8th Grade > Mathematics
PLAYING WITH NUMBERS MCQs
:
A
All the even numbers and the numbers ending with 0 are divisible by 2.
Among the given options, 123123 has its last digit as 3 which makes it an odd number. Hence, this number is not divisible by 2.
:
C
The numbers 12345, 1230 and 1234560 have last digits as 5, 0 and 0 respectively. Hence, all three numbers are divisible by 5.
On the other hand, 1232 ends with a number other than 0 or 5. Hence, it is not divisible by 5.
:
A
We know that A + 2 is a two digit number with unit digit as 0. So, A = 8
A + B is a single digit number and is equal to 9.
A + B = 9
B = 9 - A
B = 9 - 8 = 1
Therefore, A = 8 and B = 1
1 2 8+ 6 8 1 8 0 9
:
100×7+10×5+6=700+50+6=756
:
C
Let the number be abc.
Then the general form of abc will be:
100×a+10×b+1×c
Thus, 123=(100×1)+(10×2)+(1×3)
:
D
1000×2+100×3+10×2+1×4=2000+300+20+4=2324.
:
A
Let's consider a two digit number ab.Its general form will be 10×a+1×b.
If we reverse the digits, its general form will be 10×b+1×a. If we subtract both of the numbers, we get:
(10×a+1×b)−(10×b+1×a)
= (10a + b) - (10b + a)
= 10a + b -10b - a
= 9a - 9b
= 9 (a-b)
Observe that the result is a multiple of 9 and hence is exactly divisible by 9.
Similarly, for 3 digit numbers, the difference of the number and number obtained by reversing the digits (if abc is the number) will be given by:
(100×a+10×b+c)−(100×c+10×b+a)
=100a + 10b + c - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99 (a-c)
= 9×11(a−c)
which is also divisible by 9.
Note that this happens with any 2 or 3 number.
:
D
In the tens place, the numbers being added are 1 and 1. The sum given in the tens place is 2, which means there is no carry over throughout the summation.
A + 3 = 5 i.e A = 5 - 3 = 2 and
2 + B = 5 i.e B = 5 - 2 = 3
Pizza 212Coke + 313 525
:
Price of one Pen =1×10+A
Number of Pens = 7
Total price =(10+A)×7
(10+A)×7=(100+A)
70 + 7A = 100 + A
6A = 30
A = 5
:
C and D
A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9 respectively.
Here, sum of digits = 3 + 9 + 2 = 14 which is not divisible by either 3 or 9. Hence, the number is not divisible by 3 and 9.
Being an even number, the given number is divisible by 2.
A number is divisible by 4 if the number formed by the last two digits are divisible by 4.
In 392, 92 which is divisible by 4. Hence, the number is also divisible by 4.