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RPF Constable 2023 Aptitude Test - 43
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RPF Constable 2023 Aptitude Test - 43
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  • Question 1/10
    1 / -0.33

    A number is increased by 10% and then the increased number is decreased by 20%. The net increase or decrease is:

    Solutions

    Given:

    Increment in number = 10% of the number

    Decrement in number = 20% of the new number

    Concept used:

    Net increase or decrease percent = {(Original number or Final number) - (Final number or original number) ÷ Original number} × 100

    Calculation:

    Let the original number be 100

    When the number is increased by 10%

    New Number = 100 + 10% of 100

    ⇒ 100 + 10

    ⇒ 110

    Now, the number is decreased by 20%

    Final number = New number – 20% of new number

    ⇒ 110 – 20% of 110

    ⇒ 110 – 22

    ⇒ 88

    ∴ Net Decrease in percentage = { (100 – 88) / 100} × 100

    ⇒ {12/100} × 100

    ⇒ 12%

    ∴ The net decrease percentage is 12%.

  • Question 2/10
    1 / -0.33

    The ratio of new expenditure to that of the initial expenditure is 15 : 16. If the price of petrol is increased by 25% then the consumption of petrol is decreased by x%. Find the value of x.

    Solutions

    Shortcut Trick

    Formula used: (It is only valid for this type of question)

    Where,

    +x% is the percentage increased in price

    -x% is the percentage decreased in price

    +y% is the percentage increased in consumption

    -y% is the percentage decreased in consumption 

    According to the question,

    The percentage increase in price is 25%

    The percentage decreased in consumption is x%

    Therefore,

    Therefore, the consumption of petrol is decreased by 25%.

    The required answer is 25%.

    Alternate Method

    Given: 

    The ratio of new expenditure to that of the initial expenditure is 15 : 16. 

    If the price of petrol is increased by 25% then the consumption of petrol is decreased by x%.

    Concept used:

    Expenditure = Price × Consumption

    Calculation:

    Let the initial price of petrol = ₹100

    Let the initial consumption of petrol = 100 L

    Therefore, 

    The initial expenditure = ₹100 × 100 = ₹10,000

    In the second case:

    The increased price of the petrol = (100% + 25%) of ₹100 = ₹125

    The decreased consumption of the petrol = (100% - x%) of 100 L

    The new expenditure = ₹125 × (100% - x%) of 100 L

    According to the question,

    ⇒ 3 = 4(1 - x%)

    ⇒ 4x% = 1

    ⇒ x = 25%

    Therefore, the consumption of petrol is decreased by 25%.

    ∴ The required answer is 25%.

  • Question 3/10
    1 / -0.33

    A lady buys some product at Rs.5400, 2/3 of them sold at 8% profit at what profit did she sell remaining articles that she gain 18% on whole.

    Solutions

    Given:

    C.P = Rs. 5400

    C.P = Cost Price

    Calculation:

    According to the question,

    ⇒ (3600 × 8%) + (1800 × x%) = (5400 × 18%)

    ⇒ 288 + 18x = 54 × 18

    ⇒ 288 + 18x = 972

    ⇒ 18x = 972 - 288

    ⇒ x = 684/18

    ⇒ x = 38

     The correct answer is 38%.

  • Question 4/10
    1 / -0.33

    Find the least value of (A + B) if the number 27B58A4 is completely divisible by 88.

    Solutions

    Given:

    Number = 27B58A4

    The number is divisible by 88

    Concept Used:

    The divisibility rule of 8 is the last 3-digit number must divisible by 8

    The divisibility rule of 11 is the difference between a sum of alternative numbers that must be divisible by 11 or can be 0.

    Calculations:

    According to the question,

    ⇒ Number = 27B58A4

    ⇒ last 3-digit number = 8A4

    ⇒ when we put A = 2

    ⇒ then the last 3-digit becomes = 824, which is divisible by 8

    ⇒ and when we put A = 6

    ⇒ then the last 3-digit becomes = 864, which is also divisible by 8

    But the least number is 2

    ⇒ So, The value of A = 2

    ⇒ Now, The number becomes = 27B5824

    ⇒ Sum of numbers in even place = 7 + 5 + 2 = 14

    ⇒ Sum of numbers in odd places = 2 + B + 8 + 4 = 14 + B

    ⇒ Difference = 14 + B - 14 = B

    here B must be divisible by 11

    ⇒ so, 0 is the only number that can divisible by 11

    ⇒ So, the Value of B = 0

    ⇒ now, Value of A + B = 2 + 0 = 2

    ⇒ Thus, The least value of A + B is 2.

    Hence, the correct answer is option 2).

  • Question 5/10
    1 / -0.33

    A number is such that the product of its digit is 12. When 36 is added to the number, the digits get reversed. What is the number?

    Solutions

    Given:

    Product of digits = 12      ---- (1)

    Calculation:

    Let the number be ab.

