Exams › IBPS PO › Quantitative Aptitude

Sum of 4 consecutive even + 3 consecutive odd = 175. Smallest odd = smallest even + 15. Find smallest even.

1. 18
2. 14
3. 20
4. 16

Correct answer: 16

Solution

Even: n, n+2, n+4, n+6. Sum=4n+12. Odd: n+15, n+17, n+19. Sum=3n+51. Total=7n+63=175 → n=16.

Related IBPS PO Quantitative Aptitude questions

• Set A contains 3 consecutive prime numbers while set B contains 2 consecutive odd numbers. Sum of largest number of set A and set B is 64 while largest number of set A is 4 less than the 3 times of largest number of set B. Find the smallest number of set A.
• Two-digit number xy where (10x+y) > 3(10y+x). i.e., 7x > 29y. How many such numbers satisfy an additional condition?
• 5 consecutive even numbers: first+last=24. Smallest odd=smallest_even+13. 3 consecutive odds: highest?
• Number divided by 119 leaves remainder 19. Same number divided by 17: remainder?
• Two-digit number > 3×(number with digits interchanged). Ignore numbers with 0 in units. How many such numbers?
• Directions (51-55): There are five shops P, Q, R, S and T, and they sell two different items, A and B. The following pie chart shows the total number of items sold by different shops in a particular month. Total number of items sold = 500. What is the central angle corresponding to the total number of items sold by shop S?

⚔️ Practice IBPS PO Quantitative Aptitude free + battle 1v1 →