1. The selling price and cost price of an table differs by Rs.4000. What is the selling price in case the profit percentage is 20%?
   20,000
   12,000
   25,000
   22,000
   21,000
   Option A
   SP – CP = 4000
   1.2x – x = 4000
   ⇒0.2x = 4000
   ⇒x = 20,000

    

2. If a person sells his car at 8% profit against 8% loss, then he could had earned Rs12000 more. What is the cost price of car?
   
   Rs.85000
   Rs.75000
   Rs.45000
   Rs.55000
   Rs.65000
   Option B
   Let Cost Price (CP) = x
   Then Selling Price(SP) if he sells at profit = x(1+8%) = 1.08x
   and SP, if he sells at loss = x(1-8%) = 0.92%
   Given, 1.08%-0.92% = Rs12000
   0.16x = Rs.12000
   ⇒x = Rs.75000

    

3. Find the amount of water which should be added to 60% solution of citric acid to make it a 40% solution?
   
   35cc
   45cc
   40cc
   50cc
   60cc
   Option D
   Concentration of citric acid = 60%
   So in a 100cc, citric acid = 80cc and water = 40cc
   Let x be the amount of water added to make concentration of 40%
   ⇒60/(100+x) = 40/100
   ⇒ 60/(100+x) = 2/5
   ∴ x = 50cc

    

4. A bangle contains 20mg copper, 20mg gold and 40mg silver.Then find out the percentage gold in the bangle?
   22%
   20%
   35%
   25%
   15%
   Option D
   Total weight of bangle = 20+20+ 40 = 80mg
   Weight of gold = 20g
   Percentage of tin = 20g/80g x 100
   =25%

    

5. If the price of petrol is increased by 10% and subsequently by 20%. If the original price was Rs70 per litre then what is then the final price per litre?
   
   Rs. 92.4
   Rs. 96.4
   Rs. 95.4
   Rs. 72.4
   Rs. 82.4
   Option A
   Final Price = 70 x (1 + 10/100) x (1 + 20/100)
   = Rs. 92.4

    

6. The price of a onion is increased by 40%. If the original price is Rs. 300 per kg , what is the final price of the onion?
   Rs.120
   Rs.420
   Rs.220
   Rs.320
   Rs.200
   Option C
   Final Price = Initial price + Initial price x 40/100
   = Initial price x (1+40/100)
   = Rs.300x(1.4)
   = Rs.420

    

7. If 15% of a number is 4500, then the number is
   
   30,000
   50,000
   40,000
   33,000
   35,000
   Option A
   Let the number be y.
   Then, 15% of y = 4500
   ( 15 / 100 )y = 4500
   y= 30,000

    

8. 25 % of a number when added to 25 is equal to that number. What is the number?
   
   33.3
   35.3
   36.3
   37.3
   38.3
   Option A
   Let ‘x’ be the required number
   0.25x + 25 =x
   ∴ 0.75x=25
   ∴ x=33.3

    

9. Lavanya income is 20% more than prem’s. How much less is Prem’s income than that of Lavanya’s?
   
   17.6%
   16.6%
   13.6%
   14.6%
   15.6%
   Option B
   Prem’s income is less than Lavanya’s = [(20 / (100 + 20)] × 100 %
   = 16.6%

    

10. The total population of a village is 3000, out of which 1500 are males. Out of every 100 males, 98 can read and write, but only 53% of the total population can do so. Find the percentage of women who can read and write.
    
    9
    7
    5
    8
    6
    Option D
    Number of men who can read and write = 1500 × (98 ÷ 100) =1470

    Number of men and women who can read and write =3000 × (53 ÷ 100) =1590

    Number of women who can read and write 1590 – 1470 = 120

    Thus, out of 1500women 120 can read and write.

    Required percentage = ( 120× 100) ⁄ 1500 %= 8%

     

1. An amount A is invested in scheme 1 for 2 years at 10% p.a compounded annually and an amount B is invested in scheme 2 for 3 years at 16% p.a simple interest. If the interest earned by scheme 1 is 87.5% of interest earned by scheme 2 and the total sum of amount invested in both scheme is Rs 9000. Then find the amount invested in scheme 1.
   6000
   5000
   4000
   7000
   8000
   Option A
   A+ B = 9000 ——– i

   In scheme 1,

   Net effective rate = 10 + 10 + (10 * 10)/100 = 21%

   Interest = 21x/100

   Net effective rate = 3 * 16 = 48%

   Interest = 48y/100

   According to question,

   48y/100 * 87.5/100 = 21x/100

   48y/100 * 87.5 = 21x

   4200y/100 = 21x

   2y = x —— ii

   Putting (ii) in (i)

   x + y = 9000

   2y + y = 9000

   y = 3000

   x = 9000 – 3000 = 6000

    

2. Lallu alone can do a piece of work in 25 days and Mallu is 16.66% inefficient less than Lallu. Mallu and Lallu started working together for 6 days then Shalu alone completed the rest work in 21 days. In how many days Shalu alone can complete the whole work?
   41.5 days.
   40.5 days.
   36.5 days.
   35.5 days.
   37.5 days.
   Option E
   Time taken by Lallu alone = 25 days

   Ratio of efficiency of Mallu and Lallu = 5 : 6

   Ratio of time taken by Mallu and Lallu = 6 : 5

   Therefore, time taken by Mallu alone = 30 days

   Total work = LCM of 30 and 25 = 150 units

   1 day work of Mallu = 150/30 = 5 units

   1 day work of Lallu = 150/25 = 6 units

   1 day work of (Lallu+ Mallu) = 5 + 6 = 11 units

   6 days work of (A + B) = 7 * 11 = 66 units

   Left work = 150 – 66 = 84 units

   Now, Shalu can complete 84 units = 21 days

   150 units = 21 * 150/84 = 37.5 days.

