Number Solved Questions For IBPS, SBI and Other Bank Exams - Page 2
Sum of seven consecutive odd numbers is 133. Find the sum of the first and the third number in the series.
Answer : b. 38
Since difference between any two successive odd numbers is 2, we can assume 7 consecutive odd numbers to be x-6,x-4,x-2,x,x+2,x+4,x+6. (In our assumed series, difference between any two numbers is 2.)
Sum of the numbers in the series = 133 = x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 7x
Therefore, x = 133 / 7 = 19
Sum of first term and last term = x - 6 + x + 6 = 2x
Substituting x = 19 (which we found earlier), we can get the answer to be 2 x 19 = 38.
Sum of six consecutive numbers that are divisible by 3 is 99. Find the sum of the first two numbers.
Answer : b. 21
Difference between successive numbers divisible by 3 will be 3. Therefore, the series takes the form as below.
Sum of all terms in the series = 99 = x-6 + x-3 + x + x+3 + x+6 + x+9 = 6x+9
6x + 9 = 99
Or 6x = 90
Or x = 15
Sum of first two numbers = x-6 + x-3 = 2x-9 = 2(15) - 9 = 21
Sum of five consecutive numbers that are divisible by 5 is 275. Find the last number in the series.
Answer : a. 65
Difference between two adjacent numbers divisible by 5 is 5.
Series takes the form x-10,x-5,x,x+5,x+10
Sum = 275 = x-10 + x-5 + x + x+5 + x+10 = 5x
5x = 275
Or x = 55
Last number in the series = x+10 = 55 + 10 = 65
Sum of 7 consecutive natural numbers is 91. Find the last but one number.
Answer : a. 15
Difference between two successive natural numbers is 1.
Therefore, series takes the form, x-3,x-2,x-1,x,x+1,x+2,x+3
Sum = 91 = x-3 + x-2 + x-1 + x + x+1 + x+2 + x+3 = 7x
7x = 91
Or x = 13
Last but one number = x+2 = 13 + 2 = 15