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Given :

The ratios of acid and water in vessels A and B are 4 : 5 and 7 : 5, respectively.

Calculations :

The ratio of acid to water in vessel A is 4 : 5 

Let the total volume of acid and water be 4x and 5x 

The ratio of acid to water in vessel B is 7 : 5 

Let the total volume of acid and water be 7y and 5y

According to the question volume of water and acid must be equal 

So,

(4x + 7y) : (5x + 5y) = 1 : 1 

⇒ x = 2y 

Putv value of x = 2y to get volume 

Volume of vessel A = 4x + 5x = 9x 

⇒ 18y 

Volume of vessel B = 7y + 5y 

⇒ 12y 

Ratio of vessel A to B 

Volume of vessel A : Volume of vessel B = 18y : 12y 

⇒ 3 : 2 

∴ 3 : 2 will be the ratio at which vessel A and B must be mixed  

Solution:
Let the minimum length of the shortest side be x.

Then the longest side of a triangle is 3x and the third side is 3x – 2.

Perimeter of the triangle ≥ 61

x + 3x + 3x – 2 ≥ 61

7x – 2 ≥ 61

7x ≥ 61 + 2

7x ≥ 63

x ≥ 9

Thus, 9 cm is the required minimum length of the shortest side.

Solution:
Let x be the shortest board.

Then x + 3 is the second piece and 2x is the third piece.

x + (x + 3) + 2x ≤ 91
and 2x ≥ (x + 3) + 5

x + (x + 3) + 2x≤ 91
4x + 3 ≤ 91
4x ≤ 91 – 3
4x ≤88
x ≤ 44/2

2x ≥ (x + 3) + 5
2x ≥ x + 8
x ≥ 8

i.e, 8 ≤ x < 22

Thus the possible lengths of the shortest board is at least 8cm long but not more than 22cm long.

Given differential equation is

\(x^3\frac{d^3y}{dx^3}+3x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=xlog x\) which is

homogeneous linear differential equation.

Let x = e^z

⇒ log x  = z

Then x \(\frac{dy}{dx}=\frac{dy}{dz}=Dy\)

and x²\(\frac{d^2y}{dx^2}=D(D-1)y\)

and x³\(\frac{d^3y}{dx^3}=D(D-1)(D-2)y\)

Then given differential equation reduces to

D(D - 1)(D - 2)y + 3D(D - 1)y + Dy + y = ae^z

⇒ (D³ + 3D² + 2D)y + (3D²- 3D)y + Dy + y = ze^z

⇒ (D³ - 3D² + 3D² + 2D - 3D + D + 1)y = ze^z

⇒ (D³ + 1) y = ze^z

Its auxiliary equation is

m³ + 1 = 0

⇒ (m + 1)(m² - m + 1) = 0

⇒ m = -1 or m = \(\frac{1\pm\sqrt{1-4}}2=\frac{1\pm\sqrt3i}2\)

C.F. = C₁e^-z + e^z/2(C₂ cos(\(\frac{\sqrt3}2\)z) + C₃(\(\frac{\sqrt3}2\)z))

\(=\frac{C_1}x+x^{1/2}(C_2cos(\frac{\sqrt3}2log x)+C_3(\frac{\sqrt3}2log x))\)

(\(\because\) z = log x and e^z = x)

P. I. = \(\frac{1}{D^3+1}ze^z=e^z\frac{1}{(D+1)^3+1}z\)

 = e^z \(\frac{1}{D^3+3D^2+3D+1+1}z\)

 =   e^z \(\frac{1}{D^3+3D^2+3D+2}z\) 

 =  \(\frac{e^z}2\cfrac{1}{1+\frac{D^3+3D^2+3D}2}z\)

=   \(\frac{e^z}2(1+\frac{D^3+3D^2+3D}2)^{-1}\) z

= \(\frac{e^z}2(1-\frac{D^2+3D^2+3D}2+....)z\) (\(\because\) (1 + x)^-1 = 1 - x + x² - x³ + 0)

 = \(\frac{e^z}2(z-\frac32Dz)\)

= \(\frac{e^z}2(z - \frac32)\)

 = \(\frac{x}2(log x-\frac32)\) (\(\because\) z = log x and e^z = x)

\(\therefore\) Complete solution is

y = C.F. + P. I.

 = \(\frac{C_1}x+\sqrt x[C_2cos(\frac{\sqrt3}2log x)C_3sin(\frac{\sqrt3}2log x)]\) + \(\frac{x}2(log x-\frac32)\)

Given:

A invested = 50,000 for 6 months

The ratio of sum of profit of A and C to the profit of B = 3 ∶ 1

The profit of B and C are equal

Formula:

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

Calculation:

Let the profit of B and C be x.

