IBPS Clerk Prelims: Quants Day 24

Hello Aspirants. Welcome to Online Quant Section in AffairsCloud.com. We are starting IBPS Clerk course 2015 and we are creating sample questions in Quantitative Aptitude section, type of which will be asked in IBPS Clerk Prelims Exam.

Stratus – IBPS Clerk Course 2015 

Stratus - IBPS Clerk - Daily Test - Quants
[flipclock]

  1. x2 = 4, 2y2 – 13y + 20 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    Answer & Explanation
    B) x < y
    Explanation:

    x2 = 4
    x = 2 , -2
    2y2 – 13y + 20 = 0
    2y2 – 8y – 5y +20 = 0
    2y(y-4) – 5(y-4) = 0
    y = 4, 5/2
    put on number line
    -2              2                   5/2                 4
  2. 2x + 5y = 11, x – 20y = -8
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    Answer & Explanation
    A) x > y
    Explanation:

    To solve the equations, multiply 2nd equation by 2, and subtact both equations:
    2x + 5y = 11
    2x – 40y = -16
    On subtraction – 0x + 45y = 27
    y = 3/5
    now put y = 3/5 in any of the equations to find x, x= 4
  3. 2x2 + x – 36 = 0, y2 + y – 6 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    Answer & Explanation
    E) x = y or relationship cannot be determined
    Explanation:

    2x2 + x – 36 = 0
    2x2 – 8x + 9x – 36 = 0
    x = 4 , -9/2
    y2 + y – 6 = 0
    y = -3, 2
    put on number line
    -9/2              -3             2               4
  4. x2 – 5x = 0, 2y2 – y – 3 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    Answer & Explanation
    E) x = y or relationship cannot be determined
    Explanation:

    x2 – 5x = 0
    x(x-5) = 0
    x = 0, 5
    2y2 – y – 3 = 0
    2y2 +2y – 3y – 3 = 0
    y = -1, 3/2
    put on number line
    -1             0                 3/2                  5
  5. 2x2 – 9x + 4 = 0, 2y2 + 7y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    Answer & Explanation
    C) x ≥ y
    Explanation:

    2x2 – 9x + 4 = 0
    2x2 – 8x – x + 4 = 0
    x = 4 , 1/2
    2y2 + 7y – 4 = 0
    2y2 + 8y – y – 4 = 0
    y = -4, 1/2
    put on number line
    -4                   1/2                  4
  6. A mixture contains A and B in the ratio 5 : 3. 16 litres of the mixture is taken out and 16 litres of liquid B is poured in and after this the ratio of the amount of both liquids in the mixture is 15 : 17. Find the amount of liquid A present in the mixture.
    A) 24 litres
    B) 40 litres
    C) 30 litres
    D) 32 litres
    E) None of these
    Answer & Explanation
    B) 40 litres
    Explanation:

    Method of doing these type of questions when same quantity is taken out and poured in:
    Initial ratio = 5 : 3, 16 litres taken out
    Let total quantity = 5x+3x + 16 = 8x+16
    After pouring 16 litres of B, ratio is 15 : 17. So
    5x/(3x+16) = 15/17
    Solve, x = 6
    So total quantity = 8*6 + 16 = 64
    So A = 5/8 * 64 = 40 litres
  7. A and B can complete a work in 10 days and 14 days respectively. Both start this work and A left after 4 days. B completes the remaining work in how many days?
    A) 9 days
    B) 9 4/5 days
    C) 5 3/5 days
    D) 4 2/5 days
    E) None of these
    Answer & Explanation
    D) 4 2/5 days
    Explanation:

    A+B)’s 1 day’s work = 1/10 + 1/14 = 12/70
    So work done in 4 days = 4 * 12/70 = 24/35
    Now A left.
    Work left = 1 – 24/35 = 11/35
    B does 1 work in 14 days so 11/35 work in 14 * 11/35 = 22/5 days
  8. P and Q starts from point A to B 21 km away at the same time. P travels at 3 km/hr and Q at 4 km/hr. Q reaches point B and then returns towards point A and he meets P in between. Find the distance travelled by Q when he meets P.
    A) 20 km
    B) 24 km
    C) 22 km
    D) 18 km
    E) 26 km
    Answer & Explanation
    B) 24 km
    Explanation:

    Since both start at same time, when they meet they have travelled for same time
    Let both meet at point C and let BC = x, then A has travelled (21-x) km, and B has travelled (21+x) km
    Now times are same, so
    (21-x)/3 = (21+x)/4
    Solve, x = 3
    So B has travelled = 21+3
  9. A man buys a certain number of marbles at 2 for a Rupee and the same number of marbles at 4 for a Rupee. He sells all of them at 6 for Rs 2. What is his gain/loss percent in the whole transaction?
    A) 11% loss
    B) 11 1/9% gain
    C) 13 1/9% gain
    D) 11 1/9% loss
    E) None of these
    Answer & Explanation
    D) 11 1/9% loss
    Explanation:

    Buys certain number at 2/Re, let b1 = 2. Buys certain number at 4/Re, let b2 = 4
    He sells 6 marbles for Rs 2, so this means he sells 3 marbles for Re 1, let s1 = 3
    Next we have a direct formula for it:
    Gain/loss % = [(2*b1*b2/s1(b1+b2s)) – 1] * 100
    Gain/loss % = [(2*2*4/3(2+4)) – 1] * 100 = [(8/9) – 1] * 100 = -100/9 %
  10. At a certain rate of interest Rs 1000 becomes Rs 1320 in 4 years. Now the rate of interest is increased by 4% per annum. What will Rs 1000 amount to now?
    A) Rs 1440
    B) Rs 1480
    C) Rs 1280
    D) Rs 1400
    E) None of these
    Answer & Explanation
    B) Rs 1480
    Explanation:

    When rate increase by 4% fir 4 years, Increase in interest = (1000*4*4)/100 = Rs 160
    So increased amount = 1320+160