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IBPS PO 2020 Aptitude Test - 33
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IBPS PO 2020 Aptitude Test - 33

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  • Question 1/10
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    In the following questions two quadratic equations are given in variables X and Y. You have to solve these equations and determine the relation between X and Y.

    I. X2 – 21X + 110 = 0

    II. Y2 = 81

    Solutions

    X2 – 21X + 110 = 0

    ⇒ X2 – 10X – 11X + 110 = 0

    ⇒ X(X – 10) – 11(X – 10) = 0

    ⇒ (X – 10) (X – 11) = 0

    ⇒ X = 10 & 11

    Y2 = 81

    ⇒ Y2 = (± 9)2

    ⇒ Y = +9 & –9

    Hence, X > Y

     

  • Question 2/10
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    In the following question two equations are given in variables x and y. You have to solve these equations and determine the relation between x and y.

    I. x2 + 2x – 24 = 0

    II. y2 + 13y + 40 = 0

    Solutions

    I. x2 + 2x – 24 = 0

    ⇒ x2 + 6x – 4x – 24 = 0

    ⇒ x(x + 6) – 4(x + 6) =0

    ⇒ (x + 6)(x – 4) = 0

    ⇒ x = −6, 4

    II. y2 + 13y + 40 = 0

    ⇒ y2 + 8y + 5y + 40 = 0

    ⇒ y(y + 8) + 5(y + 8) = 0

    ⇒ (y + 8) (y + 5) = 0

    ⇒ y = −8, −5

     

  • Question 3/10
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    In the following question two equations are given in variables X and Y. You have to solve these equations and determine the relation between X and Y.

    I. X2 + 9X – 22 = 0

    II. Y2 + 11Y – 26 = 0

    Solutions

    X2 + 9X – 22 = 0

    ⇒ X2 + 11X – 2X – 22 = 0

    ⇒ X(X + 11) – 2(X + 11) = 0

    ⇒ (X + 11)(X – 2) = 0

    ⇒ X = –11 & 2

    Y2 + 11Y – 26 = 0

    ⇒ Y2 + 13Y – 2Y – 26 = 0

    ⇒ Y(Y + 13) – 2(Y + 13) = 0

    ⇒ (Y + 13)(Y – 2) = 0

    ⇒ Y = –13 & 2

    Hence, no relation can be established between X and Y.

     

  • Question 4/10
    1 / -0

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

    I. 4X2 – 13X – 12 = 0

    II. Y2 – 7Y – 60 = 0

    Solutions

    I. 4X2 – 13X – 12 = 0

    ⇒ 4X2 – 16X + 3X – 12 = 0

    ⇒ 4X(X – 4) + 3(X – 4) = 0

    ⇒ (X – 4)(4X + 3) = 0

    ⇒ X = 4, – 3/4

    II. Y2 – 7Y – 60 = 0

    ⇒ Y2– 12Y + 5Y – 60 = 0

    ⇒ Y(Y – 12) + 5(Y – 12) = 0

    ⇒ (Y – 12)(Y + 5) = 0

    ⇒ Y = 12, –5

    Hence, no relation can be established between X and Y.

     

  • Question 5/10
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    In the following question two equations are given in variables X and Y. You have to solve these equations and determine the relation between X and Y.

    I. X2 – 4X + 4 = 0

    II. Y2 – 12Y + 27 = 0

    Solutions

    X2 – 4X + 4 = 0

    ⇒ X2 – 2X – 2X + 4 = 0

    ⇒ X(X – 2) – 2(X – 2) = 0

    ⇒ (X – 2)(X – 2) = 0

    ⇒ X = 2 & 2

    Y2 – 12Y + 27 = 0

    ⇒ Y2 – 12Y + 27 = 0

    ⇒ Y2 – 9Y – 3Y + 27 = 0

    ⇒ Y(Y – 9) – 3(Y – 9) = 0

    ⇒ (Y – 9)(Y – 3) = 0

    ⇒ Y = 9 & 3

    Hence, Y > X

     

  • Question 6/10
    1 / -0

    Directions For Questions

    Direction: Study the following table carefully and answer the questions that follow.

    The table given below shows the cost price and profit percent of two shops A and B for 4 products – P, Q, R and S.

    ...view full instructions

    If the successive discounts offered on marked price of product S at shop A are 20% and 20%, then find the marked price of product S at shop A.

    Solutions

    Let the marked price of product S at shop A = Rs. 100a, then

    ⇒ 64a = 576

    ⇒ a = 9

    Hence, the marked price of product S at shop A = 100a = 100 × 9 = Rs. 900

     

  • Question 7/10
    1 / -0

    Directions For Questions

    Direction: Study the following table carefully and answer the questions that follow.

    The table given below shows the cost price and profit percent of two shops A and B for 4 products – P, Q, R and S.

    ...view full instructions

    What is the difference between the selling prices of product Q at shops A and B?

    Solutions

     

  • Question 8/10
    1 / -0

    Directions For Questions

    Direction: Study the following table carefully and answer the questions that follow.

    The table given below shows the cost price and profit percent of two shops A and B for 4 products – P, Q, R and S.

    ...view full instructions

    If the marked prices of product P at shop B and product R at shop A are Rs. 528 and Rs. 435 respectively, find the difference between the discount offered (in Rs.) on product P at shop B and product R at shop A.

    Solutions

    Discount offered = Marked Price – Selling Price

    Discount offered on product P at shop B = 528 − 264 = Rs. 264

    Discount offered on product R at shop A = 435 − 348 = Rs. 87

    Hence, the required difference = 264 – 87 = Rs. 177.

     

  • Question 9/10
    1 / -0

    Directions For Questions

    Direction: Study the following table carefully and answer the questions that follow.

    The table given below shows the cost price and profit percent of two shops A and B for 4 products – P, Q, R and S.

    ...view full instructions

    Profit earned on another product T at shop A is Rs. 78 more than the profit earned on product S at shop B. If the cost price of product T is 25% more than the cost price of product P, find the profit percent of product T at shop A.

    Solutions

    Profit on product T at shop A = 78 + 16% of 450 = 78 + 72 = Rs. 150

    Cost price of product T  = 240 + 25% of 240 = 240 + 60 = Rs. 300

    Hence, the required percentage = 150/300 × 100 = 50%

     

  • Question 10/10
    1 / -0

    Directions For Questions

    Direction: Study the following table carefully and answer the questions that follow.

    The table given below shows the cost price and profit percent of two shops A and B for 4 products – P, Q, R and S.

    ...view full instructions

    What is the approximate average selling price of all the given 4 products at shop A?

    Solutions

     

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