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The ratio between the LCM and HCF of 5,15, 20 is:- a)9 : 1
- b)4 : 3
- c)11 : 1
- d)12 : 1
Correct answer is option 'D'. Can you explain this answer?
The ratio between the LCM and HCF of 5,15, 20 is:
a)
9 : 1
b)
4 : 3
c)
11 : 1
d)
12 : 1
|
Ananya Das answered |
Factors are following:
5 = 5 x 1
15 = 5 x 3
20 = 2 x 2 x 5
5 = 5 x 1
15 = 5 x 3
20 = 2 x 2 x 5
LCM = 5 x 3 x 2 x 2 = 60
HCF = 5
Ratio = LCM/HCF = 60/5 = 12/1 = 12:1
HCF = 5
Ratio = LCM/HCF = 60/5 = 12/1 = 12:1
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√7 is- a)an integer
- b)an irrational number
- c)a rational number
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
√7 is
a)
an integer
b)
an irrational number
c)
a rational number
d)
none of these
|
Baishali Kulkarni answered |
Lets assume that √7 is rational number. ie √7=p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =a/b co- prime number
√7= a/b
a=√7b
squaring
a^2=7b^2 ....1
a� is divisible by 7
a=7c
substituting values in 1
(7c)^2=7b^2
49c^2=7b^2
7c^2=b^2
b^2=7c^2
b^2 is divisible by 7
that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.
√7 is irrational.
Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9 respectively.- a)99921
- b)99931
- c)99941
- d)99951
Correct answer is option 'C'. Can you explain this answer?
Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9 respectively.
a)
99921
b)
99931
c)
99941
d)
99951
|
|
Avinash Patel answered |
Greatest 5-Digit number = 99999
LCM of 8 and 9,
8 = 2 × 2 × 2
9 = 3 × 3
LCM = 2 × 2 × 2 × 3 × 3 = 72
Now, dividing 99999 by 72, we get
Quotient = 1388
Remainder = 63
So, the greatest 5-digit number divisible by 8 and 9 = 99999 - 63 = 99936
Required number = 99936 + 5 = 99941
If two positive integers p and q can be expressed as p = ab2 and q = a3b; where a, b being prime numbers, then LCM (p, q) is equal to- a)ab
- b)a2b2
- c)a3b2
- d)a2b3
Correct answer is option 'C'. Can you explain this answer?
If two positive integers p and q can be expressed as p = ab2 and q = a3b; where a, b being prime numbers, then LCM (p, q) is equal to
a)
ab
b)
a2b2
c)
a3b2
d)
a2b3
|
Meera Rana answered |
As per question, we have,
p = ab2 = a × b × b
q = a3b = a × a × a × b
So, their Least Common Multiple (LCM) = a3 × b2
p = ab2 = a × b × b
q = a3b = a × a × a × b
So, their Least Common Multiple (LCM) = a3 × b2
How many whole numbers are there between 42 and 62 ?
- a)20
- b)21
- c)23
- d)none of the above
Correct answer is option 'D'. Can you explain this answer?
How many whole numbers are there between 42 and 62 ?
a)
20
b)
21
c)
23
d)
none of the above
|
|
Aarav murthy answered |
Explanation:
There are infinite whole numbers between two given whole numbers. In the case of the question, the range is between 42 and 62. Therefore, there are infinitely many whole numbers between 42 and 62.
Here are some key points to consider:
Key Points:
- Whole numbers are the set of numbers that include all positive integers from 0, 1, 2, 3, ...
- The numbers between 42 and 62 are inclusive of both 42 and 62.
- Therefore, the count of whole numbers between 42 and 62 is infinite.
It is important to note that the concept of whole numbers does not have a specific endpoint, as they continue indefinitely in both directions on the number line. Thus, the correct answer is option 'D' - none of the above.
Which of the following rational numbers have a terminating decimal expansion?- a)125/441
- b)77/210
- c)15/1600
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following rational numbers have a terminating decimal expansion?
a)
125/441
b)
77/210
c)
15/1600
d)
|
Krishna Iyer answered |
The denominator 26 x 52 is of the form 2m x 5n, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.
The largest number which divides 245 and 1029 leaving remainder 5 in each case is- a)4
- b)8
- c)12
- d)16
Correct answer is option 'D'. Can you explain this answer?