ATQ,

The sum of profit of A and C/ Profit of B = 3/1

⇒ (300000 + x)/x = 3/1

⇒ 300000 = 2x

⇒ X = 150000

∴ The total amount invested by C is 150000.

Given:

C.P of fan = Rs. 1200

Expenditure on transportation = Rs. 16

S.P of fan = Rs. 1300

Formula used:

Profit% = (Profit/C.P) × 100

Calculation:

Total cost = Rs. (1200 + 16)

⇒ Rs. 1216

Profit = Rs. (1300 – 1216)

⇒ Rs. 84

Profit% = (84/1216) × 100

⇒ 6.90%

∴ The profit percentage is 6.90%

Given:

Investment of A = 12000 for 5 months

Investment of B = 8000

The ratio of profit at the end of the year = 5 ∶ 6

Concept used:

The ratio of the product of time and investment = Ratio of profit

If X₁, X_2, X₃ … is the ratio of investment and P₁, P_2, P₃ … is the ratio of profits then the ratio of time periods of investment is calculated as

P₁/X₁∶ P₂/X₂∶ P₃/X₃ …

Calculation:

Let ratio of time period be 5 ∶ t (Assume, B invested for t months)

Ratio of investment = 12000 ∶ 8000 = 3 ∶ 2

Ratio of product of time and investment = 3 × 5 ∶ 2t = 15 ∶ 2t

Ratio of product of time and investment = Ratio of profit

Ratio of profit = 15 ∶ 2t = 5 ∶ 6 (Given)

⇒ 15/2t = 5/6

⇒ t = (15 × 6)/10 = 90/10 = 9

Given:

Investment of A = 12000 for 5 months

Investment of B = 8000

Ratio of profit at the end of the year = 5 ∶ 6

Concept used:

Ratio of product of time and investment = Ratio of profit

If X₁, X_2,X₃ … is the ratio of investment and P₁, P_2,P₃ … is the ratio of profits then the ratio of time periods of investment is calculated as P₁/X₁∶ P₂/X₂∶ P₃/X₃ …

Calculation:

Let ratio of time period be 5 ∶ t (Assume, B invested for t months)

Ratio of investment = 12000 ∶ 8000 = 3 ∶ 2

Ratio of product of time and investment = 3 × 5 ∶ 2t = 15 ∶ 2t

Ratio of product of time and investment = Ratio of profit

Ratio of profit = 15 ∶ 2t = 5 ∶ 6 (Given)

⇒ 15/2t = 5/6

⇒ t = (15 × 6)/10 = 90/10 = 9

∴ B invested for 9 months.

Given:

Investment of Anil = Rs. 2500 for 12 months

Investment of Virat = Rs. 4000

Ratio of profit at the end of year = 3 ∶ 2

Concept used:

Ratio of product of time and investment = Ratio of profit

Calculation:

Let, Virat joined after t months

Investment of Anil for 1 year = Rs. 2500 for 12 months = Rs. (2500 × 12) = Rs. 30000

Investment of Virat for 1 year = Rs. 4000 for T months = Rs. 4000T, where T = (12 – t)

Ratio of product of time and investment = Ratio of profit

Ratio of profit = 30000 ∶ 4000T = 15 ∶ 2T = 3 ∶ 2 (Given)

Equating two ratios, 15/2T = 3/2

⇒ T = 15/3 = 5

⇒ t = 12 – 5 = 7

∴ Virat joined after 7 months.

Given:

Investment of Anil = Rs. 2500 for 12 months

Investment of Virat = Rs. 4000

Ratio of profit at the end of year = 3 ∶ 2

Concept used:

Ratio of product of time and investment = Ratio of profit

Calculation:

Let, Virat joined after t months

Investment of Anil for 1 year = Rs. 2500 for 12 months = Rs. (2500 × 12) = Rs. 30000

Investment of Virat for 1 year = Rs. 4000 for T months = Rs. 4000T, where T = (12 – t)

Ratio of product of time and investment = Ratio of profit

Ratio of profit = 30000 ∶ 4000T = 15 ∶ 2T = 3 ∶ 2 (Given)

Equating two ratios, 15/2T = 3/2

⇒ T = 15/3 = 5

⇒ t = 12 – 5 = 7

Given:

Ratio of investment of X and Y= 3 : 5

Investment of Z = 1/4 × (investment of X + investment of Y)

