Explore topic-wise InterviewSolutions in .

Correct option is (D) 4

Given that \(\sqrt{15-x\sqrt{14}}=\sqrt8-\sqrt7\)

\(\Rightarrow\) \(15-x\sqrt{14}=(\sqrt8-\sqrt7)^2\)    (By squaring both sides)

\(\Rightarrow\) \(15-x\sqrt{14}=8+7-2\sqrt8\times\sqrt7\)   \((\because(a-b)^2=a^2+b^2-2ab\) Here \(a=\sqrt8\,\&\,b=\sqrt{7})\)

\(\Rightarrow\) \(15-x\sqrt{14}=15-4\sqrt{14}\)

\(\therefore\) x = 4   (By comparing)

Correct option is D) 4

So to get full packs of both colour pencils and crayons and also of same numbers in quantity, we need to find the number of packets each need to be bought. 

It’s given that, 

Number of colour pencils in a pack = 24 

Number of crayons in a pack = 32. 

So, the least number of both colour pencils and crayons need to be purchased is given by their LCM. 

L.C.M of 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 = 96 

∴ the number of packs of pencils to be bought = 96 / 24 = 4 packs 

And, the number of packs of crayon to be bought = 96 / 32 = 3 packs

Number of cartons of coke cans = 144
Number of cartons of pepsi cans = 90
∴ The greatest number of cartons in one stock = HCF of 144 and 90
By applying Euclid’s division lemma
144 = 90 × 1 + 54
90 = 54 × 1 + 36
54 = 36 × 1 + 18
36 = 18 × 2 + 0
∴ HCF = 18
Hence the greatest number cartons in one stock = 18

To find greatest number of cartons each stack would have, we should find HCF of 144 and 90

Prime factors of 144 = 2 × 2 × 2 × 2 × 3 × 3

Prime factors of 90 = 2 × 3 × 3 × 5

Therefore HCF of 144 and 90 is:

2 × 3 × 3 = 18

Therefore the greatest number of cartons each stack would have is: 18

The correct option is: (A)

Explanation:

(a) H.C.F. (28, 16, 12) = 2 x 2 = 4

.'. Number of books each student got = 4

(b) No. of students who got Maths books = 28/4 = 7

No. of students who got Science books = 14/4 = 4

No. of students who got Social Science books = 12/4 = 3

.'. Total no. of studentswho got books = 7 + 4 + 3 = 14

Let √2 is a rational number.

∴ √2 = \(\frac{p}q\), where p and q are some integers and HCF(p,q) = 1 ….(1)

⇒ √2q = p 

⇒ (√2q)² = p² 

⇒2q² = p² 

⇒ p² is divisible by 2 

⇒ p is divisible by 2 …..(2)

Let p = 2m, where m is some integer. 

∴ √2q = p 

⇒ √2q = 2m 

⇒ (√2q)² = (2m)² 

⇒2q² = 4m² 

⇒ q² = 2m² 

⇒ q² is divisible by 2 

⇒ q is divisible by 2 …..(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1). 

Hence, our assumption is wrong. 

Thus, √2 is an irrational number. 

Hence, the correct answer is option (b).

(c) an irrational number 

It is an irrational number because it is a non-terminating and non-repeating decimal

(c) an irrational number 

It is an irrational number because it is a non-terminating and non-repeating decimal.

Clearly, 1.732 is a terminating decimal. 

Hence, a rational number. 

Hence, the correct answer is option (b).

(c) an irrational number. 

1/√2 is an irrational number.

(c) an irrational number 

2 + √2 is an irrational number. 

if it is rational, then the difference of two rational is rational. 

∴ (2 + √2) – 2 = √2 = irrational

We have to find the least number that is divisible by all numbers from 1 to 10. 

∴ LCM (1 to 10) = 2³ × 3² × 5 × 7 = 2520 

Thus, 2520 is the least number that is divisible by every element and is equal to the least common multiple.

(c) 2520 

We have to find the least number that is divisible by all numbers from 1 to 10. 

∴ LCM (1 to 10) = 2³ × 3² × 5 × 7 = 2520 

Thus, 

2520 is the least number that is divisible by every element and is equal to the least common multiple.

