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Aptitude Test - 87
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Aptitude Test - 87

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  • Question 1/10
    1 / -0

    If sin x + cot x = a and cosec x + tan x = b, find the value of cos x.

    Solutions

    Sin x + cot x = a

    1/sin x + 1/cot x = b

    (Cot x + sin x)/(sin x cot x) = b

    a/ (sin x cot x) = b

    a/b = sin x cot x

    a/b = sin x cos x/sinx

    a/b = cos x

     

  • Question 2/10
    1 / -0

    A company produces 26678 toys in each of the first two months of a year. Due to man power shortage the number of toys produced reduces in the third month so that the average production in the three months falls to 25490. Find the reduction the production of toys in the third month with respect to the second month.

    Solutions

    Average production in 3 months = 25490

    Total production in 3 months = 25490×3 = 76470

    Production in first two months = 26678 + 26678 = 53356

    Production in third month = 76470 – 53356 = 23114

    Reduction in production = 26678 – 23114 = 3564

     

  • Question 3/10
    1 / -0

    The cost of 10 vegetables is equal to the cost of 15 fruits. The cost of 9 fruits is equal to that of 10 sweets. The cost of 5 sweets is equal to the cost of 6 chocolates. If the cost of 10 vegetables is 300, find the cost of 10 fruits, 5 sweets and 5 chocolates.

    Solutions

    Let V, F, S, C represent cost of vegetable, fruit, sweet, chocolate respectively.

    10V = 15F

    So, 30V = 45F

    9F = 10S

    So, 45F = 50S

    5S = 6C

    So, 50S = 60C

    30V = 45F = 50S = 60C

    Ratio of cost of V : F : S : C = 1/30 : 1/45 : 1/50 : 1/60

    = 30 : 20 : 18 : 15

    Cost of 10V = 300

    Cost of 1 V = 30

    Cost of 1 F = 20

    Cost of 1 S = 18

    Cost of 1 C = 15

    Cost of (10F + 5S + 5C) = 10×20 + 5×18 + 5×15

    =200 + 90 + 75

    = 365

     

  • Question 4/10
    1 / -0

    The mean of 25 values is found out to be 40. Upon observation it is found that a value has been written as 25 instead of 50. Find the original mean.

    Solutions

    Mean=40

    Number of observation = 25

    Total = 40×25 = 1000

    50 has been wrongly entered as 25

    So, sum of observation=1000 – 25 + 50 = 1025

    Mean = 1025/25 = 41

     

  • Question 5/10
    1 / -0

    A man sells an item for Rs. 2400 at 10% discount. For how much does he need to sell 2 items to gain 5 %?

    Solutions

    Let CP for man be x

    SP = 0.9x = 2400

    x = 8000/3

    For 5% profit, SP = 1.05 × 8000/3

    = 2800

    SP of 2 items = 2800 × 2 = 5600

     

  • Question 6/10
    1 / -0

    A man buys an item at Rs. 42000. If he promises to pay the amount at 8% Simple interest in 12 equal monthly installments, what is the amount that he pays every month as installment?

    Solutions

    SI for 12 months = SI for 1 year = 42000×8/100

    = 3360

    Amount = 42000 + 3360 = 45360

    Monthly installment = 45360/12 = 3780

     

  • Question 7/10
    1 / -0

    A man sells basmati rice of first quality at which costs Rs. 250/kg with inferior variety that costs Rs. 120/kg. If he sells 10 kg of the mixture at Rs. 2235.2 and earns 10% profit, find the ratio of first quality rice to that of inferior quality rice.

    Solutions

    Let there be x kg of first quality rice and 10 – x kg of inferior rice

    Cost of 10kg = 250x + 120(10 – x)

    = 130x + 1200

    SP of 10 kg = 2235.2

    This gives 10% profit. Let CP = y

    1.1y = 2235.2

    Y = 2032

    But, this is equal to 130x + 1200

    130x + 1200 = 2032

    130x = 832

    x = 6.4kg (First quality rice)

    Inferior quality rice = 3.6 kg

    Ratio = 6.4/3.6 = 64/36 = 16 : 9

     

  • Question 8/10
    1 / -0

    27 men can do a piece of work in 4 days less than that of the number of days which 9 men take to do the work. Find the time taken by 6 men to do the work.

    Solutions

    M1T1/W1 = M2T2/W2

    Let the Time taken by 9 men = x days

    Time taken by 27 men to complete the work be x - 4 days

    27×(x – 4)/1 = 9x

    27x – 108 = 9x

    18x = 108

    x = 6 days

    Similarly, 9×6/1 = 6×y/1(y is time taken by 6 people to do the work)

    y = 9 days

     

  • Question 9/10
    1 / -0

    Find the last digit of the following expression.

    276×83×2323×1151

    Solutions

    Cyclicity of 7 is 4

    So, last digit of 276 is 9

    Last digit of 83 = 2

    Cyclicity of 3 = 4

    So, last digit of 2323 = 7

    Last digit of expression = Last digit of 9×2×7 = 6

     

  • Question 10/10
    1 / -0

    The ratio of ages of A and B before ten years was 4 : 9. If the sum of their ages now is 72, find the ratio of their ages after 4 years.

    Solutions

    Let their ages be x and y

    x + y = 72

    Before 10 years, (x – 10)/(y – 10) = 4/9

    9x – 90 = 4y – 40

    9x – 4y = 50

    4x + 4y = 288

    Adding these two equations, 13x = 338

    x = 26

    So, y = 36

    After 4 years, ages are 30, 50

    Ratio = 3 : 5

     

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