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the pattern in the given series is (previous number + 1, 3, 5, 7, 9)

7 + 1 = 8

8 + 3 =11

11 + 5 = 16

16 + 7 = 23

23 + 9 = 32

∴ the answer is 16

EXPLANATION:

the pattern in the given series is (previous number + 2, × 2, + 2, × 2, + 2)

5 + 2 = 7

7 × 2 = 14

14 + 2 = 16

16 × 2 = 32

32 + 2 = 34

∴ the answer is 34

The logic behind the given series:

20 + 10 = 30

30 + 12 = 42

42 + 14 = 56

56 + 16 = 72

72 + 18 = 90

∴ The next value is 90.

Given:

Gain% = 13%

Discount = 20%

Calculations:

According to the question we get the equation,

⇒ 100/(100 + gain%) × (100 – discount%)/100 × M.P

⇒ 100/(100 + 13) × (100 – 20)/100 × 2260

⇒ 100/113 × 80/100 × 2260

⇒ 1600

The Cost Price will be Rs. 1600.

Formula used:

Average speed = (total distance)/(total time)

Calculation:

Average speed = (20 + 15 + 37)/(2 + 3 + 4)

⇒ Average speed = 72/9 = 8

∴ The average speed of the train is 8 km/h

Given:

Trip one X to Y with the speed of v km/h

Trip one Y to X with the speed of 2v km/h

Trip one X to Y with the speed of 3v km/h

Trip one Y to X with the speed of 4v km/h

Formula used:

Average speed = Total distance covered/Total time taken

Speed = Distance/Time

Calculation:

Let the distance between X and Y is a km

Total distance covered = 4a

⇒ time = distance/speed

Time is taken on the first trip

⇒ time = a/v

Time is taken on the second trip

⇒ time = a/2v

Time is taken on the third trip

⇒ time = a/3v

Time is taken on the fourth trip

⇒ time = a/4v

Total time = (a/v) + (a/2v) + (a/3v) + (a/4v)

⇒ (12a + 6a + 4a + 3a)/12v = 25a/12v

Average speed = Total distance covered/Total time taken

⇒ 4a/(25a/12v) = 48v/25

⇒ Average speed = 1.92v km/h

∴ Average speed of the car lies between v km/h and 2v km/h.

Given:

Speed of car = 80 km/hr

Formula used:

1 km/hr = 1000/3600 = 5/18 m/s

Calculation:

Speed of car = 80 km/hr

⇒ 80 × 5/18

⇒ 400/18

∴ Speed of car is 22.22 m/s

GIVEN:

 3.004

CALCULATION:

 3.004 

⇒ 3004/1000

⇒ 751/250

∴ 3.004 = 751/250

GIVEN:

1/3.718 = 0.2689

CALCULATION:

1/3.718 = 0.2689

⇒1/.0003718

⇒ 10000/3.718

⇒ 10000 × 1/3.718

⇒ 10000 × .2689

⇒ 2689

∴ 1/.0003718 = 2689

GIVEN:

A and B start from the same point and travel in the same direction around a circular track 3 km in circumference. Their speeds are 2 meter / sec and 1.5 meter / sec.

CALCULATION:

Time taken by A for 1 round = 3000/2 = 1500 sec

And

Time taken by B for 1 round = 3000/1.5 = 2000 sec

Now,

LCM of 1500 and 2000 sec = 6000 sec = 100 min

Hence,

Calculation:

Let the number be x

According to question,

⇒ 2x + 5 = 6

⇒ 2x = 1

⇒ x = 0.5

∴ The number is 0.5

Concept used:

Meter per second can be changed in km per hour by multiplying with 18/5

Calculation:

10 m/s = 10 × (18/5) km/hr

⇒ 36 km/hr

∴ The required answer is 36 km/hr

Concept used:

To convert meter/sec to km/hr, we will have to multiply it by 18/5

Calculation:

35 m/s = 35 × (18/5) km/hr

⇒ (7 × 18) km/hr

⇒ 126 km/hr

∴ The required answer is 126 km/hr

Given:

P alone can fill = 30 min

Inlet P is 4 times faster than inlet Q to fill a drum

Calculation:

Time taken by P to fill the drum = 30 min

∴ time taken by Q to fill the drum

⇒ 30 × 4

⇒ 120 min

∴ required time taken

⇒ (30 × 120)/(30 + 120)

⇒ (30 × 120)/150

⇒ 24 min

Concept used:

Perfect square should have a pair of each prime factor

Calculation:

1935 = 3 × 3 × 5 × 43

1520 = 2 × 2 × 2 × 2 × 5 × 19

730 = 2 × 5 × 73

1681 = 41 × 41

Here, only 1681 has a pair of each prime factor

∴ The perfect square is 1681

Given:

We have to find the HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\)

Concept Used:

HCF of fraction = (HCF of numerator)/(LCM of denominator)

Calculation:

HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\)

⇒ \({HCF\ (9,\ 3,\ 6,\ 12) \over LCM\ (10,\ 25,\ 15,\ 5)}\)

HCF of numerators is 3

LCM of denominators is 150

Required HCF is 3/150

∴ HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) is (3/150).

We can make (3/150) fraction into a small form i.e (1/50), But we have to see options also.

In the given option (1/50) is not present. If (1/50) will be present in place of (3/150), then we can mark (1/50) as the answer. Which is totally correct.

Sometimes we have to check options also to mark the correct answers. In this case, we are marking (3/150) according to the options provided.

Given:

450 + 750 divided by 65

Concept used:

aⁿ + bⁿ is divided by (a + b) when n is an odd number

Calculation:

450 + 750 

⇒ (4²)²⁵ + (7²)²⁵

⇒ (16)²⁵ + (49)²⁵

Here, n = 25 is an odd number

⇒ 16 + 49 = 65

So, (450 + 7⁵⁰) is divisible by 65 or the multiples of 65.