    According to question

    (10a + b) + 36 = 10b + a

    ⇒ 9b - 9a = 36

    ⇒ b - a = 4      ---- (1)

    Squaring both sides, we get

    b2 + a2 - 2ab = 16

    ⇒ b2 + a2 - 2ab + 4ab = 16 + 4ab

    ⇒ (b + a)2 = 16 + 4 × 12

    ⇒ (b + a)2 = 64

    ⇒ b + a = 8 ---- (2)

    Adding (1) and (2), we get

    2b = 12

    ⇒ b = 6

    Now, a = 8 - b = 8 - 6 = 2

    ∴ Required number = 26

    Shortcut Trick

    By option method

    As we know that in question it is given that product of digit of number = 12 

    According to this given part all option are correct

    Now go with other given part

    (I) - 34

    34 + 36 = 70 not equal to 43 (reverse digit)

    (ii) 43

    43 + 36 = 79 not equal to 34 (reverse digit)

    (iii) 62

    62 + 36 = 98 not equal to 26 (reverse digit)

    (iv) 26

    26 + 36 = 62 = 26 (reverse digit)

  • Question 6/10
    1 / -0.33

    Two tourist buses start from the same point and move along two roads at right angles at speeds of 48 km/h and 36 km/h, respectively. The distance between the buses after 15 seconds is ________.

    Solutions

    Shortcut Trick

    Alternate Method

    Given:

    The two road is at right angle 

    Speed of bus A = 48 km/h

    Speed of bus B = 36 km/h

    Formula Used:

    Speed = Distance / Time

    By Pythagoras theorem

    Hypotenuse2 = Base2 + Height2

    Calculations:

    ⇒ Speed of Bus A = 48 km/h = 40/3 m/s

    ⇒ Speed of bus B = 36 km/h = 10 m/s

    Bus A travel distance in 15 sec

    ⇒ Distance = 40/3 x 15 = 200 m

    Bus B travel distance in 15 sec

    ⇒ Distance = 10 x 15 = 150 m

    as the two road is at right angle, so

    ⇒ Let Height = 200 m

    ⇒ Base = 150 m

    By Pythagoras theorem

    Hypotenuse2 = Base2 + Height2

    ⇒ Distance between two buses = 2002 + 1502 = 40000 + 22500 = 62500 = 250 m

    ⇒ Hence, Distance between the two buses is 250 m

  • Question 7/10
    1 / -0.33

    A sum of ₹3,596 is divided between A, B, and C in the ratio  What is the difference (in ₹) between the shares of A and C?

    Solutions

    Multiply 72 in equation (1)

    A : B : C = 12 : 9 : 8

    According to question,

    A + B + C = Rs. 3596

    ⇒ 12x + 9x + 8x = Rs. 3596

    ⇒ 29x = Rs. 3596

    x = 124

    Then, A = 1488, B = 1116, C = 992

    Now, The difference (in ₹) between the shares of A and C

    A - C = 1488 - 992 = 496

    ∴ The difference (in ₹) between the shares of A and C = Rs. 496

  • Question 8/10
    1 / -0.33

    A train is running at a speed of 72 km/h. If the length of the train is 240 meters and it crosses a platform in 40 seconds, what is the length of the platform?

    Solutions

    Given:

    Speed of train =  72 km/h

    Length of train = 240

    Crosses a platform in 40 seconds

    Formula used:

    Speed x Time = Distance

    Calculation:

    Speed = 72 x 5/18 = 20m /sec

    Speed x Time = Distance

    ⇒ 20 x 40 = 240 +x

    ⇒ 800 = 240 + x

    ⇒ x = 560 m

    hence, option 2 is correct answer.

  • Question 9/10
    1 / -0.33

    A dishonest dealer save 10% while buying goods and than increase the price by 20% & gives 10% discount on marked price but gives 10% less in quantity also find his gain percentage?

    Solutions

    Given:

    Profit (P) = 10%

    Increase in price = 20%

    Discount (D) = 10%

    Dishonesty = 10%

    Formula used:

    S.P = M.P - D

    S.P = Selling Price

    M.P = Marked Price

    Effective Profit = x + y + (x × y)/100

    {(S × De)/R - 100}

    S = Situation of Shopkeeper

    De = Demand of Quantity by Customer

    R = Received Quantity to Customer

    Calculation:

    Let the Cost price (C.P) by 100

    ⇒ C.P = 100

    Increased price = 120

    ⇒ S.P = 120 - 12 

    ⇒ S.P = 108

    Effective Profit = x + y + (x × y)/100

    ⇒ 10 + 8 + (10 × 8)/100

    ⇒ 18.8

    The situation of shopkeepers is 18.8 profit 

    ⇒ {(S × De)/R - 100}

    ⇒ {(108.8 × 100)/90 - 100}

    ⇒ 132 - 100

    ⇒ 32

    ∴ The net gain is 32%.

    Shortcut Trick

    Calculation:

    We can use successive method.

    C.P : S.P = 4500 : 5940 = 25 : 33

    Profit% = {(33 - 25)/25} × 100 = 32%

    ∴ The correct answer is 32%.

  • Question 10/10
    1 / -0.33

    In an election, the candidate who secured 64% of the votes polled, won by 504 votes. What is the total number of voters on the voting list, if 90% voters cast their votes and there were no invalid votes?

    Solutions

    Given:

    In an election, the candidate who secured 64% of the votes polled, won by 504 votes.

    90% voters cast their vote.

    Calculations:

    Let the number of voters on the voting list be x.

    The candidate winning be = 64% of x.

    The candidate loosing be = (100 - 64)% = 36% of x.

    Now, 64% of x - 36% of x = 504

    0.64x - 0.36x = 504

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