1. 23 ½ % of 800 +12 ¾ % of 2400 + 452 = 1125 ÷ 25 + 4 *?
   608.5
   620.5
   418.5
   618.5
   558.5
   Option D
   (47/200)*800 + (51/400)*2400 + 2025 = (1125/25) + 4x
   306 + 188 + 2025 – 45 = 4x
   2474 = 4x
   X = 2474/4 = 618.5
   

    

2. (12/75) of (48/1136) of (32/96) of 10650 =? % of 500
   5.1
   4.9
   4.8
   4.7
   4.6
   Option C
   (12/75)* (48/1136)*(32/96) * 10650 = (x/100)*500
   24 = 5x
   X = 24/5 = 4.8
   

    

3. 15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 =? + 2 7/12
   9 1/10
   9 1/7
   12 1/3
   6 1/3
   9 1/3
   Option E
   15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 – 2 7/12 = x
   (15 – 4 + 8 – 7 – 2) (1/3 – 1/6 + 5/12 – 2/3 – 7/12) = x
   10 [(4 – 2 + 5 – 8 – 7)/12] = x
   X = 10 (-8/12) = 10 (-2/3)
   X = (30 – 2)/3 = 28/3 = 9 1/3
   

    

4. (6390 ÷ 15) + (7182 ÷ 19) + (10224 ÷ 8) =? – 24 % of 750
   2262
   2160
   2346
   3422
   2211
   Option A
   (6390 ÷ 15) + (7182 ÷ 19) + (10224 ÷ 8) = x – (24/100)*750
   426 + 378 + 1278 = x – 180
   426 + 378 + 1278 + 180 = x
   X = 2262
   

    

5. ∛157464 + √8836 – 45 % of 800 =? + (1188 ÷ 9) – 362
   567
   952
   678
   389
   890
   Option B
   54 + 94 – (45/100)*800 = x + (1188/9) – 1296
   54 + 94 – 360 + 1296 = x + 132
   54 + 94 – 360 + 1296 – 132 = x
   X = 952
   

    

6. 80.0989 % of 600.099 + 33.199 % of 399.9009 – 47.7065 % of 150.298 = ?
   530
   540
   430
   420
   400
   Option B
   80/100 * 600 +33/100 * 400 – 48/100 * 150 =?
   480+132-72=?
   540=?
   

    

7. √4759.1998 × 3.11995 +50.098 % of 25.909 + 14.9086 × 2.09998 = ?
   220
   230
   240
   250
   260
   Option D
   √4761 * 3 + 50/100 * 25 + 15 * 2=?
   69*3 + 13 + 30 = ?
   207 + 43 = ?
   250 = ?
   

    

8. ? + (5.902)^3 – 28.503 * 8.080 = 2.098 * 1.998
   10
   20
   30
   40
   50
   Option B
   ? + 216 – 29 * 8 = 2 * 2
   ? + 216 – 232 = 4
   ? = 4 + 232 – 216
   ? = 20
   

    

9. 14.178 % of 3047.7 – (4.99/9.21) of (6.99/2.5) of 71 = 22.82 + ?
   230
   220
   292
   340
   450
   Option C
   14 % of 3050 – (5/9) of (7/2.5) of 72 = 23 + ?
   (14/100) * 3050 – (5/9) * (7/2.5) * 72 = 23 + ?
   427 – 112 – 23 = ?
   ? = 292
   

    

10. √286.6 × 13.99 ÷ 7.092 = ?^2 + (8.13)^1/3 + 28.089 ÷ 3.99
    3
    4
    5
    6
    7
    Option C
    17 * (14/7) = ?^2 + 2 + (28/4)
    34 – 2 – 7 = ?^2
    ?2 = 25
    ? = 5
    

     

1. Nithish can complete 60% of the work in9 days and Nithish and Kamalesh together can complete half of the work in 4.5 days. If Nithish, Kamalesh and Vijilesh together can complete 40% of the work in 3 days and they are gets the total wages is 1200, then what is the wage of Kamalesh?
   100
   200
   300
   400
   500
   Option D
   N = 100/60 * 9 = 15 days

