6th Grade > Mathematics
PLAYING WITH NUMBERS MCQs
Total Questions : 100
| Page 2 of 10 pages
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Steps: 3 Marks
Answer: 1 Mark
The greatest number of 4-digits is 9999.
The prime factors of the given numbers are:
15=3×5
25=5×5
40=2×2×2×5
75=3×5×5
The maximum number of times the prime factor 2 occurs is 3.
The maximum number of times the prime factor 3 occurs is 1.
The maximum number of times the prime factor 5 occurs is 2.
∴ The LCM of the given numbers=2×2×2×3×5×5
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
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Steps: 1 Mark
Answer: 1 Mark
Oshin can distribute the chocolates by finding the factors of 36: Each of the factors will divide the chocolates in an equal number of parts, So each child will get an equal number of chocolates.
We can find the factors of 36 in the following way:
36=1×36
36=2×18
36=3×12
36=4×9
36=6×6
Stop here, because both the factors (6) are same.
Thus, the factors are 1, 2, 3, 4, 6, 9, 12, 18 and 36. So she can distribute in 9 possible sets, depending on how many children form one set.
Answer: Option C. -> 9
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C
Factors of 45 = 1, 3, 5, 9, 15, 45
Factors of 81= 1, 3,9, 27, 81
Factors of 27= 1, 3, 9, 27
∴ The common factors of 45, 81 and27 are 1, 3 and9.
So, the H.C.F of 45, 81 and27 = 9.
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C
Factors of 45 = 1, 3, 5, 9, 15, 45
Factors of 81= 1, 3,9, 27, 81
Factors of 27= 1, 3, 9, 27
∴ The common factors of 45, 81 and27 are 1, 3 and9.
So, the H.C.F of 45, 81 and27 = 9.
Answer: Option D. -> It is the greatest 4 digit number.
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D
On simplification of the given expression, we get that
3×3×11×101 = 9999.
We can observe that it is the greatest 4 digit number.
Also, we can see that this is an odd number.
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D
On simplification of the given expression, we get that
3×3×11×101 = 9999.
We can observe that it is the greatest 4 digit number.
Also, we can see that this is an odd number.
Answer: Option A. -> 90
:
A
Multiples of 90 will be 90, 180, ...... and
Multiples of 1 will be 1, 2, 3...90 and so on.
From the above, we can say that, the least common multiple of 1 and 90 will be90.
The L.C.M. of 1 and any other number, say 'x' is always 'x'.
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A
Multiples of 90 will be 90, 180, ...... and
Multiples of 1 will be 1, 2, 3...90 and so on.
From the above, we can say that, the least common multiple of 1 and 90 will be90.
The L.C.M. of 1 and any other number, say 'x' is always 'x'.
Answer: Option B. -> 1
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B
Numbers, which do not have any common factor between them other than one, are calledco-prime numbers.
For example, 3 and 7 are co-prime numbers. Theyonly have 1 as a common factor.
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B
Numbers, which do not have any common factor between them other than one, are calledco-prime numbers.
For example, 3 and 7 are co-prime numbers. Theyonly have 1 as a common factor.
Answer: Option B. -> 4
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B
It is given that LCM = 12;HCF=2
Product of 2 numbers = HCF× LCM of the respective numbers
⇒ 6 ×the other number = 12 ×2
⇒ 6 ×the other number = 24
⇒the other number = 246= 4
Hence, theother number = 4
:
B
It is given that LCM = 12;HCF=2
Product of 2 numbers = HCF× LCM of the respective numbers
⇒ 6 ×the other number = 12 ×2
⇒ 6 ×the other number = 24
⇒the other number = 246= 4
Hence, theother number = 4
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