Explore topic-wise InterviewSolutions in Current Affairs.

Given:

An 8- digit number: 4252746B

Remainder = 0 when divided by 3

Calculation:

As per the divisibility tests, the sum of all the digits of the number should be divisible by 3

So, 

4 + 2 + 5 + 2 + 7 + 4 + 6 + B

= 30 + B

30 is completely divisible by 3. So, for remainder 0, the possible values are 0, 3, 6, 9.

∴ The number of possible values of B is 4

Given:

√33489 = ?

Calculation:

√33489 = √(3 × 3 × 61 × 61)

= 3 × 61 = 183

∴ √33489 = 183. 

GIVEN:

A number between when divided by 7, 9 and 11 leaves remainder 5 in each case.

CALCULATION:

Smallest possible number which can be divided by 7, 9 and 11 = LCM of 7, 9 and 11 = 693

Number just greater than 9000 which is also a multiple of 693 = 9009

Required number = 9009 + 5

= 9014

Given:

Numbers are 720 & 600.

Concept used:

First, break the numbers into prime factors.

Pick the least common power of each prime number from both the numbers.

Explanation:

720 = 2⁴ × 3² × 5¹

600 = 2³ × 3¹ × 5²

Least power of '2' is 3.

Least power of '3' is 1.

Least power of '5' is 1.

⇒ HCF = 23 × 31 × 5¹

∴ The largest common factor of 720 & 600 is 23 × 31 × 5¹

 You have to find the largest common factor and LCM stands for Lowest common multiple. HCF stands for Highest/largest common factor.

Given:

S.P = Rs. 54/kg

Loss = 10%

Formula used:

C.P = S.P × [100/(100 – Loss%)]

Calculation:

C.P = Rs. 54 × (100/90) = Rs. 60

To earn 20% profit

S.P = Rs. 60 × (120/100) = Rs. 72

Calculation:

(1/2)³ × (1/2)³

⇒ (1/8) × (1/8)

⇒ 1/64

∴ The answer is 1/64.

Given:

Investment of A = Rs. 9000

Investment of B = Rs. 7500

Ratio of their profit = 12 : 5

Concept used:

Partnership & Ratio method

Calculation:

Let the time period for B's investment be x

So, (9000 × 12)/(7500 × x) = 12/5

(90 × 12)/(75 × x) = 12/5

X = (90 × 12 × 5)/(12 × 75)

X = 6 months

∴ Time for which B invested in the business is 6 months

Concept Used:

The total number of factors of any natural number (N) = a^p×b^q×c^r×………, will be

Factors = (p + 1)(q + 1)(r + 1)………, Where a,b,c……. are prime numbers and p,q,r….. are natural numbers.

 Calculation:

Now as the given number is (30)⁴ represents this in the terms of factors.

⇒ (30)⁴ = (5×6)⁴ = (5×2×3)⁴ = 2⁴×3⁴×5⁴

Now apply the above-mentioned formula

⇒ The total number of factors = (4 + 1)(4 + 1)(4 + 1)

⇒ The total number of factors = 5×5×5 = 125

∴ The total number of factors is 125

Prime Number - Numbers which are divisible by 1 and itself only are called as prime numbers.

Calculation

Prime numbers between 30 and 42 are 31, 37 and 41.

∴ Sum = 31 + 37 + 41 = 109

Given:

a = \(\frac{{2\; - \;\sqrt 5 }}{{2\; + \;\sqrt 5 }}\)

b = \(\frac{{2\; + \;\sqrt 5 }}{{2\; - \;\sqrt 5 }}\)

Formula Used: 

a² – b² = (a + b) (a – b)

(a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

Calculation:

a + b = \(\frac{{2\; - \;\sqrt 5 }}{{2\; + \;\sqrt 5 }}\) + \(\frac{{2\; + \;\sqrt 5 }}{{2\; - \;\sqrt 5 }}\)

⇒ a + b = \(\frac{{{{\left( {2\; - \;\sqrt 5 } \right)}^2}\; + \;{{\left( {2\; + \;\sqrt 5 } \right)}^2}}}{{\left( {2\; + \;\sqrt 5 } \right)\left( {2\; - \;\sqrt 5 } \right)}}\)

⇒ a + b = \(\frac{{4\; + \;5\; - \;4\sqrt 5 \; + \;4\; + \;5\; + \;\;4\sqrt 5 }}{{4\; - \;5}}\)

⇒ a + b = 18/(–1)