Time period for investment of X = time period for investment of Y = 12 months

Time period for investment of Z = 8 months

Profit of Y = Rs. 22500

Concept used:

Ratio of product of time and investment = Ratio of profit

Calculation:

Let investments of X and Y be 3X and 5X respectively

Investment of Z = 1/4 × (3X + 5X) = 1/4 × 8X = 2X

Ratio of investment of time and investment = 3X × 12 : 5X × 12 : 2X × 8 = 36X : 60X : 16X = 9 : 15 : 4

Ratio of product of time and investment = Ratio of profit

Ratio of profit = 9 : 15 : 4

Profit of Y = Rs. 22500 (Given)

⇒ 15R/28 = 22500

⇒ R/28 = 1500

⇒ R = 1500 × 28 = 42000

Given

A starts a taxi service by investing Rs. 25 lakhs

After 3 months, B joins the business by investing Rs 40 lakhs

4 months after B joined, C to joins by investing Rs. 50 lakhs

Concept used

The ratio of investing money by three is same as the ratio of profit they earned

Calculation

A start business by investing 25 lakh rupees for 12 month

So total invest = 2500000 × 12

B invest  40 lakh money after 3 months of A

B invest 40 lakh for 9 months

B invest = 4000000 × 9

C invest money 5 months after B 

So C invest 5000000 for 5 months

The ratio of A, B and C investing = 2500000 × 12 : 4000000 × 9 : 5000000 × 5

⇒ 30 : 36 : 25

Profit ratio is also same i.e 30 : 36 : 25

Total profit  is Rs.273000

Profit share of C is 25/(30 + 36 + 25){273000}

∴ C's share of the Profit is Rs.75,000

Given:

Investment of A = Rs. 3000

Investment of B = Rs. 2000

Profit = Rs. 36000

Concpet used:

Profit = Investment × Time

Calculations:

⇒ A × 6 + 2A × 6 : B × 12 : C × 6

⇒ 3000 × 6 + 6000 × 6 : 2000 × 12 : 5000 × 6

⇒ A : B : C = 9 : 4 : 5

Let the ratio be 9x, 4x and 5x respectively

⇒ 9x + 4x + 5x = 36000

⇒ 18x = 36000

⇒ x = 2000

⇒ Difference between A and C = 9x – 5x = 4x

⇒ Difference between A and C = 4 × 2000

⇒ Difference between A and C = Rs. 8000

∴ The difference between the profit of A and C is Rs. 8000

Given∶

Their investing amount = Rs. 10,000

After 5 months, Sita withdrew = Rs. 2000, Geeta withdrew = Rs. 3000 and Reeta invest = Rs. 6000

Total profit at the end of the year = Rs. 36700.

Formula Used∶

Share = Amount × Time

Calculation∶

The ratio of investment of Sita, Geeta and Reeta = Sita ∶ Geeta ∶ Reeta

= (10000 × 5 + 8000 × 7) ∶ (10000 × 5 + 7000 × 7) ∶ (10000 × 5 + 16000 × 7)

= (50000 + 56000) ∶ (50000 + 49000) ∶ (50000 + 112000)

= (106000) ∶ (99000) ∶ (162000)

= 106 ∶ 99 ∶ 162

Sum of the ratio = 106 + 99 + 162 = 367

According to question∶

When, total share is 367 Rs. then profit = 36700

When share is Rs. 1 then profit = 36700/367 = Rs. 100

And when share is Rs. 99, then profit = 100 × 99 = Rs. 9900

∴ Geeta share is Rs 9900.

The given quadratic equation is 12x² + 4kx + 3 = 0, comparing it with ax² + bx + c = 0.

⇒  We get, a = 12,b = 4k and c = 3

⇒  It is given that roots are real and equal.

∴  b²−4ac = 0

⇒  (4k)²−4(12)(3)=0

⇒  16k²−144 = 0

⇒  16k² = 144

⇒  k²= 144/16​

⇒  k² = 9

∴   k = ±3

A’s capital = B’s capital + 1/3^rd of B’s capital = 4/3 × B’s capital

Time period of A for 100% capital = 5 months and remaining 50% = 7 months, (50% withdrew)

Time period of B for 100% capital = 8 months and remaining 66.66% = 4 months, (1/3^rd withdrew)

Concept used:

Ratio of product of time and investment = Ratio of profit

Calculation:

Let B’s capital be X, then A’s capital = X + 1/3 × X = 4/3 × X = 4X/3

Investment of A for 1 year = 4X/3 for 5 months + (4X/3 – 50% of 4X/3) for 7 months = 4X/3 × 5 + (4X/3 – 2X/3) × 7

⇒ 20X/3 + 14X/3 = 34X/3

Investment of B for 1 year = X for 8 months + (X – 1/3 × X) for 4 months = 8X + 2X/3 × 4

⇒ 8X + 8X/3 = 32X/3

Ratio of profit = 34X/3 ∶ 32X/3 = 34 ∶ 32 = 17 ∶ 16

Concept:

Let us consider the standard form of a quadratic equation, ax2 + bx + c =0

Discriminant = D = b2 – 4ac

• If the Discriminant > 0 then the roots are real and distinct.
• If the Discriminant = 0 then the roots are real and equal.
• If the Discriminant < 0 then the roots are Imaginary.

