## 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

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

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.

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

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.

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

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

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.**

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.

## 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%

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

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

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%

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

## 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

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

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

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=m6….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=mn
• (2)4=mn  (because 24=16)
• But from eq1 we can plug the value of “2”
• (m6)4=mn
• m(6*4)=mn
• m(24)=mn
• So we can say n=24 years.

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)

1. 3%
2. 4%
3. 4.5%
4. 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%

## 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.

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

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

## 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

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)

1. (6, 14)
2. (9, 21)
3. (15, 35)
4. (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

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

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