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SSC Selection Post-XIV 2026 (Graduation) Aptitude Test - 2
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SSC Selection Post-XIV 2026 (Graduation) Aptitude Test - 2
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  • Question 1/10
    2 / -0.5

    In an examination, the average marks of 50 students is 70. Afterwards, it is found that the marks of three students are misread as 68, 65, and 73 instead of 70, 62, and 84 respectively. Find the correct average.

    Solutions

    The correct answer is Option 4 i.e. 70.2.

    Given,

    The number of students = 50 

    Average marks = 70

    Total marks of 50 students = 50 × 70 = 3500

    Incorrect marks of 3 students = 68 + 65 + 73 = 206

    Correct marks of 3 students = 70 + 62 + 84 = 216

    Marks required to add = 216 - 206 = 10

    New total marks = 3500 + 10 = 3510 

    ∴ Correct average = 3510/50 = 70.2

  • Question 2/10
    2 / -0.5

    If x% of half of 84 is 7 times the y% of one-fourth of 48, then x is what percentage more/less than y?

    Solutions

    The correct answer is Option 4 i.e. 100%.

    x% of half of 84 = x/100 × 1/2 × 84 = 0.42x

    y% of one fourth of 48 = y/100 × 1/4 × 48 = 0.12y

    ⇒ 0.42x = 7 × 0.12y

    ⇒ 6x = 12y

    ⇒ x/y = 2/1

    Required percentage = (2 - 1)/1 × 100

    ⇒ 100%

  • Question 3/10
    2 / -0.5

    A, B and C are angles of a triangle, then what is the value of 1/(tan A tan B) + 1/(tan B tan C) + 1/(tan C tan A)?

    Solutions

    The correct answer is Option 2 i.e. 1.

    ⇒ 1/(tan A tan B) + 1/(tan B tan C) + 1/(tan C tan A) = (tan A + tan B + tan C)/(tan A tan B tan C)

    ⇒ {tan A + tan B + tan (π – A – B)}/(tan A tan B tan C) [∵ Sum of angles in triangle is π]

    ⇒ {tan A + tan B – tan (A + B)}/(tan A tan B tan C)

    [We know that ⇒ tan A + tan B = tan (A + B)/(1 – tan A tan B)]

    ⇒ {tan (A + B)/(1 – tan A tan B) – tan (A + B)}/(tan A tan B tan C)

    ⇒ - tan (A + B) (tan A tan B)/(tan A tan B tan C)

    ⇒ (tan A tan B tan C)/(tan A tan B tan C) = 1 [∵ tan (A + B) = tan (π – C) = - tan C]

  • Question 4/10
    2 / -0.5

    If the area of a rhombus and the length of the diagonal are 1176 cm2 and 56 cm respectively, then what is the length of altitude of the rhombus? (in cm)

    Solutions

    The correct answer is option 2 i.e. 33.6.

    Let the length of the other diagonal be x cm

    The Area of rhombus = 1/2 × d1d2

    ⇒ 1176 = 1/2 × 56 × x

    ⇒ 1176 = 28x

    ⇒ x = 42 cm

    Its diagonals bisect each other at 90°. So by using the Pythagoras theorem

    Side2 = Sum of squares of lengths of semi-diagonals

    ⇒ Side2 = (56/2)2 + (42/2)2

    ⇒ Side2 = 1225

    ⇒ Side = 35 cm

    Area = Side × Height

    ⇒ 1176 = 35 × Height

    ⇒ Height = 33.6 cm

  • Question 5/10
    2 / -0.5

    If a + b = c then find the value of (a + b - c)3 + 3abc.

    Solutions

    The correct answer is option 2 i.e. c- a3 - b3.

