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    4Apr, 2018
    Quantitative Aptitude: Percentage Questions Set 9
    1. The population of a city increases at the rate of 5% per annum. There is an additional annual increase of 2 % in the population due to the influx of job seekers. What will be the % increase in the population after 2 years.
      12.24%
      14.49%
      10.15%
      14.19%
      15%
      Option B
      The net annual increase = 7% Let the initial population be 100. Population after 2 years , 100*1.07*1.07 = 114.49 % change in the population = 114.49 – 100 = 14.49%

       

    2. An Athlete decided to run a certain distance in ¼ less time than he usually takes. By what per cent must be increase his average running speed to reach his goal?
      33.33%
      25%
      32.5%
      20.05%
      11.15%
      Option A
      Let he takes t minutes to cover a distance of x m. Let his usual speed is x/t m/s. Since, he decided to take 1/4 less time than he takes usually so, he takes 3t/4 minutes to cover x m. In this case his speed has to be x/(3t/4) = 4x/3t m/s. Change in speed = (4x/3t – x/t) m/s = x/3t m/s. So, the percentage of speed = 100 = (1/3)*100 = 33.33 % (approx.)

       

    3. The population of a town increases 30% and 40% respectively in two consecutive years. After the growth the present population of the town is 18200. Then what is the population of the town 2 years ago?
      11,150
      12,000
      32,500
      25,000
      10,000
      Option E
      Required Population in the town = (18200*100*100)/(130*140) = 10,000

       

    4. The price of Basmati rice is sold at Rs. 60/ kg. The rate was Rs. 45 / kg few months before. At what percent should a family reduce its consumption so as to keep the expenditure fixed?
      20%
      25%
      13%
      18%
      34%
      Option B
      Let the quantity of rice consumed when it was 45/kg be x kg. Total monthly amount = 45x To keep the expense 45x, quantity to be consumed = 45x/60 = 3x/4 Decrease in concumption = x/4 = 0.25x Change in % = (0.25x/x)*100 = 25%

       

    5. The Seller increased the price of a product by 20% so that buyers finds it difficult to purchase the required amount.Though somehow the customer managed to purchase only 60% of the required amount. Find the net percentage change in the expenditure on that product?
      60%
      55%
      72%
      70%
      80%
      Option C
      Let the initial price be Rs.1 Increased price = 1.2 * 0.6 = 0.72 Decrease in price = (1-0.28)/1*100 = 72%

       

    6. A man is working in a firm and earn to support his family , he spends 40% of his salary and saves the remaining for the future purpose . After some years his salary is increased by 30% and he increased his expenditure by 25%. By what % does his saving increase?
      33.33%
      22.11%
      10.10%
      25.14%
      20.00%
      Option A
      Let the salary be 100. Expenditure = 40 savings = 60 New salary = 130 New expense= (40 * 125)/100 = 50 New savings = 130 – 50 = 80 Required increase in savings = (80-60)*100/60 = 33.33%

       

    7. In a XYZ school, 20% of students are below 10 years of age. The number of students above 10 years of age is 2/3 of the number of students of 10 years of age which is 54. What is the total number of students in the school?
      60
      80
      50
      112.5
      100
      Option D
      Let the total number of schools be x . Number of students above 10 years of age = 80% of x. 80% of x = 54 + 2/3×54 => x = 112.5

       

    8. An ore of iron, 10% of the ore gets wasted,out of the total production of iron hematite.From the remaining iron, only 15% is pure iron, if the pure iron obtained in a year from a mine of hematite was 2700 kg. Find the quantity of hematite mined in the year .
      20,000
      30,000
      15,000
      10,000
      18,000
      Option A
      Amount of pure iron obtained = (15/100)*(90/100)*x = 27x/200 => x = (2700*200)/27 = 20,000

       

    9. Agriculture Department is predicting the area-wise sales of sugarcane for the next year. Area I with current yearly sales to Rs. 92.5 lakhs is expected to achieve a sales growth of 6.25%; Area II with current sales of Rs. 81.3 lakhs is expected to grow by 7.3% and Area III with sales of Rs. 60.5 lakhs is expected to increase sales by 8.15%. What is the expected sales growth for the next year?
      6.5%
      5.4%
      7.3%
      8.0%
      7.10%
      Option E
      Total sales = (92.5 * 6.25%) + (81.3 * 7.3%) + (60.5 * 8.15%) = 5.78 + 5.93 + 4.93 = 16.64 (approx.) Total sales from all the areas = 234.3lakhs Overall percentage growth = {16.64/234.3*100} = 7.10%

       

    10. The total car manufactured in a State is 294000, out of which 150000 are made by TATA. Out of every 1000 TATA cars, 98 are white in colour. But only 5.3% of the total car production is white . Find percentage of non-TATA cars that are white.
      0.303%
      0.550%
      0.612%
      0.464%
      0.125%
      Option C
      Number of cars that are white = (98/1000) * 150000 = 14700 Total number of white cars = 294000*(5.3/100) = 15582 Number of non-TATA cars = 15582 – 14700 = 882 Percenatge of non-TATA cars that are white = {882/(294000-150000)}*100% = 0.612%

       


    16Jul, 2017
    Quantitative Aptitude: Data Interpretation Questions Set 43 (Missing DI)

    Missing DI, Data Interpretation Questions for IBPS PO, IBPS Clerk, NIACL, NICL, SBI PO, RBI Grade B, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams.

