- Staff Selection Commission Mathematics - Percentage (1999-2017) Part 5

Staff Selection Commission Mathematics - Percentage (1999-2017)


TYPE–XI (Contd.)





Ans .

7.00


  1. Explanation :

    Original price of article = x per kg.

    New price = \(\frac{79x}{100}\)per kg

    ∴ \(\frac{100}{\frac{79x}{100}}\) - \(\frac{79x}{100}\) = 3

    \(→\) 79x = 700 \(→\) x = \(\frac{700}{79}\)

    ∴ New price = \(\frac{79x}{100}\) = \(\frac{79}{100} \times \frac{700}{79}\)

    = 7 per kg







Ans .

Rs. 4


  1. Explanation :

    Let the original price of sugar be Rs. x per kg.

    Reduced price = Rs. \(\frac{80x}{100}\) = Rs. \(\frac{4x}{5}\)per kg.

    According to the question,

    \(\frac{160}{\frac{4x}{5}}\) - \(\frac{4x}{5}\) = 8

    \(→\) 8x = 40

    \(→\) x = \(\frac{40}{8}\) = 5 per kg.

    Reduced Price = Rs. \(\frac{4 \times 5}{5}\) = Rs. 4 per kg







Ans .

12% decrease


  1. Explanation :

    Required percentage change = 10 - 20 + \(\frac{10 \times -20}{100}\) % = -12%

    Negative sign shows decrease.







Ans .

\( 37 \frac{1}{2}\)%


  1. Explanation :

    Required per cent = \(\frac{x}{100 + x} \times 100\)

    where x = 60% = \(\frac{60}{160} \times 100\) = \(\frac{75}{2}\) = \( 37 \frac{1}{2}\)%







Ans .

12 kg.


  1. Explanation :

    Let original price of sugar be Rs. x per kg.

    New price = Rs.\(\frac{120x}{100}\) = Rs.\(\frac{6x}{5}\)per kg

    According to the question,

    \(\frac{50}{x}\) - \(\frac{50}{\frac{6x}{5}}\) = 2

    \(→\) 6x = 25

    \(→\) x = Rs \(\frac{25}{6}\)kg

    ∴ Required quantity of sugar = \(\frac{50}{x}\)kg

    = \(\frac{50}{\frac{25}{6}}\)kg = \(\frac{50 \times 6}{25}\) = 12kg.







Ans .

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


  1. Explanation :

    Required per cent = \(\frac{Decrease%}{100-Decrease%} \times 100\)

    = \(\frac{10}{100-10} \times 100\) = \( 11 \frac{1}{9}\)%







Ans .

25%


  1. Explanation :

    Required percentage increase = \(\frac{x}{100-x} \times 100\)

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







Ans .

Rs. 5 per kg


  1. Explanation :

    Original price of sugar = Rs. x/kg. (let)

    New price = Rs. \(\frac{120x}{100}\) per kg.

    = Rs. \(\frac{6x}{5}\) per kg.

    According to the question,

    \(\frac{120}{x}\) - \(\frac{120}{\frac{6x}{5}}\) = 4

    \(→\) x = Rs. 5 per kg.







Ans .

25%


  1. Explanation :

    Original price of building = Rs. 100 (let)

    ∴ Its price in 2001 = Rs. 80

    Its price in 2002 = Rs. 60

    Required percentage decrease = \(\frac{80-60}{80} \times 100\) = 25%




TYPE–XII





Ans .

1700


  1. Explanation :

    Percentage of boys = 100% – 70% = 30%

    Let total no. of students be x

    \ According to question,

    30% of x = 510

    x = \(\frac{510}{30} \times 100\) = 1700







Ans .

1458


  1. Explanation :

    40% of students = 972

    ∴ 60% of students = \(\frac{972}{40} \times 60\) = 1458







Ans .

1176


  1. Explanation :

    Number of boys = \(\frac{70}{30} \times 504\) = 1176







Ans .

95


  1. Explanation :

    Required sum = 0.5% of 19000 = \(19000 \times \frac{0.5}{100}\) = 95







Ans .

9 m


  1. Explanation :

    Remaining height = 192 - \(\frac{125}{2}\)% of 192

    = 192 – 120 = 72 m

    ∴ Required distance (distance covered in second hour) then,

    = \(\frac{25}{2}\)% of 72

    = \(\frac{25 \times 72}{2 \times 100}\) = 9m







Ans .