   N + K = 100/50 * 4.5 = 9 days

   N = 1/9 – 1/15 = 1/22.5 days

   N + K + V = 100/40 * 3 = 7.5 days

   V = 1/7.5 – 1/9 = 1/45

   Work ratio of N, K and V = 1/15:1/22.5:1/45

   = 3:2:1

   K’s wage = 2/6 * 1200 = Rs.400

    

2. Arun bought a Bed and the ratio of the marked to cost price of the Bed is 6:5. If the shop offers a discount of 15% on the marked price of the Bed, then what is the profit percentage?
   3
   2
   4
   5
   6
   Option B
   MP = 6x

   CP = 5x

   SP = 6x * 85/100 = 5.1x

   Profit percentage = (5.1x – 5x)/5x * 100 = 2%

    

3. What is the speed of stream if the sum of the speed of boat in still water and stream is 20 kmph and the speed of the boat against stream is 10 kmph.
   5
   10
   15
   20
   25
   Option A
   Downstream speed = 20 kmph

   Upstream speed = 10 kmph

   Stream speed = (20 – 10)/2 = 5 kmph

    

4. Ratio of the ages of Prem and Prajesh is 3:8 and after 5 years the ratio of the ages of Prajesh to Pranshu is 7:5. If the sum of the ages of Prem, Prajesh and Pranshu is 32 years, then what is the present age of Prem?
   5
   6
   7
   8
   9
   Option B
   prem/prajesh = 3/8

   Prem + Prajesh + Pranshu = 32

   (Prajesh + 5)/(Pranshu + 5) = 7/5

   7Pranshu+ 35 = 5Prajesh + 25

   3/8 Prajesh + prajeash + (5Prajesh – 10)/7 = 32

   117Prajesh = 1872

   Prajesh = 16 years

   Prem = 3/8 * 16 = 6 years

    

5. A Ship left 2 h later than the scheduled time. But to reach its destination 2400 km away in time, it had to increase its speed by 33.33% over its usual speed. Find the new increased speed?
   400
   300
   200
   100
   50
   Option A
   Let the speed of the plane be 3x

   Then new speed = 3x * (133.33/100) = 4x

   Then

   2400 = (2 * 3x * 4x)/x

   x = 100

   New speed = 4 * 100 = 400 kmph

    

6. Two buses leave Delhi for Jammu at 7.00 pm and 8.00 pm respectively and travel at 48 kmph and 63 kmph respectively. How many kilometers from Delhi will the two buses meet?
   201.6
   209
   205
   220.7
   209.6
   Option A
   Relative distance = 48

   Relative speed = 63 – 48 = 15

   Then,

   48 = 15 * time

   3.2 h = time

   Required distance = 3.2 * 63 = 201.6 km

    

7. Find the sum of speed of boat and stream if the ratio of speed of stream and boat is 1: 7. If a boat covers 120 km upstream and downstream in 3.5 h.
   70
   80
   90
   100
   110
   Option B
   The ratio of speed of stream and boat = 1: 7 (x, 7x)

   [120/(7x – x)] + [120/(7x + x)] = 3.5
   35/x = 3.5

   x = 10

   Required sum = 8 * 10 = 80 kmph

    

8. Person1 lend Rs.X and obtained Rs. 780 at 13% per annum for 3 years.Person2 lend Rs.Y and obtained Rs. 1020 at 17% per annum for 2 years. Then find the simple interest obtained on x + y for 6 years at 7% per annum.
   2000
   2100
   2200
   2300
   2400
   Option B
   SI = pnr/100

   780 = x * 3 * 13/100

   x = 2000

   1020 = y * 2 * 17/100

   y = 3000

   Required interest = (2000 + 3000) * 7/100 * 6

   = 2100

    

9. If Rs. 2500 is invested at 20% p.a. compound interest compounded quarterly.Find the amount after one year.
   4500
   3450
   2390
   3210
   3039
   Option E
   A = p (1 + r/100)n

   A = 2500 (1 + 20/400)4

   A = 2500 * 105/100 * 105/100 * 105/100 * 105/100

   A = 3039( approx)

    

10. A water heater can heat the bucket of water in 40 minutes while another takes 45 minutes to heat under. If both heaters are switched on at same instance then how long will it take to heat approximately?
    21
    22
    23
    24
    25
    Option B
    Let the two heaters be X and Y . ‘X’ heatss at 40min

    ‘Y’ at 45min

    Together = (x x y)/(x + y)
    = (45 x 40)/(45 + 40)
    = 45 x 40/85
    = 21.1764
    = 22 min (approx).
    

