Class 6 Natural and Whole Numbers

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1 ID : in-6-natural-and-whole-numbers [1] Class 6 Natural and Whole Numbers For more such worksheets visit Answer t he quest ions (1) A rectangular courtyard with length 3 m 95 cm and breadth 2 m 0 cm is to be paved with square stones of the same size. Find the least number of such stones required. (2) What is the largest number that divides 555 and 105 leaving remainder 3? (3) Divya and Rajesh are f riends and swimming coach too. Divya goes to Martyr Day school every 3 days and Rajesh goes to Martyr Day every 5 days to deliver coaching classes. If they both delivered the coaching sessions today, how many days in the next 105 days they both would take the session on the same day? (4) Two tankers contain 1176 litres and 280 litres of petrol respectively. What is the capacity of the largest measuring container which can measure the petrol of either tanker exactly? Choose correct answer(s) f rom given choice (5) The largest perf ect number between 1 and 100 is a. 28 b. 75 c. 6 d. 45 (6) In whole numbers, the associative property is satisf ied with the f ollowing operations a. Only Addition b. Only Subtraction c. Only Multiplication d. Addition and Multiplication (7) There are 196 white balls in a basket. Ismail takes out 7 white balls f rom the basket and replaces them with 14 blue erasers. He continues to do so till all white balls are replaced by blue erasers. The total number of blue erasers in the basket will be a. 28 b. 393 c. 392 d. 391 Fill in the blanks (8) 80 cola cans and 192 f ruit juice cans need to be stacked in a school canteen. If each stack is of the same height and is to contain cans of the same type, the greatest number of cans each stack can have is. (9) The largest number which divides 1809 and 588 leaving remainder 9 and 8 respectively is. (10) Find the successor of each of the f ollowing numbers :

2 A) = B) = ID : in-6-natural-and-whole-numbers [2] C) = D) = E) = F) = Check True/False (11) 0 is the smallest natural number. True False (12) The sum of two even numbers is an even number. True False (13) 1 is the smallest natural number. True False (14) Between any two whole numbers there is a whole number. True False (15) The smallest 5 digit number is the successor of the largest 4 digit number. True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

Answers ID : in-6-natural-and-whole-numbers [3] (1) 3160 According to the question, the length and breadth of the courtyard is 3 m 95 cm and 2 m 0 cm respectively.

3 Answers ID : in-6-natural-and-whole-numbers [3] (1) 3160 According to the question, the length and breadth of the courtyard is 3 m 95 cm and 2 m 0 cm respectively. Let us f irst convert each dimension into centimeters. Since we know that 1m = 100 cm, the length and breadth of the courtyard in centimeters is 395 cm and 200 cm respectively. The pavement has to be done with square stones. To perf ectly f it the courtyard, the size of the square stones should be a f actor of both the length and breadth of the courtyard. Please ref er to this f igure to see why it is so. Since we need to minimise the number of stones needed, the size of the stones should be the largest possible. This means the size of the stones should be the highest common f actor (HCF) of 395 cm and 200 cm, the length and breadth of the courtyard in centimeters. The HCF of 395 and 200 is 5. Thus, the needed size of square stones = 5 cm The number of stones needed = Area of the couryard Area of a single stone = 5 5 = 3160 Step 5 The least number of such stones required are 3160.

4 (2) 6 ID : in-6-natural-and-whole-numbers [4] We have to f ind the largest number that divides 555 and 105 leaving remainder 3. In other words, we have to f ind the largest number that divides (555-3) and (105-3) leaving no remainder. Such number is the Highest Common Factor (HCF) of : 552 [i.e., 555-3] and 102 [i.e., 105-3]. Let us now f ind the HCF of 552 and 102. All prime f actors of 552: is a factor of is a factor of is a factor of is a factor of is a factor of 23 Theref ore, 552 = All prime f actors of 102: is a factor of is a factor of is a factor of 17 Theref ore, 102 = Step 5 The HCF of 552 and 102 is = 2 3 = 6. Step 6 Thus, the largest number which divides 555 and 105 leaving remainder 3 is 6.

5 (3) 7 ID : in-6-natural-and-whole-numbers [5] Divya goes to Martyr Day every 3 rd days and Rajesh goes to Martyr Day every 5 th days. So like this they will both will take session on every 15 days. In order to f ind the number of days af ter which both of them goes together, we need to f ind a number which is as small as possible and divisible by both 3 and 5. Theref ore we need to f ind LCM of 3 and 5 which is 15 days. Now since both of them goes to school every 15 th days. Theref ore number of session in next 105 days they will take = 105 / 15 = 7

6 (4) 56 litres ID : in-6-natural-and-whole-numbers [6] The container which can measure petrol of both tanks, should be such that its volume in litres should f ully divide 1176 and 280. T heref ore, capacity of the largest measuring container which can measure the petrol of either tanker exactly is the HCF of 1176 and 280. Let us f ind all prime f actors of 1176: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of = Let us now f ind all prime f actors of 280: is a factor of is a factor of is a factor of is a factor of is a factor of = The HCF of 1176 and 280 is = = 56 Step 5 T heref ore, the largest measuring container which can measure the petrol of either tanker exactly will have a capacity of 56 liters.

