- Staff Selection Commission Mathematics - Percentage (1999-2017) Part 4

Staff Selection Commission Mathematics - Percentage (1999-2017)


TYPE–VII (Contd.)





Ans .

1218


  1. Explanation :

    Percentage of boys = 60%

    ∴ Percentage of girls = 40%

    Boys : Girls = 60 : 40 = 3 : 2

    Number of girls = 812

    ∴ Number of boys = \(\frac{3}{2} \times 812\) = 1218







Ans .

75.5%


  1. Explanation :

    Marks scored by A :

    First subject \(→\) \(\frac{900 \times 72}{100}\) = 648

    Second subject \(→\) \(\frac{700 \times 80}{100}\) = 560

    Total marks scored = 648 + 560 = 1208

    Total maximum marks = 900 + 700 = 1600

    \ Required per cent = \(\frac{1208}{1600} \times 100\) = 75.5%







Ans .

40%


  1. Explanation :

    Percentage of failures either in 1 subject or both subjects = (35 + 45 – 20)% = 60%

    ∴ Percentage of the successful = (100– 60)% = 40%







Ans .

80%


  1. Explanation :

    Total marks of 50 students = 50 × 70 = 3500

    Total marks of 25 students = 25 × 60 = 1500

    Total marks of 24 students = 24 × 80 = 1920

    ∴ Marks obtained by last student = 3500 – 1500 – 1920

    = 80 i.e, 80%







Ans .

42 and 33


  1. Explanation :

    Let marks obtained by the first student be x.

    ∴ Marks obtained by the second student = x – 9

    According to the question,

    x = 56% of (x + x – 9)

    x = \(\frac{(2x - 9) \times 56}{100}\)

    \(→\) 100x = 112x – 504

    \(→\) 112x – 100x = 504

    \(→\) 12x = 504

    \(→\) x = \(\frac{504}{12}\) = 42

    ∴ Marks obtained by the second student = 42 – 9 = 33







Ans .

360 marks


  1. Explanation :

    Maximum marks of examination = x (let)

    According to the question,

    25% of x = 47 + 43

    \(\frac{x \times 25}{100}\) = 90

    \(\frac{x}{4}\) = 90

    \(→\) x = 4 × 90 = 360




TYPE–VIII





Ans .

4% decreased


  1. Explanation :

    Change in his salary = 20 - 20 - \(\frac{20 \times 20}{100}\) %

    = -\(\frac{400}{100}\)% = -4%

    i.e. 4% decrease







Ans .

8% increase


  1. Explanation :

    Net % change = A + B + \(\frac{AB}{100}\) %

    Here, A = 20%, B = – 10%

    ∴ Net % change

    = 20 - 10 - \(\frac{200}{100}\)

    = 10 – 2= 8%

    + ve sign shows increase







Ans .

1% decrease


  1. Explanation :

    Required change = \(\frac{(10)^2}{100}\) %decrease

    = 1% decrease







Ans .

19%


  1. Explanation :

    A single equivalent reduction to reduction series of x%, y%

    = x + y - \(\frac{xy}{100}\) %

    = 10 + 10 - \(\frac{10 \times 10}{100}\) %

    = 20 - 1 % = 19%







Ans .

10% decrease


  1. Explanation :

    The net change in price = -25 +20 - \(\frac{-25 \times 20}{100}\) % = (–25 + 20 – 5)% = – 10% Negative sign shows decrease.







Ans .

\( 2 \frac{1}{2}\)% decrease


  1. Explanation :

    Let the price of the article be 100 and the daily sale be 100

    units.

    ∴ Revenue day = 100 × 100 = 10000

    New receipts = 75 × 130 = 9750

    Decrease = (10000 – 9750) = 250

    ∴ % decrease = \(\frac{2500}{10000} \times 100\) = \( 2 \frac{1}{2}\)% decrease







Ans .

21%


  1. Explanation :

    Single equivalent percentage increase in price = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%







Ans .

