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IBPS Clerk 2020 Aptitude Test - 2
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IBPS Clerk 2020 Aptitude Test - 2
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  • Question 1/10
    1 / -0

    A man sells an article at 20% profit. If he had bought it at 20% less and sold it for Rs. 4 less, he would have gained 40 %. The cost price of the article is?

    Solutions

    Let the CP of article is Rs 100

    Then, SP = Rs. 120

    In 2nd case: CP = Rs. 80

    New SP = Rs. 112 (140% profit on 80)

    If he sells it for Rs. (120-112=8) less then , CP = Rs. 100

    If he sells it for Rs. 4 less then CP = (100/8) × 4 = Rs 50

     

  • Question 2/10
    1 / -0

    Yuvraj and Dhoni sold their bats at Rs. 7749 each but Yuvraj incurred a loss of 16%, while Dhoni gained 8%. What is the ratio of the cost price of the bat of Yuvraj to that of Dhoni?

    Solutions

    CP of Yuvraj’s bat = (7749 / 84) × 100

    = Rs. 9225

    CP of Dhoni’s bat = (7749 / 108) × 100

    = Rs. 7175

    Ratio = 9225 / 7175 = 369 : 287.

     

  • Question 3/10
    1 / -0

    Ashutosh goes from Vizag to Kolkata by sea route. The speed of the boat in still water is 45 km/h and speed of the current is 15 km/h. After reaching Kolkata he stayed there for 30 minutes and after that come back by same boat. The time taken by him in this journey is 29 hours 30 minutes, find the distance travel by him in one side.

    Solutions

    Let the distance of one side be x

    Total time taken in journey = (29.5 – 0.5) hours = 29 hours

    ATQ-

    x/(45 – 15) + x/(45 + 15) = 29

    3x/60 = 29

    x = 580 km

     

  • Question 4/10
    1 / -0

    In Cochin the pink taxi charges consist of fixed charges and additional charges per mile. The fixed charges are for a distance of up to 8 mile and additional charges are applicable per mile thereafter. The charge for a distance of 12 mile is Rs. 300 and for 22 mile is Rs. 750 . The charge for a distance of 30 mile is?

    Solutions

    let the fixed price charged by pink taxi be ‘p’

    And the additional charges per mile be ‘x’

    ATQ-

    p + 4x = 300 ….(1)

    p + 14x = 750 ….(2)

    on equating eq. (1) & (2)

    x = Rs.45

    and, p = Rs. 120

    The charge for a distance of 30 mile is = 120 + (22 × 45) = Rs.1110

     

  • Question 5/10
    1 / -0

    Kaleen left Mirzapur for Jaunpur in his car at 9:45 am. On his way to Jaunpur at 11:30 am where he met with Munna, who was heading to Mirzapur. If Munna reached Mirzapur at 1 pm, then find the ratio of speed of both the car.

    Solutions

    Kaleen takes 1 hour 45 minutes to cover the point from Mirzapur where he met with Munna ,

    while Munna takes 1 hour 30 minutes to reach Mirzapur from that point.

    Since, speed is inversely proportional to time

    So, ratio of speed of kaleen to speed of Munna = 90mins / 105mins = 6 : 7

     

  • Question 6/10
    1 / -0

    In the following question two equations are given in variables Xand Y. You have to solve these equations and determine relation between X and Y.

    I. x2 + 30x - 675 = 0

    II. y2 + 30y - 864 = 0

    Solutions

     

  • Question 7/10
    1 / -0

    In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

    Solutions

     

  • Question 8/10
    1 / -0

    In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

    I. 2x² – 21x + 34 = 0

    II. 2y² – 19y + 44 = 0

    Solutions

    I. 2x ²-21x+34=0

    2x ²-4x-17x+34=0

    2x(x-2)-17(x-2)=0

    (2x-17)(x-2)=0

    x = 17/2 = 8.5,2

    II. 2y ²-19y+44=0

    2y ²-8y-11y+44=0

    2y(y-4)-11(y-4)=0

    (2y-11)(y-4)=0

    y = - 11/2 = 5.5,4

    So, relationship cannot be established.

     

  • Question 9/10
    1 / -0

    In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

    I. 15x2 + x - 6 = 0

    II. 5y2 - 23y + 12 = 0

    Solutions

     

  • Question 10/10
    1 / -0

    In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

    I. 18x2 - 42x + 28 = 13x2 - 24x + 12

    II. 23y2 + 22y + 17 = 19y2 + 10y + 8

    Solutions

     

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