Explore topic-wise InterviewSolutions in .

The correct answer is Jupiter.

• Jupiter, the most massive planet in the solar system.
• It is fifth in distance from the Sun.
• It takes nearly 12 Earth years to orbit the Sun, and it rotates once about every 10 hours, more than twice as fast as Earth.

• Mars, fourth planet in the solar system in order of distance from the Sun and seventh in size and mass. Mars are also known as a red planet.
• Uranus is the seventh planet in distance from the Sun and the least massive of the solar system’s four giant, or Jovian planets, which also include Jupiter, Saturn, and Neptune.
• Saturn, second largest planet of the solar system in mass and size and the sixth nearest planet in distance to the Sun.

​ 

• Total eight planets orbiting the Sun, in order of increasing distance, are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.

The correct answer is Venus.

• Venus is the hottest planet in the solar system.
• Although Venus is not the planet closest to the sun, its dense atmosphere traps heat due to the greenhouse effect makes it our solar system's hottest planet.
• Venus is known as the twin sister of earth.
• Venus is also known as the Morning star or Evening star.
• Hence, option 1 is correct.

​

• There are eight planets in the Solar System.
• From the closest to farthest from the Sun, they are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.
• The first four planets are called terrestrial planets.
  • They are mostly made of rock and metal, and they are mostly solid.
• The last four planets are called gas giants.
  • This is because they are much larger than other planets and are mostly made of gas.
• The Solar System also contains other things.
  • There are asteroid belts, mostly between Mars and Jupiter.
  • Further out than Neptune, there is the Kuiper belt and the scattered disc.
  • These areas have dwarf planets, including Pluto, Makemake, Haumea, Ceres, and Eris.
  • There are thousands of very small objects in these areas. There are also comets, centaurs, and there is interplanetary dust.

Aquatic organisms like fishes obtain oxygen from water present in dissolved state through their gills. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms.

Crustose lichens

Let I = \(\int\frac{2cosx}{(sin^2x+1)(sin^2x+2)}dx\)

Let sin x = t

⇒ cos x dx = dt

\(\therefore\) I = \(\int\frac{2dt}{(t^2+1)(t^2+2)}\) 

\(=2\int\frac{(t^2+2)-(t^2+1)}{(t^2+1)(t^2+2)}dt\) 

\(=2\int(\frac1{t^2+1}-\frac1{t^2+2})dt\) 

\(=2[\int\frac1{t^2+1}dt-\int\frac1{t^2+2}dt)\) 

\(=2[tan^{-1}t-\frac1{\sqrt2}tan^{-1}\frac t{\sqrt2})+c\)

(\(\because\) \(\int\frac{1}{a^2+x^2}dx=\frac1atan^{-1}\frac xa\))

\(=2tan^{-1}t-\sqrt2tan^{-1}\frac t{\sqrt2}+c\)

\(=2tan^{-1}(sin x)-\sqrt2tan^{-1}(\frac{sin x}{\sqrt2})+c\)

(\(\because\) t = sin x)

y = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+....\infty\)

 = e^x (\(\because\) e^x = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+....\))

\(\because\) \(\frac{dy}{dx}=\frac d{dx}e^x=e^x=y\) 

or Alternate method:

y =  1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}....\infty\) 

\(\therefore\) \(\frac{dy}{dx}\) = 1 + \(\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}....\infty\)

 = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+...\infty\) 

 = y

Answer is:

b) will

Correct answer is: (c) 38

LCM of two prime numbers = product of the numbers 221= 13 x 17.

So p= 17 & q= 13

\(\therefore\) 3p - q= 51-13 =38

Correct answer is: (b) 11/36

Outcomes when 5 will come up at least once are-

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6)

Probability that 5 will come up at least once = 11/36

Correct answer is: (a) 11/13

Probability that the card drawn is neither a king nor a queen

= (52-8)/52

= 44/52 = 11/13

Given:

M.P = 28% above C.P

Gain = 6%

Formula Used:

Discount% = (M.P – S.P)/M.P × 100

Calculations:

Let the C.P of the shoes be 100.

∴ M.P = 28% of 100

⇒ 128

S.P = 6% of 100

⇒ 106

∴ Discount% = (M.P – S.P)/M.P × 100

⇒ (128 – 106)/128 × 100

⇒ 17.18 ≈ 17%

(i). Let event H and E denotes that students read Hindi newspaper and English newspaper, respectively. 

