoo33
77

Posted 16 Jul '11

> Can someone elaborate the approach as to how to solve such questions :
Q-Find the remainder when 2^89 is divided by 89 ?
A)2 B)1 C)87 D)88
OA-A
Use Eulers theorm Euler no of 89 = 88
(2^8 :: *2/89 = 2/89
So remainder = 2

Can someone elaborate the approach as to how to solve such questions :

Q-Find the remainder when 2^89 is divided by 89 ?

A)2 B)1 C)87 D)88

OA-A

Use Eulers theorm Euler no of 89 = 88

(2^8*2/89 = 2/89

So remainder = 2

Pratik Sahoo
@pratiksahoo33
77

Posted 16 Jul '11

> MM is rite!! ;)
*Two ants Amy and Pamy are running on a graph paper. Amy is at a distance 2 cm from the X-axis and 3 cm right of Y-axis. Pamy is directly 2 cm below the X-axis and 4 cm to the right side of Y-axis. They spot a food particle dropped on the first quadrant of the paper. The fo...

MM is rite!! ;)Two ants Amy and Pamy are running on a graph paper. Amy is at a distance 2 cm from the X-axis and 3 cm right of Y-axis. Pamy is directly 2 cm below the X-axis and 4 cm to the right side of Y-axis. They spot a food particle dropped on the first quadrant of the paper. The food particle's shortest distance from the X-axis is 6 cm and 8 cm from the Y-axis. If Amy can run at a speed of 1 cm/sec and Pamy at speed of 1.5 cm/sec and, Pamy is further away from the food particle than Amy, then in approximately how many seconds would the food particle be grabbed by either of the ants?

5.2 seconds

7 seconds

6.4 seconds

5.9 seconds

Suja

Is it (d ) 5.9 sec?

Pratik Sahoo
@pratiksahoo33
77

Posted 13 Jul '11

> Bhai log, you have ignored the bold part.
by going with that condition. After 4th round, a-b,2b-c,2c-d,2d-a+b
Now, a-b=48, 2b-c=48, 2c-d=48, 2d-a+b=48
2d-a+b+a-b = 96, but here we'll get d = 48. Please correct me if/where I am wrong?
If d gives a-b coins to a, then a wi...

Bhai log, you have ignored the bold part.

by going with that condition. After 4th round, a-b,2b-c,2c-d,2d-a+b

Now, a-b=48, 2b-c=48, 2c-d=48, 2d-a+b=48

2d-a+b+a-b = 96, but here we'll get d = 48. Please correct me if/where I am wrong?

If d gives a-b coins to a, then a will have 2a-2b coins

- 1 Like

Pratik Sahoo
@pratiksahoo33
77

Posted 13 Jul '11

> plz see the ques below:
find the sum of the first 20 terms of the series:
(1*2^2)+(2*3^2)+(3*4^2)+........
i am getting answer as 50050 but the sol says 47180. where am i going wrong ? ::
Gen Term = nv=> n(n^2 + 2n + 1)=>n^3 +2*n^2 + n
S = ^2 + n(n+1)/2 + 2*n*(n+1)*(2n...

plz see the ques below:

find the sum of the first 20 terms of the series:

(1*2^2)+(2*3^2)+(3*4^2)+........

i am getting answer as 50050 but the sol says 47180. where am i going wrong ?

Gen Term = nv=> n(n^2 + 2n + 1)=>n^3 +2*n^2 + n

S = ^2 + n(n+1)/2 + 2*n*(n+1)*(2n+1)/6

Put n = 20

S = 50050

Pratik Sahoo
@pratiksahoo33
77

Posted 13 Jul '11

> Father gives a certain amount of one rupee Coins to his four sons a,b,c,d. now, a gives to b as many coins as b has,b gives to c as many coins as c has,c gives to d as many coins as d has,and d gives to a as many coins a has. now, all of them has got 48 coins. what is the maximum amount of coin...

Father gives a certain amount of one rupee Coins to his four sons a,b,c,d. now, a gives to b as many coins as b has,b gives to c as many coins as c has,c gives to d as many coins as d has,and d gives to a as many coins a has. now, all of them has got 48 coins. what is the maximum amount of coins that anyone has at any point of time?

1.60

2.54

c.90

d.72

Let initially a,b,c,d

After 1st rnd, a-b,2b,c,d

After 2nd rnd,a-b,2b-c,2c,d

After 3rd rnd, a-b,2b-c,2c-d,2d

2d=48=>d=24

2c-d=48=>c=36

2b-c=48=>b=42

a-b=48=>a=90

So at any time we see that a has highest amount

So (c) 90

Whats the OA?

- 3 Likes

Pratik Sahoo
@pratiksahoo33
77

Posted 13 Jul '11

> akkitheaviator Says
>
> find the ratio of the 17th terms of two AP if the ratio of the sum upto n terms of two APs is (3n+2):(4n-13)
Is it 101:119??
:=(3n+2):(4n-13)
=>: =(3n+2):(4n-13)
=>* = (3n+2):(4n-13)
Put n = 33 and get the ans !!*

akkitheaviator Saysfind the ratio of the 17th terms of two AP if the ratio of the sum upto n terms of two APs is (3n+2):(4n-13)

Is it 101:119??

:=(3n+2):(4n-13)

=>: =(3n+2):(4n-13)

=>

Put n = 33 and get the ans !!