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RRB Group D 2025 Mix Test - 92
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RRB Group D 2025 Mix Test - 92
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  • Question 1/10
    1 / -0.33

    Three friends decided to run to the school gate. Speed of A : B : C are 5 : 4 : 6. If the distance between the starting point and the school gate is 1500 m and C beats A by 5 minutes, then find the time taken by B to complete the race.

    Solutions

    The correct answer is option 3 i.e. 37 minutes 30 seconds

    Let the speed of A, B, and C be 5x, 4x, and 6x respectively.

    Time difference = 1500/5x - 1500/6x

    ⇒ 5 = 1500(1/5x - 1/6x)

    ⇒ 1/300 = (6 - 5)/30x

    ⇒ x = 10

    Speed of B = 4(10) = 40 m/min

    Time taken by B = 1500/40

    ⇒ 37.5 minutes or 37 minutes 30 seconds

  • Question 2/10
    1 / -0.33

    The present age of the boy is two-fifth of his father. 6 years before, the age of the boy was half of his mother. If the present age of his mother is 34 years, find the present age of his father.

    Solutions

    The correct answer is option 2 i.e. 50 years.

    The present of the boy = 2/5 of his father's age

    The present age of his mother = 34 years

    Before 6 years, the age of the boy = 1/2 of his mother's age.

    The present age of his mother = 34 years

    Before 6 years, the age of his mother = 34 - 6 = 28 years.

    Before 6 years Boy's age = 1/2 of his mother's age

    = 1/2 × 28 = 14 years

    The present age of the boy = 14 + 6 = 20 years

    The present age of the boy = 2/5 of his father's age

    20 = 2/5 × His father's age

    His father's age = 20 × 5/2 = 50 years

  • Question 3/10
    1 / -0.33

    The average temperature of a city for the first 3 weeks in a month is 35 degrees, and for the last 3 weeks, it is 39 degrees. If the average temperature of the first and fourth weeks is in the ratio 9 : 13, find the average temperature of the month.

    Solutions

    The correct answer is Option 1 i.e. 36.

    Let the average temperature for the first week and the following weeks be x1, x2, x3, and x4. Therefore, we get

    Now, from this, we can get:

    ⇒ (x2 + x+ x) - ( x1 + x2 + x) = 117 - 105

    ⇒ x4 - x1 = 12

  • Question 4/10
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    The father divides his property between his son and daughter. The ratio of father to son and son to daughter is 5 : 2, such that the daughter gets Rs 400. Find the sum of the share of father and son.

    Solutions

    The correct answer is option 3 i.e. 3500.

    Given:

    The ratio of father to son and son to daughter is 5 : 2

    The daughter gets Rs. 400.

    Calculations:

    The ratio of share of father to son and son to daughter

    Father : Son = 5 : 2 ------ (1)

    Son : Daughter = 5 : 2 ------ (2)

    Multiply equation (1) with 5 and equation (2) with 2 to balance the share of son

    Father : Son = 25 : 10

    Son : Daughter = 10 : 4

    Father : Son : Daughter = 25 : 10 : 4

    Let the share be 25x, 10x and 4x

    If the value of 4x = 400

    ⇒ x = 100

    Sum of share of father and son = 25x + 10x

    Put the value of x

    = 35x

    = 35 × 100

    = 3500

  • Question 5/10
    1 / -0.33

    A sum of Rs. 64000 is invested at the rate of 25% per annum compounded annually for some time. If the interest obtained from it is Rs. 61000 then, find the time for which the sum is invested.

    Solutions

    The correct answer is option 2 i.e. 3 years.

    Principal = Rs. 64000

    Rate = 25% per annum

    CI = Rs. 61000

    So,

    Amount = 64000 + 61000 = Rs. 125000

    Applying the formula for CI:

    125000 = 64000 × (1 + 25/100)T

    ⇒ (1 + 25/100)T = 125/64

    ⇒ (5/4)T = 125/64

    ⇒ (5/4)T = (5/4)3

    So,

    T = 3

    Hence,

    Time for which the sum is invested = 3 years

  • Question 6/10
    1 / -0.33

    If A is 25% more than 36 and B is 36% more than 25, then which of the following is true?

    Solutions

    The correct answer is Option 1 i.e. A > B.

    A = 25% more than 36

    ⇒ A = 36 × (100 + 25)/100 = 45

    ⇒ B = 36% more than 25

    ⇒ B = (100 + 36)/100 × 25 = 34

    ∴ A > B

  • Question 7/10
    1 / -0.33

    Table shows income (in Rs) received by 4 employees of a company during the month of December 2020 and all their income sources.

    What is the ratio of salary of Varun to his income other than salary?

    Solutions

    The correct answer is option 4 i.e. 525 : 128

    The ratio of salary of Varun to his income other than salary = 42000: 10240

    =  525 : 128.

  • Question 8/10
    1 / -0.33

    If the sides of a triangle MNO are 15 cm, 112 cm, and 113 cm then, find the area of the triangle. (in cm2)

    Solutions

    The correct answer is option 3 i.e. 840.

    Given sides - 15cm, 112cm, and 113 cm (forms a triplet)

    Hence,

    Area of triangle = (1/2) × B × H

    ⇒ (1/2) × 112 × 15

    ⇒ 56 × 15 = 840 cm2

  • Question 9/10
    1 / -0.33

    If the total surface area of a cylinder is 704 cm2 and the ratio of height to base radius of the cylinder is 3 : 4, then what will be the volume of that cylinder?

    Solutions

    The correct answer is option 1 i.e. 384π cm3.

    The height and base radius of the cylinder are 3x and 4x, respectively.

    Total surface area of the cylinder = 2πr(h + r)

    704 = 2 × (22/7) × 4x × (3x + 4x)

    ⇒ 704 = 176x/7 × (3x + 4x)

    ⇒ 704 = 176x2

    ⇒ x= 4

    ⇒ x = 2

    Base radius = 4x = 8 cm

    Height = 3x = 6 cm

    Volume of the cylinder = πr2h = π × 64 × 6 = 384π cm3

  • Question 10/10
    1 / -0.33

    Find the value of Cos180° + Cos135°- Cos240°+ Cos315°.

    Solutions

    The correct answer is option 2 i.e. -1/2.

    Cos(90 + Ø) = -SinØ

    Cos(180 + Ø) = -CosØ

    Cos(360 - Ø) = CosØ

    Given,

    Cos180° + Cos135°- Cos240°+ Cos315°

    ⇒ Cos(90 + 90) = -Sin90 = -1

    ⇒ Cos(90 + 45) = -Sin45 = -1/√2

    ⇒ Cos(180 + 60) = - Cos60 = -1/2

    ⇒ Cos(360 - 45) = Cos45 = 1/√2

    Now,

    Cos180° + Cos135°- Cos240°+ Cos315° =  -1 - 1/√2 + 1/2 + 1/√2 = -1/2

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