RS Aggarwal Class 8 Math Fifth Chapter Playing with Numbers Exercise 5D Solution
EXERCISE 5D
OBJECTIVE QUESTIONS
Tick (√) the correct answer in each of the following:
(1) If 5×6 is exactly divisible by 3, then the least value of x is
Ans: (b) 1
5 + x + 6 = (11 + x) must be divisible by 3.
This happens when x = 1 or 4 or 7.
Since x is digit, it cannot be more than 9.
∴ x = 1
(2) If 64y8 is exactly by 3, then the least value of y is
Ans: (a) 0
6 + 4 + y + 8 = 18 + y
This is divisible by 3 as y is equal to 0.
(3) If 7×8 is exactly divisible by 9, then the least value of y is
Ans: (c) 3
7 + x + 8 = 15 + x
18 is divisible by 9.
Therefore, 15 + x = 18
⇒ x = 3
(4) If 37y4 is exactly divisible by 9, then the least value of y is
Ans: (d) 4
3 + 7 + y + 4 = 14 + y
∴ 14 y = 18
⇒ y = 18 – 14 = 4
(5) If 4xy7 is exactly divisible by 3, then the least value of (x + y) is
Ans: (a) 1
4 + x + y +7 = 11 + (x + y)
⇒ 11 + (x + y) = 12
⇒ (x + y) = 12 – 11 = 1
(6) If x7y5z is exactly divisible by 3, then the least value of (x + y) is
Ans: (d) 3
x + 7 + y + 5 = (x + y) + 12
This sum is divisible by 3 is x + y + 12 is 12 or 15.
∴ x + y + 12 = 12
⇒ x + y = 12 – 12 = 0
But x + y cannot be 0 because x and y will habe to be 0.
∴ x + y + 12 = 15
⇒ x + y = 15 – 12 = 3
(7) If x4y5z exactly divisible by 9, then the least value of (x + y + z) is
Ans: (c) 9
X + 4 + y + 5 + z = 9 + (x + y + z)
This equation is equal to 0 for the number x4y5z to be divisible by 9.
But x is the first digit, so it can’t be 0.
∴ x + 4 + y + 5 +z = 18
⇒ x + y + z = 18 – 9 = 9
(8) If 1A2B5 is exactly divisible by 9, then the least value of (A + B) is
Ans: (b) 1
1 +A + 2 + B + 5 = (A + B) + 8
The number is divisible by 9 is (A + B) = 1
(9) If the 4-digit number x27y is exactly divisible by 9, then the least value of 9x + y) is
Ans: (d) 9
X + 2 + 7 + y = (x + y) + 9
This sum will be divisible by 9, if (x + y) is 0.
Since, x is the first digit it can never be 0.
∴ x + y + 9 = 18
⇒ x + y = 9
Leave a Reply