Number System
- The sum of all the 3-digit numbers, each of which on division by 5 leaves remainder 3, is
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According to the question,
First number = a = 103
Last number = l = 998
∴ If the number of such numbers be n, then,
998 = 103 + (n – 1) × 5
⇒ (n – 1) × 5= 998 – 103 = 895⇒ n − 1 = 895 = 179 5
⇒ n = 180∴ S = n (a + 1) 2 = 180 (103 + 998) 2
= 90 × 1101 = 99090Correct Option: D
According to the question,
First number = a = 103
Last number = l = 998
∴ If the number of such numbers be n, then,
998 = 103 + (n – 1) × 5
⇒ (n – 1) × 5= 998 – 103 = 895⇒ n − 1 = 895 = 179 5
⇒ n = 180∴ S = n (a + 1) 2 = 180 (103 + 998) 2
= 90 × 1101 = 99090
- The sum of first 50 odd natural numbers is
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S = 1 + 3 + 5 + ..... to 50 terms Here, a = 1
d = 3 – 1 = 2
n = 50∴ S = n [2a + (n − 1)d] 2 = 50 [2 × 1 + (50 − 1) × 2] 2
= 25 (2 + 98) = 25 × 100
= 2500
Correct Option: D
S = 1 + 3 + 5 + ..... to 50 terms Here, a = 1
d = 3 – 1 = 2
n = 50∴ S = n [2a + (n − 1)d] 2 = 50 [2 × 1 + (50 − 1) × 2] 2
= 25 (2 + 98) = 25 × 100
= 2500
- The sum of all the natural numbers from 51 to 100 is
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As we know that the sum of N natural Numbers 1 + 2 + 3 ........ + n
S = n ( n + 1) 2
Given in the question,
∴ 51 + 52 + ..... + 100
We can write as below
⇒ 51 + 52 + ..... + 100 = (1 + 2 + 3 + 4 + 5 + ............ + 100) – ( 1 + 2 + 3 +..........+ 50)Correct Option: D
As we know that the sum of N natural Numbers 1 + 2 + 3 ........ + n
S = n ( n + 1) 2
Given in the question,
∴ 51 + 52 + ..... + 100
We can write as below
⇒ 51 + 52 + ..... + 100 = (1 + 2 + 3 + 4 + 5 + ............ + 100) – ( 1 + 2 + 3 +..........+ 50)= 100 × 101 − 50 × 51 2 2
= 5050 – 1275 = 3775
- The sum of three consecutive numbers is 87. The middle number is
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Let us assume the three consecutive natural numbers be N, N + 1, N + 2.
According to the question,
N + N + 1 + N + 2 = 87Correct Option: B
Let us assume the three consecutive natural numbers be N, N + 1, N + 2.
According to the question,
N + N + 1 + N + 2 = 87
⇒ 3N + 3 = 87⇒ 3N = 84 ⇒ N = 84 = 28 3
∴ Middle number = 28 + 1 = 29
- Sum of three consecutive even integers is 54. Find the least among them.
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Let us assume the 3 consecutive even integers be 2A, 2A + 2 and 2A + 4 respectively.
According to given question,
∴ 2A + 2x + 2 + 2x + 4 = 54Correct Option: D
Let us assume the 3 consecutive even integers be 2A, 2A + 2 and 2A + 4 respectively.
According to given question,
∴ 2A + 2x + 2 + 2x + 4 = 54
⇒ 6A + 6 = 54
⇒ 6A = 54 – 6 = 48
⇒ A = 8
∴ The least even number = 2A = 2 × 8 = 16