     

The following table shows the total number of students completed their school in different years in a certain school and the ratio of boys and girls. The percentage of students went to college and the remaining students quit their studies.

1. Find the ratio between the total number of students went to college in the year 2004 and the total number of peoples who completed their school in 2008 ?
   
   306 : 12
   306 : 125
   36 : 125
   30 : 125
   306 : 11
   Option B
   The total number of students employed in MNC companies in the year 2014
   = > 1800 * (68/100)
   The total number of students who completed their UG degree in 2018 and doing business
   = > 2500 * (20/100)
   Required ratio = [1800 * (68/100)] : [2500 * (20/100)]
   = > 306 : 125
   

    

2. Find the average number of students who completed school in all the given years together?
   2050
   2060
   2070
   2090
   2000
   Option B
   The average number of students who completed their UG degree in all the given years together
   = > (1800 + 2400 + 1600 + 2000 + 2500) / 5
   = > (10300/5) = 2060
   

    

3. Find the difference between the total number of boys who completed their school in the year 2005 to that of total number of girls who completed their school in the year 2007?
   400
   600
   100
   500
   700
   Option D
   The total number of males who completed their degree in the year 2015
   = > 2400 * (2/3) = 1600
   The total number of females who completed their degree in the year 2017
   = > 2000 * (11/20) = 1100
   Required difference = 1600 – 1100 = 500
   

    

4. The total number of students went to college in the year 2005 is approximately what percentage of total number ofwent to college in the year 2007?
   44
   77
   88
   101
   103
   Option E
   The total number of peoples employed in MNC companies in the year 2015
   = > 2400 * (72/100) = 1728
   The total number of peoples employed in MNC companies in the year 2017
   = > 2000 * (84/100) = 1680
   Required % = (1728/1680) * 100 = 102.85 % = 103 %
   

    

5. The total number of students who completed their school in 2006 and what percentage more/less than the total number of girl students who completed their school in the year 2004?
   50
   51
   52
   53
   54
   Option C
   The total number of peoples who completed their UG degree in 2016 and doing business
   = > 1600 * (24/100) = 384
   The total number of female students who completed their UG degree in the year 2014
   = > 1800 * (4/9) = 800
   Required % = [(800 – 384) / 800] * 100 = 52 % less
   

    

The following pie chart shows the percentage distribution of total number of face masks imported by Micco lab in 6 different months to china.



6. Find the difference between the total number of face masks imported by Micco lab in Oct and Feb together to that of total number of face masks imported by company B in Nov and Dec together, if the ratio between the total face masks imported by Micco lab to that of company B is 3 : 4 and the percentage distribution of total face masks imported by company B is same as Micco lab.
   3850
   1290
   2390
   3450
   1200
   Option A
   The total number of face masks imported by Micco lab in Oct and Feb together
   = >15000 * (29/100) = 4350
   The total number of face masks imported by company B in Nov and Dec together
   = >[15000 * 4/ 3] * (41/100) = 8200
   Required difference = 8200 – 4350 = 3850
   

    

7. Find the central angle corresponding to the total number of face masks imported by Micco lab in Jan and Mar together
   1120
   1130
   1080
   1909
   1000
   Option C
   The central angle corresponding to the total number of face masks imported by Micco lab in Jan and Mar together in the year 2018= (30 / 100) * 360 = 1080

    

8. Find the difference between the average number of face masks imported by Micco lab in Nov and Dec together to that of the average number of face masks imported by Micco lab in Feb and Mar together.
   1345
   1275
   4555
   6565
   1455
   Option B
   The average number of face masks imported by Micco lab in Nov and Dec together
   = > 15000 * (41/100) * (1/2) = 3075
   The average number of face masks imported by Micco lab in Feb and Mare together
   = > 15000 * (24/100) * (1/2) = 1800
   Required difference = 3075 – 1800 = 1275
   

    