Calculation:

\(\rm \sqrt 2x^2 -x + \dfrac{1}{\sqrt{2}}=0\)

⇒ 2x2 - √2x + 1 = 0

Comparing this with the standard form ax2 + bx + c = 0, we get a = 2, b = -√2 and c = 1.

∴ D = b2 - 4ac

= (-√2)2 - 4 × 2 × 1

= 2 - 8 = - 6 

D < 0

Hence roots are imaginary. 

Given:

Ratio of Priya Capital = 5

Ratio of Mahi Capital = 13

Mahi's time period = 6 month

Ratio of Priya and Mahi profit = 25 : 26

Calculation:

Let Priya capital used for X month. Then,

⇒ (5 × X)/( 13 × 6) = 25/26

⇒ 26 × 5X = 13 × 6 × 25

⇒ X = 15months

∴ Priya Capital was used for 15 months.

Given∶

In year 2010 A invests Rs. 75000

In year 2011 A’s capital value is same for the whole year = Rs. 75000

In year 2012 A invests an additional amount of Rs. 25000 so total amount invested is Rs. 100000

In case of B; B joined in year 2011 with Rs. 45000

In year 2012 he also invests an additional amount of Rs. 25000 so total amount invested = is Rs. 70000

Formula used∶

(Profit)_A ∶ (Profit)_B

(share × Time)_A ∶ (share × Time)_B

Calculations∶      

(Profit)_A ∶ (Profit)_B

[75000 × 12 + 75000 × 12 + (75000 + 25000) × 12] ∶ [45000 × 12 + (25000 + 45000) ×12]

 (1800 + 1200) ∶ (540 + 840)

 50 ∶ 23

A’s share = 50 / (50 + 23) × 43800

Concept:

Let us consider the standard form of a quadratic equation, ax2 + bx + c =0

Discriminant = D = b2 – 4ac

• If the Discriminant > 0 then the roots are real and distinct.
• If the Discriminant = 0 then the roots are real and equal.
• If the Discriminant < 0 then the roots are Imaginary.

Calculation:

Given:

The roots of the quadratic equation x2 - kx + 4 = 0 are equal.

For equal roots, Discriminant = 0

⇒ D = b2 – 4ac = 0

⇒ (-k)² - 4 × 1 × 4 = 0

⇒ k² - 16 = 0

⇒ k2 = 16

∴ k = ± 4

Concept:

1, ω and ω² are cube root of unity, where, ω = \(\rm \frac{-1+\sqrt3i}{2}\)  and ω² = \(\rm \frac{-1-\sqrt3i}{2}\), 

• ω³ = 1
• 1 + ω + ω2 = 0

In quadratic equation, \(\rm ax^2+bx+c=0\), \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Consider a quadratic equation: ax² + bx + c = 0.

Let, α and β are the roots.

• Sum of roots = α + β = -b/a
• Product of the roots = α × β = c/a

Calculation:

Here, α and β are the roots of the equation x2 + x + 1 = 0

α  \(\rm ={-1 +\sqrt{1^2-4} \over 2}\)=\(\rm \frac{-1+\sqrt3i}{2}\) and β = \(\rm \frac{-1-\sqrt3i}{2}\)......(Using \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\))

α = ω and β =  ω2 

In equation x2 - x + 1 = 0 

Roots = \(\rm {1 \pm\sqrt{1^2-4} \over 2}\)\(=\rm \frac{1\pm\sqrt3i}{2}\)

So, roots are -β and -α, i.e., - ω2 and  -ω

Sum of roots in quadratic equation x2 - x + 1 = 0, is 1

Assume roots are α7 and β13  

α⁷ + β13  = ((-ω)⁷ + (- ω2 )¹³)

= -((ω³)² + (ω³)⁸ω²)

= - (ω + ω2 )

= -(-1)

= 1

Hence, option (1) is correct.

