Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in the ‘BRACKETS’ must be solved first

Step - 2 - Any mathematical ‘OF’ or ‘EXPONENTS’ must be solved next

Step - 3 - Next the part of the equation that contains ‘DIVISION; and ‘MULTIPLICATION’ are calculated

Step - 4 - Last but not least, the parts of the equation that contains ‘ADDITION’ and ‘SUBTRACTION’ should be calculated

Now, the given expression:

⇒ 410 × ? × 26 = 67250 + 50010

⇒ 10660 × ? = 117260

⇒ ?=117260/10660=11

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in the ‘BRACKETS’ must be solved first

Step - 2 - Any mathematical ‘OF’ or ‘EXPONENTS’ must be solved next

Step - 3 - Next the part of the equation that contains ‘DIVISION; and ‘MULTIPLICATION’ are calculated

Step - 4 - Last but not least, the parts of the equation that contains ‘ADDITION’ and ‘SUBTRACTION’ should be calculated

Now, the given expression:

⇒ 207 ÷ 3 × 2.25 + 43.5 – 9 = ?

⇒ 69 × 2.25 + 43.5 – 9 = ?

⇒ 155.25 + 43.5 – 9 = ?

⇒ 198.75 – 9 = ?

⇒ 189.75 = ?

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in the ‘BRACKETS’ must be solved first

Step - 2 - Any mathematical ‘OF’ or ‘EXPONENTS’ must be solved next

Step - 3 - Next the part of the equation that contains ‘DIVISION; and ‘MULTIPLICATION’ are calculated

Step - 4 - Last but not least, the parts of the equation that contains ‘ADDITION’ and ‘SUBTRACTION’ should be calculated

Now, the given expression:

⇒ (25)³ × (4)³ – (800)² = ?²

⇒ 15625 × 64 – (800)² = ?²

⇒ 1000000 – (800)² = ?²

⇒ 1000000 – 640000 = ?²

⇒ 360000 = ?²

⇒ 600 = ?

{As we know that we are not required to find exact value so we can replace the given by there nearest places}

Given expression,

We can write,

⇒ 10.897 ≈ 11

⇒ 11.005 ≈ 11

⇒ 4.99 ≈ 5

Now,

⇒ ? ≈ √(11 × 11) × 5

⇒ ? ≈ 11 × 5 = 55

Replacing the given values by their nearest perfect values to make calculation easy, we get,

⇒ ? = 89 × 69 ÷ 26

⇒ ? = 236.19 ≈ 230

Given expression:

⇒ 17% of 2100 + ?% of 840 = 735

⇒ 357+x/100×840=735

⇒ 357 + 8.4x = 735

⇒ 8.4x = 735 – 357

⇒ 8.4x = 378

⇒ x=378/8.4

⇒ x = 45

Follow BODMAS rule to solve this question, as per the order is given below,

Step-1-Parts of an equation enclosed in 'Brackets' must be solved first,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3-Next, the parts of the equation that contains 'Division' and 'Multiplication' are calculated,

Step-4-Last but not least, the parts of the equation that contains 'Addition' and 'Subtraction' should be calculated.

Now the given expression:

⇒ ? ÷ 2 = 225.3001 × 27

⇒ ? ÷ 2 = 6083.1027

⇒ ? = 6083.1027 × 2

⇒ ? = 12166.2054

Given expression:

22² + √? = 516

⇒ √? = 516 – 484 = 32

∴ ? = (32)² = 32 × 32 = 1024

Given equation is:

⇒ 61² – 33² + (?)² = 3656

⇒ 3721 – 1089 + (?)² = 3656

⇒ (?)² = 3656 – 3721+1089 = 1024

⇒ ? = √(1024) = ± 32

As -32 is not in the option, so +32 is the answer.

We know that,

Arithmetic mean = (Sum of all terms)/(Number of terms)

According to the given information, the arithmetic mean given 12 numbers is 13.

