Quant Quiz
1. Veer calculates his profit percentage on the selling price whereas Vipin calculates his profit percentage on the cost price.
They find that the difference of their profits is Rs 100. If the selling price of both of them are the same and both of them get 25 % profit,
find their selling price?
a) Rs.1800
b) Rs.2000
c) Rs. 2500
d) Rs. 2867
e) None of these.
2. A textile vendor sells shirts at Rs. 800 per shirt, however he forced to give two successive discounts of 10 % and 5 % respectively.
However he charges sales tax on net sales price from the customer at 5 %. What price does customer has to pay to buy the shirt ?
a) Rs. 730.40
b) Rs. 690.30
c) Rs. 718.20
d) Rs. 960.50
3. A printer manufacturer initially makes a profit of 12 % by selling a particular model of printer for Rs 6500.
If the cost of manufacturing of that printer increases by 25 %, then price paid by customer is increased by 15 %.
Find the % profit made by the manufacturer ?
a) 5%
b) 6%
c) 4.5%
d) 9 %
e) None of these.
4. Mr. Kartik has taken loan from a bank in terms, he will donate 50 % of total capital at the end of the year to the bank.
He earned a profit of 50 % and donated 50 % of total to bank , next year he earned 50 % profit and donated 50 % of total to bank,
in third he again earned 50 % profit, after donating 50 % of total money he left with Rs.16,875, then find the amount donated by him at the end of 2nd year.
a) Rs. 40,000
b) Rs. 22,500
c) Rs. 16,000
d) Rs. 18,500
e) None of these.
5. Manufacturing of cement is made up with four components P, Q, R and S in ratio 2 : 3 : 4 : 5 . Due to seasonal variation-
if there are respective changes of + 10 % , - 20 % , - 30 % , + 40 % , then what would be the % change in the cost ?
a) 3.76%
b) 2.85%
c) 4.62%
d) 6.67%
e) None of these.
6. The population of a village is 5500. If the number of males increases by 11 % and the number of females increases by 20 %, then the population becomes 6330.
Find the population of females in the town?
a) 2500
b) 3000
c) 3500
d) 2250
e) None of these.
7. A man can row 8 km/hr in still water. If the river is running at 3 km/hr, it takes 3 hours more in upstream than to go downstream for the same distance. How far is the place?
a) 26.67 km
b) 32.5 km
c) 27.5 km
d) Can't determined
e) None of these.
8. A man borrows Rs. 20,000 at 10% compound interest. At the end of every year he pays Rs. 2000 as part repayment. How much does he still owe after three such installments?
a) Rs. 25000
b) Rs. 20000
c) Rs. 24000
d) Rs. 30000
e) None of these.
9. A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
a) 30 hour
b) 70 hour
c) 35 hour
d) 40 hour
e) None of these.
10. Three taps A,B and C can fill a tank in 20,30and 40 minutes respectively. All the taps are opened -
simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .
At the moment a leak developed which can empty the full tank in 60 minutes. What is the total time taken for the completely full?
a) 24 minutes
b) 25 minutes
c) 35 minutes
d) 40 minutes
e) None of these.
11) The Simple interest on a sum of money is 1/9 of the Principal and the no. of years is equal to the rate %pa. The rate % pa is?
a) 9 %
b) 3.33%
c) 10%
d) 16.67%
e) None of these
12) A sum of money is accumulating at compound interest at a certain rate of interest. If Simple interest instead of compound interest were reckoned,
the interest for the first two years would be diminished by Rs. 20 and that for the first three years by Rs. 61. Find the Sum?
a) 8000
b) 6000
c) 10,000
d) 12,000
e) None of these
13) A certain sum is interested at compound. The interest occurred in the first two years is Rs. 272 and that in the First three years is Rs. 434. Find the rate %?
a) 33.33%
b) 25%
c) 12.5%
d) 16.67%
e) None of these
14) Saurabh opened a restaurant, with a initial investment of Rs. 30,000. In that first year, he incurred a loss of 5%. However, during the second year,
he earned the Profit of 10% which in third year rose to 12.5%. Calculate his net profit for the entire period of three years?
a) 5300
b) 5620
c) 6000
d) 6600
e) None of these
15) A bottle is full of Dettol. 1/3 of it is taken out and then an equal amount of water is poured into bottle to fill it. This operation is done 4 times.
