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MathematicsClass 6CBSE

Simplify: 18 + {1 + (5 – 3) x 5}

Given that: 18 + {1 + (5 – 3) × 5} (Using BODMAS)= 18 + {1 + 2 × 5} = 18 + {1 + 10}= 18 + 11 = 29.

MathematicsClass 6CBSE

Simplify: 32 + 96 ÷ (7 + 9)

Given that: 32 + 96 ÷ (7 + 9)= 32 + 96 ÷ 16 (Using BODMAS)= 32 + 6 = 38

MathematicsClass 6CBSE

Write pairs of twin prime numbers less than 20.

Pairs of twin prime numbers are: (3, 5), (5, 7), (11, 13), (17, 19).

MathematicsClass 6CBSE

Write first three multiples of 11.

First three multiples of 11 are:11 x 1 = 11; 11 x 2 = 22; 11 x 3 = 33Hence, the required multiples are: 11,22 and 33.

MathematicsClass 6CBSE

What are the possible factors of (a) 12 (b) 18?

(a) Possible factors of 12 are:12 = 1 × 12; 12 = 2 × 6; 12 = 3 × 4Hence, the factors of 12 are 1, 2, 3, 4, 6 and 12.(b) Possible factors of 18 are:18 = 1 × 18; 18 = 2 × 9; 18 = 3 × 6Hence, the factors of 18 are 1, 2, 3, 6, 9 and 18.

MathematicsClass 6CBSE

Write first 3 multiples of 25.

We have 25 × 1 = 25; 25 × 2 = 50; 25 × 3 = 75

MathematicsClass 6CBSE

Find the HCF of 5 and 7.

Given numbers are 5 and 7. We observe that 5 and 7 are co-prime numbers.Hence, the HCF is 1.

MathematicsClass 6CBSE

If the LCM and HCF of any two numbers are 15 and 4 respectively, find the product of the numbers.

We know that the product of the number = LCM × HCF = 15 × 4 = 60Hence, the product of the given numbers = 60

MathematicsClass 6CBSE

Which of the following numbers is divisible by 3?(a) 1212(b) 625 

(a) Given number = 1212Sum of the digits = 1 + 2 + 1 + 2 = 6, which is divisible by 3.Hence, 1212 is also divisible by 3.(b) Given number = 625Sum of the digits = 6 + 2 + 5 = 13, which is not divisible by 3.Hence, 625 is not divisible by 3.

MathematicsClass 6CBSE

What is the sum of any two?(a) even numbers(b) odd numbers

(a) The sum of any two even numbers is even.Example: 4 (even) + 6 (even) = 10 (even)(b) The sum of any two odd numbers is even.Example: 5 (odd) + 7 (odd) = 12 (even)

MathematicsClass 6CBSE

Find all the multiples of 9 upto 100.

The multiples of 9 are:9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 × 4 = 36, 9 × 5 = 45, 9 × 6 = 54, 9 × 7 = 63, 9 × 8 = 72, 9 × 9 = 81, 9 × 10 = 90, 9 × 11 = 99, 9 × 12 = 108Since 108 is greater than 100 therefore all the multiples of 9 upto 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

MathematicsClass 6CBSE

Find the H.C.F of the following numbers. 27, 63

Factors of 27 are 1, 3, 9 and 27.Factors of 63 are 1, 3, 7, 9, 21 and 63.∴ Common factors of 27 and 63 are 1, 3 and 9.Highest of these common factors is 9.∴ H.C.F. of 27 and 63 is 9.

MathematicsClass 6CBSE

Find the H.C.F of the following numbers. 70, 105, 175

Factors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70.Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.Factors of 175 are 1, 5, 7, 25, 35 and 175.∴ Common factors of 70, 105 and 175 are 1,5,7 and 35.Highest of these common factors is 35.∴ H.C.F of 70, 105 and 175 is 35.

MathematicsClass 6CBSE

List all the multiples of 7 that lie between 125 and 142

By checking all the numbers from 125 to 142 which are divisible by 7, we get 126, 133, 140.So, the multiples of 7 that lie between 125 to 142 are 126, 133 and 140.

MathematicsClass 6CBSE

Write all the factors of the following number: 27

27 = 1 × 2727 = 3 × 9Thus, all the factors of 27 are 1, 3, 9 and 27.

MathematicsClass 6CBSE

Find LCM of 60 and 40.

LCM = 23 × 3 × 5 = 120

MathematicsClass 6CBSE

A school principal places orders for 85 chairs and 25 tables with a dealer. Each chair cost ₹180 and each table cost ₹140. If the principal has given ₹2500 to the dealer as an advance money, then what amount to be given to the dealer now?

Number of chairs = 85Cost of one chair = ₹ 180Cost of 85 chairs = ₹ (85 × 180)Number of tables = 25Cost of one table = ₹ 140Cost of 25 tables = ₹ (25 × 140)Total cost of all chairs and tables = ₹(85 × 180 + 25 × 140)= ₹ (15300 + 3500) = ₹ 18800Money given in advance = ₹ 2500∴ Balance money to be paid to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300

MathematicsClass 6CBSE

A housing complex built by DLF consists of 25 large buildings and 40 small buildings. Each large building has 15 floors with 4 apartments on each floor and each small building has 9 floors with 3 apartments on each floor. How many apartments are there in all?

Number of large buildings = 25Number of floors = 15Number of apartments on each floor = 4∴ Total number of apartments in large buildings = 25 × 15 × 4Number of small building = 40Number of floors = 9Number of apartments on each floor = 3∴ Total number of apartments in small buildings = 40 × 9 × 3Hence, the number of apartments in all = 25 × 15 × 4 + 40 × 9 × 3 = 1500 + 1080 = 2580.

MathematicsClass 6CBSE

Ramesh buys 10 containers of juice from one shop and 18 containers of the same juice from another shop. If the capacity of each container is same and the cost of each of the container is ₹150, find the total money spend by Ramesh.

Ramesh buys 10 containers from one shop Cost of 1 container = ₹150He buys 18 containers of the same capacity from another shop.Cost of 1 container = ₹150∴ Total money spent by Ramesh= ₹ [10 × 150 + 18 × 150]= ₹150 × (10 + 18)= ₹150 × 28= ₹4200

MathematicsClass 6CBSE

Write 10 such numbers which can be shown only as line.

2, 5, 7, 11, 13, 17, 19, 23, 29 and 31 are such numbers which can be shown only as line.123 × 9 + 4 = 1111.

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