We know that for any two positive integers a and b,

HCF (a, b) × LCM (a, b) = a × b.

Here HCF = 13, a = 26 and b = 169

Then,

13 × LCM = 26 × 169

LCM = (26 × 169) / 13

LCM = 338

(i) Prime factors of 26 = 2 × 13

Prime factors of 91 = 7 ×13

Hence LCM of 26 and 91 = 2 × 7 × 13 = 182

HCF of 26 and 91 is = 13

LCM × HCF = 182 × 13 = 2366

Product of two numbers = 26 × 91 = 2366

Hence from above result we got LCM × HCF = Product of the numbers

(ii) Prime factors of 510 = 2 × 3 × 5 × 17

Prime factors of 92 = 2 × 2 × 13

Hence LCM of 92 and 510 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

HCF of 92 and 510 is = 2

LCM × HCF = 23460 × 2 = 46920

Product of two numbers = 92×510 = 46920

Hence from above result we got LCM × HCF = Product of the numbers

(iii) Prime factors of 336 = 2 × 2 × 2 × 2 × 3 × 7

Prime factors of 54 = 2 × 3 × 3 × 3

Hence LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 3024

HCF of 54 and 336 is = 2 × 3 = 6

LCM × HCF = 3024 × 6 = 18144

Product of two numbers = 336 × 54 = 18144

Hence from above result we got LCM × HCF = Product of the numbers

Given integers are: 26 and 91 

First, find the prime factors of 26 and 91. 

26 = 2 × 13 

91 = 7 × 13 

∴ L.C.M (26, 91) = 2 × 7 × 13 = 182 

And, H.C.F (26, 91) = 13 

Verification: 

L.C.M x H.C.F = 182 x 13

=  2366 

And, product of the integers = 26 x 91 

= 2366 

∴ L.C.M x H.C.F = product of the integers 

Hence verified.

92 = 2² x 23 

570 = 2 x 3 x 5 x 17

HCF (570,92) = 2

LCM (510,92) = 2² x 23 x 3 x 5 x 17

= 23460

HCF (510, 92) x LCM (510,92)

= 2 x 23460 = 46920 

Product of two numbers =  510 x 92 = 46920

Hence, HCF x LCM = Product of two numbers

Answer is (c) 36

Product of both numbers 

= H.C.F. x L.C.M.

1080 = 30 x L.C.M

L.C.M = 36

Solution :

If the number 6ⁿ ends with the digit zero,then it is divisible by 5.Therefore the prime factorization of 6ⁿ contains the prime 5.This is not possible because the only prime in the factorization of 6ⁿ is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorization of 6ⁿ

Hence, it is very clear that there is no value of n in natural number for which 6ⁿ ends with the digit zero.

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6ⁿ = (2 ×3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

We know that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

Here a × b = 1080 and HCF = 30

∴ LCM = (a × b)/ HCF

⇒ LCM = 1080/30

⇒ LCM = 36

The LCM of given product of two numbers is 36.

Solution :

We have,

7×11×13+13

=13(7×11×1+1)

=13×78

=13×3×2×13

Hence, it is a composite number .

We have,  7×6×5×4×3×2×1+5

=5(7×6×4×3×2×1+1)

=5(1008+1)

=5×1009

Hence, it is a composite number .

Prime factorization: 

4620 = 2 × 2 × 3 × 5 × 7 × 11 

= 2² × 3 × 5 × 7 × 11

Prime factorization: 

396 = 2² × 3² × 11 

1080 = 2³ × 3³ × 5 

HCF = product of smallest power of each common prime factor in the numbers = 2² × 3² = 36 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3³ × 5 × 11 = 11880

Solution:
3 × 5 × 7 + 7 = (3 × 5 + 1) × 7
= (15 + 1) × 7
= 16 × 7 = 7 × 16
Since (3 × 5 × 7 + 7) can be expressed as a
product of primes, it is a composite number.

p also divides k

8 = 2 × 2 × 2 = 2³ 

9 = 3 × 3 = 3² 

25 = 5 × 5 = 5² 

HCF = product of smallest power of each common prime factor in the numbers = 1 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5² = 1800