52 kgs.


  1. Explanation :

    Water in 100 kg fresh fruit = 68%

    Water in dry fruit = 20%

    Decrease = 48%

    ∴ Dry fruit obtained = 100 – 48 = 52 kg







Ans .

33%


  1. Explanation :

    The net tax rate = 30 + \(30 \times \frac{10}{100}\) % = 33%







Ans .

180


  1. Explanation :

    Let z have x

    ∴ Money with Y = \(\frac{3}{2}\)x and Money with X = 3x

    ∴ 3x + \(\frac{3x}{2}\) + x = \(3 \times 110\)

    \(→\) \(\frac{6x+3x+2x}{2}\) = 330

    \(→\) 11 x = 2 × 330

    x = \(\frac{2 \times 330}{11}\) = 60

    ∴ Money with X = 3x = (3×60) = 180







Ans .

\( 83 \frac{1}{3}\)%


  1. Explanation :

    If a number is x% more than other, then the other number is less than the first number by

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

    \ Required answer = \(\frac{500}{100+500} \times 100\) = \( 83 \frac{1}{3}\)%







Ans .

48.8%


  1. Explanation :

    Let x = 10 and y = 10

    ∴ x\(y^2\) = 10 × 10 × 10=1000 units

    Decreasing values of x and y by 20%,

    Expression = x\(y^2\)

    = 8 × 8 × 8

    = 512

    Decrease= 1000–512 = 488 units

    Percentage decrease = \(\frac{488}{1000} \times 100\) = 48.8%







Ans .

88%


  1. Explanation :

    If two numbers are respectively x% and y% more than a third number, the first as a per cent of

    second is = \(\frac{100+x}{100+y} \times 100\) = \(\frac{110}{125} \times 100\) = 88%







Ans .

140


  1. Explanation :

    Let sum of money be x.

    \(\frac{11}{2}\)% of x = 220

    \(→\) x = \(\frac{220 \times 200}{11}\) = 4000

    \( 3 \frac{1}{2}\)% of 4000 = \(\frac{7}{2} \times \frac{4000}{100}\) = 140







Ans .

3000


  1. Explanation :

    Let the total number of workers in the factory be x

    \(x \times \frac{60}{100} \times \frac{75}{100}\) = 1350

    \(→\) x = \(\frac{1350 \times 100^2}{60 \times 75}\) = 3000







Ans .

10%


  1. Explanation :

    Let the third number = 100.

    ∴ First number = 70

    Second number = 63

    ∴ Required per cent = \(\frac{70-63}{70} \times 100\) = 10%







Ans .

160%


  1. Explanation :

    Let Rani’s weight be x kg.

    ∴ Meena’s weight = 4x kg.

    Tara’s weight = \(\frac{5x}{4}\)kg

    ∴ Required percentage = \(\frac{4x}{\frac{5x}{2}} \times 100\) = 160%







Ans .

570


  1. Explanation :

    Number of people who have the saving habit = \(\frac{2500 \times 60}{100}\)

    ∴ Number of shareholders = (100 – 62)% of 1500

    = \(\frac{1500 \times 38}{100}\) = 570







Ans .

8000


  1. Explanation :

    Let the original price of the article be x.

    According to the question,

    5832 = \(x(1-\frac{1}{100})^3\)

    5832 = \(x(\frac{9}{10})^3\)

    x = 8000







Ans .

9%


  1. Explanation :

    Increase in AC = 6%

    ∴ Increased AC = \(\frac{106}{100} \times 3\) = 3.18 cm

    ∴ Decreased CB = 5 – 3.18 = 1.82 cm.

    ∴ Decrease = 2 – 1.82 = 0.18 cm

    ∴ Percentage decrease = \(\frac{0.18}{2} \times 100\) = 9%







Ans .

\( 91 \frac{2}{3}\)%


  1. Explanation :

    ∵ 24= 100% ∴22 = \(\frac{100}{24} \times 22\) = \( 91 \frac{2}{3}\)%







Ans .

40%


  1. Explanation :

    Let the number of boys be x and that of girls be y.

