If x and y are two consecutive positive numbers, then is x (1) x+1 and y-1 are consecutive positive numbers.
(2) x is an even number.
Can be ans using both the statement correct or not???????????
@vbhvgupta said:Assuming sales are positive, did Spiritex Enterprises have higher sales in 1998 or in 1999? i. In 1998 the sales were twice the arithmetic mean of the sales in 1998, 1999 and 2000. ii. In 2000, the sales were three times those in 1999.Can the ans be using 1st alone????
1 alone is enough. 1988 is a typo here
@vbhvgupta said:Assuming sales are positive, did Spiritex Enterprises have higher sales in 1998 or in 1999? i. In 1998 the sales were twice the arithmetic mean of the sales in 1998, 1999 and 2000. ii. In 2000, the sales were three times those in 1999.Can the ans be using 1st alone????
yes it can be answered using 1 alone....
say sale in 2000 is 0 then sale in 1999 has to be arithmetic mean for sale of 1998 to be twice the arithmetic mean....
and the sale can never be negative...or 0 so in all the cases sale in 1998 will be greater than sale in 1999.
@bhatkushal said:what is this 140..............?
Total area of the outer strip 1m width but end corners are counted twice thats why subtracted 4
@vbhvgupta said:If x and y are two consecutive positive numbers, then is x (1) x+1 and y-1 are consecutive positive numbers. (2) x is an even number.Can be ans using both the statement correct or not???????????
@pari19O1 yar Is this correct?
@vbhvgupta said:@pari19O1 yar Is this correct?
let the numbers be x, x+1
so on adding one to x and subtracting 1 from x+1 we will get two consecutive numbers...
so x is always smaller than y for the first condition to be true...
from second, we cannot say anything about whether x is greater than y or not....
so only first...
@vbhvgupta said:If x and y are two consecutive positive numbers, then is x (1) x+1 and y-1 are consecutive positive numbers. (2) x is an even number.Can be ans using both the statement correct or not???????????
Can be answered by statement 1 alone only i guess...
What is the probability of A losing the bet?
I. The probability of throwing heads using a special coin is 2/3
II. A has bet on the event that the first 3 throws of a coin will turn heads
Is the ans shd be 2nd alone.
I. The probability of throwing heads using a special coin is 2/3
II. A has bet on the event that the first 3 throws of a coin will turn heads
Is the ans shd be 2nd alone.
@vbhvgupta said:If x and y are two consecutive positive numbers, then is x (1) x+1 and y-1 are consecutive positive numbers. (2) x is an even number.Can be ans using both the statement correct or not???????????
from statement 1
|(x+1)-(y-1)| = 1
=>|x-y+2| =1
=>(x-y)=-3 or (x-y) = -1
=>x must be less than y.
x being even or not is irrelevant.
ATDH.
@vbhvgupta said:What is the probability of A losing the bet? I. The probability of throwing heads using a special coin is 2/3 II. A has bet on the event that the first 3 throws of a coin will turn headsIs the ans shd be 2nd alone.
No it should be both together.
@vbhvgupta said:What is the probability of A losing the bet? I. The probability of throwing heads using a special coin is 2/3 II. A has bet on the event that the first 3 throws of a coin will turn headsIs the ans shd be 2nd alone.
II statement.
@vbhvgupta said:What is the probability of A losing the bet? I. The probability of throwing heads using a special coin is 2/3 II. A has bet on the event that the first 3 throws of a coin will turn headsIs the ans shd be 2nd alone.
2nd alone
@12sahil12 said:A,B,C inherited few gold coins. If they share those coins in the ratio 1:5:8, then 1 coin remains left out . But if they share the coins in the ratio 1:3:7 then no coin is left out. Comparing both the cases what is the minimum possible difference in number of coins that C can get??????Detailed Solution please!!!!!
we need to find smallest number which leaves a remainder of 1 when divided by (1+5+8) = 14 but is completely divisible by (1+3+7) =11
hence the number must be of form 14x+1 and also of form 11y. So we seek integral solutions for 14x+1 = 11y
=>y = (14x+1)/11 = {(11x+11) +(3x-10)}/11 = (x+1) + (3x-10)/11
hence (3x-10)/11 must be integer, say I
=>I = (3x-10)/11
=>x = (11*I+10)/3 = (4I+3) + (1-I)/3
=>I must be of form 3k-1
now replace I =3k+1
=>x= 4*(3k+1)+3 -k = (11k+7)
=>y = {14*(11k+7)+1}/11 = 14k+9
so the general solution for the eqn is x=11k+7 and y = 14k+9
=>no. of coins = 14*(11k+7) + 1 = 154k +99
for k=0 we get 99
Hence 99 is the answer.
if you interested in details of how to find integral solutions for eqns (ax+by=c) then you may please refer to my video (Challenge: There is small calc error in the last 2-3 mins of video, try and figure that out)
ATDH.
@vbhvgupta said:What is the probability of A losing the bet? I. The probability of throwing heads using a special coin is 2/3 II. A has bet on the event that the first 3 throws of a coin will turn headsIs the ans shd be 2nd alone.
II alone
N = 11 × 22 × 36 × 412 × 520 ..... 25 terms. Find the highest power of 75 that can divide N.
plz help..
1. 1950
2. 1330
3. 950
4. none
@badwal.aman said:N = 11 × 22 × 36 × 412 × 520 ..... 25 terms. Find the highest power of 75 that can divide N. plz help.. 1. 19502. 13303. 9504. none
it should rather be
1^1 .2^2. 3^6. 4^12.5^20 .............25^600
so in total no of 5's = 20 + 90 +210 +380 + 1200
1900
power of 75 in this = 3.5^2 = 1900/2 = 950
1^1 .2^2. 3^6. 4^12.5^20 .............25^600
so in total no of 5's = 20 + 90 +210 +380 + 1200
1900
power of 75 in this = 3.5^2 = 1900/2 = 950
@adwaitjw said:Can u please tell me why II alone?
Look for the second statement alone:
"A has bet on the event that the first 3 throws of a coin will turn heads"
=> probability=1/2*1/2*1/2=1/8
@vijay_chandola said:Look for the second statement alone:"A has bet on the event that the first 3 throws of a coin will turn heads"=> probability=1/2*1/2*1/2=1/8
But It is not mentioned normal coin... Wont that affect?
@badwal.aman said:N = 11 × 22 × 36 × 412 × 520 ..... 25 terms. Find the highest power of 75 that can divide N. plz help.. 1. 19502. 13303. 9504. none
just look for highest power of 5 and then divide by 2. (for 25).