# Free Whole Numbers Subjective Test 02 Practice Test - 6th grade

### Question 1

Is it possible to find the largest whole number on the number line? [1 MARKS]

#### SOLUTION

Solution :

No, it is not possible to find the largest number on the number line because every number has a successor in the number line.

### Question 2

What are whole numbers? How are they different from natural numbers? Are all natural numbers whole numbers? Explain. [2 MARKS]

#### SOLUTION

Solution :Whole numbers: 1 Mark

Explanation: 1 Mark

All the counting numbers are called natural numbers (i.e. 1,2,3,4...). These natural numbers along with 0 form the group of whole numbers. All the natural numbers are also whole numbers but all the whole numbers are not natural numbers.

### Question 3

State the property of closure of whole numbers. Do all the arithmetic operations follow this property? If not, why? [2 MARKS]

#### SOLUTION

Solution :Property: 1 Mark

Reason: 1 Mark

A set of numbers is said to be closed for a specific mathematical operation if the result obtained when an operation is performed on any two numbers in the set, is itself a member of the set.

Only addition and multiplication operations on whole numbers satisfy the property of closure because when we perform this operation on whole numbers, the result also lies in the set of whole numbers.But when we perform division on whole numbers, the result may be a whole number or a fraction.

So, it doesn't satisfy the property of closure. Similarly, for subtraction of two whole numbers, the result which we get maybe a positive number or negative number which clearly does not satisfy the property of closure.

Example:-

1) 3 + 4 = 7

Here 3 and 4 are whole numbers.

The addition of 3 and 4 which is 7 is also a whole number.

So, property of closure is true for addition.

(2) 4 - 3 = 1

Here, 4 and 3 are whole numbers and 1 is also a whole number.

So the property is true.

But 3 - 4 = -1

Here 3 and 4 are whole numbers.

The subtraction of 3 and 4 is -1 which is not a whole number.

(3) 1) 30 x 7 = 210

Here 30 and 7 are whole numbers.

The multiplication of 30 and 7 which is 210 is also a whole number.

So property of closure for multiplication is true.

(4)Note : Property of closure is not always true for division.

Example :45 ÷ 0 = not defined

As division with zero is not possible.

### Question 4

Does division follow the associative law? Explain it with help of examples. [3 MARKS]

#### SOLUTION

Solution :Answer: 1 Mark

Example: 2 Marks

No, the "Associative Law" states that it doesn't matter how we group the numbers (i.e. which we calculate first) we will get same result.

a÷ (b÷c) ≠ (a÷b) ÷c(except in a few special cases)Let assume a = 48, b = 16 and c = 2

48 ÷ (16 ÷ 2) = 48 ÷ 8 = 6;

But (48 ÷ 16) ÷ 2 = 3 ÷ 2 = 1.5

This example illustrates how division doesn't follow the associative property.

Regrouping the numbers resulted in two different answers.

So division does not follow the associative law.

### Question 5

Michel owes Rs. 3, Steve owes Rs. 5 but Cooper doesn't owe anything. Each of them has Rs. 13 in their pocket presently. Place these people on the number line and find who is poorest and who is richest. [3 MARKS]

#### SOLUTION

Solution :Number Line: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Money in the pocket can be taken as positive. Then the money owed will be negative.Presently Steve has Rs.13 but he owes Rs. 5. Hence, money he actually has =13-5 = 8. The given number line shows the money Steve has.Similarly, Michel owes Rs. 3, Hence actual money Michel has =Rs. 13 - Rs. 3 = Rs. 10

Cooper has Rs.13.

As we know that the number on the left side is smaller than the number on the right side, we can hence conclude that Steve is poorest and Cooper is richest.

### Question 6

Using suitable properties, find the value of following:

Simplify:

a) 126×55+126×45

b) 89×48

c) 49×12

[3 MARKS]

#### SOLUTION

Solution :Each option: 1 Mark

By using distributive property we can simplify the problem

a×(b+c)=(a+b)×(a+c)

a×(b−c)=(a−b)×(a−c)a)126×55+126×45

126(55+45)=126×100=12600.b) 89×48=(90−1)×(50−2)

=4500−180−50+2=4272.

c) 49×12=(50−1)×(10+2)

=500+100−10−2=588.

### Question 7

A milkman sold milk at two different rates depending on how much water he added to them. One of the types in which he added no water, he sold it for **₹ **101 per kilogram. The one in which he added water he sold it for **₹ **81 per kilogram. He sold 4 kilograms of pure milk and 9 kilograms in which water was added. Find his net monthly income. [4 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given that :

The cost of pure milk =₹101 per kilogram

The cost of milk in which water was added=₹81 per kilogram

Amount of pure milk he sold = 4 kilograms

Amount of impure milk he sold = 9 kilograms

∴ His net income = 101×4 + 81×9 = (100+1)×4 + (80 + 1)×(10-1)

= 400 + 4 + 800 - 80 + 10- 1 =₹1133

### Question 8

A restaurant provides a buffet lunch and dinner for Rs. 650 & Rs. 850 respectively. For a day, the maximum handling capacity of the restaurant is 1000 people and for the lunch, it is 450. Find total income of the restaurant per day. [4 MARKS]

#### SOLUTION

Solution :Steps: 3 Marks

Answer: 1 Mark

The total capacity of restaurant = 1000.Lunch income =450 × 650

= 292500No. of people in dinner =1000 - 450 = 550

Dinner income =550 × 850

= 467500So, total income for a day = 292500 + 467500 = Rs.760000.

### Question 9

John used the calculator to find the product 1499 and 319. He multiplied 319 with the successor of 1499 instead. How much should be subtracted to get the correct answer? [4 MARKS]

#### SOLUTION

Solution :Steps: 3 Marks

Answer: 1 Mark

John had to find the product of the numbers 1499 and 319.

Instead, he multiplied 319 with the successor of 1499 which is 1500

Now,1500×319 = 1500×(300+20-1) = 450000+30000-1500=478500

1499×319 = (1500-1)×(300+20-1) = 450000 + 30000 - 1500 - 300 - 20 + 1=478181

∴ The difference = 478500-478181= 319

Hence 319 should be subtracted from product to get the correct answer.

### Question 10

Determine the sum of the four number as given below:

(a) successor of 32

(b) predecessor of 49

(c) predecessor of the predecessor of 56

(d) successor of the successor of 67.

[4 MARKS]

#### SOLUTION

Solution :Each Option: 0.5 Marks

Answer: 2 Marks

a) Successor of 32=32+1 =33

b) Predecessor of 49 = 49 -1 =48

c) Predecessor of 56 = 56 -1 =55

Predecessor of 55 = 55 -1 =54

∴ Predecessor of predecessor of 56 =54

d) Successor of 67 = 67 +1 = 68

and successor of 68 = 68 +1 = 69.

∴ Successor of successor of 67 = 69.

∴ Required sum = 33 + 48 + 54 + 69 = 204