Explore topic-wise InterviewSolutions in .

LET the numbers are x and y

Then, x + y = 66

We know that,

LCM × HCF = Product of TWO numbers

⇒ xy = 6 × 120 = 720

We know that ONE to the power of any NUMBER will always be one

UNITS DIGIT in (4211)¹⁰² = 1

Units digit in (361)⁵² = 1

∴ Units digit in (4211)¹⁰² × (361)⁵² is ALSO 1

Let there are X pages in the book. 

REMAINING pages = 160

x - 3x/11 = 160

⇒ 8x/11 = 160

⇒ x = 160 × 11/8 = 220

Hence, total NUMBER of pages = 220

From the equation,

Dividend = DIVISOR × QUOTIENT + Remainder

Here, Dividend = x² + 6

Divisor = 4

Remainder = 3

Quotient = 7

⇒ x² + 6 = 4 × 7 + 3

⇒ x² + 6 = 28 + 3

⇒ x² = 31 – 6 = 25

∴ x = 5

Let the first NUMBER be 5a.

So, the second number = 60% of 5a = 3/5 × 5a = 3a

HCF × LCM = 1^st no. × 2^nd no.

⇒ 2535 = 5a × 3a

⇒ 2535 = 15a²

⇒ a² = 169

⇒ a = 13

SOLVING the expressions,

⇒ (17/3 – 5/3) = 12/3 = 4

⇒ (22/3 – 11/3) = 11/3

⇒ (13/5 – 2/5) = 11/5

⇒ (11/5 – 3/5) = 8/5

∴ As FRACTIONAL NUMBERS are not integers, (17/3 – 5/3) = 4 is an integer

The LCM of FRACTIONAL number = LCM of numerator/HCF of denominator

Let the given number be ‘x’

ACCORDING to the given conditions,

3x/4 = 3x/14 + 135

15x/28 = 135

x = 252

∴ The given number is 252

LCM of 540, 400, and 360 = 2⁴ × 3³ × 5² = 16 × 27 × 25 = 10800 SEC

⇒ 42 = 2 × 3 × 7

Hence if a number is divisible by 42 then it must be divisible by 2, 3 and 7.

For a number to be divisible by 2, last DIGIT of a number should be 2, 4, 6 or 8.

2646, 1008, 1470, 656, 2478 and 1826 all numbers are divisible by 2.

For a number to be divisible by 3 if the sum of the digits is divisible by 3.

⇒ 2646 = 2 + 6 + 4 + 6 = 18(Divisible by 3)

⇒ 1008 = 1 + 0 + 0 + 8 = 9(Divisible by 3)

⇒ 1470 = 1 + 4 + 7 + 0 = 12(Divisible by 3)

⇒ 656 = 6 + 5 + 6 = 17 (Not divisible by 3)

⇒ 2478 = 2 + 4 + 7 + 8 = 21(Divisible by 3)

⇒ 1826 = 1 + 8 + 2 + 6 = 17(Not divisible by 3)

Divisibility rule of 7(Subtract twice the last digit from the number formed by the remaining digit)

⇒ 2646 = 264 - (2 × 6) = 264 - 12 = 252(Divisible by 7)

⇒ 1008 = 100 - (2 × 8) = 100 - 16 = 84 (Divisible by 7)

⇒ 1470 = 147 - (0 × 2) = 147 (Divisible by 7)

⇒ 656 = 65 - (6 × 2) = 65 - 12 = 53 (Not divisible by 7)

⇒ 2478 = 247 - (2 × 8) = 247 - 16 = 231(Divisible by 7)

⇒ 1826 = 182 - 6 × 2 = 182 - 12 = 170(Not divisible by 7)

So, the numbers which are divisible by 42 are 2646, 1008, 1470 and 2478.

We know that dividend = DIVISOR × QUOTIENT + remainder

= (x + 2) × (4x - 5) + 12

= 4x² + 8X - 5x - 10 + 12

∴ Dividend = 4x² + 3x + 2

The ratio of two NUMBER is4:3. Such that numbers are 4x and 3x, where ‘X’ is the constant

Product of HC.F. and L.C.M. is 2028 .

