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    27Jun, 2021
    Quantitative Aptitude: Number Series Set 150
    1. Directions : What will be come in place of questions ( ? ) marks in the following number series ?
      20, 45, 100, 215, ?, 925
      320
      445
      245
      380
      450
      Option E
      The series is (*2 + 5), (*2 + 10), (*2 + 15), (*2 + 20), (*2 + 25)

       

    2. 48, 65, ?, 101, 120
      81
      82
      80
      75
      72
      Option C
      The series is (7^2 – 1), (8^2 + 1), (9^2 – 1), (10^2 + 1 ), (11^2 – 1)

       

    3. 8, 72, 172, 316, 512, ?
      620
      768
      680
      540
      820
      Option B
      The series is, (+8^2), (+10^2), (+12^2), (14^2), (+16^2)

       

    4. 14, 42, 168, ?, 5040, 35280
      840
      720
      680
      320
      580
      Option A
      The series is (*3), (*4), (*5), (*6), (*7)

       

    5. 26, 43, ?, 85, 114, 145
      60
      75
      48
      62
      54
      Option D
      The series is (+17), (+19), (+23), (+29), (+31)

       

    6. Directions : What will be wrong number in the following wrong number series ?
      5, 10, 30, 150, 600, 3600
      600
      150
      3600
      5
      30
      Option B
      The series is (*2), (*3), (*4), (*5), (*6)

       

    7. 24, 16, 20, 32, 72, 184
      20
      32
      72
      184
      24
      Option B
      The series is (*.5 + 4), (*1 + 4), (*1.5 + 4), (*2 + 4), (*2.5 + 4)

       

    8. 14, 17, 27, 42, 68, 98
      14
      42
      68
      27
      98
      Option E
      The series is +(2^2 – 1), +(3^2 + 1), +(4^2 – 1), +(5^2 + 1), (6^2 – 1)

       

    9. 4, 10, 42, 254, 2034, 20362
      4
      42
      2034
      20362
      254
      Option D
      The series is, (*2 + 2), (*4 + 2), (*6 + 2), (*8 + 2), (*10 + 2)

       

    10. 34, 48, 76, 118, 174, 248
      34
      248
      76
      48
      174
      Option B
      The series is (+14), (+28), (+42), (+56), (+70)

       

    01
    26Jun, 2021
    Quantitative Aptitude: Simplification Questions Set 111
    1. Directions : What should be come in place of question ( ? ) marks in the following given below questions ?
      32% of 480 + 25% of 680 – 20% of 380 = ?
      245
      240
      150
      247.6
      246.4
      Option D
      153.6 + 170 – 76 = ?
      ? = 247.6

       

    2. 120% of 300 + (15)^2 – 290/2 = ?
      230
      140
      440
      324
      520
      Option C
      360 + 225 – 145 = ?
      ? = 440

       

    3. (18)^2 – 8 * (12)^2 + 320 * 15 = ?
      3425
      3972
      2875
      3245
      4972
      Option B
      324 – 8 * 144 + 4800 = ?
      ? = 5124 – 1152
      ? = 3972

       

    4. 536 + 325 + 420 – 525 – 215 = ?
      342
      240
      541
      640
      540
      Option C
      1281 – 740 = ?
      ? = 541

       

    5. √4096 + √5776 – √4761 = ?
      82
      32
      78
      71
      46
      Option D
      64 + 76 – 69 = ?
      ? = 71

       

    6. Directions : What should be come in place of question ( ? ) marks in the following given questions ? ( you are not calculate exact value )
      24.99% of 599.98 + 178.01 = ? + 75.99
      302
      528
      120
      208
      252
      Option E
      25% of 600 + 178 = ? + 76
      150 + 178 = ? + 76
      ? = 328 – 76
      ? = 252

       

    7. 128.04 + (14.02)^2 – (8)^2 = ?
      250
      260
      125
      234
      360
      Option B
      128 + (14)^2 – 8^2 = ?
      128 + 196 – 64 = ?
      ? = 324 – 64
      ? = 260

       

    8. 563.99 ÷ 12 + 242.01 + √1225 = ?
      322
      228
      324
      422
      328
      Option A
      564 ÷ 12 + 242 + 35 = ?
      47 + 242 + 35 = ?
      ? = 322

       

    9. √1024.02 * 20.04 – 120 = ?
      320
      480
      340
      520
      280
      Option D
      √1024 * 20 – 120 = ?
      ? = 32 * 20 – 120
      ? = 640 – 120
      ? = 520

       

    10. 255.84 + (16.80 * 4) – (14)^2 = ?
      128
      224
      250
      128
      256
      Option D
      256 + (17 * 4) – (14)^2 = ?
      ? = 256 + 68 – 196
      ? = 324 – 196
      ? = 128

       

    02
    25Jun, 2021
    Partnership Questions Quantitative Aptitude: Partnership Set 21
    1. A ,B and C started a partnership business with investment in the ratio of 6 : 3 : 4. After one year A withdrawal half of his investment and B doubled his investment. Total profit of the business at the end of the two years is rs. 52000, then what is the sum of profit share of B and C ?
      42000
      34000
      38000
      51000
      40000
      Option B
      Let the investment of A,B and C = 6x, 3x and 4x respectively.
      profit sharing ratio of A,B and C = (6x * 1 + 3x * 1) : (3x * 1 + 6x * 1) : (4x * 2)
      = 9x : 9x : 8x = 9 : 9 : 8
      Total profit share of B and C = 52000 * 17/26 = 34000

