ten coins are tossed up simultaneously.In how many outcomes will the third coin turn up a head?
from a bag containing 100 balls, one ball weighs9 grams and all the other weigh 10 grams each.Using a simple balance where balls can be kept on either pan, what is the minimum weighs required to identify the defective ball?
Ans. A.)3 B.)4 C.)5 D.)7
Parul09 Saysten coins are tossed up simultaneously.In how many outcomes will the third coin turn up a head?
is it 7 and 2^9 ??
Six positive numbers are taken at random and are multiplied together.Then what is the probability that the product ends in an odd digit ohter than 5?
Kindly solve and explain........
is it 4^6/(9*10^5) ??? 6 numbers should be odd and none of them should be 5..
hi,
Please explain the following:
1) A student is allowed to select at most n books from a collection of (2n+1) books. If the total no of ways in which he can select at least one book is 63, find n.
a) 4 b) 3 c) 5 d) 6
actually i thought that this should be solved by this way 2^(2n+1)-1 = 63 , which gives n as 2.5. Answer is b.
2) Find the no of ways in which the letters of the word MACHINE can be arranged so that the vowels may occupy only odd positions.
a) 3!*4! b) 7P3 * 4! c) 7P4*3! d) none
my ans is 4c3*3!*4! = 4!*4!
Please explain where i went wrong. Answer is given as a
Q) a watch gains 2% per hour when the temp is in range of 40-50c and it losses at the same rate when the temp is in the range of 20-30c. however the watch owner is fortunate since it runs on time in all other temp ranges. on a sunny day the temp started soaring up from 8am at uniform rate of 2c/hr and sometime during the afternoon it stared coming down at same rate. find what time will it be by the watch at 7pm if at 8am the temp was 32c and at 4pm it was 40c.
a) 6:55pm
b)6:55:12pm
c)6:55:24pm
d)none of these
plz tell answer and give detail solution
From 4 gentleman and 4 ladies a committee of 5 is to be formed. The committee of 5 comprises of 3 secretaries, 1 president and one vice president.What will be the no of ways of selecting the committee with atleast 3 women such that atleast one woman holds the post of either a president or vice president?
sol is given as : 4c4*4c1*5!/3! + 4c2*4c3*5!/ 3! -4c2*4c3*2c2*2!
can anyone pls explain
In how many ways one white and one black rook can be placed on a chessboard so that they are never in an attacking position?
a) 64*50 b)64*49 c) 63*49 d) none
Hello guies :p
I m stuck with the following problem. Please help me.
Q: If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,
what can be the maximum number of elements in the subset.
(a) 1668 (b) 1332 (c) 1333 (d)1334
Hello guies :p
I m stuck with the following problem. Please help me.
Q: If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,
what can be the maximum number of elements in the subset.
(a) 1668 (b) 1332 (c) 1333 (d)1334
My solution:
Every no. can be expressed in the form of 9k or 9k+1 or 9k-1 or 9k+2 or 9k-2 or 9k+3 or 9k-3 or 9k+4 or 9k-4, where k is an integer.
2998 = 9 * 337 + 1
Thus, 1 to 3000 (Excluding 1 and 3000) has
334 nos. of type 9k+2 (because of no. 2999) and
333 nos. of type 9k+1, 9k, 9k+3, 9k+4, 9k-1, 9k-2, 9k-3 and 9k-4.
Now, if we chose a no. of the type 9k+2 (eg. 29), we cant chose any no. of type 9k-2 (eg. 70); else addition would be divisible by 9.
To get the max. nos. of such elements, we add 334 (Type: 9k+2) + 333 (Type: 9k+1) + 333 (Type: 9k+3) + 333 (Type: 9k+4) = 1333.
Hope it helps.
Thanks Viraj, thanks a lot.
I got the logic but u solved it in a very easy way. thanks again.... 😃
for question on burgers burger king in progressions....series is 2,9,16... remainder of 2 when divided by 7... therefore last term is 660 only... counting would be (660-2)/7 + 1 isnt it? i.e 95
as explained before... number of common burgers is 24... using a=9 and LCM(4,7).. therefore a) 660-165-95+24 and b) 24 ??
Hello guies :p
I m stuck with the following problem. Please help me.
Q: If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,
what can be the maximum number of elements in the subset.
