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Given:

S.P = 112% of C.P.

New C.P. =  90% of C.P.

New S.P. = 130% of 90% of C.P.

Concept used:

S.P. = (100 + Profit)% of C.P.

Calculation:

Let the C.P. be Rs. x.

S.P = 112% of C.P.

⇒ S.P. = 112/100 × x

New C.P. =  90% of C.P.

⇒ New C.P. = 90/100 × x

New S.P. = 130% of 90% of C.P.

⇒ New S.P. = 130/100 × 90/100 × x

⇒ New S.P. = 117/100 × x

According to the question,

New S.P. = S.P. + 6.75

⇒ 117/100 × x = 112/100 × x

⇒ (117/100 × x) – (112/100 × x) = 6.75

⇒ 5/100 × x = 6.75

⇒ x = 135

Required S.P. = (100 + 40)% of C.P.

⇒ Required S.P. = 140/100 × 135

⇒ Required S.P. = 189

∴ Rs. 189 make profit of 40%.

Given-  

Total cost of A and B = Rs. 10000

Profit% on article A = 15%

Loss% on article A = 10%

Concept Used-

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

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

Calculation-

Let the CP of two articles be A and B be x and y respectively.

According to Condition -

15% of x = 10% of y                    [∵ Profit on A = Loss on B]

⇒ x : y = 2 : 3

x = 2/5 × 10000       ∵ x + y = 10000

⇒ x = 4000

SP of article A = 4000 × (1 + 15%)

⇒ 4600

∴ The selling price of the article A is Rs. 4600

Given:

A is sold at 20% profit, B is sold for Rs. 54.60 more than the selling price of A

The overall profit earned on selling A and B is 28%

Formula used:

Profit = Selling price – Cost price

SP = [(100 + profit %)/100 × CP]

Calculation:

Let C.P of A nd B be Rs. x

According to the question:

S.P of A = (x × 120/100) = 6x/5

S.P of B = (6x/5 + 54.60)

Total C.P = Rs. (x + x) = Rs. 2x

Total S.P = (6x/5 + 54.60 + 6x/5)

⇒ (12x/5 + 54.60)

Profit = (12x/5 + 54.60 – 2x)

⇒ 2x/5 + 54.60

Now,

⇒ 28% of 2x = 2x/5 + 54.60

⇒ [(28/100) × 2x] = 2x/5 + 54.60

⇒ [(28/100) × 2x] = [2x/5 + (5460/100)]

⇒ [(28/100) × 2x] = (4x + 546)/10

⇒ 28x = 20x + 2730

⇒ 8x = 2730

⇒ x = 2730/8 = Rs. 341.25

∴The cost price of each article is Rs. 341.25.

Given: 

The cost of 30 books is Rs. 150

Concept: 

SP = [MP - (MP × Discount%)]

Calculation: 

According to the question

The cost of 30 books is 150 rupees, then

The cost of one book is 

⇒ 150/30

⇒ 5 rupees

The cost of 70 books is 

⇒ 70 × 5

⇒ 350 rupees

Now, 

The cost of 70 books after getting a 10% discount is 

⇒ [350 - (350 × 10%)]

⇒ 350 - 35

⇒ 315 rupees

∴ The cost of 70 books is 315 rupees.

Given:

An article was sold at a profit of 20%.

If the cost price would be 10% less and selling price would be Rs. 24 more, there would be profit of 40%.

Formula Used:

Profit% = Profit/CP × 100

Loss% = Loss/CP × 100

Reduced CP × profit% = New SP

Calculation:

Let the cost price of article = 100x

SP of article = 120/100 × 100x = 120x

From the question we get,

9/10 × 100x × 140/100 = 120x + 24

⇒ 126x = 120x + 24

⇒ x = 4

Original CP = 100x = 100 × 4 = Rs. 400

To gain 60%,

SP = Rs. (160/100 × 400) = Rs. 640

∴ It should be sold at Rs. 640

Given:

Cost price of shoes lies between = Rs. 500 to Rs. 700

Selling price of shoes lies between = Rs. 1000 to Rs. 1300

Concept Used:

To gain maximum profit the cost price should be minimum and the selling price should be maximum.

