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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
√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 p is a positive prime integer, then √p is –- a)A rational number
- b)An irrational number
- c)A positive integer
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If p is a positive prime integer, then √p is –
a)
A rational number
b)
An irrational number
c)
A positive integer
d)
None of these
|
Gaurav Kumar answered |
Let us assume, to the contrary, that √p is
rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √p = a/b
⇒ √p b = a
⇒ pb2 = a2 ….(i) [Squaring both the sides]
⇒ a2 is divisible by p
⇒ a is divisible by p
So, we can write a = pc for some integer c.
Therefore, a2 = p2c2 ….[Squaring both the sides]
⇒ pb2 = p2c2 ….[From (i)]
⇒ b2 = pc2
⇒ b2 is divisible by p
⇒ b is divisible by p
⇒ p divides both a and b.
⇒ a and b have at least p as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have
assumed that √p is rational.
Therefore, √p is irrational.
rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √p = a/b
⇒ √p b = a
⇒ pb2 = a2 ….(i) [Squaring both the sides]
⇒ a2 is divisible by p
⇒ a is divisible by p
So, we can write a = pc for some integer c.
Therefore, a2 = p2c2 ….[Squaring both the sides]
⇒ pb2 = p2c2 ….[From (i)]
⇒ b2 = pc2
⇒ b2 is divisible by p
⇒ b is divisible by p
⇒ p divides both a and b.
⇒ a and b have at least p as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have
assumed that √p is rational.
Therefore, √p is irrational.
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
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)
|
Dr Manju Sen 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.
If x = 
and y = 
, then the value of (x2 + y2) is- a)14
- b)- 14
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
If x = and y = , then the value of (x2 + y2) is
and y = , then the value of (x2 + y2) isa)
14
b)
- 14
c)
d)
|
Pooja Shah answered |
The correct option is Option A.
x = √3 + 1 / √3 - 1
y = √3 - 1 / √3 + 1
Now,
x² + y²
= (√3 + 1 / √3 - 1)² + (√3 - 1 / √3 + 1)²
= (3 + 2√3 + 1 / 3 - 2√3 + 1) + (3 - 2√3 + 1 / 3 + 2√3 + 1)
= (4 + 2√3 / 4 - 2√3) + (4 - 2√3 / 4 + 2√3)
= (4 + 2√3)² + (4 - 2√3)² / (4 - 2√3) × (4 + 2√3)
By solving this,
= 28 + 16√3 + 28 - 16√3 / 16 - 12
= 56 / 4
= 14
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.
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
The standard form of (1 + i) (1 + 2i) is –- a)3 + i
- b)– 3 + i
- c)1 – 3i
- d)– 1 + 3i
Correct answer is option 'D'. Can you explain this answer?
The standard form of (1 + i) (1 + 2i) is –
a)
3 + i
b)
– 3 + i
c)
1 – 3i
d)
– 1 + 3i
|
|
Amit Kumar answered |
( 1 + i ) (1 + 2i )
=1 + 2i + i + 2i^2
=1 + 3i -2 ( Because i^2 = -1)
= -1 + 3i
= a + ib = -1 + 3i
Here a = -1 and b = 3
So the standard form of (1 + i) (1 + 2i) is :
-1 + 3i
So option D is correct 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
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
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.
HCF (p,q,r) · LCM (p,q,r) =- a)pq/r
- b)qr/p
- c)p, q, r
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
HCF (p,q,r) · LCM (p,q,r) =
a)
pq/r
b)
qr/p
c)
p, q, r
d)
None of these
|
Vivek Rana answered |
Since HCF(p,q,r)*LCM(p,q,r) is not equal to pq/r, neither it is equal to qr/p and neither is p,q,r. So the correct answer is D . Also, HCF(p,q,r)*LCM(p,q,r) is not equal to p*q*r . This condition only holds for two numbers.
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
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)xy
- b)xy2
- c)x3y3
- d)x2y2
Correct answer is option 'B'. 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)
xy
b)
xy2
c)
x3y3
d)
x2y2
|
|
Ananya Das answered |
Given that, a =x3y2 = x × x × x × y × y
and b = xy3 = x × y × y × y
∴ HCF of a and b = HCF (x3y2,xy3) = x × y × y = xy2
[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]
and b = xy3 = x × y × y × y
∴ HCF of a and b = HCF (x3y2,xy3) = x × y × y = xy2
[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]
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
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.
- 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
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.
a)A irrational numberb)A whole numberc)A positive integerd)None of theseCorrect answer is option 'A'. Can you explain this answer?
b)A whole number
c)A positive integer
d)None of these
Correct answer is option 'A'. Can you explain this answer?
|
|
Ananya Das answered |
In the 1760s, Johann Heinrich Lambert proved that the number π (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is- a)10
- b)100
- c)504
- d)2520
Correct answer is option 'D'. Can you explain this answer?
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
a)
10
b)
100
c)
504
d)
2520
|
|
Vikas Kumar answered |
The least number divisible by all the numbers from 1 to 10 will be the LCM of these numbers.
We have,
1 = 1
2 = 2 * 1
3 = 3 * 1
4 = 2 * 2
5 = 5 * 1
6 = 2 * 3
7 = 7 * 1
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
So, LCM of these numbers = 1 * 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520
Hence, least number divisible by all the numbers from 1 to 10 is 2520
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
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.
