probability:
Q. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away, the probabilities of hitting the plane at 1st, 2nd, 3rd & 4th shots are 0.4, 0.3,0.2 and 0.1 respectively. What is the probability that the gun hits the plane.
a)0.7654 b)0.6976 c)0.3024 d) 0.2346
plz give the approach.
@[594309:rijz]
I think answer is b.
@[601024:crazyy_rj] Which ques. did you answer? Also, please give the solution?
@[594309:rijz]
I answered this one
2)A's salary is 1st inc. by 25% n then dec. by 20%. The result is same as B's salary increased by 20% and then reduced by 25%. Find the ratio of B's salary to that of A's?
Is my answer correct? I will explain if it is correct.
@[601024:crazyy_rj] the correct option is c.
@[522336:sharangtelkikar] can you post approach...
@[406078:pratskool] Why did you subtract it from 90.Why not 99? 
@surya the bacon said: @sharangtelkikar can you post approach...
It is given that product is less than sum. It is possible when one of the digits is 1.
Hence there are 17 no. having digit 1 between 10 & 50
@[522336:sharangtelkikar] Approach is not given but ans. is correct and ur logic also.....
@rajesh123gone said: probability: Q. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away, the probabilities of hitting the plane at 1st, 2nd, 3rd & 4th shots are 0.4, 0.3,0.2 and 0.1 respectively. What is the probability that the gun hits the plane. a)0.7654 b)0.6976 c)0.3024 d) 0.2346 plz give the approach.
Is answer B ? I'll give my approach if correct.
@abhishek-c said:Number Systems:Q. M is a two digit number which has the property that: The product of factorials of its digits > sum of factorials of its digitsHow many values of M exist?a)56 b)64 c)63 d)None of these
i think answer is c)63 ....we have all numbers from 22-29, 32-39....92-99...which gives 64 numbers, but 22 doesnt satisfy the condition..so 64-1=63
@rajesh123gone said: probability: Q. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away, the probabilities of hitting the plane at 1st, 2nd, 3rd & 4th shots are 0.4, 0.3,0.2 and 0.1 respectively. What is the probability that the gun hits the plane. a)0.7654 b)0.6976 c)0.3024 d) 0.2346 plz give the approach.
Ans is b)0.6976
Case-1: Hit in first attempt--->p=0.4
Case-2:Hit in second attempt-->p=0.6*0.3
Case-3:Hit in 3rd attempt--->p=0.6*0.7*0.2
Case-4:Hit in 4th attempt--->p=0.6*0.7*0.8*0.1
Sum=0.6976
@surya the bacon said: A triangular no. is defined as a no. which has the property of being expressed as a sum of consecutive natural no.s starting with 1. How many triangular no.s less than 1000, have the property that they are the difference of squares of two consecutive natural no.s ?(a)20 (b)21 (c)22 (d)23
Ans is b)21 i think...u have found out that we have to get the number of odd numbers from 1 to 44, which is 22....but we have to exclude 1 as 1 is not a sum of consecutive natural numbers...we must start with 3...so 22-1=21 numbers
@[521009:ricsani] Triangular no.s are in the form
1, (1+2), (1+2+3), (1+2+3+4)......
i.e. 1,3,6,10,15,21 etc.
so for every 4 no.s 2 are odd & 2 are even
44 triang. nos.
as n(n+1)/2
Also, they satisfy given property
(x+1)^2 - x^2 = n
2(x+1)=n
so 'n' is even no.
It implies that we need to find out even no.s from 44
Ans. 22 (Ans. Key is also 22)
yaa it's correct@[522336:sharangtelkikar]
Hi Puys!!
Can some1 pls. help with the solution for the below mentioned questions:-
Q1) Find the sum of all possible whole number divisors of 720?
A) 2012 B) 2624 C) 2210 D) 2418 E)2520
Q2) Find the sum of the series:-
1/1*5 + 1/5*9 + 1/9*13 +......... +1/221*225
A) 28/221 B)56/221 C)56/225 D) None of these
Q3) What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 .... + (1152!)^3 is divided by 1152?
A) 125 B)225 C)325 D)205
thanks in advance !!!
Regards,
Ayush
@abhishek-c said: @pratskool Why did you subtract it from 90.Why not 99?
because there are 90 2digit numbers
@rajesh123gone said: yaa it's correct@sharangtelkikar
Sol: Prob(1) = 0.4
Prob(2) = [1-P(1)]*0.3 = 0.6*0.3 = 0.18
similarly, Prob(3) = 0.6 *0.7* 0.2
Prob(4) = 0.6*0.7*0.8*0.1
Hence P(1+2+3+4) = 0.6976
Prob(2) = [1-P(1)]*0.3 = 0.6*0.3 = 0.18
similarly, Prob(3) = 0.6 *0.7* 0.2
Prob(4) = 0.6*0.7*0.8*0.1
Hence P(1+2+3+4) = 0.6976
There are 25 points on a plane of which 7 are col-linear , how many quadrilaterals can be formed from these points?
a)5206
b)2603
c)13015
d)none of these
Funda of calculating the number of triangles is clear to me but not of calculating the number of quadilaterals.
@mayankdangi said:There are 25 points on a plane of which 7 are col-linear , how many quadrilaterals can be formed from these points?a)5206b)2603c)13015d)none of theseFunda of calculating the number of triangles is clear to me but not of calculating the number of quadilaterals.
Im getting none of these , not sure though
Total -( all 4 points are collinear points + 3 points on collinear points&other; point is a non collinear point)
25C4-(7C4+ 7C3*18)
Solving gives none of these
Correct me if im wrong :|