The largest number which divides 245 and 1029 leaving remainder 5 in each case is
a)
4
b)
8
c)
12
d)
16
|
|
Krishna Iyer answered |
When 245 and 1029 are divided by the required number then there is left 5 as remainder. It means that 245 - 5 = 240 and 1029 - 5 = 1024 will be completely divisible by the required number.
Now we determine the HCF of 240 and 1024 by Euclid's Algorithm.
1024 = 240 x 4 + 64
240 = 64 x 3 + 48
64 = 48 x 1 + 16
48 = 16 x 3 + 0
Since remainder comes zero with last divisor 16,
required number = 16
Now we determine the HCF of 240 and 1024 by Euclid's Algorithm.
1024 = 240 x 4 + 64
240 = 64 x 3 + 48
64 = 48 x 1 + 16
48 = 16 x 3 + 0
Since remainder comes zero with last divisor 16,
required number = 16
The decimal expansion of number 
- a)terminating decimal
- b)non-terminating repeating decimal
- c)non-terminating and non-repeating decimal
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The decimal expansion of number
a)
terminating decimal
b)
non-terminating repeating decimal
c)
non-terminating and non-repeating decimal
d)
none of these
|
Naina Sharma answered |
A number with terminal decimal expansions have the denominator in the form,
2m 5n where m & n ∈ W.
The number
Which the denominator is in the form,
with m = 2 , n = 3.
Hence, it has terminal decimal expansion.
2m 5n where m & n ∈ W.
The number
Which the denominator is in the form,
with m = 2 , n = 3.Hence, it has terminal decimal expansion.
The relationship between HCF and LCM of two natural numbers is- a)HCF × LCM = Product of two natural numbers
- b)HCF × LCM = Sum of two natural numbers
- c)HCF × LCM = Difference of two natural numbers
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The relationship between HCF and LCM of two natural numbers is
a)
HCF × LCM = Product of two natural numbers
b)
HCF × LCM = Sum of two natural numbers
c)
HCF × LCM = Difference of two natural numbers
d)
None of these
|
|
Ananya Das answered |
The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.
LCM × HCF = Product of the Numbers
Suppose A and B are two numbers, then.
LCM (A & B) × HCF (A & B) = A × B
Example: Prove that: LCM (9 & 12) × HCF (9 & 12) = Product of 9 and 12.
->LCM and HCF of 9 and 12:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
LCM of 9 and 12 = 2² × 3² = 4 × 9 = 36
HCF of 9 and 12 = 3
LCM (9 & 12) × HCF (9 & 12) = 36 × 3 = 108
Product of 9 and 12 = 9 × 12 = 108
Hence, LCM (9 & 12) × HCF (9 & 12) = 108 = 9 × 12
->LCM and HCF of 9 and 12:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
LCM of 9 and 12 = 2² × 3² = 4 × 9 = 36
HCF of 9 and 12 = 3
LCM (9 & 12) × HCF (9 & 12) = 36 × 3 = 108
Product of 9 and 12 = 9 × 12 = 108
Hence, LCM (9 & 12) × HCF (9 & 12) = 108 = 9 × 12
Can you explain the answer of this question below:Find the largest number which divides 62,132,237 to leave the same reminder- A:30
- B:32
- C:35
- D:45
The answer is c.
Find the largest number which divides 62,132,237 to leave the same reminder
A:
30
B:
32
C:
35
D:
45
|
|
Anjana Khatri answered |
Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35
If two positive integers a and b are written as a = x3y2 and b = xy3 ; x, y are prime numbers, then HCF (a, b) is
- a)x4y3
- b)xy
- c)x2y3
- d) xy2
Correct answer is option 'D'. Can you explain this answer?
If two positive integers a and b are written as a = x3y2 and b = xy3 ; x, y are prime numbers, then HCF (a, b) is
a)
x4y3
b)
xy
c)
x2y3
d)
xy2
|
Diana Nair answered |
Correct option D because HCF is the highest common factor.So here in a and in b common factor are x and y and their HCF is x2y2.
The product of two consecutive integers is divisible by- a)2
- b)3
- c)5
- d)7
Correct answer is option 'A'. Can you explain this answer?
The product of two consecutive integers is divisible by
a)
2
b)
3
c)
5
d)
7
|
Raghav Bansal answered |
As the case is of 2 consecutive integers, one will be odd and the other one even,
⇒ Their product will always be even i.e. it will always be divisible by an even no., i.e. 2, here.
