10th Grade > Mathematics
REAL NUMBERS MCQs
Total Questions : 58
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Answer: Option D. -> 8
:
D
Maximum number of columns =HCF of 616 and 32
616=23×7×11
32=25
∴ HCF of 616 and 32 = 23 =8
:
D
Maximum number of columns =HCF of 616 and 32
616=23×7×11
32=25
∴ HCF of 616 and 32 = 23 =8
Answer: Option C. -> 36
:
C
The largest number that can divide both 324 and 144 isthe HCF of both the numbers.
Prime factorising the two numbers we get,
324=2×2×3×3×3×3
144=2×2×2×2×3×3
Now, taking the common factors between them will give us the HCF.
∴ HCF =2×2×3×3=36
:
C
The largest number that can divide both 324 and 144 isthe HCF of both the numbers.
Prime factorising the two numbers we get,
324=2×2×3×3×3×3
144=2×2×2×2×3×3
Now, taking the common factors between them will give us the HCF.
∴ HCF =2×2×3×3=36
Answer: Option B. -> 2
:
B
On dividing 1056 by 23, we get,
452310561035↓21
1056 = 23 × 45 + 21
Here, the remainder is 21 and divisor is 21.
The diffrenece between Divisor and remainder = 23 - 21 = 2.
⇒ If we add2 to the dividend 1056, we will get a number completely divisible by 23.
∴ The least number to be added = 2
:
B
On dividing 1056 by 23, we get,
452310561035↓21
1056 = 23 × 45 + 21
Here, the remainder is 21 and divisor is 21.
The diffrenece between Divisor and remainder = 23 - 21 = 2.
⇒ If we add2 to the dividend 1056, we will get a number completely divisible by 23.
∴ The least number to be added = 2
Answer: Option A. -> 9696
:
A
Product of two numbers = Product of their HCF and LCM.
⇒ 96 × 404 = 4 × LCM
⇒ LCM = 96×4044
⇒ LCM = 9696
∴ HCF and LCM is 4 and 9696 respectively.
:
A
Product of two numbers = Product of their HCF and LCM.
⇒ 96 × 404 = 4 × LCM
⇒ LCM = 96×4044
⇒ LCM = 9696
∴ HCF and LCM is 4 and 9696 respectively.
Answer: Option D. -> 2n5m
:
D
If pq is a rational number with terminating decimal expansion where p and q are coprimes, then theprime factorisation of q will be in the form of 2n5m.
Example:
Consider rational number 18.
Here, denominator 8=23×50
Therefore, the rational number18 will have terminating decimal expansion.
18=0.125which is terminating.
Hence verified.
:
D
If pq is a rational number with terminating decimal expansion where p and q are coprimes, then theprime factorisation of q will be in the form of 2n5m.
Example:
Consider rational number 18.
Here, denominator 8=23×50
Therefore, the rational number18 will have terminating decimal expansion.
18=0.125which is terminating.
Hence verified.
Answer: Option D. -> 364
:
D
We first find theL.C.M of 6, 9, 15 and 18, which is90.
Forgetting remainder 4, we need to add 4 to it.
i.e.,90+4 = 94.
But, 94 is not divisible by7. So, we proceed with thenext multiple of 90, which is 180.
Adding 4 to 180, we get 184, which is also not divisible by 7.
The next multiple of 90 is270.
Adding 4 to 270, we get274, againnot divisible by 7.
The next multiple of 90 is 360.
Adding 4 to it gives us 364, divisible by 7(as 3647=52).
Hence, the answer is 364.
:
D
We first find theL.C.M of 6, 9, 15 and 18, which is90.
Forgetting remainder 4, we need to add 4 to it.
i.e.,90+4 = 94.
But, 94 is not divisible by7. So, we proceed with thenext multiple of 90, which is 180.
Adding 4 to 180, we get 184, which is also not divisible by 7.
The next multiple of 90 is270.
Adding 4 to 270, we get274, againnot divisible by 7.
The next multiple of 90 is 360.
Adding 4 to it gives us 364, divisible by 7(as 3647=52).
Hence, the answer is 364.
Answer: Option A. -> 4663
:
A
Let the required number is x.
⇒(x+17) is the smallest number divisible by 520 and 468.
⇒(x+17)=LCM(520,468)
Prime factorisation of 520 and 468:
520=23×5×13
468=22×32×13
⇒LCM=23×32×5×13=4680
Thus, x+17=4680.
⇒x=4663
Therefore, the required number is 4663.
:
A
Let the required number is x.
⇒(x+17) is the smallest number divisible by 520 and 468.
⇒(x+17)=LCM(520,468)
Prime factorisation of 520 and 468:
520=23×5×13
468=22×32×13
⇒LCM=23×32×5×13=4680
Thus, x+17=4680.
⇒x=4663
Therefore, the required number is 4663.
Answer: Option C. -> 840
:
C
Given: HCF of 60 and 168 = 12
Product of two given numbers = Product of their HCF and LCM
⇒ 60 × 168 = 12 × LCM
⇒ LCM = 60×16812
⇒ LCM = 840
∴ LCM of 60 and 168 is 840.
:
C
Given: HCF of 60 and 168 = 12
Product of two given numbers = Product of their HCF and LCM
⇒ 60 × 168 = 12 × LCM
⇒ LCM = 60×16812
⇒ LCM = 840
∴ LCM of 60 and 168 is 840.
:
Two numbers 1848 and 3058, where 3058 > 1848
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638 [Using Euclid's division algorithm to the given number 1848 and 3058]
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
Therefore HCF of 1848 and 3058 is 22.
HCF (1848 and 3058) = 22
Let us find the HCF of the numbers 1331 and 22.
1331 = 22 × 60 + 11
22 = 11 × 2 + 0
HCF of 1331 and 22 is 11
HCF (22, 1331) = 11
Hence the HCF of the three given numbers 1848, 3058 and 1331 is 11.
HCF (1848, 3058, 1331) = 11
Answer: Option B. -> 270
:
B
Let the smaller number be x.
⇒Larger number = x+1365
By Euclid's division lemma, the larger number can also be written as 6x+15.
⇒x+1365=6x+15
⇒5x=1350
⇒x=270
∴ The smaller number is 270 and the larger number is270+ 1365= 1635.
:
B
Let the smaller number be x.
⇒Larger number = x+1365
By Euclid's division lemma, the larger number can also be written as 6x+15.
⇒x+1365=6x+15
⇒5x=1350
⇒x=270
∴ The smaller number is 270 and the larger number is270+ 1365= 1635.