# Quantitative Reasoning – Practice – What can you learn from this question!

PSA_LEVEL-UP_QNR18.Two consecutive numbers are removed from the progression 1,2,3, … , n. The arithmetic mean of the remaining numbers is 26¼. What is the value of n ?

A. 60
B. 81
C. 50
D. Cannot be determined

Ans: The first thing to notice is the arithmetic mean given when two of the numbers are removed.

26¼ = 105/4
We know that average multiplied by the number of numbers = Sum of all those numbers. And sum must be a natural number here. This means that whatever is the given condition the denominator ‘4’ tells that the “number of numbers” would be a multiple of 4.

Only option C. 50 suggests that when 2 numbers are removed, the “number of numbers” would be 48. No other option fulfills that.

A. 60 —> 60-2 =58 not divisible by 4.
B. 81 —> 81-2 =79 not divisible by 4.
C. 50 —> 50-2 =48 divisible by 4.
D. Cannot be determined —> Incorrect option.

What else can be learnt from the question?

(i) The sum of first ‘n’ natural numbers would be : n(n+1)/2

(ii) Average of first ‘n’ natural numbers would be : [n(n+1)/2]/n = (n+1)/2

(iii) For sometime if you assume that no numbers are removed from the series 1,2,3, … , n and the average is say 26, n would be 2*26 -1 or 51. So you can safely pick up a option close to 51 and work in a trial and error manner to understand the scenario.

(iv) Now that we know 50 is the value of n, sum of ‘n’ natural numbers = 50*51/2 = 5100/4 = 1275.
When two of the consecutive numbers are removed, the average is 105/4; the sum in this case= (105/4) * 48 = 105* 12 = 1260. Since the difference is 15, the numbers removed must be 7 & 8.