- Prologue
- Percentages: Profit Loss
- Percentages: Simple Interest
- Percentages: Compound Interest
- Percentages: Misc. Sums
- Ratio
Prologue
- Time Limit: 30 minutes. 2 Marks for correct answer; -0.66 marks for wrong answers.
- Mitron, my [T25] series contains 25 Mock MCQ sets
- I usually don’t give direct answers below the MCQs but indirect hints. However, since everyone is not equally good at maths, and sometimes more help is required so this sub-set of questions contain full length calculation and answers.
- Make sucre you’ve read / brushed up the formulas related to %, profit loss, SI-CI & Ratios before attempting this round.
- This 44th set contains Maths questions from UGC NET exams of 2018 and 2019.
- धीरे धीरे करते हुए, so far I’ve done 44 sets x 25 questions each = 1100 MCQs मुफ्त मे अपलोड हो गए, और पता भी नहीं चला!
- Given the fact that UPSC Prelims even asks contemporary topics even from 2-3 years preceding the exam, so you should go through all of them available at Mrunal.org/Prelims
Percentages: Profit Loss
Q. A shopkeeper sells a refrigerator for Rs 22,000.00 and makes a profit of 10%. If he desires to make a profit of 18% what should be his selling pace? (NET19Dec06-II)
1. Rs 23,600
2. Rs 39,600
3. Rs 36,000
4. Rs 24,600
HINT:
Cost price(CP) | +profit | =selling price (SP) | |
situationA | 100%CP | +10% of CP | =22000 |
situationB | 100%CP | +18% of CP | =?? |
As per situation A:
- 100%CP+10% of CP=22000
- 110%CP=22000
- (110/100) x CP = 22000
- CP = 22000 x (100/110)
Update the table
Cost price(CP) | +profit | =selling price (SP) | |
situationA | 100%CP | +10% of CP | =22000 |
situationB | CP=20000 | +18% of CP | =?? |
As per situation B:
- 100%CP+18% of CP=??
- 118% of CP =??
- (118/100) x 20,000=??
- Ans = 23600
Correct Answer:[1]
Q. A person buy 3 item; of Rs. 1500 each and 2 items of Rs. 1800 each from a shopping mall. On the total bill a discount of 10% is given. What is the average cost of the items? (NET19Dec06-I)
(1) 1650
(2) 1458
(3) 1658
(4) 1550
HINT:
item | price | Quantity | total |
A | 1500 | 3 | 4500 |
B | 1800 | 2 | 3600 |
total | Total → | 8100 | |
Minus Discount 10% | -810 | ||
Final bill | =7290 |
- Average cost of the items = Rs. 7290 / 5 items = Rs. 1458
Correct Answer:[2]
Q. At what percentage above the cost price must an article be marked so as to gain 8% after allowing a customer a discount of 10%? (NET19Jun25-II)
1. 18%
2. 20%
3. 28%
4.10.8%
HINT:
- Better we go by elimination method. Suppose cost price is 100
Options | Option-A | Option-B |
Cost price | 100 | 100 |
Marked up by | 18%=118 | 20%=120 |
sold@10% discount | 90% of 118=106 | 90% of 120=108 |
Profit against Rs.100 cost price | 6 so not fitting | 8 so it’s fitting. |
Correct Answer:[2]
Q. A retailer marks all his goods at 40% above the cost price and thinking that he will still make 20% profit, offers a discount of 20% on the marked price. What is his actual profit percentage on the sales? (NET19Jun25-I)
(1) 20%
(2) 18%
(3) 16%
(4) 12%
HINT:
Options | Option-A |
Cost price | 100 |
Marked up by | 40%=140 |
sold@20% discount | 80% of 140=112 |
Profit against Rs.100 cost price | 12 |
Correct Answer:[4]
Q. A person incurs 10% loss by selling a watch for Rs. 1,800. At what price should the watch be sold to earn 10% profit? (NET19Jun26-II)
(1) Rs. 1980
(2) Rs. 2160
(3) Rs. 2,200
(4) Rs. 2,360
HINT:
Cost price(CP) | profit or loss | =selling price (SP) | |
situation-A | 100%CP | -10% of CP | =1800 |
situation-B | 100%CP | +10% of CP | =?? |
Situation-A
- 100%CP-10% of CP=1800
- 90%CP=1800
- CP=1800*(100/90)= Rs. 2000
Update the table
Cost price(CP) | profit or loss | =selling price (SP) | |
situationA | 100%CP | -10% of CP | =1800 |
situationB | 2000=100%CP | +10% of CP | =?? |
Situation-B: we want to earn 10% profit then what should be the selling price?