25


  1. Explanation :

    Effective increase percentage

    = 10 + 20 + \(\frac{10 \times 20}{100}\) % = 32%

    \(\frac{x \times 132}{100}\) = 33

    \(→\) x = \(\frac{33 \times 100}{132}\) = 25







Ans .

44%


  1. Explanation :

    Effective percentage increase = 20 + 20 + \(\frac{20 \times 20}{100}\) % = 44%







Ans .

1% decrease


  1. Explanation :

    Net change = 10 - 10 - \(\frac{10 \times 10}{100}\) %

    = –1% = 1% decrease







Ans .

32%


  1. Explanation :

    Required percentage increase = 10 + 20 + \(\frac{10 \times 20}{100}\) % = 32%







Ans .

44% increase


  1. Explanation :

    Required effect = 80 - 20 - \(\frac{80 \times 20}{100}\) % = 44%

    Positive sign shows increase.







Ans .

decreased by 1%


  1. Explanation :

    Net effect = 10 - 10 - \(\frac{10 \times 10}{100}\) % = -1%

    Negative sign shows decrease







Ans .

10% decrease


  1. Explanation :

    Required per cent effect = 20 - 25 - \(\frac{20 \times 25}{100}\) % = -10%







Ans .

21


  1. Explanation :

    Percentage effect = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%







Ans .

\(\frac{8}{15}\)


  1. Explanation :

    Original fraction = \(\frac{x}{y}\)

    ∴ \(\frac{\frac{120x}{100}x}{y \times \frac{80}{100}}\) = \(\frac{4}{5}\)

    ∴ \(\frac{120x}{80y}\) = \(\frac{4}{5}\)

    ∴ \(\frac{6x}{4y}\) = \(\frac{4}{5}\)

    ∴ \(\frac{x}{y}\) = \(\frac{8}{15}\)







Ans .

\(\frac{95}{48}\)


  1. Explanation :

    Let original fraction be

    \(\frac{x}{y}\)

    According to the question,

    \(\frac{\frac{120}{100}x}{\frac{95y}{100}}\) = \frac{5}{2}

    \(\frac{120x}{95y}\) = \frac{5}{2}

    \(\frac{x}{y}\) = \frac{95}{48}







Ans .

decreased by \( 6 \frac{1}{4}\)%


  1. Explanation :

    Effective value = x + y + \(\frac{x \times y}{100}\) %

    = 25 - 25 - \(\frac{25 \times 25}{100}\) % = decreased by \( 6 \frac{1}{4}\)%







Ans .

240


  1. Explanation :

    Larger number = x and smaller

    number = 520 – x

    \(\frac{96x}{100}\) = \(\frac{520-x}{100} \times 112\)

    \(→\) 96x = 520 × 112 – 112x

    \(→\) 112x + 96x = 520 × 112

    \(→\) 208x = 520 × 112

    \(→\) x = \(\frac{520 \times 112}{208}\) = 280

    ∴ Smaller number

    = 520 – 280 = 240







Ans .

400


  1. Explanation :

    If the original price of article be x, then

    \(x \times \frac{80}{100}\times \frac{130}{100}\) = 416

    \(→\) x = \(\frac{416 \times 100\times 100}{80 \times 130}\) = 400







Ans .

80


  1. Explanation :

    If the number be x, then \(x \times \frac{245}{200}\) = 98

    \(→\) x = \(\frac{98 \times 200}{245}\) = 80







Ans .

\( 11 \frac{1}{9}\)%


  1. Explanation :

    Let the original price be 100

    New price after 10% decrease = 90

    In order to restore the price to its original value,

    it must be increased by 10% increase = \(\frac{10}{90} \times 100\) = \(\frac{100}{9}\) = \( 11 \frac{1}{9}\)%







Ans .

25%


  1. Explanation :

    Clearly, 75% of the number = 225

    ∴ Number = \(\frac{225 \times 100}{75}\) = 300

    Again, 125% of 300 = 375

    Hence, the number should be increased by 25%







Ans .