Calculation:

Minimum CP = Rs.500

Maximum SP = Rs.1300

Profit on 1 pair of shoes = SP – CP = Rs. (1300 – 500) = Rs.800

Maximum profit on 10 pair of shoes = Rs. (10 × 800) = Rs.8000

∴ Maximum profit is Rs. 8000

Given:

Cost price of both articles = Rs.3500

Profit gained from first article = 20%

Loss from 2nd article = 10%

Formula Used:

Profit% = Profit/CP × 100

Loss% = Loss/CP × 100

Calculation:

Let the cost price of first article be 100x and cost price of second article be 100y.

Selling price for first article = 120/100 × 100x = 120x

Selling price for second article = 90/100 × 100y = 90y

So, 120x = 90y

⇒x = 3y/4      ----eq. 1

And total cost price = 100x + 100y = 3500

⇒ x + y = 35      ----eq.2

From eq.1 and eq.2 by substituting values

We get, x = 15 and y = 20

CP of first article = 100 × 15 = Rs. 1500

CP of second article = 100 × 20 = Rs. 2000

∴ CP of first and second article are Rs. 1500 and Rs. 2000 respectively.

Given: 

The cost price of the article is 250 rupees and sold at a loss of 5%.

Formula used: 

S.P = [(100 - loss%)/100] × C.P

Calculation: 

The selling price of the article is 

⇒ [(100 - 5)/100] × 250

⇒ (95/100) × 250

⇒ (95 × 5)/2

⇒ 237.50 rupees 

∴ The required selling price of the article is 237.50 rupees.

Given:

Cost price (C.P.) ∶ Selling price (S.P.)  = 20 ∶ 21.

Formula used:

Profit = S.P. – C.P.

Profit % = (P/C.P.) × 100 

Where

P → Profit

S.P. → Selling price

C.P. → Cost price

Calculations:

Let the CP be 20x and SP be 21x.

Then, Profit = 21x – 20x = x

Profit % = (x/20x) × 100 = 5%

∴ The required Profit% is 5%.

Given :

ab is a 2-digit prime number

a ≠ b

To find :

The sum of ab and ba

Solution :

List of such prime numbers is

13, , 31, 17, 71, 37, 73, 79, 97.

⇒ Sum of these numbers = 13 + 31 + 17 + 71 + 37 + 73 + 79 + 97

⇒ Sum of these numbers = 418

∴ The required number is 418.

√1521 = 39 and √1225 = 35

So, √1521 - √1225 = 39 - 35 = 4

Given : 

AB × BA = BCB and A ≠ B ≠ C

Calculation

A and B cannot take the same value as A ≠ B

So, let A = 1 and B = 2

Then, AB × BA = 12 × 21 = 252

⇒ BCB = 252 

∴ C = 5 

A = 1, B = 2 and C = 5

∴ A + B + C = 1 + 2 + 5 = 8

Calculation : 

x = 2/3 and y = 3/4

x - y = 2/3 - 3/4 = (8 - 9)/12 = -1/12

∴ (x - y)^-1 = -12

x^-1 = 3/2 and y^-1 = 4/3

x^-1 - y^-1 = 3/2 - 4/3 = (9 - 8)/6 = 1/6

Then, the rational number between (x - y)^-1 and x-1 - y^-1 = (-12 + 1/6)/2 = (1 - 72)/12 = -71/12

 A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero. For example : 2, 6/5, -9/12, etc.

Rational number between two numbers a and b = (a + b)/2.

