Question

The sum of all two digit numbers each of which leaves remainder 3 when divided by 5 is

1.  1120

2.  1064

3.  999

4.  952

Answer 1

Hi there!
Here's the answer :

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¶¶¶ POINTS TO REMEMBER ¶¶¶


¶ Dividend = (Divisor × Quotient) + Remainder.

¶ In an A.P,
• Last term = First term × [(n-1)×Difference]
>>> Tn = a+(n-1)d

• Sum of the sequence = n/2(first term + last term) ,
Where n = number of terms in the sequence
>>> Sn = n/2(a+l)

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¶¶¶ SOLUTION :

The Range of Numbers = 10 to 99.

¶ The No.s exactly divisible by 5 are

10, 15, 20, 25, 30, 35, ……… 90, 95.

¶ The No.s which leaves reminder of 3.
=> 3 is to be added to each no. which are exactly divisible by 5

13, 18, 23, 28, 33, 38, ……… 93, 98.

This is a A.P.

a= 13 ; d= 5 ; l = 98 ; d = 5

• Now, To find no. of terms 'n'
Tn = a+(n-1)d

=> 98 = 13 + 5(n-1)
=> 5n - 5 = 98 - 13
=> 5n - 5 = 85
=> 5n = 90
=> n = 18

• Now, to find sum of n terms 'Sn'
Sn = (n/2)×(a+l)

=> Sn = (18/2) × (13 + 98)
=> Sn = 9(111)
=> Sn = 999

•°• Required Sum = 999.

This answer is in Option (3.)
•°• Option (3.) is the correct answer.

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:)

Hope it helps

Answer 2

Answer:

999.....................