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(b) Limiting nutrients

0, 3, 8, 15, 24, __

Series follows the following pattern:

0 + 3 = 3

3 + 5 = 8

8 + 7 = 15

15 + 9 = 24

24 + 11 = 35

∴ Next number is 35

Considering the given series

4, 6, 9, 36, 52, 177, 213

The logic of the given series can be explained as

4 + 1³ = 5

5 + 2² = 9

9 + 3³ = 36

36 + 4² = 52

52 + 5³ = 177

177 + 6² = 213

∴ Wrong term in given number series is 6.

Given:

We have to find the wrong term in the given series

93, 113, 132, 159, 185, 213

Solution:

The given series follows the following pattern

9 × 10 = 90 + 3 = 93

10 × 11 = 110 + 3 = 113

11 × 12  = 132 + 3 = 135 ≠ 132

12 × 13 = 156 + 3 = 159

13 × 14 = 182 + 3 = 185

14 × 15 = 210 + 3 = 213

∴ In the above series 132 is wrong number

The series follows the given pattern:

3 + 14 × 1 = 17

17 + 14 × 2 = 45

45 + 14 × 3 = 87

87 + 14 × 4 = 143 (X)

143 + 14 × 5 = 213 

∴ the value of X in the given series 3, 17, 45, 87, X, 213 is 143

Series follows the following pattern:

625 × (1/5) = 125

125 × (2/5) = 50

50 × (3/5) = 30

30 × (4/5) = 24

24 × (5/5) = 24

∴ The next number in the series 625, 125, 50, 30, 24, ? is 24

Formula used:

 \(HCF \ of \ fractions =\frac{HCF \ of \ numerators}{LCM \ of \ denominators}\)

Calculations:

Factors of 7 = 7 × 1

Factors of 14 = 2 × 7

Hence HCF of numerators (7, 14) = 7

Factors of 90 = 2 × 3 × 3 × 5

⇒ 2¹ × 3² × 5¹

Factors of 15 = 3 × 5

⇒ 3¹ × 51

Factors of 10= 2 × 5

⇒ 21 ×  51

LCM of denominators (90, 15, 10) = 2 × 5 × 3² = 90

∴  HCF of fractions 7/90, 14/15 and 7/10 is 7/90

Given:

The cost of a piece of stone varies with the square of its weight loss.

⇒ Price = k × weight²......(k is any constant)       

Let the weight of a piece of stone be (3a + 5a = 8a).

Calculation:

The initial value of stone = k.64a²

⇒ 64ka² = 1600

Then,

The new value of stone becomes = k (9a² + 25a²) = 34ka²

New value of stone = (34 × 1600)/64 = Rs. 850

∴ Loss occurred = 1600 - 850 = Rs. 750

Series follows the following pattern:

15 × 4 + 5 = 65

65 × 2 - 5 = 125

125 × 1 + 5 = 130

130 × 0.5 - 5 = 60

60 × 0.25 + 5 = 20

∴ The missing term is 60

Given:

(i) Cost of 4 socks and 5 mufflers = 5400

(ii) Cost of 2 socks and 7 mufflers = 4500

Calculations:

⇒ Let the cost price of one pair of socks be x.

⇒ Let the cost price of one muffler be y.

⇒ According to the question

⇒ 4x + 5y = 5400           …………………..(i)

⇒ 2x + 7y = 4500           ……………………(ii)

⇒ After solving equation (i) and (ii),

⇒ y = 400 ; x = 850

Required Ratio:

= 400 : 850

= 40 : 85

= 8 : 17.

Given:

A man saved = Rs. 600

Number of days = 30 days 

Calculations:

In one day saved money = 600 ÷ 30

⇒ Rs. 20

 The number of days needed by him to save Rs. 1,160 = 1160 ÷ 20

⇒ 58 days

∴  The number of days needed by him to save Rs. 1,160 is 58 days

Given:

Number  = 7386038

Concept used:

Divisibility rule 3:

A number is divisible by 3 if the sum of the number is divisible by 3.

Sum of numbers = 7 + 3 + 8 + 6 + 0 + 3 + 8 = 35 

As, 35 is not divisible by 3

So, the number 7386038 is also divisible by 3

Divisibility rule of 9:

A number is divisible by 9 if the sum of the digits is divisible by 9 and is exactly divisible by 3.
Sum of numbers = 7 + 3 + 8 + 6 + 0 + 3 + 8 = 35

AS, 35 is not divisible by 9

So, the number 7386038 is also divisible by 9

Divisibility rule of 11:

A number is divisible by 11, if the alternating sum of the digits in the number, read from left to right. If that is divisible by 11 the number is divisible by 11.
Sum of alternating numbers

7 + 8 + 0 + 8 = 23 

3 + 6 + 3 = 12
Difference = 23 - 12 = 11
AS, 11 (i.e. difference) is divisible by 11

So, the number 7386038 is also divisible by 11

∴ The number is divisible by 11

If the number is not divisible by 3 the number is not divisible by 9

As, 3 is the factor of 9

GIVEN :

The loss occurred is Rs. 3080

The ratio between the weights of two pieces is 2 : 7

CONCEPT :

The cost of a piece of gem stone varies with the square of its weight.