Since, given that in a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers . 

Therefore, n(E) = 40%, n(H)= 60% and n(H ∩ E) = 20%. 

Now, we know that n(H ∪ E) = n(H) + n(E) − n(H ∩ E). 

Therefore, n(H ∪ E) = 40 + 60 − 20 = 80%. 

Therefore, 80% students who reads either Hindi or English newspaper in the hostel . 

Therefore, 20% students in the hostel who reads neither Hindi nor English newspaper . 

Therefore, the probability that student reads neither Hindi nor English newspaper is \(\frac{20}{100}= \frac{2}{10} = \frac{1}{5}.\)

(By changing percentage into the probability form. ) 

(ii). Now, the probability that student reads both newspaper is P(H ∩ E) = \(\frac{20}{100}= \frac{2}{10} = \frac{1}{5}\) . 

(By changing percentage into the probability form.)

And the probability that student read Hindi newspaper is P(H) = \(\frac{60}{100}\) = \(\frac{6}{10}\) = \(\frac{3}{5}\) .

Therefore, the probability that student reads English newspaper given that he reads Hindi newspaper is P(E|H). 

Now, P(E|H) = \(\frac{P(H∩E)} {P(H)}\) = \(\frac{1/5} {2/5}\) = \(\frac{1}{3}\) . (Conditional probability)

Hence, if the student reads Hindi newspaper, then the probability that he reads English newspaper is \(\frac{1}{3}\).

(iii). The probability that student read English newspaper is P(E) = \(\frac{40}{100} = \frac{4}{10} = \frac{2}{5}.\) 

Therefore, The probability that student reads Hindi newspaper given that he reads English newspaper is P(H|E).

Now, P(H|E) = \(\frac{P(H ∩ E)} {P(E)}\)  = \(\frac{1/5} {2/5}\) = \(\frac{1}{2}.\) (Conditional probability) 

Hence, if the student reads English newspaper, then the probability that he reads Hindi newspaper is \(\frac{1}{3}\) . 

(iv). The students in the hostel who reads only English newspaper is n(E_only) = n(E) − n(H ∩ E) = 40 − 20 = 20%. 

Therefore, the probability that student reads only English newspaper is P (E_only) = \(\frac{20}{100} = \frac{2}{10} = \frac{1}{5}.\) 

(By changing percentage into the probability form.) 

(v). The students in the hostel who reads only Hindi newspaper is n(H_only) = n(H) − n(H ∩ E) = 60 − 20 = 40%. 

Therefore, the probability that student reads only English newspaper is P(H_only) = \(\frac{40}{100} = \frac{4}{10}\) = \(\frac{2}{5}\) . 

(By changing percentage into the probability form. ) 

Hence, the probability that student reads only Hindi newspaper is \(\frac{2}{5}\) .

Given that events and are two independent events.

Therefore, P(A ∩ B) = P(A)P(B). ...(1) 

Now, the probability of occurrence of at least one of A and B is P(A ∪ B).

We know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A)P(B)(By equation 1) 

⇒ P(A ∪ B) = P(A) + P(B)[1 − P(A)] = P(A) + P(B)P(A′) ( \(\because\) P(A′) = 1 − P(A)) 

⇒ P(A ∪ B) = 1 − 1 + P(A) + P(B)P(A′) = 1 − [1 − P(A)] + P(B)P(A′) 

⇒ P(A ∪ B) = 1 − P(A ′ ) + P(B)P(A ′ ) = 1 − P(A′)[1 − P(B)] ( \(\because\) P(A ′) = 1 − P(A)) 

⇒ P(A ∪ B) = 1 − P(A ′ )P(B ′ ).

Hence, if A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1 - P(A′) P(B′ ). 

(Hence prove)

Yes, I have seen idols, medals, cannon and decorative goods made up of bronze.

Concept:

Trapezoidal rule is given by:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)

\({\rm{Number\;of\;intervals(n)}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)

where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

From question

\(I = \mathop \smallint \nolimits_0^4 2.f\left( x \right).dx\)

Let f(x) = 2.f(x)

∴ \(f\left( x \right) = \frac{{f\left( x \right)}}{2}\)

\(I = \mathop \smallint \nolimits_0^4 \frac{{f\left( x \right)}}{2} \cdot dx = \frac{1}{2}\mathop \smallint \nolimits_0^4 f\left( x \right) \cdot dx\)

From the formula:

\(\mathop \smallint \nolimits_0^4 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\;\)

\(= \frac{{1}}{2}\left[ {\left( {5 + 7} \right) + 2\left( {2 + 1 + 3} \right)} \right]\)

∴ \(I = \frac{{24}}{2} = 12\)

Concept:

The letters A, E, I, O, and U are called vowels. The other letters in the alphabet are called consonants.

n ! = n(n - 1)(n - 2)....3.2.1

Calculations:

We know that, the letters A, E, I, O, and U are called vowels. The other letters in the alphabet are called consonants.

In the word 'DAUGHTER', the vowels are ‘a, u, e’ and the consonants are “d, g, h, t, r”

All the vowels should come together, so consider the bundle of vowels as one letter, then the total letters will be 6. [(aue), d, g, h, t, r]

So, no. of ways of arranging these letters = 6! × 3! = 4320

Concept:

If a, b, c are in GP than \(r = \frac{b}{a} = \frac{c}{b}\)

If first term of GP is a and common ratio is r than nth term of GP is given by ar^{n - 1}

Calculation:

Given: x, 2x + 2, 3x + 3 are the first three terms of a geometric progression than

\(\frac{{2x + 2}}{x} = \frac{{3x + 3}}{{2x + 2}}\)

\(\frac{{2x + 2}}{x} = \frac{3}{2}\)

\(4x + 4 = 3x \Rightarrow x = - 4 = a\)

\(r = \frac{c}{b} = \frac{{3x + 3}}{{2x + 2}} = \frac{3}{2}\)

4th term in the geometric progression is T₄= ar^{n - 1}

\({T_4} = ( - 4){\{ \frac{3}{2}\} ^{4 - 1}} = - 13.5\)

Hence, option A is the correct answer.

Concept:

Let the two numbers as a and b.

A.M. = \(\rm \dfrac {a+b}{2}\)

G.M. = \(\rm\sqrt{ab}\) 

Relation between AM and GM:

(G.M.)2 = (A.M.)(H.M.)

Calculations:

Given, the arithmetic mean of two numbers exceeds their geometric mean by 2 and the geometric mean exceeds their harmonic mean by 1.6.

⇒ A.M. = G.M. + 2         ....(1)

and G.M. = H.M. + 1.6

We know that, (G.M.)² = (A.M.)(H.M.)

⇒ (G.M.)2 = (G.M. + 2)(G.M. - 1.6)

⇒ (G.M.)2 = (G.M.)² + 0.4 G.M. - 3.2 

⇒ 0.4 G.M. = 3.2

⇒ G.M. = 8                  ....(2)

From equation (1), we get 

⇒A.M. = 10                  ....(3)

Consider, the two numbers as a and b.

then, G.M. = \(\rm\sqrt{ab}\) and A.M. = \(\rm \dfrac {a+b}{2}\)

From equation (2) and (3), we get

⇒ \(\rm\sqrt{ab}\) = 8 and \(\rm \dfrac {a+b}{2}\) = 10

⇒ ab = 64 and a + b = 20 

⇒ a (20 - a) = 64

⇒ a²  - 20a - 64 = 0

⇒ a =16 or a = 4

⇒ b = 4 or b = 16

Hence, the two numbers are 16 and 4.

Hence, The arithmetic mean of two numbers exceeds their geometric mean by 2 and the geometric mean exceeds their harmonic mean by 1.6. the two numbers are 16 and 4.

Answer: a) a-b=3

S_n = \(\frac{3\times1^3}{1^2}+\frac{5\times(2^3+2^3)}{1^2+2^2}+\frac{7\times(1^3+2^3+3^3)}{1^2+2^2+3^2}\) + .... + \(\frac{2n+1)(1^3+2^3+3^3+....+n^3)}{1^2+2^2+3^2+....+n^2}\)

∴ ΣT_n = \(\cfrac{(2n+1)\frac{n^2(n+1)^2}4}{\frac{n(n+1)(2n+1)}6}\) 

\(=\frac32 n(n+1)\) 

\(=\frac32(n^2+n)\)

∴ S_n = ΣT_n = \(\frac32\)(Σn² + Σn)

 = \(\frac32(\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2)\) 

 = \(\frac34\)n(n + 1)\((\frac{2n+1}3+1)\) 

 = \(\frac14\)n(n + 1) (2n + 4)

