6th Grade > Mathematics
PLAYING WITH NUMBERS MCQs
Total Questions : 100
| Page 3 of 10 pages
Answer: Option A. -> product of both numbers
:
A
Prime numbers have only two factors 1 and the number itself.
Hence,LCM of two prime numbers is always the product of the two numbers.
For example, LCM of 3 and5 = 3×5 = 15
Similarly, LCM of 7and 11 = 7×11 = 77
:
A
Prime numbers have only two factors 1 and the number itself.
Hence,LCM of two prime numbers is always the product of the two numbers.
For example, LCM of 3 and5 = 3×5 = 15
Similarly, LCM of 7and 11 = 7×11 = 77
:
Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
To check the divisibility of a number by 11, the rule is to find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is either 0 or divisible by 11, then the number is divisible by 11.
3+ 9 = 12, A + 8
A + 8 - 12 = 0
A - 4 = 0 or A -4 = 11
A = 4 or A= 15, A = 15 is not possible.
So, the number is 3498.
Question 23. Rashi is going to plant 54 oak trees and 27 pine trees. Rashi would like to plant the trees in all rows that have the same number of trees and are made up of only one type of tree. What is the greatest number of trees Rashi can have in each row? What is the total number of rows she needs to plant? [4 MARKS]
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Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
We need to understand that if Rashi has to plant the same number of trees of the same type in a row, then the number of trees in each row should be a factor of 54 and 27.
Now, the greatest common factor will give the greatest number of trees Rashi can plant in each row.
The prime factorization of the numbers is given below.
54= 2 × 3 × 3 × 3
27= 3 × 3 × 3
So, the HCF will be 3 × 3 × 3 = 27
Therefore 27 trees can be planted in each row.
Total number of rows which will be planted = 54 ÷ 27 + 27÷27 = 3
:
An odd number is anintegerwhich is not amultipleoftwo.
For example 1,3,5,7...
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Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
The LCM of 3, 4 and 5 = 3 × 4 × 5 = 60
Therefore, the required number is 2 more than 60.
Hence, the required least number = 60 + 2 = 62.
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Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
For the shortest height, we need to find the LCM of 12 and 18
The prime factorization of 12 and 18 are:
12 = 2 × 2 × 3;
18 = 2 × 3 × 3
In these prime factorizations, the maximum number of times the prime factor 2 occurs is two; this happens for 12.
Similarly, the maximum number of times the factor 3 occurs is two; this happens for 18.
The LCM of the two numbers are the product of the prime factors counted the maximum number of times they occur in any of the numbers.
Thus, in this case LCM = 2 × 2 × 3 × 3 = 36.
Hence, the shortest height for which the two stacks will have the same height is 36 inches.
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Definition: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
A number for which sum of all its factors is equal to twice the number is called a perfect number.
We have to find out if 496 is a perfect number or not.
To find if a number is a perfect number or not first we need to find out all the factors of the number.
So,
1×496=496
2×248=496
4×124=496
8×62=496
16×31=496
The various factors of 496 are 1,2,4,8,16,31,62,124,248,496.
Now sum of all the factors= 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496 = 992 = 2×496
Hence, the sum of all the factors of the of 496 is twice the number.
∴496 is a perfect number.
Answer: Option B. -> Writing a number as a product of only prime factors.
:
B
Prime factorization is finding the factors of a number that are all prime.
Eg:- 12 = 2×2×3
36 = 2 × 2×3×3 etc.
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B
Prime factorization is finding the factors of a number that are all prime.
Eg:- 12 = 2×2×3
36 = 2 × 2×3×3 etc.
Answer: Option D. -> 13
:
D
For any two numbers x and y, if x is divisibleby y, then x is also divisible by each factor of y.
For example, 12 is divisible by 6, and 2 is a factor of 6. so, 12 is also divisible by 2.
Now, factors of 26 = 1, 2, 13, 26
312 is divisible by 26, hence it is also divisible by 13.
:
D
For any two numbers x and y, if x is divisibleby y, then x is also divisible by each factor of y.
For example, 12 is divisible by 6, and 2 is a factor of 6. so, 12 is also divisible by 2.
Now, factors of 26 = 1, 2, 13, 26
312 is divisible by 26, hence it is also divisible by 13.
Answer: Option A. -> False
:
A
If a number is divisible by a number 'n', it is also divisible by the factors of 'n'.
Factors of 8 = 1, 2, 4, 8
Any number divisible by 8 will also be divisible by 4.
But vice-versa is not true. For example 20is divisible by 4 but not divisible by 8
:
A
If a number is divisible by a number 'n', it is also divisible by the factors of 'n'.
Factors of 8 = 1, 2, 4, 8
Any number divisible by 8 will also be divisible by 4.
But vice-versa is not true. For example 20is divisible by 4 but not divisible by 8