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SSC MTS 2024 Aptitude Test - 3
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SSC MTS 2024 Aptitude Test - 3
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  • Question 1/10
    3 / -0

    Solutions

    Formula Used:

    a3 + b3 = (a + b)(a2 + b2 - ab)

    Calculations:

    Now, (0.04)3 + (0.21)3 = [(0.04) + (0.21)][(0.04)2 + (0.21)2 - (0.04 × 0.21)]

    ⇒ √[(0.04)2 + (0.21)2 - (0.04 × 0.21)]/[(0.04) + (0.21)][(0.04)2 + (0.21)2 - (0.04 × 0.21)]

    ⇒ √(1/[(0.04) + (0.21)]

    ⇒ √1/0.25 = √4

    ⇒ 2

    ∴ The value of given expression is 4.

  • Question 2/10
    3 / -0

    Find the single equivalent discount for successive discounts of 12%, 18% and 25% on the marked price of a car.

    Solutions

    Given:

    Discount (d1) = 12%

    Discount (d2) = 18%

    Discount (d3) = 25%

    Formula:

    Where “d1” and “d2” are successive discounts offered.

    Calculation:

  • Question 3/10
    3 / -0

    Mr. Shubham started a business with Rs 20,000. Mr. Karan joined Mr. Shubham after one year and invested Rs 40,000. After two years since the start of the business, they earned a profit of Rs 45,000. What will be Mr. Karan’s share in the profit?

    Solutions

    Given:

    The initial investment made by Mr. Shubham is RS 20,000.

    Mr. Karan's initial investment is Rs 40,000 but after one year of business starting.

    Concept:

    The ratio of profit = Amount invested by one partner × time period of investment ⦂ Amount invested by other partner × time period of investment

    Calculation:

    ⇒ The ratio of share of profit = Shubham : Karan

    ⇒ Amount invested by shubham × time period of investment ⦂ Amount invested by Karan × time period of investment

    ⇒ 20,000 × 2 : 40,000 × 1

    ⇒ 40,000 : 40,000 ⇒ 1 : 1

    Karan’s share = (1/2) × 45,000

    ⇒ Rs 22,500

  • Question 4/10
    3 / -0

    What is the fourth proportional to 6, 24 and 83?

    Solutions

    let the fourth proportional be x

    ⇒ 6 ∶ 24 ∶∶ 83 ∶ x

    ⇒ 6/24 = 83/x

    ⇒ x = 83 × (24/6) = 332

  • Question 5/10
    3 / -0

    The following pie chart shows the production of different types of garments in Company XYZ in 2014.

    If a total of 150,000 tonnes of textiles were produced in 2014, then what was the production of woolen garments (in tonnes) in 2014?

    Solutions

    Total production of the year is 150000 tonnes

    Woolen garments = 15%

    So, 150000 × 15/100

    ⇒ 22500

    ∴ The production of woolen garments (in tonnes) in 2014 is 22500

  • Question 6/10
    3 / -0

    A and B can complete a piece of work in 7 days and 14 days respectively. They start working together but after 4 days B leaves the work. In how many days the total work would be finished?

    Solutions

    Given:

    Time taken to complete the whole work by A = 7 days

    Time taken to complete the whole work by B = 14 days

    Number of days A and B worked together = 4 days

    Concept used:

    If A and B can do a piece of work in 'x' days and 'y' days, respectively.

    They start working together and after 't' days B leaves the work,

    then the time taken to finish the whole work will be [(x/y) × (y - t)] days.

    Calculation:

    Here, x = 7 days, y = 14 days, t = 4 days

    As we know that

    Required time = [(x/y) × (y - t)] days

    ⇒ [(7/14) × (14 - 4)]

    ⇒ (1/2) × 10

    ⇒ 5 days

    ∴ Total work would be finished in 5 days.

    Alternate Method

    Calculation:

    One day work of (A + B) = (1/7) + (1/14)

    ⇒ (3/14) units

    Work done by (A + B) in 4 days = 4 × (3/14)

    ⇒ (6/7) units

    Remaining Work = 1 - (6/7)

    ⇒ (1/7) units

    Since B left after 4 days.

    Remaining work done by A = (1/7) units

    Time taken by A to do remaining work = [(1/7)/(1/7)] days

    ⇒ 1 day

    So, Total days = (4 + 1) days

    ⇒ 5 days

    ∴ Total work would be finished in 5 days.

  • Question 7/10
    3 / -0

    A and B working together complete a job in 12 days. For this job A gets paid Rs.4,980 and B gets paid Rs.3,320. If A had worked alone on this job he would have completed it in how many days?

    Solutions

    Given:

    A and B together complete a job = 12 days

    A gets for this job = Rs. 4980

    B gets for this job = Rs. 3320

    Calculation:

    A and B price ratio = 4980 : 3320 = 3 : 2

    ⇒ A and B efficiency ratio = 3 : 2

    ⇒ Total work = 12 × (2 + 3) = 60 units

    ⇒ A one day work efficiency = 3 unit/day

    ⇒ A alone completed the work = 60/3 = 20 days

    ∴ A alone completed the job in 20 days.

  • Question 8/10
    3 / -0

    A person lent a certain sum of money at the rate of 30% per annum at simple interest. The interest in 5 years exceeds the amount lent by Rs.3200. What is the loan amount?

    Solutions

    Given:

    The rate of interest (r) = 30% per annum

    The time period (t) = 5 years

    The interest in 5 years exceeds the amount lent by Rs.3200

    Formula Used:

    The formula for simple interest is I = PRT/100

    Where:

    I is the interest, P is the principal amount 

    R is the rate of interest, T is the time period

    Calculation:

    From the given problem,

    I = P + 3200

    (P × 30 × 5)/100 = P + 3200

    1.5P = P + 3200

    0.5P = 3200

    P = 3200/0.5 

    ⇒ P = 6400

    Hence, the loan amount (Principal) is Rs. 6400.

  • Question 9/10
    3 / -0

    Two vessels A and B contain spirit and water in the ratio 3 : 8 and 6 : 5. In what ratio should their ingredients be mixed to obtain a solution of spirit and water in the ratio 5 : 6?

    Solutions

    Given:

    A and B contain spirit and water in the ratio 3 : 8 and 6 : 5 respectively

    Calculation:

    ∴ Required ratio = (1/11) : (2/11) = 1 : 2

    Alternate Method

    Let the total quantity in A be 11x and in B be 11y, which is to be mixed

    According to the question:

    (3x + 6y)/(8x + 5y) = 5/6

    ⇒ 18x + 36y = 40x + 25y

    ⇒ 22x = 11y

    ⇒ x/y = 1/2

    ∴ They should be mixed in the ratio 1 : 2

  • Question 10/10
    3 / -0

    Find the value of ‘?’ in the following question

    Solutions

    Concept used:

    Follow BODMAS rule to solve this question, as per the order given below:

    Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

    Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

    Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

    Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

    Now,

    Calculation:

    Considering the given equation

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