    Then, 71x + 73y = 71.8 (x + y)

    \(→\) 71.8x – 71x = 73y – 71.8y

    \(→\) 0.8x = 1.2y

    \(\frac{x}{y}\) = \(\frac{1.2}{0.8}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)

    ∴ \(\frac{x}{y}\) + 1 = \(\frac{3}{2}\) + 1

    \(\frac{x + y}{y}\) = \(\frac{5}{2}\)

    ∴ Percentage of girls = \(\frac{y}{x+y} \times 100\) = \(\frac{2}{5} \times 100\) = 40%







Ans .

50%


  1. Explanation :

    Let the number of books in shelf B be 100.

    ∴ Number of books in shelf A = 80

    On transferring 25% i.e \(\frac{1}{4}\) of books of shelf A to shelf B.

    B = 100 + 20 = 120

    Again, on transferring \(\frac{1}{4}\) of books of shelf B to shelf A.

    A = 60 + \(\frac{120}{4}\) = 90

    ∴ Required percentage = \(\frac{90}{180} \times 100\) = 50%







Ans .

178200


  1. Explanation :

    Total revenue earned = \(9900 \times \frac{20}{100} \times 10 \) + \(9900 \times \frac{80}{100} \times 20 \)

    = (19800 + 158400)

    = 178200







Ans .

\( 77 \frac{7}{9}\)


  1. Explanation :

    Let Tina’s weight = 1 kg

    Lina’s weight = 2 kg

    Neha’s weight = 1.4kg

    Mina’s weight = 1.8 kg.

    \(\frac{1.8x}{100}\) = 1.4

    \(→\) x = \( 77 \frac{7}{9}\)







Ans .

37.5%


  1. Explanation :

    Let the number of seats initially in the cinema hall be 100 and the cost of each ticket be

    100.

    ∴ Total revenue = 100 × 100 = 10000

    In second condition,

    Number of seats = 125

    Cost of each ticket = 110

    ∴ New revenue = 125 × 110 = 13750

    Increase in revenue collection = . (13750 – 10000) = 3750

    ∴ Percentage increase = \(\frac{3750}{10000} \times 100\) = 37.5%







Ans .

80


  1. Explanation :

    Total amount = x

    ∴ x - \(\frac{x}{5}\) - \(\frac{4x}{5} \times \frac{5}{100}\) - 120 = 1400

    \(\frac{19x}{25}\) = 1520

    \(→\) x = 2000

    ∴ Expenditure on transport = \(\frac{1}{25} \times 2000\) = 80







Ans .

30,000


  1. Explanation :

    Let the total number of employees be x.

    \(x \times \frac{69}{100}\) = 20700

    \(→\) x = \(\frac{20700 \times 100}{69}\) = 30,000







Ans .

700 apples


  1. Explanation :

    Let the fruit seller had originally x apples.

    According to the question;

    x – 40% of x = 420

    x - \(\frac{40}{100} \times x\) = 420

    \(\frac{3x}{5}\) = 420

    x = 700







Ans .

80%


  1. Explanation :

    Let third number = 100

    First number = 120

    Second number = 150

    Required percentage = \(\frac{120}{150} \times 100\) = 80%







Ans .

\( 45 \frac{5}{11}\)%


  1. Explanation :

    The batsman scored 3 × 4 + 8 × 6 = 60 runs by boundariesand sixes respectively.

    Then, Runs scored by running

    = 110 – 60 = 50

    \ Required percentage = \(\frac{50}{110} \times 100\) = \(\frac{500}{11}\) =\( 45 \frac{5}{11}\)%







Ans .

123.75


  1. Explanation :

    y = \(\frac{110}{100} \times 125\) = 137 5

    ∴ x = 90% of y

    = \(\frac{90 \times 137.5}{100}\) = 123.75







Ans .

2.5%


  1. Explanation :

    Error = 5.5 minutes

    ∴ Error per cent = \(\frac{5.5}{3 \times 60 + 40} \times 100\) = 2.5 per cent







Ans .

24


  1. Explanation :

    Per cent of families having either a cow or a buffalo or both = 60 + 30 – 15 = 75

    It means 25 per cent of families do not have a cow or a buffalo.

    ∴ Required number of families = 25% of 96 =\(96 \times \frac{25}{100}\) = 24







Ans .

