6.4 Mathematical Aptitude: Fractions, Percentages, Profit & Loss

Key Takeaways

  • Fraction-to-percentage conversion equivalence (1/2=50%,1/3=33.33%,1/6=16.67%,1/7=14.28%,1/8=12.5%,1/12=8.33%1/2 = 50\%, 1/3 = 33.33\%, 1/6 = 16.67\%, 1/7 = 14.28\%, 1/8 = 12.5\%, 1/12 = 8.33\%) enables rapid mental computation of complex data values.

  • The product constancy rule dictates that if the price of a commodity increases by 1/n1/n, consumption must decrease by 1/(n+1)1/(n+1) to maintain a constant total expenditure.

  • Successive percentage changes follow the net change formula a+b+ab100%a + b + \frac{ab}{100}\%, while two successive discounts d1d_1 and d2d_2 yield an equivalent single discount of d1+d2−d1d2100d_1 + d_2 - \frac{d_1 d_2}{100}.

  • Profit and loss percentages are universally calculated on the Cost Price (CP) unless explicitly stated otherwise; Marked Price (MP) serves as the sole base for commercial discounts.

  • If two articles are sold at the exact same Selling Price (SP), one at a profit of x%x\% and the other at a loss of x%x\%, the overall transaction always results in a net loss of x2100%\frac{x^2}{100}\%.

Last updated: August 2026

Mathematical Aptitude: Fractions, Percentages, Profit & Loss

Quick Answer: Percentages, profit, and loss form the mathematical core of UGC NET quantitative questions. Convert percentages to fractions instantly (1/6=16.67%,1/8=12.5%1/6 = 16.67\%, 1/8 = 12.5\%). Master the product constancy rule: if price rises by 1/n1/n, consumption must fall by 1/(n+1)1/(n+1) for constant expenditure. Successive changes combine as a+b+ab100a + b + \frac{ab}{100}, while successive discounts combine as d1+d2−d1d2100d_1 + d_2 - \frac{d_1 d_2}{100}. Always compute profit/loss on Cost Price (CPCP) and discounts on Marked Price (MPMP).


1. Fraction-to-Percentage Mastery Table

Memorizing fraction-to-percentage equivalents eliminates tedious manual division during exam calculations.

FractionPercentageDecimal MultiplierFractionPercentageDecimal Multiplier
1/11/1100%100\%1.001.001/91/911.11%=1119%11.11\% = 11\frac{1}{9}\%0.11110.1111
1/21/250%50\%0.500.501/101/1010%10\%0.100.10
1/31/333.33%=3313%33.33\% = 33\frac{1}{3}\%0.33330.33331/111/119.09%=9111%9.09\% = 9\frac{1}{11}\%0.09090.0909
1/41/425%25\%0.250.251/121/128.33%=813%8.33\% = 8\frac{1}{3}\%0.08330.0833
1/51/520%20\%0.200.201/131/137.69%=7913%7.69\% = 7\frac{9}{13}\%0.07690.0769
1/61/616.67%=1623%16.67\% = 16\frac{2}{3}\%0.16670.16671/141/147.14%=717%7.14\% = 7\frac{1}{7}\%0.07140.0714
1/71/714.28%=1427%14.28\% = 14\frac{2}{7}\%0.14280.14281/151/156.67%=623%6.67\% = 6\frac{2}{3}\%0.06670.0667
1/81/812.5%=1212%12.5\% = 12\frac{1}{2}\%0.12500.12501/161/166.25%=614%6.25\% = 6\frac{1}{4}\%0.06250.0625

Key Fraction Multiples

  • 2/3=66.67%2/3 = 66.67\%
  • 3/4=75%3/4 = 75\%
  • 2/5=40%2/5 = 40\%, 3/5=60%3/5 = 60\%, 4/5=80%4/5 = 80\%
  • 3/8=37.5%3/8 = 37.5\%, 5/8=62.5%5/8 = 62.5\%, 7/8=87.5%7/8 = 87.5\%
  • 2/7=28.57%2/7 = 28.57\%, 3/7=42.86%3/7 = 42.86\%, 4/7=57.14%4/7 = 57.14\%
  • 2/9=22.22%2/9 = 22.22\%, 4/9=44.44%4/9 = 44.44\%, 7/9=77.78%7/9 = 77.78\%
  • 2/11=18.18%2/11 = 18.18\%, 3/11=27.27%3/11 = 27.27\%, 5/11=45.45%5/11 = 45.45\%

