@Aman.Malhotra said:Four identical bags are distributed among four boys. If each boy can get any number of bags then what is the probability that no boy gets more than two bags?(a)18/35(b)2/7(c)19/35(d)16/35
19/35?
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
the 3digit number should be divisible by 11
1000/11=90
100/11=9
so 90-9=81
1000/11=90
100/11=9
so 90-9=81
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
81?
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
a number written six times is always divisble by 11
as sum of terms at odd positions is equal to sum of numbers at even positions.
check the criterion for divisibility by 11 rule
so if the inserted 3 digit number is divisible by 11, the entire number is divisible by 11
such numbers are 11(10)..11(90)=81
as sum of terms at odd positions is equal to sum of numbers at even positions.
check the criterion for divisibility by 11 rule
so if the inserted 3 digit number is divisible by 11, the entire number is divisible by 11
such numbers are 11(10)..11(90)=81
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
Every Consecutive six digit no. is divisible by 11
Student entered no. after 4 but before 7. i.e 4_ _ _ 7
So digot before 4 are divisible by 11 and digit after 7 are also divisible by 11
In short find all three digit no. that are divisible by 11
990-110/11 +1 =81
Student entered no. after 4 but before 7. i.e 4_ _ _ 7
So digot before 4 are divisible by 11 and digit after 7 are also divisible by 11
In short find all three digit no. that are divisible by 11
990-110/11 +1 =81
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
Any six digit no containing same digit is divisible by 11
So, the no abc777777 will be divisible by 11
Hence, the no abc * 10^6 + 777777 will be divisible by 11
Hence, abc is divisible by 11
There are 90 - 9 = 81 three digit numbers divisible by 11
@Cat.Aspirant123 said:A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.82928182
81
number of 3 digit numbers which are divisible by 11
extension :- A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.
A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.
75
65
50
60
75
65
50
60
@Aman.Malhotra said:Four identical bags are distributed among four boys. If each boy can get any number of bags then what is the probability that no boy gets more than two bags?(a)18/35(b)2/7(c)19/35(d)16/35
a+b+c+d=4, ways 7C3=35
P(A boy gets more than 2 bags)=
Select 1 and give 3 bags, 3+k1+k2+k3=1 = 4c1*3c2=12
Select 1 and give 4 bags= 4c1=4
P(X')=(12+4)/35
P(X)=1-(16/35)=19/35
P(A boy gets more than 2 bags)=
Select 1 and give 3 bags, 3+k1+k2+k3=1 = 4c1*3c2=12
Select 1 and give 4 bags= 4c1=4
P(X')=(12+4)/35
P(X)=1-(16/35)=19/35
@jain4444 said:81 number of 3 digit numbers which are divisible by 11 extension :- A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.
No is 1234N7
4N7123
14N723
124N73
1234N7
here 123 can be arranged in 3!=6 ways and possible value of n is 81
6*81*4=1944 ??
@jain4444 bhai What is the OA ???
4N7123
14N723
124N73
1234N7
here 123 can be arranged in 3!=6 ways and possible value of n is 81
6*81*4=1944 ??
@jain4444 bhai What is the OA ???
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
(125-250/3)/(250/3)*100
=125/250*100
=50
=125/250*100
=50
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
1000 rs = 1000 apples = cost price
he receives = 1250 apples
A = number of apples pretend to sell to customer
A*83.333% = 1250
=> A = 1500
so , profit %age = 50%
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
50 % profit
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
50% profit
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
Profits while buying = 25%
Profits while selling = 1/0.833 => 20%
Overall profit => 1.25 * 1.2 = 1.5
So 50%
Profits while selling = 1/0.833 => 20%
Overall profit => 1.25 * 1.2 = 1.5
So 50%
@Cat.Aspirant123 said:A treacherous fruit vendor, while purchasing apples from the wholesaler, manages to receive 25% more apples than the quantity for which he pays. He cheats his customers as well and while selling the apples to them he delivers 16 2/3% less apples than the quantity for which the customers pay. If he claims to sell the apples at their cost price, find his net profit percentage.75655060
Profit-1 = 25%
Profit-2 = 1/(1-0.1777) = 1/0.8333 = 20%
(1+25%)(1+20%) = (1+P)
P = 50%
Using colour red blue & green, hw many diff ways can we paint edges of a regular pentagon sch tat no 2 edges meeting at a common vertex are of same colour ?
abc a three digit number such that a>= b>= >=c. hw many no feasible?
@hiteshkhurana82 said:Using colour red blue & green, hw many diff ways can we paint edges of a regular pentagon sch tat no 2 edges meeting at a common vertex are of same colour ?
198??
3^5 - 3*(1+2+3+4+5) = 198
a cube is to be painted wid 2 colours, each colour being used for 3 faces of cube. hw many diff cubes painted possible?