## Quants Profit & Loss A: 2

 Question 1
```An electronic machine was sold for Rs. 2500. Had it been sold for
Rs. 4000 there would have been an additional profit of 15%.
What will be the cost price of the electronic machine?```
 A Rs. 9000 B Rs. 10000 C Rs. 15000 D Rs. 25000
Question 1 Explanation:
Let the cost price be Rs. x Original Selling Price = Rs. 2500 Original Profit = (2500 – x) Rs. Profit % = (2500 – x)*100/CP Now, new Selling Price = Rs. 4000 Original CP = Rs. x New profit = (4000 – x) Rs. New profit % = (4000 – x)*100/CP (4000 – x)*100/CP – (2500 – x)*100/CP = 15 (400000 – 100x)/CP – (250000 + 100x)/CP = 15 150000/CP = 15 15CP = 150000 CP = 150000/15 CP = 10000 So, the original Cost price is Rs. 10000
 Question 2
``` Mukesh buys a worth Rs. 15,000. He sells it to Ramesh at a profit
of 5%. If after sometime Ramesh sells it back to Mukesh at a loss
of 15%, then completely:```
 A Mukesh gains Rs. 1500 B Mukesh loses Rs. 150 C Mukesh loses Rs. 1500 D Mukesh gains Rs. 150 E None of these
Question 2 Explanation:
Mukesh’s cost price (CP) = 15,000 Profit on selling price (SP) = 5% SP= (100+5%/100) * CP SP= 105/100 *15,000 SP of Mukesh = Rs. 15750 profit of Mukesh is 15750-15,000= Rs.750 Cost Price of Ramesh= Mukesh’s selling price= 750 loss % = 20% using the formula, SP= (100-20%/100)*CP SP= 80/100 * 750 SP =Rs.600 loss= CP-SP Loss= 750-600 = Rs. 150
 Question 3
```The list price of an article is Rs. 400. If two consecutive discounts
of 25% and 15% are allowed, its selling price will be:```
 A Rs. 150 B Rs.250 C Rs.255 D Rs. 350 E None of these
Question 3 Explanation:
Discount for 25%=(25/100)×400=Rs100 S.P=400-100=Rs 300. discount of further 15%=(15/100)×300=Rs 45 S.P=300-45=Rs 255
 Question 4
```The rate of compound interest at which a sum of Rs. 600 amounts
to Rs. 800 in 1 years?```
 A 55.5% B 33.3% C 67.7% D None of these
Question 4 Explanation:
Let the rate be R% p.a Then, 600 x(1+R/100)1 =800 (1+R/100)1 =800/600 ==4/3 (1+R/100)1=(4/3)1 1+R/100=4/3 R=33.3%
 Question 5
```Anamika bought 50 kg of sugar at the rate of Rs. 6.50 per kg and 30 kg
of wheat at the rate of Rs. 8.00 per kg. She mixed the two. At
what amount (App.) per kg should she sell the mixture to get 10% profit?```
 A Rs. 7.76 B Rs. 7.50 C Rs. 8.86 D Rs. 9.00 E none of these
Question 5 Explanation:
Total quantity of the mixture = 50 + 30 = 80 kg Cost price of 80 kg mixture = 50×6.50 + 30×8 = Rs. 565 The selling price of 80 kg mixture to earn 10% profit = 565×110/100= Rs. 621.5 i.e. Selling price of 1 kg mixture to earn 10% profit = 621.5/ 80= Rs. 7.76
 Question 6
```The cash price of a radio is Rs. 3000. A buyer paid Rs. 1200 in
cash and promised to pay the remaining money in 2 monthly equal
installments at the rate of 5% per annum compound interest.
What is the value of each installment?```
 A Rs. 650 B Rs. 623.45 C Rs. 750 D Rs. 660.97 E None of these
Question 6 Explanation:
Cash price of the radio = Rs. 3000 Payment made by the buyer in cash, in the beginning, = Rs. 1200 Remaining amount to be paid = 3000 – 1200 = Rs. 1800 Let each installment be of Rs. x. Then the principals for the amount of Rs. x for the 1st year and 2nd year will be as follows. => [ X / ( 1+(r/100)) ] + [ X / ( 1+(r/100))^2 ] + [ Y / ( 1+(r/100))^3 ] ………. upto N terms [ X / ( 1+(r/100)) ^N ] => x (20/21) + x(20/21)2 = 1800 => x (20/21) (1261/ 441)= 1800 => x= 1800 x 21 x 441)/ (20 x1261) => Rs. 660.97
 Question 7
```The residents of a city decrease at the rate of 10% per annum.
If its population two years ago was 9000, what is its present population?```
 A 5000 B 7200 C 7290 D 7500 E None of these
Question 7 Explanation:
Present Population = 9000 (1- 10/100)2 = 9000 x 9/10 x 9/10 = 7290
 Question 8
A certain sum of money at simple interest becomes Rs. 2800 in 3 years and Rs. 3180.50 in 1 5/2 years. What is the rate of interest per annum?
 A 11% B 12% C 12.4% D 11.4% E None of these
Question 8 Explanation:
Let the sum=Rs x and rate%= r, then Sum after 3 years = x + x*r*3/100 = 2800 —-(i) and Sum after 1 7/2 or 9/2 years = x + x*r*(9/2)/100 = 3180.50 —-(ii) Subtracting (i) from (ii), 9xr/200 – 3xr/100 = 380.5 => 3xr = 380.5 x 200 => xr= 25366 Substituting value of xr=25366 in (i), x + 25366*3/100 = 2800 => x = 2800 – 760.98 = 2039.02 So r= 25366/2039.02= 12.4
 Question 9
``` If the simple interest on a sum at 5% per annum for two years is
Rs. 70, what is the compound interest on the same sum for the same time?```
 A Rs. 71.50 B Rs. 71.75 C Rs. 70.65 D Rs. 81.75 E None of these
Question 9 Explanation:
Let take x as principal SI = (PRT/100) 70 = (x × 5× 2 /100) (70/100)x 10= x x = 700 Now CI = P (1+ r/n)nt =[700(1+5/100)^2] =700×21/20×21/20 =771.75 CI=(771.75-700)= Rs 71.75
 Question 10
```A woman covers a distance of 1500 km in 50 days resting 5 hours a day,
if she rests 15 hours a day and walks with speed 7⁄2 times of the
previous in how many days will she cover```
 A 31 B 31.25 C 30 D 33 E None of these
Question 10 Explanation:
The woman’s previous speed = 1500/(50*(24-5))km/h = 10/7 km/h New speed = 10/7 * 7/2 km/h = 5 km/h Per day she walks for (24-15)= 9 hours Each day she covers = 5*9 Kms = 45 kms The total time taken for covering 720 Kms = 720/24 days = 30 days
There are 10 questions to complete.