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Q1. Find whether the number 3948 is divisible by 6 or not.

Solution

A number is divisible by 6 only if it is divisible by 2 and 3. Consider the number 3948. Since its units place digit is 8, which is even. So the number 3948 is divisible by 2. Again, sum of digits = 3 + 9 + 4 + 8 = 24. Since, 24 is divisible is 3. So the number 3948 is divisible by 3. Thus, the number is divisible by both 2 and 3. Hence, 3948 is divisible by 6.

Q2. In the following addition, find A.

Solution

Consider: At one's place, A + 3 = _1 So think of a number which when added to 3 gives one's place as 1. Such a number is 8 as 8 + 3 = 11 Taking, A = 8, we obtain the addition as below:

Q3. Using the digits 4, 0 and 5, form all possible 3 digit numbers without repeating the digits. Also, classify them in a multiple of 2 and 3.

Solution

Digits are: 4, 0, 5 3 digit numbers are: 405, 450, 540, 504 A number is multiple of 2 if its unit place is even. Therefore, 450, 540 and 504 are multiples of 2. A number is multiple of 3 if the sum of digits is divisible by 3. For all the above 3 digit numbers, sum of digits is 9, which is divisible by 3. Hence all numbers are divisible by 3.

Q4. In the following product, find A, B and C.

Solution

Here, at unit's place A x 5 = _A That is we need to search for a number, which when multiplied to 5 gives the same unit place. Such a number is 5 or 0, as 5 x 5 = 25 or 0 x 5 = 0 Case (1) Taking 5 in place of A, we get Here, B can take the value 2, which satisfy the condition, taking, B = 2, we get Thus, A = 5, B = 2 and C = 1 Case (2) Taking 0 in place of A, we get Here B can take the value 5, which satisfy the condition, taking B = 5, we get Thus, A = 0, B = 5 and C =2.

Q5. Write the following numbers in expanded form.  i) 548  ii) 6985  iii) 85  iv) 356 

Solution

  i) 548 = 500 + 40 + 8 = 5 × 100 + 4 × 10 + 8 × 1 (ii) 6985 6985 =  6000 + 900 + 80 + 5 6985 = 6 × 1000 + 9 × 100 + 8 × 10 + 5 × 1 (iii) 85 85 = 80 + 5 85 = 8 × 10 + 5 × 1 (iv) 356  365 = 300 + 60 + 5 365 = 3 × 100 + 6 × 10 + 5 × 1