    (x + y)3 = x3 + y3 + 3xy(x + y)

    If a + b = c ..........(I)

    a + b - c = 0

    So, (a + b - c)3 = 0

    Cubing eq. (I) On both sides we get 

    (a + b)3 = c3

    ⇒ a3 + b3 + 3ab(a + b) = c3

    ⇒ 3abc = c3 - a3 - b3

    So,

    (a + b - c)3 + 3abc

    = 0 + c3 - a3 - b3 

    = c3 - a3 - b3

  • Question 6/10
    2 / -0.5

    A train running at 58 km/hr is trying to overtake a train running at 49 km/hr. The faster train overtakes the slower train in 3 minutes. If train with faster speed is shorter than the other train by 10 m, find the length of the longer train.

    Solutions

    The correct answer is Option 3 i.e. 230 m.

    Speed of faster train = 58 km/hr

    Speed of slower train = 49 km/hr

    Relative speed (same direction) = 58 - 49 = 9 km/hr = 9 × (5/18) = 2.5 m/s

    Time = 3 minutes = 3 × 60 = 180 seconds.

    Distance travelled = Sum of lengths of both trains = Relative speed × time

    Let length of trains be x and x + 10.

    ⇒ x + x + 10 = 2.5 × 180 

    ⇒ 2x + 10 = 450

    ⇒ 2x = 440

    ⇒ x = 220

    Length of longer train = x + 10 = 220 + 10 = 230 m

  • Question 7/10
    2 / -0.5

    Directions For Questions

    Direction: The line graph below shows the number of youth addicted to Drugs and the number of youth inclined towards Yoga in 5 different states A, B, C, D and E. Study the graph carefully and answer the questions that follow. All the values in the graph are in lakhs.

    ...view full instructions


    The total number of youth addicted to drugs in states A and E is approximately what percentage of the total number of youth inclined towards yoga in states C and D?

    Solutions

    The correct answer is option 3 i.e. 88.5%.

    Total number of youth addicted to drugs in states A and E

    = 12 + 15 = 27 lakhs

    And

    Total number of youth inclined towards Yoga in states C and D

    = 18 + 12.5 = 30.5 lakhs

    Hence,

    Required percentage = [27/30.5] × 100 = 88.5% (Approx.)

  • Question 8/10
    2 / -0.5

    In a cyclic quadrilateral PQRS, SR, and PQ are extended to meet at point T. If TS = 16 cm, TR = 12 cm, and TQ = 8 cm, then the length of the TP in cm is?

    Solutions

    The correct answer is option 1 i.e. 24 cm.

    TS × TR = TP × TQ

    ⇒ 16 × 12 = TP × 8

    ⇒ TP = (16 × 12)/8

    ⇒ TP = (2 × 12) cm = 24 cm

    Hence, the length of TP is 24 cm

  • Question 9/10
    2 / -0.5

    Rs. 5000 was divided into two parts, if one part is deposited at 8% and another at 10%, the whole annual interest sums up to Rs. 450. How much was lent at 8%?

    Solutions

    The correct answer is option 1 i.e. Rs. 2500.

    Let the amount lent at an interest rate of 8% be Rs. x

    Amount lent at 10% interest rate = Rs. (5000 - x)

    Simple Interest = (P × R × T)/100

    Where, P = Principal , R = Rate of interest, T = time period in years

    Simple Interest from amount lent at 8% = x × 8 × 1/100 = 0.08x

    Simple Interest from amount lent at 10% = (5000 - x) × 10 × 1/100 = 500 - 0.1x

    According to the question -

    ⇒ 0.08x + 500 - 0.1x = 450

    ⇒ 0.02x = 50

    ⇒ x = 50/0.02

    ⇒ x = Rs. 2500

  • Question 10/10
    2 / -0.5

    If 450 apples are distributed among Sita, Reeta, and Geeta in the ratio of 27 : 11 : 7. How many more apples does Sita have than Reeta?

    Solutions

    The correct answer is option 1 i.e. 160.

    Total apples = 450

    ⇒ Number of apples with Sita = [450/(27 + 11 + 7)] × 27

    ⇒ 450 × 27/45

    ⇒ 270

    ⇒ Number of apples with Reeta = [450/(27 + 11 + 7)] × 11

    ⇒ 450 × 11/45

    ⇒ 110

    Required number of apples = 270 - 110 = 160

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