Let the cost price of hydrochloric acid = Rs. 10 per litre

∴ Selling price of mixture = Rs. 10 per litre

Given,

Profit = 25%

∴ Cost Price of mixture = (10 × 100)/125 = 8

∴ Proportion of water and hydrochloric acid = 2 : 8 = 1 : 4

There are 5 letters in the word with M repeated 3 times.

So, total no. of different ways possible = 5!/3!

If two M are kept fixed at ends, there are 3 remaining letters which can be arranged in 3! Ways.

So, required probability = 3!/(5!/3!)

 = (3!)²/5!

Let, the bigger digit = x and smaller digit = y

According to problem,

⇒ x + y = 12       .… (1)

⇒ x – y = 4       .… (2)

From (1) and (2) we get, x = 8 and y = 4

∴ The product of the two digits of the  number = xy = 8 × 4 = 32

99.8/9.98 = 499/?

⇒ 10 = 499/?

⇒ ? = 499/10 = 49.9

= 31/9 + 41/9 - (5 × 29/9)

= 31/9 + 41/9 - 145/9

= (31 + 41 - 145)/9

= -73/9

Follow BODMAS rule to solve this question as per order 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 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:

9800 ÷ {(769 ÷ 13 + 19) of 23/9} = ?

⇒ 9800 ÷ {(769/13 + 19) of 23/9} = ?

⇒ 9800 ÷ {(769/13 + 19) of 23/9} = ?

⇒ 9800 ÷ {((769 + 247)/13) of 23/9} = ?

⇒ 9800 ÷ {(1016/13) of 23/9} = ?

⇒ 9800 ÷ {(1016/13) × 23/9} = ?

⇒ 9800 ÷ {(1016/13) × 23/9} = ?

⇒ 9800 ÷ {(23368/117)} = ?

⇒ 9800 × {(117/23368)} = ?

∴ ? = 114660/23368 = 49

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

The given expression is :

⇒ 87.60 + 99.81 + ? = 74.91% of 300

We can write the given values as :

⇒ 87.60 ≈ 88 and 99.81 ≈ 100

⇒ 74.91 ≈ 75

⇒ 88 + 100 + ? = 75% of 300

⇒ 88 + 100 + ? = (75/100) × 300

⇒ 88 + 100 + ? = 75 × 3

⇒ 88 + 100 + ? = 225

⇒ 188 + ? = 225

∴ ? ≈ 37

Given expression is,

⇒ 69.709 - 32.94 × 1.98 = 149.86 ÷ ?

We can write the given values as 

⇒ 69.709 ≈ 70 and 32.94 ≈ 33

⇒ 1.98 ≈ 2 and 149.86 ≈ 150

⇒ 70 - 33 × 2 = 150 ÷ ?

⇒ 70 - 66 = 150 ÷ ?

⇒ 4 = 150 ÷ ?

⇒ ? = 150/4

∴ ? ≈ 37

Let Rs. 1270 be divided into two parts x and y

Given,

We can write now,

(12% of x) + (7% of y) = 8% of (x + y)

⇒ 0.12x + 0.07y = 0.08x + 0.08y

⇒ 0.04x = 0.01y

⇒ x/y = ¼

So, x = ¹/₅ × 1270 = 254, y = ⁴/₅ × 1270 = 1016

∴ The required two parts of the amount = Rs. 254 and Rs. 1016.

We know that, total cost = price of raw materials +labor cost

Assume that, initial price of raw materials = R

Initial labor cost = 25% of the cost of raw material = 0.25 × R

∴total cost =R + 0.25 × R = 1.25 × R

Now, the cost of raw material increases by 15%.

Thus, new price of raw materials = R + (15% of R) = 1.15 × R

New labor cost = 30% of 1.15R = 0.3 × 1.15R = 0.345R

∴New total cost = new price of raw materials + new labor cost

⇒New total cost = 1.15R + 0.345R = 1.495R

Total increase in total cost = 0.245 × R