2. Percentage Change, Base Shifts, and Product Constancy

The Multiplier Method

  • An increase of x%x\% corresponds to a multiplying factor of (1+x100)(1 + \frac{x}{100}). Example: Increase of 25%  ⟹  New Value=Original×1.25\text{Example: Increase of } 25\% \implies \text{New Value} = \text{Original} \times 1.25
  • A decrease of x%x\% corresponds to a multiplying factor of (1−x100)(1 - \frac{x}{100}). Example: Decrease of 15%  ⟹  New Value=Original×0.85\text{Example: Decrease of } 15\% \implies \text{New Value} = \text{Original} \times 0.85

The Product Constancy Rule (AB = Constant)

When the product of two variables remains constant (A×B=KA \times B = K), such as Price×Consumption=Expenditure\text{Price} \times \text{Consumption} = \text{Expenditure} or Speed×Time=Distance\text{Speed} \times \text{Time} = \text{Distance}:

If A increases by 1n, then B must decrease by 1n+1 to keep the product constant.\text{If } A \text{ increases by } \frac{1}{n}, \text{ then } B \text{ must decrease by } \frac{1}{n+1} \text{ to keep the product constant.} If A decreases by 1n, then B must increase by 1n−1 to keep the product constant.\text{If } A \text{ decreases by } \frac{1}{n}, \text{ then } B \text{ must increase by } \frac{1}{n-1} \text{ to keep the product constant.}
Increase in PriceFraction IncreaseRequired Decrease in ConsumptionPercentage Decrease
+100%+100\%+1/1+1/1−1/(1+1)=−1/2-1/(1+1) = -1/2−50%-50\%
+50%+50\%+1/2+1/2−1/(2+1)=−1/3-1/(2+1) = -1/3−33.33%-33.33\%
+33.33%+33.33\%+1/3+1/3−1/(3+1)=−1/4-1/(3+1) = -1/4−25%-25\%
+25%+25\%+1/4+1/4−1/(4+1)=−1/5-1/(4+1) = -1/5−20%-20\%
+20%+20\%+1/5+1/5−1/(5+1)=−1/6-1/(5+1) = -1/6−16.67%-16.67\%
+16.67%+16.67\%+1/6+1/6−1/(6+1)=−1/7-1/(6+1) = -1/7−14.28%-14.28\%

Successive Percentage Changes

When a quantity undergoes two consecutive percentage modifications, a%a\% followed by b%b\% (where increases are positive and decreases are negative):

Net Percentage Change=a+b+a×b100(%)\text{Net Percentage Change} = a + b + \frac{a \times b}{100} \quad (\%)
  • Example: A salary is increased by 20%20\% and subsequently decreased by 10%10\%: Net Change=20+(−10)+20×(−10)100=10−2=+8% (Net increase of 8%)\text{Net Change} = 20 + (-10) + \frac{20 \times (-10)}{100} = 10 - 2 = +8\% \text{ (Net increase of } 8\%).
Loading diagram...
Commercial Pricing Architecture: CP, MP, Discount, and SP

3. Profit, Loss, and Discount Fundamentals

Core Commercial Definitions

  • Cost Price (CPCP): The total expenditure incurred to acquire or manufacture an article.
  • Selling Price (SPSP): The actual price at which the article is sold to a customer.
  • Marked Price (MPMP / List Price): The price labeled on the article before discounts.

Fundamental Formulas

Profit (P)=SP−CP(when SP>CP)\text{Profit } (P) = SP - CP \quad (\text{when } SP > CP) Loss (L)=CP−SP(when CP>SP)\text{Loss } (L) = CP - SP \quad (\text{when } CP > SP) Profit %=(SP−CPCP)×100=(ProfitCP)×100\text{Profit } \% = \left(\frac{SP - CP}{CP}\right) \times 100 = \left(\frac{\text{Profit}}{CP}\right) \times 100 Loss %=(CP−SPCP)×100=(LossCP)×100\text{Loss } \% = \left(\frac{CP - SP}{CP}\right) \times 100 = \left(\frac{\text{Loss}}{CP}\right) \times 100