⇒ Price = k × weight²......(k is any constant)       

ASSUMPTION :

Let the weight of a piece of gem stone be (2a + 7a = 9a)

CALCULATION :

The initial value of diamond = 81ka²

Let the original price of a diamond be Ra.M.

⇒ 81ka² = M

Then,

The new value of diamond becomes = k (4a² + 49a²) = 53ka²

New value of diamond = (53 × M)/81

Then,

⇒ M - 53M/81 = 3080

⇒ M = Rs. 8910

∴ The original price of gem stone is Rs.8910.

Given:

Ratio of males and females of a village = 5 : 3

Number of males in village = 800

Calculation:

Let the number of males and females be 5x and 3x respectively

⇒ 5x = 800

⇒ x = 160

Number of females = 3x

⇒ 3 × 160

⇒ 480

∴ The number of females is 480

Given:

Product of numbers = 5400

HCF of numbers = 30

Formula required:

Product of two numbers = LCM of two number × HCF of two numbers

Calculations:

Let the numbers be '30a' '30b'

5400 = 30a × 30b 

⇒ a × b = 5400/(30 × 30) 

⇒ a × b = 6

Factors of 6 = 1, 2, 3, 6

Hence, we can write in 6 the form of  a × b = (2 × 3) or (1 × 6) 

Hence, the pair of numbers is = [(30 × 2), (30 × 3)] or [(30 × 1), (30 × 6)]

⇒ (60, 90) or (30, 180)

∴ The possible number of pairs is 2

GIVEN :

The ratio of weight of pieces is 3 : 2 : 1

The diamond of worth Rs. 9000 is accidently dropped and broke into 3 pieces.

CONCEPT :

The cost of a piece of diamond varies with the square of its weight.

⇒ Price = k × weight²......(k is any constant)       

ASSUMPTION :

Let the weight of a piece of diamond be (3a + 2a + a = 6a)

CALCULATION :

The initial value of diamond = 36ka²

⇒ 36ka² = 9000

Then,

The new value of diamond becomes = k (9a² + 4a² + a²) = 14ka²

New value of diamond = (14 × 9000)/36 = Rs. 3500

∴ Loss occurred = 9000 - 3500 = Rs. 5500

(D) None of these

Correct option:

(A) 1

Solution: For a rational number p/q to have terminating decimal representation, the prime factorisation of q should be of the form 2^m x 5ⁿ, where m and n are non-negative integers.

Total distance is S = 480 km.

Let the speed of the train is x km/hour.

∴ Time taken to cover the distance 480 km is t = \(\frac{480}{x}\) hours … (1)

Given that if speed of the train is decreasing by 8 km/hour, it take 3 more hours to cover the same distance.

∴ t +3 = \(\frac{480}{x\, - \,8}\)

⇒ \(\frac{480}{x}+3=\frac{480}{x\,-\,8}\) (By putting the value of t)

⇒ 480(x – 8 – x) + 3x (x –8) = 0 

⇒ 3x (x – 8) = 480 × 8 

⇒ x (x – 8) = 160 × 8 = 1280. 

⇒ x² – 8x – 1280 = 0 

⇒ x² – 40x + 32x – 1280 = 0 

⇒ x (x – 40) + 32 (x – 40) = 0 

⇒ (x + 32)(x – 40) = 0 

⇒ x = –32 or x = 40 

⇒ x = 40. (∵ Speed never be negative) 

Hence, usual speed of the train is 40 km / hour.

Solution:
HCF(420, 130) will give the maximum number of barfis that can be placed in each stack.
By Euclid’s division algorithm,
420 = 130 x 3 + 30
130 = 30 x 4 + 10
30 = 10 x 3 + 0
So, HCF(420, 130) = 10
Hence, the sweetseller can make stacks of 10 for both kinds of barfi.

Given that \(\alpha,\beta\) are zeros of kx² – 2x + 3k such that \(\alpha + \beta\) = \(\alpha\,\beta\). 

Since, α & β are zeros of kx² – 2x + 3k. 

Therefore, sum of zeros = \(\alpha + \beta\) = \(\frac{-b}{a}\) = \(\frac{-(-2)}{k}\) = \(\frac{2}{k}\). (In kx² – 2x + 3k, a = k, b = –2, c = 3k) 

& product of zeros \(\alpha,\beta\) = \(\frac{c}{a} = \frac{3k}{k} = 3\). 