3,850


  1. Explanation :

    Let the amount invested at 6% = x

    ∴ Amount invested at 5% = (10000 – x )

    According to the question

    \(\frac{(10000 - x) \times 5}{100}\) - \(\frac{x \times 6}{100}\) = 76.50

    \(→\) 50000 – 5x – 6x = 7650

    \(→\) 50000 – 11x = 7650

    \(→\) 11x = 50000 – 7650 = 42350

    x = 3850







Ans .

3


  1. Explanation :

    Kites of 20 are available for 19.

    Hence, discount = 5%

    If one gets kites of 20 for 18, discount = 10%

    ∴ Required answer

    20 kites \(→\) 2 kites

    27 kites \(→\) \(\frac{2}{20} \times 27\) = 3







Ans .

135


  1. Explanation :

    First part = x and second part = y.

    According to question

    \(\frac{x \times 80}{100}\) = \(\frac{y \times 60}{100}\) + 3

    \(→\) 4x – 3y = 15 ...(i)

    Again

    \(→\) 8y = 9x + 60

    \(→\) 8y – 9x = 60 ...(ii)

    By equation (i) × 8 + (ii) × 3,

    32x – 24y = 120

    24y – 27x = 180



    5x = 300 \(→\) x = 60



    From equation (i)

    4 × 60– 3y = 15

    \(→\) 3y = 240 – 15 = 225

    \(→\) y = \(\frac{225}{3}\) = 75

    ∴ x + y = 60 + 75 = 135







Ans .

\( 11 \frac{31}{54}\)%


  1. Explanation :

    Value of 100 stock = 108

    ∵ Income on investing 108 = \(\frac{25}{2}\)

    ∴ Income on investment of 27000

    = \(\frac{25}{2 \times 108} \times 27000\)

    = 3125

    ∴ Gain per cent = \(\frac{3125}{27000} \times 100\) = \( 11 \frac{31}{54}\)%







Ans .

4764


  1. Explanation :

    Required mass of lead = \(8000 \times \frac{60}{100} \times (1-\frac{3}{400})\)

    = \(8000 \times \frac{60}{100} \times (\frac{397}{400})\) = 4764 kg.







Ans .

\(\frac{2}{9}\)


  1. Explanation :

    According to the question \(x + y = (x^2 + y^2) \times \frac{1}{5} \)

    Again,

    \(x + y = (x^2 - y^2) \times \frac{1}{4} \)

    \(\frac{x^2 + y^2}{5} = \frac{x^2 - y^2}{4}\)

    \(→\)\( 5x^2 -4x^2 = 5y^2 + 4y^2\)

    \(→\)\( x^2 = 9y^2\)

    \(→\) x = 3y

    \(\frac{x + y}{x^2} = \frac{x^2 + y^2}{5x^2} = \frac{2}{9}\)







Ans .

200


  1. Explanation :

    Let the total number of eggs bought be x.

    10% of eggs are rotten.

    \ Remaining eggs = \(\frac{90x}{100}\) = \(\frac{9x}{10}\)

    After giving 80% of eggs to the neighbour,

    Remaining eggs = \(\frac{9x}{50}\)

    According to the question

    \(\frac{9x}{50}\) = 36

    \(→\) 9x = 36 × 50

    \(→\) x = 200







Ans .

Rs. 4,00,000


  1. Explanation :

    Amount with man in the

    beginning = Rs. x (let).

    Amount given to son and daughter = 80%.

    Remaining amount = 20% of x = Rs.\(\frac{x}{5}\)

    Remaining amount after donations to trust = \(\frac{x}{5} \times \frac{20}{100}\) = Rs. \(\frac{x}{25}\)

    ∴ \(\frac{x}{25}\) = 16000

    \(→\) x = 16000 × 25 = Rs. 400000







Ans .

Rs. 40000


  1. Explanation :

    Let the business man’s present earning be Rs. x.

    According to the question,

    \(x \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100}\) = 72000

    \(x \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4}\) = 72000

    \(x \times \frac{9}{5}\) = 100

    x = Rs. 40000







Ans .

48025


  1. Explanation :

    Number of blood cells in first 6 hours = \(40000 (1+\frac{10}{100})^2(1-\frac{10}{100})(1+\frac{5}{100})^2\) = 48024.9 = 48025







Ans .

\( 33 \frac{1}{3}\)%


  1. Explanation :

    Let 100 pairs of shoes be bought for Rs. 100.

