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Solution: Group all 'S' together and call it X 
remaining letters have 3A, 2I, 2N, one T and one O 
so total letters = 10 
out of this 3! terms will have indistinguishable 'A' 
2! indistinguishable 'I' and same for N 
So total = 10!/(3!*2!*2!) 
=151200

The word "ASSASSINATION" has 4S , 3A , 2I , 2N , T , O , 4S are together. this is considered as one block as 1 letter. we have 3A , 2I, 2N , 4S , T , O 

Therefore , Number of words = Number of Permutations of 3A , 2I , 2N , 4S , T , O = 10! / 3!2!2! 

= 10X9X8X7X6X5X4X3X2X1 / (3X2X1) X (2X1) X (2X1)

= 10X9X8X7X5X3X2 = 151200

Calculation:

For 1st year, the population is P

For 2nd year, population P₂ = its 3 + 2P

For 3rd year, P₃ = 3 + 2P₂ = 9 + 4P

For 4th year, P₄ = 3 + 2P₃ = 3 + 2(3 + 2P₂) = 21 + 8P

For nth year, we see, the pattern or the formula for population here is: 3(2^n-1– 1) + 2^n-1P

Then, 2034 will be 16th term of the series

Population in 2034 = 3(2¹⁵– 1) + 2¹⁵ × 1000

⇒ 2¹⁵(1003) – 3

∴ The population at the beginning of 2034 will be 215(1003) – 3

Calculation:

LCM of 16, 24, 36 and 48 is 144

According to question

So, the smallest number = 144 - 6 = 138

∴ 138 is the smallest number with which 6 being added makes the sum divisible by 16, 24, 36 and 48.

The correct option is 3  i.e. 138.

Given:

Number = 0.529, 2.01

Calculation:

The product of two number = 0.529 × 2.01 = 1.06329

According to the question:

Required difference = 1.06329 – 1 = 0.06329

∴ The required answer is 0.06329.

Given:

The product of two numbers is 9216

Concept used:

The larger number is divided by the smaller number, the quotient is 16 leaving no remainder.

Larger number is 16 times of the smaller number

Calculation:

Let the smaller number be x

And larger number be 16x

According to the question:

The product of two numbers = 9216

⇒ (x × 16x) = 9261

⇒ x² = 9261/16

⇒ x2 = 576

⇒ x2 = 24²

⇒ x = 24

Now, The smaller number = x = 24

Larger number = 16x = 16 × 24 = 384

Difference = 384 – 24 = 360

∴ The difference between the numbers are 360.

Concept used:

Any number that can be divided by a, b, c, and d is always a multiply a multiple of LCM(a, b, c, and d)

Calculation:

L. C. M of 10, 5, 4 and 2 = 20

On dividing 3432 by 20, the remainder is 12

Number to be added = 20 – 12 = 8

∴ The required number is 8.

Correct answer is (D) 2520

Correct answer is (B) 2

Given:

The number is 2497

It is exactly divisible by 5, 6, 4, and 3

Concept Used:

Any numbers which are divisible by the numbers a, b, c, d,

Are always a multiple of the L.C.M of the numbers a, b, c, d.

Calculation:

The LCM of 3, 4, 5, and 6 is 60.

⇒ So the number is exactly divisible by 60

⇒ 2497/60

⇒ (2460 + 37)/60

⇒ 2460 is exactly divisible by 60

⇒ The remaining = 37/60

⇒ Remainder = 60 – 37

⇒ Remainder = 23

∴ The least number which should be added is 23

Correct answer is 13

Explanation:

Let a = 91, b = 26.

Then, a * b = HCF(a, b) * LCM (a, b)

So here, 91 * 26 = HCF (91, 26) * 182

That is, 2366 = HCF (91, 26) * 182

Therefore, HCF (91, 26) = 2366/182

So, HCF (91, 26) = 13

Solution:
327.7081 = 3277081/10000 = 3277081 / 2⁴ x 5⁴ = p/q
Here, q is of the form 2^m × 5ⁿ, where m and n are natural numbers.
The prime factors of p and q will be either 2 or 5 or both.

Given:

Profit of A = 10%

Profit of B = 5%

CP of C = Rs. 4620

Concept used:

SP = (100 + Profit)% × CP

Calculation:

Let CP of A be Rs. x.