- 100%CP+10%CP=SP
- 110%CP=SP
- (110/100)*2000=SP
- 2200=Selling price.
Correct Answer:[3]
Q. Sanjay sold an article at a loss of 25%. If the selling price had been increased by Rs. 175, there would have been a gain of 10%. What was the cost price of the article? (NET19Jun26-I)
(1) Rs. 350
(2) Rs. 400
(3) Rs. 500
(4) Rs. 750
HINT:
Cost price(CP) | profit or loss | =selling price (SP) | |
situationA | 100%CP | -25% of CP | =SP |
situationB | 100%CP | +10% of CP | =SP+175 |
We can make two equations
100%CP-25% of CP=SP
75%CP=SP….eq1
From situation-B
100%CP+10% of CP=SP+175
110%CP=SP+175….eq2
Now plug value from eq1 to eq2
110%CP=75%CP+175
(110-75)%CP=175
35%CP=175
(35/100)xCP=175
CP=175*(100/35)
CP=500
Correct Answer:[3]
Q. Manoj’s commission is 10% on all sales upto Rs. 10,000 and 5% on all sales exceeding this. He remits Rs. 75,500 to his company after deducting his commission. The total sales will come out to be (NET19Jun20-II)
1. Rs 78,000
2. Rs 80,000
3. Rs 85,000
4. Rs 90,000
HINT:
- Let, total sales be Rs. x.
- His commission on total sales is:
- Total commission = 10% of 10000 + 5% of (x-10000)
- Total commission = 1000 + 5x/100 – 500
- Total commission = x/20+500
- Remits the amount:
- 75500 = total sales minus his commission
- 75500 = x- ((x/20) + 500))
- 75500 + 500 = 19x/20
- 76000 = 19x/20
- x = (76000 × 20) / 19
- x = 4000 × 20
- x = 80000
Correct Answer:[2]
Q. By selling 70 pencils for Rs.90- a person makes a loss of 25%. How many pencils should be sold for Rs.234 to have a profit of 30%? (NET 19 Dec 04-II)
(1) 95
(2) 100
(3) 105
(4) 110
HINT:
Cost price(CP) | profit or loss | =selling price (SP) | |
situationA | 100%CP | -25% of CP | =SP=90 |
situationB | 100%CP | +30% of CP | =234 |
SituationA
100%CP-25% of CP=90
75%CP=90
CP=90* (100/75)
CP=120 for which 70 pencils were sold. So, cost price of 1 pencil = 120/70=1.71
Cost price(CP) | profit or loss | =selling price (SP) | |
situationA | 100%CP | -25% of CP | =SP=90 |
situationB | 100%CP | +30% of CP | =234 |
SituationB: we want to earn 30% profit by selling @₹234
130%CP=234
CP=234* (100/130)=180
Since the cost price of 1 pencil is 1.71 then in ₹180 cost price, we can accomodate 180/1.71=105.**
Correct Answer:[3]
Q. What is the exact equivalent discount of three successive discounts of 10%,15%, and 20% by sale of an article? (NET19Dec04-I)
A. 35.8%
B. 38.5%
C. 35.5%
D. 38.8%
HINT:
Original | |
Selling price | 100 |
discount1 | 10%=90 |
discount2 | 15%=85%(90) |
discount3 | 20%=80%of(85%(90)) |
Final price | (80/100)*(85/100)*(90)=61.20 |
Total discount | 100-61.20=38.8 |
Alternate method is through formula, but if something can be solved through common logic then unnecessarily we need not waste time memorizing formula.