Solution:
Since, 2520 = 2 x 2 x 2 x 3 x 3 x 5 x 7
= 2³ x 3² x 5 x 7

Therefore,  p = 2 and q = 5

Prime factorization: 

1152 = 2⁷ × 3² 

1664 = 2⁷ × 13 

HCF = product of smallest power of each common prime factor in the numbers = 2⁷ = 128 

LCM = product of greatest power of each prime factor involved in the numbers = 2⁷× 3² × 13 = 14976

Find Prime factors of 1152 and 1664

1152 = 2⁷ × 3²

1664 = 2⁷ × 13

LCM (1152, 1664) = 2⁷ × 3² × 13 = 14976

HCF (1152, 1664) = 2⁷ = 128

Verification:

HCF x LCM = 14976 x 128 = 1916928

Product of given numbers = 1152 x 1664 = 1916928

HCF x LCM = Product of given numbers

Verified.

Solution:
90 = 2 x 3 x 3 x 5 = 2 x 3² x 5
144 = 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 3²
HCF (90, 144) = 2 x 3² = 18
LCM (90, 144) = 2⁴ x 3² x 5 = 720

Find Prime factors of 144 and 198

144 = 2⁴ × 3²

198 = 2 × 3² × 11

LCM (144, 198) = 2⁴ × 3² × 11 = 1584

HCF (144, 198) = 2 x 3² = 18

Verification:

HCF x LCM = 1584 x 18 = 28512

Product of given numbers = 144 x 198 = 28512

HCF x LCM = Product of given numbers

Verified.

First, find the prime factors of the given integers: 8, 9 and 25 

For, 8 = 2 × 2 x 2 

9 = 3 × 3 

25 = 5 × 5 

Now, L.C.M of 8, 9 and 25 = 2³ × 3² × 5² 

∴ L.C.M (8, 9, 25) = 1800 

And, H.C.F (8, 9 and 25) = 1

Find Prime factors of 96 and 404

96 = 2 x 2 x 2 x 2 x 2 x 3

404 = 2 x 2 x 101

LCM (96 and 404) = 2⁵ × 3 × 101 = 9696

HCF (96 and 404) = 2² = 4

Verification:

HCF x LCM = 9696 x 4 = 38784

Product of given numbers = 96 x 404 = 38784

HCF x LCM = Product of given numbers

Verified.

Prime factorization: 

96 = 2⁵ × 3 

404 = 2² × 101 

HCF = product of smallest power of each common prime factor in the numbers = 2² = 4 

LCM = product of greatest power of each prime factor involved in the numbers = 2⁵ × 3 × 101 = 9696 

First, find the prime factors of the given integers: 17, 23 and 29 

For, 17 = 1 × 17 

23 = 1 × 23 

29 = 1 × 29 

Now, L.C.M of 17, 23 and 29 = 1 × 17 × 23 × 29 

∴ L.C.M (17, 23, 29) = 11339 

And, H.C.F (17, 23 and 29) = 1

17 = 17 

23 = 23 

29 = 29 

HCF = product of smallest power of each common prime factor in the numbers = 1 

LCM = product of greatest power of each prime factor involved in the numbers = 17 × 23 × 29 = 11339

30 = 2 × 3 × 5 

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3² 

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2⁴ × 3³ 

HCF = product of smallest power of each common prime factor in the numbers = 2 × 3 = 6 

LCM = product of greatest power of each prime factor involved in the numbers = 2⁴ × 3³ × 5 = 2160

Let a is a positive odd integer and apply Euclid’s division algorithm

a = 6q + r, Where 0 ≤ r < 6

for 0 ≤ r < 6 probable remainders are 0, 1, 2, 3, 4 and 5.

a = 6q + 0

or a = 6q + 1

or a = 6q + 2

or a = 6q + 3

or a = 6q + 4

or a = 6q + 5 may be form

Where q is quotient and a = odd integer.

This cannot be in the form of 6q, 6q + 2, 6q + 4.