SP of A = (100 + Profit)% × CP of A

⇒ SP = (100 + 10)% × x

⇒ SP = 110/100 × x

⇒ SP = 11x/10

SP of A = CP of B 

⇒ CP of B = 11x/10

SP of B = (100 + Profit)% × CP of B

⇒ SP = (100 + 5)% × 11x/10

⇒ SP = 105/100 × 11x/10

SP of B = CP of C

⇒ 105/100 × 11x/10 = 4620

⇒ x = Rs. 4000

∴ The cost price of A is Rs. 4000.

We know that the product of the numbers is equal to the product of their HCF and LCM.

Let the other number is x.

Then 54x = 27× 162.

(∵ LCM(54, x) = 162 & HCF(54, x) = 27 and one number is 54.)

⇒ x = \(\frac{27\times162}{54}\) = \(\frac{162}{2}\) = 81.

Hence, the other number is 81.

The given number is 3.24636363…. = \(3.24\bar{63}.\)

(This number is non-terminating and repeating number which is a rational number.

Because, a number is irrational only if the number has non-terminating and non-repeating decimal)

second method :- Let x = 3.246363……….

⇒ 100x = 324.636363……….

Now, 100x – x = 324.636363…. – 3.246363…..

⇒ 99x = 321.39

⇒ x = \(\frac{321.39}{99}=\frac{32139}{9900}=\frac{3571}{1100}\)

⇒ x = \(\frac{3571}{1100}\) which is a rational number.

Hence, 3.246363….. is a rational number.

Prime factorization of 4620 is 

4620 = 2 × 2 × 3 × 5 × 7 × 11 

= 2² × 3 × 5 × 7 × 11. 

Hence, the prime factorization of 

4620 is \(2^2 \times 3 \times 5 \times 7 \times 11\).

Solution:
117 = 65 x 1 + 52
65 = 52 x 1 + 13
65 = 13 x 5 + 0
=> HCF(65, 117) = 13
Since, HCF(65, 117) = 65m - 117
=> 13 = 65m - 117
=> 65m = 117 + 13 = 130
=> m = 130/65 = 2
Now, 65 = 5 x 13
and 117 = 3 x 3 x 13 = 3² x 13
=> LCM(65, 117) = 3² x 5 x 13 = 585

Given:

Investment of A = Rs. 1.5 lakh

Investment of B = Rs. 1 lakh

Total profit after 1 year = Rs. 24,000

Concept used:

Share of profit is directly proportional to the amount of investment multiplied by time of investment

Calculation:

Total investment of A and B = Rs. (1.5 + 1) lakh

⇒ Rs. 2.5 lakh

Share of A = Rs. (1.5/2.5) × 24,000

⇒ Rs. 14,400

∴ The share of A is Rs. 14,400

Given:

Three partners invested in the ratio 4 : 6 : 9 for 3 years, 1 year and 4 years respectively. The total profit was 180000.

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Let the investments by A , B and C be 4x, 6x and 9x respectively

Ratio of profit = amount × time

Ratio of profits of A, B and C = 4x × 3 : 6x × 1 : 9x × 4 = 2x : x : 6x

Profit of B = x/9x × profit = 1/9 × 180000 = 20000

Given:

Variance = 361

Concept used:

The positive square root of the variance is called Standard deviation

S.D = √Variance

Calculation:

S.D = √Variance

⇒ S.D = √361

⇒ S.D = 19

∴ The standard deviation of the data is 19.

Given:

The standard deviation of a population is 5

Concept used:

standard deviation = √variance

Calculation:

Variance = (standard deviation)²

Variance = 5² = 25 

∴ The variance is 25.

Given:

Mean of the distribution = 24

Standard deviation = 6

Concept used:

Coefficient of variance = standard deviation/mean ×  100

Calculation:

Coefficient of variance = 6/24 × 100 = 100/4

Coefficient of variance = 25%

∴ The coefficient of variance is 25%.

Given: work together how many days will it take them to complete the work?

R can do work = 15 days

K can do work = 21 days

Concept used:

Total work = LCM

Formula used:

Efficiency = Work/Time

Calculations:

Total work = LCM of 15 and 21 = 105

Then efficiency of R+K = 7 + 5 = 12

The efficiency of R ∶ K  = 7 ∶ 5

∴ Working together they completed the work in 8(3/4) days.