⇒ Their product will always be even i.e. it will always be divisible by an even no., i.e. 2, here.
What is the number x The LCM of x and 18 is 36. The HCF of x and 18 is 2.- a)2
- b)3
- c)1
- d)4
Correct answer is option 'D'. Can you explain this answer?
What is the number x The LCM of x and 18 is 36. The HCF of x and 18 is 2.
a)
2
b)
3
c)
1
d)
4
|
|
Anjana Khatri answered |
LCM * HCF = product of the integers
36 * 2 = 18 + p
36 * 2 / 18 = p
72 / 18 = p
Therefore p = 4
Hope you liked my answer. If so mark me as brainliest.
on simplification gives an irrational or a rational number.
Every positive odd integer is of the form 2q + 1, where ‘q’ is some- a)whole number
- b)natural number
- c)integer
- d)none of these
Correct answer is 'C'. Can you explain this answer?
Every positive odd integer is of the form 2q + 1, where ‘q’ is some
a)
whole number
b)
natural number
c)
integer
d)
none of these
|
|
Rohan Kapoor answered |
Let a be any positive integer and b = 2.
Applying Euclid’s algorithm, we have:
a = 2q + r, for some integer q ≥ 0, and 0 ≤ r < 2
a = 2q or 2q + 1
If a = 2q, then a is an even integer.
Now, a positive integer can either be even or odd. Thus, any positive odd integer is of the form 2q + 1.
The HCF of 2472, 1284 and a third number N is 12. If their LCM is 23 x 32 x 5 x 103 x 107, then the number N is :
- a)22 x 32 x 7
- b)22 x 33 x 103
- c)22 x 32 x 5
- d)24 x 32 x 11
Correct answer is option 'C'. Can you explain this answer?
The HCF of 2472, 1284 and a third number N is 12. If their LCM is 23 x 32 x 5 x 103 x 107, then the number N is :
a)
22 x 32 x 7
b)
22 x 33 x 103
c)
22 x 32 x 5
d)
24 x 32 x 11
|
Bhargavi Pillai answered |
Given that, LCM (2472, 1284, and n) is 23 × 32 × 5 × 103 × 107
Let us express the numbers 2472, and 1284 as a product of prime numbers.
2472 = 2 × 2 × 2 × 3 × 103 (23 × 3 × 103)
1284 = 2 × 2 × 3 × 107 (22 × 3 × 107)
HCF (2472, 1284)= 2 × 2 × 3 (22 × 3)
'n' should also have one of the factors as HCF, and another factor as the missing element of other numbers from the LCM (i.e., 3 × 5)
Therefore,
n = 22 × 3 × (3 × 5)
n = 22 × 32 × 5
n = 4 × 9 × 5
n = 180
Therefore, if the HCF of 2472 and 1284 and a third number 'n' is 12 and if their LCM is 23 × 32 × 5 × 103 × 107, then the number 'n' is 180 (22 × 32 × 5)
Which of the following is a rational number?- a)√15
- b)√9
- c)√10
- d)√12
Correct answer is option 'B'. Can you explain this answer?
Which of the following is a rational number?
a)
√15
b)
√9
c)
√10
d)
√12
|
Niharika Mehta answered |
√9 is a rational number because √9 = 3 and 3 is a rational number.
All non-terminating and non-recurring decimal numbers are- a)irrational numbers
- b)natural numbers
- c)rational numbers
- d)integers
Correct answer is option 'A'. Can you explain this answer?
All non-terminating and non-recurring decimal numbers are
a)
irrational numbers
b)
natural numbers
c)
rational numbers
d)
integers
|
|
Krishna Iyer answered |
All non-terminating and non-recurring decimal numbers are irrational numbers.
The HCF of 867 and 255 is- a)51
- b)25
- c)55
- d)35
Correct answer is option 'A'. Can you explain this answer?
The HCF of 867 and 255 is
a)
51
b)
25
c)
55
d)
35
|
|
Amit Sharma answered |
As, 867=255 × 3 +102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
So, HCF (867,255) = 51
255 = 102 × 2 + 51
102 = 51 × 2 + 0
So, HCF (867,255) = 51
If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then LCM (a, b) is- a)xy
- b)xy2
- c)x3y3
- d)x2y2
Correct answer is option 'C'. Can you explain this answer?