    

3. A box contains x green marbles, 4 red marbles and 5 white marbles 3 marbles are picked up randomly one after the another without replacement from the box. Find the value of x if probability of 3 marbles being green is 1/16.
   9
   8
   7
   6
   5
   Option C
   Probability of getting 1st green marble = x/(x + 4 + 5) = x/x + 9

   Probability of getting 2nd green marble without replacement = (x – 1)/(x + 8)

   Probability of getting 3rd green marble without replacement = (x – 2)/(x + 7)

   By question,

   [x/x + 9] * [(x – 1)/(x + 8)] * [(x – 2)/(x + 7)] = 1/16
   Now we will check through option and hence, option A = 7 satisfied the equation

   So, value of x = 7

    

4. Kiran purchased an old car for Rs 30000 and spent 3500 in servicing, 1500 in engine repair. If he wants to earn 20% profit then at what price he must have to sell the car?
   Rs 12000
   Rs 52000
   Rs 32000
   Rs 42000
   Rs 40000
   Option D
   Total C.P of car = (30000 + 3500 + 1500) = Rs 35000

   Required Profit = 20%

   So, required S.P = 120/100 * 35000 = Rs 42000

    

5. Man 1 and 2 started a business by investing Rs 2500 and Rs 2200 respectively. After 7 months they had invested 500 and 800 more amounts respectively. If the difference between the share of 1 and 2 is Rs 3486, find the total share earned after 1 year.
   Rs 103314
   Rs 204414
   Rs 104414
   Rs 103489
   Rs 105000
   Option C
   Ratio of shares of 1 and 2 = (2500 * 7 + 3000 * 5) + (2200 * 7 + 3000 * 5)

   = 325: 304

   Let the total profit earned after 1 year = x

   So, (325 – 304)/(325 + 304) * x = 3486

   21x/629 = 3486

   x = Rs 104414

    

6. 8 years ago the average ages of Lalith and Brijesh is 31 years and the present age of Lalith is 12 years less than the average of the ages of Brijesh and Simon 6 years hence. If Brijesh is 24 years elder than Simon, then find Simon’s age after 2 years?
   26
   27
   21
   22
   25
   Option A
   Lalith + Brijesh = 31 * 2 + 16 = 78———-(1)

   (Brijesh + 6 + lalith + 6)/2 – A = 12

   Brijesh + Simon – 2Lalith = 12——–(2)

   Brijesh – Simon = 24———–(3)

   (2) + (3)

   2Brijesh – 2Lalith = 36

   Brijesh – Lalith = 18——-(4)

   (4) + (1)

   2Brijesh = 96

   Brijesh = 48 years

   Simon= 48 – 24 = 24 years

   Simon’s age after 2 years = 24 + 2 = 26 years

    

7. If the ratio of the radius to slanting height of the conical vessele is 3:5 and volume of the cylindrical vessele is equal to the volume of the conical vessele. If the radius of the cylinder is 4 cm and the height of the cylindrical vessele is equal to the side of the square whose perimeter is 192 cm, then what is the curved surface area of the conical vessele?
   200∏
   140∏
   210∏
   240∏
   540∏
   Option D
   Height of the conical vessele = √(5×2 – 3×2) = 4x

   Volume of the conical vessele = 1/3 * 22/7 * r2 * h

   Volume of the cylindrical vessele = 22/7 * r2 * h

   Side of the square = 192/4 = 48

   Height of the cylindrical vessele = 12

   1/3 * 22/7 * 3x * 3x * 4x = 22/7 * 4 * 4 * 48

   x = 4 cm

   Radius of the conical vessele = 3 * 4 = 12 cm

   Slating height of the conical vessele = 5 * 4 = 20 cm

   CSA of the conical vessele = 22/7 * r * l

   = 22/7 * 12 * 20 = 240∏

    

8. Mayur and Suraj started the business with the investment in the ratio of 5:2 and after 6 months Suraj withdrew half of his initial investment. At the end of one year Mayur and Suraj gets Rs.4000 and Rs.1500 from the total profit obtained for managing the business and the profit ratio of Mayur and Suraj is 3:1. What is the total profit obtained at the end of year?
   14500
   16000
   14500
   13000
   12000
   Option E
   Profit ratio of Mayur and Suraj = 5x * 12:(2x * 6 + x * 6)

   = 10:3

   Total profit = (13x + 4000 + 1500) = 13x + 5500

   (10x + 4000)/(3x + 1500) = 3/1

   9x + 4500 = 10x + 4000

   x = 500

   Total profit = 13 * 500 + 5500 = 12000

    