7 (5) a. 28 ID : in-6-natural-and-whole-numbers [7] A perf ect number is a positive integer that is equal to the sum of its proper positive divisors, excluding the number itself. For example, all positive divisors of 6 excluding itself are 1, 2 and 3, and = 6. Theref ore, we can say that 6 is a perf ect number. All positive divisors of 28, excluding the number itself are 1, 2, 4, 7 and 14, and = 28. There is no other number between 28 and 100 that is a perf ect number. Theref ore, we can say that 28 is the largest perf ect number between 1 and 100. (6) d. Addition and Multiplication T he Associative Property states that if we are adding or multiplying three or more numbers, it does not matter where we put the parenthesis. It is applicable f or addition and multiplication. For example, 3 + (5 + 7) = (3 + 5) + 7, 3 (5 7) = (3 5) 7. Theref ore, we can say that, in whole numbers the associative property is satisf ied with Addition and Multiplication. (7) c. 392 Each time, Ismail replaces 7 white balls with 14 blue erasers. In other words, we can say that 1 white ball is replaced by Ismail with 14 7 blue erasers. Since there are 196 white balls in the basket, the number of blue erasers they will be replaced with by Ismail = = = 392 Thus, the total number of blue erasers in the basket will be: 392.

8 (8) 16 ID : in-6-natural-and-whole-numbers [8] Since each stack needs to be of the same height and is to contain cans of the same type, the number of cans in each stack should be a f actor of both 80 and 192. Since the number of cans in each stack should additionally be greatest possible, it should be the Highest Common Factor(HCF) of 80 and 192. Lets f ind the HCF of 80 and 192. All prime f actors of 80: is a factor of is a factor of is a factor of is a factor of is a factor of 5 Thus, 80 = All prime f actors of 192: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 3 Thus, 192 = Step 5 The HCF of 80 and 192 is = = 16 Step 6 The greatest number of cans each stack can have 16.

9 (9) 20 ID : in-6-natural-and-whole-numbers [9] We need to f ind the largest number that divides 1809 leaving remainder 9 and divides 588 leaving remainder 8. Such number is the H.C.F of (1809-9) and (588-8), that is, the HCF of 1800 and 580. Let us now f ind the H.C.F of 1800 and 580. All prime f actors of 1800: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 5 Thus, 1800 = Step 5 All prime f actors of 580: is a factor of is a factor of is a factor of is a factor of 29 Thus, 580 = Step 6 The H.C.F of 1800 and 580 is = = 20. Step 7 Theref ore, the largest number which divides 1809 and 588 leaving remainders 9 and 8 respectively is 20.

10 (10) A) ID : in-6-natural-and-whole-numbers [10] The successor of is = B) The successor of is = C) The successor of is = D) The successor of is = E) The successor of is = F) The successor of is = (11) False Natural numbers are the numbers used f or counting: 1, 2, is not a natural number. Theref ore, the answer is f alse.

11 (12) True ID : in-6-natural-and-whole-numbers [11] We know that every even number is divisible by 2. Consider two even numbers x and y. Since they are even, they can be written as x=2a and y=2b respectively, f or some numbers a and b. T heref ore, the sum x+y=2a+2b=2(a+b). It can be clearly seen that x+y is divisible by 2 and theref ore is even. We have seen that the sum of two even numbers is an even number. Theref ore, the answer is true. (13) True Natural numbers are the numbers used f or counting: 1, 2, 3... The smallest natural number is 1. Theref ore, the answer is true. (14) False Whole numbers are the numbers 0, 1, 2, 3,... It is not necessary that there is a whole number between any two whole numbers. For example, between the two whole numbers 2 and 3, there is no other whole number. Theref ore, the answer is f alse. (15) True The smallest 5 digit number is The largest 4 digit number is The successor of 9999 is = Theref ore, the smallest 5 digit number is the successor of the largest 4 digit number. The answer is true.

Grade 6 LCM and HCF. Answer t he quest ions. Choose correct answer(s) f rom given choice. For more such worksheets visit

Grade 6 LCM and HCF. Answer t he quest ions. Choose correct answer(s) f rom given choice. For more such worksheets visit ID : gb-6-lcm-and-hcf [1] Grade 6 LCM and HCF For more such worksheets visit www.edugain.com Answer t he quest ions (1) Ingravio-Tage is a comet that orbits around the sun once in 56 years, and Sunerva-Primo