Explore topic-wise InterviewSolutions in .

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520. 

Prime factorization of 468 and 520 is: 

468 = 2² × 3² × 13 

520 = 2³ × 5 × 13 

LCM = product of greatest power of each prime factor involved in the numbers = 2² × 3² × 5 × 13 = 4680 

The required number is 4680 – 17 = 4663. 

Hence, 

the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

The required number is determine with the help of LCM of 468 and 520.

Using prime factorization, find the LCM:

468 = 2² × 3² × 13

520 = 2³ × 5 × 13

LCM = 4680

The required number = LCM(468 and 520) – 17

= 4680 – 17 = 4663

The required number is 4663.

First let’s find the smallest number which is exactly divisible by both 520 and 468. 

That is simply just the LCM of the two numbers. 

By prime factorisation, we get 

520 = 2³ × 5 × 13 

468 = 2² × 3² × 13 

∴ LCM (520, 468) = 2³ × 3² × 5 × 13 = 4680 

Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468 i.e. we will get a remainder of 0 in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468. 

So that is found by, 4680 – 17 = 4663 

∴ 4663 should be the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

(b) 2 

Let q be the quotient. 

It is given that: 

Remainder = 7 

On applying Euclid’s algorithm, i.e. dividing n by 9, we have 

n = 9q + 7 

⇒ 3n = 27q + 21 

⇒ 3n – 1 = 27q + 20 

⇒ 3n – 1 = 9 × 3q + 9 × 2 + 2

⇒ 3n – 1 = 9 × (3q + 2) + 2 

So, when (3n-1) is divided by 9, we get the remainder 2.

No, every positive integer cannot be of the form 4q + 2, where q is an integer.

Justification:

All the numbers of the form 4q + 2, where ‘q’ is an integer, are even numbers which are not divisible by ‘4’.

For example,

When q=1,

4q+2 = 4(1) + 2= 6.

When q=2,

4q+2 = 4(2) + 2= 10

When q=0,

4q+2 = 4(0) + 2= 2 and so on.

So, any number which is of the form 4q+2 will give only even numbers which are not multiples of 4.

Hence, every positive integer cannot be written in the form 4q+2

Two irrational numbers between 0.7 and 0.77 can take the form 

0.70101100111000111…………. and 

0.70200200022……………

The decimal forms of  5/7 and 7/9 are

\(\frac 57 = 0.\overline{714285}....,\frac 79 = 0.7777.....= 0.\overline7\)

∴ An irrational between 5/7 and 7/9 is 0.727543…………

There are infinitely many irrational numbers between 

5/7 and 7/9

1. b) 13915

2. c) 11

3. b) 23

4. a) composite number

5. c) 5 x 11² x 23

The given number ends in 5, Hence it is a multiple of 5. therefore it is a  composite number.

Prime factorization: 

a = a 

b = b 

LCM = product of greatest power of each prime factor involved in the numbers = a × b 

Thus, LCM (a, b) = ab

The correct answer is

P³

HCF of two numbers = 145 

LCM of two numbers = 2175 

Let one of the two numbers be 725 and other be x. 

Using the formula, product of two numbers = HCF × LCM 

we conclude that 

725 × x = 145 × 2175

x = \(\frac{145\times2175}{725}\)

= 435 

Hence, the other number is 435.

Given: HCF = 145

LCM = 2175

One of the two numbers = 725

Let another number be m.

To Find: The value of m

We know, product of two numbers = HCF × LCM

Substitute the values, we get

725 x m = 145 x 2175

m = (145 x 2175)/725 = 435

The number is 435. Answer!

Given, 

HCF of two numbers = 16 

And, their product = 3072 

We know that, 

LCM x HCF = Product of the two numbers 

LCM x 16 = 3072 

⇒ LCM = \(\frac{3072}{16}\) = 192

∴ The LCM of the two numbers is 435.

True

Let a = bq + r. b = 2, q = m

r = 0.

a = 2m

We know that LCM × HCF = Product of the numbers

Therefore Other Number = \(\frac{LCM\times HCF}{One\,number}\) = \(\frac{2175\times 145}{725}\) = 435

True, because n(n+1)(n+2) will always be divisible by 6, as least one of the factors will be divisible by 2 and at one of the factors will be divisible by 3.

True

Let a = bq + r: b = 2, q = m

- 2 < r < 0

i.e., r = -1

a = 2m – 1 for odd integer.

On dividing 380 by 16 we get, 

23 as the quotient and 12 as the remainder. 

Now, since the LCM is not exactly divisible by the HCF its can be said that two numbers cannot have 16 as their HCF and 380 as their LCM.

True,because n(n+1) will always be even, as one out of the n or (n+1) must be even.

We know that any odd positive integer n can be written in form 4q + 1 or 4q + 3.