\( 30 \frac{5}{9}\)%


  1. Explanation :

    Let the number be 100.

    After 20% increase, number = 120

    After 20% increase of 120, number = \(120 \times \frac{120}{100}\) = 144

    ∴ Per cent decrease

    = \(\frac{44}{144} \times 100\)

    = \(\frac{275}{9}\) = \( 30 \frac{5}{9}\)%







Ans .

\(\frac{25}{4}\)%


  1. Explanation :

    Let the original number of employees be 100 and wages per

    head be 100.

    Total wages = (100 × 100) = 10000

    New number of employees = 125

    New wages per head = 75

    Total new wages = (125 × 75) = 9375

    Decrease = (10000 – 9375) = 625

    ∴ Percentage decrease = \(\frac{625}{10000} \times 100\)

    = \(\frac{625}{100}\) = \(\frac{25}{4}\)%







Ans .

5000


  1. Explanation :

    Let original number be x

    \(\frac{90}{100}x \times \frac{110}{100}\) = x - 50

    \(\frac{99x}{100}\) = x-50

    x - \(\frac{99x}{100}\) = 50

    \(\frac{x}{100}\) = 50

    \(→\) x = 5000







Ans .

5%


  1. Explanation :



    Let the income be x and the rate of income tax be y %

    According to the question,

    \(\frac{xy \times 1.19}{100}\) - \(\frac{xy}{100}\) = \((x - \frac{xy}{100}) \times \frac{1}{100}\)

    \(→\) 1.19 xy – xy = - \(\frac{xy}{100}\)

    \(→\) 0.19 y = - \(\frac{y}{100}\)

    \(→\) \(\frac{y}{100}\) + 0.19y = 1 \(→\) y\(\frac{1+19}{100}\) = 1

    \(→\) = \(\frac{100}{20}\) = 5%







Ans .

50%


  1. Explanation :

    Man’s income = 100 (let)

    Expenditure = 75

    Savings = 25

    New income = \(\frac{100 \times 120}{100}\) = 120

    New expenditure = \(\frac{75 \times 110}{100}\) = 82.5

    Savings = 120 – 82.5 = 37.5

    Increase in savings = 37.5 – 25 = 12.5

    \ Increase per cent = \(\frac{12.5}{25} \times 100\) = 50%







Ans .

33.1%


  1. Explanation :

    Single equivalent increase for 10% and 10% = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%

    Again, single equivalent increase for 21% and 10% = 21 + 10 + \(\frac{21 \times 10}{100}\) % = 33.1%







Ans .

increased by 8.9%


  1. Explanation :

    Increase in first year = 10%

    Decrease in 2nd year = 10%

    Effective result = 10 - 10 - \(\frac{10 \times 10}{100}\) % = –1%

    Increase in 3rd year = 10%

    ∴ Effective result = 10 - 1 - \(\frac{10 \times 1}{100}\)%

    = (9 – 0.1)% = 8.9% (increase)







Ans .

80


  1. Explanation :

    Let the number be x.

    ∴ (20 + 25)% of x = 36

    \(\frac{45x}{100}\) = 36

    \(→\) x = \(\frac{36 \times 100}{45}\) = 80







Ans .

500


  1. Explanation :

    Effective percentage = -20 + 20 - \(\frac{20 \times 20}{100}\) % = -4%

    If the number be x, then 4% of x = 20

    \(x \times \frac{4}{100}\) = 20

    x = \(\frac{20 \times 100}{4}\) = 500







Ans .

\(\frac{100x}{100+x}\)%


  1. Explanation :

    \(\frac{B \times 30}{100}\) \(→\) increased value \(→\) P + \(\frac{Px}{100}\)

    = P\(\frac{100+x}{100}\)

    ∴ Required answer =\(\frac{x}{100+x} \times 100\)







Ans .

28%


  1. Explanation :

    Effective percentage decrease = x + y + \(\frac{x \times y}{100}\) %

    = - 10 - 20 + \(\frac{-10 \times -20}{100}\) %

    = (– 30 + 2)% = –28%







Ans .