       

    2. P started a business with the investment of rs.5500. After four months Q and R joined the business with investment of rs.6000 and rs.4500 respectively. At the end of the one year they earned total profit rs.22500. Find profit share of R ?
      5400
      7500
      5000
      4850
      3250
      Option A
      profit sharing ratio of P,Q and R = (5500 * 12) : (6000 * 8) : (4500 * 8) = 11 : 8 : 6
      profit share of R = 22500 * 6/25 = 5400

       

    3. Amit and Sumit invested rs.5000 and rs.7500 in a partnership business. After one year Amit added rs.2500 and sumit added rs.2000 in the business. If profit of Sumit at the end of two years is rs.13600, then find the total profit of the business ?
      20800
      14500
      10000
      23600
      28400
      Option D
      Profit sharing ratio of Amit and Sumit = (5000 * 1 + 7500 * 1) : (7500 * 1 + 9500 * 1) = 25 : 34
      profit share of sumit = 13600
      total profit of the business = 13600 * 59/34 = 23600

       

    4. X,Y and Z started a business with investment ratio of 4 : 3 : 5. If they invested for 5 months, 8 months and 4 months respectively, then what is profit sharing ratio of X,Y and Z ?
      4 : 2 : 5
      2 : 5 : 3
      10 : 9 : 5
      3 : 4 : 8
      5 : 6 : 5
      Option E
      Profit sharing ratio of X, Y and Z = (4 * 5) : (3 * 8) : (5 * 4) = 20 : 24 : 20 = 5 : 6 : 5

       

    5. A and B investment rs.8000 and rs.6000 for 12 months and 14 months respectively in the business. If 25% of total profit is given to B as salary for managing business and remaining profit is distributed in their profit sharing ratio and profit share of A is rs.3200, then find total profit of the business ?
      8400
      6900
      4800
      8000
      4800
      Option D
      Profit sharing ratio of A and B = (8000 * 12) : (6000 * 14) = 8 : 7
      Let total profit of the business = 100x
      salary given to B = 100x * 25/100 = 25x
      remaining profit = 100x – 25x = 75x
      profit share of A = 75x * 8/15 = 40x
      40x = 3200
      x = 80
      total profit of the business = 100x = 100 * 80 = 8000

       

    6. Lipu and Papu start a business with initial investments in the ratio of 9 : 11 and their annual profit were in the ratio of 3 : 1. If Lipu invested the money for 11 months, then Papu invested his money for how many months ?
      3 months
      8 months
      4 months
      14 months
      7 months
      Option A
      Let Papu invested his money for x months.
      9 * 11/ 11 * x = 3/1
      x = 3 months

       

    7. A , B and C enter into partnership business. A invests some money at the beginning, after 4 months B invests double the money of A’s invests and after 6 months C invests thrice the amount of A. If total profit of the business is rs.9200 at the end of one year, then how much profit B get ?
      4600
      2800
      3400
      3200
      3800
      Option D
      Let A invests = rs.x
      B invests = 2x
      C invests = 3x
      profit sharing ratio of A , B and C = (x * 12 ) : (2x * 8) : (3x * 6)
      =6 : 8 : 9
      profit share of B = 9200 * 8 /23 = 3200

       

    8. X , Y and Z entered into a partnership farm with investing rs. 12000, rs. 18000 and rs. 20000 respectively. X being a working partner gets 20% of the annual profit for the work. Find the profit share of X in an annual profit of rs. 25000 ?
      8600
      7300
      9800
      5820
      8750
      Option C
      Profit sharing ratio of X , Y and Z = 12000 : 18000 : 20000 = 6 : 9 : 10
      X gets profit for being a working partner = 25000 * 20 /100 = 5000
      remaining profit = 25000 – 5000 = 20000
      profit share of X of the remaining profit = 20000 * 6 /25 = 4800
      total profit of X = 5000 + 4800 = 9800

       

    9. A and B started a business by investing rs.15000 and rs.20000 respectively. By mistake they divided the total profit in the ratio of 1 : 3, so that A loses rs.500. What is total profit of the business ?
      1400
      2400
      4200
      2800
      5200
      Option D
      Profit sharing ratio of A and B = 15000 : 2000 = 3 : 4
      Let total profit of the business = rs. x
      profit sharing ratio of A = 3x/7
      By mistake profit share of A = x/4
      A losses = 500
      = 12x – 7x/28
      x = 500 * 28/5 = 2800
      total profit of the business = rs.2800

       

    10. A and B started a business with capital of rs.28000 and rs.35000. After 6 months, A invests rs.4000 more, after 8 months B withdrawal rs. 5000 while C joins with capital of rs.40000. Annual profit of the business is rs.9200, then how much profit A get ?
      4300
      3600
      4800
      5200
      5500
      Option B
      profit sharing ratio of A, B and C =(28000* 6 + 32000 * 6) : (35000 * 8 + 3000 * 4) : (40000 * 4) = 9 : 10 : 4
      profit share of A = 9200 * 9/23 = 3600

       

    03
    24Jun, 2021
    Quantitative Aptitude: Quadratic Equations Questions Set 74
    1. I. x^2 – 13x – 198 = 0
      II. y^2 – 21y + 108 = 0
      X > Y
      X < Y
      X ≤ Y
      X ≥ Y
      X= Y or no relation.
      Option E
      I. x^2 – 22x + 9x – 198 = 0
      x = 22, -9
      II. y^2 – 12y – 9y + 108 = 0
      y = 12, 9

