সোমবার, ১৪ সেপ্টেম্বর, ২০১৫

Important GMAT Arithmetic and Multiplicative Progressions for Bank Recruitment Exam (Part 1)

Question 1
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
  1. 897
  2. 164,850
  3. 164,749
  4. 149,700
  5. 156,720
The correct choice is (B) and the correct answer is 164,850.
Explanatory Answer
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, the given series is an AP with the first term being 101 and the last term being 998 and thhe common difference being 3.
Sum of an AP = Ap Summation Formula

We know that in an A.P., the nth term an = a1 + (n - 1)*d
In this case, therefore, 998 = 101 + (n - 1)* 3
i.e., 897 = (n - 1) * 3
Therefore, n - 1 = 299
Or n = 300.Sum of the AP will therefore, be 101+998/2 * 300 = 164,850


Question 2
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
  1. 128
  2. 142
  3. 143
  4. 141
  5. 129
The correct choice is (E) and the correct answer is 129.
Explanatory Answer
The smallest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 103.
The largest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 999.
The series of numbers that satisfy the condition that the number should leave a remainder of 5 when divided by 7 is an A.P (arithmetic progression) with the first term being 103 and the last term being 999 having a common difference of 7.
We know that in an A.P, 'l' the last term is given by l = a + (n - 1) * d, where 'a' is the first term, 'n' is the number of terms of the series and 'd' is the common difference.
Therefore, 999 = 103 + (n - 1) * 7
Or 999 - 103 = (n - 1) * 7
Or 896 = (n - 1) * 7
Or n - 1 = 128
Or n = 129

Question 3
The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m+2?
  1. m + 4
  2. n + 6
  3. n + 3
  4. m + 5
  5. n + 4
The correct choice is (E) and the correct answer is n + 4.
Explanatory Answer
The fastest way to solve problems of this kind is to take numerical examples.
The average of 5 consecutive integers from 1 to 5 is 3. Therefore, the value of m is 1 and the value of n is 3.
Now, the average of 9 consecutive integers starting from m + 2 will be average of integers from 3 to 11.
The average of numbers from 3 to 11 is 7.
Now look at the answer choices. Only choice (E) satisfies this condition.

Question 4
The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?
  1. 300
  2. 120
  3. 150
  4. 170
  5. 270
The correct choice is (C) and the correct answer is 150.
Explanatory Answer
The sum of the 4th and 12th term = 20.
Let t1 be the first term, t4 be the fourth term, and t12 be the 12th term
Then t4 + t12 = 20
t4 can be expressed as t1 + 3d
Similarly, t12 can be expressed as t1 + 11d
Then t4 + t12 = 20 can be expressed as   t1 + 3d + t1 + 11d = 20
=>      2t1 + 14d = 20
=>      t1 + 7d =10
=>      t8 = 10
The sum of the first 15 terms = 15/2 ( t1 + t15) 
In an arithmetic progression t1 + t15 = t1 + t1 + 14d = 2t1 + 14d = 2(t1 + 7d) = 2(t8).
Therefore, the sum of the first 15 terms =15/2 (2 *10) = 150 

Question 5
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?
  1. 3
  2. 1/3
  3. 2
  4. 9
  5. 1/9
The correct choice is (C) and the correct answer is 2.

Question 6
Set A contains all the even numbers between 2 and 50 inclusive. Set B contains all the even numbers between 102 and 150 inclusive. What is the difference between the sum of elements of set B and that of set A?
  1. 2500
  2. 5050
  3. 11325
  4. 6275
  5. 2550
The correct choice is (A) and the correct answer is 2500.
Explanatory Answer
SET 1 - (2, 4, 6, 8,...., 50). A total of 25 consecutive even numbers.
SET 2 - (102, 104, 106,....., 150). Another set of 25 consecutive even numbers.
Difference between 1st term of set 1 and set 2 is 100. Difference between 2nd term of set 1 and set 2 is 100 and so on.
So total difference is (100 + 100 + 100 + ....) = 25*100 = 2500.

Question 7
A set S contains the following elements: {7, 11, 15, 19, 23, x}. What is the value of x?
(1)    The elements are in arithmetic progression
(2)    x is prime

The correct choice is (C). Statements (1) and (2) Taken TOGETHER are sufficient to answer the question asked.
Explanatory Answer
Statement 1 : The elements are in arithmetic progression.
So, x could either be 3 or 27.

INSUFFICIENT.

Statement 2 : x is prime

x could be any prime number.
INSUFFICIENT
Combining the two statements, we know that x could take only two values from statement 1. viz., 3 or 27; from statement 2 we know that x is prime. 3 is the only value that satisfies both the conditions.
Hence, choice(C) is the correct answer.

Question 8
In the first 1000 natural numbers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?
  1. 11
  2. 14
  3. 12
  4. 13
  5. 10
The correct choice is (D). The correct answer is 13.
Explanatory Answer
Finding the first number that satisfies both the conditions is an iterative process.
The first number that satisfies both the conditions is 53.
Every 77th number after 53 will satisfy both the conditions.
However, after that the periodicity with which the numbers appear will be the LCM of the two divisors. The divisors in this question are 7 and 11 and their LCM is 77.
The terms that satisfy the conditions can be expressed as 77k + 53, where k takes values from 0 to 12.
i.e., a total of 13 numbers.
Choice (D) is correct.