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Given:

Selling price of car = Rs. 400

Loss % = 20%

Formula used:

CP = SP × 100 / (100 + P%)

SP = CP × (100 + P %) / 100

Where P = Profit

Calculations:

At 20% loss

⇒ SP = Rs. 400

⇒ CP × 80% = 400 

⇒ CP = Rs. 500

For 30% profit

⇒ SP = CP × 130%

⇒ SP = 500 × 130/100 = Rs. 650

∴ The selling price of car will be Rs. 650.

Given:

Old profit = 30%

New profit = 45%

Formula used:

\(\;Profit\% \; = \;\frac{{SP\; - \;CP}}{{CP}}\; \times 100\)

Calculation:

Let the original cost price of the article be X.

Old selling price at 30% profit will be = \(\frac{{130}}{{100}}\;x\)

⇒ 1.3x

both cost price and selling price of the article are decreased by Rs.100

∴ New CP = x -100 and new SP = 1.3x - 100

\(\;Profit\% \; = \;\frac{{SP\; - \;CP}}{{CP}}\; \times 100\)

\(45\; = \;\frac{{\left( {1.3x\; - \;100} \right)\; - \;\left( {x\; - \;100} \right)}}{{x\; - \;100}}\; \times \;100\;\)

⇒ 0.45x - 45 = 1.3x - 100 - x + 100

⇒ 0.45x - 0.3x = 45

⇒ 0.15x = 45

x = 300

c The original cost price is 300

Given:

List price of an article = Rs. 8500 and two discount offered first one is 12% and second one we have to find. Also the selling price = Rs. 6358

Calculation:

List price = Rs. 8,500 and Discount = 12% (12% = -3/25 = 22/25)

⇒ After first discount price came down by = (8500) × 22/25 = 85 × 88 = Rs. 7480

⇒ As given selling price = Rs. 6,358

⇒ Discount given = Rs. (7480 - 6358) = Rs. 1122

⇒ Discount is on the given List price = (1122/7480) × 100 = 15%.

Given:

Selling Price = Rs. 1350

Calculation:

Let the cost price of one ball be x

Now selling 26 balls for Rs. 1,350, there is a loss equal to the cost price of eight balls

Cost price of 26 balls = 26x, loss = 8x(8 balls)

Loss = Cost Price – Selling Price

⇒ 8x = 26x – 1350

⇒ 18x = 1350

⇒ x = 75 

∴ The cost price of a ball is Rs.75

Given:

MP = Rs. 150

Discount% = 20%

Formula used:

SP = MP - Discount

Discount = MP × Discount%

Calculation:

Discount = MP × Discount%

⇒ Discount = 150 × 20/100

⇒ Discount = 30

SP = MP - Discount

⇒ SP = 150 - 30

⇒ SP = Rs. 120

∴ The selling price is Rs. 120.

Given:

Initial Selling Price = Rs.400

Final Selling Price = Rs.350

Formula Used:

Loss = CP – SP

Calculation;

Let the Cost Price of the table be Rs.x

Let the initial loss be y

So, the final loss = [(100 + 5)/100] × y = 1.05y

Hence, we get:

y = x – 400                  ----(i)

⇒ 1.05y = x – 350      ----(ii)

On combining the equations (i) and (ii), we get:

1.05 × (x – 400) = x – 350

⇒ 1.05x – 420 = x – 350  

⇒ x = 1400

∴ The cost price of table is Rs.1400

Given:

profit = 300/47%

Formula Used:

Profit% = Error weight/Actual weight given by shopkeeper × 100

Calculations:

Let error = x gms

Then,

⇒ x/(1000 – x) × 100 = 300/47

⇒ 100x/1000 – x = 300/47

⇒ 47x = 3 (1000 – x)

⇒ 47x = 3000 – 3x

⇒ 50x = 3000

⇒ x = 3000/50

⇒ x = 60

∴ Weight used = (1000 – 60) = 940 gms.

Given:

Purchase price = Rs.18,000 

Transportation charges = Rs.2,000

Selling price = Rs. 25,000

Formula Used:

Profit = Selling price – Cost price

Profit% = (Profit/ Cost price) × 100

Calculation

⇒ Cost price = Purchase price + Transportation charges

⇒ Cost price = 18,000 + 2,000 = Rs.20,000

⇒ Selling price = Rs. 25,000

Cost price < Selling price

Profit = Selling price – Cost price

⇒ Profit = Rs.25,000 – Rs. 20,000

⇒ Profit = Rs. 5,000

Profit% = (Profit/ Cost price) × 100

⇒ Profit% = (5,000/ 20,000) × 100

⇒ Profit%  = 25%

∴ Manish gains 25% in this transaction.                           

