If the sum of one-half and one-fifth of the number exceeds one-third of that number by 7 ¹/3 , the number is

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
? n - 4n/7 = 24
? (7n - 4n)/7 = 24
? 3n/7 = 24
? 3n = 24 x 7
? n = 24 x 7/3
? n = 8 x 7
? n = 56
Sum of the digits of the number = 5 + 6 = 11

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t₁ = a + ( n - 1 ) x d
t₁ = a + (1 - 1) x d = a + 0 x d = a
Fifth term t₅ = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10
? a + a + 4d = 10
? 2a + 4d = 10
? a + 2d = 5.............................(1)

Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
? S₁₀ = 10/2[ 2a + ( 10 - 1 ) x d ]
? S₁₀ = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99
? 10a + 45d - a = 99
? 9a +45d = 99
? a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
? a + 5d - a - 2d = 11 - 5
? 3d = 6
? d = 2
Put the value of d in equation (1), we will get
? a + 2 x 2 = 5
? a + 4 = 5
? a = 5 - 4
? a = 1
Since t_n = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t₃ = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
? a + ( m - 1 ) x d = t_m
? a + ( m - 1 ) x d = 1/n
? a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for n^th term,
? a + ( n - 1 ) x d = t_n
? a + ( n - 1 ) x d = 1/m
? a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - a - md + d = 1/m - 1/n
? nd - md = 1/m - 1/n
? d(n - m) = 1/m - 1/n
? d(n - m) = (n - m)/mn
? d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
? a + md - d = 1/n
? a + ( m x 1/mn ) - 1/mn = 1/n
? a + 1/n - 1/mn = 1/n
? a = 1/n - 1/n + 1/mn
? a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
S_mn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
S_mn = mn/2 [2/mn + 1 - 1/mn ]
S_mn = mn/2 [ 1 + 1/mn ]
S_mn = mn/2 [ ( mn + 1 )/mn ]
S_mn = mn/2 (1 + mn)/mn
S_mn = 1/2 (1 + mn)
S_mn = (1 + mn)/2 = (mn + 1)/2

According to question,
First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like ?10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
? a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
? 10 + (n - 1) x 7 = 94
? 10 + 7n - 7 = 94
? 7n + 3 = 94
? 7n = 94 - 3 = 91
? n = 91/7
? n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676

According to given question,
First day 1 rs, second day 2 rs, third day 4 rs.....
1 , 2 , 4............ and so on
The given series is in geometric progression (GP).
1, 2¹, 2²,.............
Sum of first n terms in a geometric progression(GP)
S_n = a (rⁿ ? 1) /(r ?1)
if r > 1 and where a= the first term, r = common ratio,n = number of terms
As per the given question a = 1 , r = 2 and n = 20;
S_n = 1(2²⁰ ? 1) /(2 ?1)
S_n = (2²⁰ ? 1) /1
S_n = (2²⁰ ? 1)
S_n = 2²⁰ ? 1

Let us assume the salary of driver be ? R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
? Total Income during one week = R + (5R/4)
? Total Income during one week = (4R + 5R)/4 = 9R/4

Required Fraction = Tips in a week/Total Income in a week
? Required Fraction = 5R/4 / 9R/4
? Required Fraction = 5R/4 x 4/9R
? Required Fraction = 5/9

Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15

The Total number of Guests = 2a = 3b = 4c = 60

Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
? R + 10 = 3M - 30
? 3M - R = 10 + 30
? 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
? 6M - M = 170
? 5M = 170
? M = 170/5
? M = 34
Put the value of M in equation (1), we will get
? 2R - 34 = 90
? 2R = 90 + 34
? 2R = 124
? R = 124/2
? R = 62

Let us assume the numbers are a and b.
According to question,
sum of two numbers = 25
a + b = 25......................(1)
difference of two numbers = 13
a - b = 13........................(2)
add the Equation (1) and (2)
a + b+ a - b = 25 + 13
? 2a = 38
? a = 19
Put the value of a in equation in (1)
19 + b = 25
? b = 25 - 19
? b = 6
Product of the numbers = a x b
put the value of a and b,
? Product of the numbers = 19 x 6
? Product of the numbers = 114

Let us assume the numbers are a and b.
According to question,
a² - b² = 256000 .......................(1)
and a + b = 1000 .........................(2)
On dividing the Equation (1) with Equation (2),
(a² - b²)/(a + b) = 256000/1000
(a² - b²)/(a + b) = 256
(a + b)(a - b)/(a + b) = 256
a - b = 256...............................(3)
Add the equation (2) and (3), we will get
a + b + a - b = 1000 + 256
2a = 1256
a = 628
Put the value of a in equation (2), we will get
658 + b = 1000
b = 1000 - 628
b = 372
So answer is 628 , 372.