⇒ a + b = (–18)

a – b = \(\frac{{2\; - \;\sqrt 5 }}{{2\; + \;\sqrt 5 }}\) – \(\frac{{2\; + \;\sqrt 5 }}{{2\; - \;\sqrt 5 }}\)

⇒ a – b = \(\frac{{{{\left( {2\; - \;\sqrt 5 } \right)}^2}\; - \;{{\left( {2\; + \;\sqrt 5 } \right)}^2}}}{{\left( {2\; + \;\sqrt 5 } \right)\left( {2\; - \;\sqrt 5 } \right)}}\)

⇒ a – b = \(\frac{{4\; + \;5\; - \;4\sqrt 5 \; - \;\left( {4\; + \;5\; + \;4\sqrt 5 \;} \right)}}{{4\; - \;5}}\)

⇒ a – b = (–8√ 5 )/(–1)

⇒ a – b = 8√ 5 

a² – b² = (–18) × (8√5)

⇒ a² – b² = (–144√5)

∴ The value of a² – b² is –144√5

(b) 38² = 1444 

39² = 1521 

Required number = 1521 – 1500 = 21

Given:

The HCF and LCM of polynomial p(m, n) and q(m, n) is 4m²(m² - n²) and (m³ – n³)

Formula used:

Product of polynomial = Product of HCF and LCM of polynomials

p(x) × q(x) = LCM of (p(x) and q(x)) × HCF of (p(x) and q(x))

Calculation:

By using the given formula Product of polynomial = Product of HCF and LCM of polynomials

∴ p(m, n) × q(m, n) = LCM of (p(m, n) and q(m, n)) × HCF of (p(m, n) and q(m, n))

⇒ p(m, n) × q(m, n) = 4m²(m² - n²) × (m³ – n³)

⇒ p(m, n) × q(m, n) = 4m² {m⁵ + n⁵ – m²n² (m + n)}

⇒ p(m, n) × q(m, n) = 4m⁷ + 4m²n⁵ – 4m⁴n² (m + n)

Hence, option (1) is correct

Number of toothpicks to built the top trapezoid = 5

Number of toothpicks to built 2^nd row trapezoid = 9

Number of toothpaste to built 3^rd row trapezoid = 13

Here, we have to find how many trapezoids will be built by using 1000 toothpicks.

(i) \(\because\) 5, 9, 13,...... are A.P.

whose first term is a = 5

common difference id d = a₂ - a₁ = a₃ - a₃ = 4

Here, we have to find for which value of n.

the sum of A.P. is equal to 1000 or just less than or greater than 1000.

Sum of A.P. = \(\frac{n}{2}[2a + (n-1)d]\)

\(=\frac{n}{2}[10 + (n-1)7]\)      (\(\because\) a = 5, d = 4)

\(=\frac{n}{2}\times2[5 + 2n- 2]\)

= n (3 + 2n)

\(\because\) At n = 21, sum of A.P. 

= 21 (3 + 42)

= 21 x 45 

= 545 < 1000

But at n = 22, sum of A.P. 

= 22 (3 + 44)

= 22 x 47

= 1034 > 1000

So, we can built 21 trapezoids in last complete row with the help of 945 toothpicks and there will be 55 toothpicks still left.

To built 22 trapezoids we will need 34 more toothpick

(ii) \(\because\) a₁ = 5, a₂ = 9, a₃ = 13,.....

\(\therefore \) a_n = a + (n-1)d

\(\therefore \) a₂₁ = 5 + (21 - 1)4       (a = a₁ = 5, d = 4, n = 21)

= 5 + 20 x 4

= 85

Hence, in last complete row (21^throw) we have to use 85 toothpicks to construct trapezoids in the last complete row.

Answer: (x+5)(x+10)=x²+9x+20

Explanation:(x+5)(x+4)

                = x(x+4)+5(x+4)

               = x²+4x+5x+20

               = x²+9x+20

1. The king gave the innocent prisoner a sum of money and released him from jail.
2. After his release from the jail, the man went straight to the market, where some birds were kept for sale.
3. The shopkeeper was surprised because the man set all the birds free which he had purchased.
4. The man told the shopkeeper that he too would have done the same thing if he had been in prison like him for no fault.
5. Set him free = released him.
surprised = astonished.
6. Change of heart.

