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Aptitude Test 236
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Aptitude Test 236
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  • Question 1/10
    1 / -0.33

    Find the next term in the series.

    841, 890, 1011, 1236, 1597, ?

    Solutions

    The series is as follows:

    841 + 72 = 890

    890 + 112 = 1011

    1011 + 152 = 1236

    1236 + 192 = 1597

    1597 + 232 = 2126

     

  • Question 2/10
    1 / -0.33

    In the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.

    I. x2 +21x +110 = 0

    II. y2 +7y +10 = 0

    Solutions

    x2 +21x +110 = (x + 11)(x + 10)= 0

    => x = -11, -10

    From equation II:

    y2 +7y +10 = (y + 5)(y + 2) = 0

    => y = -5, -2

    So, x < y

     

  • Question 3/10
    1 / -0.33

    Mohan and Sohan entered into a partnership with investment of Rs.80000 and Rs.60000 respectively. After one year, Mohan invested Rs.5000 more. After one more year, Sohan invested Rs.20000 more. At the end of three years, they earned a total profit of Rs.108000. 10% of the profit goes to a charity. Find the share of Sohan in the profit.

    Solutions

    Ratio of shares in the profit:

    Mohan : Sohan = (80000 + 85000 x 2) : (60000 x 2 + 80000)

    = (80000 + 170000) : (120000 + 80000)

    = 250000 : 200000

    = 5:4

    Share of Sohan in the profit = 4/(5 + 4) x (100 - 10)/100 x 108000

    = 4/9 x 90/100 x 108000

    = Rs.43200

     

  • Question 4/10
    1 / -0.33

    Find the next term in the series.

    26, 78, 273, 1092, 4914, ?

    Solutions

    The series is as follows:

    26 x 3 = 78

    78 x 3.5 = 273

    273 x 4.0 = 1092

    1092 x 4.5 = 4914

    4914 x 5.0 = 24570

     

  • Question 5/10
    1 / -0.33

    When a boy swims with the stream, his speed is increased by 12.5% whereas when he swims against the stream, his speed is decreased by 17.5 km per hour. How much time, the boy will take to swim 672 km in still water?

    Solutions

    Let in still water, the speed of the boy = u km per hour

    With stream, its speed = 112.5% of u = 1.125u km per hour

    Let the rate of stream = v km per hour then, u + v = 1.125u

    0.125u = v

    u/v = 1/0.125 = 8/1

    let u = 8a then v = a

    Against the stream, his speed = 8a - a = 7a km per hour

    Decrease in his speed = 8a - 7a = a km per hour = 17.5 km per hour

    Therefore, the speed of the boy in still water = 17.5*8 = 140 km per hour

    The time taken by the boy to travel 672 km in sill water @ 140 km per hour = 672/140 = 4.8 hours = 4 hours 48 minutes

     

  • Question 6/10
    1 / -0.33

    When a cow was tethered in a grass field then it grazed the total area (maximum area) of 1886.5 sq. cm. what is the perimeter of the total area grazed by the cow?

    Solutions

    When the cow was tethered in the grass field, it would graze the maximum area in circular shape

    Let the length of rope = r cm = radius of the area grazed by the cow

    The area = πr2 = 1886.5 sq. cm

    (22/7)*r2 = 1886.5

    R2 = 1886.5*7/22 = 85.75*7 = 600.25 sq. cm

    R = √600.25 = 24.5 cm

    The perimeter = 2πr = 2*(22/7)*24.5 = 154 cm

     

  • Question 7/10
    1 / -0.33

    4 years ago, ratio of ages of Umar and Vinit was 5:8 and after 4 years, the age of Umar will be 70% of the age of Vinit. What is the ratio of the present ages of Umar and Vinit?

    Solutions

    Let the present ages of Umar and Vinit be 'u' years and 'v' years respectively.

    So, (u - 4)/(v - 4) = 5/8

    => 5v - 20 = 8u - 32

    => 8u - 5v = 12..(i)

    Also, (u + 4) = (70/100)*(v + 4)

    => (u + 4) = (7/10)*(v + 4)

    => 10u + 40 = 7v + 28

    => 7v - 10u = 12..(ii)

    Solving (i) and (ii), we get

    u = 24 and v = 36

    Required ratio = 2:3

     

  • Question 8/10
    1 / -0.33

    Find the next term in the series.

    1002, 1334, 1732, 2202, 2750, ?

    Solutions

    The series is as follows:

    103 + 2 = 1002

    113 + 3 = 1334

    123 + 4 = 1732

    133 + 5 = 2202

    143 + 6 = 2750

    153 + 7 = 3382

     

  • Question 9/10
    1 / -0.33

    Directions For Questions

    Directions : The table given below gives the information about the total number of students participated in a singing competition from five different schools. It also gives the partial information about the percentage of students who participated in singing competition from class 9th and from class 10th.

    Total number of students who participated in singing competition = The total number of students who participated in singing competition from class 9th + The total number of students who participated in singing competition from class 10th

    ...view full instructions

    From the all five schools together, the total number of class 9th participants were how much more/less than that of class 10th participants?

    Solutions

    The total number of class 9th participants = 580 + 819 + 962 + 441 + 738 = 3540

    The total number of class 10th participants = 870 + 441 + 518 + 539 + 902 = 3270

    The required difference = 3540 - 3270 = 270

     

  • Question 10/10
    1 / -0.33

    Rs 10000 is lent to A and B in some proportion at 6% simple interest. The total amount owed by A at the end of 8 years is equal to the total amount owed by B at the end of 10 years. Find the approximate amount lent individually to A and B.

    Solutions

    Let's suppose A's and B's shares be 'a' and 'b' respectively.

    According to the question; a + b = 10000... (1)

    Also, a + (a X 6 X 8)/100 = b + (b X 6 X 10)/100

    => a + (12a/25) = b + (15b/25)

    => 37a/25 = 40b/25

    => 37a = 40b... (2)

    From (1) and (2) we will get, a = Rs 5200 and b = Rs 4800

     

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