Given:

Sum of three numbers = 98

Ratio of first and second number = 2 ∶ 3

Ratio of second and third number = 5 ∶ 8

Calculation:

Let the first number be 2x and second number be 3x

As, the ratio of second and third number = 5 ∶ 8

3x ∶ Third number = 5 ∶ 8

⇒ Third number = 24x/5

Sum of all numbers = 98

⇒ 2x + 3x + 24x/5 = 98

⇒ 49x/5 = 98

⇒ x = 10

Second number = 3x = 3 × 10 = 30

∴ Second number is 30

Given :

A can do a work in ‘x’ days

B can do the work in 15 days

The efficiency of ‘A’ is 75% of the efficiency of ‘B’

Concept used :

The efficiency of a person × Number of days he took to complete the work = Total Work

Number of days to complete the work = Total work/Work done every day

Calculations:

Let the working efficiency of B be ‘b’

Let the working efficiency of A be ‘a’

Work for A = Work for B

The efficiency of A × x = Efficiency of B × 15    

⇒ a × x = b × 15

⇒ 75% of b × x = b × 15

⇒ (3/4) b × x = b × 15

⇒ x = 20 days

Now, total work = LCM of 20 and 15 = 60 units

A’s per day work = (Total work/Total days he took) = 60/20

⇒ 3 units

B’s per day work = 60/15

⇒ 4 units

A and B total work for everyday = 4 + 3 = 7 units

Total number of days taken by them = Total work/work done by them every day

⇒ 60/7

∴ They both will complete the work in 60/7 days 

Given:

Sum of three numbers = 98.

Ratio of First number to Second number is 2 : 3.

Ratio of Second number to Third number is 5 : 8.

Calculation:

Let the three numbers be A, B and C. Then,

A : B = 2 : 3 and B : C = 5 : 8

⇒ \( 5 \times \frac{3}{5}\) : \(8 \times \frac{3}{5}\) = 3 : \(\frac{{24}}{5}\)

⇒ A : B : C = 2 : 3 : \(\frac{{24}}{5}\)

⇒ 10 : 15 : 24

⇒ B = \(98 \times \frac{{15}}{{49}}\) = 30.

∴ Second number is 30.

Let x, y and z be the efficiencies of Sachin, Saurav and Rahul respectively.

According to the question,

⇒  4 × (x + y + z) = 46%      ---- (1)

After 4 days when Rahul left,

⇒  6 × (x + y) = 54%

⇒  (x + y) = 9%      ---- (2)

As we know, the ratio of efficiency of Sachin and Saurav

⇒  x : y = 5 : 4      ---- (3)

So, x = 5% and y = 4%      {from eq. (2) and (3)}

⇒  4 × (9% + z) = 46%      {from eq. (1) and (2)}

⇒  36% + 4z = 46%

⇒ 4z = 10%

⇒  z = 2.5%

Since, Rahul has the least efficiency, so he’s the slowest worker among the three.

Days required by Rahul to complete his work alone = 1/(2.5%) = 40 days

Given :

Ram can do the work in 20 days

Shyam can do the work in 25 days

Concept used :

Total days took to complete the work = Total work/work done each day

Calculations :

 Total work = LCM of 20 and 25 = 100 units

Ram’s per day work = 100/20 = 5 units

Shyam’s per day work = 100/25 = 4 units

Ram’s and Shyam’s each day work = 5 + 4  = 9 units

Total work done by them in 8 days = 8 × 9 = 72 units

Now Shyam got an injury  

Remaining work = 100 – 72 = 28 units

Shyam’s efficiency after 8 days = 50% of 4 = 2 units

Now Ram’s and Shyam’s each day work = 5 + 2 = 7 unit

Days required to complete the remaining work = 28/7

⇒ 4 days

 Total days = 8 + 4 = 12 days

∴ Total number of days taken by them is 12 

Given :

Ram can do the work in 16 days 

Shyam can do the work in 24 days 

Rohan can do the work in 32 days 

Concept used :

Total days to complete the work = Total work/Work done per day 

Calculations :

Total work = LCM of 16, 24 and 32 

⇒ 96 units 

Ram's per day work = 96/16 = 6 units 

Shyam's per day work = 96/24 = 4 units 

Rohan's per day work = 96/32 = 3 units

Ram, Shyam and Rohan per day work = 6 + 4 + 3 = 13 units

Their total work in 4 days = 4 × 13 = 52 units

Remaining work = 96 - 52 = 44 units 

Days taken by Shyam to complete the remaining work = 44/4 

⇒ 11 

∴ Shyam work alone for 11 days to complete the remaining work   

Given:

20 labourers, working 7 hours a day can finish a piece of work in 30 days

The next set of labourers work for 5 hours a day, finishes the work in 40 days.

Concept used:

M₁ × D₁ × H₁ ÷ W₁ = M2 × D2 × H₂ ÷ W2

Where M → number of people, D → number of days, H → hours W → wages, work 

Calculation:

Let the number of men working be x.

