Integer Review: Fundamentals of Integers , Rules for Multiplication , Rules for divisibility, Division Terminology, Even & Odd Integer, Rules about even & odd integer, Prime Number, Prime Factorization , Distinct Prime Number, LCM & HCF, Composite Numbers , Consecutive Integers
Class Questions : Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5
Practice Questions : Solutions
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Q.3 Five consecutive integers have a sum of -116.
Quantity A Quantity B
Median of five integers -20
A. If Quantity A is greater
B. If Quantity B is greater
C. If Quantity A and Quantity B are equal
D. The relationship cannot be determined from given information
Solution;
Take 1st integer as ‘a’ so five Consecutive integers will be
a + a+1 + a+2 + a+3 + a+4 = -116
6a + 10 = -116
6a = – 116 – 10
6a = -126
a =
a = – 21
Median of integers will be middle number of list of Consecutive Integers, which is = a+2
So, a+2 = -21+2 = -19
As -19 is greater than -20 the correct answer choice will be ‘A’.
Q.4 x>1
If x is divisible by 15 then what is the least number by which x is divisible?
A.3
B. 2
C. 5
D. 9
E. 15
Solution;
As x is divisible by 15 that means, x can be divided by those numbers also which are factors of 15.
Factors of 15 = 3 x 5
= 15 x 1
Here the least number is 1 but the condition is already given that x > 1 so the least number will be 3.
The correct answer is ‘A’.