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:

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 ‘EXPONEMTS’ 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:

⇒ √7056 + 13 × 24 – 1157 ÷ 13 = ?

⇒ 84 + 13 × 24 – 1157 ÷ 13 = ?

⇒ 84 + 13 × 24 – 89 = ?

⇒ 84 + 312 – 89 = ?

⇒ 84 + 312 – 89 = ?

⇒ 396 – 89 = ?

⇒ 307 = ?

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

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

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

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

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

Given expression is,

⇒ {(81 + 17 × 2) ÷ 5} + 2 = (?)²

⇒ (?)² = {(81 + 34) ÷ 5} + 2

⇒ (?)² = (115 ÷ 5) + 2

⇒ (?)² = 23 + 2

⇒ (?)² = 25

⇒ (?)² = 5²

⇒ ? = 5

Total amount of calorie intake =Calories from Pie A + Calories from Pie B + Calories from Pie C

⇒ Total amount of calorie = (20.5)(0.25 × 3) + (30.5)(0.1) + (10)(0.25)

⇒ Total amount of calorie = (20.5)(0.75) + (30.5)(0.1) + (10)(0.25)

⇒ Total amount of calorie = 15.375 + 3.05 + 2.5 = 20.925

∴ The man had 20.925 calories in total.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, the BODMAS rule must be followed,

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

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

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

Given expression is,

2.75 of 3.25 – 1.75 + 2.5 - 3.25 = (?) × 5

⇒ 8.9375 – 1.75 + 2.5 - 3.25 = (?) × 5

⇒ 8.9375 + 2.5 – 1.75 - 3.25 = (?) × 5

⇒ 11.4375 – 1.75 - 3.25 = (?) × 5

⇒ 9.6875 - 3.25 = (?) × 5

⇒ 6.4375 = (?) × 5

⇒ ? = 6.4375 ÷ 5

⇒ ? = 1.2875

9³ = (3²)³ = 3^{2 × 3} = 3⁶

81² = (3⁴)² = 3^{4 × 2} = 3⁸

27³ = (3³)³ = 3^{3 × 3} = 3⁹

∵ [ (a^m)ⁿ = a^mn]

∴ 9³ × 81² ÷ 27³ = 3⁶ × 3⁸ ÷ 3⁹

= 3⁶ × 3^{(8 – 9)} = 3⁶ × 3^-1

∵ [a^m ÷ aⁿ = a^{m – n}]

= 3^{(6 – 1)} = 3⁵

∵ [a^m × aⁿ = a^(m+n)]

∴ 3⁵ = 3^? 

⇒ ? = 5

First of all, we use given quantities and place them in the formula, to know the required quantities.

Cost of 1 litre oil of first type = Cost price of dearer = d = Rs. 95

Cost of 1 litre oil of second type = Cost price of cheaper = c = Rs. 89

Desired cost of 1 litre of the mixture = Mean price = m = Rs. 92.50

Given, average marks in Hindi subject of a class of 54 students is 76.

∴ Total marks scored by 54 students = 54 × 76

⇒ Total marks scored by 54 students = 4104

Given, two marks 36 and 47 were misread as 60 and 77

∴ Actual total marks scored by 54 students = 4104 + 36 + 47 – 60 – 77

⇒ Actual total marks scored by 54 students = 4050

∴ Actual average marks = 4050/54 = 75

To avoid any wastage minimum length that must be purchased would be equal to the least common factor of 7 m and 1 m 70 cm

7 m = 700 cm

And 1 m 70 cm = 170 cm

LCM of 700 and 170 = 11900 cm

11900 cm = 119 m

∴ To avoid any wastage, he should purchase minimum length of 119 m.

CP = Cost Price & SP = Sell Price

Let the CP of each consignment be Rs. x.

For first consignment,

CP = Rs. x, Loss = 15%

MP = CP(1 − loss%) = x(1 − 0.15) = 0.85x

For second consignment,

CP = Rs. x, Profit = 35%

MP = CP(1 + profit%) = x(1 + 0.35) = 1.35x

For third consignment,

CP = Rs. x, Loss = 20%

MP = CP(1 − loss%) = x(1 − 0.20) = 0.80x

Total CP = Rs. 3x

Total MP = Rs. 3x

∴ Profit or loss = Rs. 0