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CDS I 2026 Mathematics Test - 5
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CDS I 2026 Mathematics Test - 5
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  • Question 1/10
    1 / -0.33

    If a2 + 2(b - 1) (c - 1) = 73, b+ 2(c - 1)(a - 1) = 75.5, c+ 2(a - 1) (b - 1) = 78.5 then find a+b+c  where  a, b, c are positive numbers?

    Solutions

    The correct answer is option 2  i.e. 17

    a2 + 2(b - 1) (c - 1) = 73

    b+ 2(c-1)(a-1) = 75.5

    c+ 2(a - 1) (b - 1) = 78.5

    Adding all equations:

    (a2+b2+c2) + 2 (ab+bc+ca) - 4 (a+b+c) = 227 - 6

    (a+b+c)2  - 4 (a+b+c) - 221 = 0

    a+b+c = 17 (a,b,c are positive)

  • Question 2/10
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    Solutions

  • Question 3/10
    1 / -0.33

    The value of cot 13° cot 27° cot 60° cot 63° cot 77° is:

    Solutions

    According to the question;

    ⇒ cot 13° cot 27° cot 60° cot (90° - 27°) cot (90° - 13°)

    ⇒ cot 13° cot 27° cot 60° cot (90° - 27°) cot (90° - 13°)

    ⇒ cot 13° cot 27° cot 60° tan 27° tan 13°

    ⇒ cot 13° × cot 27° × cot 60° × 1/cot 27° × 1/cot 13°

    ⇒ cot 60° = 1/√3

  • Question 4/10
    1 / -0.33

    A bus during its journey travels 60 minutes at a speed of 80 km/h, another 90 minutes at a speed of 100 km/h, and 4 hours at a certain speed. If the average speed of the bus during its whole journey is 70 km/h, what is the speed of the bus during the 4 hours?

    Solutions

    The correct answer is Option 1 i.e. 56.25 km/h.

    60 minutes = 1 hour and 90 minutes = 1.5 hours

    First distance = 80 × 1 = 80 km

    Second distance = 100 × 1.5 = 150 km

    Let speed for the last segment = x km/h

    (80 + 150 + 4x)/(1 + 1.5 + 4) = 70

    ⇒ (230 + 4x)/6.5 = 70

    ⇒ 230 + 4x = 455

    ⇒ 4x = 225

    ⇒ x = 225/4 = 56.25 km/h

  • Question 5/10
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    There exists a natural number N which is 50 times its own logarithm to the base 10, then the sum of the digits in N is

    Solutions

    The correct answer is option 2 i.e. 1

    N = 50 log10N

    Log10N = N/50

    ⇒ N = (10)N/50

    N1/N = (10)1/50 = (10)2/100 = (100)1/100

    ∴ N = 100.

  • Question 6/10
    1 / -0.33

    Let x be the least number which when divided by 12, 18, 20, 27 and 30, the remainder in each case is 2 and x is divisible by 47. If the HCF of x and 1932 is y, then the sum of the digits of y is?

    Solutions

    The correct answer is Option 2 i.e. 10.

    X = LCM(12, 18, 20, 27)K + 2

    X = 540K + 2

    X = 47 × 11K + 23K + 2

    X is divisible by 47 if K = 4

    X = 540 × 4 + 2 = 2162

    HCF(2162, 1932) = 46 = y

    Sum of digits of y = 4 + 6 = 10

  • Question 7/10
    1 / -0.33

    In the given figure, O is the centre of the circle ∠BCA = 50°. The value of ∠BDA is:

    Solutions

    The correct answer is option 3 i.e. 50°.

    Angles in the same segment are equal.

    ∠BDA = ∠BCA = 50°

  • Question 8/10
    1 / -0.33

    If the radius of a sphere is increased by 11%, then what is the percentage (correct to two decimal places) increase in its volume?

    Solutions

    The correct answer is option 4 i.e. 36.76%.

    Volume of sphere (V) = (4/3)πr3

    New radius (R) = 1.11r

    New volume = (4/3)π(R)3 = (4/3)π(1.11r)3 = (4/3)πr3 × 1.367631 = 1.367631V

    Increase = (1.367631V - V) = 0.367631V

    Percentage increase = (0.367631V)/V × 100 ≈ 36.76%

  • Question 9/10
    1 / -0.33

    (2310 - 1024) is not divisible by?

    Solutions

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

    (2310 - 1024) = (2310 - 210)

    an - bn is divisible by (a - b) and (a + b) if n = even

    So, (2310 - 210) is divisible by 21 and 25

    Hence, it is not divisible by 4

  • Question 10/10
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    Solutions

    tan 3x + cot(2x + π/3) = 0

    ⇒ tan 3x = -cot(2x + π/3)

    ⇒ tan 3x = tan[(2x + π/3) + π/2]

    ⇒ tan 3x = tan(2x + 5π/6)

    For tan A = tan B, the general solution is A = nπ + B, where 'n' is an integer

    So, 3x = nπ + (2x + 5π/6)

    ⇒ 3x - 2x = nπ + 5π/6

    ⇒ x = nπ + 5π/6

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