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Dillesh4096
Bunuel
Andy solves problems 74 to 125 inclusive in a Math exercise. How many problems does he solve?

A. 53
B. 52
C. 51
D. 50
E. 49

Numbers form an AP series {74, 75, 76, . . . . . . .125}
--> Last term = a + (n - 1)d
--> 125 = 74 + (n - 1)1
--> 125 = 73 + n
--> n = 52

IMO Option B




Hi! How can we deal with such a question, to find the numbers, when it is not in an A.P? Bunuel
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Bunuel
Andy solves problems 74 to 125 inclusive in a Math exercise. How many problems does he solve?

A. 53
B. 52
C. 51
D. 50
E. 49

We are counting 74 to 125. If we subtract 73 on both ends we are counting 1 to 52 which is 52 numbers. Hence the formula for the number of consecutive terms is the range + 1.

Let's assume we were counting all the even-numbered questions between 74 and 125 (inclusive) instead, then we would start by finding the effective range first which is 74 to 124. Then we apply the same formula, but the range will be cut in half first because count only one number for every 2 numbers in this range, and finally don't forget to add 1. So (124-74)/2 + 1 = 26 would be the amount of even-numbered questions in between 74 and 125 inclusive.

Ans: B
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Most of the series on GMAT will be arithmetic. Otherwise, it could be manually counting or finding patterns. Take this question for example:

How many three-digit numbers have a tens digit of 7?

We start by writing a three-digit number, leaving the hundreds and units digit blank: _ 7 _. Notice that the hundreds digit can range from 1 to 9, and the units digit can range from 0 to 9. So that's 9 possibilities for the hundreds digits and 10 possibilities for the units digits. The seven is fixed so it's just 1 possibility, finally we multiply all those numbers to get 9*1*10 = 90 possibilities.

So there we go, that's pretty much all you need to know for GMAT in terms of counting numbers in a range! Also please see check out my 1st response as I include the general method of counting numbers within a particular range.

Roosh18
Dillesh4096
Bunuel
Andy solves problems 74 to 125 inclusive in a Math exercise. How many problems does he solve?

A. 53
B. 52
C. 51
D. 50
E. 49

Numbers form an AP series {74, 75, 76, . . . . . . .125}
--> Last term = a + (n - 1)d
--> 125 = 74 + (n - 1)1
--> 125 = 73 + n
--> n = 52

IMO Option B




Hi! How can we deal with such a question, to find the numbers, when it is not in an A.P? Bunuel
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Solution



Given:
    • Andy solves math problems from question number 74 to 125, both inclusive

To find:
    • The number of problems solved by Andy

Approach and Working Out:
    • The total number of problems solved by Andy = 125 – 73 = 52

Hence, the correct answer is Option B.

Answer: B
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For inclusive counting remember the equation:

(High - Low)/increment + 1

(125-74)/1 + 1 = 52
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Bunuel
Andy solves problems 74 to 125 inclusive in a Math exercise. How many problems does he solve?

A. 53
B. 52
C. 51
D. 50
E. 49

When we include both the first and last items, as in this question, we must add 1 to the difference.

Thus, the number of problems Andy solves is 125 - 74 + 1 = 52.

Answer: B
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