Online Mock Test 1: Number System

1) When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

10

7

192

None of these

When a number is divided by 36 and leaves a remainder of 19, it means that the number can be expressed as 36n + 19, where n is an integer. To find the remainder when this number is divided by 12, we substitute the expression into the division: (36n + 19) ÷ 12. Simplifying this expression, we get 3n + 1 with a remainder of 7. Therefore, the correct answer is 7.

Explanation

When a number is divided by 36 and leaves a remainder of 19, it means that the number can be expressed as 36n + 19, where n is an integer. To find the remainder when this number is divided by 12, we substitute the expression into the division: (36n + 19) ÷ 12. Simplifying this expression, we get 3n + 1 with a remainder of 7. Therefore, the correct answer is 7.

2) 10²⁵ - 7 is divisible by

9

2

3

None of these

A number is divisible by 3 if the sum of its digits is divisible by 3.

10^25 - 7 can be written as 9999999999999999999999993.

The sum of the digits is 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 3 = 225.

225 is divisible by 3, so 10^25 - 7 is also divisible by 3.

Explanation

A number is divisible by 3 if the sum of its digits is divisible by 3.

10^25 - 7 can be written as 9999999999999999999999993.

The sum of the digits is 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 3 = 225.

225 is divisible by 3, so 10^25 - 7 is also divisible by 3.

3) Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.

99840

99960

99900

99990

To find the greatest number of five digits that is divisible by 7, 10, 15, 21, and 28, we need to find the least common multiple (LCM) of these numbers. The LCM of 7, 10, 15, 21, and 28 is 420, which means that any multiple of 420 is divisible by all these numbers. To find the greatest number of five digits, we need to find the largest multiple of 420 that is less than 100,000. The largest multiple of 420 that is less than 100,000 is 99,960. Therefore, the correct answer is 99960.

Explanation

To find the greatest number of five digits that is divisible by 7, 10, 15, 21, and 28, we need to find the least common multiple (LCM) of these numbers. The LCM of 7, 10, 15, 21, and 28 is 420, which means that any multiple of 420 is divisible by all these numbers. To find the greatest number of five digits, we need to find the largest multiple of 420 that is less than 100,000. The largest multiple of 420 that is less than 100,000 is 99,960. Therefore, the correct answer is 99960.

4) The sum of the first 100 numbers, 1 to 100 is divisible by

2, 4 and 8

2 and 4

2 only

None of these

The sum of the first 100 numbers can be calculated by using the formula for the sum of an arithmetic series, which is (n/2)(first term + last term). In this case, the first term is 1 and the last term is 100, so the sum is (100/2)(1 + 100) = 50(101) = 5050. Since 5050 is an even number, it is divisible by 2. However, it is not divisible by 4 or 8, as these numbers require the sum to be divisible by 2 twice and three times respectively. Therefore, the correct answer is 2 only.

Explanation

The sum of the first 100 numbers can be calculated by using the formula for the sum of an arithmetic series, which is (n/2)(first term + last term). In this case, the first term is 1 and the last term is 100, so the sum is (100/2)(1 + 100) = 50(101) = 5050. Since 5050 is an even number, it is divisible by 2. However, it is not divisible by 4 or 8, as these numbers require the sum to be divisible by 2 twice and three times respectively. Therefore, the correct answer is 2 only.

5) Find the G.C.D of 12x²y³z², 18x³y²z⁴, and 24xy⁴z³

The G.C.D (Greatest Common Divisor) is the highest power of each variable that divides all the given terms. In this case, the highest power of x that divides all the terms is x^2, the highest power of y is y^2, and the highest power of z is z^2. Therefore, the G.C.D of 12x^2y^3z^2, 18x^3y^2z^4, and 24xy^4z^3 is 1x^2y^2z^2.

Explanation

The G.C.D (Greatest Common Divisor) is the highest power of each variable that divides all the given terms. In this case, the highest power of x that divides all the terms is x^2, the highest power of y is y^2, and the highest power of z is z^2. Therefore, the G.C.D of 12x^2y^3z^2, 18x^3y^2z^4, and 24xy^4z^3 is 1x^2y^2z^2.

