Successive Discounts
Concept 1: Basic Formulae
If there are two discounts:
If the first discount is x% and second discount is y%, then
Total discount = (x + y − $\frac{xy}{100}$)%
SP = MP × (1−$\frac{x}{100}$) (1−$\frac{y}{100}$)
If there are three discounts x%, y% and z%, then:
SP = MP × (1−$\frac{x}{100}$) (1−$\frac{y}{100}$) (1−$\frac{z}{100}$)
Q. What is the single discount equivalent to a series of discounts of 10%, 20% and 50%?
Explanations :
Let MP = Rs. 100
So, $SP_1$ = 100 – 10% of 100 = Rs. 90
$SP_2$ = 90 – 20% of 90 = 90 – 18 = Rs. 72
$SP_3$ = 72 – 50% of 72 = 72 – 36 = Rs. 36
So, discount percentage = [(100 – 36)/100] × 100 = 64%
Let MP = Rs. 100
If there are three discounts x%, y% and z%, then:
SP = MP × (1−$\frac{x}{100}$) (1−$\frac{y}{100}$) (1−$\frac{z}{100}$) = 100 × (90/100) × (80/100) × (50/100) = 100 × (9/10) × (4/5) × (1/2) = Rs. 36
So, discount percentage = [(100 – 36)/100] × 100 = 64%
Q. A shopkeeper gives successive discounts of 20%, 10% and 5% on an item. What will be the selling price of the item if its cost price is Rs. 800?
(a) Rs. 523.6
(b) Rs. 547.2
(c) Rs. 576
(d) Rs. 640.5
Explanation:
Selling price after first discount = 800 - 800 × (20/100) = 800 – 160 = Rs. 640
Selling price after second discount = 640 × (90/100) = Rs. 576
Selling price after third discount = 576 × (95/100) = Rs. 547.2
Answer: (b)
Q. A trader allows two successive discounts of 20% and 10%. If he sells the article for Rs. 108, then the marked price of the article is:
(a) Rs. 120
(b) Rs. 135
(c) Rs. 140
(d) Rs. 150
Explanation:
Net discount = -20 - 10 + (200/100) = 28%
SP = MP × [(100 - Discount percentage)/100]
or 108 = MP × [(100 - 28)/100]
or 108 = MP × (72/100)
or MP = 150
Answer: (d)
Concept 2
In case of Successive Discounts, the discount is lower if the two discount values are closer.
Let’s see how.
We already know that, Net discount percentage = (x + y − $\frac{xy}{100}$)%
When x + y = constant, then the value of xy is the largest when x = y. The closer the values of x and y will be the higher will be the value of xy, and so lower will be the value of net discount percentage.
E.g. if x + y = 20, then maximum value of xy = 10 × 10 = 100 (i.e. when x = y = 10)
So, Net discount percentage = 10 + 10 – 1 = 19% (minimum possible value)
If x = 1, and y = 19, then Net discount percentage = 1 + 19 – 0.19 = 19.81%
Q. To maximize his profit on his product, should a shopkeeper provide two discounts of 40% and 30% or 50% and 20%?
Explanations :
Let MP be Rs. 100
Case 1:
After 40% discount, SP = Rs. 60
And after 30% discount, SP = Rs. 42
Case 2:
After 50% discount, SP = Rs. 50
And after 20% discount, SP = Rs. 40
So, the shopkeeper should opt for discounts of 40% and 30%. This way the net discount offered by him will be less.
Case 1:
40% discount = 2/5; So, SP1/MP = 3/5
30% discount = 3/10; So, SP2/SP1 = 7/10
So, SP2/MP = (3/5) × (7/10) = 21/50 = 42/100
So, net discount = 58%
Case 2:
50% discount = 1/2; So, SP1/MP = 1/2
20% discount = 1/5; So, SP2/SP1 = 4/5
So, SP2/MP = (1/2) × (4/5) = 4/10 = 40/100
So, net discount = 60%
So, the shopkeeper should opt for discounts of 40% and 30%. This way the net discount offered by him will be less.
Case 1:
Net discount after 40% and 30% discount = (x + y − $\frac{xy}{100}$)% = (40 + 30 − $\frac{(40)(30)}{100}$)% = 70 – 12 = 58%
Case 2:
Net discount after 50% and 20% discount = (x + y − $\frac{xy}{100}$)% = (50 + 20 − $\frac{(50)(20)}{100}$)% = 70 – 10 = 60%
So, the shopkeeper should opt for discounts of 40% and 30%. This way the net discount offered by him will be less.