So, according to the question,

When n = 4q + 1,

Then n² – 1 = (4q + 1)² – 1 = 16q² + 8q + 1 – 1 = 8q(2q + 1), is divisible by 8.

When n = 4q + 3,

Then n² – 1 = (4q + 3)² – 1 = 16q² + 24q + 9 – 1 = 8(2q² + 3q + 1), is divisible by 8.

So, from the above equations, it is clear that, if n is an odd positive integer

n² – 1 is divisible by 8.

Hence Proved.

Correct option is (C) positive irrational number

Assume contrary that \(\sqrt3 +\sqrt5 \) is a rational number.

\(\therefore\) \(\sqrt3 +\sqrt5 \) \(=\frac ab\)  (Every rational number can be written as \(\frac pq\) form)

\(\Rightarrow\) \((\sqrt3 +\sqrt5 )^2\) \(=\frac{a^2}{b^2}\)   (By squaring both sides)

\(\Rightarrow\) \(3+5+2\sqrt{15}\) \(=\frac{a^2}{b^2}\)

\(\Rightarrow\) \(2\sqrt{15}\) \(=\frac{a^2}{b^2}-8\) \(=\frac{a^2-8b^2}{b^2}\)

\(\Rightarrow\) \(\sqrt{15}\) \(=\frac{a^2-8b^2}{2b^2}\)   ___________(1)

\(\because\) a & b both are integers & \(b\neq0\)

\(\therefore\) \(a^2-8b^2\) is an integer.

\(\Rightarrow\) \(\frac{a^2-8b^2}{2b^2}\) is a rational number.

But \(\sqrt{15}\) is an irrational number.

This is a contradiction.   (From (1))

\(\therefore\) Our assumption is wrong.

\(\Rightarrow\) \(\sqrt3 +\sqrt5 \) is a positive irrational number.

\((\because\) Sum of two positive number is always positive)

Correct option is C) positive irrational number

Correct option is (C) 11/2

Rational number between 5 and 6 is \(\frac{5+6}2=\frac{11}2.\)

Correct option is C)11/2

Let us consider two odd positive numbers be x and y where

x = 2p + 1 and y = 2q + 1

From question,

x² + y² = (2p + 1)² +(2q + 1)²

= 4p² + 4p + 1 + 4q² + 4q + 1

= 4p² + 4q² + 4p + 4q + 2

= 4 (p² + q² + p + q) + 2

Form above result, we can conclude that x and y are odd positive integer, then x² + y² is even but not divisible by four.

A rational number between √3 and √5 is √(324)=1.8=18/10=9/5

Correct option is (B) left side of the zero

\(\because\) Negative numbers are lie on left side of zero on number line.

Since, \(\frac{-2}{3}\) is a negative number.

\(\therefore\) \(\frac{-2}{3}\) lies on left side of the zero.

B) left side of the zero

Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) 

We have n³ – n = n(n²-1)= (n-1)n(n+1), 

For n= 6q, 

⇒ (n-1)n(n+1)= (6q-1)(6q)(6q+1) 

⇒ (n-1)n(n+1)= 6[(6q-1)q(6q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q-1)q(6q+1)] 

For n= 6q+1, 

⇒ (n-1)n(n+1)= (6q)(6q+1)(6q+2) 

⇒ (n-1)n(n+1)= 6[q(6q+1)(6q+2)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)] 

For n= 6q+2, 

⇒ (n-1)n(n+1)= (6q+1)(6q+2)(6q+3) 

⇒ (n-1)n(n+1)= 6[(6q+1)(3q+1)(2q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)] 

For n= 6q+3, 

⇒ (n-1)n(n+1)= (6q+2)(6q+3)(6q+4) 

⇒ (n-1)n(n+1)= 6[(3q+1)(2q+1)(6q+4)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)] 

For n= 6q+4, 

⇒ (n-1)n(n+1)= (6q+3)(6q+4)(6q+5) 

⇒ (n-1)n(n+1)= 6[(2q+1)(3q+2)(6q+5)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)] 

For n= 6q+5, 

⇒ (n-1)n(n+1)= (6q+4)(6q+5)(6q+6) 

⇒ (n-1)n(n+1)= 6[(6q+4)(6q+5)(q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+4)(6q+5)(q+1)] 

Hence, for any positive integer n, n³ – n is divisible by 6.

Let n be any positive integer.

Thus, the three consecutive positive integers are n, n+1 and n+2.

We know that any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) 

So, 

For n= 6q, 

⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2) 

⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)] 

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)] 

For n= 6q+1, 

⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3) 

⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)] 

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)] 

For n= 6q+2, 

⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4) 

⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)] 

For n= 6q+3, 

⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5) 

⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)] 

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)] 

For n= 6q+4, 

⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6) 

⇒ n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)] 

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+2)(3q+1)(2q+2)] 

For n= 6q+5, 

⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7) 

⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)] 

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+5)(q+1)(6q+7)] 

Hence, the product of three consecutive positive integers is divisible by 6.

