model 3 successive discount Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Equivalent discount for successive discounts of 20% and 10%

= $[20 + 10 - {20 × 10}/100]%$ = 28%

Net selling price = 72% of 2000

= Rs.${72 × 2000}/100$ = Rs.1440

Using Rule 3,

Here, M.P. = Rs.2000, $D_1 = 20%, D_2$ = 10%

S.P. = M.P.$[({100 - D_1}/100)({100 - D_2}/100)]$

= M.P.$[2000 × ({100 - 20}/100) × ({100 - 10}/100)]$

= $2000 × {80 × 90}/10000$ = Rs.1440

Answer: (b)

Using Rule 5,

Successive discounts of x% and y%

= $(x + y - {x + y}/100)%$

Required discount

= $(20 + 10 - {20 × 10}/100)$%

= 30 - 2 = 28%

Answer: (d)

Price after 10% first discount

= $1000 × {100 - 10}/100$

=$1000 × 90/100$ = Rs.900

Given :

Price after second discount = Rs.810

Second discount

= 900 - 810 = Rs.90

Percentage of second discount

= ${90 × 100}/900$ = 10%

Answer: (b)

Single of discount for successive discounts 10% and 20%

= $(20 + 10 - {20 × 10}/100)$% = 28%

Equivalent discount for discounts 28% and 25%

= $(28 + 25 - {28 × 25}/100)$%

= 53 - 7 = 46%

Single equivalent discount

= 100 - $[({100 - D_1}/100)({100 - D_2}/100)({100 - D_3}/100) × 100]$

= 100 - $[({100 - 10}/100)({100 - 20}/100)({100 - 25}/100) × 100]$

= $100 - 90/100 × 80/100 × 75/100 × 100$

= 100 - 54 = 46%

Answer: (b)

Let the second discount be x%.

Then, 90 % of (100 - x) % of 800 = 612

$90/100 × {100 - x}/100 × 800 = 612$

100 - x = ${612 × 100}/{90 × 8}$ = 85

x = 100 - 85 = 15%

Using Rule 3,

Here, M.P. = Rs.800, S.P. = Rs.612, $D_1 = 10%, D_2$ = ?

S.P. = M.P.$({100 - D_1}/100)({100 - D_2}/100)$

612 = 800 × $({100 - 10}/100) × ({100 - D_2}/100)$

612 = $800 × 90/100 × {100 - D_2}/100$

$6120/72 = 100 - D_2$

$D_2 = {100 - 6120}/72$

= ${7200 - 6120}/72$ = 15%