[all divides by 2]

Hence, any positive odd integer is of the form 6q + 1, or 6q + 3 or (6q + 5)

24 = 2 × 2 × 2 × 3 = 2³ × 3 

36 = 2 × 2 × 3 × 3 = 2² × 3² 

40 = 2 × 2 × 2 × 5 = 2³ × 5 

HCF = product of smallest power of each common prime factor in the numbers = 2² = 4 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5 = 360

Prime factors of 30, 72, 432

30 = 2 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2⁴ × 3³

HCF(30, 72, 432) = 2 x 3 = 6

LCM (30, 72, 432) = 2⁴ × 3³× 5 = 2160

We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m. 

Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5 

Thus we have: 

(6 m +1)² = 36 m² + 12 m + 1 = 6 (6 m² + 2 m) + 1 = 6 q + 1, q is an integer 

(6 m + 3)² = 36 m² + 36 m + 9 = 6 (6 m² + 6 m + 1) + 3 = 6 q + 3, q is an integer 

(6 m + 5)² = 36 m² + 60 m + 25 = 6 (6 m² + 10 m + 4) + 1 = 6 q + 1, q is an integer. 

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

According to Euclid’s division Lemma,

Let the positive integer = n

And b=3

n =3q+r, where q is the quotient and r is the remainder

0<r<3 implies remainders may be 0, 1 and 2

Therefore, n may be in the form of 3q, 3q+1, 3q+2

When n=3q

n+2=3q+2

n+4=3q+4

Here n is only divisible by 3

When n = 3q+1

n+2=3q=3

n+4=3q+5

Here only n+2 is divisible by 3

When n=3q+2

n+2=3q+4

n+4=3q+2+4=3q+6

Here only n+4 is divisible by 3

So, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3.

Hence Proved

Prime factors of 24, 36, 40

24 = 2 × 2 × 2 × 3 = 2³ × 3

36 = 2 × 2 × 3 × 3 = 2² × 3²

40 = 2 × 2 × 2 × 5 = 2³ × 5

HCF(24, 36, 40) = 2² = 4

LCM (24, 36, 40) = 2² × 3²× 5 = 360

Prime factors of 21, 28, 36, 45

21 = 3 × 7

28 = 2 × 2 × 7 = 2² × 7

36 = 2 × 2 × 3 × 3 = 2² × 3²

45 = 5 × 3 × 3 = 5 × 3²

HCF(21, 28, 36, 45) = 1

LCM (21, 28, 36, 45) = 2² × 3² x 5 × 7 = 1260

Find Prime factors of 36 and 84

36 = 2 x 2 x 3 x 3

84 = 2 x 2 x 3 x 7

LCM (36, 84) = 2 x 2 x 3 x 3 x 7 = 252

HCF (36, 84) = 2 x 2 x 3 = 12

Verification:

HCF x LCM = 12 x 252 = 3024

Product of given numbers = 36 x 84 = 3024

HCF x LCM = Product of given numbers

Verified.

First, find the prime factors of the given integers: 24, 15 and 36 

For, 24 = 2 × 2 x 2 x 3 

15 = 3 × 5 

36 = 2 × 2 × 3 × 3 

Now, LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5 

= 2³ x 3² x 5 

∴ LCM (24, 15, 36) = 360 

And, HCF (24, 15 and 36) = 3

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer. Then corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that

a = 6q + r where 0 ≤ r < 6

Cubing both the sides using (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ a³ = (6q + r)³ = 216q³ + r³ + 3 × 6q × r(6q + r)

⇒ a³ = (216q³ + 108q²r + 18qr²) + r³

where 0 ≤ r < 6

Case I When r = 0 we get a³ = 216q³

⇒ a³ = 6(36q³) ⇒ a³ = 6m + 0 where m = 36q³ is an integer

Case II When r = 1 we get

a³ = (216q³ + 108q² + 18q) + 1

⇒ a³ = 6(36q³ + 18q² + 3q) + 1

⇒ a³ = 6m + 1, where m = (36q³ + 18q² + 3q) is an integer.

Case III When r = 2 we get

⇒ a³ = 6(36q³ + 36q² + 12q + 1) + 2

⇒ a³ = 6m + 2

where m = (36q³ + 36q² + 12q + 1) is an integer.