We can write 1.2 as 12/10 = 120/100

1.2 = 120/100​,    0.12 = 12/​100

HCF of the numerators 120 and 12 is 12.

HCF of denominators 100 and 100 is also 100.

So, HCF of 120/100 and 12/100 is 12/100 i.e., 0.12.

∴ HCF of 1.2 and 0.12 is 0.12.

Hence the correct option is (c) 0.12

Correct option is C) 0.12

9√7 - 6√7

= (9 – 6)√7 = 3√7

The answer is irrational.

Given:

The length, breadth and height of a room are 8 m and 25 cm, 6 m 75 cm and 4 m 50 cm.

To find:

the longest rod which can measure the three dimensions of the room exactly.

Solution:

To find the length of largest rod,

we should find HCF of 8m and 25 cm, 6 m 75 cm and 4 m 50 cm

As 1 m = 100 cm ⇒ 8m and 25 cm

= 8 × 100 cm + 25 cm

= 800 cm + 25 cm

= 825 cm ⇒ 6 m and 75 cm = 6 × 100 cm + 75 cm

= 600 cm + 75 cm

= 675 cm

⇒ 4 m and 50 cm = 4 × 100 cm + 50 cm

= 400 cm + 50 cm

= 450 cm

Length = 825 cm; Breadth = 675 cm; Height = 450 cm

Prime factors of 825 = 3 × 5 × 5 × 11

Prime factors of 675 = 3 × 3 × 3 × 5 × 5

Prime factors of 450 = 2 × 3 × 3 × 5 × 5

Therefore HCF of 825, 675 and 450 is: 3 × 5 × 5 = 75

Hence, the length of rod is: 75 cm 

NOTE:

Always find the HCF of the given values to find their maximum.

Length of room = 8m 25cm = 825 cm
Breadth of room = 6m 75m = 675 cm
Height of room = 4m 50m = 450 cm
∴ The required longest rod
= HCF of 825, 675 and 450
First consider 675 and 450
By applying Euclid’s division lemma
675 = 450 × 1 + 225
450 = 225 × 2 + 0
∴ HCF of 675 and 450 = 825

Now consider 625 and 825
By applying Euclid’s division lemma
825 = 225 × 3 + 150
225 = 150 × 1 + 75
150 = 75 × 2 + 0
HCF of 825, 675 and 450 = 75

Length of the courtyard = 18 m 72 cm = [18(100) + 72] cm [As, 1 m = 100 cm] = 1872 cm

The breadth of the courtyard = 13 m 20 cm = [13(100) + 20] cm = 1320 cm

To find the maximum edge of the tile we need to calculate HCF of length and breadth,

Using Euler’s division lemma:

a = pq + r where 0 ≤ r ≤ p

1872 = 1320 × 1 + 552

As 'r' is not equals to 0,

So apply Euler's division on 1320 and 552,

1320 = 552 × 2 + 216

As 'r' is not equals to 0,

So apply Euler's division on 552 and 216,

552 = 216 × 2 + 120

As 'r' is not equals to 0,

So apply Euler's division on 216 and 120,

216 = 120×1 + 96

As 'r' is not equals to 0,

So apply Euler's division on 120 and 96,

120 = 96 ×1 + 24

As 'r' is not equals to 0,

So apply Euler's division on 96 and 24,

96 = 24 × 4 + 0

Therefore HCF of 1872 and 1320 is 24

Maximum edge can be 24 cm.

Number of tile = \(\frac{Area\,of\,courtyard}{Area\,of\,one\,\,tile}\) = \(\frac{1872\times1320}{24\times 24}\) = 4290 tiles

A rational numbers may has its decimal expansion either terminating or non-terminating. repeating An irrational numbers has its decimal expansion non-repeating and non-terminating.

If any number ends with the digit 0 or 5, it is always divisible by 5. If 12n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5.

12 = 2 × 2 × 3 = 2² × 3

⇒ 12n = (2² × 3)n = 2²n × 3n

Since its prime factorisation does not contain 5.

Hence, 12n cannot end with the digit 0 or 5 for any natural number n.

C. a³b²

Let p = ab² = a × b × b

And q = a³b = a × a × a × b

⇒ LCM of p and q = LCM (ab², a³b) = a × b × b × a × a = a³b²

[Since, LCM is the product of the greatest power of each prime factor involved in the number]

As per Euclid’s division lemma: For any two positive integers, say a and b, there exit unique integers q and r, such that a = bq + r; where 0 ≤ r < b.