If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then LCM (a, b) is
a)
xy
b)
xy2
c)
x3y3
d)
x2y2
|
|
Wahid Khan answered |
A=x3y2=x(x)(x)(y)(y)
b=xy3=x(y)(y)(y)
LCM of a and b=xy2(x2×y)
=x3y3
so option C is the correct answer
b=xy3=x(y)(y)(y)
LCM of a and b=xy2(x2×y)
=x3y3
so option C is the correct answer
If ‘p’ is a prime number, then √p is- a)Irrational
- b)Rational
- c)Integer
- d)prime number
Correct answer is 'A'. Can you explain this answer?
If ‘p’ is a prime number, then √p is
a)
Irrational
b)
Rational
c)
Integer
d)
prime number
|
|
Namita rane answered |
√p is an irrational number because square root of every prime number is an irrational number.
Any ____________ is of the form 4q + 1 or 4q + 3 for some integer ‘q’.- a)composite number
- b)positive odd integer
- c)prime number
- d)positive even integer
Correct answer is option 'B'. Can you explain this answer?
Any ____________ is of the form 4q + 1 or 4q + 3 for some integer ‘q’.
a)
composite number
b)
positive odd integer
c)
prime number
d)
positive even integer
|
|
Neha Patel answered |
Let a be any positive integer
by EDL a = bq +r
0 ≤ r < b
possible remainders are 0, 1, 2 , 3
this shows that a can be in the form of 4q, 4q+1, 4q+2, 4q+3 q is quotient
as a is odd a can't be the form of 4q or 4q+2 as they are even
so a ill be in the form of 4q + 1 or 4q+3
hence proved
The HCF of two consecutive even numbers is- a)1
- b)2
- c)0
- d)3
Correct answer is 'B'. Can you explain this answer?
The HCF of two consecutive even numbers is
a)
1
b)
2
c)
0
d)
3
|
|
Rajiv Gupta answered |
Let's take two consecutive even numbers i.e, 2x and 2x+2.
Method 1 : Factorisation
Now factor both numbers
2 * x and 2 * (x+1). Therefore, the only common factor we get here is one and only 2, which turns out to be HCF itself.
Hence, HCF of two even consecutive number is 2.
Example take 10 and 12 or 4 and 6 HCF always turns out to be 2 as
10=2*5 and 12=2*6
Method 2:
If we calculate through simple division process. Let's say two consecutive even numbers are n and n+2.
If n+2 is divided by n, we get the remainder as 2.
And n is divisible by 2 as it is an even number.
Hence, HCF is 2.
If 112 = q×6+r, then the possible values of r are:
- A:2, 3, 5
- B:0, 1, 2, 3, 4, 5
- C:1, 2, 3, 4
- D:0, 1, 2, 3
The answer is b.
A:
2, 3, 5
B:
0, 1, 2, 3, 4, 5
C:
1, 2, 3, 4
D:
0, 1, 2, 3
|
|
Ananya Das answered |
For the relation x = qy+r, 0 ⩽ r < y So, here r lies between 0 ⩽ r < 6. Hence r = 0, 1, 2, 3, 4, 5
n2 - 1 is divisible by 8 if n is
- a)An integer
- b)A natural number
- c)An odd integer
- d)An even integer
Correct answer is option 'C'. Can you explain this answer?
n2 - 1 is divisible by 8 if n is
a)
An integer
b)
A natural number
c)
An odd integer
d)
An even integer
|
Pooja Shah answered |
Option A: If n is integer similarly n²- 1 is not divisible by 8.
Option B: If n is a natural number then it is not possible for n=1,2.
Option C: If n is odd the possible for n is 3,5,7...
Option D: If n is even then it is not possible for n=2.
Option B: If n is a natural number then it is not possible for n=1,2.
Option C: If n is odd the possible for n is 3,5,7...
Option D: If n is even then it is not possible for n=2.
Thus, the option C is correct.
The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is- a)900
- b)1200
- c)2500
- d)3600
Correct answer is option 'D'. Can you explain this answer?
The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is
a)
900
b)
1200
c)
2500
d)
3600
|
|
Naina Sharma answered |
L.C.M. of 3, 4, 5, 6, 8 = 2 × 2 × 2 × 3 × 5 = 120
Pair of 2, 3 and 5 is not completed.