9. If the total number of face masks imported by Micco lab in Aug is 45% more than the total number of face masks imported by the same company in Jan and the total number of face masks imported by Micco lab in Sep 2018 is 60% more than the total number of face masks imported by the same company in Feb then find the total number of face masks imported by the Micco lab in Aug and Sep together.
   7710
   3490
   5590
   7800
   7000
   Option A
   The total number of face masks imported by the Micco lab in Aug and Sep together
   = >15000 * (20/100) * (145/100) + 15000 * (14/100) * (160/100)
   = > 4350 + 3360 = 7710
   

    

10. Find the ratio between the total number of face masks imported by Micco lab in Nov and Jan together to that of total number of face masks imported by Micco lab in Oct and Mar together.
    38: 25
    18: 25
    38: 15
    38: 23
    8: 25
    Option A
    The total number of face masks imported by Micco lab in Nov and Jan together
    = > 15000 * (38/100)
    The total number of face masks imported by Micco lab in Oct and Mar together
    = > 15000 * (25/100)
    Required ratio = [15000 * (38/100)]: [15000 * (25/100)]
    =38: 25
    

     

1. Three members started a business by investing in the ratio of 6 : 7 : 10. After 5 months, P1 invested Rs. 10000 more but P2 withdraw Rs. 20000. Find the initial investment of P2, if the share of P1, P2 and P3 is in the ratio of 43 : 42 : 46?
   Rs. 15000
   Rs. 20000
   Rs. 25000
   Rs. 30000
   Rs. 35000
   Option E

   [6x * 5 + (6x + 10000) * 7] : [7x * 12] : [10x * 5 + (10x – 20000) * 7] = 43 : 42 : 46
   = > [30x + 42x + 70000] : [84x] : [50x + 70x – 140000] = 43 : 42 : 46
   = > [72x + 70000] : [84x] : [120x – 140000] = 43 : 42 : 46
   by qs,
   (72x + 70000) / 84x = (43/42)
   72x + 70000 = 86x
   70000 = 14x
   x = 70000 / 14 = 5000
   The initial investment of P2= 7x = Rs. 35000
   

    

2. The difference between the ages of A and B, 2 years ago is 15 years. If the ratio of their ages, 7 years hence will be 3 : 2, then find the present age of B’s Father, who is 25 years older than B?
   42
   44
   46
   48
   50
   Option D
   2 years ago, A– B = 15
   7 years hence, The ratio of A and B = 3 : 2 (3x, 2x)
   Present age of A and B = 3x – 7, 2x – 7
   From Question,
   3x – 2x = 15
   x = 15
   Present age of B = 2x – 7 = 23 years
   Present age of B’s mother = 23 + 25 = 48 years
   

    

3. 30 Male workers can complete the work in 12 days. After 5 days from the start of the work, some workers left the job and the remaining work will be completed by remaining workers in 10 days. Then find the total number of workers left the job after 5 days of the work?
   9
   10
   11
   12
   13
   Option A
   Total work = 30 * 12 = 360 units
   5 days work = 30 * 5 = 150 units
   Remaining work = 360 – 150 = 210 units
   Given,
   210 / (30 – x) = 10
   21 = 30 – x
   x = 30 – 21 = 9
   The total number of workers left the job after 5 days of the work = 9
   

    

4. The length of Train 1 is 150 m which crosses the train 2 of length 200 m in 42 sec, running in same direction. Find the speed of second train (In km/hr), if the speed of train 1 is 90 km/hr and the first train is faster than second one?
   40
   50
   60
   80
   100
   Option C
   T = D / S
   According to the question,
   42 = (150 + 200) / [(90 – x) * (5/18)]
   42 = (350 * 18) / [[(90 – x) * 5]
   (350 * 18) / (42 * 5) = 90 – x
   30 = 90 – x
   x = 90 – 30 = 60
   The speed of second train = 60 km/hr
   

    

5. In a total monthly salary of Guru, he spends 15 % on Rent, 9 % on travelling expenses and 20 % for his children’s education. In the remaining monthly salary, he spends 45 % for medicines and other expenses and the remaining salary is saved by him, which is equal to Rs. 21560. Find the monthly salary of Guru?
   Option
   Let the monthly salary of Shyam be x,
   x * (56/100) * (55/100) = 21560
   x = 21560 * (100 / 56) * (100 / 55) = 70000
   The monthly salary of Guru = Rs. 70000
   

    

