Olympiad Test: Integers - Class 6 MCQ

From above number line we cans see that A is 7 marks right of 0 and B is 4 marks to the left of 0

So, A = 7 and B = −4

Hence, Option (D) is correct

Option (c) is correct; in the figure given below, each unit has a value of 5.

The given integers are 10, 0, 5, 5, 7.

- Descending order: 10, 7, 5, 5, 0

- Ascending order: 0, 5, 5, 7, 10

In both arrangements, the middle integer is 5.

Therefore, the integer that always remains in the middle of the arrangement is 5.

Option A is right 

If a number side to be an even number is must be divisible by 22. Therefore the even number between 5858 and 8080 are

60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

So, there are 1010 even numbers between 58 and 80

Starting from −4, move 7 unit right. So, from the above figure, we can conclude that integer 3 is 7 more than −4.

The number line represents all the real numbers. The positive numbers are on the right of 0 and the negative numbers are on the left of 0, as shown in the above figure.

Since 5 is a positive number, it lies to the right of 0.

Hence, on the number line, the integer 5 is located to the right of 0.

Let −a be the negative integer.
Thus its additive inverse becomes −(−a) which is equal to a.
Hence additive inverse of a negative integer is always positive.
Option B is right.

Write given integers in addition expression and we get: 

(+5) + (−6)

Subtract smaller integer 5 from bigger integer 6 as follows:

6 − 5 = 1

Since, the bigger integer −6 is a negative integer, so the result should also be negative that is −1

Hence, +5 − 6 = −1

3rd alphabet to the left of C is O.

O lies on 0 on number line. Hence, its coordinate is 0.

When a positive integer and a negative integer are added, we can get a positive or negative integer.
Therefore, statement 1 is false.

Fro example: −4 + 3 = −1 or 4 + (−3) = 1
And when two negative integers are added, we get a negative integer.
∴ Statement 2 is false.

For example: −3 + (−4) = −7

We have,

A = 8 and B = −4

The numbers on the number line increase on moving from left to right on the number line. The point B is left to the number 0, so B is less than 0.

A. − 4 > −10 ⇒ B > −10  True

B is right to −10. So, B is greater.

B. 8 > 0 ⇒ A > 0 True

A is right to 0. So, A is greater.

C. A > −4 ⇒ A > B  True

A is right to B. So, A is greater.

D. As − 4 < 0 ⇒ B < 0. So, B > 0 is false. 

0 is right to B. So, 0 is greater.

Option A is right.
The integer 5 units to the right of 0 on the number line' is 0 + 5 = +5.

To find the predecessor of a number we move from right to the left on the number line

And , −17 lies on left side of −16

Hence, −17 is the predecessor of −16

Option (B) is the correct answer

Let the number be x.

Then according to the problem we get,

x + 18 = −34

or x = −34 − 18

or x = −52.

If the magnitude of positive number is greater than the negative number then there will be some positive remainder after adding the negative number.Hence,it will be the sign of a large number.

Additive inverse of number is that when added to a number it results in 0

Additive inverse of itself = a + 0 = 0

Therefore a = 0

Hence the correct answer is option C.

This is a sample number line.

We can observe that: The numbers on the left of 0 are negative numbers and the numbers on the right of 0 are positive numbers.

Given that 

We have to find the  value of given expression

9 − (−6)

we know that (−) × (−) = +

9 + 6

= 15

So option C is correct

If we write the given numbers in words well be easily able to solve this.

One lakh two and the second number is ten lakh two.

Obviously the second number is greater then the first.

So option B is the correct answer.

The smallest positive integer is 1. Positive integers are the numbers greater than 0, and the smallest of these is 1.

Thus, the smallest positive integer is 1Thus, the smallest positive integer is 1