∴ 13 = (2 + 9 + 13 + 17 + 11 + 5 + 12 + 18 + 14 + 8 + 15 + x)/12

⇒ 13 = (124 + x)/12

∴ 124 + x = 13 × 12

∴ x = 156 – 124 = 32

Let ‘a’ be the number of litres of milk to be added. Given the ratio of milk and water before and after mixing with milk as 3 : 2 and 4 : 1 respectively.

Equating the quantity of milk present in the ingredients to the final mixture we have,

0.6 × 40 + a = 0.8 × (40 + a)

⇒ 0.2 × a = 0.2 × 40

⇒ a = 40

∴ Quantity of milk to be added is 40 litres.

We have,

3x + 7 = 7x + 5.

⇒ 4x = 2

⇒ x = 1/2 = 0.5

We also know,

3x + 7 = x² + M

⇒ 3 × 0.5 + 7 – (0.5)² = M

⇒ M = 1.5 + 7 - 0.25 = 8.25

120% of M will be = (120/100) × 8.25 = 9.9

The pattern of the given number series is

⇒ 3 × 0² = 0

⇒ 3 × 1² = 3

⇒ 3 × 2² = 12

⇒ 3 × 3² = 27

⇒ 3 × 4² = 48

⇒ 3 × 5² = 75

The pattern of the series :(Prime No's. used)

64 = 2⁶

243 = 3⁵

625 = 5⁴

343 = 7³

121 = 11² 

13 = 13¹

The Pattern of the given series :

1²  + 1 = 2 

1³ + 1² = 2 

2² + 2 = 6

2³ + 2² = 12

3² + 3 = 12

3³ + 3² = 36

4² + 4 = 20 = ?

The pattern of the given number series is:

⇒ 2

⇒ 2 × 1 + 1 = 3

⇒ 3 × 2 + 2 = 8

⇒ 8 × 3 + 3 = 27 = ?

⇒ 27 × 4 + 4 = 112

⇒ 112 × 5 + 5 = 565

Area of circle = πr² (where π = 22/7)

Let the radius = r

New Radius = r + 1

Old area = πr²

New area = π(r+1)²  = π(r² + 1 + 2r)

New Area - Old area = 22

π(r² + 1 + 2r) - πr² = 22

⇒ π(2r + 1) = 22

⇒ 2r + 1 = 7

⇒ r = 3

The pattern of given series is :

⇒ 6 × 10 – 9 = 51

⇒ 51 × 8 – 7 = 401

⇒ 401 × 6 – 5 = 2401

⇒ 2401 × 4 – 3 = 9601

⇒ 9601 × 2 – 1 = 19201

⇒ 19201 × 0 – (-1) = 1 = ?

Let the first number be x and second number be y

If 25% of x is subtracted from y i.e. y−25/100x, then y becomes 5/6y

Concept: If M₁ Persons can do a work W₁ in H₁ hours and M₂ persons can do a work W₂ in H₂ hours then we have a general formula in this case.

If 12 furnaces consume 13 tonnes of coal in 612612 hours, then for 7 furnaces for consuming 14 tonnes

We have, M₁ = 12 furnaces

W₁= 13 tones

H₁ = 

M₂ = 7 furnaces

W₂ = 14 tones

Let H₂ = H hours

By putting all these values in the above general formula

From 2 : 20 pm to 3 pm only Train A travels.

∴ Only Train A travels for 40 min.

∴ In this 40 min. distance travelled by Train A = 90 × 40/60 km = 60 km [∵ Speed of Train A = 90 km/h]

∴ Remaining distance between Kolkata and Siliguri = 360 - 60 = 300 km

After 3pm Train A and Train B both travel.

Relative Speed between Train A and Train B = 60 + 90 = 150 km/h [∵ Train A and Train B travel towards opposite direction]

Let, T hours needed before both the trains meet.

∴ 150 × T = 300

⇒ T = 2

∴ Both the trains will meet at (3 + 2) = 5 pm.