0


  1. Explanation :

    Cost of edible oil = 100 per kg.

    Consumption = 1 kg.

    Again, New price = 125 per kg.

    Consumption = 0.8 kg.

    Expenditure = Rs. (125 × 0.8) = Rs. 100

    Percentage effect = 25 - 20 - \(\frac{20 \times 25}{100}\) % = 0%




TYPE - IX





Ans .

27500


  1. Explanation :

    Let the total number of votes be 100.

    Number of uncast votes = 8

    ∴ Number of votes polled = 92

    Number of votes obtained by the winner = 48

    ∴ Number of votes obtained by the loser = 92 – 48 = 44

    If the difference of win be 4 votes, total voters = 100

    ∴ When the difference be 1100 votes, total voters

    = \(\frac{100}{4} \times 1100\) = 27500







Ans .

16800


  1. Explanation :

    Let the total number of voters enrolled be x.

    Number of votes polled = = 75% of x = \(\frac{3x}{4}\)

    Number of valid votes = \(\frac{3x}{4}\) - \(\frac{2}{100} \times \frac{3x}{4}\)

    = \(\frac{3x}{4}\) - \(\frac{2}{100} \times \frac{3x}{4}\)

    = \(\frac{147x}{200}\)



    Now, 75% of = \(\frac{147x}{200}\) = 9261

    or \(\frac{3}{4}\) of \(\frac{147x}{200}\) = 9261

    or x = \(\frac{9261 \times 4 \times 200}{3 \times 147}\) = 16800







Ans .

42,000


  1. Explanation :

    Difference of percentage of

    votes = 60% – 40% = 20%

    ∴ 20% of total votes = 14000

    ∴ 60% of total votes

    = \(\frac{14000}{20} \times 60\) = 42000







Ans .

52%


  1. Explanation :

    Let total employees = 100

    ∴ Required percentage = \(\frac{40 \times 40}{100}\) + \(\frac{60 \times 60}{100}\)

    = 16 + 36 = 52%







Ans .

1490


  1. Explanation :

    Let votes polled = x

    \(x \times \frac{60-40}{100}\) = 298

    \(→\) x ×\(\frac{1}{5}\) = 298

    \(→\) x = 298 × 5 = 1490







Ans .

56056


  1. Explanation :

    Number of valid votes = \(104000 \times \frac{98}{100}\) = 101920

    ∴ Valid votes received by the candidate = \(\frac{101920 \times 55}{100}\) = 56056







Ans .

8000


  1. Explanation :

    If the number of votes polled be x, then \(\frac{x \times 20}{100}\) = 1600

    \(→\) x = \(\frac{1600 \times 100}{20}\) = 8000







Ans .

8640


  1. Explanation :

    Vote percentage of third candidate

    = 100 – 40 – 36 = 24%

    ∴ Votes got by third candidate = \(\frac{36000 \times 24}{100}\) = 8640







Ans .

3,00,000


  1. Explanation :

    Total votes polled = x

    ∴ (57 – 43)% of x = 42000

    \(x \times \frac{14}{100}\) = 42000

    x = \(\frac{42000 \times 100}{14}\) = 300000







Ans .

700


  1. Explanation :

    Total number of votes polled = x

    \(\frac{x \times 84}{100}\) - \(\frac{x \times 16}{100}\) = 476

    \(\frac{68x}{100}\) = 476

    \(→\) x = \(\frac{476 \times 100}{68}\) = 700







Ans .

30000


  1. Explanation :

    Number of valid votes = x (let)

    \ (62 – 38)% of x = 7200

    \(x \times \frac{24}{100}\) = 7200

    x = \(\frac{7200 \times 100}{24}\) = 30000







Ans .

600


  1. Explanation :

    Total number of votes polled = x (let)

    According to the question

    \(\frac{x \times 62}{100}\) - \(\frac{x \times (100 - 62)}{100}\) = 144

    \(\frac{x \times 62}{100}\) - \(\frac{x \times 38}{100}\) = 144

    x = 600







Ans .