Correct Answer:[4]
Percentages: Simple Interest
Q. A sum of Rs. 2.000 yields Rs 180 with simple interest in nine (9) months. The annual rate of interest is (NET19Dec02-I)
(1) 9%
(2) 10%
(3) 11%
(4) 12%
HINT:
Principal | 2000 |
Months (T) | 9/12 |
interest | r? |
Final amount A = Principal+interest | 2000+180=2180 |
A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods. Where r is in decimal form; r=R/100
A = P(1 + rt)
- 2180=2000(1+r(9/12))
- r=0.12 which is in decimal format.
- 0.12=R/100
- R=0.12*100=12%
Correct Answer:[4]
Q. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 9 years and that for 12 years? (NET19Jun20-I)
(1) 3:4
(2) 2:3
(3) 4:3
(4) Data Inadequate
HINT:
Accumulated Amount A= | A |
Principal P = | 100 |
Simple Rate of interest= | 10% assume = 0.10 |
Time period | 9 years and 12 years |
Now we know the simple interest formula is A = P(1 + rt)
A9YR=100(1+(0.10*9))=190, so simple interest earned is 190-100=₹90
A12YR=100(1+(0.10*12))=220, so simple interest earned is 220-100=₹120
Ratio simple interest for 9 years to 12 years= 90/120=3:4
Correct Answer:[1]
Q. The simple interest on a certain sum of money at the rate of 5% p.a. for four years is Rs. 420. At what rate of interest, the same amount of interest can be received on the same sum after two and half years? (NET19Jun21-II)
(1) 4%
(2) 6%
(3) 8%
(4) 10%
HINT:
Accumulated Amount A= | A=P+I |
Principal P = | P |
Simple Rate of interest R%= | 5% |
Time period | 4 years |
Simple Interest amount | 420 |
Simple INTEREST Amount = PRT/100
420=P*5*4/100
P=(420*100)/(5*4)=2100
But, we want to earn, ₹420 within 2.5 years =
24+6=30 months divided by 12 month per year = 30/12=again it’s 2.5 so we can simply take 2.5
Simple INTEREST Amount = PRT/100
420=2100*(R/100)*(2.5)
R=8%
Correct Answer:[3]
Q. If the simple interest on a certain sum for 1 year 3 months at 8.5% per annum exceeds the simple interest on the same sum for 8 months at 7.5% per annum by Rs. 45, then the sum is (NET19Jun21-I)
(1) Rs.600
(2) Rs.800
(3) Rs.6,000
(4) Rs.8,000
HINT:
situation 1 | situation 2 | ||
Accumulated Amount | A | ||
Principal | P | ||
Rate of Interest | R | 8.5% | 7.5% |
Time | T | 1 year 3 months =(15/12) years | 8 months = 8/12 years |
Simple interest in Situation1 minus situation2 = 45
Simple Interest = PRT/100, so,
45= ((P*8.5*(15/12))/100)-((P*7.5*(8/12))/100)
P=800
Correct Answer:[2]
Percentages: Compound Interest
Q. The compound interest on a certain sum of money for 3 years at a rate of 10% per annum is Rs. 993.00. The simple interest on the same sum at the rate of 8% per annum for 4 years would be equal to which of the following? (NET19Dec05-I)
(1) Rs. 900.00
(2) Rs. 930.00
(3) Rs. 985.00
(4) Rs. 990.00
HINT:
Accumulated Amount A= | 993 |
Principal P = | P |
Compound Rate of interest= | 10% |
Time period | 3 years |
Compound interest accumulated Amount = Principal * [1 + Rate of Interest/100]^{Time period}
- 993=P(1+10/100)^3
- 993=P(1.10)^3
- 993=P(1331/1000)
- P=746
Now’ve to find simple interest
Accumulated Amount A= | A |
Principal P = | 746 |
Simple Rate of interest= | 8% = 0.08 |
Time period | 4 years |
- A = P(1 + rt)
- A=746*(1+(0.08*4))
- A=approx 985
Correct Answer:[3]
Q. A sum of money with compound interest becomes Rs. 2,400 in one year and Rs. 3,000 in two years. Find out the principal amount. (NET19Dec03-I)
1) Rs. 1,900
2) Rs. 1,915
3) Rs. 1,910
4) Rs. 1,920
HINT:
- Let’s go by elimination method by plugging values from each option.