(a) 1668 (b) 1332 (c) 1333 (d)1334
See,the methods of making 9 can be:
1+8=9 10+8=18
2+7=9 11+7=18
3+6=9 12+6=18
4+5=9 13+5=18
5+4=9 14+4=18
6+3=9 15+3=18
7+2=9 16+2=18
8+1=9 17+1=18
9+0=9 18+0=18 (Just to show elimination of 9k types)
Thus if u eliminate 5 digits out of 9,sum of 9k cannot be made....Hence left no. of digits = 4
Therefore 2997 multiplied by 4/9 = 1332
Now '1' is not included,hence subtract it from 1332
hence total 1332-1+2(for 2998,2999)=1333
My solution:
Every no. can be expressed in the form of 9k or 9k+1 or 9k-1 or 9k+2 or 9k-2 or 9k+3 or 9k-3 or 9k+4 or 9k-4, where k is an integer.
2998 = 9 * 337 + 1
Thus, 1 to 3000 (Excluding 1 and 3000) has
334 nos. of type 9k+2 (because of no. 2999) and
333 nos. of type 9k+1, 9k, 9k+3, 9k+4, 9k-1, 9k-2, 9k-3 and 9k-4.
Now, if we chose a no. of the type 9k+2 (eg. 29), we cant chose any no. of type 9k-2 (eg. 70); else addition would be divisible by 9.
To get the max. nos. of such elements, we add 334 (Type: 9k+2) + 333 (Type: 9k+1) + 333 (Type: 9k+3) + 333 (Type: 9k+4) = 1333.
Hope it helps.
Hey,tell me why here u are you not including the number 9 ??? coz if u include it,it will not make any difference as there is no other 9k type number and hence this 9 cannot combine with any other number to form a number than is a multiple of 9 ???
Please help me with the following problem:
Arun, Bikas & Chetakar have a total of 80 coins among them. Arun triples the number of coins with others by giving them
some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the
number of coins he had at the beginning?
(a) 11 (b) 10 (c) 9 (d) 12
Please help me with the following problem:
Arun, Bikas & Chetakar have a total of 80 coins among them. Arun triples the number of coins with others by giving them
some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the
number of coins he had at the beginning?
(a) 11 (b) 10 (c) 9 (d) 12
Let A,B,C be the number of coins with Arun,Bikas,Chetakar resp.
Hence initially , A+B+C=80;
After first distribution,
A-2B-2C,3B,3C are the number of coins with Arun,Bikas,Chetakar resp.
After second distribution,
3A-6B-6C,7B-2A-2c,9C are the number of coins with Arun,Bikas,Chetakar resp.
Hence we have three equations,
A+B+C=80;
A+B+C=80;
7B-2A-2C=20
Solving this you will get B = 20
Hence Bikas had 20 coins initially....
This solution looks flawless...thanks. but 20 is not in the options. I was wandering if the options are wrong!!!
arijit_medya SaysThis solution looks flawless...thanks. but 20 is not in the options. I was wandering if the options are wrong!!!
hey , the answer is 20 . but the equation given above is wrong
the equations should be
a+b+c=80
7b-2a-2c=20
multiplying the first equation by 2
gives 2a+2b+2c=160 so , we get 9b=180 therefore b=20
hope it helps
regards
Yah, how about the following...... :)
x+y+z = 80;
after first distribution: x1 + 3y + 3z = 80;
after 2nd distribution: 3x1 + y1 + 9z = 80;
now y1 = 20 given.
So from the third equation we get, 3x1 + 9z = 60 or x1 + 3z = 20; substituting this value in the
2nd equqtion gives : 3y = 60 or y = 20
hop its also correct and easy...
A set S is formed by including some of the first 1000 natural numbers where S contains the maximum if numbers satisfying the following:
1. No number of the set S is prime.
2. When two numbers of the set are selected at a time, we always see a co prime number.
What is the number of elemnts in the set S?
a) 11 (b) 12 (c) 13 (d) 7
I got the answer as 11, takin the square of the prime numbrs up to 31. The answer given in the book is 12..... Please give me a clue 
A set S is formed by including some of the first 1000 natural numbers where S contains the maximum if numbers satisfying the following:
1. No number of the set S is prime.
2. When two numbers of the set are selected at a time, we always see a co prime number.
What is the number of elemnts in the set S?
a) 11 (b) 12 (c) 13 (d) 7
I got the answer as 11, takin the square of the prime numbrs up to 31. The answer given in the book is 12..... Please give me a clue
Counted 1 or not ?
The numbers are : 1,4,9,25,49,121,169,289,361,529,841,961 (basically the squares of prime number )
Total =12 such elements