9. The ratio of the monthly salary of person 1 to 2 is 4:3 and the ratio of the monthly salary of 2 to 3 is 2:1. If the difference between the 3’s monthly income and his savings is Rs.12000 and the saving’s of 3 is half of his expenditure, then what is the average of the income of person 1, 2 and 3?
   24000
   14000
   34000
   30000
   20000
   Option C
   person 3’s salary = x

   Expenditure = y

   Savings = s

   x = y + s

   y + s – s = 12000

   y = 12000

   Savings = 12000 * ½ = 6000

   x = 12000 + 6000 = 18000

   person 2’s income = 2/1 * 18000 = 36000

   person 1’s income = 4/3 * 36000 = 48000

   Average income of 1, 2 and 3 = (48000 + 36000 + 18000)/3 = 34000

    

10. If the marked price of the chair is Rs.6000 more than the money Kunal had but the merchant offers two successive discounts 10% and 5% respectively. Now he left with him Rs.9370. What is the marked price of the chair?
    116000
    105000
    106000
    100000
    105550
    Option C
    Kunal have = x

    MP of the mobile = x + 6000

    (x + 6000) * 90/100 * 95/100 + 9370 = x

    0.855x + 5130 + 9370 = x

    x = 100000

    MP of chair = 100000 + 6000 = 106000

     

1. 75 % of 320 – 24 % of 150 + 80 % of 280 = ?
   414
   234
   457
   456
   428
   Option E
   75 % of 320 – 24 % of 150 + 80 % of 280 = ?

   240 – 36 + 224 = ?

   ? = 428

    

2. 2030 ÷ √841 + 12.5 * 24 + ? = 47 * 11
   123
   147
   156
   134
   176
   Option B
   2030 ÷ √841 + 12.5 * 24 + ? = 47 * 11

   70 + 300 + ? = 517

   ? = 147

    

3. 6300 ÷ 35 + 15 * 28 – 60 % of 420 = ?
   548
   348
   788
   578
   678
   Option B
   6300 ÷ 35 + 15 * 28 – 60 % of 420 = ?

   180 + 420 – 252 = ?

   348 = ?

    

4. 21^2 + ?^2 – 32= 45 % of 180 + 80 * 8
   664
   221
   225
   289
   144
   Option D
   21^2 + ?^2 – 32= 45 % of 180 + 80 * 8

   441 + ?^2 – 9 = 81 + 640

   ?^2 = 289

   ? = 17

    

5. 315 ÷ √225 – 6.5 * 12 + 55% of 240 = ?
   75
   76
   77
   78
   79
   Option A
   315 ÷ √225 – 6.5 * 12 + 55% of 240 = ?

   21 – 78 + 132 = ?

   ? = 75

    

6. 52 % of 1200 – 112 + (11/12) of 1440 = ? + 850
   980
   675
   789
   456
   973
   Option E
   52 % of 1200 – 112 + (11/12) of 1440 = x + 850

   (52/100) * 1200 – 121 + (11/12) * 1440 – 850 = x

   x = 624 – 121 + 1320 – 850

   x = 973

    

7. 132 ÷ 6 × 12.5 – ? – 12 % of 50 + 93 = 152
   500
   846
   678
   768
   456
   Option B
   132 ÷ 6 × 12.5 – x – 12 % of 50 + 93 = 152

   (132 / 6) * (25/2) – (12/100) * 50 + 729 – 152 = x

   275 – 6 + 729 – 152 = x

   x = 846

    

8. ∛148877 – √5184 + 18 % of 150 + 133 = ? + 28
   2345
   5643
   2345
   2177
   5436
   Option D
   ∛148877 – √5184 + 18 % of 150 + 133 = x + 28

   53 – 72 + (18/100) * 150 + 2197 – 28 = x

   x = 53 – 72 + 27 + 2197 – 28

   x = 2177

    

9. 18 × 3 7/9 of (? ÷ 104) of 728 = 3808
   7
   6
   8
   5
   4
   Option C
   18 × 3 7/9 of (x ÷ 104) of 728 = 3808

   18 * (34/9) * (x/104) * 728 = 3808

   x = 8

    

10. (51120 ÷ 9) + (3/11) of 1155 = 25 % of ?
    24980
    13980
    23900
    23880
    23980
    Option E
    (51120 ÷ 9) + (3/11) of 1155 = 25 % of x

    (51120/9) + (3/11) * 1155 = (25/100) * x

    5680 + 315 = (x/4)

    5995 * 4 = x

    x = 23980

     

Find the number in ?:-

1. 1… 4… 20… 120… 840… ?
   4590
   6720
   4320
   5560
   7690
   Option B
   1 * 4 = 4
   4 * 5 = 20
   20 * 6 = 120
   120 * 7 = 840
   840 * 8 = 6720
   

    

2. ?… 8776… 1096… 136… 16… 1
   73216
   70288
   70716
   75216
   70216
   Option E
   (8776 – 8)/8 = 1096
   (1096 – 8)/8 = 136
   (136 – 8)/8 = 16
   (16 – 8)/8 = 1
   8776 * 8 + 8 = 70216
   

    