H.C.F. ×L.C.M. = product of numbers

⇒ 2028 = 4x × 3x = 12x²

⇒ x² = 2028/12 = 169

⇒ x = 13

The numbers are 3, 6, 9…., 150

Nth TERM of an AP = a + (n - 1)d

3 + (n - 1)3 = 150 ⇒ n = 50

Sum of n terms of an AP = n/2(2a + (n - 1)d)

Where a is first term, n is NUMBER of terms and d is common difference

a = 3, n = 50 and d = 3

Sum = 50/2 × (6 + 49 × 3) = 3825

TOTAL cost of the TWO pieces of cloth = 1.2 × 330 + 1.3 × 270 = Rs. 747

Given that he paid an amount of Rs. 1000 at the counter

Amount he get back = 1000 – 747 = Rs. 253

Let, 11 CONSECUTIVE numbers are,

⇒ (X - 5), (x - 4), (x - 3), (x - 2), (x - 1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5)

ACCORDING to problem,

⇒ (x - 5) + (x - 4) + (x - 3) + (x - 2) + (x - 1) + x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 11

⇒ 11x = 11

⇒ x = 1

∴ The lowest NUMBER = 1 - 5 = -4

a³b - ab³ = ab(a² – b²) = ab(a + b)(a – b)- - - (1)

a³b² + a²b³ = a²b²(a + b)- - - (2)

ab(a + b)- - - (3)

From EQUATIONS (1), (2) and (3), we get,

LCM = ab(a + b) × ab × (a – b) = a²b²(a² – b²)

∴ LCM of a³b - ab³, a³b² + a²b³ and ab(a + b) = a²b²(a² – b²)

GIVEN,

LET other NUMBER be M.

Then,

$(\RIGHTARROW {\rm{}}\frac{y}{x} = \frac{x}{{{y^2}}} \times M)$

As we KNOW,

⇒ Sum of first ‘N’ terms of square series = (1² + 2² + … + n²) = n(n + 1)(2n + 1)/6

⇒ Sum of first ‘n’ terms of CUBE series = (1³ + 2³ + … + n³) = [n(n + 1)/2]²

We can write,

Let the first term of AP be a and common difference be d

From the problem’s statement

⇒ 3^rd term = a + (3 - 1) d = a + 2d = 13----(i)

⇒ 5^th term = a + (5 - 1) d = a + 4D = 21----(ii)

Now (ii) - (i)

⇒ 2d = 8

⇒ d = 4, PUT this in (i)

⇒ a = 13 - 8 = 5

Now the 13^th term can be given as a + 12d

⇒ 13^th term = 5 + 12 × 4 = 53

The LEAST number divisible by 8, 12 and 28 is LCM of these numbers i.e. 168. Clearly, any multiple of 168 would be EXACTLY divisible by each of the numbers 8, 12 and 28.

But the required number is not to EXCEED 900. We can find the required number USING following steps-

Step-1-

Divide 900 by 168,

⇒ (900/168) ⇒ we find quotient 5.

Step-2-

Multiply 168 by the quotient 5, which is our required number.

⇒ Required number = 168 × 5 = 840

GIVEN that, 1² + 2² + 3² + … n² $(= \;\frac{{n\;\LEFT( {n\; + \;1} \right)\left( {2n\; + \;1} \right)}}{6})$

⇒ 1² + 2² + 3² + … 17² $(= \;\frac{{17\;\left( {17\; + \;1} \right)\left( {2\; \times 17\; + \;1} \right)}}{6}\; = \;1785)$

And,

1² + 2² + 3² + … 7² $(= \;\frac{{7\;\left( {7\; + \;1} \right)\left( {2\; \times 7\; + \;1} \right)}}{6} = 140)$

LET the numbers p and q be 4X and 3X respectively.

4x = 4 × x

3x = 3 × x

LCM (4x, 3x) = 4 × 3 × x = 36

∴ x = 36/12 = 3

The numbers p and q are 12 and 9 respectively.

∴ Their sum = 12 + 9 = 21 

⇒ 13³? = (13³)¹² = (2197)¹² = (2196 + 1)¹²

⇒ When (2196 + 1)¹² is divided by 2196 all the TERM is divisible by 2196 EXCEPT 1¹²

⇒ HCF MUST be a factor of LCM.

⇒ Factor of 120 = 2 × 2 × 2 × 3 × 5

⇒ From above we SEE that 8, 12 and 24 are the factor but not 35

Given,

NUMBER is DIVISIBLE by 7 and when divided by 5, 10, 12 and 15 LEAVES remainder 2 in each case

∴ Smallest number which is divisible by 5, 10, 12 and 15

= LCM of 5, 10, 12 and 15

Finding LCM

5 = 1× 5

10 = 2 × 5

12 = 2 × 2 × 3

15 = 3 × 5

∴ LCM = 2 × 2 × 3 × 5 = 60

∴ The number which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case is (60n + 2) where, n is an integer.

Now, we have to find the smallest value of n for which (60n + 2) is divisible by 7.