6) What is the value of M and N respectively? If M39048458N is divisible by 8 & 11; Where M & N are single digit integers?

7, 8

8, 6

6, 4

5, 4

The value of M and N respectively is 6 and 4. This is because for a number to be divisible by 8, the last three digits must be divisible by 8. In this case, the last three digits are 458, which is divisible by 8. Additionally, for a number to be divisible by 11, the difference between the sum of the digits in odd positions and the sum of the digits in even positions must be divisible by 11. In this case, the sum of the digits in odd positions is 3+0+4+8=15, and the sum of the digits in even positions is 9+4+5=18. The difference between these sums is 18-15=3, which is divisible by 11. Therefore, M=6 and N=4.

Explanation

The value of M and N respectively is 6 and 4. This is because for a number to be divisible by 8, the last three digits must be divisible by 8. In this case, the last three digits are 458, which is divisible by 8. Additionally, for a number to be divisible by 11, the difference between the sum of the digits in odd positions and the sum of the digits in even positions must be divisible by 11. In this case, the sum of the digits in odd positions is 3+0+4+8=15, and the sum of the digits in even positions is 9+4+5=18. The difference between these sums is 18-15=3, which is divisible by 11. Therefore, M=6 and N=4.

7) If both 11² and 3³ are factors of the number a * 4³ * 6² * 13¹¹, then what is the smallest possible value of 'a'?

121

363

3267

33

To find the smallest possible value of 'a', we need to find the common factors of 112 and 33. The prime factorization of 112 is 2^4 * 7, and the prime factorization of 33 is 3 * 11. The common factors are 1 and 11. Since 'a' needs to be a factor of the given number, it must be divisible by 11. Among the given options, 363 is the smallest number that is divisible by 11. Therefore, the smallest possible value of 'a' is 363.

Explanation

To find the smallest possible value of 'a', we need to find the common factors of 112 and 33. The prime factorization of 112 is 2^4 * 7, and the prime factorization of 33 is 3 * 11. The common factors are 1 and 11. Since 'a' needs to be a factor of the given number, it must be divisible by 11. Among the given options, 363 is the smallest number that is divisible by 11. Therefore, the smallest possible value of 'a' is 363.

8) Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

1050

540

1440

1590

Anita initially multiplied 35 with another number to get a product. However, she mistakenly used 53 instead of 35. As a result, the product increased by 540. To find the new product, we need to subtract 540 from the incorrect product she obtained. So, the correct answer is 1590.

Explanation

Anita initially multiplied 35 with another number to get a product. However, she mistakenly used 53 instead of 35. As a result, the product increased by 540. To find the new product, we need to subtract 540 from the incorrect product she obtained. So, the correct answer is 1590.

9) 48 students have to be seated such that each row has the same number of students as the others. If at least 3 students are to be seated per row and at least 2 rows have to be there, how many arrangements are possible?

4

10

8

7

In order to find the number of arrangements, we need to find the factors of 48 that are greater than or equal to 3. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. However, since we need at least 2 rows, we can eliminate the factors 1 and 48. Additionally, the number of students per row cannot exceed 16, so we can eliminate the factors 24 and 48. Thus, the possible arrangements are 2, 3, 4, 6, 8, 12, and 16, giving us a total of 7 arrangements.

Explanation

In order to find the number of arrangements, we need to find the factors of 48 that are greater than or equal to 3. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. However, since we need at least 2 rows, we can eliminate the factors 1 and 48. Additionally, the number of students per row cannot exceed 16, so we can eliminate the factors 24 and 48. Thus, the possible arrangements are 2, 3, 4, 6, 8, 12, and 16, giving us a total of 7 arrangements.

10) Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

Is even

Is odd

Is even

The given answer "z is even" cannot be true because it contradicts the statement that z is odd. Since x and y are odd and positive, their sum will always be even. Therefore, the sum of x and y cannot be odd. The only statement that cannot be true is "z is even."

Explanation

The given answer "z is even" cannot be true because it contradicts the statement that z is odd. Since x and y are odd and positive, their sum will always be even. Therefore, the sum of x and y cannot be odd. The only statement that cannot be true is "z is even."