Find the final ratio of Dettol and water in the bottle?
a) 43: 65
b) 17: 43
c) 16:63
d)16:65
e) None of these
Answers:
1. b
2. c
3. c
4. b
5. b
6. a
7. c
8. b
9. c
10. a
11. b
12. a
13. c
14. b
15. d
Solution
1. Since veer calculates his profit percentage on Selling price so ,
CP = SP-25/100 SP = 75/100 SP = 3/4 SP
&vipin calulates his profit percentage on Cost price, So
SP = CP+25/100 CP= 5/4 CP or CP = 4/5 SP
As, Difference in Profit =100
SP/4 - SP/5 = 100
So, Selling Price = 2000
2. Final Price Paid by customer = 800*(90/100) *(95/100) * (105/100) = 718.20
3. New SP= 6500+ 15% of 6500 = 7475
Old Cost price = 6500 -12% of 6500 =5720
New Cost Price = 125% of 5720 = 7150
% Profit = (7475-7150)/ 7150 *100 = 4.5%
4. Let the initial capital is X
After one year 3/4 of X will be donated aur utna hi hath me bachega
After 2nd yr 9/16 of X will be donated ...& same
After 3rd yr 27/64 of X will be donated and the same in hand
27/64 of X = 16875
calculate x= 40000
Amount denoted in 2nd year =9/ 16 of X= 22500
5. Ratio of material's 2 : 3 : 4 : 5
Let is value be 20 , 30, 40, 50 so total sum = 140
Now, there is + 10 % increase in cost of P, 20 + 2 = 22
- 20 % decrease in the cost of Q , 30 - 6 = 24
- 30 % decrease in the cost of R , 40 - 12 = 28
+ 40 % increase in the cost of S , 50 + 20 = 70
Now, total sum of latest cost = 22+ 24 + 28 + 70 = 144
% change of Cost = (144-140)/140 * 100 = 2.85%
6. Let X is the initial population of Male and Y be the initial population of Female
X + Y = 5500 .........(1)
If the number of males increases by 11 % and the number of females increases by 20 %, then
(111/100 )X+ (120/100) Y = 6330
1.11 X + 1.2 Y = 6330 , On putting the value of X from equeation (1)
1.11 ( 5500 - Y ) + 1.2 Y = 6330
Y = 2500, Population of females in town.
7. Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr.
1. b
2. c
3. c
4. b
5. b
6. a
7. c
8. b
9. c
10. a
11. b
12. a
13. c
14. b
15. d
Solution
1. Since veer calculates his profit percentage on Selling price so ,
CP = SP-25/100 SP = 75/100 SP = 3/4 SP
&vipin calulates his profit percentage on Cost price, So
SP = CP+25/100 CP= 5/4 CP or CP = 4/5 SP
As, Difference in Profit =100
SP/4 - SP/5 = 100
So, Selling Price = 2000
2. Final Price Paid by customer = 800*(90/100) *(95/100) * (105/100) = 718.20
3. New SP= 6500+ 15% of 6500 = 7475
Old Cost price = 6500 -12% of 6500 =5720
New Cost Price = 125% of 5720 = 7150
% Profit = (7475-7150)/ 7150 *100 = 4.5%
4. Let the initial capital is X
After one year 3/4 of X will be donated aur utna hi hath me bachega
After 2nd yr 9/16 of X will be donated ...& same
After 3rd yr 27/64 of X will be donated and the same in hand
27/64 of X = 16875
calculate x= 40000
Amount denoted in 2nd year =9/ 16 of X= 22500
5. Ratio of material's 2 : 3 : 4 : 5
Let is value be 20 , 30, 40, 50 so total sum = 140
Now, there is + 10 % increase in cost of P, 20 + 2 = 22
- 20 % decrease in the cost of Q , 30 - 6 = 24
- 30 % decrease in the cost of R , 40 - 12 = 28
+ 40 % increase in the cost of S , 50 + 20 = 70
Now, total sum of latest cost = 22+ 24 + 28 + 70 = 144
% change of Cost = (144-140)/140 * 100 = 2.85%
6. Let X is the initial population of Male and Y be the initial population of Female
X + Y = 5500 .........(1)
If the number of males increases by 11 % and the number of females increases by 20 %, then
(111/100 )X+ (120/100) Y = 6330
1.11 X + 1.2 Y = 6330 , On putting the value of X from equeation (1)
1.11 ( 5500 - Y ) + 1.2 Y = 6330
Y = 2500, Population of females in town.
7. Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr.
If he takes t hours more in upstream than to go downstream for the same distance,
the distance = (x^2 - y^2)t/ 2y Km
x = 8 km/hr
y = 3 km/hr
t = 3 hours
As per the formula,
Distance = 27.5 Km.
