Mathematics Paper 1 (Divisibility and Remainder) - CTET & State TET MCQ

Given: 

The divisor is 5 times the quotient

The divisor is 4 times the remainder

Remainder = 5

Formula Used:

Dividend = Divisor × Quotient + Remainder

Calculation:

Divisor = 4 × Remainder = 4 × 5 = 20

Divisor = 5 × Quotient 

⇒ Quotient = 20/5 = 4

Dividend = Divisor × Quotient + Remainder

⇒ 20 × 4 + 5 = 85

Calculation

Let the number be 893k + 193

⇒ (893k + 193)/47

⇒ 19k + 193/47

⇒ 19k + 4 + 5/47

The remainder will be 5.

Concept used:

For a number to be divided by 11 the difference of the sum of digits in odd and even places must be the multiple of either 11/0.

Calculation:

Let be the number is 07X6

⇒ (0 + X) = (7 + 6)

⇒ X = 13

∴ X = 13 - 11

⇒ X = 2

∴ The correct answer is 2.

Concept used:

Divisibility law of 6 ⇒ A number is divisible by 6 if the sum of its digit is divisible by 3 and the last digit is even.

Divisibility law of 7: Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

The divisibility rule of 13 states that:  Add the unit place digit after multiplication with 4 to the remaining number to the left of the digit at units place. If the result of the addition is divisible by 13, then the complete number is also divisible by 13.

Calculation

⇒ 10373 + 24871 = 35244

Option 1:

⇒ 35244 - 4 × 2 = 35236 ⇒ 35236 - 4 × 6 = 35212

35212 is not divisible by 7, so 35244 is also not divisible by 7.

Option 2:

last three digits of the number i.e. 244 is not divisible by 8, hence 35244 is also not divisible by 8.

Option 3:

⇒ 3 + 5 + 2 + 4 + 4 = 18

The sum of its digit is divisible by 3 and the last digit is even, hence number is divisible by 6.

Option 4:

⇒ 35244 + 4 × 4 = 3540 

3540 is not divisible by 13, so 3540 is also not divisible by 13.

The answer is option 3.

CONCEPT : 

A number is divisible by 15 if it is divisible by both 3 and 5. 

CALCULATION :

We solve this type of question by option

1025 = 1 + 0 + 2 + 5 = 8 = is not divisible by 3 but divisible by 5

2045 = 2 + 0 + 4 + 5 = 11 is not divisible by 3 but divisible by 5

2015 = 2 + 0 + 1 + 5 = 8 is not divisible by 3 but divisible by 5

1440 = 1 + 4 + 4 + 0 = 9 is divisible by 3 and 5

∴ The required answer is 1440.

Concept used

Divisibility of 9: The sum of digits of the number should be divisible by 9

Calculation

1231 = 1 + 2 + 3 + 1 = 7, not divisible by 9.

2367 = 2 + 3 + 6 + 7 = 18, divisible by 9.

6462 = 6 + 4 + 6 + 2 = 18, divisible by 9.

5354 = 5 + 3 + 5 + 4 = 17, not divisible by 9.

7020 = 7 + 0 + 2 + 0 = 9, divisible by 9.

1341 = 1 + 3 + 4 + 1 = 9, divisible by 9.

There are 4 such numbers that are divisible by 9.

Concept used

Divisibility of 9: The sum of digits of the number is divisible by 9.

Divisibility of 4: The last two digits of the number is divisible by 4.

Divisibility of 3: The sum of digits of the number is divisible by 3.

Divisibility of 13: Multiply its unit place digit by 4, then add the product obtained to the number formed by the rest of the digits of the number.

Calculation

76050 = 7 + 6 + 0 + 5 + 0 = 18

Since 18 is divisible by 9 it is also divisible by 3.

76050 is not divisible by 4 because the last two digits of the number is not divisible by 4

Calculation:

Let the two digit number be 10x + y.

Number obtained by reversing the digit 10y + x.

Now adding the numbers we get,

⇒ 10x + x + 10y + y

⇒ 11x + 11y

⇒ 11(x + y)

We can see that the result is divisible by 11.

∴ The correct answer is 11.

Given:

The largest 5-digit number that is exactly divisible by 88.

Concept used:

Reminder = dividend - quotient x divisor

Largest five-digit number = 99999

Calculation :

Divide the largest five-digit number by 88, and we got

remainder = 31

this means if the number was 31 less, it would be completely divisible by 88 After dividing 99999 with 88, we get 31 as the remainder.

so largest 5-digit number divisible by 88 = 99999-31 = 99968

Hence, '99968' is the correct answer.

Calculation:

6 is itself a composite number which is a product of two prime numbers that are 2 and 3.

So, any number that is divisible by 6 is always divisible by 2 and 3.

∴ A number that is divisible by 6 will be divisible by 2 and 3.