Find the value of x then 
- a)x = 2
- b)x = -2
- c)x = 1
- d)x = -1
Correct answer is option 'A'. Can you explain this answer?
Find the value of x then
a)
x = 2
b)
x = -2
c)
x = 1
d)
x = -1
|
|
Krishna Iyer answered |
(3/5)2x-3= (5/3)x-3
(3/5)2x-3=(3/5)3-x
2x-3=3-x
2x+x=6
3x=6
x=2
(3/5)2x-3=(3/5)3-x
2x-3=3-x
2x+x=6
3x=6
x=2
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.
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
|
Rohini Seth answered |
Let a be a given positive odd integer.
Applying Euclid’s Division Lemma to aa and b = 4,,
We have, a = 4q+r where 0 ⩽ r < 4 ⇒ r = 0, 1, 2, 3 ⇒ a = 4q or 4q+1 or 4q+2 or 4q+3
But a = 4q and 4q+2 = 2(2q+1)are clearly even.
Also a = 4q, 4q+1, 4q+2, 4q+3 are consecutive integers, therefore any positive odd integer is of the form 4q+1 and 4q+3 where q is some integer.
Applying Euclid’s Division Lemma to aa and b = 4,,
We have, a = 4q+r where 0 ⩽ r < 4 ⇒ r = 0, 1, 2, 3 ⇒ a = 4q or 4q+1 or 4q+2 or 4q+3
But a = 4q and 4q+2 = 2(2q+1)are clearly even.
Also a = 4q, 4q+1, 4q+2, 4q+3 are consecutive integers, therefore any positive odd integer is of the form 4q+1 and 4q+3 where q is some integer.
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.
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
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
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.
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 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
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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
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
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Clifford Academy answered |
52n −22n is of the form a2n − b2n which is divisible by both (a + b) and (a – b).
So, 52n − 22n is divisible by both 7, 3.
So, 52n − 22n is divisible by both 7, 3.
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
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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.
The value of i37+ 1/i67 is –- a)1
- b)-1
- c)-2i
- d)-2
Correct answer is option 'C'. Can you explain this answer?
The value of i37+ 1/i67 is –
a)
1
b)
-1
c)
-2i
d)
-2
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Arun Yadav answered |
i^37+1/i^67
=(i^2)^18.i + 1/(i^2)^33.i
=1.i + 1/-1.i (Since i^2=-1)
=i -1/i
=(i^2 - 1)/i
=(-1-1)/i
=-2/i
Now, multiplying by i/i,
-2/i . i\i
=-2i/i^2
=-2i/-1
=2i
A rational number can be expressed as a terminating decimal if the denominator has the factors- a)2 or 5 only
- b)2 or 3 only
- c)2 and 5 only
- d)2, 3 or 5 only
Correct answer is option 'A'. Can you explain this answer?
A rational number can be expressed as a terminating decimal if the denominator has the factors
a)
2 or 5 only
b)
2 or 3 only
c)
2 and 5 only
d)
2, 3 or 5 only
|
|
Vikas Kumar answered |
If the denominator of rational number contains no prime factors other then 2 or 5 or both, then this rational number can be expressed as terminating decimal.
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.
The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:- a)215
- b)220
- c)225
- d)235
Correct answer is option 'C'. Can you explain this answer?
The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:
a)
215
b)
220
c)
225
d)
235
|
|
Rahul Kapoor answered |
Let the lcm be x and hcf be y and the other number be z.
Given lcm of 2 numbers is 45 times their hcf, the sum of HCF + LCM is 1150.
y = 45x. ---- (1)
x + y = 1150 --- (2)
Substitute equation (1) in (2), we get
46x = 1150
x = 25.
Substitute x = 25 in (1), we get
y = 45 * 25 = 1125.
We know that product of two numbers = LCM * HCF
125 * z = 25 * 1125
z = 25 * 1125/125
= 225.
The other number = 225.
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
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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)..
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
Correct answer is option 'B'. Can you explain this answer?
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
|
Athira Yadav 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
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.
0.737373...=- a)(0.73)3
- b)73/100
- c)73/99
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
0.737373...=
a)
(0.73)3
b)
73/100
c)
73/99
d)
None of these
|
|
Arun Sharma answered |
⇒ a = 0.737373...
⇒ 100a = 73.737373
= 73 + a
⇒ a = 73/99 = p/q
⇒ p=73 and q = 99 are co-prime.
Here, q=32 X 11.
Which of the following is false:- a)H.C.F.(p,q,r) × L.C.M.(p,q,r) = p×q×r
- b)H.C.F.(p,q,r) = 1; if p,q,r are prime numbers
- c)H.C.F.(a,b) × L.C.M.(a,b) = a×b
- d)L.C.M.(p,q,r) = p×q×r; if p,q,r are prime numbers
Correct answer is option 'A'. Can you explain this answer?
Which of the following is false:
a)
H.C.F.(p,q,r) × L.C.M.(p,q,r) = p×q×r
b)
H.C.F.(p,q,r) = 1; if p,q,r are prime numbers
c)
H.C.F.(a,b) × L.C.M.(a,b) = a×b
d)
L.C.M.(p,q,r) = p×q×r; if p,q,r are prime numbers
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Raghav Bansal answered |
H.C.F.(p,q,r)× L.C.M.(p,q,r) ≠ p×q×r. This condition is only applied on HCF and LCM of two numbers.
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