Case IV When r = 3 we get

a³ = (216q³ + 324q² + 162q) + 27

⇒ a³ = (216q³ + 324q² + 162q + 24) + 3

⇒ a³ = 6(36q³ + 54q² + 27q + 4) + 3

⇒ a³ = 6m + 3

where m = (36q² + 54q² + 27q + 4) is an integer.

Case V When r = 4 we get

a³ = (216q² + 432q² + 288q) + 64

⇒ a³ = 6(36q³ + 72q² + 48q) + 60 + 4

⇒ a³ = 6(36q³ + 72q² + 48q + 10) + 4

⇒ a³ = 6m + 4

where m = (36q³ + 72q² + 48q + 10) is an integer.

Case VI When r = 5 we get

a³ = (216q³ + 540q² + 450q) + 120 + 5

⇒ a³ = 6(36q³ + 90q² + 75q + 20) + 5

⇒ a³ = 6m + 5

where m = (36q³ + 90q² + 75q + 20) + 5

Hence, the cube of a positive integer of the form 6q + r, q, is an integer and r = 0,1,2,3,4,5 is also of the forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4 and 6m + 5i.e., 6m + r.

Let the positive integer = a

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6

a² = (6q + r)² = 36q² + r² + 12qr [∵(a+b)² = a² + 2ab + b²]

a² = 6(6q² + 2qr) + r²   …(i), where,0 ≤ r < 6

When r = 0, substituting r = 0 in Eq.(i), we get

a² = 6 (6q²) = 6m, where, m = 6q² is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

a² + 6 (6q² + 2q) + 1 = 6m + 1, where, m = (6q² + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

a² = 6(6q² + 4q) + 4 = 6m + 4, where, m = (6q² + 4q) is an integer.

When r = 3, substituting r = 3 in Eq.(i), we get

a² = 6(6q² + 6q) + 9 = 6(6q² + 6a) + 6 + 3

a² = 6(6q² + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

When r = 4, substituting r = 4 in Eq.(i) we get

a² = 6(6q² + 8q) + 16

= 6(6q² + 8q) + 12 + 4

⇒ a² = 6(6q² + 8q + 2) + 4 = 6m + 4, where, m = (6q² + 8q + 2) is integer.

When r = 5, substituting r = 5 in Eq.(i), we get

a² = 6 (6q² + 10q) + 25 = 6(6q² + 10q) + 24 + 1

a² = 6(6q² + 10q + 4) + 1 = 6m + 1, where, m = (6q² + 10q + 1) is integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Hence Proved

Given, 6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5 

Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube on L.H.S and R.H.S, 

For 6q, 

(6q)³ = 216 q³ = 6(36q)³ + 0 

= 6m + 0, (where m is an integer = (36q)³) 

For 6q+1, 

(6q+1)³ = 216q³ + 108q² + 18q + 1 

= 6(36q³ + 18q² + 3q) + 1 

= 6m + 1, (where m is an integer = 36q³ + 18q² + 3q) 

For 6q+2, 

(6q+2)³ = 216q³ + 216q² + 72q + 8 

= 6(36q³ + 36q² + 12q + 1) +2 

= 6m + 2, (where m is an integer = 36q³ + 36q² + 12q + 1) 

For 6q+3, 

(6q+3)³ = 216q³ + 324q² + 162q + 27 

= 6(36q³ + 54q² + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q³ + 54q² + 27q + 4) 

For 6q+4, 

(6q+4)³ = 216q³ + 432q² + 288q + 64 

= 6(36q³ + 72q² + 48q + 10) + 4 

= 6m + 4, (where m is an integer = 36q³ + 72q² + 48q + 10) 

For 6q+5, 

(6q+5)³ = 216q³ + 540q² + 450q + 125 

= 6(36q³ + 90q² + 75q + 20) + 5 

= 6m + 5, (where m is an integer = 36q³ + 90q² + 75q + 20) 

Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Let n be any arbitrary positive odd integer. 

On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have 

n = 6m + r, where 0 ≤ r ˂ 6. 

As 0 ≤ r ˂ 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5. 

⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5 

But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 ( ∵ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even integer whereas n is an odd integer) 

⇒ n = 6m + 1 or n = 6m + 3 or n = 6m + 5 

Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer