Correct option  (1) d a b c 

Explanation:

a → isobaric 

d → isochoric 

 b → isothermal 

 c → adiabatic (d a b c)

The Correct option is (c) 5P₁V₁

(a) 150 K, 900 cal 

Given 

Cost price of 100 mangoes = Rs. 125

Number of mangoes purchased = 400

Total profit = Rs. 100

Formula used

Selling price(SP) = Cost price(CP) + Profit(P)

Calculation

CP of 100 mangoes = Rs. 125

CP of 400 mangoes = 125 × 4 = Rs.500

SP = CP + P

⇒ SP = 500 + 100 = 600

SP of 400 mangoes = Rs. 600

⇒ SP of 1 dozen(12) mangoes = 600 × 12/400 = Rs. 18

Given:

Cost price of 400 chairs = Selling price of 320 chairs

Formula Used:

Profit percentage = {(SP – CP)/CP} × 100

Calculation:

Let CP of 1 chair be Rs. x

CP of 320 chairs = Rs. 320x

CP of 400 chairs = Rs. 400x

SP of 320 chairs = Rs. 400x

Profit percentage = (400x – 320x)/320x × 100 = 25%

Given:

Cost price of 320 books = Selling price of 400 books

Formula Used:

Loss% = {(C.P – S.P)/C.P} × 100

Calculation:

Let C.P of 1 book = Rs. x

C.P of 320 books = Rs. 320x

S.P of 400 books = Rs. 320x

C.P of 400 books = Rs. 400x

Loss = Rs. (400x – 320x) = Rs. 80x

Loss% = (80x/400x) × 100 = 20%

15th of August 

15 AUGUST

15 th August is our independence day

\(9a^2 - 9b^2 + 6a + 1\)

\(= 9a^2 + 6a + 1 - 9b^2\)

\(= (3a + 1)^2 -(3b)^2 \)    \((\because a^2 + 2ab + b^2 = (a + b)^2)\)

\(= (3a + 1 - 3b)(3a + 1 + 3b)\)   \((\because a^2 - b^2 = (a + b) (a - b))\)

\(= (3 (a - b) + 1) (3(a + b) + 1)\)

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Add the following expressions:

1.
   (i) 5x+7x+(-6x) = 5x+7x-6x = (5+7-6)x = 6x
   (ii) 3x/5 + 2x/3 + (-4x/5)
        taking x as common
(3/5+2/3-4/5)x
(9+10-12)/15 x
7x/15
(vii) 5x-2x²-8, 8x²-7x-9, 3+7x²-2x
5x-2x²-8+8x²-7x-9+3+7x²-2x

13x²-4x-14

Similarly you can do all the question

The teacher can select a student for monitor in two exclusive ways
(i) Select a boy among 10 boys, which can be done in 10 ways OR
(ii) Select a girl among 8 girls, which can be done in 8 ways. 

Hence, by the fundamental principle of addition, either a boy or a girl can be selected in 10 + 8 = 18 ways.

Given that,

α and β are zeros of the polynomial x² – x – 12. 

The sum of zeros = α + β = \(\frac{-b}{a}\) 

= \(\frac{−(−1)}{1}\) 

= 1. 

The product of zeros = αβ 

= \(\frac{c}{a}\) 

= \(\frac{−12} {1}\) 

= −12. 

Now,

(α – β)² = (α + β)² – 4αβ 

= 1² – 4 × –12 

= 1 + 48 

= 49. 

⇒ α − β = ±7. 

Case I :-  α − β = 7 and 

α + β = 1. 

⇒ 2α = 8 

⇒ α = 4. 

Then,

4 + β = 1 

⇒ β = 1 − 4 

= −3. 

i.e., α = 4 and 

β = −3 .

Case II :- α − β = −7 and 

α + β = 1 

⇒ 2α = −6 

⇒ α = −3. 

Then,

−3 + β = 1 

⇒ β = 1 + 3 = 4. 

i.e., α = −3 and 

β = 4. 

Hence, 

–3 and 4 are zeros of given polynomial x² – x – 12. 

We want to find a quadratic equation whose zeros are 2α and 2β 

i.e., quadratic equation is (x – 2α) (x – 2β) = 0 

⇒ (x − 8)(x + 6) = 0 

(∵ either α = −3 & β = 4 or α = 4 and β = −3) 

⇒ x² − 2x − 48 = 0. 

∴ Required quadratic equation is x² − 2x − 48 = 0.

We have five cards – ten, jack, queen, king and ace of diamonds. 

Total cards = n(S) = 5. 

(a) Option (iv)

Total queen cards in given set of cards n(E₁) = 1. 

∴ Probability that drawn card is a queen = \(\frac{n(E_1)}{ n(S)}\) = \(\frac{1}{5}\). 

Hence,

Option (iv) is correct. 

(b) Option (iii)

If the queen is drawn and put aside. 

Then total number of remaining cards = n(S') = 4. 

And total ace cards in remaining set of cards = n(E₁) = 1. 

∴ The probability that 2nd card is an ace = \(\frac{n(E_1)}{ n(S')}\) = \(\frac{1}{4}\) . 

Hence, 

Option (iii) is correct. 