- Interest in second year is 3000-2400=₹600.
- So, if ₹600 interest is paid on ₹2400 then % interest is (600/2400) x 100 = 25%
- So, now we’ve compound interest % 25%
Accumulated Amount A= | 3000 |
Principal P = | P |
Compound Rate of interest= | 25% |
Time period | 2 years |
Compound interest formula is
Amount = Principal * [1 + Rate of Interest/100]^{Time period}
3000=P(1+(25/100))]^2
3000=P(1.25)^2
3000=P(1.5625)
P=3000*10000/15625
P=1920
Correct Answer:[4]
Q. The simple interest on a certain sum of money for 3 years at 4% per annum is half the compound interest on Rs. 2,000 for 2 years at 10% per annum. The sum invested on simple interest is (NET19Jun20-II)
1. Rs 8,750
2. Rs 1,750
3. Rs 2,500
4. Rs 3,500
HINT:
situation 1 (SI) | situation 2 (CI) | ||
Accumulated Amount | A | ||
Principal | P | find? | 2000 |
Rate of Interest | R | 4% | 10% |
Time | T | 3 years | 2 years |
Interest earned | (CI)/2 | CI |
Situation 2
Compound interest formula is
Accumulated Amount = Principal * [1 + Rate of Interest/100]^{Time period}
A=2000*(1+(10/100))^{2}
- A=2420
- CI=A-P
- CI=2420-2000=420
- SI is half of CI
- SI=½ of CI
- SI=½*420= 210
- SI=PRT/100
- 210=P*4*3/100
- 21000/12=P
- P= Rs. 1750
Correct Answer:[2]
Q. The sum of money doubles at compound interest in 6 years. In how many years will it become 16 times? (NET19Jun24-I)
(1) 16 years
(2) 24 years
(3) 48 years
(4) 96 years
HINT:
Method1: arithmetic and geometric progression series
Increase in year (every +6 years_ | Increase in money (2x) |
0 (originally) | 1x |
+6 years than before | 2 times before (1)=2 |
+6 +6=12 years | 2 times before (2)=4 |
12 years +6 =18 years | 2 times before (4)=8x |
18 years +6= 24 years | 2 times before (8)=16x |
Hence the answer is 24 years.
Method2: rule of 72
- In compound interest rate, the time taken to double the money is years=72/%(of interest rate)
- So, if money doubles in 6 years
- 6 years = 72/(R%)
- R%=12%
- We want the accumulated amount 16 times, so assume the original principal was 100. Then
- Accumulated Amount = Principal * [1 + Rate of Interest/100]^{Time period}
- 16*(100)=100(1+(12/100))^{t}
- But this solution will require logarithm, but answer will be t= approx 24.46 years.
Method3: make two equations
- Principal= P
- The sum doubles in 6 years. That is 2P
- According to compound interest formula, 2P=P(1+r/100)^{6}
- 2=(1+r/100)^{6}
- Let’s call (1+r/100)=m
- 2=m^{6}….eq1
afterwards,
- The sum becomes 16 times in a number of years.