3. 13…15… 27… ?… 143… 293
   45
   54
   63
   36
   69
   Option C
   13 + 12 + 13 =15
   15 + 22 + 23 = 27
   27 + 32 + 33 = 63
   63 + 42 + 43= 143
   143 + 52 + 53 = 293
   

    

4. 41… 52… 67… 89… 121… ?
   146
   156
   166
   176
   186
   Option C
   41 52 67 89 121 166
   11 15 22 32 45
   4 7 10 13
   3 3 3
   

    

5. 6… 3… 3… 4.5… ?… 22.5
   9
   11
   13
   15
   17
   Option A
   6 * 1/2 = 3
   3 * 2/2 = 3
   3 * 3/2 = 4.5
   4.5 * 4/2 = 9
   9 * 5/2 = 22.5
   

    

Find the wrong term:-


6. 22… 36… 54… 76… 102… 130
   22
   36
   54
   76
   130
   Option E
   22 36 54 76 102 132
   14 18 22 26 30

    

7. 37… 75… 227… 911… 4559… 27358
   27358
   4559
   911
   227
   75
   Option A
   37 * 2 + 1 = 75
   75 * 3 + 2 = 227
   227 * 4 + 3 = 911
   911 * 5 + 4 = 4559
   4559 * 6 + 5 = 27359
   

    

8. 107… 118… 131… 148… 168… 190
   190
   167
   148
   131
   118
   Option B
   107 + 11 = 118
   118 + 13 = 131
   131 + 17 = 148
   148 + 19 = 167
   167 + 23 = 190
   

    

9. 17… 18… 34… 54… 68… 161
   161
   68
   54
   34
   18
   Option A
   17 * 2 = 34 ,,,, 18 * 3 = 54
   34 * 2 = 68 ,,,,,, 54 * 3 = 162
   

    

10. 47… 54… 66… 80…. 97…. 112
    54
    66
    80
    97
    112
    Option C
    47 + 7 * 1 = 54
    54 + 6 * 2 = 66
    66 + 5 * 3 = 81
    81 + 4 * 4 = 97
    97 + 3 * 5 = 112
    

     

Directions (1-5): Study the following information carefully to answer the following:

Rajaji college has five different types of groups, Maths, Physics , Chemistry, Commerce and computer . The total number of students are 2000. One-fourth of the total number of students are Commerce . 10% of the total students are Chemistry students . 25% of the total students are Physics. Remaining Students are either Computer or Physics students . The number of Physics Students are 120 more than the number of students of computer.

1. If 20% of commerce students are failed , what is the number commerce students are passed?
   500
   400
   300
   200
   100
   Option B
   Number of Maths students =25 x 2000/100 = 500
   Number of Commerce students = 2000 x1/4 = 500
   Number of Chemistry students =10 x2000 /100 = 200
   Number of computer + Physics students = 2000 – ( 500 +500 + 200) = 800
   Number of computer students + no.of computer students +120 =800
   2( number of computer students ) = 800 – 120 = 680
   Number of computer students = 680/ 2= 340.
   Number of physics students = 120 + 340 =460

   Number of commerce students = 500
   Failed = 20% of 500 = 20/100 x 500 = 100
   So, number of students passed = 500 – 100 = 400.
   

    

2. What is the ratio of the total number of Commerce students to the total number of Students in computer and physics students together?
   
   1:5
   3:4
   5:8
   6:3
   5:9
   Option C
   Ratio of commerce and ( computer + physics)= 500 🙁 340 +460 )
   = 500: 800 = 5:8

    

3. What is average of the total number of students in computer, Chemistry and Commerce students together?
   456.8
   234.8
   346.7
   123.6
   236.9
   Option C
   Average = (500 + 200 + 340 ) / 3 = 346.7

    

4. What is the difference between the total number of students in computer and maths students together and the number of Physics students?
   
   280
   380
   480
   580
   680
   Option B
   Difference = ( computer + maths) – physics
   = ( 340 + 500 ) – 460 = 380
   

    

5. The number of Chemistry students are approximately what percent of the total number of maths students and commerce students together?
   
   10
   20
   30
   40
   50
   Option B

   Number of maths + commerce = 500 + 500 =1000.
   Number of Chemistry students = 200
   percentage = 200/ 1000 x 100 = 20%
   

    

Directions (6-10): Study the following information carefully to answer the following:

The subsequent tables show the quantity of 3 kinds of books Social, Science and English sold by Venus Publication in 5 totally different colleges of India, within the year 2019 and the other table shows the profit % earned by Venus Publication from these 5 colleges on the 3 books given above. Study each the graphs fastidiously and answer the queries that follow:

The following table shows the no of books sold in thousands:



6. Total no. of Social books sold in B, D and E is what percentage of total English books sold in these schools?
   
   56.5%
   34.90%
   65%
   45%
   78.47%
   Option E
   Total Social books sold in B, D and E together
   = 40,000 + 5,000 + 10,000 = 55,000
   Total English books sold in these schools together
   = 30,000 + 20,000 + 20,000 = 70,000
   Required percentage = (55,000/70,000) × 100 = 78 .47%
   

    