       

    2. I. x^2 – 9x + 20 = 0
      II. y^2 – 11y + 30 = 0
      X > Y
      X < Y
      X ≥ Y
      X ≤ Y
      X= Y or no relation.
      Option D
      I. x^2 – 5x – 4x + 20 = 0
      x = 5, 4
      II. y^2 – 6y – 5y + 30 = 0
      y = 6, 5

       

    3. I. x^2 – 30x + 216 = 0
      II. y^2 – 34y + 168 = 0
      X > Y
      X < Y
      X ≥ Y
      X= Y or no relation.
      X ≤ Y
      Option D
      I. x^2 – 18x – 12x + 216 = 0
      x = 18, 12
      II. y^2 – 28y – 6y + 168 = 0
      y = 28, 6

       

    4. I. x^2 = 1024
      II. y^2 – 38y + 192 = 0
      X > Y
      X < Y
      X ≥ Y
      X ≤ Y
      X= Y or no relation.
      Option E
      I. x^2 = 1024
      x = 32, -32
      II. y^2 – 32y – 6y + 192 = 0
      y = 32, 6

       

    5. I. x^2 – 27x + 92 = 0
      II. y^2 + 22y + 112 = 0
      X > Y
      X < Y
      X ≤ Y
      X ≥ Y
      X= Y or no relation.
      Option A
      I. x^2 – 23x – 4x + 92 = 0
      x = 23, 4
      II. y^2 + 14y + 8y + 112 = 0
      y = -14, -8

       

    6. I. 4x^2 – 2x – 12 = 0
      II. 3y^2 + 20y + 32 = 0
      X > Y
      X < Y
      X= Y or no relation.
      X ≤ Y
      X ≥ Y
      Option A
      I. 4x^2 – 8x + 6x – 12 = 0
      4x = 8, -6
      x = 2, – 1.5
      II. 3y^2 + 12y + 8y + 32 = 0
      3y = -12, -8
      y = -4, -2.67

       

    7. I. x^4 = 4096
      II. y^2 – 28y – 160 = 0
      X > Y
      X < Y
      X ≤ Y
      X= Y or no relation.
      X ≥ Y
      Option C
      I. x^4 = 4096
      x = 8, -8
      II. y^2 – 20y – 8y – 160 = 0
      y = 20, 8

       

    8. I. x^2 – 20x + 96 = 0
      II. y^2 + 3y – 28 = 0
      X > Y
      X < Y
      X ≤ Y
      X ≥ Y
      X= Y or no relation.
      Option A
      I. x^2 – 12x – 8x + 96 = 0
      x = 12, 8
      II. y^2 + 7y – 4y – 28 = 0
      y = -7, 4

       

    9. I. x^2 – x – 6 = 0
      II. y^2 + 5x + 4 = 0
      X > Y
      X < Y
      X ≥ Y
      X= Y or no relation.
      X ≤ Y
      Option D
      I. x^2 – 3x + 2x – 6 = 0
      x = 3, -2
      II. y^2 + 4y + y + 4 = 0
      y = -4, -1

       

    10. I. x^2 – 5x – 84 = 0
      II. y^2 – 32y + 252 = 0
      X > Y
      X < Y
      X ≥ Y
      X= Y or no relation.
      X ≤ Y
      Option B
      I. x^2 – 12x + 7x – 84 = 0
      x = 12, -7
      II. y^2 – 14y – 18y + 252 = 0
      y = 14, 18

       

    04
    23Jun, 2021
    Quantitative Aptitude: Ratio and Proportion Questions Set 10
    1. The ratio of monthly income of A , B and C is 5 : 8 : 3. If annual income of B is rs.192000, then what is the sum of the monthly income of A and C ?
      18000
      40000
      14000
      16000
      52000
      Option D
      Let monthly income of A, B and C = 5x, 9x and 3x respectively.
      Annual income of B = 12 * 8x = 96x
      96x = 192000
      x = 2000
      sum of the monthly income of A and C = 5x + 3x = 8x = 8 * 2000 = 16000

       

    2. Three are two numbers in the ratio of 12 : 17. If 8 is substracted from each number the new numbers are in the ratio of 2 : 3. If 12 is added in each number then what will be the new ratio of the numbers ?
      4 : 5
      5 : 2
      3 : 4
      1 : 4
      4 : 15
      Option C
      Let two numbers are 12x and 17x respectively.
      12x – 8/ 17x – 8 = 2/3
      36x – 24 = 34x – 16
      x = 4
      two numbers are = (12 * 4), (17 * 4) = 48, 68
      If we added 12 in each number, new ratio = (48 + 12) : (68 + 12) = 3 : 4

       

    3. There are two numbers in the ratio of 5 : 8. If 20 is added in first number and 6 is added in second number, then ratio of two numbers become 6 : 7. Find the larger of the two numbers ?
      50
      48
      32
      64
      28
      Option D
      Let 1st numbers = 5x
      2nd numbers = 8x
      5x + 20/ 8x + 6 = 6/7
      48x + 36 = 35x + 140
      x = 8
      Langer number = 8 * 8 = 64

       

    4. Total income of A, B and C is rs.57000. If A, B and C spend 60%, 80% and 82% of their income respectively and ratio of their saving is 8 : 5 : 9, then what is the income of B ?
      18000
      20000
      25000
      33000
      15000
      Option E
      Let saving of A,B and C = 8x, 5x and 9x respectively.
      income of A = 8x/40 * 100 = 25x
      income of B = 5x/20 * 100 = 25x
      Income of C = 9x/18 * 100 = 50x
      Total income = 57000
      income of B = 57000 * 25x/95x = 15000