To make it a perfect square, the number should be multiplied by 2, 3, 5.
Pair of 2, 3 and 5 is not completed.
To make it a perfect square, the number should be multiplied by 2, 3, 5.
Required number = 120 x 2 x 3 x 5 = 3600.
Two natural numbers whose difference is 66 and the least common multiple is 360, are:
- a)120 and 54
- b)90 and 24
- c)180 and 114
- d)130 and 64
Correct answer is option 'B'. Can you explain this answer?
Two natural numbers whose difference is 66 and the least common multiple is 360, are:
a)
120 and 54
b)
90 and 24
c)
180 and 114
d)
130 and 64
|
|
Aditya Shah answered |
Let the two numbers be x and x+66.
Since, the LCM is 360, there must be 5,2 and 3 as the prime factors of the two numbers.
So, one pair is 24 & 90, because 90−24=66.
Also, 24=2×2×2×3
90=2×3×3×5
∴ LCM=2×2×2×3×3×5=360.
So, the numbers are 24 and 90.
The HCF of two consecutive numbers is- a)2
- b)3
- c)0
- d)1
Correct answer is option 'D'. Can you explain this answer?
The HCF of two consecutive numbers is
a)
2
b)
3
c)
0
d)
1
|
Akshara Basu answered |
The HCF of two consecutive numbers is always 1.
- a)m = 2 and n = 5
- b)m = 4 and n = 5
- c)m = 5 and n = 3
- d)m = 3 and n = 2
Correct answer is option 'C'. Can you explain this answer?
a)
m = 2 and n = 5
b)
m = 4 and n = 5
c)
m = 5 and n = 3
d)
m = 3 and n = 2
|
Anisha Mukherjee answered |
Comparing the denominators of both fractions, we have m = 5 and n = 3
Can you explain the answer of this question below:The product of two numbers is -20/9. If one of the numbers is 4, find the other. - A:–5/9
- B:3/11
- C:12/39
- D:–9/11
The answer is a.
The product of two numbers is -20/9. If one of the numbers is 4, find the other.
A:
–5/9
B:
3/11
C:
12/39
D:
–9/11
|
|
Ananya Das answered |
We have two numbers such that their product is equal to -20/9.
So we have x*y=-20/9
One no. is given 4, so
x*4=-20/9
x=-20/9 = 4
x=-20/9 x 1/4=-5/9
So we have x*y=-20/9
One no. is given 4, so
x*4=-20/9
x=-20/9 = 4
x=-20/9 x 1/4=-5/9
If ‘a’ and ‘b’ are both positive rational numbers, then 
- a)neither rational nor rational number
- b)an irrational number
- c)a rational number
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If ‘a’ and ‘b’ are both positive rational numbers, then
a)
neither rational nor rational number
b)
an irrational number
c)
a rational number
d)
none of these
|
|
Avinash Patel answered |
= (a−b) Since a and b both are positive rational numbers, therefore difference of two positive rational numbers is also rational.
If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is- a)150
- b)90
- c)900
- d)1800
Correct answer is option 'A'. Can you explain this answer?
If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is
a)
150
b)
90
c)
900
d)
1800
|
|
Anisha Mukherjee answered |
Using the result, HCF × LCM = Product of two natural numbers ⇒ LCM (a, b) = 1800/12 = 150
The HCF of two consecutive numbers is- a)2
- b)3
- c)0
- d)1
Correct answer is 'D'. Can you explain this answer?
The HCF of two consecutive numbers is
a)
2
b)
3
c)
0
d)
1
|
Anshu Shah answered |
The HCF of two consecutive numbers is always 1.
If two positive integers a and b are written as a and x3y2 and b = xy3, where x, y are prime numbers, then HCF(a, b) is- a)xy
- b)xy2
- c)x3y3
- d)x2y2
Correct answer is option 'C'. Can you explain this answer?
If two positive integers a and b are written as a and x3y2 and b = xy3, where x, y are prime numbers, then HCF(a, b) is
a)
xy
b)
xy2
c)
x3y3
d)
x2y2
|
Gaurav Kumar answered |
Here, a = x3y2 and b = xy3.
