@ishu1991 said:yes u ar ryt plss explain
gimme some time.. will upload a diagrm.. it'll be easier to look and understand
@joyjitpal said:find the sum of the following seriesS = 7/6 + 13/12 +21/20 +31/30 +.........+ 9901/9900
S=1+1/6 +1+1/12 + 1+1/20 +..... + 1+1/9900
S=1+1/2.3 +1 +1/3.4+....+1+1/99.100
S=98*1 + 1/2 -1/3 +1/3-1/4 + .....+1/99 -1/100
S=98+1/2-1/100
S=98+49/100
S=98.49
@joyjitpal said:find the sum of the following seriesS = 7/6 + 13/12 +21/20 +31/30 +.........+ 9901/9900
9849/100
@joyjitpal said:find the sum of the following seriesS = 7/6 + 13/12 +21/20 +31/30 +.........+ 9901/9900
98.49
Nth term=n2+3n+3/ n2+3+2
=1+1/n2+3n+2
=1+1/(n+1) -1/(n+2)
Sum of series till 98th term=98+1/2-1/100=98.49
@joyjitpal said:find the sum of the following seriesS = 7/6 + 13/12 +21/20 +31/30 +.........+ 9901/9900
s = 7/6 + 13/12 + 21/20 + ..........9901/9900
= 1 + 1/6 + 1 + 1/12 + ............1 + 1/9900
= 98 + 1/2 - 1/3 + .........1/99 - 1/100
= 98 + 1/2 - 1/100
= 98 + 49/100
= 98.49 ?
@saurabhlumarrai said:in an attempt to find the prime number between(and including) 101 and 1000,all the numbers from 101 to 1000 are written on a sheet of paper.the first three steps of the process involve crossing out all multiplesof 2,3 and 5.After these three steps,how many numbers are left uncrosed?a)230b)235c)240d)245
2k form -> 102,104,...,1000 = 450 numbers
3k form -> 102, 105,...,999 = 300 numbers
5k form -> 105,110,...,1000 = 180 numbers
(2*3)k form -> 102,108,...,996 = 150 numbers
(2*5)k form -> 110,120,...,1000 = 90 numbers
(3*5)k form ->105,120,...,990 = 60 numbers
(2*3*5)k form -> 120,150,...,990 = 30 numbers
hence, total multiples of 2,3 and 5 = 450+300+180-(150+90+60)+30 = 660...
numbers that are left -> 900-660 = 240....
110) What is the sum of all positive rationals p/30 (in lowest terms) which are less than 10?(a) 360 (b) 400 (c) 450 (d) none of the foregoing
colonel major and general started a work for rs 816. colonel and major did 8/17 of the total work , while major and general did 12/17 of work. what is the amount of the least efficient person?
@saurabhlumarrai said:in an attempt to find the prime number between(and including) 101 and 1000,all the numbers from 101 to 1000 are written on a sheet of paper.the first three steps of the process involve crossing out all multiplesof 2,3 and 5.After these three steps,how many numbers are left uncrosed?a)230b)235c)240d)245
multiples of 2 = 450
multiples of 3 = 300
multiples of 5 = 180
multiples of both 2 and 5 = 90
multiples of 3 and 5 = 60
multiples of 2 and 3 = 150
multiples of 2 , 3 and 5 = 30
no.s left uncrossed = 900 - 450 - 300 - 180 + 150 + 90 + 60 - 30 = 240 ?
@vbhvgupta said:colonel major and general started a work for rs 816. colonel and major did 8/17 of the total work , while major and general did 12/17 of work. what is the amount of the least efficient person?
1/C+1/M+1/G = 1
1/c+1/M = 8/17
1/m+1/g = 12/17
adding above 2 and sub in the first
1/m = 20/17-1 = 3/17
and he is the least efficient
bhak sala 
@vbhvgupta said:colonel major and general started a work for rs 816. colonel and major did 8/17 of the total work , while major and general did 12/17 of work. what is the amount of the least efficient person?
3/17?
Q.
There are two containers each having balls of two colours, green and red. The ratio of the green balls to the red balls in the first container is 2 : 3 and in the second container it is 2 : 1. N balls are randomly picked from the first container and transferred to the second container. The ratio of green balls to the red balls in the second container now becomes 1 : 1. However, the ratio of balls in the first container remains 2 : 3. If the number of balls in the first container is 25, then the value of N for which the final number of balls in the second container becomes 24 is
a)5
b)10
c)15
d)20
There are two containers each having balls of two colours, green and red. The ratio of the green balls to the red balls in the first container is 2 : 3 and in the second container it is 2 : 1. N balls are randomly picked from the first container and transferred to the second container. The ratio of green balls to the red balls in the second container now becomes 1 : 1. However, the ratio of balls in the first container remains 2 : 3. If the number of balls in the first container is 25, then the value of N for which the final number of balls in the second container becomes 24 is
a)5
b)10
c)15
d)20
@ishu1991 said:110) What is the sum of all positive rationals p/30 (in lowest terms) which are less than 10?(a) 360 (b) 400 (c) 450 (d) none of the foregoing
none of these?
1/30 + 2/30+ 3/30 + ... 299/30
1/30 (1+2+3+... 299)
1/30*((299* 300)/2)
= 299*5= 1495??
1/30 + 2/30+ 3/30 + ... 299/30
1/30 (1+2+3+... 299)
1/30*((299* 300)/2)
= 299*5= 1495??
@joyjitpal said:(990*1/2*2/3*4/5 +2 ) - (90*1/2*2/3*4/5 +2) gives 240
can u please elaborate this relationship-990*1/2*2/3*4/5...i mean to say why multiplied only with 1/2*2/3*4/5.can it be something else for different digit, like the same question changes to 2000 then what will be the multiplicative factor...??
@joyjitpal said:Q. There are two containers each having balls of two colours, green and red. The ratio of the green balls to the red balls in the first container is 2 : 3 and in the second container it is 2 : 1. N balls are randomly picked from the first container and transferred to the second container. The ratio of green balls to the red balls in the second container now becomes 1 : 1. However, the ratio of balls in the first container remains 2 : 3. If the number of balls in the first container is 25, then the value of N for which the final number of balls in the second container becomes 24 isa)5b)10c)15d)20
15??
options se karlo bhai ye wala toh...
btw joyjit bhai...V ke fan ho aap 😁 