       

    5. The ratio of salaries of P, Q and R is 5 : 4 : 6. If the salary of P, Q and R are increased by 20%, 25% and 32% respectively. Find the new ratio of their salaries.
      5 : 8 : 3
      3 : 8 : 9
      3 : 5 : 6
      30 : 25 : 42
      30 : 28 : 15
      Option D
      Let salary of P, Q and R are 5x, 4x and 6x respectively.
      new salary ratio = (5x * 120/100) : (4x * 125/100) : (6x * 140/100)
      = 6x : 5x : 42x/5
      = 30 : 25 : 42

       

    6. In what ratio should rice at rs.13 per kg and rs.20 per kg be mixed. So that the mixture be worth of rs.17 per kg ?
      5 : 2
      3 : 4
      2 : 1
      5 : 4
      4 : 3
      Option B
      By alligation method,
      13 20
      17
      3 4
      ratio = 3 : 4

       

    7. A bag has coins in the denominations of 50p, 25p and 20p in the ratio of 4 : 2 : 1. If the total value of the coins is rs.108. Find the number of 20p coins in the bag ?
      50
      80
      45
      40
      240
      Option D
      Let the number of 50p, 25p, and 20p coins in the bag are 4x, 2x and x respectively.
      (50 * 4x) + (25 * 2x) + (20 * x) = 10800p
      200x + 50x + 20x = 10800
      270x = 10800
      x = 40
      number of 20p coins = 40

       

    8. First number and second number are 32% and 48% more than the third number. Find the ratio of first number, third number and second number.
      33 : 25: 37
      33 : 37 : 25
      33 : 38 : 25
      5 : 7 : 8
      6 : 5 : 3
      Option A
      Let third number = 100
      first number = 100 * 132/100 = 132
      second number = 100 * 148/100 = 148
      ratio of first, third and second number = 132 : 100 : 148 = 33 : 25 : 37

       

    9. A marriage party is attended by a total of 80 persons. If the ratio of men and women in the party is 3 : 2, how many more women must join the party such that the ratio becomes 6 : 5 ?
      4
      12
      14
      8
      19
      Option D
      Total persons = 80
      men = 80 * 3/5 = 48
      women = 80 * 2/5 = 32
      Let the number of women will join = x
      48/32 + x = 6/5
      192 + 6x = 240
      x = 8

       

    10. The students of three class are in the ratio of 5 : 6 : 4. If 8 students of each class increased so that, the ratio becomes 6 : 7 : 5, then Find the total number of students in the class before increased ?
      140
      120
      165
      200
      170
      Option B
      Let the number of students in 1st, 2nd and 3rd class are 5x, 6x and 4x respectively.
      5x + 8/6x + 8 = 6/7
      36x + 48 = 35x + 56
      x = 8
      total students of class before increased= 5x + 6x + 4x = 15x = 15 * 8 = 120

       

    05
    22Jun, 2021
    Quantitative Aptitude: Data Sufficiency Questions Set 15
    1. what is distance between city A and city B ?
      I. Sanju and Manju started their journey simultaneously from city A to city B with their speeds in the ratio of 3 : 2.
      II. Manju takes 2 hours more than Sanju to reach city B.
      II. The difference between speeds of Sanju and Manju is 25 km/hr.
      Only I and II is sufficient.
      Only II and III is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      All I, II and III are sufficient.
      Option E
      From I, Let speed of Sanju = 3x
      speed of Manju = 2x
      From II, we can not calculate distance between city A and B.
      From III, difference between speed of Sanju and Manju is given, so we can not calculate the answers.
      from statement, I, II, III together,
      Difference between their speeds = 3x – 2x = 25
      x = 25
      speed of Sanju = 3 * 25 = 75
      speed of Manju = 2 * 25 = 50
      Let distance = D
      D/50 – D/75 = 2
      3D – 2D/150 = 2
      D = 300

       

    2. In how many days will B alone completed the work ?
      I. A and B together can complete the work in 16 days.
      II. B and C together can complete the work in 12 days.
      III. A and C together can complete the work in 18 days.
      Only I and II is sufficient.
      Only II and III is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      All I , II and III are sufficient.
      Option E
      From I, only from statement I we can not get the answer.
      From I and II,
      LCM of 16 and 12 = 144
      Total work = 144
      efficiency of A and B = 144/16 = 9
      efficiency of B and C = 144/12 = 12
      Here, we can not get answers.
      From I, II and III,
      efficiency of A and C = 144/18 = 8
      A + B + B + C + A + C = 9 + 12 + 8 = 29
      A + B + C = 29/2 = 14.5
      Here, we can get the answers.

       

    3. What will be the difference between two numbers ?
      I. the square of the first numbers is 9 times the second number.
      II. The ratio of the first number to the second number is 3 : 4.
      Only I is sufficient.
      Only II is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option E
      From I, we can not calculate the answers.
      From I and II, Let first numbers = 3x
      second number = 4x
      (3x)^2 = 9 * 4x
      9x^2 = 36x
      x = 4
      Difference = (4 * 4) – (3 * 4) = 4

       

    4. What was the percent profit or loss incurred by selling an article for rs.14000 ?
      I. The ratio of selling price to the cost price of the article is 7 : 5.
      II. The marked price of the article is 60% more than the cost price.
      Only I is sufficient.
      Only II is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option A
      From I,
      Ratio of SP and CP of the article = 7 : 5
      SP = 14000
      CP = 14000/7 * 5 = 10000
      profit % = (14000 – 10000)/10000 * 100 = 40%
      only statement I is sufficient.

