Number System
- The number 1, 2, 3, 4, ...., 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be :
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Zeros are obtained if there is any zero at the end of any multiplicand and if 5 or multiple of 5 are multiplied by any even number. i.e. (5)n (2)m has n zeros if n < m or m zeros if m < n
Now, we obtain the index of 5 as follows :index = 
1000 
+ 1000 + 1000 + 1000 5 52 53 54
= 200 + 40 + 8 + 1 = 249.
Certainly, n will be less than m.
∴ Number of zeros = 249Correct Option: D
Zeros are obtained if there is any zero at the end of any multiplicand and if 5 or multiple of 5 are multiplied by any even number. i.e. (5)n (2)m has n zeros if n < m or m zeros if m < n
Now, we obtain the index of 5 as follows :index = 1000 + 1000 + 1000 + 1000 5 52 53 54
= 200 + 40 + 8 + 1 = 249.
Certainly, n will be less than m.
∴ Number of zeros = 249
- How many numbers less than 1000 are multiples of both 10 and 13 ?
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The number of multiples of 130 are obtained by dividing 1000 by 130. The quotient i.e. 7 gives the result.
Correct Option: D
The number of multiples of 130 are obtained by dividing 1000 by 130. The quotient i.e. 7 gives the result.
- A two digit number is five times the sum of its digits. If 9 is added to the number, the digits interchange their positions. The sum of digits of the number is :
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Let the two digit number be 10y + x.
According to the question,
10y + x = 5 (x + y)
⇒ 10y + x – 5x – 5y = 0
⇒ 5y – 4x = 0 ....(i)
And,
10y + x + 9 = 10x + y
⇒ 9x – 9y = 9
⇒ x – y = 1 ... (ii)
From equation (i),
⇒ 5y – 4(1 + y) = 0
[From (ii)]
⇒ 5y – 4 – 4y = 0
⇒ y = 4
∴ From equation (ii),
x = 4 + 1 = 5
∴ Number = 10 × 4 + 5 = 45
∴ Sum of digits = 4 + 5 = 9Correct Option: B
Let the two digit number be 10y + x.
According to the question,
10y + x = 5 (x + y)
⇒ 10y + x – 5x – 5y = 0
⇒ 5y – 4x = 0 ....(i)
And,
10y + x + 9 = 10x + y
⇒ 9x – 9y = 9
⇒ x – y = 1 ... (ii)
From equation (i),
⇒ 5y – 4(1 + y) = 0
[From (ii)]
⇒ 5y – 4 – 4y = 0
⇒ y = 4
∴ From equation (ii),
x = 4 + 1 = 5
∴ Number = 10 × 4 + 5 = 45
∴ Sum of digits = 4 + 5 = 9
- Of the three numbers, the second is twice the first and it is also thrice the third. If the average of three numbers is 44, the difference of the first number and the third number is :
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Let the first number be x.
∴ Second number = 2xand third number = 2x 3 Now, x + 2x + 2x = 44 × 3 3 ⇒ 3x + 6x + 2x = 132 3
⇒ 11x = 132 × 3⇒ x = 132 × 3 = 36 11
∴ Required difference= x − 2x = x = 36 = 12 3 3 3 Correct Option: C
Let the first number be x.
∴ Second number = 2xand third number = 2x 3 Now, x + 2x + 2x = 44 × 3 3 ⇒ 3x + 6x + 2x = 132 3
⇒ 11x = 132 × 3⇒ x = 132 × 3 = 36 11
∴ Required difference= x − 2x = x = 36 = 12 3 3 3
- A number consists of two digits such that the digit in the ten’s place is less by 2 than the digit in the unit’s place. Three times the number added to 6/7 times the number obtained by reversing the digits equals 108. The sum of digits in the number is :
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Let the unit’s digit be x.
∴ Ten’s digit = x – 2
∴ Number = 10 (x – 2) + x
= 10x – 20 + x = 11x – 20
New number obtained after reversing the digits
= 10x + x – 2 = 11x – 2
According to the question3(11x − 20) + 6 (11x – 2) = 108 7 ⇒ (11x − 20) + 2 (11x – 2) = 36 7
⇒ 77x – 140 + 22x – 4 = 252
⇒ 99x = 252 + 144⇒ x = 396 = 4 99
∴ Number = 11x – 20
= 11 × 4 – 20 = 24
∴ Sum of digits = 2 + 4 = 6Correct Option: D
Let the unit’s digit be x.
∴ Ten’s digit = x – 2
∴ Number = 10 (x – 2) + x
= 10x – 20 + x = 11x – 20
New number obtained after reversing the digits
= 10x + x – 2 = 11x – 2
According to the question3(11x − 20) + 6 (11x – 2) = 108 7 ⇒ (11x − 20) + 2 (11x – 2) = 36 7
⇒ 77x – 140 + 22x – 4 = 252
⇒ 99x = 252 + 144⇒ x = 396 = 4 99
∴ Number = 11x – 20
= 11 × 4 – 20 = 24
∴ Sum of digits = 2 + 4 = 6
