model 3 reducing & exceeding prices Practice Questions Answers Test with Solutions & More Shortcuts

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)

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