Given :  x = 597 × 1022 + 73.5 × 1021 

Calculation

 x = 597 × 1022 + 73.5 × 1021

= 597 × 10 × 10²¹ + 73.5 × 10²¹

= 5970 × 10²¹ + 73.5 × 10²¹

= (5970 + 73.5) × 10²¹ = 6043.5 × 10²¹ = 6.0435 × 10²⁴

On comparing it with the standard form of x which is x = 6.0435 × 10^m  we get

m = 24

Given:

Raju bought 50 kg of rice at a discount of 10%

The shopkeeper offered 10 kg rice for free  

Formula used:

Profit % = (Profit/Cost price) × 100

Calculation:

Let the cost of rice be Rs. x per kg

⇒ Price of 50 kg rice = Rs. 50x

But he bought it at a discount of 10%

⇒ Cost Price of Rice for Raju = Rs. (50x × 90/100 ) = Rs. 45x

⇒ The price of extra 10 kg rice = Rs. 10x

Raju didn’t paid Rs. 10x though he got (50 + 10) kg = 60 kg of rice

But he sold the 60 kg rice at a price Rs. 60x

⇒ Profit gained by him = (60x – 45x) = 15x

⇒ Profit percentage = (15x/45x) × 100

⇒ 33.33 %

∴ Raju made 33.33% profit on the whole transaction.

Concept used :

Number of parallelograms when n parallel lines intersect n parallel line = ⁿc₂ × nc₂

Calculations :

Here 4 parallel lines are intersecting 4 parallel lines so n will be 4 

Total number of parallelogram = ⁴c₂ × ⁴c₂

⇒  (4!/(2! × (4 - 2)!) × (4!/(2! × (4 - 2)!)

⇒ (4!)/(2! × 2!) × (4!)/(2! × 2!)                      (n! = n.(n - 1).(n - 2)..........1) 

⇒ 36 

∴ Option 4 will be the correct choice.

Given:

x : y = 3 : 2

x + y = 90

Calculation:

Let us take the value of x and y be 3a and 2a

x + y = 3a + 2a 

⇒ 3a + 2a = 90

⇒ 5a = 90 

⇒ a = 18

⇒ x = 3a = 54 

⇒ y = 2a = 36

⇒ x - y = 54 - 36 = 18

 ∴ The value (x - y) is 18

Given:

The ratio of milk and water in the 300 L of the mixture is 1:5.

Calculation:

Let x Litre of milk is added in the mixture so that the ratio becomes 2 ∶ 3

⇒ (50 + x)/250 = 2/3

⇒ (50 + x)× 3 = 250×2

⇒ 150 + 3x = 500

⇒ x = 350/3

∴ 350/3 Litre of milk is added

Given:

Initial ratio of milk and water in the given mixture = 4 : 5

Adding 30 liters of water in the mixture then ratio of milk and water = 1 : 2

Calculation:

Let the initial amount of mixture be p liters

Initial amount of milk in the mixture = 4/9 × p liters

⇒ 4p/9 liters

Initial amount of water in the mixture = 5/9 × p liters

⇒ 5p/9 liters

Now, 30 liters of water is added

Final amount of water = 30 + 5p/9 liters

Ratio of milk and water after adding 30 liters of water = (4p/9)/(30 + 5p/9)

⇒ (4p/9)/(30 + 5p/9) = 1/2

⇒ 2 × (4p/9) = 30 + 5p/9

⇒ 8p/9 = (270 + 5p)/9

⇒ 8p = 270 + 5p

⇒ 3p = 270

⇒ p = 270/3 

⇒ p = 90 liters

Initial amount of milk in the given mixture = 4/9 × 90 liters

⇒ 40 liters

∴ The initial quantity of milk in the mixture is 40 liters

Given:

Total mixture = 65 litres

Ratio of water and milk in a mixture = 5 : 8

Final ratio of water and milk after adding more water = 3 : 4

Calculation:

Amount of water in mixture = (5/13) × 65 litre

⇒ 25 litres

Amount of milk in mixture = (8/13) × 65

⇒ 40 litres

Let the amount of water added be x litre

According to the question,

(25 + x)/40 = 3/4

⇒ 100 + 4x = 120

⇒ 4x = 20

⇒ x = 5

∴ The amount of water to be added is 5 litres

Given:

Initial concentration = 60 litre

Final concentration = 48.6 litre

Quantity removed in each cycle = x litre

The process is repeated two times means: n = 2

Formula:

FC = IC × (1 – Quantity removed in each cycle/total IC)ⁿ

Here,

FC = final concentration

IC = Initial concentration

n = The process repeated how many times

Calculation:

