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Concept used:

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

Calculation:

Our equation 55 – [13 + {12 ÷ (8 – 2)}] × 10 ÷ 3

⇒ 55 – [13 + {12 ÷ 6}] × 10 ÷ 3

⇒ 55 – [13 + 2] × 10 ÷ 3

⇒ 55 – 15 × 10 ÷ 3

⇒ 55 – 15 × 10 × 1/3

⇒ 55 – 50 = 5

∴ The correct answer is 5

Calculation:

Value of the total stock = 25 × 15000 + 80 × 16000 + 54 × 55000 + 41 × 27000 + 64 × 24000

375000 + 1280000 + 2970000 + 1107000 + 1536000

7268000 = 72.68 lakh

∴ Option 3 is the correct answer.

Given:

Present population of village = 28000

Rate at which population increases = 5%

Time period = 2 years

Formula used:

\({P_n} = P{\left( {1 + \frac{R}{{100}}} \right)^n}\)

Where, P_n = Population after n years

P = Present population

R = Rate of growth of population

n = time period (in years)

Calculations:

Population after 2 years,

\( \Rightarrow {P_2} = 28000{\left( {1 + \frac{5}{{100}}} \right)^2}\)

\( \Rightarrow {P_2} = 28000{\left( {\frac{{21}}{{20}}} \right)^2}\)

\( \Rightarrow {P_2} = 28000\left( {\frac{{441}}{{400}}} \right)\)

⇒ P₂ = 70 × 441

⇒ P₂ = 30870

∴ The population after 2 years will be 30870

Given:

If 24 is subtracted from a number then it reduces by 40%.

Calculation:

Let the number be x

On subtracting 24 the number reduces by 40%

⇒ 40% of x = 24

⇒ x = (24 × 100)/40

⇒ x = 60

∴ required number is 60.

Tip: We can find the answer quickly by going through the options.

ALTERNATE METHOD:

Let the number be x

If is substracted from a number then it reduces by 40%

⇒ x - 24 = 60x/100

⇒ x - 24 = 3x/5

⇒ 5x - 120 = 3x

⇒ 2x = 120

⇒ x = 60

∴ required number is 60.

Shortcut Trick

Identify the highest powers of prime factors across all numbers.

Numbers: 45 = 9 × 5 = 3² × 5¹

25 = 5²

35 = 5 × 7

LCM = 3² × 5² × 7 = 9 × 25 × 7 = 1575.

∴ The correct answer is 1575.

Alternate Method

Given: Three numbers 45, 25, and 35.

Formula Used: LCM (Least Common Multiple) is the smallest number which is exactly divisible by each of the given numbers.

⇒ Prime factorization of 45 = 3 × 3 × 5 = 3² × 5

⇒ Prime factorization of 25 = 5 × 5 = 5²

⇒ Prime factorization of 35 = 5 × 7

⇒ LCM = 3² × 5² × 7

⇒ LCM = 9 × 25 × 7

⇒ LCM = 225 × 7 = 1575

∴ The correct answer is 1575.

Additional Information

HCF and LCM Relation

For two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a, b) × LCM(a, b).

LCM of Fractions

LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators).

HCF of Fractions

HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators).

Given:

P can finish the work in 15 days and Q can finish the work in 20 days.

Concept used:

If time taken finish the work = x days

∴ Part of work completed in one day = 1/x

Calculation:

P can finish the work in 15 days.

∴ One day work of P = 1/15

 And, Q can finish the work in 20 days.

∴ One day work of Q = 1/20

Let P and Q together complete the work in x days.

