Quant Test for SBI PO 2018 Prelim Exam Set – 12

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

Click here to know the details of the Examination

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under:

Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam. 


 

Directions(1-5): Find the relationship between x and y and choose an option accordingly.

  1. I. 2x² + x – 300 = 0
    II. y² – 45y + 324 = 0

    y>x
    x>=y
    x>y
    x=y or relation cannot be established.
    y>=x
    Option D
    I. 2x² + x – 300 = 0
    2x² + x – 300 = 0
    x = 12, -12.5
    II. y² – 45y + 324 = 0
    y² – 45y + 324 = 0
    y = 36, 9
    x = y or relation cannot be established

     

  2. I. 12/√x – 23/√x = 5√x
    II. √y/12 – 5√y/12 = 1/√y

    x>y
    y>x
    x>=y
    y>=x
    x=y or relation cannot be established.
    Option A
    I. 12/√x – 23/√x = 5√x
    =>12/√x – 23/√x = 5√x
    =>-11 = 5x
    => x = -2.2
    II. √y/12 – 5√y/12 = 1/√y
    => √y/12 – 5√y/12 = 1/√y
    => √y[1/12 – 5/12]= 1/√y
    =>y = -3
    x > y

     

  3. I.[48 / x(4/7)] – [12 / x(4/7)] = x(10/7)
    II. y³ + 783 = 999

    x>=y
    x>y
    x=y or relation cannot be established.
    y>x
    y>=x
    Option E
    I.[48 / x4/7] – [12 / x4/7] = x10/7
    => (48 – 12) / x4/7 = x10/7
    => 36 = x(10/7 + 4/7)
    => 36 = x2
    =>x = ± 6
    II. y³ + 783 = 999
    =>y3 + 783 = 999
    =>y3 = 999 – 783
    => y3 = 216
    => y = 6
    x≤ y

     

  4. I. 6×2 + 13x + 5 = 0
    II. 9y2 + 22y + 8 = 0

    x>y
    x=y or relation cannot be established.
    y>x
    x>=y
    y>=x
    Option B
    I. 6×2 + 13x + 5 = 0
    => 6×2 + 10x + 3x + 5 = 0
    => 2x (3x + 5) +1 (3x + 5) = 0
    => (3x + 5) (2x + l) = 0
    => x = (5/3) or (1/2)
    II. 9y2 + 22y + 8 = 0
    => 9y2 + 18y + 4y + 8 = 0
    => 9y (y + 2) + 4 (y + 2) = 0
    => (y + 2) (9y + 4) = 0
    => y = -2 or -(4/9)
    x=y or relation cannot be established.

     

  5. I. 14x -25 = 59 – 7x II. (y + 222)^(1/2) – (36)^(1/2) = (81)^(1/2)
    x=y or relation cannot be established.
    y>x
    x>=y
    y>=x
    x>y
    Option E
    I. 14x + 7x = 59 + 25
    =>21x = 84
    => x = 4
    II. (y+ 222)^(1/2) = (36)^(1/2) + (81)^(1/2)
    => (y+222)^(1/2) = 6 + 9 = 15
    =>y + 222 = 225
    => y = 225 – 222 = 3
    x > y

     

  6. A water tank has three taps A, B and C. Tap A, when opened, can fill the water tank alone in 4 hours. Tap B, when opened, can fill the water tank alone in 6 hours and tap C, when opened, can empty the water tank alone in 3 hours. If taps A, B and C are opened simultaneously how long will it take to fill the tank completely ?
    15 hours
    5 hours
    12 hours
    10 hours
    8 hours
    Option C
    Part of the tank filled in 1 hour when all the taps are opened =(1/4)+(1/6)- (1/3) = 1/12
    Hence, 12 hours required to fill the tank.

     

  7. A person invested some amount at the rate of 12% simple interest and a certain amount at the rate of 10% simple interest. He received yearly interest of Rs 130. But if he had interchanged the amounts invested, he would have received Rs 4 more as interest. How much did he invest at 12% simple interest ?
    Rs.200
    Rs.400
    Rs.100
    Rs.500
    Rs.700
    Option D
    Amount invested at 12% = Rs x .
    Amount invested at 10% = Rs y
    130 = (x*12*1)/100 + (y*10*1)/100
    =>13000 = 12x + 10y—–(1)
    and 134 =(x*10*1)/100 + (y*12*1)/100
    => 13400 = 10x + 12y—–(2)
    Solving Eqs. (1) and (2), we get x = Rs 500

     

  8. A video magazine distributor made 3500 copies of the March issue of the magazine at a cost of Rs 350000. He gave 500 cassettes free to some key video libraries. He also allowed a 25% discount on the market price of the cassette and gave one extra cassette free with every 29 cassettes bought at a time. In this manner, he was able to sell all the 3500 cassettes that were produced. If the market price of a cassette was Rs150, then what is his gain or loss per cent for the March issue of video magazine ?
    4.5%
    3.12%
    5.4%
    2.15%
    6.78%
    Option E
    According to the question,
    CP of 3500 cassettes = Rs 350000
    The SP of one cassette after discount is Rs 112.50
    SP of each set of 30 cassettes = (29 +1) = Rs 29 × 112.50 = Rs 3262.50
    SP of 3500 cassettes including 500 free cassettes = Rs 326250
    Overall loss on Rs 350000 is Rs 23750
    Loss percent = (23750/350000) × 100 = 6.78%

     

  9. A car driver travels from the plains to the hill station, which are 200 km a part at an average speed of 40 km/h. In the return trip he covers the same distance at an average speed of 20 km/h the car. Find the average speed of the car over the entire distance of 400 km .
    26.67km/hr.
    20.25km/hr.
    32.4km/hr.
    45km/hr.
    None of these
    Option A
    Average speed
    =(2* 40 *20)/(20*40) = 26.67 km/hr.

     

  10. Raksha invested an amount of Rs. 60,000 to start a software business. After six months, Kamal joined her with an amount of Rs. 90,000. After one year from the commencement of the business, Raksha put in an additional amount of Rs. 20,000. At the end of three years, they earned a proflt of Rs. 71,20,000. What is Raksha’s share in the proflt ?
    Rs. 40,20,000
    Rs. 35,20,000
    Rs. 11,15,000
    Rs. 72,40,000
    Rs. 30,00,000
    Option B
    Equivalent capital of Raksha for 3 year
    = Rs. (60000 × 1 + 80000 × 2) = Rs. (60000 + 160000) = Rs. 220000
    Equivalent capital of Kamal for 3 year = Rs. 90000 × (5/2) =Rs 225000
    Ratio of capitals = 220000 : 225000 = 44 : 45
    Sum of ratios = 44 + 45 = 89
    Total profit = Rs. 7120000
    Raksha’s share = (44/89)*7120000 = Rs. 35,20,000

     


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