Number System
- In a three-digit number, the digit at the hundred’s place is two times the digit at the unit’s place and the sum of the digits is 18. If the digits are reversed, the number is reduced by 396. The difference of hundred’s and ten’s digit of the number is
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Let the number be 100 (2x) + 10y + x = 201x + 10y ....(i)
∴ 2x + y + x = 18
⇒ 3x + y = 18 .....(ii)
When the digits are reversed, number
= 100(x) + 10y + 2x
= 102x + 10y .....(iii)
∴ 201x +10y – 102x – 10y = 396
⇒ 99x = 396 ⇒ x = 4
∴ From equation (i)
3 × 4 + y = 18
⇒ y = 18 – 12 = 6
∴ Required difference = 2x – y = 2 × 4 – 6 = 2Correct Option: B
Let the number be 100 (2x) + 10y + x = 201x + 10y ....(i)
∴ 2x + y + x = 18
⇒ 3x + y = 18 .....(ii)
When the digits are reversed, number
= 100(x) + 10y + 2x
= 102x + 10y .....(iii)
∴ 201x +10y – 102x – 10y = 396
⇒ 99x = 396 ⇒ x = 4
∴ From equation (i)
3 × 4 + y = 18
⇒ y = 18 – 12 = 6
∴ Required difference = 2x – y = 2 × 4 – 6 = 2
- A number consists of two digits and the digit in the ten’s place exceeds that in the unit’s place by 5. If 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is
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Let unit’s digit be x.
Ten’s digit = x + 5
Number = 10 (x + 5) + x
= 11x + 50
Again,
11x + 50 – 5 (2x + 5)
= 10x + x + 5
⇒ 11x + 50 – 10x – 25 = 11x + 5
⇒ 10x = 20 ⇒ x = 2
∴ Required sum
= 2x + 5 = 2 × 2 + 5 = 9Correct Option: C
Let unit’s digit be x.
Ten’s digit = x + 5
Number = 10 (x + 5) + x
= 11x + 50
Again,
11x + 50 – 5 (2x + 5)
= 10x + x + 5
⇒ 11x + 50 – 10x – 25 = 11x + 5
⇒ 10x = 20 ⇒ x = 2
∴ Required sum
= 2x + 5 = 2 × 2 + 5 = 9
- Of the three numbers, the sum of the first two is 55, sum of the second and third is 65 and sum of third with thrice of the first is 110. The third number is
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Let the numbers be x, y and z.
&there4 x + y = 55 ....(i)
y + z = 65 ....(ii)
3x + z = 110 ....(iii)
By equation (iii) – (ii),
3x – y = 110 – 65 = 45 ....(iv)
By equation (i) + (iv),
4x = 45 + 55 = 100
⇒ x = 25
From equation (iii),
75 + z = 110
⇒ z = 110 – 75 = 35Correct Option: C
Let the numbers be x, y and z.
&there4 x + y = 55 ....(i)
y + z = 65 ....(ii)
3x + z = 110 ....(iii)
By equation (iii) – (ii),
3x – y = 110 – 65 = 45 ....(iv)
By equation (i) + (iv),
4x = 45 + 55 = 100
⇒ x = 25
From equation (iii),
75 + z = 110
⇒ z = 110 – 75 = 35
- Instead of multiplying a number by 0.72, a student multiplied it by 7.2. If his answer was 2592 more than the correct answer, then the original number was
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Let the original number be x.
According to the question,
7.2 × x – 0.72 × x = 2592
⇒ x (7.2 – 0.72) = 2592
⇒ x × 6.48 = 2592⇒ x = 2592 6.48 = 2592 × 100 = 400 648
Correct Option: A
Let the original number be x.
According to the question,
7.2 × x – 0.72 × x = 2592
⇒ x (7.2 – 0.72) = 2592
⇒ x × 6.48 = 2592⇒ x = 2592 6.48 = 2592 × 100 = 400 648
- A man ate 100 grapes in 5 days. Each day, he ate 6 more grapes than those he ate on the earlier day. How many grapes did he eat on the first day ?
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Let number of grapes eaten on the first day be x.
∴ x + x + 6 + x + 12 + x + 18 + x + 24 = 100
⇒ 5x + 60 = 100
⇒ 5x = 100 – 60 = 40⇒ x = 40 = 8 5 Correct Option: A
Let number of grapes eaten on the first day be x.
∴ x + x + 6 + x + 12 + x + 18 + x + 24 = 100
⇒ 5x + 60 = 100
⇒ 5x = 100 – 60 = 40⇒ x = 40 = 8 5
