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RBI Grade B 2022 Quantitative Aptitude Test - 3
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RBI Grade B 2022 Quantitative Aptitude Test - 3

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  • Question 1/10
    1 / -0.25

    In the following questions an equation followed by some information is given. Read the equation and information carefully to answer the given questions.

    x2 +  18x + P = 0.

    Roots of the above equation are m and n and where m2 – n2 = –108 .

    Find the value of P.

    Solutions

    Given equation: x2 + 18x + P = 0

    After comparing the given equation with x2 – (Sum of roots)x + (Products of roots) = 0

    Sum of the roots (m and n) = –18

    So, m + n = –18 …..(1)

    And m2 – n2 = –108

    (m + n)(m – n) = –108

    –18 × (m – n) = –108

    m – n = 6 …..(2)

    After solving both the equations, we get:

    m = –6, n = –12

    The value of P = (–6) × (–12) = 72

     

  • Question 2/10
    1 / -0.25

    In the following questions an equation followed by some information is given. Read the equation and information carefully to answer the given questions.

    x2 +  18x + P = 0.

    Roots of the above equation are m and n and where m2 – n2 = –108 .

    Find the value of P + 2m – n2.

    Solutions

    Given equation: x2 + 18x + P = 0

    After comparing the given equation with x2 – (Sum of roots)x + (Products of roots) = 0

    Sum of the roots (m and n) = –18

    So, m + n = –18 …..(1)

    And m2 – n2 = –108

    (m + n)(m – n) = –108

    –18 × (m – n) = –108

    m – n = 6 …..(2)

    After solving both the equations, we get:

    m = –6, n = –12

    Required value = P + 2m – n2

    = 72 + 2 × (–6) – (–12)2

    = 72 – 12 – 144 = –84

     

  • Question 3/10
    1 / -0.25

    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

    Hence, x > y

     

  • Question 4/10
    1 / -0.25

    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. 4x2 + 15x + 9 = 0

    II. 15y2 + 11y + 2 = 0

    Solutions

    I. 4x2+ 15x + 9 = 0
    ⇒ 4x2+ 12x + 3x + 9 = 0
    ⇒ 4x(x + 3) + 3(x + 3) = 0
    ⇒ (4x + 3)(x + 3) = 0
    ⇒ x = –3 or –3/4

    II. 15y2+ 11y + 2 = 0
    ⇒ 15y2+ 5y + 6y + 2 = 0
    ⇒ 5y(3y + 1) + 2(3y + 1) = 0
    ⇒ (5y + 2)(3y + 1) = 0
    ⇒ y = –1/3 or –2/5

    Hence, x < y

     

  • Question 5/10
    1 / -0.25

    Area of a square is 64 cm2 and the perimeter of a rectangle is 4 cm more than half of the perimeter of square. If the breadth of the rectangle is 4 cm and length of the rectangle is equal to the radius of a circle. Find the difference between the area and the circumference of the circle.

    Solutions

    Area of square = 64 cm2

    Side of square = 8 cm

    Perimeter of square = 4 × 8 = 32 cm

    According to the question ,

    Perimeter of the rectangle = 4 + 32/2 = 20

    2 × (length + breadth) = 20 cm

    Length = 10 – 4 = 6 cm

    Length = radius of circle = 6 cm

    Area of circle – circumference of circle = Πr2 – 2Πr = 36Π – 12Π = 24Π

     

  • Question 6/10
    1 / -0.25

    Two persons A and B start simultaneously from same point in the same direction and distance between them after 30 minutes becomes 16 km, then what will be the distance between them when they travel in opposite direction for the same time period and ratio of their speeds is 3 : 2?

    Solutions

    Let the speed of A and B is '3x' and '2x' respectively.

    Relative speed of both the persons when travelling in same direction = 3x − 2x = x

    relative speed of both the persons when travelling in opposite direction = 3x + 2x = 5x

    According to the data in the question,

    Also, we can observe that if in half an hour 16 km is covered with speed x. Then in the same with speed 5x, 80km will be covered.

     

  • Question 7/10
    1 / -0.25

    In a monthly test of maths total marks scored by Satish, Varun and Saumya is 245. The ratio of the marks scored by Satish and Saumya is 5 : 6 and that of the marks scored by Saumya and Varun is 9 : 8. Find the total marks scored by Satish in the maths.

    Solutions

    Ratio of marks of Satish and Saumya = 5 : 6

    Ratio of marks of Saumya and Varun = 9 : 8

    Ratio of the marks scored by Satish, Saumya and Varun respectively

    = 5 × 9 : 6 × 9 : 6 × 8

    = 45 : 54 : 48 = 15 : 18 : 16

    ∴ Marks scored by Satish

     

  • Question 8/10
    1 / -0.25

    In how many different ways, can the letters of the word ‘FLAVOUR’ be arranged in such a way that the vowels always come together?

    Solutions

    In the word FLAVOUR:

    Vowels are – A, O, U (3)

    consonants are – F, L, V, R (4)

    Required number of ways = (4 + 1)! × 3! = 120 × 6 = 720

     

  • Question 9/10
    1 / -0.25

    The ratio of milk and water in a mixture of 90 liter milk and water is 7 : x. If 15 liter of water is added in the mixture, the ratio of water to milk becomes 1 : x, then find the quantity of milk in the initial mixture.

    Solutions

    ⇒ (105x + 105)x = 1 × 630

    ⇒ (x + 1)x = 6

    ⇒ x2 + x − 6 = 0

    ⇒ (x + 3)(x − 2) = 0

    ⇒ x = 2

     

  • Question 10/10
    1 / -0.25

    The MRP of a article is 50% more than its manufacturing cost. The article is sold through a retailer, who earns 25% profit on his purchased price. What is the profit or loss percentage of the manufacturer who sells his article to the retailers, If the retailers give 20% discount on MRP?

    Solutions

    Let manufacturing cost be Rs. 100.

    Then, MRP = Rs. 150

    Discount = 20%

    Customer's price = Selling price of retailer = 0.8 × 150 = Rs. 120

    Profit of retailer = 25% on purchased price

    Let purchasing price of the retailer = Rs. a

    According to question,

    1.25a = 120
    ⇒ a = Rs. 96

    Selling price for manufacturer = Purchasing price of retailer = Rs. 96

     

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