Let the distance travelled be x km

Let the speed of current be v km/hr

Let the speed of boat in still water be b km/hr

∴ Speed of boat upstream = (b – v) km/hr

⇒ Speed of boat downstream = (b + v) km/hr

∴ Time taken to travel x distance upstream = x/(b – v) hr

⇒ Time taken to travel x distance downstream = x/(b + v) hr

⇒ b + v = 3(b – v)

⇒ b + v = 3b – 3v

⇒ 4v = 2b

⇒ b/v = 2 : 1

∴ Ratio of speed in still water to speed of current is 2 : 1

Let the amount = P

∵ He invested 1/3 part of capital at 7%,

∴ S.I. on P/3 part at 7% =

∵ He invested 2/5 part at 10%,

∴ S.I. on 2P/5 part at 10% =

∴ Remaining part = P – P/3 – 2P/5

∵ Remaining part, he invested at 12%,,

∴ S.I. on (P - P/3 – 2P/5) at 12% =,

⇒ 143P/750 = 1430

⇒ P = 7500

Let the cost price of the goods to be Rs. 100

He had initially marked his goods up by 40%.

∴ After40% mark-up,

⇒ MP = Rs. 100 + 40% of Rs. 100

⇒ Rs.100 + Rs. 40 = Rs. 140

So,He offers a discount of Rs. 40 on his MP of Rs. 140

∴ The % discount offered by him = (Discount)/(Marked Price) × 100

⇒ (40/140) × 100 = 28.5%

∴ The % discount offered by him is 28.5%

Let the cost price be Rs. 100

Gain required = 15% 

∴ Selling price = Rs. 115 

Let the marked price be Rs. x 

Then, discount = 20% of Rs. x

⇒ Rs. (x × 20/100) = Rs. x/5

∴ selling price = (Marked Price) - (discount) 

⇒ Rs. {x - (x/5)} = Rs. 4x/5

∴ 4x/5 = 115 

⇒ x = {115 × (5/4)} = 143.75 ≈ 144 

∴ Marked price = Rs. 144 

Hence, the marked price is 44% above cost price.

Let the digit in the ten’s place be ‘x’.

Digit in unit’s place = (x + 2)

∴ Required number = 10(x) + x + 2 = 11x + 2

Now,

Sum of digits = 4/19 × (11x + 2)

⇒ x + x + 2 = 4/19 × (11x + 2)

⇒ 2x + 2 = 4/19 × (11x + 2)

⇒ 38x + 38 = 44x + 8

⇒ 44x – 38x = 38 – 8

⇒ 6x = 30

⇒ x = 30/6 = 5

Hence, the number = 11(5) + 2 = 55 + 2 = 57

Let the unit digit of number be x and tenth digit be y

Then,

Number = 10y + x

Number obtained by reversing its digits = 10x + y

Number obtained by interchanging the digits of a two digit number is less than the original number by 54

∴ (10y + x) – (10x + y) = 54

⇒ 9y – 9x = 54

⇒ y – x = 6      ----eq (1)

Sum of the digits of the number = 8

∴ y + x = 8      ----eq (2)

By adding eq (1) and eq (2), we get

2y = 14

⇒ y = 7

By putting this value of y in eq(2), we get

7 + x = 8

⇒ x = 1

∴ Original two digit number = (10 × 7) + 1 = 71

According to question -

⇒ The no. of bottle sizes possible = Factors of HCF (170, 102, 374)

⇒ HCF (170, 102, 374) = 34

⇒ Factors of 34 = 1, 2, 17, 34

∴ Total no. of bottle sizes possible = 4

On dividing 28614 by 89, we get

⇒ 89 × 321 + 45 = 28614

The required number to be added = 89 - 45 = 44

We can also verify as:

⇒ 28614 + 44 = 28658 = 89 × 322

∴ 44 should be added to 28614 to make it exactly divisible by 89