For, n = 1, 60n + 2 = 62, not divisible by 7

For, n = 2, 60n + 2 = 122, not divisible by 7

For, n = 3, 60n + 2 = 182, which is divisible by 7

Product of the digits should be LESS than or EQUAL to the sum of the digits;

This is possible only if the NUMBER has 0 or 1.

All such numbers more than 10 and less than 40 are 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 30, 31

The number 22 will also satisfy the given condition (As the Product of the digits is equal to the sum of the digits)

The NUMBER obtained is greater than the number obtained on reversing the digits, thus the TEN’s digit is greater than the unit’s digit

Let ten’s digit and unit’s digit be 2x and x respectively

Then, [10(2x) + x] + (10x + 2x) = 132

⇒ 21x + 12x = 132

⇒ 33X = 132

⇒ x = 4

The product between the sum and difference of the digits of the number = (2x + x)(2x – x)

⇒ [2(4) + 4][2(4) – 4]

⇒ 12 × 4 = 48

∴ The product is 48

Any number can be expressed in the form (6k + n), where k is an integer, and n varies from 0 to 5.

Now, when the square of this number is divided by 6, the REMAINDER will be the same as the remainder obtained by dividing n² by 6.

For square of (6k + 1), remainder will be the same as remainder obtained by dividing 1² by 6, i.e. 1.

For square of (6k + 2), remainder will be the same as remainder obtained by dividing 2² by 6, i.e. 4.

For square of (6k + 3), remainder will be the same as remainder obtained by dividing 3² by 6, i.e. 3.

For square of (6k + 4), remainder will be the same as remainder obtained by dividing 4² by 6, i.e. 4.

For square of (6k + 5), remainder will be the same as remainder obtained by dividing 5² by 6, i.e. 1.

Numbers with H.C.F 15 must contain 15 as a factor.

Multiples of 15 between 40 and 100 = 45, 60, 75 and 90

Number of pairs with H.C.F 15 are (45, 60), (45, 75)(60, 75) and (75, 90)

H.C.F of (60, 90) is 30 and that of (45, 90) is 45.

Let the tens and UNITS DIGITS of the NUMBER be X and y respectively

x × y = 72

⇒ y = 72/x

10x + y + 9 = 10y + x

⇒ 9Y – 9x = 9

⇒ 72/x – x = 1

⇒ x² + x – 72 = 0

⇒ (x + 9)(x – 8) = 0

⇒ x = 8

y = 72/8 = 9

∴ Number is 89

To be divisible by all three of 5, 15 and 20, the NUMBER must be divisible by LCM of 5, 15 and 20

LCM of 5, 15 and 20 is 60 (= 2 × 3 × 2 × 5).

Thus, the number must be a multiple of 60.

⇒ 9999/60 = 166.65

∴ Largest multiple of 60 less than 9999 is: 60 × 166 = 9960

LET the number be y

According to the PROBLEM statement

⇒ y + 20/y = 9

⇒ y² - 9y + 20 = 0

⇒ (y - 5) (y - 4) = 0

the REQUIRED number is 5 which satisfies the above given condition

We can WRITE,

⇒ √128 = √(2⁶ × 2) = 2³√2 = 8√2

⇒ √72 = √(2² × 3² × 2) = 2 × 3√2 = 6√2

⇒ √32 = √(2⁴ × 2) = 2² × √2 = 4√2

Hence,

⇒ (√128 + √72)/√32

= (8√2 + 6√2)/4√2

= 14√2/4√2

= 7/2

For any number to be completely divisible by 80, it must be divisible by 8 and 10 both

So, for the number 347XY to be divisible by 10 the LAST digit must be a 0

This means, Y = 0 and also for any number to be divisible by 8 the last three digits of the number must be divisible by 8.

When 347X0/(10 × 8)

THUS gives 347X divides exactly by 8

So, the last three digits ie 47X must be divisible by 8

This means X = 2 is the only possible value

⇒ (2³¹ + 2³² + 2³³ + 2³⁴⁾

⇒ 2³¹ (2⁰ + 2¹ + 2² + 2³⁾

⇒ 2³¹ (1 + 2 + 4 + 8)

⇒ 2³¹ × 15

⇒ 2³⁰ × 2 × 15

⇒ 3 ? 10 × 2³⁰

∴ Given number will be divisible by 10

Considering STATEMENT I,

⇒ √324 = 18

⇒ √3.24 = √(324/100) = 18/10 = 1.8

⇒ √0.0324 = √(324/10000) = 18/100 = 0.18

⇒ √324 + √3.24 + √0.0324 = 18 + 1.8 + 0.18 = 19.98

Considering statement II,

⇒ √(129 + √121 + √361 + √100)

= √(129 + 11 + 19 + 10)

= √169 = 13

A number M is divisible by 25 if it ends with 00, 25, 50, or 75.