Applying the formula here, 

20 × 7 × 30 = x × 5 × 40

⇒ x = (20 × 7 × 30)/(5 × 40) = 21

∴ 21 labourers will work to finish the work. 

Proportionate capital of Ram, Rohan and Ravi
= 10,000 × 6 : 7500 × 12 : 7500 × 12
= 60,000 : 90,000 : 90,000
or
ratio = 2 : 3 : 3

Ram's share = 2500x2/8= Rs.625

(i) Two balls are selected from 17 balls in ¹⁷C₂w ways Ways of choosing 2 Red balls from 9 red is ⁹C₂ ways. 

∴ (P both red) = \(\frac{^9C_2}{^{17}C_2} = \frac{9}{34}\)

(ii) Without replacement, 

probability of 1st red ball = \(\frac{9}{17}\)

And also probability of 2nd red ball is = \(\frac{8}{16}\)

(iii) With replacement Probability of 1st red ball = \(\frac{9}{17}\)

And also probability of 2nd red ball is = \(\frac{9}{17}\)

∴ P(both red) = \(\frac{9}{17}\) x \(\frac{9}{17}\)= \(\frac{81}{289}\)

Sum of the ratios 7 : 8 = 7 + 8 = 15

Total share Kunal has = Rs.360 × (7/15) = Rs.168 

Total share Mohit has = Rs.360 × (8/15) = Rs.192

i. The north east trade winds blowing in the northern hemisphere and the south east trade winds blowing in the southern hemisphere lead to the formation of monsoon winds. Monsoon winds are winds that change direction in accordance with season. Monsoon is the seasonal reversal of wind in a year.

The factors responsible for the formation of monsoon winds are the apparent movement of the sun, Coriolis force and difference in heating. During the summer in the Northern Hemisphere, high temperature is experienced along the region through which the Tropic of Cancer passes. The pressure belts shift slightly northwards. The south east trade winds also cross the equator and move towards the north.

As the trade winds cross the equator, they get deflected and transfer into south west monsoon winds under the influence of the Coriolis effect. The low pressure formed over the land due to the intense day temperature attracts these sea winds and further contributes to the formation of south west monsoon winds. As a result of the formation of high pressure zones over the Asian landmass during winter and low pressure zones over the Indian Ocean, the north east trade winds get strengthened. These are the north east monsoon winds

ii. Global pressure belts in the northern hemisphere.

• Equatorial low pressure belt: between 5°N and 5°S. 
• Subtropical high pressure belt: 30°N 
• Subpolar low pressure belt: 60°N 
• Polar high pressure belt: 90°N 
• Planetary winds in the Northern hemisphere: 
• Trade winds: Blow from subtropical high pressure belt to equatorial low pressure belt. 
• Westerlies: Blow from subtropical high pressure belt to subpolar low pressure belt. 
• Polar winds: Blow from polar high pressure belt to subpolar low pressure belt.

Winds that change their direction according to change in season are called monsoon winds.

Monsoon winds are formed due to the apparent movement of the sun, Coriolis force and differences in heating. 

Sun’s rays fall vertically to the north of the equator during certain months due to the tilt of the earth’s axis. This leads to an increase in temperature along the region through which Tropic of Cancer passes. The pressure belts also shift slightly northwards in accordance with this.

The south east trade winds cross the equator and move towards the north. As the trade winds cross the equator, they get deflected and transform into south west monsoon winds under the influence of the Coriolis effect. The low pressure formed over the land due to the intense day temperature attracts these sea winds and further contributes to the formation of south west monsoon winds. As a result of the formation of high pressure zones over the Asian landmasses during winter and low pressure zones over the Indian Ocean, the north east trade winds get strengthened. These are the north east monsoon winds.

a. Equatorial low pressure belt 

b. Subtropical high pressure belt 

c. Subpolar low pressure belts 

d. Polar high pressure belts

Land Breeze:

• Blows during the night. 
• Blows from land to sea.

Sea Breeze:

• Blows during the daytime.
• Blows from sea to land.

During day time, the land gets heated up quickly. As a result, the air in contact with the land also gets heated. This leads to the formation of low pressure over the land which causes comparatively cooler air to blow from the sea. This is known as sea breeze.

As the land cools faster than the sea during the night, it would be high pressure over the land and low pressure over the sea. This results in the movement of air from the land to sea. This is the land breeze.

• Characteristic features of land and sea breezes 
• Characteristic features of mountain and valley breezes