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how many negative integer sol exists for x^2-y^2 =400?
If x is a real no then what is the no of solutions for the eq:
((x^4-16)(x^2-4))^1/2?
the set of all real nos x for which x^2-|x+2|+x >0

how many negative integer sol exists for x^2-y^2 =400?

If x is a real no then what is the no of solutions for the eq:

((x^4-16)(x^2-4))^1/2?

the set of all real nos x for which x^2-|x+2|+x >0

- 1 Like 20Comments
- @saphira1 I think ur 3rd answer is wrong it can not be -i.... 29 Jul.
- i think the answer for third one should be .. from 2 to i.... 29 Jul.

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swati garg
@saphira1
34

Answer1 - 4 ways (-101, -99), (-52,-48) (-29, -21) and (-25, -15).

Answer2 - infinite solutions.

Answer 3 (-inf, -rt2) (rt2, inf)

nitin nandan
@nitinknandan
3

@saphira1 exactly my point, if u see the number of positive integral solutions will also be the same

swati garg
@saphira1
34

yeah. That's how I got it too. Only takeaway from the word "negative" is that |x|>|y|.

Anirban Kar
@Anirban_kar
14

Yes, @swati The total answers will be 8 only. We just have to take the even factors that make up 400 only. Total factors = 15, of which (5, 80), (25,16), & (1,400) will be taken off. So, it seems that the answer is 15 -6 = 9. But, again (20,20) will have only 1 solution instead of 2. So, 8 is the final answer. Thanks for the tips.

mayank asd
@mack01
48

@saphira1 I think ur 3rd answer is wrong it can not be -inf to -rt2?

_rininir_ _dev_
@asaktidev
1

i think the answer for third one should be .. from 2 to infinity

Here are few ques...
A speaks truth in 60% cases and B in 70% cases. Wat is the prob that they will do the same thng while describing a single event. If they both lie, they tell the same thing?
A nd B toss coin alternatively till one of them gets a head. If A starts find the prob that B will wi...

Here are few ques...

A speaks truth in 60% cases and B in 70% cases. Wat is the prob that they will do the same thng while describing a single event. If they both lie, they tell the same thing?

A nd B toss coin alternatively till one of them gets a head. If A starts find the prob that B will win the game.

the prob of picking a spade or an ace not of spade from a pack of 52 cards is?

- 9Comments
- third one is 4/13..plz explain other two also 2 & 3. 27 Jul.
- @webby the second one is P(B winning)= P(A gets tails)*.... 28 Jul.

alok k
@allok
2

@webby the second one is

P(B winning)= P(A gets tails)*P(B gets head) + P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets head) +P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets head) +. so on upto infinity

the probability of P(A gets tails)= P(B gets head) = P(B gets tails) = 1/2

So P(B winning) is a GP with infinite terms with first term a = 1/4 and r= 1/4

So P(B winning) =a/(1-r) = 1/4/(1-1/4)=1/3

For the 3rd one in a pack of cards there are 13 spades + 3 ace except the ace of spades

So we have probability =16/52 =4/13

how to find the 30th term of this series

1,3,6,10,15...

plz explain

- 2 Likes 6Comments
- 465. 25 Jul.

In a group of newly married couples, 10% of husbands and 45% of wives are unemployed. What is the probability that a person selected at random is employed?

- 10Comments
- 29/100. 25 Jul.
- 29/40. 25 Jul.

sagar vaid
@sagarvaid
8

Let there be 100 couples.. 90 husbands and 55 wives arr emplyed I.e. 145 out of 200 ppl are employed.. hence probability =145/200=29/40