Now, given that \(\alpha + \beta\) = \(\alpha\,\beta\) 

⇒ \(\frac{2}{k}\) = 3. 

Hence, k = \(\frac{2}{3}\) .

If α,β are the zeroes of kx² – 2x + 3k

then α+β=2/k and αβ=3k/k=3

again we have α+β=αβ

so 2/k =3

=>k=2/3

We have function f(x) = x² + 1 on set R.

We can clearly seen that the function is defined for all real values of x.

Therefore, the domain of the function f(x) is R.

Let x ∈ . Then, x² ≥ 0.

⇒ x² + 1 ≥ 1.(By adding 1 both sides of inequality)

⇒ f(x) ≥ 1.

Therefore, the range of the function f(x) is [1, ∞).

Hence, the domain of function f(x) is R and the range of the function f(x) is [1, ∞).

Let one zero of the given polynomial is α.

Given that one zero of given polynomial is reciprocal of the other.

∴ The other zero of given polynomial is \(\frac{1}{a}.\)

Hence, the zeros of the given polynomial (α² + 9)x² + 13x + 6 are and \(\frac{1}{a}.\)

Now, the product of the zeros is α. \(\frac{1}{a}\) = \(\frac{6a}{a^2+9}\)

⇒ \(\frac{6a}{a^2+9}=1\)

⇒ α² + 9 = 6a

⇒ α² − 6 + 9 = 0

⇒ (α − 3)² = 0 (∵ (α − b)² = α² − 2ab + b²)

⇒ α = 3.

Hence, the value of is 3.

Given system of equation is x + 2y = 3 

⇒ x + 2y – 3 = 0 

And 5x + ky + 7 = 0. 

By comparing given system of equations with a₁x + b₁y + c₁ = 0 & a₂x + b₂y + c₂ = 0, 

We get a₁ = 1, b₁ = 2, c₁ = –3 and a₂ = 5, b₂ = k, c₂ = 7. 

Since, given that system of equations has no solution.

\(\therefore \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.\)

Therefore, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) & \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ \(\frac{1}{5}\) = \(\frac{2}{k}\)⇒ k = 2× 5 = 10. (By cross multiplication) 

And \(\frac{2}{k}\) ≠ \(\frac{-3}{7}\) 

⇒ –3k ≠14 

⇒ k ≠ \(\frac{-14}{3}\) .

Hence, for k = 10, the given system of equations has no solution.

Given system of linear equations is

3x – 2y = 5 … (1) 

2x + y = 7 … (2) 

Putting x = 3 and y = 2 in equation (2), 

L.H.S = 2 × 3 + 2 = 6 + 2 = 8 ≠ 7 = R.H.S

Hence, x = 3 and y = 2 not satisfying equation (2) which implies that x = 3 and y = 2 is not a solution of the given system of equations.

Hence Proved

Given that HCF of two numbers is 27 and their LCM is 162. 

One number is 81. 

Let the other number is a. 

We know that for two positive integer a & b, HCF (a, b) × LCM (a,b) = a × b.

Therefore, HCF (81, a) × LCM (81, a) = 81 × a

⇒ a = \(\frac{HCF(81,a) \times LCM(81,a)} {81}\)

⇒ a = \(\frac{27 \times 162}{81} = 27 \times 2 = 54.\)

Hence, the other number is 54.

24 × 1.5 = 36

36 ÷ 2 = 18

18 × 1.5 = 27

Hence, option 4 is correct.

Given:

We have to find the missing term in the given series

16, 54, 128, ____, 432

Calculation:

The given series followes the following pattern

2 × 2³ = 16

2 × 3³ = 54

2 × 4³ = 128

2 × 5³ = 250

2 × 6³ = 432

∴ Required number is 250

Considering the above series,

8, 12, 18, 26, 36, ?

The logic of the series can be explained as,

12 - 8 = 4

18 - 12 = 6

 26 - 18 = 8

36 - 26 = 10

By observing the pattern, we can write

? - 36 = 12

⇒ ? = 48

∴ The value of ? is 48

The pattern of the given series :

⇒ 1 + 2 = 3

⇒ 3 + 4 = 7

⇒ 7 + 6 = 13 

⇒ 13 + 8 = 21 

⇒ 21 + 10 = 31 

∴ Answer will be 21 

We know that a rational number has a terminating decimal expansion, if the prime factorization of the denominator is of the form of 2^m5ⁿ, where m and n are non negative integers.

Now solving one by one.

A) 31/3125 \(=\frac{31}{5\times5\times5\times5\times5}=\frac{31}{5^52^0}\)

So, given number has terminating decimal expansion.

B) 17/512 \(=\frac{17}{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}=\frac{17}{2^95^0}\)

So, given number has terminating decimal expansion.

