Olympiad Test: Integers - Free MCQ with solutions Class 6 Mathematics

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.

We are given the integers 10, 0, 4, 5, 7. We need to arrange them in both ascending order (smallest to largest) and descending order (largest to smallest). Then we must check which integer always comes in the middle.

Step 1: Arrange in ascending order
Ascending means from the smallest to the greatest: 0, 4, 5, 7, 10.
Here, the middle number is the third one (since there are 5 numbers). So, the middle is 5.

Step 2: Arrange in descending order
Descending means from the greatest to the smallest: 10, 7, 5, 4, 0.
Here again, the middle number is the third one. So, the middle is still 5.

Step 3: Compare both
Middle in ascending order = 5.
Middle in descending order = 5.
So, 5 always remains in the middle, no matter how we arrange.

Final Answer: The integer that always stays in the middle is a) 5

Option A is right 

If a number side to be an even number is must be divisible by 2. Therefore the even number between 58 and 80 are

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

So, there are 10 even numbers between 58 and 80

Starting from −4, move 7 unit right on the number line 

.
So, we can conclude that integer 3 is 7 more than −4.
Correct Answer: Option C

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.

When a positive integer and a negative integer are added, we subtract their absolute values.
The sign of the result depends on the number with the greater absolute value.
Therefore, the sum has the same sign as the larger number.

The additive inverse of a number is the number which, when added to it, gives 0.
Only 0 satisfies this condition because:

0 + 0 = 0

Hence, 0 is the only integer that is its own additive inverse.

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.

Subtracting a negative integer is the same as adding its positive value.

So,
9 − (−6) = 9 + 6 = 15

Therefore, the correct answer is 15.

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