- According to compound interest formula, 16P=P(1+r/100)^{n}
- 16=(1+r/100)^{n}
- Here also Let’s call (1+r/100)=m
- 16=m^{n}
- (2)^{4}=m^{n} (because 2^{4}=16)
- But from eq1 we can plug the value of “2”
- (m^{6})^{4}=m^{n}
- m^{(6*4)}=m^{n}
- m^{(24)}=m^{n}
- So we can say n=24 years.
Correct Answer:[2]
Q. A certain principal invested at compound interest payable yearly amounts to Rs. 10816.00 in 3 years and Rs. 11248.64 in 4 Years. What is the rate of interest? (NET19Dec04-I)
- 3%
- 4%
- 4.5%
- 5.5%
HINT:
Year3 | 10816.00 |
Year4 | 11248.64 |
Interest: year4-year3 | 432.64 earned on the previous amount 10816.00 |
So, % interest is = (432.64/10816.00)*100=4%
Correct Answer:[2]
Percentages: Misc. Sums
Q. A student obtains overall 78% marks in his examination consisting of Physics, Chemistry, Mathematics, Computer Science and General English. Marks obtained in each subject and the maximum marks are indicated in the following table (NET19Dec06-II)
Subject | Physics
(200 max marks) |
Chemistry
(100 max marks) |
Mathematics
(200 max marks) |
Computer
Science (200 max marks) |
General
English (100 max marks) |
Marks
Obtained |
155 | 80 | 165 | 140 | X |
The marks (X) obtained by the student in General English would be
1. 68
2. 74
3. 84
4. 90
HINT:
- Total marks of all subjects= 200+100+200+200+100= 800 maximum marks
- Marks scored by student= 155+80+165+140+X=(540+X)
- % = marks scores / total marks
- 78%=(540+x)/800
- 78/100=(540+x)/800
- 78*8=540+x
- 624=540+x
- x=624-540=84 marks secured in English paper.
Correct Answer:[3]
Q. During one year, the population of a town increased by 5% and during the Next year, the population decreased by 5%. If the total population is 29925 at the end of the second year, and then what was the population size in the beginning of the first year? (NET19Jun21-II)
(1) 25000
(2) 30000
(3) 33000
(4) 36000
HINT:
Let’s go by trial-error-elimination method,
- Original Population can’t be 25000 because if it increases by 5% in year1 it’ll be in the range of 26,*** something so in year2 if it decreased by 5% it couldn’t be 29,****
- Let’s try option2
Option2 | |
Starting pop | 30000 |
Year1 increase by 5%
100%+5%=105%=1.05 |
30000*1.05 |
Year2 decrease by 5%
100%-5%=95%=0.95 |
0.95*(30000*1.05) |
Is population @end of year2 29925? | yes! |
Another method (but more cumbersome):
- Let the population be X
- Increase by 5%
- Hence it will be=X+(5/100)X= 1.05X
- Next year it is decreased by 5%
- 5% of 1.05X = 1.05×(5/100)X = 0.0525X
- 1.05X – 0.0525X= 0.9975X
- But, the population at the end of second year is 29925.
- 0.9975X = 29925
- X=30000
Correct Answer:[2]
Q. If x and y are positive numbers and x is 25 % greater than y, what is the value of the ratio y/x? (NET19Jun21-I)
(1) 0.75
(2) 0.80
(3) 1.20
(4) 1.25
HINT:
- Let’s take, y is 100. So, x is 25% more than y, then x will be 125.