(M + 5)(M + 1) = M² + 6M + 5

When a number (X + Z) is always divisible by y, the remainder of (x + z) when divided by y will be zero

When M² + 6M + 5 is divided by 25, the remainder will be due to sum of that.

M is divisible by 25, M² is also divisible by 25

6M will also be divisible by 25

5 when divided by 25 leaves the remainder of 5

So the sum M² + 6M + 5 is divided by 25, the remainder will be 5

When a number is divisible by both 4 and 7, then the number is ALSO divisible by 4 × 7 = 28.

The numbers from 33 to 143 which are divisible by 28 are 56, 84, 112 and 140

Hence, there are 4 numbers between 33 and 143 which are divisible by 4 and 7.

Let the 3 numbers be n - 2, n, n + 2

⇒ (n - 2)(n)(n + 2) = 1287

⇒ (n² - 4)(n) = 1287

⇒ n³ - 4n - 1287 = 0

⇒ n = 11 is the only odd NUMBER that is the root of above equation

H.C.F of two NUMBERS is the highest factor present in both the numbers.

So we can say, if the HCF of two numbers is 9 then the two numbers can be 9x & 9y where x & y are co-prime.

The LCM of these two numbers = PRODUCT of these two numbers

[Because co-prime numbers don’t have any common factors other than 1]

So, LCM of 9x & 9y = 9xy [Only one 9 in the LCM because it’s a common factor of both the numbers]

∴We can conclude that the LCM must be an integral multiple of 9 as x & y are integers.

Given the expression: $(\frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }})$

Multiplying the numerator and DENOMINATOR by (7 - 4√3), we get

$(\Rightarrow \frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }} = \frac{{\LEFT( {4 + 3\sqrt 3 } \right) \TIMES \left( {7 - 4\sqrt 3 } \right)}}{{\left( {7 + 4\sqrt 3 } \right) \times \left( {7 - 4\sqrt 3 } \right)}})$

We know that, (a + b) (a - b) = a² – b²

Hence,

$(\Rightarrow \frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }} = \frac{{\left( {4 + 3\sqrt 3 } \right) \times \left( {7 - 4\sqrt 3 } \right)}}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}})$

$(= \frac{{\left( {28 + 21\sqrt 3- 16\sqrt 3- 36} \right)}}{{49 - 48}})$

Formula:

(a + B)² = a² + b² + 2ab

(a – b)(a + b) = a² – b²

Given,

x = 3 + 2√2

$(\RIGHTARROW \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }})$

Multiplying and dividing by 3 – 2√2

$(\Rightarrow \frac{1}{x} = \frac{{3 - 2\sqrt 2 }}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}\;)$

⇒ 1/x = 3 – 2√2

$({\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} + 2)$

$(\Rightarrow {\left( {3 + 2\sqrt 2+ 3 - \;2\sqrt 2 } \right)^2} - 2 = {x^2} + \frac{1}{{{x^2}}})$

As per the given DATA,

We know that DIVIDEND = DIVISOR × quotient + remainder

⇒ 11509 = divisor × 71 + 7

⇒ 11502 = divisor × 71

⇒ Divisor = 162

For this to be divisible by 11, sum of numbers at even places - sum of numbers at odd places MUST be zero or divisible by 11

⇒ (4 + 1) - (3 + 0) = 5 - 3 = 2

instead of 2 there must be 0, so that it will be divisible by 11

Let the number be 3x and 4x.

LCM of 3x and 4x = 3 × 4 × x = 240.

⇒ 12X = 240.

⇒ x = 20.

LET the NUMBERS be 2x and 3x

LCM of 2x and 3x = 6X

Given that LCM = 66

⇒ 6x = 66

⇒ x = 66/6 = 11

∴ The numbers are 22 and 33

∴ SUM of the numbers = 22 + 33 = 55

Time interval between 3^rd and 1^st alarm = 88.88% of 180 = 8/9 × 180 = 160 secs

∴ REQUIRED time = LCM of (220, 180, 160) = 15840 SECONDS = 15840/60 = 264 min

Given, if your allowance was a Re.1 and it is DOUBLED every day

Thus, after (n + 1) DAYS, the allowance = 2ⁿ × 1 = 2ⁿ

⇒ 2ⁿ > 35

⇒ 2⁶ = 64

⇒ n = 6