We know that –

FC = IC × (1 – Quantity removed in each cycle/total IC)ⁿ …….(1)

Put all the given values in equation (1) then we get

48.6 = 60 × (1 – x/60)²

⇒ 48.6/60 = (1 – x/60)²

⇒ (486/600) = (1 – x/60)²

⇒ (81/100)² = (1 – x/60)²

⇒ (9/10)² = (1 – x/60)²

⇒ 9/10 = 1 – x/60

⇒ x/60 = 1 – 9/10

⇒ x/60 = 1/10

⇒ 10x = 60

⇒ x = 60/10

⇒ x = 6

∴ The Value of x = 6 litres

Given:

The initial and the final ratio of milk and water in a given mixture are 4∶9 and 1∶3. 20 L of water is added to this mixture.

Calculation:

Let the initial quantity of milk and water in the mixture be 4x, 9x

After adding 20 L of water the ratio becomes 1:3

⇒ (4x)/(9x + 20) = 1/3

⇒ 12x = 9x + 20

⇒ 3x = 20, x = 20/3

Now the initial quantity of water in the mixture = 9×(20/3) = 60 L

∴ Initial quantity of water in the mixture is 60L.

Mental, Physical, Social, Financial.

Correct option: (c) Clarity of speech

a. This type of suspension gives the maximum room in the engine compartment due to the absent of upper control arm. 

b. It is simple in construction and light in weight. 

c. Due to its light weight, road irregularities are easily countered and hence provide increased road safety. 

d. It improves the ride comfort and gives a light and self-stabilizing steering, also the wheel camber is more stable. 

e. In addition to its relatively low initial cost, its maintenance, repair or replacement is less expensive.

Both A and B are correct :

(a) Mental or Physical

(b) Social or Financial

Correct option: (d) Spiritual

According to the Bohr Atomic model, a small positively charged nucleus is surrounded by revolving negatively charged electrons in fixed orbits. He concluded that electron will have more energy if it is located away from the nucleus whereas electrons will have less energy if it located near the nucleus.

The stress causal agents

• Mental 
• Physical 
• Social 
• Financial

The stress management are

It helps people lead and healthy and happier life

Two options of source of funds from banks

1.Cash credit: the facility given to the Industrial / Business customers is known as ‘Cash Credit’ (CC) account in which the stock (raw material / work in process / finished goods) lying in the go down is pledged or hypothecated as the security by the bank. 

2. Overdraft: An overdraft facility is an open-ended facility. Normally the limit is initially sanctioned for a period of one year and rolled over after a review by the bank of the facility utilised by the borrower. Bank charges interest on the actual amount utilised by the borrower.

The ecological imbalance.

1. Destruction of forests

2. industrialization  

3. urbanization

Correct answer is Frequency.

Green Economy

Correct option: d. All of the above.

Correct option: C) Oxford English Dictionary

1. Individuals feel more empowered and responsible. 

2. It provides flexibility to choose and define working hours and working mechanisms.

a. Physical Barriers 

b. Language Barriers 

c. Gender Barriers 

d. Attitudinal Barriers ( ½*4=2)

The correct answer is (i), (ii) and (iii).

• Social responsibility is a moral obligation on a company or an individual to make decisions or actions that is in favour and useful to society.
• Moral values that are inherent in society create a distinction between right and wrong.
• In this way, social fairness is believed (by most) to be in the “right”, but more frequently than not this “fairness” is absent.
• Every individual has a responsibility to act in manner that is beneficial to society and not solely to the individual.

(a) 1 and 2 only 

Throttling process involves the passage of a higher pressure fluid through a narrow constriction. This process is adiabatic, and there is no work interaction. Hence, 

Q = 0 & W = 0

\(\Delta\)PE = 0 (Inlet and outlet are at the same level) 

\(\Delta\)KE = 0 (KE does not change significantly) 

\(\therefore\) Applying the SFEE, h₁ = h₂ 

Thus, enthalpy remain constant Further, the velocity of flow is kept low and any difference between the kinetic energy upstream and downstream is negligible. The effect of the decrease in pressure is an increase in volume.