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SSC CHSL 2025 Aptitude Test - 3
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SSC CHSL 2025 Aptitude Test - 3
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  • Question 1/10
    2 / -0.5

    From the given graph. find the difference in the number of Laptops and mobiles sold in comparison to the number of other products sold by a shop dealing in electronic items in the month of March 2019.

    Solutions

    The correct answer is option 3 i.e. 500.

    Total number of mobile and laptop sold = 500 + 600 = 1100

    Total number of Radio, TVs, and watches sold = 100 + 300 + 200 = 600

    ATQ:-

    Difference = 1100 - 600 = 500

  • Question 2/10
    2 / -0.5

    The parallel sides of a trapezium and its height are in an arithmetic progression with a common difference of 4. If the height is the highest term and the area of the trapezium is 160 sq. mits, find the ratio of the length of greatest parallel side to that of the smallest parallel side.

    Solutions

    The correct answer is Option 4 i.e. 3 : 2.

    Let the lengths of the parallel sides of the trapezium be a (smallest) and a + 4 (greatest), and let the height be a + 8 (highest term).

    The area of a trapezium = 1/2 × sum of parallel sides × height

    ⇒ 160 = 1/2 × (a + a + 4) × (a + 8)

    ⇒ 160 = 1/2 × (2a + 4) × (a + 8)

    ⇒ 160 = (a + 2)(a + 8)

    ⇒ 160 = a2 + 8a + 2a + 16

    ⇒ 160 = a2 + 10a + 16

    ⇒ a2 + 10a - 144 = 0

    Solving this quadratic equation using the formula

    ⇒ a = 16/2, -36/2

    ⇒ a = 8, -18

    So, a = 8

    Smallest side = a = 8

    Greatest side = a + 4 = 12

    Thus,

    Required ratio = 12/8 = 3/2 = 3 : 2

  • Question 3/10
    2 / -0.5

    Find the amount of water contained in a cylindrical tank of radius 7 m and height 20 m. It is known that the tank is completely filled.

    Solutions

    The correct answer is option 4 i.e. 3080m3

    Radius r = 7 m

    Height h = 20 m

    the volume of a cylinder V = πr2h

    ⇒ 22/7 × 72 × 20 = 3080m3

  • Question 4/10
    2 / -0.5

    If Tan A = 15/8 and Tan B = 7/24, then Cos A + Cos B =?

    Solutions

    The correct answer is option 1 i.e. 608/425.

    Given,

    Tan A = 15/8 and Tan B = 7/24

    we know that,

    Tan θ = P/B and Cos θ = B/H

    For Tan A = 15/8

    ⇒ HA = √(PA2 + BA2) = √(225 + 64) = √289 = 17

    Now,  for Tan B = 7/24

    ⇒ HB = √(PB2 + BB2) = √(49 + 576) = √625 = 25

    Now,

    Cos A + Cos B

    = BA/HA + BB/HB

    = 8/17 + 24/25

    = (200 + 408)/425

    = 608/425

  • Question 5/10
    2 / -0.5

    Solutions

    The correct answer is Option 3 i.e. 76.

    x - 1/x = k

    x3 - 1/x3 = k3 + 3k

  • Question 6/10
    2 / -0.5

    A shop keeper purchased 10 kg of rice and 20 kg of sugar at Rs.75 and Rs. 85 per kg respectively. On selling, he gained 20% on rice and 10% on sugar. What was the total sale price?

    Solutions

    The correct answer is option 2 i.e. Rs 2770

    Understanding/Application

    20% gain on Rice

    Total SP of rice = (Rate/100) (100 + Gain%) (weight) = (75/100)(100 + 20)(10)

    = (.75)(120)(10) = Rs 900

    Now 10% gain on Sugar

     Total SP of Sugar = (Rate/100) (100 + Gain%) (weight) = (85/100)(100 + 10)(20)

    = (.85)(110)(20) = Rs 1870

    Total SP = 900 + 1870 = Rs 2770

  • Question 7/10
    2 / -0.5

    If the price of a book is first decreased by 30% and then increased by 20%, then the net change in the price will be:

    Solutions

    The correct answer is option 3 i.e. 16%.

    Let the initial number be 100 then,

    First, it decreases by 30 i.e. it becomes 70

    Then it increases by 20% i.e,

    ⇒ 70 + 20% of 70 = 84

    Net decrease = (100 - 84)/100 × 100 = 16%

  • Question 8/10
    2 / -0.5

    Simplify: 8 ÷ [(9 - 5) ÷ {(7 - 2) ÷ (3 + 9 ÷ 24)}]

    Solutions

    The correct answer is Option 1 i.e. 80/27.

    8 ÷ [(9 - 5) ÷ {(7 - 2) ÷ (3 + 9 ÷ 24)}]

    ⇒ 8 ÷ [(9 - 5) ÷ {(7 - 2) ÷ (3 + 9/24)}]

    ⇒ 8 ÷ [(9 - 5) ÷ {(5) ÷ (3 + 3/8)}]

    ⇒ 8 ÷ [(4) ÷ {(5) ÷ 27/8}]

    ⇒ 8 ÷ [(4) ÷ {40/27}]

    ⇒ 8 ÷ [4 × 27/40]

    ⇒ 8 ÷ [27/10]

    ⇒ 8 × 10/27

    ⇒ 80/27

  • Question 9/10
    2 / -0.5

    The LCM of 48, 72, and another number, x, is 576. Which of the values given below can be the value of x?

    Solutions

    The correct answer is option 2 i.e. 192.

    The LCM of 48 = 24 × 3

    LCM of 72 =  23 × 32

    LCM of 576 =  26 × 32

    Let's assume the third number is x.

    The third number should be the multiple of x = 26 × 3= 64 × 3k

    if k = 0 then x = 64

    k = 1 then x = 192

    k = 2 then x = 576

    The value of the third number cannot be more than 576.

    So, From the options, the third number will be 192.

  • Question 10/10
    2 / -0.5

    Atul can do a piece of work in 18 days, and Aaravya is 50% more efficient than Atul. How much more time does Atul take than Aaravya to do the same piece of work?

    Solutions

    The correct answer is option 2 i.e. 6 days.

    Aaravya is 50% more efficient than Atul

    Ratio of efficiency (Aaravya : Atul) = 3 : 2

    Let the efficiency of Aaravya and Atul be 3x and 2x.

    Total work = Number of days taken by Atul × Efficiency of Atul = 18 × 2x

    ⇒ 36x

    Number of days taken by Aaravya = 36x/3x = 12 days

    Hence, the required difference = 18 - 12 = 6 days

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