 = \(\frac12\)n (n + 1) (n + 2)

Let P(n) = \(\cfrac{(1^4+\frac14)(3^4+\frac14)...((2n-1)^4+\frac14)}{(2^4+\frac14)(4^4+\frac14)....((2n)^4+\frac14)}\)

Put n = 1, we get P(1) = \(\cfrac{1^4+\frac14}{2^4+\frac14}=\frac{1+\frac14}{16+\frac14}\) = \(\cfrac{\frac54}{\frac{65}4}\)

= \(\frac5{65}=\frac1{13}\)

From option (a), \(\frac{1}{4n^2+2n+1}\) at n = 1 = \(\frac1{4+2+1}=\frac17\)

Hence, (a) is wrong.

From option (b), \(\frac1{8n^2+4n+1}\) at n = 1 = \(\frac1{8 + 4 + 1}=\frac1{13}\)

Hence, option(b) may be true.

From option (c), \(\frac1{4(2n^2+n+1)}\) at n = 1 = \(\frac1{4(2+1+1)}=\frac1{16}\)

Hence, option (c) is wrong.

From option (d), \(\frac1{4(8n^2-4n+1)}\) at n = 1 = \(\frac{1}{8-4+1}=\frac15\)

Hence, option(d) is wrong.

Only option (b) may be true.

\(\therefore\) P(n) = \(\frac1{8n^2+4n+1}\) which can be proved by mathematical induction.

Given sequence is :

3,-1,\(\frac{1}{3}\),\(\frac{-1}{9}\),....

∴ a₁ = 3, a₂ = -1, a₃ = \(\frac{1}{3}\), a₄ = \(\frac{-1}{9}\),....

Common ration = r = \(\frac{a_4}{a_3}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_2}{a_1}\) = \(\frac{-1}{3}\) < 1

n^th term of given sequence is,

a_n = ar^n-1

= 3 x (\(\frac{1}{3}\))^n-1

= (-1)^n-1 x 3 x \(\frac{1}{3^{n-1}}\)

= \(\frac{(-1)^{n-1}}{3^{n-2}}\)

Sum upto 13 terms = S₁₃ = \(\frac{a(1-r^{13})}{1-r}\)

= \(\frac{3(1-(\frac{-1}{3})^{13})}{1-(\frac{-1}{3})}\)

= \(\frac{3(1+\frac{1}{3^{13}})}{1+\frac{1}{3}}\)

= \(\frac{3(1+\frac{1}{3^{13}})}{\frac{4}{3}}\)

= \(\frac{9}{4}\)(1 + \(\frac{1}{3^{13}}\))

Correct option is (C) The reminder is 1 when 931^th term is divided by 2

Given sequence is

2,3,5,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26......

2003^rd term = ?

1892^th term = 1937

1980^th term = 2026

2003^rd term = 2026 + (2003 - 1980)

 = 2026 + 23 = 2049

870^th term = 901

930^th term = 962

\(\therefore\) 931^st term = 963

(a, b) \(\frac{T_{2003}}{2048}=\frac{2049}{2048}\) the reminder left 1

option (a), and (b) are false

(c, d) \(\frac{T_{931}}2=481+\frac12\)

Thus, reminder left 1

option (c) is correct

Given

Number of boys = 5

Number of girls= 4

Concept

 ⁿC_n = ⁿC₀ = 1

Calculation

As there is at least 2 girls from the committee of 4 members. Therefore there could be 2,3 and 4 girls in a committee.

So,

⁵B₀  ×  ⁴G ₄+ ⁵B₁×  ⁴G ₃+ ⁵ B ₂ × ⁴ G ₂

⇒ 1 + (5 × 4!/1! × 4!) × (4 × 3!/3! × 1!) + (5 × 4 × 3!/2 × 1 × 3!) × (4 × 3 × 2 × 1)/(2 × 1 × 2 × 1)

⇒ 1 + 20 + 60

⇒ 81

Given:

The number is 75 × 73 × 78 × 76.

Concept used:

Divide each number individually by 34 and break it into a small number and find a number smaller than 34.

Calculation:

75 × 73 × 78 × 76 is divided by 34.

⇒ (75 × 73 × 78 × 76)/34

⇒ (75/34) × (73/34) × (78/34) × (76/34)

⇒ 7 × 5 × 10 × 8

⇒ (35 × 80)/34

⇒ (35/34) × (80/34)

⇒ 1 × 12

⇒ 12

∴ The remainder is 12.

