Quant Questions – Profit and Loss Problems Set-1

Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.
1).If A saves Rs 220 by getting a discount of 8.33% then find the amount paid by him?
a)  Rs 2000
b)  Rs 2200
c)  Rs 2420
d)  Rs 2500
2).Three friends A, B& C went to the market to buy a saree at 3 different stores. A went to Kala Mandhir which was offering two successive discounts of 10% and 40%, B went to Kala Sangam, which was offering two successive discounts of 20% and 30%, C went Kala Niketan, which was offering two successive discounts of 24% and 26%. Out of the three friends, who got the least discount, if the Marked Price of all the three sarees, was the same?
a)  A
b)  B
c)  C
d)  B&C
3).A readymade garment store marks up his merchandise in such a way that a profit of 16% is earned even after giving a discount of 6% to the customer. If a customer is wrongly given a discount of 10% instead of 6%, then what is the profit % in this car ?
a)  15.67%
b)  12.76%
c)  14.76%
d)  Cannot be determined
4).A dishonest trader gives only 900 gm for 1 kg. Although he claims to be selling at a profit of x% above cost Price, his actual Profit (Owing to faulty measure) is 30%. What is the value of x?
a)  15
b)  17
c)  21
d)  14
5).A sold a cycle to B at a profit of 16.67%, B sold it to C, at a loss of 7.14%. C spent 5% of his cost price on spare parts and further sold it to A at a Profit of 10%. Find the loss incurred by A (in Rs).
a)  50.75
b)  150.75
c)  100.5
d)  120.75
6).Two friends A& B invest in two different schemes of XYZ mutual funds in the ratio of 3 : 2. The ratio of % return on a& B’s investment is 4 : 3. What are the combined earning of A& B taken together if A earns more than B by Rs 300?
a)  Rs 1200
b)  Rs 900
c)  Rs 500
d)  Rs 600
7).Kajol had the exact amount of money required to buy a packet of sweets. However, after bargaining she got a discount of 10% from which she bought a chocolate for Rani worth Rs 8. Find the amount of money paid by Kajol for the packet of sweets.
a)  Rs 80
b)  Rs 70
c)  Rs 75
d)  None of these
8).”True Value” a shop for second hand cars bought a cars for Rs 1,00,000 each. They spent a total of Rs 1,25,000 on repair of all the three cars and then sold them to 3 different customers. If one of the cars was sold for Rs.1,45,000 and it is known that a total profit of 40%? Was made on all the 3 cars, then find price at which the other 2 cars were sold if both the remaining cars were sold for same price.
a)  Rs 2,25,000
b)  Rs 1,45,000
c)  Rs 3,05,000
d)  Rs 1,55,000
9).A dishonest trader cheats while buying from wholesaler and also while selling to a customer. While he takes 1200gm per kg from wholesaler, he gives 900gm per kg to the customer. Also it is known that his mark up % and discount % is 60% and 10% respectively. Find his actual profit.
a)  72%
b)  85%
c)  92%
d)  100%
10).When a shopkeeper sells 20 pencils, he makes a profit equivalent to selling price of 6 erasers. However, when he sells 20 erasers, he incurs a loss equivalent to selling price of 8 pencils. Also it is known that % profit is twice % of loss and ratio of cost price of pencil : Eraser is 3 : 2. Find the ratio of selling price of pencil to that of Eraser.
a)  3 : 2
b)  4 : 3
c)  4 : 1
d)  None of these
Answers:
1). c)  2). c)  3). c)  4). b)  5). a)  6). b)  7). d)  8). a)  9). c)  10). d)
Detailed Explanation:
 
1.We Know 8.33% = (1 /12), so, (1/12) MP = 220
MP = 220 × 12 = 2640 and S.P =2640 – 220 = Rs 2420
Answer: (c)
 
2.Let  M.P of each of the saree = Rs 100
A paid 100 × (90 / 100) × (60 / 100) = Rs 54
B paid 100 ×(80 / 100) × (70 / 100) = Rs 56
C paid 100 × (76 / 100) × (74 / 100) = Rs 56.2
C paid the most and hence got the least discount.
Answer: (c)
 
3.To find the mark up %, use [ a + b +(ab/100)]
Formula,
16 = a – 6 +[(a)(-6) /100]
22= [a – (6a / 100)] ➾ 22 = [94/100] a  and a=(2200 / 94) = 23.4 %
Let C. P = 100, MP = 123.4.
S.P =123.4 × (90 / 100) = 114.76
Profit % = 14. 76%
Answer: (c)
 
4.Let  CP/gm = Rs.1
Actual CP is for 900gm = Rs 900
CP for 1 kg = Rs 1000, above which he claims to be selling at a profit of x%.
SP for 1 kg = Rs 1000 +[x / 100 (1000)]= 1000 + 10x
Actual Profit % = [SP – Actual CP  / Actual CP ] × 100
30 =[ [(1000 + 10x) – 900] / 900 ] × 100
30 × 9 = 100 + 10x ➾x = 17%
Answer: (b)
 
5.16.67 % = (1 / 6) , 7.14 % = (1 / 14)
Let CP of A = 600, then he sold to B for Rs 700, B sold it to C for Rs 650, amount spent by C = [(5 / 100) × 650] = 32.5
CP for C = 650 + 32.5 = 682.5 and he sells to A at profit of 10%
i.e., SP =682.5 × (110 /100) = 750.75
loss incurred by A = 750.75 – 700 = Rs. 50.75
Answer: (a)
 
6.Let investment of A& B = 3x and 2x& their percentage return = 4y% and 3y% then
[(4y / 100) × 3x] – 300 = (3y /100) × 2x
(12 / 100) xy – 300 = (6 / 100)xy
(6 / 100)xy = 300. Their combined earnings is (18 / 100)xy which is equal toRs 900.
Answer: (b)
 
7.10% of MP = 8, MP of sweets packet = 80, and amount she paid = 80 – 8 = Rs.72.
Answer: (d)
 
8.Total C.P on 3 cars = 1,00,000 × 3 + 1,25,000 = 4,25,000.
Total profit = (40/100)×425000=170000
Total S.P = 595000 out of which one of the car is sold for RS.145000
S.P for remaining two = 4,50,000
Price per car = (4,50,000/ 2)= Rs 2,25,000
Answer: (a)
 
9.Actual CP  → Let  C.P of trader be Rs 1 / gram, he pays Rs 1000 for 1200 gm, hence his actual C.P.= (10 / 12) per gm, for 900 gm = (10 / 12) × 900
Actual SP   → [He is charging according to 1000 gm]
C.P of 1000gm = Rs 1000, MP = 16000
S.P = 1600 – 10% = 1440
Actual Profit % = {[1440 – (9000 / 12) ] / [9000 / 12]} × 100
= (8280 / 12) × (12 / 9000) × 100
=92%
Answer: (c)
 
10.Ratio is 1: 4
Pencil         Eraser
S.P. / Unit                            X                 a
C.P./  Unit                           Y                  b
When 20 pencils are sold → 20 x – 20 y = 6 a
i.e., 20 (x – y) = 6a
When 20 erasers are sold → 20 b – 20 a= 8x
i.e., 20 (b – a) =  8x
Also % Profit = (6a / 20y)× 100 and % loss = (8x / 20b) × 100
Now (6a / 20y) = 2 (8x/ 20b)à 3a / y = 8x / b ……(i)
Also y : b = 3 : 2à (y / b) = (3 /2) or 2y = 3b, Putting in (i)
We get, 3a / (3/2)b = 8x / bà 2a = 8x and (a / x)  = (8/ 2)
i.e.,  x : a = 1:4
Answer: (d)
 
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