Solution: (b)

Reduced price => Rp/100y per kg

R = Rs 270 , p=10% ,y=5kg

{(270×10)/(100×5)} = Rs. 5.4

Original prices per kg= Rp/(100-p)y per kg

{(270 x 10)/ (100 – 10)5} = 2700/450 = Rs. 6

Difference will be = 6-5.4= Rs 0.6 = 60 paise

Alternate Method:

Let the original price of sugar be Rs. p.

Then, reduced price of sugar

{p-(10p/100)} = 90p/100 = Rs. 9p/10

Expenditure on any quantity = price × total consumed qty

Total Consumed Quantity = Expenditure on any quantity/price

Now, according to the question,

{270/(9p/10)- (270/p)} = 5

=> {(270 x 10)/9p} – (270/p) = 5

=> {(300/p) – (270/p)} = 5

=> {(300 – 270)/p} = 5

=> p = 30/5 = 6

∴ Original price of sugar = Rs. 6

Reduced price of sugar

(9 x p)/ 10 = (9 x 6)/10 = Rs. 5.40

∴ Required difference = (6 –5.40)= Rs. 0.60 or 60 paise

Original prices per kg= Rp/(100-p)y per kg

{(1 x 10)/(100 – 10)0.05} = Rs. 20/9 /kg

It means the man can buy 1 kg or 1000 gm in Rs 20/9 or he can buy

(1000 x 9)/20 = 450g in one rupee.

Alternate Method:

Original price of wheat = Rx x/kg

New price = Rs. 9p/10 per kg

Expenditure on any quantity = price per qty × total consumed qty

Total Consumed Quantity = (Expenditure on any quantity)/ (price per quantity)

According to the question,

{1/(9p / 10)} – 1/p = 50/1000

{(10/9p) – (1/p)} = 50/1000

(10/9p) – (1/p) = 1/20

9p = 20

p=20/9

Original Price = Rs. 20/9 per kg

=> Rs. 20/9 = 1000g

=> Rs. 1 = (1000 x 9)/25 = 450 g

Solution: (b)

In 1 kg of ore, the iron = (90/100) x (25/100) x 1000 = 225 gms

Hence for 225 gm iron, the ore required = 1 kg

So, to get 60 kg of iron, the ore required = (1/0.225) x 60 = 266.67 kg

Solution: (b)

Sugar in original solution = (75 x 30) / 100 = 22.5 g

Let x g of sugar be mixed

According to the question

{(22.5 + p)/(75 + p)} x 100 = 70

⇒ 2250 + 100p = 75 × 70 + 70p

⇒ 2250 + 100p = 5250 + 70p

⇒ 30p = 5250 – 2250= 3000

=> p = 3000/30 = 100g