       

    5. What is the amount invested in scheme B ?
      I. The amount invested in scheme A is rs.48000.
      II. The amount invested in scheme A is 60% of the total amount invested.
      III. The amount invested in scheme A and B are in the ratio of 3 : 2.
      Only I and II is sufficient.
      Only I and III is sufficient.
      Either only I and II or only I and III is sufficient.
      Neither I nor II is sufficient.
      All I, II and III are sufficient.
      Option C
      From I, invested by A = 48000
      from I and II,
      Let, total investment of A and B = 100x
      investment of A = 60x
      investment of B = 40x
      ratio of investment A and B = 3 : 2
      investment of B = 48000/3 * 2 = 32000
      From I and III,
      Investment of A = 48000
      Investment ratio of A and B = 3 : 2
      investment of B = 48000/3 * 2 = 32000
      either only statement I and II or only statement I and III is sufficient.

       

    6. What is the present age of B ?
      I. The age of B is 4 years older than A.
      II. The ratio of present age of A and C is 5 : 4.
      Only I is sufficient.
      Only II is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option D
      From I,
      We can not get answer,
      from I and II,
      Let present age of A and C = 5x and 4x respectively.
      the age of B = 5x + 4
      from I and II together we can not get answer.

       

    7. What is the speed of boat in still water ?
      I. The boat can go a distance of 320 km in 8 hours in downstream speed.
      II. The ratio of downstream speed and upstream speed is 5 : 3.
      Only I is sufficient.
      Only II is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option E
      From I,
      Downstream speed = 320/8 = 40 km/hr
      only statement I is not sufficient.
      From I and II,
      Downstream speed = 40 km/hr
      upstream speed = 40/5 * 3 = 24 km/hr
      speed of boat in still water = (40 + 24)/2 = 32 km/hr
      both statement I and II is required to get answer.

       

    8. Find the rate of interest ?
      I. The difference between CI and SI on an amount at rs.4000 for 2 years is rs.90.
      II. The simple interest incurred on certain sum is rs.2400 in 4 years.
      Only I is sufficient.
      Only II is sufficient.
      Either I or II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option A
      From I,
      CI – SI = PR^2/100^2
      90 = 4000 * R^2/10000
      R = 15%
      From II, Here principal is not given so we can not get answer.
      only statement I is sufficient.

       

    9. What is the length of rectangle ?
      I. The perimeter of square field is 56m and breadth of the rectangle filed is 4m more than the side of square field.
      II. The area of rectangular field is 360 m^2.
      Only I is sufficient.
      Either I or II is sufficient.
      Only II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option E
      From I,
      Perimeter of square filed = 56m
      side of square = 56/4 = 14m
      breadth of rectangular field = 14 + 4 = 18 m
      Here we can not get answer.
      From I, and II,
      Breadth of rectangular = 14 + 4 = 18
      Area of rectangular = 360 m^2
      length = 360/18 = 20m
      both I and II is required to get answer.

       

    10. What is the length of train B ?
      I. Length of train A is 180 m and it crosses a pole in 12 seconds.
      II. The ratio of speed of train A and train B is 3 : 5. Train B crosses a 280 m long bridge in 20 seconds.
      Only I is sufficient.
      Either I or II is sufficient.
      Only II is sufficient.
      Neither I nor II is sufficient.
      Both I and II are sufficient.
      Option E
      From I, length of train A = 180 m
      speed of train A = 180/12 = 15 m/sec
      only statement I is not sufficient.
      Both from I and II,
      speed of train B = 15/3 * 5 = 25 m/sec
      let length of train B = x
      x + 280/20 = 25
      x = 220
      both statement I and II is required to get answer.

       

    06
    21Jun, 2021
    Quantitative Aptitude: Simple/Compound Interest Set 21
    1. The difference between compound interest and simple interest on rs.42000 at same rate of interest for 2 years is rs.945. What is the rate of interest ?
      10%
      15%
      14%
      20%
      25%
      Option B
      945 = PR^2/100^2
      945 = 42000 * R^2/10000
      R^2 = 225
      R = 15%

       

    2. Rohit borrows rs.4000 at 10% on compound interest and at the end of the every year he paid rs.1000 back. How much amount should he pay at the end of the 3rd year to settlement all his debt ?
      2500
      3000
      3240
      3014
      2480
      Option D
      total amount at the end of the 1st year = 4000 * 110/100 = 4400
      After he paid rs.1000, remaining amount = 4400 – 1000 = 3400
      total amount at the end of 2nd year = 3400 * 110/100 = 3740
      After he paid rs.1000, remaining amount = 3740 – 1000 = 2740
      total amount at the end of 3rd year = 2740 * 110/100 = 3014
      At the end of 3rd year he paid 3014 to settlement his debt.