7. What is the total profit earned on Science book from B, C and E if cost price of one Science book is Rs. 120 (in Rs. lakhs)
   Rs. 52,80,000
   Rs. 22,80,000
   Rs. 12,80,000
   Rs. 70,80,000
   Rs. 2,80,000
   Option B
   Total profit earned =25,000×120×40/100+20,000×120×30/100+15,000×120×20/100
   = Rs. 22,80,000
   

    

8. If Rs.120 is the cost price of Social book , then what is the total selling price of this book earned from A, C and D together (in Rs. Lakhs )?
   79 , 20, 000
   60 , 20, 000
   59 , 20, 000
   79 , 45, 000
   79 , 40, 000
   Option A
   Selling price earned from School A = 30,000 × 120 × 140/100 = 50, 40,000
   Selling price earned from School C= 15,000 × 120 × 120/100 = 21, 60, 000
   Selling price earned from School D= 5,000 × 120 × 120/100 = 7,20,000
   Total selling price earned = 79 , 20, 000. (add all selling prices)
   

    

9. If one Social book, one Science book and one English book is costs Rs. 150, Rs. 120 and Rs. 100 respectively then total selling price of these three books in School D is what percent more or less than the total selling price of these three books in School E (Nearest integer)?
   5
   6
   7
   8
   9
   Option B
   Total selling price of all the three books in School D =5,000×150×120/100+20,000×120×125/100+20,000×100×110/100=61,00,000
   Total selling price of all the three books in School E =10,000×150×130/100+15,000×120×120/100+20,000×100×120/100=65,10,000
   Required percentage =(65,10,000-61,00,000 )/ 65,10,000 × 100 = 6% less
   

    

10. What is the average number of Science books sold in five schools together?
    5
    10
    15
    20
    25
    Option D
    Required average no. of Science books
    = (20 + 25 + 20 + 20 + 15) /5 = 100 / 5 = 20
    

     

1. Number of ways of drawing 3 flowers from a basket containing 5 pink and 7 yellow flowers if at least one pink flower is to be included in the draw.
   125
   145
   175
   185
   195
   Option D
   Total no. of ways of drawing 3 flowers
   12C3= 12*11*10/3*2=220
   No. of ways of drawing 3 flowers if no pink ball is drawn
   7C3=7*6*5/3*2=35
   Hence req. no. of ways = 220-35=185
   

    

2. There are 5 MEN and 3 WOMEN in a family. They are photographed in groups of 2 MEN and one WOMEN.
   10
   20
   30
   40
   50
   Option C
   No. of different photographs = 5C2 * 3C1
   (5*4/2)*3
   =30
   

    

3. In majority voting between three parties A, B and C. A got 30%, B got 48% vote and C got rest of the votes polled. If B got 720 more votes than A then find how many votes were polled?
   3000
   4000
   5000
   6000
   7000
   Option B
   The difference in the percentage of votes received by A and B= 48-30= 18%
   Total number of votes polled= 720*100/18= 4000
   

    

4. In Parliament election between two Parties A and B, 10% of the votes were invalid. The ratio of votes received by A and B is 7:5. If the number of votes polled in favor of A was 630, then find the number of votes polled?
   1100
   1200
   1300
   1400
   1500
   Option B
   Total number of votes received by A and B= 630*12/7= 1080
   Total number of votes polled= 1080*100/90= 1200 votes.
   

    

5. Persons X, Y, Z jogging around a circular field and complete a single round in 18, 22, 30 seconds respectively. In how much time they will meet again at the starting point?
   16minutes 30 seconds
   6minutes 30 seconds
   16minutes 10 seconds
   16minutes 20 seconds
   5minutes 30 seconds
   Option A

   Required time = LCM of 18, 22, 30
   = 990 seconds
   =990/60 minutes=16minutes 30 seconds
   

    

6. The respective ratio between present age of Kalim’s Aunty, Uncle and cousin is 16:18:3. Sum of ages of aunty and uncle is 68. Difference between the age of cousin after 6 yrs and aunty’s age before 18 yrs.
   5
   4
   3
   2
   1
   Option D
   Ratio of present age = 16:18:3
   Sum of ages 16x+18x=34x=68
   Therefore we can find uncles’s age=36, aunty’s age=32 and cousin’s age=6.
   cousin’s age after 6 yrs= 12 yrs
   aunty’s age before 18 yrs=14 yrs
   Difference=2 yrs
   

    

7. Present age of Arun and Anil are in the ratio 4:5. Five years hence ratio of their ages will be 5:6. Arun’s present age is.
   10
   20
   30
   40
   50
   Option B
   4x+5/5x+5 = 5/6
   24x+30 = 25x+25
   X= 30-25=5 years
   Arun’s present age is 20 years.
   

