model 3 reducing & exceeding prices Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Reduced price of 6.2kg of sugar

= 10% of Rs.1116 = Rs.111.6

Reduced price per kg

= Rs.$(111.6/6.2)$ = Rs.18

Using Rule 31,

New price = ${10 × 1116}/{100 × 6.2}$

= $1116/62$ = Rs.18

Answer: (d)

Let the original price of sugar be Rs.x/kg.

New price = Rs.${9x}/10$ /kg

$270/{{9x}/10} - 270/x =1$

$300/x - 270/x = 1 ⇒ 30/x = 1$ ⇒ x = Rs.30 /kg

Using Rule 31,

New price = ${10 × 270}/{100 × 1}$ = Rs.27

Let the original price be Rs. x

$x - {10x}/100 = 27 ⇒ {90x}/100 = 27$

$x = 2700/90$ ⇒ x = 30

∴ Original price = Rs.30 per kg.

Answer: (d)

Let the original price of sugar = Rs.x /kg.

Reduced price of sugar = 80% of x

= ${x × 80}/100 = Rs.{4x}/5$kg

$36/{{4x}/5} - 36/x = 1/2$

$45/x - 36/x = 1/2$

$9/x = 1/2$ ⇒ x = 9 × 2 = Rs.18/Kg

Using Rule 31,

New price = ${20 × 36}/{100 × 500/100}$ = Rs.14.4

Let the original price be Rs x A.T.Q.

$x - {20x}/100$ = 14.4

${80x}/100$ = 14.4

$x = 144/8$ ⇒ x =18

∴ Original price = Rs.18

Answer: (c)

Let original price of rice per kg = Rs.x (let)

New price of rice per kg = Rs.${3x}/4$

$600/{{3x}/4} - 600/x = 10$

$600(4/{3x} - 1/x) = 10$

$600({4 - 3}/{3x})$ = 10

$600/{3x} = 10 ⇒ x = 600/30 = Rs.20$

New price = ${3x}/4 = {3 × 20}/4$ = Rs.15/kg

Method 2: Simple Approach,

If the price of an article is reduced by a% and buyer gets c kg more for some Rs.b, the new price per kg of article = ${ab}/{100 × c}$

= ${25 × 600}/{100 × 10}$ = Rs.15

Using Rule 31,

Reduced price per kg = ${25 × 600}/{100 × 10}$ = Rs.15

Answer: (a)

Original price of sugar = Rs. x/kg. (let)

New price = Rs.${120x}/100$ per kg.

= Rs.${6x}/5$ per kg.

According to the question,

$120/x - 120/{{6x}/5} = 4$

$120/x - {120 × 5}/{6x}$ = 4

$120/x - 100/x$ = 4

$30/x - 25/x$ = 1

$5/x$ = 1 ⇒ x = Rs.5 per kg.

Answer: (d)

Required percentage increase = $x/{100 - x} × 100$

= $(20/{100 - 20}) × 100 = 20/80 × 100 = 25%$