@bhatkushal said:also till 50 we will have 15 primes
i removed 15 primes
1 will be there as 0! /1 is taken
so 35
what have i missed???
@bhatkushal said:also till 50 we will have 15 primes
total primes=15
1 extra =4
==> total required numbers=50-16=34
@techsurge said:i removed 15 primes1 will be there as 0! /1 is takenso 35 what have i missed???
answer should be 34 by this logic
out of the 50 numbers we are subtracting 15 primes and 4.
therefore 34 is the answer
am Icorrect?
@maddy2807 said:Statements: 1. Some forests are zoos. 2. All zoos are lions. 3. No lion is panther. 4. All panthers are amphibians.Conclusions: I. Some zoos are panthers. II. Some panthers are forests. III. Some amphibians are panthers. IV. All lions are forests.
3?
@vijay_chandola said:One question from my side:[x] = Greatest integer less than or equal to x{x} = x €“ [x]How many real values of x satisfy the equation 5[x] + 3{x} = 6 + x?(a) 0 (b) 1 (c) 2 (d) More than 2
0?
there are totally 12 cards in two suits with six cards in each of them . All six digits are distinct . now 4 cards are chosen among them. wat is the probability that there is no pair in the chosen cards ??
a.16/33
b.1/33
c.1/2
d.none
@vijay_chandola said:OA of the previous question: a) 0Two different solutions of honey, milk and water are mixed with each other three times in varying proportions. The concentration of honey and milk in the three resulting solutions are found to be (10%, 16%), (12%, 12%) and (16%, x%) respectively. What is the value of x?(a) 4 (b) 7 (c) 8 (d) 10
4?
@rubikmath said:there are totally 12 cards in two suits with six cards in each of them . All six digits are distinct . now 4 cards are chosen among them. wat is the probability that there is no pair in the chosen cards ??a.16/33b.1/33c.1/2d.none
12c4 - 6c4 / 12c4 ?
@mailtoankit yaar koi iss waley answer ki approach bataooo.... question hi samjh nahi aaraha yahah toh
Q)How many pairs of previous integers (m,n) satisfy the equation:-
1/m!+1/n!=1/20?
a 0
b 1
c 2
d 3
@gs4890 said:12c4 - 6c4 / 12c4 ?
i think dis would be the prob. for all the cards are not pairs .. !
all 4 are pairs --> 6c4/12c4 ryt ??
all 4 are pairs --> 6c4/12c4 ryt ??
@bhatkushal said:Q)How many pairs of previous integers (m,n) satisfy the equation:-1/m!+1/n!=1/20?a 0b 1c 2d 3
0 ??
@rubikmath said:there are totally 12 cards in two suits with six cards in each of them . All six digits are distinct . now 4 cards are chosen among them. wat is the probability that there is no pair in the chosen cards ??a.16/33b.1/33c.1/2d.none
16/33