    

8. In how many different ways can the letters of the word MONICOT can be arranged so that vowels occupy odd positions?
   220
   350
   470
   576
   668
   Option D
   Required no. of ways =4P3* 4p4
   = 576 ways
   

    

9. In how many different ways can the letters of the word HEADING can be arranged so that vowels always come together?
   230
   450
   540
   660
   720
   Option E
   HEADING = 7 letter
   3vowels (EAI) supposed to form one letter
   Now remaining letter arranged in 5!
   Vowels arranged in 3!
   Therefore 5!*3! = 720 ways

    

10. The ratio of men and women in a office is 3: 5. If 10% of men and 15% of women speak Tamil, the difference between men and women who did not speak Tamil is 155, find the total number of employee in the office.
    880
    270
    740
    630
    540
    Option

    Men: Women = 3: 5
    Unspeak Tamil = 3 * 90/100 : 5 * 85/100
    = 2.7 : 4.25
    Total = 155/1.55 * 8 = 800
    

     

1. Kiran saves 9% of his income. Two years after his income hiked by 19% but his savings remain the same.Then find the expenditure percent of him?
   20.25%
   22.56%
   24.56%
   34%
   20.87%
   Option E
   Let kiran income be = Rs.100
   Savings =Rs.9
   Expenditure= Rs.91
   New income= Rs.119
   New expenses= 119-9 = 110
   Required %= 110-91/ 91 *100 = 20.87%
   

    

2. The salary of two persons A and B are 15% and 23% less than that of a third person C. Find What percentage of Salary B is less than that of salary A.?
   10%
   10.5%
   9%
   9.4%
   11%
   Option D
   Let the C salary be Rs.100
   Then A’s = 85 and B’s= 77
   Req. % = 85-77/ 85 * 100 = 9.4%

    

3. To give continuous three discount of 10% is equal to giving one discount of —%
   23%
   24%
   26%
   27%
   28%
   Option E
   continuous discount = 10 + 10 – 10 * 10/ 100
   = 20%
   Continuous discount = 20 %+ 10% – 20 * 10/ 100 = 28%

    

4. If a TV is sold at Rs. 8000 which is marked at Rs.9500.then what is the discount percentage?
   15.78%
   11.78%
   12.78%
   13.78%
   17.78%
   Option A
   MP *( 100-D) / 100 =SP
   => 9500 * ( 100 -D) / 100 = 8000
   = 15.78%
   

    

5. The ratio of man and woman in a city is 7:6. If 73% of men and 87% of women are educated the what is the percentage of people who are not educated?
   45.89%
   19.46%
   20.54%
   21%
   22%
   Option C
   Men = 700
   Women=600
   Educated men = 511
   educated women = 522
   Req % = 1300 – 1033/ 1300 = 20.54%
   

    

6. Today is Saturday. After 64 days it will be ?
   
   monday
   wednes day
   sunday
   friday
   tuesday
   Option C
   Each day of the week is repeated after 7 days
   ∴ After 63 days, it would be saturday
   So, After 64 days, it would be Sunday

    

7. Three students height are in the ratio 3 :4 : 5 . The sum of the tallest and the shortest equals the sum of the third and 52. The shortest height is ?
   29
   21
   30
   39
   47
   Option D
   Let the heights be 3H, 4H and 5H.
   Then, 5H + 3H = 4H + 52
   ⇒ 4H = 52
   ⇒ H = 13
   ∴ Then smallest number = 3H = 3 x13 = 39

    

8. a train crosses a platform in 30 sec and a pole in 18 sec. If a speed of train is 45 km/hr. Then find the length of the train?
   150 m
   100m
   200m
   250m
   300m
   Option A
   length of the plat form is x m
   length of the train = 45 * 5/18 * 18 = 225 m
   So x+ 225 / 30 = 45 * 5/18
   => x= 375-225= 150m
   

    

9. Mr. Gosh can print 40 pages in 10 minutes, Mr. Giri and Mr. Gosh both working together can print 250 In 25 minutes. In how many minutes Mr.Giri can print 36 pages?
   2 min
   3 min
   4 min
   6 min
   5 min
   Option D
   Efficiency ( per minute) of Gosh = 4 copies/min
   Efficiency of Gosh and Giri together = 10 pages/min
   ∴ Efficiency of Giri alone = 10 – 4 = 6 pages/min
   ∴ Mr. Giri needs 6 min to print 36 pages.

    

10. kailash, on race crosses first 100 km, at 64 km/hr. and the next 150km at 80km/hr. the average speed for the first 250 km of the race is ?
    
    71.11km/hr.
    81.11km/hr.
    61.11km/hr.
    70.11km/hr.
    69.11km/hr.
    Option A
    Average speed = 2t1 x t2/( t1 + t2)
    = (2x64x80)/(64+80)km
    = (2 x 64 x 80 ) / 144 km/hr.
    =71.11km/hr.

     

Find the wrong number in the series

1. 157..164..175..188..203..224..
   203
   157
   164
   175
   224
   Option A
   Clue: +7…+11……+19

    

2. 301..308..628..1899..7610..
   301
   308
   628
   1899
   7610
   Option E
   Clue: +7 * 1 ..+6* 2….+4*4

    

3. 21..26..62..201..842..4145
   21
   26
   62
   201
   842
   Option E
   Clue: *1+5..*2+10…*5+25

    

4. 63.. 67.. 78.. 92..117…
   63
   67
   78
   92
   117
   Option C
   Clue +2^2, +3^2…..