       

    3. A man borrows rs.8000 for 3 years and he gave at 15%, 20% and 30% rate of interest for 1st year, 2nd year and 3rd year respectively. Then how much amount he paid at the end of 3rd years in compound interest ?
      14435
      14352
      17250
      19540
      20450
      Option B
      Amount in 1st year = 8000 + (8000*15/100) = 9200
      Amount in 2nd year = 9200 + (9200*20/100) = 11040
      Amount in 3rd year = 11040 + (11040 * 30/100) = 14352

       

    4. What is the compound interest on rs.8600 at 20% p.a. for 1 years, compounded half yearly.
      2240
      2420
      1440
      1806
      1425
      Option D
      Rate of interest for 1 year = 20%
      rate of interest for half year = 10%
      by applying successive,
      10 + 10 + 10 * 10/100 = 21%
      compound interest = 8600 * 21/100 = 1806

       

    5. Shivam invested a certain amount at the rate of 14% for 4 years in simple interest and after 4 years he obtained a total simple interest of rs.5600. If invested the same amount at the same rate of interest for 2 years in compound interest then how much interest would he have obtained ?
      3220
      2996
      2834
      2450
      2830
      Option B
      Let principal be rs. x
      = x * 14/100 * 4 = 5600
      x = 5600 * 100/56 = 10000
      by applying successive, rate of compound interest = 14 + 14 + 14 * 14/100 = 19.96%
      compound interest = 10000 * 29.96/100 = 2996

       

    6. The difference between simple interest and compound interest on a certain sum at 20% rate of interest for 2 years is 500. What is difference between simple interest and compound interest for 1st year, if interest is compounded half-yearly ?
      120
      180
      125
      150
      190
      Option C
      500 = PR^2/100^2
      500 = P * 400/10000
      P = 12500
      simple interest for 1st year = 12500 * 20/100 = 2500
      Rate of interest for compound half yearly = 20/2 = 10%
      by apply successive method,
      rate of CI = 10 + 10 + 10 * 10/100 = 21%
      compound interest = 12500 * 21/100 = 2625
      Difference = 2625 – 2500 = 125

       

    7. On a certain sum of money simple interest occurred after 4 years is rs.4800 and compound interest occurred after 2 years is rs.2580. If rate of interest in both cases are same, find the principal amount ?
      12000
      8000
      14000
      18000
      20000
      Option B
      SI after 4 years = 4800
      SI for 1 year = 4800/4 = 1200
      SI for 2 years = 2400
      CI for 2 years = 2580
      by applying SI and CI ratio formula,
      SI/CI = 200/200 + r
      2400/2580 = 200/200 + r
      r = 15%
      Let principal = rs. x
      x * 15/100 * 4 = 4800
      x = 4800*100/60 = 8000

       

    8. What is difference between compound interest and simple interest for 2 years at the rate of 20% on a principal of rs.8500 ?
      240
      520
      340
      624
      125
      Option C
      SI for 2 years = 8500 * 20/100 * 2 = 3400
      by applying successive
      rate of compound interest after 2 years = 20 + 20 + 20 * 20/100 = 44%
      CI = 8500 * 44/100 = 3740
      Difference = 3740 – 3400 = 340

       

    9. If a money get tripled in 5 years on a certain rate of simple interest, then find the rate of interest ?
      20%
      40%
      48%
      45%
      25%
      Option B
      let principal = rs. x
      amount = 3 * x = 3x
      simple interest = 3x – x = 2x
      2x = x * R/100 * 5
      R = 40%

       

    10. Shiva invested rs. X for 4.5 years at the rate of 20% p.a. on simple interest. If he received rs.2520 as interest, then what is the value of X ?
      2800
      1700
      3400
      4600
      3800
      Option A
      2520 = X * 20/100 * 4.5
      X = 2800

       

    07
    20Jun, 2021
    Quantitative Aptitude: Probability Questions – Set 20
    1. A bag contains 8 green balls and 12 blue balls. If a ball is drawn randomly, then find the probability of getting either green ball or blue ball.
      2
      3
      2/5
      2/3
      1
      Option E
      required = (8C1) + (12C1) = 20
      total = (20C1) = 20
      probability = 20/20 = 1

       

    2. If two cards are drawn randomly from a pack of 52 cards. Then what is the probability of both cards are red cards ?
      28/91
      23/108
      25/102
      8
      17
      Option C
      required = (26C2) = 13 * 25
      Total = (52C2) = 26 * 51
      probability = (13 * 25)/(26 * 51) = 25/102

       

    3. How ,any 4-digit numbers can be formed from the digits 6, 2, 5, 7, 3, which are divisible by 5 and none of the digits is repeated ?
      140
      220
      28
      24
      96
      Option D
      A number to be divisible by 5, unit digit must be 0 or 5.
      So, in 4-digit numbers unit place is required to be filled by digit 5 and rest three places of required 4-digit numbers will be filled by rest 4 digit.
      numbers of ways = 4! = 4 * 3 * 2 * 1 = 24

       

    4. Out of 52 playing cards two cards is picked randomly, then what is the probability of getting that one red king and one black queen ?
      2/663
      5/13
      8/58
      5
      14
      Option A
      required = (2C1) * (2C1) = 4
      Total = (52C2) = 26 * 51
      probability = 4/(26*51) = 2/663

       

    5. A bag contains 2 red pens, 6 blue pens and 4 green pens. Find the probability of getting two same color pens from the bag ?
      5/8
      2/4
      1/3
      5/6
      4
      Option C
      required = (2C2) + (6C2) + (4C2)
      = 1 + 15 + 6 = 22
      total = (12C2) = 66
      probability = 22/66 = 1/3

       