8. Amount after 3 years on Rs.20000 at 10% compound interest
= 20000( 1+ 10/100)^3= 26620.............(1)
He paid Rs.2000 after 1st year.
Hence Rs.2000 and its compound interest for 2 years (i.e., amount on 2000 after 2 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 2nd year.
Hence Rs.2000 and and its compound interest for 1 year (i.e., amount on 2000 after 1 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 3rd year.
Hence this Rs.2000 also need to be reduced from (1)
Hence, remaining amount= 26620 - 2000(1+10/100)^2 - 2000(1+10/100) - 2000 = Rs. 20000
i.e, he still owes Rs.20000 even after three installments.
9. Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4hours respectively to fill the tank.
1/x+2/x+4/x = 1/5
on Solving,
x = 35 hrs.
10. A, B, C do it in 20, 30 , 40 min respectively Let total work be 120
Efficiency of A,B,C = 6,4,3 respectively
First 5 min work = 13* 5 = 65
remaining = 120-65 = 55
next 6 min work = 7*6 = 42
Remaining = 13
Now, 120 work is done by leak in 60 minutes
Leak's efficiency = 120/60 = -2(as it empties the tank)
Leak +C = 3-2 = 1
time taken = 13/1 = 13 minutes
Total time taken = 5+6+13 = 24 Minutes
11. R=(100/9)^1/2 =3.33%
12. P(r/100)^2= 40
P(r/100)^2 ×(300+r)/100 =61
On Solving P= 8000
13. P[(1+r/100)^2 -1]= 272 ----(1)
P[(1+r/100)^3 -1]= 434-----(2)
Consider 1+r/100 =a
On dividing eqn (2) by (1)
So (a3-1)/ (a2-1) = 434/272
On solving a=9/8
So 1+r/100 = 9/8
So r= 12.5%
14. 1st year loss = 1500
Price after 1st year = 28500,
2nd year profit = 2850
Price after 2nd year =31350
Profit in 3rd year = 3918.75
So net profit during 3 years = 3918.75+ 2850 – 1500= 5268.75
So option E is true
15. Final ratio of Dettol and water = (1-1/3)^4/{1-(1-1/3)^4} = 16: 65
the distance = (x^2 - y^2)t/ 2y Km
x = 8 km/hr
y = 3 km/hr
t = 3 hours
As per the formula,
Distance = 27.5 Km.
8. Amount after 3 years on Rs.20000 at 10% compound interest
= 20000( 1+ 10/100)^3= 26620.............(1)
He paid Rs.2000 after 1st year.
Hence Rs.2000 and its compound interest for 2 years (i.e., amount on 2000 after 2 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 2nd year.
Hence Rs.2000 and and its compound interest for 1 year (i.e., amount on 2000 after 1 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 3rd year.
Hence this Rs.2000 also need to be reduced from (1)
Hence, remaining amount= 26620 - 2000(1+10/100)^2 - 2000(1+10/100) - 2000 = Rs. 20000
i.e, he still owes Rs.20000 even after three installments.
9. Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4hours respectively to fill the tank.
1/x+2/x+4/x = 1/5
on Solving,
x = 35 hrs.
10. A, B, C do it in 20, 30 , 40 min respectively Let total work be 120
Efficiency of A,B,C = 6,4,3 respectively
First 5 min work = 13* 5 = 65
remaining = 120-65 = 55
next 6 min work = 7*6 = 42
Remaining = 13
Now, 120 work is done by leak in 60 minutes
Leak's efficiency = 120/60 = -2(as it empties the tank)
Leak +C = 3-2 = 1
time taken = 13/1 = 13 minutes
Total time taken = 5+6+13 = 24 Minutes
11. R=(100/9)^1/2 =3.33%
12. P(r/100)^2= 40
P(r/100)^2 ×(300+r)/100 =61
On Solving P= 8000
13. P[(1+r/100)^2 -1]= 272 ----(1)
P[(1+r/100)^3 -1]= 434-----(2)
Consider 1+r/100 =a
On dividing eqn (2) by (1)
So (a3-1)/ (a2-1) = 434/272
On solving a=9/8
So 1+r/100 = 9/8
So r= 12.5%
14. 1st year loss = 1500
Price after 1st year = 28500,
2nd year profit = 2850
Price after 2nd year =31350
Profit in 3rd year = 3918.75
So net profit during 3 years = 3918.75+ 2850 – 1500= 5268.75
So option E is true
15. Final ratio of Dettol and water = (1-1/3)^4/{1-(1-1/3)^4} = 16: 65
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