- Ratio of y/x = 100/125 = ⅘ = 0.8
Correct Answer:[2]
Ratio
Q. If income of X is 20% more than that of Y and income of Y is 25% less than that of Z, then the total income of X, Y and Z are in which of the following ratios? (NET19Dec05-II)
(1) 18:15:20
(2) 18:15:10
(3) 18:10:15
(4) 18:12:15
HINT:
Assume income of Y is 100
Income of Y | 100 |
Income of X is 20% more than Y | 120 |
Income of Y is 25% less than Z | 400/3
=> Y=0.75% of Z => 100=0.75(z) =>400/3=Z |
So, now we’ve the three numbers as following
x | y | z | |
originally | 120 | 100 | 400/3 |
Multiply with 3 | 360 | 300 | 400 |
Now let’s try to reduce the number as much as we can by dividing it with common figures like 10 | 36 | 30 | 40 |
Again divide by 2 | 18 | 15 | 20 |
Correct Answer:[1]
Q. The ratio of two numbers a and b is 3:7. After adding 9 to each number, the ratio becomes 9:17. The numbers a and b are: (NET19Dec02-II)
- (6, 14)
- (9, 21)
- (15, 35)
- (18, 42)
HINT:
Method-1: trial error
- For (a,b) → ratio of two numbers a and b is 3:7. This condition is fulfilled in all four options.
- After adding 9 to each number, the ratio becomes 9:17. So, we should try to find a number which comes in a table of 17. Such as 17,34,51
- If denominator number was “9” less than table figures of 17, then it could be
- 8,25,42
- So, only fouth option is fitting. Let’s check
Option4 | |
(numerator, denominator) | (18, 42) |
After adding 9 to each | (27, 51)= which is same as 9/17. |
Method-2: assume “X”
- Suppose, the ratio of numbers are 3x and 7x = 3x:7x
- 9 is added to the ratios and the new ratio obtained is 9:17.
- (3x+9)/(7x+9)=9/17
- 17(3x+9)=9(7x+9)
- 51x+153=63x+81
- 72=12x
- x=6
- Now, input the value of x in the original ratios. We’ll get the values of a and b.
- 3x:7x=(3*6):(7*6)
- 18:42
Correct Answer:[4]
Q. Two numbers are in the ratio 3:7. If 8 is added to both the numbers, their ratio becomes 5:9. The numbers are (NET19Dec02-I)
(1) 12, 28
(2) 6, 14
(3) 15, 35
(4) 24, 56
HINT:
Method-1: trial error
After adding 8 to the numerator, the next figure must end with number that comes in table of 5.
option | numerator | After adding “+8”, is it likely to be in the table of 5 |
1 | 12 | 20 |
2 | 6 | 14 |
3 | 15 | 23 |
4 | 24 | 32 |
Only option 1 is matching. Just to crosscheck
Option1 | |
(numerator, denominator) | (12, 28) |
After adding 8 to each | (20, 36)= which is same as 5/9 |
Method-2: assume “X”
- 3:7 can be written as = 3x:7x
- 8 is added to both the numbers and new ratio obtained is 5:9.
- (3x+8)/(7x+8)=5/9
- 9(3x+8)=5(7x+8)
- 27x+72=35x+40
- 32=8x
- x=4
- Input the value of x in the ratios.
- 3x:7x=(3*4):(7*4)
- 12:28
Correct Answer:[1]
Q. In a class, there are 60% female students and the remaining are male. 40% of the female students and 70% of the male students have opted for the research methodology course. If in all, 60 students have not opted for the research methodology course, which of the following is the total number of students in the class? (NET19Dec05-II)
(1) 120
(2) 125
(3) 130
(4) 140
HINT:
Assume class population is “c”
60% female = 0.6c
Remaining are male so 0.4c
Male: 0.4c | Female: 0.6c | Total students in absolute figure | |
Opted research methodology
(40% female and 70% male) |
0.7*0.4c | 0.4*0.6c | ?? |
Not Opted research methodology
(60% female and 30% male) |
0.3*0.4c | 0.6*0.6c | 60 |
0.3*0.4c+0.6*0.6c=60
0.12c+0.36c=60
0.48c=60
c=125
Correct Answer:[2]
Visit Mrunal.org/prelims for Previous Rounds of [T25] Mock MCQ Question / Answers.
One Comment on “[T25] Aptitude Mock Round#44: Maths: %, Profit-Loss, Simple & compound Interest, Ratios”
Sir 1 st question of compund interest wrong he apka please correct krlo