∴ One day work of P and Q = 1/20 + 1/15 = 1/x

⇒ 1/x = 7/60

So, 5 days work of P and Q will be = 5 × 7/60 = 7/12

Now, remaining work = 1 – 7/12 = 5/12

∴ Number of days required by Q to finish the work = (5/12)/(1/20) = \(8\frac{1}{3}\) days

Given:

Marker Price (MP) = Rs.1280

Profit % = 12

Discount % = 30

Formulas used:

Discount % = Discount/MP × 100

Profit % = Profit/Cost Price × 100

Selling Price = Marked Price - Discount

Profit = Selling Price - Cost Price

Cost Price = (Selling price × 100)/(100 + profit)

Calculation:

Discount % = Discount/1280 × 100 

⇒ 30 = Discount/1280 × 100

⇒ Discount = 128 × 3 = Rs.384

Selling Price = 1280 - 384 =  Rs.896

Cost Price = (Selling price × 100)/(100 + profit)

⇒ (896 × 100)/(100 + 12)

∴ CP = 89600/112 = Rs.800 

Alternate Method

Cost Price = MP × (100 - D)/100 × 100/(100 + P)

⇒ Rs.1280 × 70/100 × 100/112 = Rs.800

Shortcut Trick

For two pipes P and Q filling a tank in x and y hours respectively, time taken together = (x × y) ÷ (x + y).

Time taken = (7 × 13) ÷ (7 + 13) = 91 ÷ 20 hours.

⇒ 91 ÷ 20 = 4.55 hours = 4 hours + (0.55 × 60) minutes.

⇒ 4 hours 33 minutes.

∴ The correct answer is 4 hr 33 min.

Alternate Method

Given: Time for Pipe P = 7 hours, Time for Pipe Q = 13 hours.

Formula Used: Total Work = LCM of individual times; Efficiency = Total Work ÷ Time.

⇒ Combined Efficiency of P + Q = 13 + 7 = 20 units/hour.

⇒ Total Time taken = Total Work ÷ Combined Efficiency = 91 ÷ 20 hours.

⇒ 91 ÷ 20 = 4 (11/20) hours.

⇒ Convert fraction to minutes: (11 ÷ 20) × 60 = 33 minutes.

⇒ Total Time = 4 hours 33 minutes.

∴ The correct answer is 4 hr 33 min.

Additional Information

Inlet and Outlet Pipes

Inlet pipes have positive efficiency (fill tank), while outlet/leakage pipes have negative efficiency (empty tank).

Net Work Done

If pipe A fills in x hours and pipe B empties in y hours, net work in 1 hour = (1/x) − (1/y).

Alternate Day/Hour Work

If pipes operate alternately, calculate the combined work for one full cycle (e.g., 2 hours) before finding total time.

Spend on food = 10%

⇒ Spend on fuel = (100 – 10) × 20% = 18%

⇒ Remaining salary of man = 100% – (10 + 18)% = 72%

⇒ 72% → Rs. 4680

⇒ 100% → Rs. 6500

∴ His salary is Rs. 6500.

Shortcut Trick

When a solid object is melted to form another, the total volume remains conserved.

⇒ Volume of Sphere = Volume of Cone

⇒ (4/3) × π × (3)³ = (1/3) × π × r² × (r / 2)

⇒ 4 × 27 = r³ / 2

⇒ r³ = 216 ⇒ r = 6 cm

∴ The correct answer is 6 cm.

Alternate Method

Given:

Radius of the sphere (R) = 3 cm

Let the radius of the cone be r cm.

Height of the cone (h) = r / 2

Formula Used:

Volume of Sphere = (4/3) × π × R³

Volume of Cone = (1/3) × π × r² × h

Calculations:

⇒ Volume of Cone = Volume of Sphere

⇒ (1/3) × π × r² × h = (4/3) × π × R³

Substituting the given values into the equation:

⇒ (1/3) × π × r² × (r / 2) = (4/3) × π × 3³

Canceling out (π / 3) from both sides of the equation:

⇒ r³ / 2 = 4 × 27

⇒ r³ / 2 = 108

⇒ r³ = 216

⇒ r = (216)^1/3

⇒ r = 6 cm

∴ The correct answer is 6 cm.

Additional Information

Principle of Melting and Recasting

Whenever a 3D solid figure is melted and converted into another solid, the initial volume is exactly equal to the final volume regardless of the shape.

Total Surface Area of Cone

The total surface area of a cone involves the curved area plus the circular base area: TSA = π × r × l + π × r², where l is the slant height.

Surface Area of Sphere

A complete sphere has a continuous curved surface with an area given by the formula: SA = 4 × π × R².