Also written as: Dividend = (Divisor x Quotient) + Remainder

Given: Dividend = 1365, Quotient = 31, Remainder = 32 

Let the divisor be x. 

Dividend = Divisor × Quotient + Remainder 

1365 = x × 31 + 32 

⇒ 1365 – 32 = 31 x 

⇒ 1333 = 31 x 

⇒ x = \(\frac{1333}{31}\) = 43 

Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

Given: Divisor = 61, Quotient = 27, Remainder = 32

To Find: Dividend

As per Euclid’s lemma, we have

Dividend = (divisor x quotient) + remainder

Dividend = (61 × 27) + 32 = 1679

Therefore, dividend is 1679.

Step 1: Choose bigger number: 2520 > 405

On dividing 2520 by 405, we get

Quotient = 6, remainder = 90

2520 = (405 x 6) + 90

Step 2: On dividing 405 by 90, we get

Quotient = 4 and Remainder = 45

405 = 90 x 4 + 45

Step 3: On dividing 90 by 45, we get

Quotient = 2 and Remainder = 0

90 = 45 x 2 + 0

Step 4: Since remainder is zero, stop the process and write your answer.

Therefore, H.C.F. of 405 and 2520 is 45.

Correct option is (A) \(\sqrt\frac{a}{b}-\sqrt\frac{b}{a}\)

\(\sqrt{\frac{a}{b}+\frac{b}{a}-2} \) \(=\sqrt{\left(\sqrt{\frac ab}\right)^2+\left(\sqrt{\frac ba}\right)^2-2\sqrt{\frac ab}\times\sqrt{\frac ba}} \)

\(=\sqrt{\left(\sqrt{\frac ab}-\sqrt{\frac ba}\right)^2} \)   (By assuming \(\sqrt\frac{a}{b}=A\,\&\,\sqrt\frac{b}{a}=B\) and \((A-B)^2=A^2+B^2-2AB)\)

\(=\left(\left(\sqrt{\frac ab}-\sqrt{\frac ba}\right)^2\right)^\frac12\)

\(=\sqrt\frac{a}{b}-\sqrt\frac{b}{a}\)

Correct option is  A) \(\sqrt{\cfrac{a}{b}}\) - \(\sqrt{\cfrac{b}{a}}\)

Correct option is (B) 3/4

Rational number which lies between \(\frac{1}{2}\) and \(\sqrt1\)

\(=\frac{\frac12+\sqrt1}2\) \(=\frac{\frac12+1}2\) \(=\frac{\frac32}2\)

\(=\frac3{2\times2}\) \(=\frac34\)

Alternative :-

\(\because\) \(\frac{1}{2}\) = 0.5 and \(\sqrt1\) = 1

Now, (A) \(\frac{9}{4}=2+\frac14\) = 2.25 > 1

(B) \(\frac{3}{4}\) = 0.75 which lies between 0.5 and 1.

\(\therefore\) \(\frac{1}{2}\) < \(\frac{3}{4}\) < \(\sqrt1\)

(C) \(\frac{5}{4}\) = 1 + \(\frac14\) = 1.25 > 1

(D) \(\frac{7}{4}\) = 1 + \(\frac{3}{4}\) = 1.75 > 1

Correct option is B) 3/4

Correct option is (D) All

The multiplicative inverse of 2 is \(\frac12.\)

2 is a natural number, a whole number and also an integer.

But \(\frac12\) is neither a natural number, nor a whole number nor an integer.

Thus, the multiplicative inverse does not exist in the set of natural numbers, whole numbers and integers.

Correct option is (D) All

Correct option is (B) 2 and 3

\(\because 4<5<9\)

\(\Rightarrow\) \(\sqrt4<\sqrt5<\sqrt9\)

\(\Rightarrow\) \(2<\sqrt5<3\)

\(\therefore\) \(\sqrt{5}\) will lie between 2 and 3.

Correct option is  B) 2 and 3

Correct option is (A) 8 + 5√2

Area of park \(=(3+\sqrt{2})(2+\sqrt{2})\)

\(=6+3\sqrt{2}+2\sqrt{2}+2\)

= \(8+5\sqrt{2}\)

Correct option is  A) 8 + 5√2 

Correct option is (B) rational

\((2+\sqrt{2})(2-\sqrt{2})\) \(=2^2-(\sqrt2)^2\) = 4 - 2 = 2 is a rational number.

C) can’t be determined

Correct option is (D) real

\(\because\) S is the set of all irrational numbers and Q is the set of all rational numbers.

\(\therefore\) The combination of Q and S is the set of all real numbers.

Correct option is  D) real

Correct option is (A) \(0.\bar{6}\)

\(\frac23=2\times0.333.....\)

= 0.666 ....... = \(0.\bar6\)

Correct option is   A) \(0.\overline{6}\)