    Click on images to open in new tab

    Directions (1-5): In the given table data is given about the shop.Some electronic goods are  sold by a shop. Some data is missing. Study the table carefully and answer the questions based on data in table and particular questions.

    1. What is the ratio of the cost price of mobile and cost price of laptop?
      A)  4:2
      B)  5:3
      C)  6:4
      D)  5:4
      E) None of these
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option D
      Solution:
      Cost Price of mobile  = 25000 * 100/ (100+25) = 20000
      Cost Price of laptop = 20000 * 100/ (100+25) = 16000
      Required Ratio = 20000 : 16000 = 5 : 4[/su_spoiler]
    2. Market Price of T.V is how much percent more than the cost price of T.V ?
      A)  37.5 %
      B)  35.5%
      C)  36.0%
      D)  40.7%
      E)  32.6%
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
      Solution:
      Amount of profit on T.V  = 2000
      Cost Price = 2000* 100/10 = 20000
      Selling Price = 20000 + 2000 = 22000
      Market Price = Selling Price  * 100 / (100 – Discount%)
                                   =  20000 * 100 /80 = 27500
      Required Percentage = [(27500 -20000) /20000] * 100 = 37.5%
      [/su_spoiler]
    3. What is the ratio between the percentage of profit and percentage of discount of Mobile ?
      A)  3:2
      B)  4:3
      C)  2:3
      D) 4:5
      E) 4:3
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option C
      Solution:
      Percentage of profit on Mobile  = 25%
      Percentage of discount on Mobile = [40000 – 25000/40000]*100
      .                                                                    = 37.5%
      Required Ratio = 25 : 37.5
      .                               = 2 : 3
      [/su_spoiler]
    4. Selling Price of D.V.D is what percent of cost price of A.C ?
      A)  5.6%
      B)  7.2%
      C)  5.3%
      D)  3.4%
      E) None of these
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
      Solution:
      Market Price of D.V.D = 500 * 100/20 = 2500
      Selling Price of D.V.D = 2500 – 500 = 2000
      Selling Price of A.C = 50000 * 60 / 100 = 30000
      Cost Price of A.C = 30000 * 100/80 = 37500
      Required Percentage = (2000/37500)*100 = 5.3%
      [/su_spoiler]
    5. What is the amount (in Rupees) of profit /loss  of shop by selling all electronic goods ?
      A)  5000
      B)  5700
      C)  5600
      D) 5200
      E)  5500
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option B
      Solution:
      Selling Price of A.C = 50000 * 60/100 = 30000
      Loss on A.C = 30000 * 100/80 – 3000 = Rs.7500
      Market Price of D.V.D  = 500 * 100 /20 = 2500
      Selling Price of D.V.D = 2500 – 500 = 2000
      Profit on D.V.D =2000 – 1800 = Rs.200
      Profit on T.V = Rs. 2000
      Profit on Mobile = 25000 – 25000 * 100/25 = Rs. 5000
      Profit on Laptop = 20000 – 20000 * 100 / 125 = Rs. 4000
      Profit on Radio = 10,000 * 20 /100 = Rs.2000Total Profit =  -7500 +  200 + 2000 + 5000 + 4000 + 2000
      = Rs. 5700 [/su_spoiler]

    Directions (6-10): The given tables show the time taken by filling and emptying pipes to fill and empty a tank respectively. Some data is missing. Study the table carefully and answer the questions based on data in table and particular questions.