    6. A basket contains 3 green, 4 orange and 5 pink balls. If two balls are drawn randomly from the basket, then what the probability of selecting that both balls are of different colors ?
      4/65
      57/68
      8/45
      47/66
      4/15
      Option D
      possibility of both balls are different colors = ( 1 green and 1 orange) , (1 orange and 1 pink ), (1 green and 1 pink)
      required = (3C1) * (4C1) + (4C1) * (5C1) + (3C1) * (5C1)
      = 12 + 20 + 15 = 47
      total = (12C2) = 66
      probability = 47/66

       

    7. Sunil threw two dices together. What is the probability of getting that the sum of the two outcomes is 4 ?
      1/8
      1/12
      4
      5
      2
      Option B
      Possibility to get = (2, 2), (3, 1), (1, 3)
      required = 3
      total = 36
      probability = 3/36 = 1/12

       

    8. When three coins are tossed simultaneously, then find the probability of getting at least one tale.
      7/8
      5/8
      1/4
      4/15
      2
      Option A
      possibility = 1 tale or 2 tale or 3 tale
      required = (3C1) + (3C2) + (3C3)
      = 3 + 3 + 1 = 7
      total = 2^3 = 8
      probability = 7/8

       

    9. Four boys and 2 girl sit in a row for presenting their project paper, then what is the probability that they will sit in alternate position ?
      1/6
      2/3
      1/8
      1/15
      2/5
      Option D
      probability = 4! * 2!/6! = 1/15

       

    10. Probability of a question solved by A and B is 1/5 and 1/4 respectively, then find the probability that at least one of them will solve the questions ?
      5/6
      1/4
      11/25
      2/5
      5/9
      Option D
      possibility = 1 – probability of question not solved by anyone
      probability = 1 – (1 – 1/5) * (1 – 1/4)
      = 1-(4/5) * (3/4)
      = 2/5

       

    08
    19Jun, 2021
    Quantitative Aptitude: Mensuration Questions Set 15
    1. The ratio of length and breadth of rectangular field is 3 : 2. If the perimeter of rectangular field is 40cm, then what is the area of square whose side 5cm more than the length of rectangular field ?
      216
      289
      324
      196
      400
      Option B
      Let length and breadth or rectangular field = 3x and 2x
      perimeter = 40
      2(3x + 2x) = 40
      10x = 40
      x = 4
      length = 3 * 4 = 12cm
      side of square = 12 + 5 = 17
      Area of square = 17 * 17 = 289 cm^2

       

    2. Length of two rectangles A and B is 14 cm and 20 cm respectively and sum of their breadth 14 cm. If ratio of perimeter of A and B is 5 : 7, then what is the difference between area of these rectangles ?
      84
      54
      42
      62
      76
      Option E
      Let breadth of rectangular A = x
      and breadth rectangular B = 14 – x
      perimeter of rectangular A = 2 (14 + x)
      perimeter of rectangle B = 2 (20 + 14 – x)
      = 2 (34 – x)
      2 (14 + x)/2 (34 – x) = 5/7
      340 – 10x = 196 + 14x
      x = 6
      breadth of rectangle A = 6 cm
      breadth of rectangle B = 14 – 6 = 8 cm
      Area of rectangle A = 14 * 6 = 84 cm^2
      Area of rectangle B = 20 * 8 = 160 cm^2
      Difference = 160 – 84 = 76 cm^2

       

    3. If the diagonal of a cube is of length ‘x’, then what is the lateral surface area of that cube ?
      2x^2/3
      5x/3
      4x/3
      4x^2/3
      8x
      Option D
      Diagonal of cube = a√3
      a√3 = x
      a = x/√3
      lateral surface area = 4a^2 = 4*(x/√3)^2 = 4x^2/3

       

    4. If the total surface area of a cube is 8664 cm^2, the what is the length of its diagonal ?
      46√3
      34√3
      38√3
      24√3
      19√2
      Option C
      T.S.A = 8664
      6a^2 = 8664
      a^2 = 1444
      a = 38
      diagonal = a√3 = 38√3

       

    5. If the ratio of radius of two spheres A and B is 3 : 5, then what is the ratio of their volumes ?
      27 : 125
      3 : 5
      12 : 25
      9 : 25
      6 : 15
      Option A
      Ratio of volume of A and B = 4/3π(3)^3 : 4/3π(5)^3
      = 27 : 125

       

    6. The ratio of length and breadth of rectangles is 5 : 3 and its area is 240 m^2. If the perimeter of square is 8m less than the perimeter of rectangles, then what is the area of square ?
      324
      196
      1024
      1444
      1296
      Option B
      Let length and breadth of rectangles = 5x and 3x
      Area = 240
      15x^2 = 240
      x = 4
      length = 5 * 4 = 20
      breadth = 3 * 4 = 12
      perimeter of rectangle = 2 (20+ 12) = 64
      perimeter of square = 64 – 8 = 56
      side of square = 56/4 = 14
      Area of square = 14 * 14 = 196

       

    7. The non – parallel sides of a trapezium are equal to side of perimeter of square 40cm and the parallel sides are 8cm and 20cm respectively, find the perimeter of trapezium ?
      38
      24
      20
      36
      40
      Option D
      perimeter of square = 40
      side of square = 40/4 = 10cm
      non-parallel side of trapezium = 10
      one parallel side = 8cm
      other parallel side = 20cm
      breadth of tringle = 20 – 8/2 = 6
      let height of trapezium = h
      10^2 = h^2 + 6^2
      h^2 = 64
      h = 8
      perimeter = 10 + 10 + 8 + 8 = 36 cm

       

