Quant Test for IBPS PO 2018 Prelim Exam Set – 27

Directions(1-5): Find the relation between x and y and choose a correct option.

  1. I. 12/√π‘₯ + 8/√π‘₯ = 8√π‘₯
    II. βˆšπ‘¦/4 + 5βˆšπ‘¦/12 = 1/√y

    y>x
    y>=x
    x>=y
    x>y
    x=y or relation cannot be established.
    Option D
    I.12/√π‘₯ + 8/√π‘₯ = 8√π‘₯
    =>20/√π‘₯ = 8√π‘₯
    =>8x = 20
    =>x = 20/8
    x = 5/2
    II.βˆšπ‘¦/4 + 5βˆšπ‘¦/12 = 1/βˆšπ‘¦
    => 8βˆšπ‘¦/12 = 1/βˆšπ‘¦
    y = 3/2
    x > y

     

  2. I. 11x + 13y = 157
    II. 10x – 7y = 11

    y>=x
    x>=y
    x>y
    x=y or relation cannot be established.
    y>x
    Option E
    I. 11x + 13y = 157
    II. 10x – 7y = 11
    Y = 7
    x = 6
    y>x

     

  3. I. 25/π‘₯^2 – 12/π‘₯ + 9/π‘₯^2 = 4/π‘₯^2
    II. 9.84–2.64 = 0.95 + y^2

    x>y
    y>=x
    x>=y
    y>x
    x=y or relation cannot be established.
    Option C
    I. 25/π‘₯^2 – 12/π‘₯ + 9/π‘₯^2 = 4/π‘₯^2
    => [25+9βˆ’4]/ π‘₯^ 2 = 12/π‘₯
    x = 2.5
    II. 9.84–2.64 = 0.95 + y^2
    => y^2 = 6.25
    y = Β±2.5
    x >= y

     

  4. I. x^ 2–43x+ 462 = 0
    II. y^ 2–37y+342 = 0

    y>x
    x>y
    y>=x
    x=y or relation cannot be established.
    x>=y
    Option B
    I.x^ 2 – 22x–21x+ 462 = 0
    x = 22,21
    II.y^2 – 37y + 342 = 0
    =>y^2 – 19y – 18y + 342 = 0
    y = 19,18
    x>y

     

  5. I. 4x^2–25x+25 = 0
    II. 2y^2–13y+21 = 0

    x=y or relation cannot be established.
    x>=y
    y>x
    y>=x
    x>y
    Option A
    I.4x^2–25x+25 = 0
    =>4x^2–20x–5x+25 = 0
    x = 5, 1.25
    II.2y^2–13y+21 = 0
    => 2y^2–7y–6y+21 = 0
    y = 3.5, 3
    No relation

     

  6. The work done by a woman in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day 9 men can complete a work in 6 days then in how many days can 12 men, 12 women and 12 boys together finish the same work working 8 hours per day?
    2(5/2) days
    3(1/2) days
    2(5/7) days
    4(3/2) days
    1(1/2) days
    Option E
    8 Women = 6 Men = 12 Boys
    12M + 12W + 12B = 12M + 9M + 6M = 27M
    Now, applying the above formula, we have
    9 Γ— 6 Γ— 6 = 27 Γ— 8 Γ— D2
    D2 = [9 Γ— 6 Γ— 6]/[27 Γ— 8] = 1(1/2) days

     

  7. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, then find the number of valid votes that the other candidate got.
    2200
    2800
    2700
    3200
    2000
    Option C
    Valid votes = 80%
    Total valid votes = 7500*(80/100)
    1st candidate got 55% of the total valid votes.
    Hence the 2nd candidate should have got 45% of the total valid votes
    => Valid votes that 2nd candidate got
    = total valid votes x (45/100) 7500*(80/100)*(45/100)
    = 2700

     

  8. Graham lent sum of Rs.840 to Rajesh in the beginning of the year at a certain rate of interest. After 6 months Rs.420 is lent to the same person but the rate of interest is twice the former. At the end of the year Rs.80 is earned as total interest by graham then what is the original rate of interest.
    6.34%
    5.68%
    2.22%
    4.37%
    5.35%
    Option A
    Since the second loan is given after 6 months
    n= 1/2 [840 Γ— 𝑅 Γ— 1]/100 + [420 Γ— 2𝑅 Γ— 1]/[100 Γ— 2] = 80
    =>8.4 R + 4.2 R = 80
    =>12.6 R = 80
    R= 80/12.6 = 6.34%

     

  9. A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of C respectively (in Rs):
    Rs.100
    Rs.300
    Rs.420
    Rs.250
    Rs.400
    Option D
    A’s 5 days work = 50%
    B’s 5 days work = 33.33%
    C’s 2 days work = 16.66% [100- (50+33.33)] Ratio of contribution of work of A, B and C = 50 : 33(1/3) : 16(2/3)
    = 3 : 2 : 1
    A’s total share = Rs. 1500
    B’s total share = Rs. 1000
    C’s total share = Rs. 500
    A’s one day’s earning = Rs.300
    B’s one day’s earning = Rs.200
    C’s one day’s earning = Rs.250

     

  10. If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word β€˜SACHIN’ appears at serial number :
    601th
    589th
    610th
    590th
    615th
    Option A
    If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
    If it started with c then the remaining 5 positions can be filled in 5! Ways.
    Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
    If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
    The required word SACHIN can be obtained after the 5X5!= 600 Ways i.e. SACHIN is the 601th letter.

     


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