Calculating SP from CP and Vice-Versa

SP=CP×(1+P%100)orSP=CP×(1−L%100)SP = CP \times \left(1 + \frac{P\%}{100}\right) \quad \text{or} \quad SP = CP \times \left(1 - \frac{L\%}{100}\right) CP=SP1+P%100=SP×100100+P%orCP=SP×100100−L%CP = \frac{SP}{1 + \frac{P\%}{100}} = \frac{SP \times 100}{100 + P\%} \quad \text{or} \quad CP = \frac{SP \times 100}{100 - L\%}

Commercial Discounts and Markup

  • Discount: Calculated strictly on Marked Price (MPMP): Discount=MP−SP\text{Discount} = MP - SP Discount %=(DiscountMP)×100\text{Discount } \% = \left(\frac{\text{Discount}}{MP}\right) \times 100 SP=MP×(1−Discount %100)SP = MP \times \left(1 - \frac{\text{Discount } \%}{100}\right)
  • Single Equivalent Discount for Successive Discounts (d1%d_1\% and d2%d_2\%): Dequivalent=d1+d2−d1×d2100(%)D_{\text{equivalent}} = d_1 + d_2 - \frac{d_1 \times d_2}{100} \quad (\%)
    • Example: Two successive discounts of 20%20\% and 10%10\%: Deq=20+10−20×10100=30−2=28%D_{\text{eq}} = 20 + 10 - \frac{20 \times 10}{100} = 30 - 2 = 28\%
  • Linking CP, MP, Profit, and Discount: MPCP=100+Profit %100−Discount %\frac{MP}{CP} = \frac{100 + \text{Profit } \%}{100 - \text{Discount } \%}

4. Advanced Profit & Loss Theorems

Theorem 1: Equal Selling Price with Identical Profit and Loss Percentages

When two distinct articles are sold at the exact same Selling Price (SPSP), one at a profit of x%x\% and the other at a loss of x%x\%:

  • The overall transaction always results in a net loss.
  • The net loss percentage is given by: Net Loss %=(x10)2=x2100(%)\text{Net Loss } \% = \left(\frac{x}{10}\right)^2 = \frac{x^2}{100} \quad (\%)
  • Example: A merchant sells two laptops for ₹24,00024,000 each. On one he gains 20%20\% and on the other he loses 20%20\%. Net Loss %=202100=400100=4%\text{Net Loss } \% = \frac{20^2}{100} = \frac{400}{100} = 4\%

Theorem 2: Dishonest Dealer and False Weight Logic

When a dishonest trader claims to sell goods at cost price (CPCP) but uses a false weight smaller than the true weight:

Gain %=(ErrorTrue Value−Error)×100=(Claimed Weight−Actual WeightActual Weight)×100\text{Gain } \% = \left(\frac{\text{Error}}{\text{True Value} - \text{Error}}\right) \times 100 = \left(\frac{\text{Claimed Weight} - \text{Actual Weight}}{\text{Actual Weight}}\right) \times 100
  • Example: A shopkeeper claims to sell sugar at cost price but uses a 900 g900\text{ g} weight instead of a 1 kg (1000 g)1\text{ kg } (1000\text{ g}) weight.
    • Error=1000−900=100 g\text{Error} = 1000 - 900 = 100\text{ g}
    • Gain %=(100900)×100=19×100=11.11%=1119%\text{Gain } \% = \left(\frac{100}{900}\right) \times 100 = \frac{1}{9} \times 100 = 11.11\% = 11\frac{1}{9}\%.
Test Your Knowledge

Due to inflation, the price of petrol increases by 25%. By what percentage must a motorist reduce petrol consumption so that the total monthly expenditure on petrol remains unchanged?

A

25%

B

16.67%

C

20%

D

33.33%

Test Your Knowledge

A retail outlet advertises a clearance sale offering two successive discounts of 30% and 10% on all apparel. What is the single equivalent discount percentage?

A

40%

B

38%

C

35%

D

37%

Test Your Knowledge

A dishonest grocer professes to sell pulses at cost price but uses a fraudulent weight of 800 grams in place of a 1 kilogram standard measure. What is the grocer's true profit percentage?

A

25%

B

20%

C

15%

D

12.5%

Test Your Knowledge

A real estate investor sold two residential plots for ₹18,00,000 each. On the first plot, the investor realized a profit of 15%, while on the second plot, the investor incurred a loss of 15%. What was the overall percentage outcome of the combined sale?

A

No profit, no loss (0%)

B

Loss of 2.25%

C

Profit of 2.25%

D

Loss of 1.5%

Sections you finish are checked off in the contents.