The correct option is 1 i.e. Profit, 25%

Given:

False weight = 680 gram

Formula used:

When false weight is used, then

Gain% = [error/(true value – error)] × 100

Where error is the difference between the true weight and false weight.

Calculation:

Error = 1000 gram – 680 gram

⇒ 320 gram

Gain% = [320/(1000 – 320)] × 100

⇒ [320/680] × 100

⇒ \(47\frac{1}{{17}}\)%

Given:

Cost price of 15 mangoes = Selling price of 10 mangoes

Formula Used:

Profit percentage = {(S.P – C.P)/C.P} × 100

Calculation:

Let the C.P. of each mango is Rs. 1

C.P of 15 mangoes = Rs. 15

S.P. of 10 mangoes = Rs. 15

C.P of 10 mangoes = Rs. 10

Profit = Rs. (15 – 10) = Rs. 5

Profit percentage = (5 × 100)/10 = 50%

Given:

CP of 15 apples = Rs. 2

SP of 5 apples = Rs. 3

Formula used:

Gain% = (Gain/CP) × 100%

Calculation:

SP of 15 apples = Rs. (3/5) × 15

⇒ Rs. 9

Gain = Rs. (9 – 2)

⇒ Rs. 7

Gain% = (7/2) × 100%

⇒ 350%

∴ The gain% is 350%

Given:

SP of 33 mangoes = CP of 44 mangoes

Formula used:

P = S.P. – C.P.

L = C.P. – S.P.

P% = (P/C.P.) × 100

L% = (L/C.P.) × 100

Where,

P → Profit 

L → Loss

SP → Selling price

CP → Cost price

Calculations:

Let the CP of 1 mangoe be Rs.y and its SP be x.

So the SP of 33 mangoes = 33x

And the CP of 44 mangoes = 44y

According to the question,

33x = 44y

⇒ x ∶ y = 4 ∶ 3

Now let x and y be 4k and 3k respectively.

Since x > y

So P = 4k – 3k = k

P% = (P/CP) × 100

⇒ P% = (k/3k) × 100 = \(33{{1} \over 3}\)%

∴ The required Profit% is \(33{{1} \over 3}\)%.

Given equations is 2x² – 3x + 1 = 0

⇒ 2x² – 2x – x + 1 = 0

⇒ 2x(x – 1) – 1(x – 1) = 0

⇒ (2x – 1)(x – 1) = 0

⇒ 2x – 1 = 0 or x – 1 = 0

⇒ x \(=\frac{1}{2}\) or x = 1

Hence, roots of given quadratic equation is 1 and \(\frac{1}{2}\).

But given that \(\alpha\) and \(\beta\) are roots of given equation.

Let \(\alpha \) = 1 and \(\beta=\frac{1}{2}\)

Then \(\frac{\alpha}{2\beta}+3\) \(=\frac{1}{2\times \frac{1}{2}}+3\) = 1 + 3 = 4

And \(\frac{\beta}{2\alpha}+3\) \(=\frac{\frac{1}{2}}{2\times 1}+3\) \(=\frac{1}{4}+3\) \(=\frac{13}{4}\)

The equations whose roots are 4 & \(\frac{13}{4}\) is 

(x – 4)(x – \(\frac{13}{4}\)) = 0

⇒ x² – 4x – \(\frac{13}{4}\mathrm x\) + \(4\times \frac{13}{4}\) = 0

⇒ x² – \(\frac{29}{4}\mathrm x\) + 13 = 0

Hence, the required equation is x² – \(\frac{29}{4}\mathrm x\) + 13 = 0.

Given:

The Sumit paid = Rs. 67200

Formula used:

\(GM\% =\ {GM\over TM}× 100\)

(Where GM = Gain marks and TM = Total marks)

Calculation:

Let us assume Mohan bought a car at price P

⇒ Shyam purchase the car from Mohan at = P × (120/100) = 6P/5

⇒ Sumit purchase the car from Sumit at = (6P/5) × (70/100) = 84P/100

⇒ 84P/100 = 67200

⇒ P = \({67200\ \times\ 100 \over 84 }\) = 80000

∴ The required result will be 80000.

