Search Previous Year Questions

Given:

The cost price of an article is 20% less than its marked price.

Discount% = 10%

Concept:

We assume total percentage will 100%.

Formula used:

S.P = M.P × [(100 – D%)/100]

P = S.P – C.P

P% = (P/C.P) × 100

Where, C.P = Cost Price, S.P = Selling Price, P = Profit, D = Discount, M.P = Marked Price

Calculation:

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

Then C.P = x × (100 – 20)% = x × 80% = 4x/5

After 20% discount,

S.P = M.P × [(100 – D%)/100]

⇒ x × [(100 – 10)/100]

⇒ x × (90/100)

⇒ 9x/10

Since S.P > C.P

P = S.P – C.P

⇒ (9x/10) – (4x/5)

⇒ (9x – 8x)/10

⇒ x/10

P% = (P/C.P) × 100

⇒ [(x/10)/(4x/5)] × 100

⇒ [(x/10) × (5/4x)] × 100

⇒ 1/8 × 100

⇒ 25/2

⇒ 12.5

∴ The Profit percentage is 12.5%

Alternate Method

Let M.P be 100.

C.P = 100 – 20 = 80

S.P = 100 – 10 = 90

P = S.P – C.P = 90 – 80 = 10

P% = (P/C.P) × 100 = (10/80) × 100 = 12.5

∴ The Profit percentage is 12.5%

According to the question,

⇒ The present age of Rohan = 8 + 10 = 18 yrs

⇒ The present age of Rajiv = 18 - 3 = 15 yrs

Similarly

The present age of  sister = 10 yrs

⇒ Present age of Father = 18 + 15 +10 = 43 yrs

Sum of Present age of Parents = 84 years

⇒ Present age of Mother = 84 - 43 = 41 yrs

∴ The required Difference = 43 - 41 = 2 yrs

We know that,

Area of an equilateral triangle $= {{\sqrt 3 } \over 4}{\left( {side} \right)^2}$

∵ Side of the triangle $= {{48} \over 3} = 16\ cm$

∴ Area of the triangle ​$= {{\sqrt 3 } \over 4} \times {\left( {16} \right)^2}\ c{m^2}$

$= {\rm{ }}64\sqrt 3\ c{m^2}$

Shortcut Trick

Initial milk in 80 L (ratio 2:3) = 80 × (2/5) = 32 L.

Mixture replaced = 25% of 80 L = 20 L.

Milk removed = 25% of 32 L = 8 L.

Milk added = 20 L (equal to the mixture removed).

Final Milk = 32 − 8 + 20 = 44 L.

∴ The correct answer is 44 litres.

Alternate Method

Given: Total Mixture = 80 L, Ratio (M:W) = 2:3.

⇒ Initial Milk = (2/5) × 80 = 32 L.

⇒ Initial Water = (3/5) × 80 = 48 L.

⇒ Replacement amount = 25% of 80 = 20 L.

⇒ Milk in 20 L mixture = (2/5) × 20 = 8 L.

⇒ Milk left after removal = 32 − 8 = 24 L.

⇒ Milk added = 20 L.

⇒ Final amount of Milk = 24 + 20 = 44 L.

∴ The correct answer is 44 litres.

Additional Information

Mixture Replacement Rule

When a part of a mixture is replaced by one of its components, the ratio of the components in the remaining part stays the same as the original ratio.

Formula for Remaining Quantity

Remaining amount of original liquid = Initial amount × (1 − x/V)ⁿ, where x is quantity replaced, V is total volume, and n is number of operations.

Alligation in Replacement

Replacement problems can also be solved by treating the added pure component as a 100% concentration mixture.

Let the total work = 360 units

Efficiency of A $= \frac{360}{60} = 6$ units/day

Efficiency of B $= \frac{360}{45} = 8$ units/day

Efficiency of C $= \frac{360}{30} = 12$ units/day

According to question

Amount of work done 

$= 6 \times 5 + (6 + 8) \times 7.5 + (6 + 8 + 12) \times 5\ = 30 + 105 + 130 = 265: units$

Remaining work = 360 – 265 = 95 unit

Required time $= \frac{(600\% \times 95)}{6} = 95: days$

The average price of per kg of rice sold by all the shopkeepers = (43.5 + 42 + 40 + 44 + 42.5)/5 = Rs. 42.4

The average price of per kg of jowar sold by all the shopkeepers = (38 + 39.5 + 37 + 36.5 + 39)/5 = Rs. 38

 ∴ The difference = 42.4 – 38 = Rs. 4.4

The total price (per kg) of all the products sold by C = (37.5 + 37 + 40 + 41.5 + 34 + 28) = 218

The total price (per kg) of all the products sold by E = (35 + 39 + 42.5 + 39 + 33.5 + 31) = 220

∴ Required ratio = 109 : 110

Total earning of shop keeper D = 15 × 36.5 + 18 × 41 + 31 × 30.5 = Rs. 2231

∴ Total earning of shop keeper D = Rs. 2231

Amount earned by shopkeeper B = 66 × 39.5 = Rs. 2607

Amount earned by shopkeeper D = 75 × 34 = Rs. 2550

∴ The difference = 2607 – 2550 = Rs. 57

Amount earned by shopkeeper A = 88 × 43.5 + 66 × 40 = Rs. 6468

Amount earned by shopkeeper B = 88 × 42 + 66 × 43 = Rs. 6534

Amount earned by shopkeeper C = 88 × 40 + 66 × 41.5 = Rs. 6259

Amount earned by shopkeeper D = 88 × 44 + 66 × 41 = Rs. 6578

Amount earned by shopkeeper E = 88 × 42.5 + 66 × 39 = Rs. 6314

∴ Shopkeeper D will earn the highest.