Formula used∶

Ratio of their profits - (share × Time)_A ∶ (share × Time)_B ∶ (share × Time)_C

Calculations∶

Let C invested Rs. x

A invested Rs. 2x

B invested Rs. 5x

A and B invested their capitals for 12 months

C invested his initial capital for first 7 months and then increases his capital by 20%

Ratio of their share -

(share × Time)_A ∶ (share × Time)_B ∶ (share × Time)_C

(2x × 12) ∶ (5x × 12) ∶ (x × 7 + 120% × x × 5)

24 ∶ 60 ∶ 13

Share of \(B = \frac{{60}}{{97}} \times 4850\)

Concept:

For an equation ax2 + bx +c = 0

• Sum of the roots = \(\rm-b\over a\)
• Product of the roots = \(\rm c\over a\)
• Roots of the equation = \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
• If α and β are the roots of the equation the equation can be represented as (x - α)(x - β) = 0

The roots of a quadratic equation ax2 +bx + c = 0 are real if:

b2 - 4ac ≥ 0

Calculation:

For a and b are the roots of 4x2 - 18x + 20 = 0 then, 

a + b = \(\rm 18\over 4\) = 4.5 

ab = \(\rm 20\over 4\) = 5

(a + b)² = a2 + b2 + 2ab

⇒ 4.5 × 4.5 = a2 + b2 + 2(5)

⇒ a2 + b2  = 20.25 - 10

⇒ a2 + b2  = 10.25

Now a³ + b³  = (a + b)(a² + b² - ab)

⇒ a3 + b3  = (4.5)(10.25 - 5)

⇒ a3 + b3  = (4.5) × (5.25)

⇒ a3 + b3   = 23.625

Concept:

Consider a quadratic equation: ax2 + bx + c = 0.

Let, α and β are the roots.

• Sum of roots = α + β = -b/a
• Product of the roots = α × β = c/a

Calculation:

Given quadratic equation: 4x2 + 8x – β = 0 and roots are -5α and 3

Now, sum of roots:

⇒ -5α + 3 = -(8)/4 = -2 

⇒ -5α = -5

⇒ α = 1

Now, Product of the roots:

⇒ (-5α)(3) = -β/4

⇒ 4 × (-5) × (3) = -β

⇒ β = 60

Hence, option (2) is correct. 

Calculation∶

MP = Rs. 23,000

Then, SP = MP - D% of MP

SP = 80% of MP = 80% of 23000

SP = 18,400

\(CP = \left( {\frac{{SP}}{{1 + g\%}}} \right)\)

⇒ Rs. (18400/115) × 100

⇒ Rs. 16,000 

∴ The cost price is Rs. 16,000

We know that quadratic equation ax² + bx + c = 0 has equal roots if D = b² – 4ac = 0. 

Therefore, given quadratic equation x² – 4kx + k = 0 has equal roots if D = 0. 

⇒(−4k)² – 4 × 1 × k = 0. (∵ In equation x² – 4kx + k = 0, a = 1, b = – 4k & c = k) 

⇒ 16k² – 4K = 0 

⇒ 4k (4k – 1) = 0

⇒ k = 0 or 4k – 1 = 0 

⇒ k = 0 or k = \(\frac{1}{4}\) .

Hence, for k = 1 and k = \(\frac{1}{4}\) , the quadric equation x² – 4kx + x =0 has equal roots.

How many multiples between 10 and 250. 60 

Given:

SP of the article = Rs. 546

Discounts are 20% and 25% respectively

Concept used:

Equivalent discount = (x + y – xy/100) %

MP = {100/(100 – value of discount%)}  × SP

Calculation:

Equivalent discount = (20 + 25 – 500/100)% = 40%

So marked price = MP = Rs. (100/60 × 546) = Rs. 910

∴ Marked price is Rs. 910

The probability of six on both if two dice will throw at once 1/36

Given:

⇒ Selling price of an item = Rs.4420

⇒ Profit percentage = 30%

Formula:

⇒ Selling price = Cost price × (profit percentage + 100)/100

Calculation:

⇒ 4420 = Cost price × (100 + 30)/100

⇒ Cost price = 3400

∴ The cost price of an item is Rs.3400.

Given:

Loss% = 5%

Gain% = 8%

Concept used:

SP = CP × (100 – loss%)/100

Calculation:

Let initial CP be Rs. x

Loss% = 5%

SP = x × (100 – 5)/100

⇒ SP = 95x/100

⇒ SP = 19x/20

If it is sold at Rs. 650 more,

SP = (19x/20) + 650

⇒ (19x/20) + 650 = 108x/100

⇒ (19x/20) + 650 = 27x/25

⇒ 650 = 27x/25 – 19x/20

⇒ x = 100 × 650/13

⇒ x = 5000

∴ The initial CP is Rs. 5000

Given:

Sp of an item is increased from Rs. 2252 to Rs. 2360

Concept:

Increase in profit is proportional to increase in selling price.

