model 3 successive discount Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 4 EXERCISES
The following question based on discount topic of quantitative aptitude
(a) 20%
(b) 15%
(c) 10%
(d) 12%
The correct answers to the above question in:
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%
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Read more problems on successive discount Based Quantitative Aptitude Questions and Answers
Question : 1
The difference between a discount of 40% on Rs.500 and two successive discounts of 36% and 4% on the same amount is
a) Rs.7.20
b) Rs.2.00
c) zero
d) Rs.1.93
Answer »Answer: (a)
Using Rule 5,
Single equivalent discount of two successive discounts of 36% and 4%
= 36 + 4 - ${36 × 4}/100$
= 40 - 1.44 = 38.56
Percentage difference
= 40 - 38.56 = 1.44
Required difference
= 500 × ${1.44}/100$ = Rs.7.20
Question : 2
A bicycle, marked at Rs.2,000, is sold with two successive discount of 20% and 10%. An additional discount of 5% is offered for cash payment. The selling price of the bicycle at cash payment is
a) Rs.1,668
b) Rs.1,568
c) Rs.1,368
d) Rs.1,468
Answer »Answer: (c)
Using Rule 5,
Single equivalent discount for two successive discounts of 20% and 10%
= $(20 + 10 - {20 × 10}/100)%$ = 28%
Now , single discount for 28% and 5%
= $(28 + 5 - {28 × 5}/100)%$
= (33 - 1.4) % = 31.6%
Required selling price of bicycle at cash payment
= (100 - 31.6) % of Rs.2000
= ${2000 × 68.4}/100$ = Rs.1368
Question : 3
A dealer buys a car listed at Rs.200000 at successive discounts of 5% and 10%. If he sells the car for 179550, then his profit is
a) 4%
b) 5%
c) 10%
d) 9%
Answer »Answer: (b)
Equivalent discount
= $10 + 5 - {10 × 5}/100 = 14.5%$
CP (for buyer)
= 85.5% of Rs.200000
= Rs.$({85.5 × 200000}/100)$ = Rs.171000
SP = Rs.179550
Gain = Rs.(179550 –171000) = Rs.8550
Gain % = $8550/171000 × 100 =5%$
Using Rule 3,
Here, M.P. = 200000,
S.P. is C.P. byer for $D_1 = 5%, D_2$ = 10%
S.P.= M.P.$({100 - D_1}/100)({100 - D_2}/100)$
= 200000$({100 - 5}/100)({100 - 10}/100)$
= 20 × 95 × 90
C.P. for buyer =171000
S.P. = 179550
Profit =S.P. - $\text"C.P."/\text"C.P" ×100%$
= $8550/171000$ × 100 = 5%
Question : 4
A single discount equivalent to the successive discounts of 10%, 20% and 25% is
a) 60%
b) 46%
c) 55%
d) 45%
Answer »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%
Using Rule 4,If $D_1, D_2, D_3$ are successive discounts, then equivalent discount/overall discount is (in percentage)100 - $[({100 - D_1}/100)({100 - D_2}/100)({100 - D_3}/100) × 100]$
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%
Question : 5
The marked price of a watch is Rs.1000. A retailer buys it at Rs.810 after getting two successive discounts of 10% and another rate which is illegible. What is the second discount rate?
a) 6.5%
b) 8%
c) 15%
d) 10%
Answer »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%
Question : 6
Successive discounts of 10% and 20% are equivalent to a single discount of :
a) 12%
b) 28%
c) 30%
d) 15%
Answer »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%
GET discount PRACTICE TEST EXERCISES
model 1 profit x after discount y
model 2 hiking & discounting
model 3 successive discount
model 4 mixed discount problems of marked price
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