When 65 or the multiples of 65 divided by 65 then the remainder will come to zero.

∴ The remainder will be 0.

Given:

Ratio of two numbers = 5 : 6

HCF of given numbers = 6

Formula used:

Product of two numbers = HCF × LCM

Calculation:

Let the numbers be 5x and 6x respectively

HCF of 5x and 6x = x

Now, according to the question

x = 6

Now, Numbers = 5x = 5 × 6 = 30

Second number = 6x = 6 × 6 = 36

Now, LCM = (30 × 36)/6

⇒ 180

∴ The LCM of these two numbers is 180

Given:

We have to find the difference between largest and smallest number of six digit

Calculation:

Largest number of six digit is 999999

Smallest number of six digit is 100000

Difference between this two number is (999999 - 100000) = 899999

∴ The difference between largest and smallest number of six digit is 899999.

Calculation:

Smallest number of six digits = 100000

Greatest number of five digits = 99999

The required difference = 100000 - 99999 = 1

∴ The required difference is 1.

Calculation:

496 = 6 × 82 + 4

It means remainder will be 4

∴ The remainder when 496 is divided by 6 is 4

Given:

Smallest 4 digit number = 1000

Greatest 4 digit number = 9999

Calculation:

Sum = 1000 + 9999

⇒ 10999

∴ The sum of smallest and greatest four digits number is 10999

Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10. 

Without decimal places, these numbers are 63, 105 and 210.

Now, H.C.F. of 63, 105 and 210 is 21. 

H.C.F. of 0.63, 1.05 and 2.1 is 0.21. 

L.C.M. of 63, 105 and 210 is 630. 

L.C.M. of 0.63, 1.05 and 2.1 is 6.30. 

Let the required numbers be 15.x and 11 x. 

Then, their H.C.F. is x. So, x = 13. 

The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

Given:

 \((1^2 + 2^2 + 3^2 + ...+10^2)=385\)

Calculation:

\(2^2(1^2 +2^2 +3^2 + ...+ 10^2)\)

⇒ 4 × 385 = 1540

∴ The value of \((2^2 +4^2 +6^2 + ...+ 20^2)\) is equal to 1540.

Other number = 11 x 693/77 = 99

Given:

HCF of two numbers = 11

LCM of two numbers = 693

Formula used:

LCM × HCF = Product of two numbers

Calculation:

Let the other number be x

⇒ 11 × 693 = 77 × x

⇒ x = 693/7

⇒ 99

∴ The other number is 99

Ozone layer absorbs the ultra voilet radiations. Depletion of ozone layer may lead to skin cancer in human beings, damage eyes, decrease crop yield and disturb the global rainfall pattern. Therefore the damage to the ozone layer is a cause for concern.

Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catanation. These compounds may have long chains of carbon or even carbon atoms arranged in rings.

Eg: Butane, Pentane, Hexane etc,.

A lens is a transparent material bound by two surfaces, of which one or both surfaces are spherical.

V = 2 cm/s

v = 0.02m/s

\(\eta\) = 4.8 x 10^-4

Density of oil = 0.8gl cm³

Density of air = 18/cm³ J

We know that

r² = \(\frac{9\eta v}{29(p-\sigma)}\)

r² = \(\frac{9\times4.8\times10^{-5}\times0.02}{2\times9.8(1000-800)}\) 

r² = \(\frac{0.864\times10^{-5}}{3920}\)

r² = 0.00022 x 10^-5

r² = 2.2 x 10^-9

r = \(\sqrt{2.2\times10^{-9}}\) m

hence diameter of spherical drop

d = 2 x \(\sqrt{2.2\times10^{-9}}\) m

For a real and inverted image, the height of the image is negative and the height of the object is positive. So, the magnification of the concave mirror is negative.

For an object at infinity, all the paraxial rays after reflective from the concave mirror will pass through the focus, thus the image is formed at focus, Hence the focal length at the concave mirror can be determined do by measuring he distance b/w the screen and the mirror.

Work done W = qΔv

 = 2 x (50 - 20)

W = 2 x 30

W = 60 N

\(2N_2O_5 \longrightarrow 4NO_2 + O_2\)

Overall rate of reaction r = \(\frac{-1}2 \frac{d[N_2O_5]}{dt}\)

\(= +\frac14\frac{d[NO_2]}{dt}\)

\(= + \frac{d[O_2]}{dt}\)

∵ Given, rate of appearance of NO₂ = \(= \frac{d[NO_2]}{dt} = 2 mol/L\)

(i)  ∴ rate of reaction r = \(+\frac14\frac{d[NO_2]}{dt}\)

\(= \frac14\times2 \)

\(= \frac12 mol/L.s\)

(ii) rate of disappearance of N₂O₅ = \(- \frac{d[N_2O_5]}{dt}\)

∵ \(r =\frac{-1}2 \frac{d[N_2O_5]}{dt}\)

\(- \frac{d[N_2O_5]}{dt} = 2 \times\frac12\)

\(= 1 mol/L.s\)

Hence, rate of disappearance of N₂O₅ is 1 mol/L.s.

As we know 

Osmotic pressure \(\pi = iCRT\)

\(i = \) Vant's Hoff factor

For Na2SO4, \(i = 3\)

For glucose, \(i = 1\)

∴ \(\pi _{Na_2SO_4} = 3 \times 0.2 \times 0.082 \times 300 \,atm\)

\(= 14.76 \, atm\)

\(\pi _{glucose} = 1 \times 0.5 \times 0.082 \times 300 \,atm\)

\(= 12.3 \, atm\)

Hence, difference in osmotic pressure = 14.76 - 12.3 = 2.46 atm.