Formula:

y% of Cp = increase in Sp

Where,

y% = increase in profit %

Cp = cost price

Sp = sell price

Calculation:

Increase in sell price = Rs. (2360 – 2252) = Rs. 108

Let cost price be x

10% of x = 108

10/100 × x = 108

1/10 × x = 108

x = 108 × 10

x = 1080

∴ Cost price of the item is Rs. 1080

Given:

Discount = 15%

SP of a set of pens = Rs. 1751

Formula used:

Discount% = (Discount/Marked Price) × 100

Marked Price = Selling Price + Discount

Calculation:

Marked price of a set of pens

⇒ MP × (85/100) = 1751

⇒ MP = (1751 × 100)/85

⇒ MP = 2060

∴ The Marked price of a set of pens is Rs. 2060.

GIVEN:

An article passing through two sellers is sold at a profit of 50% at the original cost price. The first seller makes a profit of 35%.

CONCEPT:
Basic concept of Profit and Loss.

CALCULATION:

Let the actual cost price of the article be x.

Then, the final selling price of the article after passing through both the sellers

= (1 + 50/100)x = 1.5x

Now, we know the profit % made by the first seller is 35%.

Let the profit percent made by the second seller be y%.

So,

The selling price by the first seller to the second seller = (1 + 35/100)x = 1.35x

The selling price by the second seller = (1 + y/100) × (1.35x)

Hence,

(1 + y/100) × (1.35x) = 1.5x

⇒ 1 + y/100 = 150/135

⇒ y/100 = 15/135

⇒ y = 100/9

Given:

Cost Price (CP) of an item = Rs.840

Profit on selling the item by the shopkeeper = 10%

Loss on selling the item by the first buyer = 5%

Formulae Used:

When an item is sold at a Profit of x%, the Selling Price becomes:

SP = [(100 + x)/100] × CP

Similarly, when the item is sold at a Loss of y%, the Selling Price becomes:

SP = [(100 – y)/100] × CP

Calculation:

Selling price of the item when sold by the Shopkeeper is obtained as;

SP = [(100 + 10)/100] × 840

⇒ SP = 1.1 × 840 = Rs.924

This SP will be the CP for the first buyer.

So, when the first buyer sells the item at a loss of 5%, the final SP becomes:

SP = [(100 – 5)/100] × 924 

⇒ Final SP = 0.95 × 924 = Rs.877.80

∴ The final Selling Price of the item is Rs.877.80

Given:

Sujal’s capital = Rs. 65,000; Mandeep’s capital = Rs. 75,000

Profit = Rs. 56,000

Calculation:

Ratio of profit = 65,000 ⦂ 75,000

⇒ 13 : 15

Mandeep’s share of profit = (15/28) × 56,000

⇒ Rs. 30,000

Given:

S.P. reduces 20%

Initial gain = 15%

Final gain = 10%

Formula used:

Gain  = (gain%/100) × C.P.

S.P. = C.P. + Gain

Calculation:

Let C.P. = 100

⇒ Initial S.P. = 100 + (0.15 × 100)

⇒ Initial S.P. = 100 +15

⇒ Initial S.P. = 115

Now, calculate final S.P.

⇒ Final S.P. = 100 + (0.10 × 100)

⇒ Final S.P. = 100 + 10

⇒ Final S.P. = 110

Change in S.P = 115 -110

⇒ 5 units difference = 2000

⇒ 1 units = 400

Hence, 100 × 400 = 40,000

∴ Cost price of suit = Rs. 40,000

Given:

(i) Total profit = 50%

(ii) 1^st buyer’s profit = 20%

Calculations:

Let CP be 100

Then, SP is 150.

Let Profit made by 2^nd dealer be x%

Then, (100 + x)% of 120% of 100 = 150

⇒ (100 + x)/100 × 120/100 × 100 = 150

⇒ 6(100 + x) = 750

⇒ 600 + 6x = 750

⇒ 6x = 750 – 600

⇒ 6x = 150

⇒ X = 25%

∴ the profit of the 2^nd dealer is 25%.