    8. The circumference of a circle A is equal to the perimeter of a square whose area is 1936 m^2. Diameter of circle B is four times the radius of circle A what is the area of circle B ?
      4532
      3952
      2840
      4900
      9856
      Option E
      Area of square = 1936 m^2
      side of square = √1936 = 44m
      perimeter of square = 44 * 4 = 176m
      2πr = 176
      r = 176 *7 /44 = 28
      diameter of circle B = 28 * 4 = 112
      radius = 112/2 = 56m
      Area = πr^2 = 22/7 * 56 * 56 = 9856

       

    9. If the height and radius of a cylinder increased by 25% and 20% respectively. Then what is the change in percent in its curved surface area ?
      20%
      50%
      40%
      25%
      58%
      Option B
      Let the height of cylinder = 4h
      increased height = 4h* 125/100 = 5h
      and radius of cylinder = 5r
      increased radius = 5r * 120/100 = 6r
      initial curved surface area = 2 * π * 4h * 5r = 40πhr
      new curved surface area = 2 * π * 5h * 6r = 60πhr
      % change = 60πhr – 40πhr/40πhr * 100 = 50%

       

    10. The length , breadth and height of a cuboid is 14m, 15m, 20m respectively,. Find the lateral surface area of cuboid ?
      1225
      1160
      4250
      2420
      3840
      Option B
      L.S.A = 2 * 20 (14 + 15) = 40 * 29 = 1160 m

       

    09
    18Jun, 2021
    Mixed Quantitative Aptitude Questions Set 217
    1. Train A crosses a 400m long bridge and a pole in 32 second and 12 seconds respectively. If length of train B is 40m more then the length of train A and speed of train B is 32m/sec. Then how much time required if train A and B is running in opposite direction ?
      14 sec
      4 sec
      18 sec
      10 sec
      19 sec
      Option D
      Let length of train A = x
      x+400/32 = x/12
      32x = 12x + 4800
      20x = 4800
      x = 240
      speed of train A = 240/12 = 20m/sec
      length of train B = 240 + 40 = 280m
      speed of train B = 32m/sec
      Time = 240 + 280/32 + 20 = 520/52 = 10sec

       

    2. The ratio of number of male and female in a group is 8 : 5. If the percentage of male and female increased by 20% and 25% respectively. Then find the new ratio of male and female, if total number of male and female is 260 initially.
      8 : 3
      5 : 4
      7 : 8
      5 : 3
      4 : 11
      Option D
      Total number of male and female = 260
      number of male = 260 * 8/13 = 160
      number of female = 260 * 5/13 = 100
      increased male = 160 * 5/4= 200
      increased female = 100 * 6/5 = 120
      ratio = 200 : 120 = 5 : 3

       

    3. The ratio of age of A and B before 5 years is 2 : 3 and D is 12 years elder than B. The present age of C is one fourth of the present age of D. If the age of C after 6 years is 14 years, then find the present age of B.
      28 years
      20 years
      34 years
      38 years
      42 years
      Option B
      Let the age of A and B before 5 years = 2x and 3x
      present age of A = 2x + 5
      present age of B = 3x + 5
      present age of D = 3x + 5 + 12 = 3x + 17
      present age of C = 3x + 17/4
      present age of C = 14 – 6 = 8 years
      3x + 17/4 = 8
      3x + 17 = 32
      3x = 32 – 17
      3x = 15
      x = 5
      present age of B = 3x + 5 = 20 years.

       

    4. A bag contain 4 red pens, 6 green pens and 8 blue pens. If 2 pens are taken at randomly, then find the probability that there are no blue pens.
      8/17
      5/17
      8/15
      2/8
      4
      Option B
      required = 10C2 = 10 * 9/2 * 1 = 45
      total = 18C2 = 18 * 17 / 2*1 = 153
      probability = 45/153 = 5/17

       

    5. Shiva invested rs.4200 in a scheme at 18% per annum for 3 years at simple interest and he also invested same amount in another scheme at 20% per annum for 3 years. What is the difference between simple interest and compound interest which he earned from both the scheme ?
      880
      789.6
      435.6
      940
      840
      Option B
      simple interest = 4200*18/100*3 = 2268
      by applying successive method,
      20 + 20 + 400/100 = 44%
      44 + 20 + 880/100 = 72.8
      compound interest = 4200 * 72.8/100 = 3057.6
      Difference = 3057.6 – 2268 = 789.6

       

    6. Directions : what will be the wrong number in the following wrong number series ?
      34, 35, 43, 52, 116, 945
      116
      43
      52
      945
      34
      Option D
      The series is (+1)^2, (+2)^3, (+3)^2, (+4)^3, (+5)^2

       

    7. 18, 10, 11, 17.5, 36, 94
      17.5
      10
      18
      36
      94
      Option E
      The series is (*.5 + 1), (*1 + 1), (*1.5 +1), (*2 + 1) , (*2.5 + 1)

       

    8. 52, 56, 65, 90, 139, 260
      52
      56
      65
      90
      260
      Option B
      The series is, (+2)^2, (+3)^2, (+5)^2, (+7)^2, (+11)^2

       

    9. 5, 8, 21, 70, 290, 1461
      5
      290
      8
      1461
      70
      Option B
      The series is (*1 + 3), (*2 + 5), (*3 + 7), (*4 + 9), (*5 + 11)

       

    10. 153, 161, 177, 215, 233, 273
      161
      177
      233
      215
      233
      Option D
      The series is, (+8), (+16), (+24), (+32), (+40)

       

    10