6200


  1. Explanation :

    Total voters in the list = x

    Votes got by the winner = \(\frac{47x}{100}\)

    Votes got by the loser = x – \(\frac{x}{10}\) - 60 - \(\frac{47x}{100}\)

    = \(\frac{9x}{10}\) - 60 - \(\frac{47x}{100}\)

    = \(\frac{43x}{100}\) - 60



    According to the question,

    \(\frac{47x}{100}\) - \(\frac{43x}{100}\) + 60 = 308

    x = 6200







Ans .

1490


  1. Explanation :

    Total votes polled = x

    According to the question,

    (60 – 40)% of x = 298

    \(→\) \(x \times \frac{20}{100}\) = 298

    \(→\) \(\frac{x}{5}\) = 298

    \(→\) x = 298 × 5 = 1490




TYPE–X





Ans .

217800


  1. Explanation :

    Required population after two years = = 180000 \((1+ \frac{10}{100})^2\)

    = \(180000 \times \frac{11}{10}\times \frac{11}{100}\) = 217800







Ans .

48.8%


  1. Explanation :

    If the present worth of the equipment be 100, then its price after 3 years

    = \(100 \times (\frac{80}{100})^3\) = 51.2

    ∴ Depriciation = 48.8%







Ans .

68921


  1. Explanation :

    P = P0 \((1 + \frac{R}{100})^T\)

    P = 64000 \((1 + \frac{5}{200})^3\)

    P = 64000 \(( \frac{41}{40})^3\)

    P = \(\frac{64000 \times 41 \times 41 \times 41}{40 \times 40 \times 40}\)

    P = 68921







Ans .

10000


  1. Explanation :

    Suppose the value of property two years ago was x

    According to question

    ∴ x\((1-\frac{10}{100})^2\) = 8100

    ∴ x\((\frac{90}{100})^2\) = 8100

    \(→\) x = 10000







Ans .

56,700


  1. Explanation :

    Let the present population be P.

    ∴ P = 62500\((1 - \frac{4}{100})^2\)

    P = 57600







Ans .

54,080


  1. Explanation :

    Required population = \(( 1 + \frac{4}{100})^2\) = 54,080







Ans .

Let the man’s annual salary in 2006 be x.

∴ \(\frac{110x}{100}\) = 880000

x = = 800000


  1. Explanation :







Ans .

62500


  1. Explanation :

    Population of the village two years ago = \(\frac{P}{(1+\frac{R}{100})^2}\)

    = \(\frac{67600}{(1+\frac{4}{100})^2}\)

    = 62500







Ans .

1,80,500


  1. Explanation :

    A = \(P (1-\frac{R}{100})^T\)

    A = \(200000 (1-\frac{5}{100})^2\)

    = 1,80,500







Ans .

4,80,000


  1. Explanation :

    Value of the property 3 years ago

    = \(\frac{P}{(1-\frac{R}{100})^T}\)

    = \(\frac{411540}{(1-\frac{5}{100})^3}\)

    = 480000







Ans .

85184


  1. Explanation :

    Population of town = \(P (1-\frac{R}{100})^T\)

    = \(P (1-\frac{R}{100})^T\)

    = \(64000 (1-\frac{10}{100})^3\)

    = 85184







Ans .

6400


  1. Explanation :

    Present population = \(10000 (1-\frac{20}{100})^2\) = 6400







Ans .

5


  1. Explanation :

    Required percent = \(\frac{1}{4} \times 3\) + \(\frac{2}{3} \times 5\) + \( (1-\frac{1}{4}-\frac{2}{3}) \times 11\) = 5%







Ans .

3,150


  1. Explanation :

    Required price of the machine = = \(6250 (1-\frac{10}{100})(1-\frac{20}{100})(1-\frac{30}{100})\) = 3150







Ans .

40,500


  1. Explanation :

    Required value = \(50000 (1-\frac{10}{100})^2\) = 40,500







Ans .