⇒ a = x * x * x * y * y and b = xy * y * y
∴ LCM(a, b) = x * x * x * y * y * y = x3 * y3 = x3y3
LCM = x3y3
⇒ a = x * x * x * y * y and b = xy * y * y
∴ LCM(a, b) = x * x * x * y * y * y = x3 * y3 = x3y3
LCM = x3y3
The multiplicative inverse of zero is- a)is 1
- b)is 0
- c)is 1/0
- d)does not exist
Correct answer is option 'B'. Can you explain this answer?
The multiplicative inverse of zero is
a)
is 1
b)
is 0
c)
is 1/0
d)
does not exist
|
|
Krishna Iyer answered |
Some rules associated with Multiplicative Inverse are discussed in following ways :-
Rule 1 = If product of two Fractional Numbers is equal to 1, then each of the Fractional Numbers is the Multiplicative Inverse of other.
Rule 2 = If the product of a Fractional Number and a Whole Number is equal to 1, then each is the Multiplicative Inverse of other.
Rule 3 = Multiplicative Inverse of 1 is also 1.
Rule 4 = Multiplicative Inverse of 0 (zero) does not exists
Rule 1 = If product of two Fractional Numbers is equal to 1, then each of the Fractional Numbers is the Multiplicative Inverse of other.
Rule 2 = If the product of a Fractional Number and a Whole Number is equal to 1, then each is the Multiplicative Inverse of other.
Rule 3 = Multiplicative Inverse of 1 is also 1.
Rule 4 = Multiplicative Inverse of 0 (zero) does not exists
The answer of 1⁄3 x 8² + 2² is- a)7.1667
- b)6.1667
- c)4.1667
- d)5.1667
Correct answer is option 'C'. Can you explain this answer?
The answer of 1⁄3 x 8² + 2² is
a)
7.1667
b)
6.1667
c)
4.1667
d)
5.1667
|
|
Raghav Bansal answered |
Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70-5),
117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65,117 [for the largest number]
For this, 117 = 65 × 1 + 52 [∵ dividend = divisior × quotient + remainder]
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.
117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65,117 [for the largest number]
For this, 117 = 65 × 1 + 52 [∵ dividend = divisior × quotient + remainder]
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.
The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is - a)13
- b)65
- c)875
- d)1750
Correct answer is option 'A'. Can you explain this answer?
The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is
a)
13
b)
65
c)
875
d)
1750
|
|
Krishna Iyer answered |
The largest number by which x , y
divisible and gives the remainder a ,
and b is
the HCF of ( x - a ) and ( y - b)
According to the given problem ,
According to the given problem ,
The largest number which divides
70 and 125 leaving remainders 5 and
8 respectively are
HCF of ( 70 - 5 ) = 65 and
( 125 - 8 ) = 117
65 = 5 × 13
117 = 3 × 3 × 13
HCF ( 65 , 117 ) = 13
Required number is 13.
Which of the following numbers has terminating decimal expansion?- a)3/11
- b)3/5
- c)5/3
- d)3/7
Correct answer is option 'B'. Can you explain this answer?
Which of the following numbers has terminating decimal expansion?
a)
3/11
b)
3/5
c)
5/3
d)
3/7
|
|
Arun Sharma answered |
3/5 has terminal decimal expansion because terminal decimal expansion should have the denominator 2 or 5 only.
Every positive even integer is of the form ____ for some integer ‘q’.- a)2q
- b)2q – 1
- c)2q + 1
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Every positive even integer is of the form ____ for some integer ‘q’.
a)
2q
b)
2q – 1
c)
2q + 1
d)
none of these
|
|
Bhargavi Pillai answered |
Let a be any positive integer and b = 2
Then by applying Euclid’s Division
Lemma, we have, a = 2q + r where 0 ⩽ r < 2 r = 0 or 1
Therefore, a = 2q or 2q+1
Therefore, it is clear that a = 2q i.e.,
a is an even integer in the form of 2q
Then by applying Euclid’s Division
Lemma, we have, a = 2q + r where 0 ⩽ r < 2 r = 0 or 1
Therefore, a = 2q or 2q+1
Therefore, it is clear that a = 2q i.e.,
a is an even integer in the form of 2q
The product of a non zero rational and an irrational number is
- a)Always irrational
- b)Always rational
- c)Rational or irrational
- d)One
Correct answer is option 'A'. Can you explain this answer?