C) 23/200 \(=\frac{23}{2\times2\times2\times5\times5}=\frac{23}{2^35^2}\)

So, given number has terminating decimal expansion.

Hence option (D) has a non terminating repeating decimal expansion.

Calculation:

The series follows the following pattern

12960 ÷ 6 = 2160

2160 ÷ 5 = 432

432 ÷ 4 = 108

108 ÷ 3 = 36

36 ÷ 2 = 18

∴ The missing number is 2160.

Calculation:

The pattern of the number series is:

⇒ 5 + 3 = 8

⇒ 8 + 6 = 14 

⇒ 14 + 12 = 26 

⇒ 26 + 24 = 50 

⇒ 50 + 48 = 98 

∴ The value of ? is 50

Given :

The sum of three numbers is 280 

The ratio of 1st to 2nd number is 2 : 3 and the ratio of 2nd to third number is 4 : 5

Concept used :

If the ratio of x and y is a : b and the ratio of y and z is c : d then, 

x : y : z = a × c : c × b : b × d 

Calculations :

Let the three numbers be p, q, and r 

p : q : r = 2 × 4 : 4 × 3 : 3 × 5 

p : q : r = 8 : 12 : 15 

Let the value of p, q and r be 8x, 12x and 15x respectively 

According to the question 

Sum of all the numbers is 280 

so, 

8x + 12x + 15x = 280 

⇒ x = 8 

So, p = 64, q = 96 and r = 120 

∴ The second number will be 96 

i. 75 = 3 × 25 

= 3 × 5 × 5 

135 = 3 × 45 

= 3 × 3 × 15 

= 3 × 3 × 3 × 5 

∴ HCF of 75 and 135 = 3 × 5 

= 15 

LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3 

= 675 

ii. 114 = 2 × 57 

= 2 × 3 × 19 

76 = 2 × 38 

= 2 × 2 × 19 

∴ HCF of 114 and 76 = 2 × 19

= 38 

LCM of 114 and 76 = 2 × 19 × 3 × 2 

= 228

iii. 153 = 3 × 51 

= 3 × 3 × 17 

187 = 11 × 17 

∴ HCF of 153 and 187 = 17 

LCM of 153 and 187 = 17 × 3 × 3 × 11 

= 1683

v. 32 = 2 × 16 

= 2 × 2 × 8 

= 2 × 2 × 2 × 4 

= 2 × 2 × 2 × 2 × 2 

24 = 2 × 12 

= 2 × 2 × 6 

= 2 × 2 × 2 × 3 

48 = 2 × 24 

= 2 × 2 × 12 

= 2 × 2 × 2 × 6 

= 2 × 2 × 2 × 2 × 3 

∴ HCF of 32, 24 and 48 = 2 × 2 × 2 

= 8 

LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3 

= 96

Calculation:

The series follows the following pattern:

⇒ 6 × 0.5 = 3

⇒ 3 × 1 = 3

⇒ 3 × 1.5 = 4.5

⇒ 4.5 × 2 = 9

⇒ 9 × 2.5 = ? = 22.5

 ∴ The value of ‘?’ is 22.5.

Given:

The train ticket fare in 2nd class AC = Rs. 2500

The train ticket fare in 3rd class AC = Rs. 2000

Fare increase in 2^nd class AC = 20%

Fare increase in 3^rd class AC = 10%

Calculations:

The train ticket fare in 2^nd class AC and 3^rd class AC after the increase

New train ticket fare in 2nd class AC = (120/100) × 2500

⇒ Rs. 3000

New train ticket fare in 3rd class AC = (110/100) × 2000

⇒ Rs. 2200

Ratio of new fares of 2 and 3 class AC = 3000 ∶ 2200

⇒ 15 ∶ 11

∴ The ratio of the new fares of 2nd class AC and 3rd class AC is 15 ∶ 11

Given-   

Ratio of boys and girls in a school = 27 : 23

Difference between the number of boys and girls = 200

Calculation-

Let the number of boys and girls be 27x and 23x respectively.

According to Condition-

27x - 23x = 200

⇒ x = 50

The number of boys = 27 × 50

⇒ 1350

∴ The number of boys is 1350.

The logic behind the following series is as follows:

33 + 3² = 42

42 + 5² = 67

67 + 7² = 116

116 + 9² = 197

197 + 11² = ? = 318

 ∴ The value of ‘?’ is 318.

The series follows the following pattern:

45 + 2² = 49

49 – 3² = 40

40 + 4² = 56

56 – 5² = 31 (?)

31 + 6² = 67

∴ 31 will come in the place of question mark (?)

Explanation:

The resulting sum is added with alternative subtraction and addition of squares of consecutive numbers i.e  + 2² – 3² + 4² – 5² + 6²

Correct option is C) fasten

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