    

5. 6..7..15..51..208..
   6
   7
   15
   51
   208
   Option D
   Clue:*1+1,*2+2

    

6. 174..169..162..150..138..
   174
   169
   162
   150
   138
   Option E
   174..169..162..152..138..
   -5 -7 -10 -14
   -2 -3 -5

    

7. 193..97..47..25..13..7
   193
   97
   47
   25
   13
   Option C
   clue: 193..97..49..25..13..
   -96 -48 -24 -12
   /2 /2 /2
   

    

8. 1..3..12..48..400..6432
   1
   3
   12
   48
   400
   Option C
   Clue:*1+2,*2+4,*4+8,*8+16*16+32

    

9. 17..24..13..26..9
   Option
   Clue: +7,-11,+13,-17

    

10. 12..21..45..95..216..
    12
    21
    45
    95
    216
    Option C
    Clue:+3^2 ,+5^2, +7^2….

     

Directions (1-5): Study the following information carefully and answer the questions that follow:
The following bar graph shows the production of egg (in lakh tones) by three different farms A, B and in the given the years.

1. In which year the percentage of production of farm B to the production of farm A is maximum?
   2011
   2012
   2013
   2014
   2015
   Option A
   In 2011= (80/60)*100=133.3%
   In 2012= (45/65)*100=69.2%
   In 2013= (65/50)*100=130%
   In 2014= (65/55)*100=118.2%
   In 2015= (55/70)*100=78.5%
   

    

2. What is the difference between the production of farm A in 2012 and farm B in 2011?
   20
   15
   10
   5
   25
   Option B
   Required difference= 80 lakh tones- 65 lakh tones
   = 15 lakh tones
   

    

3. What is the ratio of the average production of farm C in the period 2011-2014 to the average production of farm A in the same period?
   49: 35
   51: 46
   49: 44
   47: 46
   49: 46
   Option E
   Total production in the given period of farm A = 60 + 65 + 50 + 55
   = 230
   Required average production of farm A = 230/4 = 57.5
   Total production in the given period of farm C = 55 + 75 + 70 + 45 = 245
   Required average production of farm C = 245/4 = 61.25
   Required ratio = 61.25: 57.5
   = 49: 46
   

    

4. The average production for five years was Maximum for which farm?
   A
   B
   C
   none
   A and B
   Option B
   Average production of farm A in the given period= (60+65+50+55+70)/5
   = 60
   Average production of farm B in the given period= (80+45+65+65+55)/5
   = 62
   Average production of farm C in the given period= (55+75+70+45+60)/5
   = 61
   Thus the farm B has the maximum average
   

    

5. What is the approximate percentage increase in the production of farm C from 2011 to 2013?
   17
   27
   37
   47
   57
   Option B
   Required percentage increase from 2011 to 2013 = [(70 – 55)/55] * 100
   = 15/55 * 100
   = 27%
   

    

Directions (6-10): Study the following information carefully and answer the questions that follow:


The table shows the discount % given by malls on different items. The Market Price of all things in all malls is the same. Some values are missing.

6. If the average SP of things II in all the malls is Rs 2568, Find the MP of things II.
   
   Rs 1600
   Rs 2600
   Rs 3600
   Rs 4600
   Rs 5600
   Option C
   SP by store A = (100-28)/100 * MP = 72% of MP, by B = 82% of MP, by C = 60% of MP
   So (72+82+60)/3 * MP/100 = 2568
   Solve, MP = Rs 3600
   

    

7. Difference between SP of things I by malls A and B is Rs 486, Find the SP of same things by store C.
   Rs 4104
   Rs 4100
   Rs 3104
   Rs 4004
   Rs 1104
   Option A
   Difference in SP = 93% of MP – 84% of MP
   So 9% of MP = 486
   Solve, MP = 5400
   So SP by store C = 76% of 5400 = Rs 4104
   

    

8. Average SP of things III by malls A and B is Rs 3608, by malls B and C is Rs 3300. Find the SP of things III by store C.
   Rs 1080
   Rs 2080
   Rs 3080
   Rs 4080
   Rs 5080
   Option C
   Let x% discount by store B
   So [84 + (100-x)]/2*100 * MP = 3608
   And [70 + (100-x)]/2*100 * MP = 3300
   Put value of (100-x) from 1 equation to another and solve for MP
   MP = Rs 4400
   So SP by store C = 70/100 * 4400 = Rs 3080
   

    

9. Mall A earned 10% profit by selling product V. If CP of things V at all mall is same then find the ratio of profits by B and C in selling V.
   9 : 14
   9 : 19
   9 : 15
   9 : 13
   9 : 17
   Option A
   MP = x
   SP by A = 88% of x
   So CP by A = 100/110 * 88/100 *x = 4x/5
   Now CP is same
   SP by B = 89x/100, so profit of B = 89x/100 – 4x/5 = 9x/100
   SP by C = 94x/100, so profit of B = 94x/100 – 4x/5 = 14x/100
   So required ratio = 9 : 14
   

    

10. What is the ratio between the selling price of artice II at mall A and mall B?
    36/41
    36/45
    36/47
    37/41
    31/41
    Option A
    So required ratio is 5 : 6
    SP of things II at store A= (100-28)%MP= 72%MP
    SP of things II at stire B= (100-18)= 82%MP
    Required ratio= 72%MP/82%MP= 72/82=36/41

     

