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

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: (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: (b)

10 percent of Rs.837

= $10/100 × 837$ = Rs.83.7

Reduced per kg price

= $83.7/6.2$ = Rs.13.50

Reduced price per kg

= ${10 × 837}/{100 × 6.2}$ = Rs.13.50

Answer: (c)

Required percentage change

= $(10 - 20 + {10 × (-20)}/100)$%

= –12% Negative sign shows decrease.

Required percentage = $(10 - 20 - {10 × 20}/100)$

= (–10 – 2) = 12% decrease.

Answer: (a)

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

Reduced price = Rs. ${80x}/100 = Rs.{4x}/5$ per kg.

According to the question,

$160/{{4x}/5} - 160/x$ = 8

${40 × 5}/x - 160/x = 8$

$200/x - 160/x = 8$

$40/x = 8 ⇒ 8x = 40 ⇒ x = 40/8 = 5$ per kg.

Reduced Price = ${4x}/5 = {4 × 5}/5$ = Rs. 4 per kg

Using Rule 31,

Reduced price per kg. = ${21 × 160}/{100 × 8}$ = Rs. 4

Answer: (b)

If the number be x, then

x – 15 = ${4x}/5$

5x – 75 = 4x ⇒ x = 75

40% of 75 = ${75 × 40}/100 = 30$