Given:

(i) Mansi invested = Rs.2,20,000

(ii) Stuti invested = Rs.2,50,000

(iii) Stuti’s profit = Rs.8500

Calculations

Ratio of profits : ratio of investments

Mansi’s share : Stuti’s share

= 220000 : 250000 = 22 : 25

Let Mansi’s share be 22x

Let Stuti’s share be 25x

According to the question,

25x = 8500

⇒ x = 340

Total profit earned = (22x + 25x) = 47x

= 47 × 340

= 15980

∴ Total profit earned by both is Rs.15980.

Concept:

The solution to the quadratic equation Ax² + Bx + C = 0 can also be given by: \(\rm x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}\).

The quantity B² - 4AC is also called the discriminant.

• If B² - 4AC ≥ 0, the roots are real.
• If B² - 4AC = 0, the roots are real and equal.
• If B² - 4AC < 0, the roots will be complex and conjugates of each other.

The sum of both the roots of the quadratic equation Ax² + Bx + C = 0 is \(\rm -\frac{B}{A}\) and the product of the roots is \(\rm \frac{C}{A}\).

Calculation:

Using the expressions for the sum and the product of the roots, we have:

α + β = -p            ... (1)

αβ = \(\rm \frac{p^2}{2}\)            ... (2)

α⁴ + β⁴ = r            ... (3)

Squaring equation (1), we get:

α² + β² + 2αβ = p²

Using equation (2), we get:

⇒ α2 + β2 = 0

Squaring again, we get:

⇒ α4 + β4 + 2α2β2 = 0

Using equations (2) and (3), we get:

⇒ r = -\(\rm \frac{p^4}{2}\)            ... (4)

The discriminant of the equation 2x2 - 4p2x + 4p⁴ - 2r = 0 is:

(-4p2)² - 4(2)(4p⁴ - 2r)

= 16p⁴ - 32p4 + 16r

Using equation (4), we get:

= 16p4 - 32p4 - 8p4

= -24p⁴, which is always negative for non-zero real p.

Since the discriminant is < 0, the roots are imaginary.

Concept:

The solution to the quadratic equation Ax2 + Bx + C = 0 is given by:

\(x = {-B \pm \sqrt{B^2-4AC} \over 2A}\)

• If B² - 4AC ≥ 0, the roots are real.
• If B2 - 4AC = 0, the roots are real and equal.
• If B2 - 4AC < 0, the roots will be complex and conjugates of each other.
• The quantity B2 - 4AC is also called the determinant.

Calculation:

Comparing the given equation 2x2 + kx + 1 = 0 with the general equation Ax2 + Bx + C = 0, we can say that:

A = 2, B = k and C = 1.

We know that for the roots to be equal, the discriminant of the quadratic equation must be equal to 0.

∴ B2 - 4AC = 0

k2 - 4(2)(1) = 0

k2 = 8

k = ±√8

k = ±2√2.

• The sum of both the roots of the quadratic equation Ax2 + Bx + C = 0 is \(\rm -\frac{B}{A}\) and the product of the roots is \(\rm \frac{C}{A}\).
• The graph of a quadratic polynomial is a parabola.

Concept:

Consider a quadratic equation: ax² + bx + c = 0.

Let, α and β are the roots.

• Sum of roots = α + β = -b/a
• Product of the roots = α × β = c/a
• tan(A + B) = \(\rm \frac{tanA+tanB}{1-tanAtanB}\)

Calculation:

Here, tanα and tanβ  are the roots of the equation 2x2 + bx + c = 0

Sum of roots =  tan α + tan β = -b/2

Product of roots = c/2

tan (α + β) = \(\rm \frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta}\)

= \(\rm \frac{\frac{-b}{2}}{1-\frac c 2}\)

= \(\rm \frac{-b}{2-c}\)

= \(\rm \frac{b}{c-2}\)

cot (α + β) = \(\rm \frac{c-2}{b}\)

Hence, option (2) is correct.

Given:

(i) Reema’s Gain = 25%

(ii) Meena’s Loss = 15%

(iii) SP of Meena = Rs.8500

Calculations:

Let Reema purchased the earring for Rs. X

Required Equation:

⇒ X × 125/100 × 85/100 = 8500

⇒ X = 8500 × 100 × 100/125 × 85

⇒ X = 8000.

Reema purchased the earring for Rs.8000.