Correct option is A) us

Correct option is C) ourselves

Correct option is B) ours

Correct option is (c) 0.35° 

If the encoder has ‘N’ tracks then the resolution = \(\cfrac{360°}{2^N}\)

In the given data N = 10

So, Resolution = \(\cfrac{360°}{2^{10}}\) = 0.35°

Correct option is A) some blue ones

Given:

As different Discount percentage is given as shown in the question

Formula used:

Discount = Discount% × Marked Price

Calculation:

⇒ 8% discount on marked price ₹400 = (400 × 8)/100 = ₹ 32

⇒12% discount on marked price ₹240 = (240 × 12)/100 = ₹ 28.8

⇒7% discount on marked price ₹500 = (500× 7)/100 = ₹ 35

⇒10% discount on marked price ₹320 = (320× 10)/100 = ₹ 32

∴ ₹ 35

Given- 

Discount given by publisher = 30%

Discount given by distributor = 20%

Formula Used-

Selling Price = Marked Price(1 - Discount%)

Profit% = (SP - CP)/CP × 100

Calculation-

Let the printed price of the book be 100

The publisher sells the book to distributor at 100(1 - 30%)

⇒ 70

Distributor sells the book to bookseller at 100(1 - 20%)

⇒ 80

∴ Profit% of distributor = (80 - 70)/70 × 100

⇒ \(14 \frac{2}{7}\)%

Given:

A girl buys chocolates at 4 for Rs. 3 and sells them at 5 for Rs. 4.

Concept:

This problem can be easily solved by equating number of chocolates.

Calculations:

So, cost price of 20 chocolate = Rs. 15

And, the selling price of 20 chocolate = Rs. 16

Then, profit = Rs. (16) - (15) = Rs. 1

Let the cost of the object be Rs x

Commission paid to first agent = 2/100 × x = 0.02x

Commission paid to second agent = (100 + 5) /100 × 0.02x = 1.05 × 0.02x

Commission paid to third agent = (100 + 5) /100 × 1.05 × 0.02x = 1.05² × 0.02x …

Commission paid to ninth agent = 1.05⁸ × 0.02x

Total commission paid = 0.02x + 1.05 × 0.02x + 1.05² × 0.02x + … + 1.05⁸ × 0.02x

⇒ 0.02x × (1 + 1.05 + 1.05² + … + 1.05⁸)

⇒ 0.02x × (1.05⁹ – 1) / (1.05 – 1) [∵ Sum of n terms of G.P. = a (rⁿ – 1) / (r – 1)]

⇒ 0.02x × (1.55 – 1) / 0.05 = 0.22x

Let the sales he made be Rs. x

If sales are less than 1 lakh, he would have received only basic pay which should be same for half the sales.

∴ Sales > 1 lakh

Let us assume 200000 > x > 100000

Half the sales will be less than 100000

∴ Salary received them = Basic pay = Rs. 15000

Salary received for sale of Rs. x = Basic pay + Commission of 10%

⇒ 50000 = 15000 + (x – 100000) × 10/100 [∵ Commission is paid on every penny above 1 lakh]

⇒ x = 450000

It contradicts the assumption

Let us assume x > 200000

Salary received = Basic pay + Commission of 10% + Bonus of 20%

⇒ 50000 = y + (x – 100000) × 10/100 + (x – 200000) × 20/100

100000 = y + 0.3x      ---- 1

Maximum value of x = 100000/0.3 = Rs. 33333 1/3

∴ Half of sales is always less than Rs. 200000

∴ No bonus is paid when sales is halved

⇒ 15000 = y + 10/100 × (x/2 – 100000)

⇒ 25000 = y + 0.05x   ---- 2

Equation 1 – Equation 2

⇒ 75000 = 0.25x

⇒ x = 300000

Substituting in equation 1

⇒ y = Rs. 10000

Given-

A sells a watch to B at a profit of 20%

B sells it to C at a loss of 10%

Formula Used-

Profit% = (CP - SP)/CP × 100       [where CP = Cost Price and SP = Selling Price]

Loss% = (SP - CP)/CP × 100      [where CP = Cost Price and SP = Selling Price]

Calculation-

Let the price at which A buys the watch is 100x.

