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:

⇒7/19 of 1064 = (?)² × 2

⇒ 7/19 × 1064 = (?)² × 2

⇒ 392 = (?)² × 2

⇒ 392 ÷ 2 = (?)²

⇒ 196 = (?)²

⇒ 14 = ?

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:

⇒ 25³ × 4³ × √810000 = ?²

⇒ (5²)³ × (2²)³ × 900 = (?)²

⇒ 5⁶ × 2⁶ × 30² = (?)

⇒ 5³ × 2³ × 30 = ?

⇒ 125 × 8 × 30 = ?

⇒ 30000 = ?

(269.88 ÷ 29.11 × 512 ÷ 31.75)^0.5

Approximating values to the closest integers and solving

= (270 ÷ 30 × 512 ÷ 32)^0.5

⇒ (9 × 16)^0.5 = 3 × 4

= 12

Original quantity of milk in the container = 40 litres

At first, the quantity of liquid removed from the container = 4 litres

Quantity of milk left in the container after first procedure = 40 – 4 = 36 litres

Let us try to write this quantity in terms of 40 litres to make it easier to do bigger calculations.

= 29.16 litres

∴ Milk now present in the container is 29.16 litres

Let the three consecutive odd numbers be x, x + 2, x + 4

We know,

It is given that Sum of three consecutive numbers = Average + 38

⇒ x + (x + 2) + (x + 4) = (x + 2) + 38

⇒ 2x + 4 = 38

⇒ 2x = 34

⇒ x = 17

⇒ x + 2 = 19

∴ The second number is 19

Let the cost price of an article be Rs. x.

For 48% profit, sell price = cost price (1 + profit %) = Rs. x (1 + 0.48) = Rs. 1.48x

But this sell price is obtained after 60% discount on Marked Price.

Sell Price = Marked Price (1 − Discount %)

⇒ 1.48x = Marked Price (1 − 0.60)

⇒ Marked Price = 1.48x / 0.4 = 3.7x

∴ The object should be marked at 270% higher price.

Cost of 1 kg pulses of 1st quality = Cost price of cheaper = c = Rs. 15

Cost of 1 kg pulses of 2nd quality = Cost price of dearer = d = Rs. 20

Cost of 1 kg of the mixture = Mean price = m = Rs. 16.50

By formula,

∴ The ratio of cheaper to dearer or 1st quality ∶ 2nd quality is 7 ∶ 3

Number of days Kamal takes to complete a work = 20 days

∴ In 1 day Kamal can do = 1/20 part of work

Given: Bimal is 20% more efficient than Kamal

∴ In 1 day Bimal can do = 1.2/20 part of work

Given: Suresh is 25% more efficient than Bimal

Let ‘l’,’b’ be the length and breadth of the rectangular plot as shown in the figure.

Area of 1 metre broad pathway = Area of outer rectangle – Area of inner rectangle

⇒ Area of 1 metre broad pathway = lb – (l – 2)(b – 2)

⇒ Area of 1 metre broad pathway = lb – lb + 2l + 2b – 4

Area of 1 metre broad pathway = 2(l + b) – 4

∵ We don’t know the values of l and b, the given data is inadequate.

Let the speed of the slower train be ‘x’ km/h, then,

Speed of the faster train = ‘x + 7’ km/h.

The trains are moving in opposite directions, so the relative speed = x + (x + 7) km/h

And, time taken = Distance/Speed

⇒ 2x + 7 = 138/6

⇒ 2x + 7 = 23

⇒ x = 16/2 = 8 km/h

∴ Speed of the slower train = 8 km/h.