(c) Option (iv) 

Total queen cards in the remaining set of cards = n(E₃) = 0.

∴ The probability that 2nd card is queen card = \(\frac{n(E_3)}{ n(S') }\)

= \(\frac{0}{4}\) 

= 0. 

Hence,

Option (iv) is correct. 

(d) Option (iv)

From five cards, 

Two cards are drawn. 

Let \(E'_4\) = Event that first card is ten and second card is ace card. 

\(E''_4\) = Event that first card is ace card and second card is ten card. 

S₄ = total possible outcomes in drawing two cards = 5 × 4 = 20. 

∴ Probability that drawn card are an ace card and ten card = P(\(E'_4\)) + P(\(E''_4\)) 

= \(\frac{1}{5\times 4}\) + \(\frac{1}{5\times 4}\) 

= \(\frac{2}{20}\) 

= \(\frac{1}{10}\)

Hence,

The correct answer is option (iv). 

(e) Option (i) 

In an ordinary year, total day = 365 

= 52 weeks and 1 day. 

Hence,

There will be 52 Mondays for sure. 

So,

In ordinary year there will be 52 Mondays and 1 day will be left This one day could be a Monday or Tuesday or Wednesday or Thursday or Friday or Saturday or Sunday. 

Hence,

Total possible outcomes = 7 

But favorable outcome = 1. 

∴ Probability of getting 53 Mondays = \(\frac{1}{7}\). 

Hence, 

Option (i) is correct.

Concept:

The relation between AM, GM and HM:

(GM)2 = (AM) × (HM)

Calculation:

Given AM = 24 and HM = 6

And we know that (GM)² = (AM) × (HM)

⇒ GM =

 \(\sqrt{24\times 6}=\sqrt{144}\\=12 \)

Hence, option (1) is correct.

Quartile deviation (Q.D.) = (Q₃ - Q₁)/2

= (121 - 61)/2

= 60/2 = 30.

Given :

(?)³/4 = (?)²

Calculation :

According to the question,

⇒ ?³/4 = ?²

⇒ ?³/?² = 4

⇒ ? = 4

∴ The required answer is 4.

Given:

The LCM and HCF of two numbers are 120 and 3

One of these numbers is 24

Formula used:

Product of two no. = product of their LCM and HCF

Calculation:

Let the other no. be p

So, 24 × p = 120 × 3

⇒ 24 × p = 360

⇒ p = 15

Therefore, the second number is 15

Concept:

The divisibility rule of 3 states that a number is completely divisible by 3 if the sum of its digit is divisible by 3 or multiple of 3.

Calculation:

⇒ 7 + A + 3 = 10 + A  (Possible value of A is 2, 5, 8)

⇒ 136 + 5B7 = 7A3    (Possible value of B is 6, 7, 8, 9)

⇒ For A = 2, B = 8

⇒ 7 + 2 + 3 = 12, (Which is divisible by 3 or multiple of 3)

⇒ The value of B = 8

∴ The required result will be 8.

Calculation:

⇒ The single-digit page (From 1 to 9) = 9 digit

⇒ Double digit page (From 10 to 99) = 2 × 90 = 180 digit

⇒ Three digit page (From 100 to 999) = 3 × 900 = 2700 digit

⇒ Total digit used = 9 + 180 + 2700 = 2889

⇒ The remaining digit = 3089 - 2889 = 200

⇒ The number of page with four digits = 200/4 = 50

⇒ The required number of total page = 999 + 50 = 1049

∴ The required result will be 1049.

Given: 

A and B can do a piece of work in 18 days. B and C do it in 30 days.

A is twice as good a workman as C.

Calculation: 

Let total work is LCM of (18, 30) equal to 90 units. 

According to the question

The efficiency of A and B together is 

⇒ 90/18

⇒ 5 units

The efficiency of B and C together is 

⇒ 90/30

⇒ 3 units

The difference between the efficiency of A and C is 

⇒ 5 - 3

⇒ 2 units

The ratio of the efficiency of A and C is 

⇒ 2 : 1

Now, We say that the efficiency of A is 4 units and C is 2 units, then 

The efficiency of B is 

⇒ 5 - 4

⇒ 1 units 

So, 

The time taken by B to complete the work is 

⇒ 90/1

⇒ 90 days

∴ B alone completes the whole work in 90 days.

Given:

Ram is thrice efficient than Shyam and Ram finishes the work in 15 days less than Shyam

Concept used:

If time taken finish the work = X days

∴ Part of work completed in one day = 1/X

Efficiency ∝ 1/Time

Calculation:

Ram is thrice efficient than Shyam

∴ Ratio of Efficiency = 3 ∶ 1

And ratio of the time taken = 1 ∶ 3

Now, Ram takes 15 days less than shyam to finish the work

∴ 3x – x = 15

⇒ x = 7.5

So, Ram takes 7.5 days and Shyam takes 22.5 days to finish the work

Now, Let both takes x days to finish the work

∴ Time taken by both to finish the work = (1/7.5 + 1/22.5) = 1/x

⇒ x = 5.625 days

So, both of them will finish the work in 5.625 days

Hence, option (1) is correct