    1. Pipes B and D are to fill the tank. They both are opened. After 8 minutes pipe R was also opened. It took 3 11/39 minutes more to fill the tank. In how much time pipe R can empty the tank?
      A) 24 minutes
      B) 48 minutes
      C) 50 minutes
      D) 60 minutes
      E) 72 minutes
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option C
      Solution:
      B+D = 1/25 + 1/40 = 13/200
      Worked for 8 minutes, so did 13/200 * 8 = 13/25 work
      So 200/13 – 8 = 96/13 minutes left to complete work,
      But after R was opened, it took 3 11/39 minutes more, means total 96/13 + 128/39 = 416/39 = 32/3 minutes
      So now  B+D – R
      13/25  + (13/200 – 1/R) * 32/3 = 1
      Solve, x = 50 minutes[/su_spoiler]
    2. Pipes A, C and D are opened to fill a tank. After 10 minutes A and D were closed and C had to fill the remaining tank. But accidentally at the same time pipe Q was also opened which was closed after 3 minutes. Now C filled the tank. If pipe Q was not opened, how much time would have been saved?
      A) 1 minute
      B) 5 minutes
      C) 3 minutes
      D) 4 minutes
      E) 2 minutes
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option E
      Solution:
      A+C+D = 1/24 + 1/48 + 1/40 = 7/80
      Worked for 10 minutes, so filled 7/80 * 10 = 7/8th of tank
      Now C and Q are opened for 3 minutes — (1/48 – 1/72)* 3 = 1/48th filled now
      So 7/8 + 1/48 + 1/48 *x = 1
      Solve,  x = 5 minutes
      So total time = 10+3+5 = 18 minutes
      If Q was not opened::
      1/8th of tank was to be filled by C
      So 1/48 * u = 1/8,
      Solve, u = 6
      So total time here = 10+6 = 16 minutes
      So time that would have been saved = 18-16 = 2 minutes[/su_spoiler]
    3. 11/40th of a tank was filled. Pipes C and D were opened. After some time pipe P was also opened. Pipe P was closed after time equal to the time it was opened at the starting . If after closing pipe P, the tank was filled in 6 minutes, find what fraction of tank was filled by pipe C?
      A) 7/8th
      B) 3/8th
      C) 2/7th
      D) 5/7th
      E) None of these
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option B
      Solution:
      C+D = 1/48 + 1/40 =  11/240
      Let after x minutes P was opened, so P was opened for x minutes [as given]
      So
      11/40 + (1/48 + 1/40) * x + (1/48 + 1/40 – 1/60)*x + (1/48 + 1/40) * 6 = 1
      OR  11/40 + (1/48 + 1/40) * 2x – 1/60 * x + (1/48 + 1/40) * 6 = 1
      Solve, x = 6 minutes
      So C was opened for 2x+6 = 18 minutes
      So filled 1/48 * 18 = 3/8 of tank
      [/su_spoiler]
    4. Pipe D was to fill a tank. To increase its efficiency, pipe A and C were opened for 2 minutes each starting with pipe A, then C and so on. This procedure ran for 16 minutes, after which pipe A and C were closed. Accidentally pipe S was opened and now it took 4 minutes more to fill the tank. Find in how much time pipe S can empty 3/40th of tank.
      A) 11 minutes
      B) 12 minutes
      C) 3 minutes
      D) 7 minutes
      E) 6 minutes
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option E
      Solution:
      For 2 minutes A and D. So
      (1/24 + 1/40)* 2= 2/15
      For next 2 minutes C and D. So
      (1/48 + 1/40)* 2= 11/120
      So in 4 minutes — 2/15 + 11/120 = 9/40
      Procedure ran for 16 minutes, so multiply by 4 both sides
      In 16 minutes, tank filled = 9/40 * 4 = 9/10
      Now 1/10 tank left which D can fill in 4 minutes, but with D and S opened, took 4 more minutes, so 8 minutes
      So (1/40 – 1/x) * 8 = 1/10
      Solve, x = 80 minutes
      Full tank in 80 minutes so 3/40th in 3/40 * 80 = 6 minutes [/su_spoiler]
    5. To fill an empty tank all filling pipes A, B, C and D are opened one by one in same order. A and C are opened for 4 minutes each while B and D for 5 minutes each. In this way, in how much time the tank will be filled (approximately)?
      A) 30 minutes
      B) 32 minutes
      C) 27 minutes
      D) 22 minutes
      E) 35 minutes
      [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option A
      Solution:
      A for 4 minutes = 1/24 * 4  = 1/6
      B for 5 minutes = 1/25 * 5  = 1/5
      C for 4 minutes = 1/48 * 4  = 1/12
      D for 5 minutes = 1/40 * 5  = 1/8
      So in 18 minutes (4+5+4+5), tank filled is 1/6 + 1/5 + 1/12 + 1/8 = 23/40
      Next A’s turn for 4 minutes = 1/6
      Next B’s turn for 5 minutes = 1/5
      Filled now = 23/40 + 1/6 + 1/5 = 113/120
      Now 7/120 is left
      Now C’s turn, full tank in 48 minutes, so 7/120 in  48 * 7/120 = 14/5 minutes
      So total = 18 + 4 + 5+ 14/5 = 27 + 3 = 30 mins approx [/su_spoiler]

    Click here for more Data Interpretation Sets

     

     

     

    14Jun, 2017
    Quantitative Aptitude: Quadratic Equations Set 19

    Quadratic Equations Practice Sets IBPS PO NICL, NIACL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

    Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

    1. I. 3x2 + 8x + 4 = 0,
      II. 2y2 – 7y – 4 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
      Solution: 
      3x2 + 8x + 4 = 0
      3x2 + 6x + 2x + 4 = 0
      Gives x = -2/3, -2
      2y2 – 7y – 4 = 0
      2y2 + y – 8y – 4 = 0
      Gives y = 4, -1/2
      Put all values on number line and analyze the relationship
      -2…. -2/3….-1/2….. 4 [/su_spoiler]
    2. I. 2x2 – 13x + 20 = 0,
      II. 3y2 + 4y – 20 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
      Solution: 
      2x2 – 13x + 20 = 0
      2x2 – 8x – 5x + 20 = 0
      Gives x = 5/2, 4
      3y2 + 4y – 20 = 0
      3y2 – 6y + 10y – 20 = 0
      Gives y = 2, -10/3
      [/su_spoiler]
    3. I. 3x2 + x – 14 = 0,
      II. 3y2 – 5y – 12 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
      Solution: 
      3x2 + x – 14 = 0
      3x2 – 6x + 7x – 14 = 0
      Gives x = 2, -7/3
      3y2 – 5y – 12 = 0
      3y2 – 9y + 4y – 12 = 0
      Gives y = -4/3, 3
      [/su_spoiler]
    4. I. 3x2 – 2x – 21 = 0,
      II. 3y2 + 19y + 28 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
      Solution: 
      3x2 – 2x – 21 = 0
      3x2 – 9x + 7x – 21 = 0
      Gives x = -7/3, 3
      3y2 + 19y + 28 = 0
      3y2 + 12y + 7y + 28 = 0
      Gives y= -7/3, -4
      [/su_spoiler]
    5. I. 4x2 + 23x + 28 = 0,
      II. 4y2 – y – 14 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
      Solution: 
      4x2 + 23x + 28 = 0
      4x2 + 16x + 7x + 28 = 0
      Gives x = -4, -7/4
      4y2 – y – 14 = 0
      4y2 – 8y + 7y – 14 = 0
      Gives y= -7/4, 2 [/su_spoiler]
    6. I. 4x2 + x – 18 = 0,
      II. 4y2 – 3y – 27 = 0
      A) If x > y
      B) If x < y
      C) If x ≥ y
      D) If x ≤ y
      E) If x = y or relation cannot be established
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
      Solution: 
      4x2 + x – 18 = 0
      4x2 – 8x + 9x – 18 = 0
      Gives x = -9/4, 2
      4y2 – 3y – 27 = 0
      4y2 – 12y + 9y – 27 = 0
      Gives y = -9/4, 3
      [/su_spoiler]
    7. I. 3x2 – 16x + 21 = 0,
      II. 2y2 – y – 6 = 0
      A) x > y
      B) x < y
      C) x ≥ y
      D) x ≤ y
      E) x = y or relationship cannot be determined
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
      Solution: 
      3x2 – 16x + 21 = 0
      3x2 – 9x – 7x + 21 = 0
      Gives x = 3, 7/3
      2y2 – y – 6 = 0
      2y2 – 4y + 3y – 6 = 0
      So y = -3/2, 2
      [/su_spoiler]
    8. I. 3x2 + x – 2 = 0,
      II. 4y2 + 13y + 10 = 0
      A) x > y
      B) x < y
      C) x ≥ y
      D) x ≤ y
      E) x = y or relationship cannot be determined
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
      Solution: 
      3x2 + x – 2 = 0
      3x2 + 3x – 2x – 2 = 0
      Gives x = -1/3, 2
      4y2 + 13y + 10 = 0
      4y2 + 8y + 5y + 10 = 0
      So y = -2, -5/4
      [/su_spoiler]
    9. I. 4x2 + 29x + 45 = 0,
      II. 4y2 – 3y – 27 = 0
      A) x > y
      B) x < y
      C) x ≥ y
      D) x ≤ y
      E) x = y or relationship cannot be determined
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
      Solution: 
      4x2 + 29x + 45 = 0
      4x2 + 20x + 9x + 45 = 0
      Gives x = -5, -9/4
      4y2 – 3y – 27 = 0
      4y2 – 12y + 9y – 27 = 0
      So y = -9/4, 3
      [/su_spoiler]
    10. I. 3x2 – 22x + 35 = 0,
      II. 3y2 – 16y + 21 = 0
      A) x > y
      B) x < y
      C) x ≥ y
      D) x ≤ y
      E) x = y or relationship cannot be determined
      [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
      Solution: 
      3x2 – 22x + 35 = 0
      3x2 – 15x – 7x + 35 = 0
      Gives x = 7/3, 5
      3y2 – 16y + 21 = 0
      3y2 – 9y – 7y + 21 = 0
      So y = 7/3, 3
      [/su_spoiler]

    Click here for more Quadratic Equations Practice Sets