1. The boy was crying because he was hungry.
2. The tree asked the boy to eat the fruit which is offered to him.
3. The boy looked worried because he was going to marry while he had no house to live in.
4. The tree and the boy became close friends due to their long mutual company and association.
5. bending down = bringing one of its branches close to the boy.
worried look = sight reflecting the anxieties or worries of the mind.
6. Best Friend

y=sin(1-x^2)/(1+x^2)

Let x=tanA

dx/dA=sec^2A=1+tan^2A

dx/dA=1+x^2………………..(1)

y=sin[(1-tan^2A)/(1+tan^2A)]

y=sin(cos2A)

dy/dA=cos(cos2A).(-2sin2A)

=-2.cos[(1-tan^2A)/(1+tan^2A)].[2tanA/(1+tan^2A)]

dy/dA=-2[2x/(1+x^2)].cos[(1-x^2)/(1+x^2)]

dy/dA=-4x/(1+x^2).cos[(1-x^2)/(1-x^2)]………(2)

Divide eq.(2)by (1)

dy/dA÷dx/dA=-4x/(1+x^2).cos[(1-x^2)/(1+x^2)] ÷(1+x^2).

Therefore, Answer = dy/dx={-4x/(1+x^2)^2}.cos{(1-x^2)/(1+x^2)} 

It depends on n,

For n = 1, the number of ways = 1 = 1!

For n = 2, the number of ways = 2 = 2!

For n = 3, the number of ways = 7 = 3! + 1

For n = 4, the number of ways = 48 = 2 x 4!

.........

Varies when n changes.

Concept used:

If N is expressed in terms of the product of prime factors

N = a^x × b^y × c^z

Where,

a, b and c = Prime factors

x, y and z = Highest power of the prime factors

Then,

Number of factors = (x + 1)(y + 1)(z + 1)

Calculations:

Expressing 42 in terms of prime factors,

⇒ 42 = 2 × 3 × 7

Number of factors = (1 + 1) × (1 + 1) × (1 + 1)  

⇒ 2 × 2 × 2

⇒ 8

∴ The number of factors of 42 is 8

Given:

A train leaves station P at 8:18 a.m.

The train reaches the station Q at 10:28 p.m. on the same day

Concept used:

1 hour = 60 minutes

We can write, 12 ∶ 00 = 11 ∶ 60

Calculation:

Time is taken by the train up to 12 ∶ 00 pm = 11 ∶ 60 - 8 ∶ 18 = 3 hours 42 minutes

Time is taken by the train from 12 ∶ 00 pm to 10 ∶ 28 pm = 10 hours 28 minutes 

Total time is taken by the train to reach station Q = 3 hours 42 minutes + 10 hours 28 minutes

⇒ 13 hours 70 minutes 

⇒ 14 hours 10 minutes

∴ The time is taken by train to reach Q is 14 hours 10 minutes

Given:

First dividend = 3488

First divisor = 12

Second dividend = 2478

Second divisor = 11

Concept used:

Dividend = Quotient × Divisor + Remainder

Calculations:

The closest multiple of 12 to 3488 = 3480

Remainder = 3488 - 3480

⇒ 8

The closest multiple of 11 to 2478 = 2475

Remainder = 2478 - 2475

⇒ 3

Difference between the remainder in both cases = 8 - 3

⇒ 5

∴ The difference between the remainders in both the cases is 5

Divisibility rule of 11 → A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11

Divisibility rule of 12 → A number is divisible by 12 if it is divisible by 3 and 4 both

Divisibility rule of 3 → A number is divisible by 3 if the sum of digits of the number is divisible by 3

Divisibility rule of 4 →  A number is divisible by 4 if the last two digits of the number is divisible by 4

Concept:

All Boolean algebra laws are shown below

Correct option is a) Assertion is correct but Reason is wrong.

Correct option is b) Assertion is wrong but Reason is correct.

b) Assertion is wrong but Reason is correct. 

Correct answer is a. Spaces between the stitches 

Correct answer is b. Blanket stitch 

Correct answer is  c. Embroidery frame

1. (D)₹13,000

2. (C)₹39,000

3. (C) 5000

4. (B) ₹13,000

Correct option is 

1 a. Issue against consideration other than cash

2 d. 50,000 shares

3 d. Excess applications can be allotted with preference shares

4 a. ₹5000

Correct option is 

1 a. 8000, ₹16000

2 a. 12000, ₹24000

3 a. 2000 shares

4 a. ₹14800

Correct option is 

1 (C) ₹12,50,000

2 (B) ₹15,000

3 (C) premium

4 (D)3500

Correct answer is a. To transfer the Design