Rs. 720 was divided among A, B, C, D, E. The sum received by them was in ascending order and in arithmetic progression. E received Rs. 40 more than A. How much did B receive?
Rs. 134
Given:-
A+B+C+D+E = Rs. 720
E - A = 40
Concept Used:-
Arithmetic Series -
a, a + d, a + 2d, a + 3d, a + 4d
nth term (Tn) = a + (n -1)d
Calculation:-
Let A receives amount a and the difference between each successive person is d rupees.
AmountE = a + 4d
According to the question,
Amount E = AmountA + 40
⇒ a + 4d - a = 40
⇒ 4d = 40
⇒ d = 10
Also,
Total amount = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)
⇒ 720 = 5a + 10d
⇒ 720 = 5a + 100
⇒ a = 124
⇒ AmountB = a + d = 124 + 10 = 134 rupees
Alternate Method
Calculation:
A, B, C, D and E
Since the received amounts are in AP,
the difference between two successive members is constant.
⇒ B – A = C – B = D – C = E – D
We have, E – A = 40,
⇒ B – A = 10, C – B = 10, D – C = 10, E – D = 10,
Let A receives x rupees,
then B, C, D and E will receive,
⇒ x + 10, x + 20, x + 30, x + 40
According to the question,
⇒ x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 720
⇒ 5x + 100 = 720
⇒ 5x = 620
⇒ x = 124
B will receive = x + 10 = 124 + 10 = 134
∴ B will receive 134 rupees
Important Questions from Integers
How many three digit whole numbers are there between 75 and 405?
Find the number of integers between $1$ and $150$ (inclusive) having $7$ as one of the digits but which are not divisible by $7$.
Integers are listed from 700 to 1000. In how many integers is the sum of the digits 10 ?
Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed ?
Consider the following statements :
1. The sum of 5 consecutive integers can be 100.
2 The product of three consecutive natural numbers can be equal to their sum.
Which of the above statements is/are correct ?
