Sample Aptitude Questions of Cairn

  1. A copper wire is bent in the form of an equilateral triangle and has area 121√3cm2 . If the same wire is bent into the form of a circle. The area (in cm2) enclosed by the wire is (take π - 22/7)
    1. 364.5
    2. 693.5
    3. 346.5
    4. 639.5
    5. None of these
    Answer: Option C
    Area of an equilateral triangle = √3/4×a2 = 121√3cm2 We get a = 22 cm So perimeter of an equilateral triangle = 3a = 66 cm As the same wire is bent into the form of a circle, 2πr = 66 cm We get r = 21/2 cm
    Area of circle = πr2 = 346.5 cm2
  1. What will be the ratio of petrol and kerosene in the final solution formed by mixing petrol and kerosene that are present in three identical vessels in the ratio 4:1,5:2 and 6 :1 respectively?
    1. 166 : 22
    2. 83 : 22
    3. 83 : 44
    4. 78 : 55
    5. None of these
    Answer: Option B
    Three identical vessels in the ratio 4:1, 5:2 and 6:1 respectively. Petrol : kerosene (4 : 1 = 5)7 (5 : 2 = 7)5 (6 : 1 = 7)5 28 : 7 =35 25 : 10 =35 30 : 5 =35 83 : 22
  2. Mrs. Sharma invests 15% of her monthly salary, i.e., Rs. 4428 in Mutual Funds. Later she invests 18% of her monthly salary on Pension Policies also she invests another 9% of her salary on Insurance Policies. What is the total monthly amount invested by Mrs. Sharma?
    1. Rs. 113356.8
    2. Rs. 12398.4
    3. Rs. 56678.4
    4. Can't determined
    5. None of these
    Answer: Option B
    15% of monthly salary = Rs 4428
    So monthly salary = Rs 29500
    Total money invested = 42% of 29500 = Rs 12398.4
  3. At present, Aanshi is five years younger to Binny. Binny’s age twenty- years hence will be equal to twice of Aanshi’s age five years ago. What will be Binny’s age eight year hence?
    1. 42 years
    2. 35 years
    3. 30 years
    4. 48 years
    5. None of these
    Answer: Option D
    Let age of Aanshi be A, Bunny be B.
    According to equation, A= B-5
    Also, B+20 = 2(B-10)
    Solving these we get, B= 40. So after 8 years, his age will be 48 years.
  4. In a bag, there are 8 red balls and 7 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green in colour?
    1. 28/65
    2. 22/65
    3. 37/65
    4. 3/13
    5. 1/13
    Answer: Option A
    Probability that two balls are red and one ball is green in colour = 8*7*7*3! / 15*14*13*2! = 28/65
  1. A vessel contains 120 litres of mixture of milk and water in the respective ratio of 11 : 4. Forty-five litres of this mixture was taken out and replaced with 5 litres of water. What is the percentage of water in the resultant mixture?
    1. 1.35
    2. 31.25
    3. 25
    4. 20
    5. 15
    Answer: Option B
    Total volume 120 litres
    Milk Water
    88 32
    if 45 litres of mixture is removed 5/8 of the mixture is left
    Milk Water
    55 20
    5 litres water is added
    Milk Water
    55 25
    % of water in the mixture= 25*100/ 80 = 31.25%
  2. What would be the compound interest accrued on an amount of Rs. 8400 at the rate 12.5% per annum at the end 3 yr? (Rounded, off to two digits after decimal)
    1. Rs. 420.62
    2. Rs. 2584.16
    3. Rs. 3560.16
    4. Rs. 3820.14
    5. None of these
    Answer: Option C
    C.I = (1+R/100)T – Amount Put R=12.5% p.a Time = 3 years Amount = Rs 8400 We get C.I = (1+12.5/100)3 – 8400 C.I = 3560.16
  3. Twice the speed of a boat downstream is equal to thrice the speed upstream. The ratio of its speed in still water to the speed of current is
    1. 1 : 5
    2. 1 : 3
    3. 5 : 1
    4. 2 : 3
    Answer: Option C
    Let the boat speed in still water be b.
    Let the stream speed be x.
    2(b+ x) = 3(b-x)
    5x=b
    b/x=5/1
  4. How many terms are there in an A.P. whose first and fifth terms are -14 and 2, respectively, and the sum of terms is 40?
    1. 15
    2. 10
    3. 5
    4. 20
    Answer: Option B
    Now the common difference of this AP is 16/4 = 4.
    The sum of an AP is n/2 {2a + (n – 1)d}
    Substituting we get, 40 = n/2 {2×-14 + (n – 1)4}
    The best way to solve this is by plugging options. Put in n = 10 and get the RHS as 40.
  5. A bottle is full of Dettol. One-third of it is taken out and then an equal amount of water is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle.
    1. 13 : 55
    2. 20 : 74
    3. 16 : 65
    4. 10 : 48
    Answer: Option C
    As in denominator we have to take 1/3 four times so, we start by assuming 81 ml of dettol in the bottle. After the first iteration you will be left with 2/3 × 81 = 54 ml. After the second iteration you will be left with 2/3 × 54 = 36 ml. After the third iteration you will be left with 2/3 × 36 = 24 ml. After the fourth iteration you will be left with 2/3 × 24 = 16 ml. So the required ratio will be 16 : (81 – 16) = 16 : 65

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