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31. The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.
According to question,
(10x + y) - (10y + x) = 54
10x - 10y + y - x = 54
Or, 9x - 9y = 54
Or, x - y = 6 -------(i)
Sum of digits,
(x + y) = 10 ------- (ii)
(i) - (ii)
So, x - y - x - y = 6 - 10
Or, -2y = -4
Or, y = 2 and, x = 8
Then, the required number is
= (10y + x)
= 10 × 2 + 8
= 28
Solution:
Let number be (10x + y) According to question,
(10x + y) - (10y + x) = 54
10x - 10y + y - x = 54
Or, 9x - 9y = 54
Or, x - y = 6 -------(i)
Sum of digits,
(x + y) = 10 ------- (ii)
(i) - (ii)
So, x - y - x - y = 6 - 10
Or, -2y = -4
Or, y = 2 and, x = 8
Then, the required number is
= (10y + x)
= 10 × 2 + 8
= 28
32. If A381 is divisible by 11, find the value of the smallest natural number A?
Hence, (A + 8) - (3 + 1) = 0 or multiple of 11.
To get the difference 0 or multiple of 11, we need 7 at the place of A.
So, sum of odd place - sum of even place
= 15 - 4 = 11. And this is divisible by 11.
Solution:
A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11.Hence, (A + 8) - (3 + 1) = 0 or multiple of 11.
To get the difference 0 or multiple of 11, we need 7 at the place of A.
So, sum of odd place - sum of even place
= 15 - 4 = 11. And this is divisible by 11.
33. If 381A is divisible by 9, find the value of the smallest natural number A?
So, (3 + 8 + 1 + A) = must be divisible by 9
Thus, smallest natural number be 6
Or, (3 + 8 + 1 + 6) = 18, this is divisible by 9
Solution:
A number is divisible by 9 when the sum of its digit is divisible by 9So, (3 + 8 + 1 + A) = must be divisible by 9
Thus, smallest natural number be 6
Or, (3 + 8 + 1 + 6) = 18, this is divisible by 9
34. A forester wants to plant 44 apples tree, 66 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of tree (i.e. only one type of tree in one row). The number of rows (minimum) that required is:
Solution:
35. The greatest number which will divides: 4003, 4126 and 4249, leaving the same remainder in each case:
Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.[Taken for Positive result].
Detailed Explanation:
The numbers can be written as,
4003 = AX + P where P = Remainder
4126 = BX + P
4249 = CX + P
(B - A) × X = 123
(C - B) × X = 246
Thus the X is factor of 123 and 246
Solution:
Rule- Greatest number with which if we divide P, Q, R and it leaves same remainder in each case.Number is of form = HCF of (P - Q), (P - R)Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.[Taken for Positive result].
Detailed Explanation:
The numbers can be written as,
4003 = AX + P where P = Remainder
4126 = BX + P
4249 = CX + P
(B - A) × X = 123
(C - B) × X = 246
Thus the X is factor of 123 and 246
36. Find the number of divisors of 1420.
Here, we count the number of terms in the expansion.
In the above expansion,
1420 contains number of terms = 3 × 2 × 2 = 12
Then, the number of factors or divisors = 12
Solution:
1420 = 2 × 2 × 5 × 71 = (20 + 21 + 22) × (50 + 51) × (710 + 711)Here, we count the number of terms in the expansion.
In the above expansion,
1420 contains number of terms = 3 × 2 × 2 = 12
Then, the number of factors or divisors = 12
37. The LCM of two numbers is 1890 and their HCF is 30. If one of them is 270, the other will be
Or, 30 × 1890 = 270 × N
Or, N =
Or, N = 210
Solution:
HCF of the numbers × LCM of the numbers = Multiplication of the numbersOr, 30 × 1890 = 270 × N
Or, N =
Or, N = 210
38. Find the least number of five digits which when divided by 40, 60, and 75, leave remainders 31, 51 and 66 respectively.
60 - 51 = 9
75 - 66 = 9
Difference between numbers and remainder is same in each case.
Then,
The answer = {(LCM of 40, 60, 75) - 9}
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
LCM = 2 × 2 × 2 × 5 × 5 × 3 = 600
But, the least number of 5 digits = 10000
we get remainder as 400
Then, the answer = 1000 - (600 - 400) - 9 = 10191
Solution:
Difference, 40 - 31 = 960 - 51 = 9
75 - 66 = 9
Difference between numbers and remainder is same in each case.
Then,
The answer = {(LCM of 40, 60, 75) - 9}
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
LCM = 2 × 2 × 2 × 5 × 5 × 3 = 600
But, the least number of 5 digits = 10000
we get remainder as 400
Then, the answer = 1000 - (600 - 400) - 9 = 10191
39. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed?
448 = 2 × 2 × 2 × 2 × 2 × 2 × 7
HCF = 2 × 2 × 2 × 2 × 2 × 2 = 64
Hence, number of classes required
= 9 + 7
= 16
Shortcut:
The numbers (3 × 3) and 7 are not a part of HCF. And sum of multiplication of these number is the required answer.
Solution:
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3448 = 2 × 2 × 2 × 2 × 2 × 2 × 7
HCF = 2 × 2 × 2 × 2 × 2 × 2 = 64
Hence, number of classes required
= 9 + 7
= 16
Shortcut:
The numbers (3 × 3) and 7 are not a part of HCF. And sum of multiplication of these number is the required answer.
40. Which of following can never be ending of a perfect square?
Solution:
A perfect square never ends with odd number of zeros.