Concept:

For an equation ax2 + bx +c = 0

• Sum of the roots = \(\rm-b\over a\)
• Product of the roots = \(\rm c\over a\)
• Roots of the equation = \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
• If α and β are the roots of the equation the equation can be represented as (x - α)(x - β) = 0

The roots of a quadratic equation ax2 +bx + c = 0 are real if:

b2 - 4ac ≥ 0

Calculation:

2x2 + (10 - a)x + 25 = 0

∵ a is the factor of the equation

⇒ 2a² + (10 - a)a + 25 = 0

⇒ a² + 10a + 25 = 0

⇒ (a + 5)² = 0

⇒ a = - 5

Putting a in the equation

2x² + (10 - (-5)) x + 25 = 0

2x² + 15x + 25 = 0

(x + 5)(2x + 5) = 0

x = -5, -2.5

Given :

A man sold his furniture at 25% gain.

Had he sold at 15% loss he would have received Rs.800 less 

Concept used :

Profit = selling price - cost price = profit % on cost price 

Loss = cost price - selling price = loss % on cost price  

Calculations :

Let the cost price of the item be 100x 

Selling price at 25% profit = cost price + profit 

⇒ cost price + 25% of 100x 

⇒ 100x + 25x = 125x 

Selling price at 15% loss = cost price - loss

⇒ 100x - 15% of 100x 

According to the question 

125x - 85x = 800 

⇒ 40x = 800 

⇒ x = 20 

⇒ cost price = 100x 

⇒ 100 × 20 = 2000 rupees 

∴ Cost price of the furniture is 2000 rupees 

Given:

The given polynomials are (x² – 3x - 28) and (x² – 10x + 3a) and their HCF is (x - 7)

Concept Used:

Concept of HCF and factor theorem

According to factor theorem If (x - a) is factor of P(x) then P(a) = 0

Calculation:

Let the polynomials are P(x) = (x² – 3x - 28) and G(x) = x² – 10x + 3a) .If (x - 7) is the HCF of the given polynomial then (x - 7) is the factor of polynomials. Then according to the factor theorem

∴ G(7) = 0

Now, put x = 7 in G(x)

∴ 49 – 10 × 7 + 3a = 0

⇒ a = 7

Hence, option (1) is correct

Concept:

For an equation ax2 + bx +c = 0

• Sum of the roots = \(\rm-b\over a\)
• Product of the roots = \(\rm c\over a\)
• Roots of the equation = \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
• If α and β are the roots of the equation the equation can be represented as (x - α)(x - β) = 0

The roots of a quadratic equation ax2 +bx + c = 0 are real if:

b2 - 4ac ≥ 0

Calculation:

For a and b are the roots of x2 - 11x + y = 0 then, 

a + b = 11 and ab = y

⇒ b = 11 - a

Given 3a - 2b = 8

⇒ 3a - 2(11 - a) = 8

⇒ 5a - 22 = 8 

⇒ 5a = 30

⇒ a = 6

As b = 11 - a = 5

Now ab = y 

⇒ y = 6 × 5

⇒ y = 30

Given:

It is given that X = LCM of (5abc, 2a² and 3b³) and Y = HCF of (ab, 3bc and - bca)

Concept Used:

Basic concept of HCF and LCM

Calculation:

X = LCM of (5abc, 2a² and 3b³)

∴ LCM of (5abc, 2a² and 3b³) = 30 a²b³c

Now, Y = HCF of (ab, 3bc and -bca)

∴ HCF of (ab, 3bc and -bca) = b

Consider the first statement, Y is linear but X is not

∴ X is 30 a²b³c and Y is b

So, statement 1 is correct and statement 2 is incorrect

Now, Consider the third statement, the value of XY is 30 a²b⁴c

So, XY = 30 a²b³c × b

⇒ 30 a²b⁴c

So, statement 3 is correct

Hence, option (2) is correct

p(x) = x³+6x²+11x+18

Let q(x) = x²+2x+3

p(x)/q(x) \(=\frac{x^3+6x^2+11x+18}{x^2+2x+3}\)

= x + \(\frac{4x^2+8x+18}{x^2+2x+3}\)

= (x+4) + 6/x²+2x+3

∴ p(x) = (x+4) (x²+2x+3) + 6

⇒ p(x) - 6 = (x+4) (x²+2x+3)

Therefore, we must add -6 to p(x) so that the resulting polynomial is exactly divisible by q(x) = x²+2x+3.

Given :

A trader marks his goods 40 % above cost price 

He gives 25 % discount on marked price 

Calculations :

Let the cost price be Rs.100 

Marked price = 100 + 40 % of 100 = 100 + 40 

⇒ 140 rupees 

Now, 

He gives 25 % discount on this marked price 

Selling price = 140 - 25 % of 140 

⇒ 140 - 35 

⇒ Rs. 105 

Profit %  = (selling price - cost price)/cost price × 100 

⇒ (105 - 100)/100 × 100 

⇒ 5 % 

∴ Profit is 5 %