    New budget = Rs. 160

    New price = Rs. 1.20 pair of shoes

    ∴ Number of shoes bought = \(\frac{160}{1.2}\) = \( 133 \frac{1}{3}\)

    ∴ Percentage increase = \( 33 \frac{1}{33}\)%







Ans .

84


  1. Explanation :

    Average of set A = \(\frac{27 + 28 + 30 + 32 + 33}{5}\) = \(\frac{150}{5}\) = 30

    Case II,

    New average = \(\frac{30 \times 130}{100}\) = 39

    ∴ 150 + k = 39 × 6 = 234

    \(→\) k = 234 – 150 = 84







Ans .

Rs. 128


  1. Explanation :

    Initial amount with the man = Rs. x (let).

    Remaining amount after first bet = Rs. \(\frac{x}{4}\)

    Remaining amount after second bet = Rs.\(\frac{x}{16}\)

    Remaining amount after third bet = Rs.\(\frac{x}{64}\)

    ∴ \(\frac{x}{64}\) = 2

    x = 128







Ans .

1000000


  1. Explanation :

    Initial number of soldiers in the army = x

    According to the question

    \(x \times (\frac{90}{100})^3\) = 729000

    \(→\) x = = 1000000







Ans .

16


  1. Explanation :

    Required percentage decrease = \(\frac{25-21}{25} \times 100\) = \(\frac{400}{25}\) = 16%







Ans .

= 11088


  1. Explanation :

    Required number of workers = \(8000 \times \frac{105}{100} \times \frac{110}{100} \times \frac{120}{100}\)

    = 11088







Ans .

2000


  1. Explanation :

    1% = 100 basis points

    ∴ 82.5% = 8250 basis points and 62.5% = 6250 basis points

    ∴ Required difference

    = 8250 – 6250

    = 2000 basis points







Ans .

\( 22 \frac{8}{11}\)%


  1. Explanation :

    Percentage increase in sales = \(\frac{51300 - 41800}{41800} \times 100\)

    = \(\frac{9500}{418}\) = \( 22 \frac{8}{11}\)%







Ans .

7


  1. Explanation :

    Defective parts of 120 machine parts = \(\frac{120 \times 5}{100}\) = 6

    Defective parts of 80 machine parts = \(\frac{80 \times 10}{100}\) = 8

    Total defective parts = 6 + 8 = 14

    ∴ Required percent = \(\frac{14}{200} \times 100\) = 7%







Ans .

\( 3 \frac{1}{3}\)%


  1. Explanation :

    Error = (1.55 – 1.5) metre = 0.55 metre

    ∴ Error per cent = \(\frac{0.05}{15} \times 100\) = \(\frac{50}{15}\) = \( 3 \frac{1}{3}\)%







Ans .

36500


  1. Explanation :

    Duty payment :

    Laptop \(→\) Rs \(\frac{210000 \times 10}{100}\) = Rs. 21000

    Mobile phone \(→\) Rs \(\frac{100000 \times 8}{100}\) = Rs. 8000

    Television set \(→\) Rs \(\frac{150000 \times 5}{100}\) = Rs. 7500

    Total Duty Payment

    = Rs. (21000 + 8000 + 7500)

    = Rs. 36500







Ans .

Rs. 1600


  1. Explanation :

    Initial amount with the person = Rs. x (let)

    After an expense of \(\frac{15}{2}\)%

    Remaining amount = \((100 - \frac{15}{2})% of x.\)

    = \((\frac{185}{2})% of x.\) = Rs. \(\frac{185x}{200}\)

    After an expense of 75% of it,

    Remaining amount = \(\frac{37x}{40 \times 4}\) = \(\frac{37x}{160}\)

    According to the question,

    \(\frac{37x}{160}\) = 370

    \(→\) 37x = 370 × 160

    \(→\) x = Rs. 1600







Ans .

60


  1. Explanation :

    Let the number of matches played between India and Pakistan in the first case be x.

    Number of wins by Pakistan = \(\frac{60x}{100}\) = \(\frac{3x}{5}\)

    \(\frac{\frac{3x}{5}}{x+30}\) = \(\frac{30}{100}\)

    \(→\) 2x = x + 30

    \(→\) x = 30

    ∴ Total number of matches = 30 + 30 = 60