1000


  1. Explanation :

    If the price of machine 3years ago be x. then 729 = \(x (1-\frac{10}{100})^3\) 729 = \(x \times (\frac{9}{10})^3\) \(→\) x = 1000







Ans .

1720


  1. Explanation :

    Required Raman’s salary = \(\frac{100}{100+5} \times 1806\) = \(\frac{100}{105} \times 1806\) = 1720







Ans .

4000


  1. Explanation :

    ∴ Men : Women = 1 : 1

    ∴ Number of men = \(\frac{1}{2} \times 8000\) = 4000







Ans .

5600


  1. Explanation :

    Let the number of males = x

    \ Number of females= 9800 – x

    According to the question,

    \(x \times \frac{108}{100}\) + \((9800-x) \times \frac{105}{100}\) = 10858

    \(→\) 108 x + 9800 ×105 – 105x

    = 1045800

    \(→\) 3x + 1029000 = 1045800

    \(→\) 3x = 1045800 – 1029000

    = 16800

    \(→\) = \(\frac{16800}{3}\) = 5600







Ans .

5120


  1. Explanation :

    Population in the beginning of the year = \(\frac{10000}{ (1+\frac{25}{100})^3}\) = 5120







Ans .

111%


  1. Explanation :

    If the number of men be 100, then Number of women = 90

    ∴ Required per cent = \(\frac{100}{90} \times 100\) = 111%







Ans .

562432


  1. Explanation :

    Required population = \(500000 (1+\frac{4}{100})^3\) = 562432







Ans .

4000


  1. Explanation :

    If the population of village two years ago be P0 , then 4410 = \(P0 (1+\frac{1}{20})^2\)

    P0 = 4000







Ans .

19,950


  1. Explanation :

    Value of TV after one year = 21000 × (100 – 5)% = \(\frac{21000 \times 95}{100}\) = = Rs. 19950







Ans .

72.8 %


  1. Explanation :

    Single equivalent increase for 20% and 20% = 20 + 20 + \(\frac{20 \times 20}{100}\) % = 44%

    Single equivalent increase for 44% and 20% = 44 + 20 + \(\frac{44 \times 20}{100}\) % = = 72.8%







Ans .

32.5


  1. Explanation :

    Population of town = 1000

    Males \(→\) 600

    Females \(→\) 400

    Literate males = \(\frac{600 \times 20}{100}\) = 20

    Total literate inhabitants = \(\frac{1000 \times 25}{100}\) = 250

    ∴ Literate females = 250 – 120 = 130

    ∴ Required percent = \(\frac{130}{400} \times 100\) = 32.5%







Ans .

10%


  1. Explanation :

    If the rate of increase per annum be R%, then

    A = \(P (1-\frac{R}{100})^T\)

    48400 = \(40000 (1+\frac{R}{100})^2\)

    R = 10% per annum







Ans .

Rs. 10,000


  1. Explanation :

    A = \(P (1-\frac{R}{100})^3\)

    7290 = \(P (1-\frac{10}{100})^3\)

    P = Rs. 10000







Ans .

8000


  1. Explanation :

    9261 = \(P0 (1+\frac{5}{100})^3\)

    P0 = 8000







Ans .

6125


  1. Explanation :

    Original population of village = x (let)

    According to the question,

    \(x \times \frac{95}{100} \times \frac{80}{100}\) = 4655

    x = 6125







Ans .

9600


  1. Explanation :

    In the village,

    Females = 3000

    Males = 9000 – 3000 = 6000

    After respective increases,

    Population of village = \(3000 \times \frac{105}{100}\) + \(\frac{6000 \times 107.5}{100}\)

    = 3150 + 6450 = 9600







Ans .

60


  1. Explanation :

    Let the population of the city be 100.

    Total illiterate people = 40

    Poor people = 60

    Rich people = 40

    Illiterate rich people = \(\frac{40 \times 10}{100}\) = 4

    ∴ Illiterate poor people = 40 – 4 = 36

    ∴ Required per cent = \(\frac{36}{60} \times 100\)% = 60%







Ans .