Concept:

Sum of elements along principle diagonal = Trace = ∑ Eigen values

Product of eigen values = determinant = det (A)

Calculation:

Given:

\(A =\left[ {\begin{array}{*{20}{c}} {1}&{2}&{-1}\\ {3}&{5}&{2}\\ {1}&{k}&{2} \end{array}} \right]\)

We know that;

Det (A) = product of eigen values

1 × (10 - 2 × k) - 2 × (6 -2) - (3 × k - 5) = -8

10 - 2k - 8 - 3k + 5 = -8

7 - 5k = -8

15 = 5k

k = 3

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

1. If λ is an Eigenvalue of a matrix A, then λn will be an Eigenvalue of a matrix An.
2. If λ is an Eigenvalue of a matrix A, then kλ will be an Eigenvalue of a matrix kA where k is a scalar.
3. Sum of Eigenvalues is equal to the trace of that matrix.
4. The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A.
5. If λ is an Eigenvalue of matrix A, then λ2 will be an Eigenvalue of matrix A2.
6. If λ1 is an Eigenvalue of matrix A, then (λ1 + 1) will be an Eigenvalue of the matrix (A + I).
7. Eigenvalues of a matrix and its transpose are the same because the transpose matrix will also have the same characteristic equation.

Calculation:

Given:

\(A = \left[ {\begin{array}{*{20}{c}} 2&3\\ x&y \end{array}} \right]\)

Sum of eigen values = Trace (A) = 2 + y

Product of eigen values = |A| = 2y – 3x

∴ 4 + 8 = 2 + y     … i)

4 × 8 = 2y – 3x      … ii)

∴ 2 + y = 12     … iii)

 2y – 3x = 32     … iv)

∴ Solving i) and ii) we get x = -4 and y = 10

Given:

A horse and a cow was sold for Rs. 12,000 each

Horse sold at a loss of 20%

Cow sold at a gain of 20%

Concept used:

Loss = CP - SP

Calculation:

Let the cost price of the horse be H and that of the cow be C.

Horse sold at a loss of 20%

⇒ H × (80/100) = 12,000 

⇒ H = 15,000

The cow sold at a gain of 20%

⇒ C × (120/100) = 12,000

⇒ C = 10,000

Total CP = H + C = Rs. (15,000 + 10,000) = Rs. 25,000

Total SP = Rs. (2 × 12,000) = Rs. 24,000

Net loss = Total CP - Total SP = Rs. (25,000 - 24,000) = Rs 1000

∴ Net loss of Rs 1000 in the entire transaction.

Concept:

If λ is an eigenvalue of a matrix A and k  is a scalar then:

1) λ^m is the eigenvalue of matrix A^m (m belongs to N).

2) kλ  is an eigenvalue of matrix kA.

3) λ + k is an eigenvalue of the matrix A + kI.

4) λ - k  is an eigenvalue of matrix A - kI.

Calculation:

Given:

3, 5 are the eigenvalues then,

Using Property 1;

A³ = λ³ = 3³ = 27.

A³ = λ³ = 5³ = 125.

Correct option is B) less

The series follows the following pattern

⇒ 362 - 1 = 361 = 19²

⇒ 762 - 362 = 400 = 20²

⇒ 1203 - 762 = 441 = 21²

⇒ 1687 - 1203 = 484 = 22²

∴ The missing number is 1687.

Given-

When x is added to each of 9, 15, 21 and 31, the numbers so obtained are in proportion.

Concept Used-

If a, b, c and d are in proportion, then a/b = c/d

Mean proportion of two numbers a and b is √ab

Calculation- 

According to Question-

(9 + x)/(15 + x) = (21 + x)/(31 + x)

⇒ (9 + x)(31 + x) = (15 + x)(21 + x)

⇒ x² + 40x + 279 = x² + 36x + 315

⇒ 4x = 36

⇒ x = 9

Mean proportion of (3x - 2) and (5x + 4) = √{(3×9 - 2)(5×9 + 4)}

⇒ √(25 × 49)

⇒ 35 

∴ Mean proportional between the numbers (3x - 2) and (5x + 4) is 35.

Given :- 

An amount of Rs. 800 is distributed between Ravi, Mohan and Govind in the proportions 2 : 5 : 3,

Concept :- 

Ratio represent amount must be divided in that value.

