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

Answer: (b)

Let 'x' be the marked price Single Discount = 15%

100 - 15 = 85

85% of x = 17,000

$x = {17,000}/85 × 100$ = Rs.20,000

Required SP

= $20,000 × 95/100 × 90/100$

= 180 × 95 = Rs.17100

M.P. = ${S.P. × 100}/{100 - D}$

= ${17000 × 100}/{100 - 15}$

= ${17000 × 100}/85$

M.P. = 20000

Also,

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

= 20000$({100 - 5}/100)({100 - 10}/100)$

= $20000 × 95/100 × 90/100$

= 180 × 95 = Rs.17100

Answer: (d)

Using Rule 5,

Let the original price be Rs.100

Increased price = Rs.130

Equivalent discount

= $(10 + 10 - {10 × 10}/100)$ = 19%

Ultimate price of the article = 81% of 130 = 105.3

i.e. increase by 5.3%.

Answer: (c)

A single discount equal to the two successive discounts

= $(10 + 5 - {10 × 5}/100)%$ = 14.5%

Selling price of the article

= 85.5% of Rs.240

= Rs.${85.5 × 240}/100$ = Rs.205.20

Using Rule 3,

Here, M.P. = Rs. 240, $D_1 = 10%, D_2$ = 5%

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

= $240({100 - 10}/100)({100 - 5}/100)$

= $240 × 90/100 × 95/100$

= Rs.205.20

Answer: (b)

Marked price = Rs.160

After 10% discount

S.P = $90/100 × 160$ = Rs.144

Let other discount = x%

${(100 - x)}/100 × 144$ = Rs.122.40

100 - x = $12240/144$

100 - x = 85

x = 100 - 85 = 15%

Using Rule 3,

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

122.40 = 160$({100 - 10}/100)({100 - D_2}/100)$

${1224000}/160 = 90 × ({100 - D_2}/1)$

$1224000/{160 × 90} = 100 - D_2$

$85 = 100 - D_2 ⇒ D_2$ = 15%

Answer: (d)

Using Rule 5,

Single equivalent discount for successive discounts of 10% and 20%.

= $(10 + 20 - {20 × 100}/100)$% = 28%

Single equivalent discount for 28% and 30%.

= $(28 + 30 - {28 × 30}/100)$% = 49.6%