IBPS Clerk Prelims: Quants Day 20

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. Two articles are bought for Rs 19,500. After this one is sold at a loss of 20% and the other at a profit of 15%. Find the respective cost prices of the two articles, if after selling it is found that the selling price of both the articles is same.
    A. 10,500 and 9,000
    B. 11,500 and 8,000
    C. 11,000 and 8,500
    D. 9,500 and 10,000
    E. None of these
    Answer & Explanation
    B. 11,500 and 8,000
    Explanation:

    Let CP of 1st article = x, then of other = 19500 – x
    One at loss of 20%, other at 15% gain. SP is same
    So, 80/100 * x = 115/100 (19500-x)
    80x + 115x = 115*19500
    x = 115*19500/195 = 11500
  2. A jar contains 3 red balls, 4 blue balls, and 6 white balls. Find the probability of at least 2 red balls, when 3 balls are drawn at random.
    A. 11/144
    B. 31/286
    C. 15/132
    D. 19/286
    E. None of these
    Answer & Explanation
    B. 31/286
    Explanation:

    Case 1 : 2 red balls:
    Prob = 3C2 × 10C1 / 13C3 = 15/143.
    Case 2 : 3 red balls:
    Prob = 3C3 / 13C3 = 1/286.
    So total prob. = 15/143 + 1/286= 31/286
  3. A and B together can complete the work in 10 days. B and C can complete the same work in 20 days. A works for 5 days and B works for 7 days, and now the remaining work is done by C in 26 days. Find the number of days for which B can alone work to complete the job.
    A. 25
    B. 26
    C. 30
    D. 32
    E. None of these
    Answer & Explanation
    C. 30
    Explanation:

    5 times A’s 1 day’s work + 7 times B’s 1 day’s work + 26 times C’s 1 day’s work = 1
    Or it can be written as
    5 times A’s 1 day’s work + (5 times B’s 1 day’s work+2 times B’s 1 day’s work) + (2 times C’s 1 day’s work+24 times C’s 1 day’s work) = 1
    (5 times A’s 1 day’s work + 5 times B’s 1 day’s work) + (2 times B’s 1 day’s work + 2 times C’s 1 day’s work) + 24 times C’s 1 day’s work) = 1
    5 (A+B)’s work + 2 (B+C)’s work + 24 C’s work = 1
    5 * 1/10 + 2 * 1/20 + 24 * 1/C = 1
    Solve, C = 60 days
    B and C complete in 20 days. So B’s 1 day’s work is 1/20 – 1/60 = 1/30
  4. Find the number of ways in which the letters of the word COMPLETE be arranged such that all the vowels of the given word are always together.
    A. 2180
    B. 2200
    C. 2060
    D. 2160
    E. None of these
    Answer & Explanation
    D. 2160
    Explanation:

    CMPLT OEE
    We have to take OEE together, so in all we have 6 letters {C, M, P, L, T, OEE}
    Ways for these 6 is 6!
    Now for OEE, ways are 3!, and since it contains 2 E’s, so ways are 3!/2!
    Total ways = 6! × 3!/2! = 2160
  5. Two articles are bought for Rs 6000. The first one is sold at 10% loss and the other at a gain of 10%. In this whole transaction a gain of Rs 100 is made on the whole. Find the cost price of the article sold at 10% gain is
    A. Rs 3870
    B. Rs 3500
    C. Rs 2850
    D. Rs 4070
    E. Rs 4225
    Answer & Explanation
    B. Rs 3500
    Explanation:

    Gain% on the whole = 100/6000 * 100 = 5/3%
    By the method of allegation:
    -10                                                 10
    .                             5/3
    10 – 5/3 = 25/3                5/3 – (-10) = 35/3
    Ratio = 25/3 : 35/3 = 5 : 7
    So CP of 2nd article = 7/(5+7) * 6000 = 3500
  6. A 100 litres solution of sugar and water consists of 20% of sugar. The solution is heated and the water evaporates. After this sugar concentration becomes 80% of the total remaining solution. Find the amount of water being evaporated.
    A. 78 litres
    B. 60 litres
    C. 75 litres
    D. 62 litres
    E. 64 litres
    Answer & Explanation
    C. 75 litres
    Explanation:

    In 100 litres solution, sugar = 20 l, water = 80 l
    Suppose x litres of water evaporated.
    So now the solution becomes (100-x) litres and water becomes (80-x) litres.
    In this (100-x) litres, 20% is water.
    So 20/100 (100-x) = 80-x
    Solve, x = 75
  7. A town has a total population 1,20,000. In this town the percentage of males is 55% and the rest are females. If only 48% of the males and 60% of the females are allowed to vote, then find the total number of voters present.
    A. 64,320
    B. 64,080
    C. 70,000
    D. 69,890
    E. 62,080
    Answer & Explanation
    B. 64,080
    Explanation:

    Males = 55/100 * 1,20,000 = 50/100 of 1,20,000 + 5/100 of 1,20,000 = 60,000 + 6000 = 66,000
    Male voters = 48/100 * 66,000 = 50% of 66,000 – 2% of 66,000 = 33,000 – 1320 = 31680
    Females = 1,20,000 – 66,000 = 54,000
    Female voters = 60/100 * 54,000 = 32400
    Total voters = 31680 + 32400
  8. A man can row 10 km/hr in still water. When the river is running at 4 km/hr, it takes him 5 hrs to row to a place and come back. How far is the place?
    A. 20 km
    B. 22 km
    C. 21 km
    D. 35 km
    E. 24 km
    Answer & Explanation
    C. 21 km
    Explanation:

    B is speed of boat in still water, R is speed of stream
    Time is total time taken for upstream and downstream
    Distance = time * [B^2 – R^2] / 2*B
    =5 * [10^2 – 4^2] / 2*10
  9. In a rectangle if its length increases by 25% and the breadth decreases by 20%, then what is the percentage change in the perimeter of the rectangle?
    A. 20%
    B. 0%
    C. -5%
    D. Cannot be determined
    E. None of these
    Answer & Explanation
    D. Cannot be determined
    Explanation:

    Perimeter = 2(l+b)
    New perimeter = 2(1.25l + 0.80b)
    Change in perimeter = 2 [(1.25l – l) – (0.80b – b)] = 2[0.25l + 0.20b] As the values of l and b are not known, change cannot be determined.
  10. The area of a square is 450 cm2. Find the length of diagonal.
    A. 30 cm
    B. 44 cm
    C. 28 cm
    D. 35 cm
    E. 32 cm
    Answer & Explanation
    A. 30 cm
    Explanation:

    1/2 d2 = 450
    d2 = 900, d = 30