The product of a non zero rational and an irrational number is
a)
Always irrational
b)
Always rational
c)
Rational or irrational
d)
One
|
|
Rising Star answered |
The product will always be irrational...
let us have an example--}
2( rational number) *√2(irrational number)=2√2(irrational)..
let us have an example--}
2( rational number) *√2(irrational number)=2√2(irrational)..
Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?- a)12.20 pm
- b)12.12 pm
- c)12.11 pm
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?
a)
12.20 pm
b)
12.12 pm
c)
12.11 pm
d)
none of these
|
|
Ananya Das answered |
LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.
The HCF and LCM of two numbers is 9 and 459 respectively. If one of the number is 27, then the other number is- a)459
- b)153
- c)135
- d)150
Correct answer is option 'B'. Can you explain this answer?
The HCF and LCM of two numbers is 9 and 459 respectively. If one of the number is 27, then the other number is
a)
459
b)
153
c)
135
d)
150
|
Girisha reddy answered |
Using the result, HCF * LCM = Product of two natural numbers ⇒ the other number = product of two no. = LCM *HCF
Let unknown no. = X
>> X * 27 = 9*459
X = 9*459/27
X = 153
Let unknown no. = X
>> X * 27 = 9*459
X = 9*459/27
X = 153
So, option B is the correct answer.
You can study all the concepts of mathematics Class 10 by going through the link:
You can study all the concepts of mathematics Class 10 by going through the link:
If A = 2n + 13, B = n + 7, where n is a natural number, then HCF of A and B is:
- a)2
- b)1
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
If A = 2n + 13, B = n + 7, where n is a natural number, then HCF of A and B is:
a)
2
b)
1
c)
3
d)
4
|
|
Ritu Saxena answered |
Taking different values of n we find that A and B are coprime
e.g. put n=1
we get A= 2(1)+13=15
B=1+7=8
∴ HCF = 1
∴ HCF = 1
put n=2
we get A= 2(2)+13=17
B=2+7=9
∴ HCF = 1
∴ HCF = 1
A rational number can be expressed as a non-terminating repeating decimal if the denominator has the factors- a)other than 2 or 5
- b)2 or 3 only
- c)2 or 5 only
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
A rational number can be expressed as a non-terminating repeating decimal if the denominator has the factors
a)
other than 2 or 5
b)
2 or 3 only
c)
2 or 5 only
d)
none of these
|
|
Pooja Shah answered |
A rational number can be expressed as a non-terminating repeating decimal if the denominator has the factors other than 2 or 5.
There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:
- a)22
- b)16
- c)36
- d)21
Correct answer is option 'B'. Can you explain this answer?
There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:
a)
22
b)
16
c)
36
d)
21
|
|
Gaurav Kumar answered |
The number 576 can be factorised as,
576 = 2×2×2×2×2×2×3×3
The number 448 can be factorised as,
448=2×2×2×2×2×2×7
Write the common factors of the given numbers.
2×2×2×2×2×22×2×2×2×2×2
Multiply the common factors to determine the highest common factor (HCF) of the given numbers.
2×2×2×2×2×2 = 642×2×2×2×2×2 = 64
Since the highest common factor (HCF) of the given numbers is 64, this implies that each section will have 64 number of students.
Now, we need to find the number of sections formed.
Let us first find the number of sections formed by the total number of boys by dividing 576 by 64.
576/64 = 9
Now, find the number of sections formed by the total number of girls by dividing 448 by 64.
448/64=7
Thus, the total number of sections formed will be 9+7=16
Hence, option B is the correct answer.
576 = 2×2×2×2×2×2×3×3
The number 448 can be factorised as,
448=2×2×2×2×2×2×7
Write the common factors of the given numbers.
2×2×2×2×2×22×2×2×2×2×2
Multiply the common factors to determine the highest common factor (HCF) of the given numbers.
2×2×2×2×2×2 = 642×2×2×2×2×2 = 64
Since the highest common factor (HCF) of the given numbers is 64, this implies that each section will have 64 number of students.
Now, we need to find the number of sections formed.
Let us first find the number of sections formed by the total number of boys by dividing 576 by 64.
576/64 = 9
Now, find the number of sections formed by the total number of girls by dividing 448 by 64.
448/64=7
Thus, the total number of sections formed will be 9+7=16
Hence, option B is the correct answer.
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