31200


  1. Explanation :

    Population of city after two years = \(P (1+\frac{R1}{100})(1+\frac{R1}{100})\) = \(20000 (1+\frac{20}{100})(1+\frac{30}{100})\) = \(20000 \times \frac{120}{100} \times \frac{130}{100}\) = 31200







Ans .

4100


  1. Explanation :

    Let the total population of village be x.

    According to the question,

    \(\frac{x \times 14}{100}\) = 574

    \(→\) x = \(\frac{574 \times 100}{14}\) = 4100




TYPE–XI





Ans .

13.50


  1. Explanation :

    10 per cent of 837 = \(\frac{10}{100} \times 837\) = 83.7

    ∴ Reduced per kg price = \(\frac{83.7}{62}\) = 13.50







Ans .

18


  1. Explanation :

    Let the original price of sugar = x /kg.

    Reduced price of sugar = 80% of x

    = \(\frac{x \times 80}{100}\) = \(\frac{4 \times x}{5}\)

    ∴ \(\frac{36}{\frac{4x}{5}}\) - \(\frac{36}{x}\) = \(\frac{1}{2}\)

    \(→\) x = 9 × 2 = 18/Kg







Ans .

18


  1. Explanation :

    Reduced price of 6.2kg of sugar

    = 10% of 1116

    = 111.6

    ∴ Reduced price per kg = \(\frac{111.6}{62}\) = 18







Ans .

30


  1. Explanation :

    Let the original price of sugar be x/kg.

    ∴ New price = \(\frac{9x}{10}\)

    \(\frac{270}{\frac{9x}{10}}\) - \(\frac{270}{x}\) = 1

    \(→\) x = 30 /kg







Ans .

12.96


  1. Explanation :

    Let the original price of apples be x/ dozen

    ∴ New price = \(\frac{4x}{5}\) /dozen

    \(\frac{54}{\frac{4x}{5}}\) - \(\frac{54}{x}\) = \(\frac{10}{2}\)

    \(\frac{4x}{5}\) = 12.96







Ans .

36


  1. Explanation :

    The original price of 1 egg = x

    Present price = \(\frac{3}{2}x\)

    \(\frac{24}{x}\) - \(\frac{24}{\frac{3x}{2}}\) = 4

    \(→\) x = 2

    ∴ Present price of eggs per dozen = \(12 \times \frac{3}{2} \times 2\) = 36







Ans .

16


  1. Explanation :

    Original price of wheat = x /kg.

    New price of wheat = \(\frac{4x}{5}\)

    \(\frac{320}{\frac{4x}{5}}\) - \(\frac{320}{x}\) = 5

    \(→\) x = 16







Ans .

28.80


  1. Explanation :

    Original rate = x per egg

    New rate = \(\frac{6x}{5}\) per egg

    \(\frac{24}{x}\) - \(\frac{24 \times 5}{6x}\) = 2

    x = 2

    ∴ New rate = \(\frac{12}{5}\) per egg.

    ∴ Rate per dozen of eggs = \(\frac{12}{5} \times 12\)= 28.80







Ans .

15


  1. Explanation :

    Let original price of rice per kg = x (let)

    ∴ New price of rice per kg = \(\frac{3x}{4}\)

    \(\frac{600}{\frac{3x}{4}}\) - \(\frac{600}{x}\) = 10

    \(→\) x = 20

    ∴ New price = \(\frac{3 \times 20}{4}\) = 15/kg







Ans .

20%


  1. Explanation :

    Percentage decrease = \(\frac{0.25}{1.25} \times 100\) = 20%







Ans .

30


  1. Explanation :

    If the number be x, then x – 15 = \(\frac{4x}{5}\)

    \(→\) 5x – 75 = 4x \(→\) x = 75

    ∴ 40% of 75 = \(\frac{75 \times 40}{100}\) = 30