Calculation :-

Let ratio be in x so,

⇒ Ravi share's = 2x

⇒ Mohan share's = 5x 

⇒ Govind share's = 3x

Now,

⇒ Total amount = 2x + 5x + 3x

⇒ 800 = 10x 

⇒ x = (800/10)

⇒ x = 80 

Now, Put the value of x 

⇒ Ravi share's = 2 × 80 = Rs. 160

⇒ Mohan's share's = 5 × 80 = Rs. 400

⇒ Govind share's = 3 × 80 = Rs. 240

⇒ Shares of Mohan and Govind sum = 400 + 240 = Rs. 640 

∴ Shares of Mohan and Govind sum is Rs. 640

Concept used :-

Trick to multiply two numbers ending in 5 with difference of 20 between them.

Method :-

• Write 125 at the end.
• Add 1 to ten’s place of the bigger number.
• Now multiply the increased digit with the ten’s digit of the smaller number.
• Write the product so obtained on the left side of 125

Calculation :-

Let the product be A125

A = (9 + 1) × 7 = 70 (step 2 and 3)

_ 1 2 5

7 0 _ _

Concept used :-

Trick to multiply two digit numbers ending in 1.

Method :-

• Multiply the left most digits of both the numbers.
• Add the left most digits of both the numbers.
• Put the result from Step(2) next to the result of Step(1)
• Put 1 next to the result of Step(3)

Calculation :-

Let the product be AB1

A = 2 × 7 = 14

B = 2 + 7 = 9

AB1 = 1491

We see mountains, rivers, valleys, ocean, also terrestrial animals and plants.

frequency * wavelength = speed

wavelength = 3*10⁸/ 98.8*10⁶

                   = 3.03 m

Correct option is (2) 191.24 J K^-1

As we know

\(\Delta S = nC_V\, ln\frac{T_2}{T_1} + nR\,ln\frac{V_2}{V_1}\)    .....(1)

∵ Expansion occurs, reversibly and isothermally.

∴ The above equation (1) becomes-

∴ \(\Delta S = nR\, ln \frac{V_2}{V_1}\)    .....(2)

Putting the values of n, R, V₂ and V₁ in equation (2) - 

\(\Delta S = 10\times 8.314 \times ln \frac{100}{10}\)

\(= 2.303 \times 10\times 8.314 \times log 10\)

\(= 191.47 JK^{-1}mol^{-1}\)

As you sight at the portion of the pencil that was submerged in the water, light travels from water to air (or from water to glass to air). This light ray changes medium and subsequently undergoes refraction. As a result, the image of the pencil appears to be broken.

(9q + 6p)/(3q - 5p)

⇒ (9 + 6p/q)/(3 - 5p/q)

⇒ (9 + 6 × 1/5)/(3 - 5 × 1/5)

⇒ [(45 + 6)/5)/[(15 - 5/5)]

⇒ (51/5)/2

⇒ 51/10

Let the two digit number is yx = x + 10y. 

(∵ Any two digit number ab is written as ab = 10a + b) 

Given that the sum of digits of the number is 12. 

∴ x + y = 12 … (1) 

The number obtained by interchanging its digits is xy = y + 10x. 

Given that, the number obtained by interchanging its digits exceeds the given number by 18. 

i.e. y + 10x = x + 10y +18 

⇒ 9x – 9y = 18 … (2) 

Now multiplying equation (1) by 9, we get 

9x + 9y = 108 … (3) 

Now, adding equation (2) and (3), we get 

(9x – 9y) + (9x + 9y) = 18 + 108. 

⇒18x = 126 

⇒ x = \(\frac{126}{18}\) = 7. 

Now putting x = 7 in equation (1), we get 7 + y =12 ⇒ y = 12 – 7 = 5. 

Hence, the number is x + 10y = 7 + 10 × 5 = 7 + 50 = 57. 

Hence, the number is 57.

• Machines and tools made of iron are used widely.
• The amount of iron used in a country determines its standard of living.

• Singhbhum – Jharkhand 
• Sundargarh – Odisha 

• Magnetite 
• Hematite 
• Limonite 
• Siderite

Option : (b).Bauxite

• Minerals with metallic content are called metallic minerals. Eg : Iron ore. 
• Minerals without metallic content are called non metallic minerals. Eg : Mica.

• Temperature : High temperature above 24°C. 
• Rainfall : Annual rainfall above 150 cm. 
• Soil : Alluvial soil.