A sells the watch to B at 100x(1 + 20%)

⇒ 120x

B sells the watch to C at 120x(1 - 10%)

⇒ 108x

According to Question -

108x = 432

⇒ x = 4

⇒ 100x = 400

∴ A paid Rs. 400 for the watch.

In original case:

Salary for first month = 10000 + 10/100 × (250000 – 150000) = Rs 20000

Salary for 2^nd month = 10000 + 10/100 × (300000 – 150000) = Rs 25000

Salary for 3^rd month = 10000 + 10/100 × (450000 – 150000) = Rs 40000

Salary for 4^th month = 10000 + 10/100 × (500000 – 150000) = Rs 45000

Total salary = 130000

In changed case

Salary received as basic pay = 4 × 12000 = Rs 48000

Salary received as commission = 130000 – 48000 = Rs 82000

Commission per month = 82000/4 = Rs 20500

Commission = (Sales – 150000) × 16/100

20500 = (Sales – 150000) × 0.16

128125 = Sales – 150000

Salary received as basic pay = 4 × 20000 = Rs. 80000

For month - I

Salary received as commission in original case = 10/100 × (250000 – 100000) = Rs 15000

Salary received as bonus in original case = 0

Total salary except basic = Rs. 15000

When 20% more sales, Sales = 120/100 × 250000 = 300000

Salary received as commission = 10/100 × (300000 – 100000) = 20000

Salary received as bonus = 0

Total salary except basic = Rs. 20000

For month - II

Salary received as commission in original case = 10/100 × (500000 – 100000) = Rs. 40000

Salary received as bonus in original case = 20/100 × (500000 – 300000) = Rs. 40000

Total salary except basic = Rs. 80000

When 20% more sales, Sales = 120/100 × 500000 = 600000

Salary received as commission = 10/100 × (600000 – 100000) = 50000

Salary received as bonus = 20/100 × (600000 – 300000) = Rs. 60000

Total salary except basic = Rs. 110000

For month - III

Salary received as commission in original case = 0

Salary received as bonus in original case = 0

Total salary except basic = Rs. 0

When 20% more sales, Sales = 120/100 × 100000 = 120000

Salary received as commission = 10/100 × (120000 – 100000) = 2000

Salary received as bonus = 0

Total salary except basic = Rs. 2000

For month - IV

Salary received as commission in original case = 0

Salary received as bonus in original case = 0

Total salary except basic = Rs. 0

When 20% more sales, Sales = 120/100 × 70000 = 84000

Salary received as commission in original case = 0

Salary received as bonus in original case = 0

Total salary except basic = Rs. 0

Total salary originally = Rs. 80000 + Rs. 15000 + Rs. 80000 = Rs. 175000

Total salary if 20% more sales = Rs. 80000 + Rs. 20000 + Rs. 110000 = Rs. 210000

% increase in salary = (210000 – 175000) /175000 × 100

 Correct Option (d) i_c is less than β_dci_b.

Explanation:

For transistor in common emitter mode the relation between collector current and collector emitter voltage is given by v_ce= v_cc-I_CR_C

cut off

 occurs when I_c= 0 

Saturation

 It occurs when there is no longer a change in collector current for a change in base current

 Active Mode

 In this mode the following relation is valid i_c =β I_β

Thus in saturation the collector current does not increase with base current

A device used to display one or more digital signals so that they can be compared to expected timing diagrams for the signals is a logic analyser.

An n stage ripple counter can count up to the binary value represented by the n-bits. = 2ⁿ - 1

Example:

if n = 3

then count is 0, 1 ,2 ,3 ,4, 5, 6, 7 (order may vary)

that is 2^n-1 = 2³ - 1 = 7 (maximum value)

The average power read by the wattmeter meter, is

\( = { \vee _{rms\;}}{I_{rms}}cos\phi = \frac{{110\;\sqrt 2 }}{{\sqrt 2 }} - \frac{{5\sqrt 2 }}{{\sqrt 2 }}{\rm{cos}}\left( {60°} \right)\)

\(= 110 \times 5 \times \frac{1}{2} = \frac{{550}}{2}\omega \)

Full scale reading of the meter = 110 × 5 = 500 ω

550 ω = 110 divisions

\(\frac{{550}}{2}\omega = \frac{{110}}{2} \Rightarrow 55\;divisions\)

The superposition theorem is used when the circuit contains number of voltage sources. 

Cryogenics refers to such work at very low temperatures, near absolute zero.