8th Grade > Mathematics
PLAYING WITH NUMBERS MCQs
Total Questions : 54
| Page 2 of 6 pages
Answer: Option A. -> 123123
:
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.
:
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.
Answer: Option D. -> 2324
:
D
1000×2+100×3+10×2+1×4=2000+300+20+4=2324.
:
D
1000×2+100×3+10×2+1×4=2000+300+20+4=2324.
Answer: Option A. -> True
:
A
Ifthe ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
For the number 244 in the ones place we find 4 which is divisible by 2. Therefore we can say that the number 244 is divisible by 2.
:
A
Ifthe ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
For the number 244 in the ones place we find 4 which is divisible by 2. Therefore we can say that the number 244 is divisible by 2.
Answer: Option D. -> A = 2, B = 3
:
D
In the tens place, the numbers being added are 1 and 1. The sum given in the tens place is2, 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
Pizza212Coke+313525
:
D
In the tens place, the numbers being added are 1 and 1. The sum given in the tens place is2, 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
Pizza212Coke+313525
Answer: Option D. -> 8
:
D
Study the addition in the ones column:
Q + 3 = 1,a number whose ones digit is 1.
If Q = 8, then 8 + 3 = 11.
So, the puzzle can be solved as shown below:
318+183501
Hence, Q = 8
:
D
Study the addition in the ones column:
Q + 3 = 1,a number whose ones digit is 1.
If Q = 8, then 8 + 3 = 11.
So, the puzzle can be solved as shown below:
318+183501
Hence, Q = 8
:
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
Answer: Option A. -> True
:
A
Let's considera two digitnumber ab.Its general form will be 10×a+1×b.
If wereverse 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 3number.
:
A
Let's considera two digitnumber ab.Its general form will be 10×a+1×b.
If wereverse 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 3number.
Answer: Option A. -> True
:
A
Let the3 digit number beabc.On rearrangingthedigits I get, abc, bac, cba, cab, acb, bca. Hence Iwill get 6 three digit numbers.
:
A
Let the3 digit number beabc.On rearrangingthedigits I get, abc, bac, cba, cab, acb, bca. Hence Iwill get 6 three digit numbers.
Answer: Option C. -> 591
:
C
On dividing 1591 by 1000, the remainder is the last 3 digits. Hence it is 591.
:
C
On dividing 1591 by 1000, the remainder is the last 3 digits. Hence it is 591.
Answer: Option C. -> fde
:
C
(f×100) + (d×10) + e is written asfde which is a three digit number.
For example, (2 × 100) + (3 × 10) + 1 is the expanded form of 231.
:
C
(f×100) + (d×10) + e is written asfde which is a three digit number.
For example, (2 × 100) + (3 × 10) + 1 is the expanded form of 231.