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10th Grade > Mathematics

REAL NUMBERS MCQs

Total Questions : 58 | Page 1 of 6 pages
Question 1. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
  1.    9
  2.    6
  3.    7
  4.    8
 Discuss Question
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
Question 2. Find the largest number which can divide both 324 and 144.
  1.    21
  2.    9
  3.    36
  4.    18
 Discuss Question
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
Question 3. What is the least number that must be added to 1056 so the number is divisible by 23?
  1.    1
  2.    2
  3.    3
  4.    0
 Discuss Question
Answer: Option B. -> 2
:
B
On dividing 1056 by 23, we get,
45231056103521
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
Question 4. For two numbers 96 and 404, the HCF is 4. What is the LCM of the two numbers?
  1.    9696
  2.    9698
  3.    9600
  4.    9700
 Discuss Question
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.
Question 5. If pq is a rational number with terminating decimal expansion where p and q are coprimes, then q can be represented as:
(Here, n and m are non-negative integers.)
  1.    3n5m
  2.    2n5m
  3.    7n5m
  4.    2n5m
 Discuss Question
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.
Question 6. The least multiple of 7 which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is ___.
  1.    74
  2.    94
  3.    184
  4.    364
 Discuss Question
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.
Question 7. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
  1.    4663
  2.    2244
  3.    7893
  4.    4416
 Discuss Question
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.
Question 8. If the HCF of 60 and 168 is 12, what is the LCM?
  1.    480
  2.    240
  3.    840
  4.    420
 Discuss Question
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.
Question 9. HCF of 1848, 3058 and 1331 is 
___
 Discuss Question

:
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
Question 10. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number ?
  1.    240
  2.    270
  3.    295
  4.    360
 Discuss Question
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.

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