1. 35m²+16m-3=0
   10n²+136n-56=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option E
   35m²+16m-3=0
   35m² -5m+ 21m-3=0
   5m(7m -1) + 3(7m -1)=0
   (7m-1) ( 5m+3)=0
   m= 1/7 , -3/5
   10n²+136n-56=0
   10n² -4n+ 140n -56=0
   n( 10n-4) + 14(10n-4)=0
   (10n-4) ( n+14)=0
   n= 4/10,- 14
   Hence the relation cannot be determined
   

    

2. 10m²-136m-14=0
   25n²-10n-8=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option C
   10m²-136m-14=0
   10m² -140m+m-14=0
   10m(m -14) + 1(m -14)=0
   (m-14) ( 10m+1)=0
   m= +14 , -1/10
   25n²-10n-8=0
   25n² -20n+ 10n -8=0
   5n( 5n-4) + 2(5n-4)=0
   (5n-4) ( 5n+2)=0
   n= 5/4, -2/5
   Hence m>n
   

    

3. 11m²+120m= 11
   8n²+74n+143=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option E
   11m²+120m= 11
   11m² +121m-m-11=0
   11m(m +11) -1(m +11)=0
   (m+11) ( 11m-1)=0
   m=- 11 , 1/11
   8n²+74n+143=0
   8n² +22n+52n+143 =
   2n( 4n+11) +13(4n+11)=0
   (4n+11) ( 2n+13)=0
   n= -11/4, -13/2
   Hence the relation cannot be determined
   

    

4. 8m²+51m+18=0
   49n²-1=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option A
   8m²+51m+18=0
   8m² +48m+3m+18=0
   8m(m +6) + 3(m +6)=0
   (m+6) ( 8m+3)=0
   m= -6 , -3/8
   49n²-1=0
   (7n-1) ( 7n+1)=0
   n= 1/7, -1/7
   Hence n>m
   

    

5. 40m²-13m+1=0
   18n²+77n-18=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option D
   40m²-13m+1=0
   40m² -8m-5m+1=0
   8m(5m -1) -1(5m -1)=0
   (5m-1) ( 8m-1)=0
   m= 1/5 , 1/8
   18n²+77n-18=0
   18n² +81n- 4n -18=0
   9n( 2n+9) – 2(2n+9)=0
   (2n+9) ( 9n-2)=0
   n= -9/2, 2/9
   Hence m ≥ n
   

    

6. 16m²+60m+14=0
   8n²– 77n-30=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option A
   16m²+60m+14=0
   16m² + 56m+ 4m+14=0
   8m(2m +7) + 2(2m +7)=0
   (2m+7) ( 8m+2)=0
   m= -7/2 , -2/8
   8n²– 77n-30=0
   8n² -80n+ 3n -30=0
   8n( n-10) + 3(n-10)=0
   (n-10) ( 8n+3)=0
   n= 10, -3/8
   Hence n>m
   

    

7. 35m²+32m+5=0
   9n²– 36n+32=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option A
   35m²+32m+5=0
   35m² + 25m+ 7m+5=0
   5m(7m +5) + 1(7m +5)=0
   (7m+5) ( 5m+1)=0
   m= -5/7 , -1/5
   9n²– 36n+32=0
   9n² -12n-24n +32=0
   3n( 3n-4) -8(3n-4)=0
   (3n-4) ( 3n-8)=0
   n= 4/3, 8/3
   Hence n>m
   

    

8. 35m²-18m-81=0
   10n²+184n-114=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option C
   35m²-18m-81=0
   35m² – 45m+ 63m-81=0
   5m(7m -9) + 9(7m -9)=0
   (7m-9) ( 5m+9)=0
   m= 9/7 , -9/5
   10n²+184n-114=0
   10n² -6n+ 190n -114=0
   n( 10n-6) + 19(10n-6)=0
   (10n-6) ( n+19)=0
   n= 6/10,- 19
   Hence m>n
   

    

9. 10m²+79m+63=0
   48n²+10n-2=0
   m<n
   m<=n
   m>n
   m>=n
   the relation cannot be determined
   Option A
   10m²+79m+63=0
   10m² +70m+9m+63=0
   10m(m +7) + 9(m +7)=0
   (m+7) ( 10m+9)=0
   m= -7 , -9/10
   48n²+10n-2=0
   48n² -6n+ 16n -2=0
   6n( 8n-1) + 2(8n-1)=0
   (8n-1) ( 6n+2)=0
   n= 1/8, -2/6
   Hence n>m
   

    

10. 2m²-35m= -17
    2n²+35n= +17
    m<n
    m<=n
    m>n
    m>=n
    the relation cannot be determined
    Option D
    2m²-35m= -17
    2m² -34m-1m+17=0
    2m(m -17) -1(m -17)=0
    (m-17) ( 2m-1)=0
    m= 17 , 1/2
    2m²-35m= -17
    2m² -34m-1m+17=0
    2m(m -17) -1(m -17)=0
    (m-17) ( 2m-1)=0
    m= 17 , 1/2
    2n²+35n= +17
    2n² +34n+1n+17=0
    2n(n +17) +1(n +17)=0
    (n+17) ( 2n+1)=0
    n= -17 , -1/2
    m>n