6. Four Partners distributed a sum of Rs 44352 among themselves. 1st got 3/8th of total amount. 2nd got 1/6 th part of the remaining amount and the remaining amount was divided between 3 and 4 in the ratio 3 : 2. The amount received by 4 is
   9240
   9040
   5240
   6740
   9200
   Option A
   1’s share = 3/8 * 44352=16632
   Remaining amount = (44352 – 16632) = 27720
   2’s share = 4620
   Remaining amount=27720-4620=23100
   share of 4=23100*2/5=9240
   

    

7. The ratio of daily salary of two labours is 7 : 5 and one gets daily Rs 120 more than the other, what are their sum of daily salaries?
   Rs.820
   Rs.520
   Rs.700
   Rs.720
   Rs.450
   Option D
   Let their daily salary are 7x and 5x. Now
   7x – 5x = 120
   2x = 120
   X = 60
   So daily salary are 420 and 300.
   Sum= 420+300= Rs.720

    

8. The ratio of income of two friends is 5:4 and their expenditure is as 3:2. If at the end of the year, each saves Rs. 3600, then the income of friend 1 is
   Rs.9000
   Rs.6000
   Rs.6400
   Rs.8000
   Rs.9500
   Option A
   Let the incomes be 5x and 4x respectively. Now
   (5x-3600)/(4x-3600) = 3/2
   x = 1800.
   Income of friend 1= 5x = Rs.9000
   

    

9. If 20% of 45% of 30% of a income of Lalitha is same as 50% of 25% of 40% of Kavitha then what is the ratio of their salaries?
   10: 27
   50: 27
   50: 15
   13: 27
   50: 13
   Option B
   Let the first salary be X and the second be Y
   20/100 * 45/100 * 30/100 * X = 50/100 * 25/100 * 40/100 * Y
   20*45*30*X = 50*25*40*Y
   X/Y = 50*25*40/(20*45*30) = 50/27
   Ratio = 50: 27
   

    

10. Silver and alluminium are in the ratio 5 : 3 in 800gm . How much of alluinium (in grams) should be added to make the ratio 5 : 4 ?
    290
    120
    100
    170
    280
    Option C
    Let 5x : 3x
    Given, 5x + 3x = 800g
    8x = 800 g
    x = 100 g
    = 500 g : 300 g
    Let a gram of Alluminium is added
    500/(300+a)=5/4
    2000 = 1500 + 5a
    500= 5a
    a = 100 g
    

     

Directions (1-10): Find the wrong term in the following series each of which follows some particular pattern:

1. 1121.., 1120.., 1109.., 998.., – 128
   1121
   1120
   1109
   998
   -128
   Option E
   1121-1=1120
   1120-11=1109
   1109-111=998
   998-1111=- 113 (not -128)
   

    

2. 20.., 27.., 32.., 35.., 28
   20
   27
   32
   35
   28
   Option E
   20+7=27
   20+7+5=32
   20+7+5+3=35
   20+7+5+3+1=36 (not 28)
   

    

3. 100 .., 909.., 999.., 1899.., 10999
   100
   909
   999
   1899
   10999
   Option E
   100*9+9=909
   100*9+99=999
   100*9+999=1899
   100*9+9999=10899 (not 10999)
   

    

4. 10.5…, 15.5.., 5.5.., 20.5… , 15.5
   10.5
   15.5
   5.5
   20.5
   15.5
   Option E
   10.5+5=15.5
   15.5-10=5.5
   5.5+15=20.5
   20.5-20=0.5 (not 15.5)
   

    

5. 10.., 7.., 6.., 6.., 6.5.., 7
   10
   7
   6
   6.5
   7
   Option E
   10/2+2= 7
   7/2+2.5= 6
   6/2+3= 6
   6/2+3.5=6.5
   6.5/2+4=7.25 (not 7)
   

    

6. 69.., 60.., 58.1.., 56.11.., 52
   69
   60
   58.1
   56.11
   52
   Option E
   69-9=60
   60-1.9=58.1
   58.1-1.99=56.11
   56.11-1.999=54.111 (not 52)
   

    

7. 8.., 19.., 61.., 244… , 1251
   8
   19
   61
   244
   1251
   Option D
   8*2+3=19
   19*3+4=61
   61*4+5=249(not 244)
   249*5+6=1251
   

    

8. 65.., 56..,49..,42..,41
   65
   56
   49
   42
   41
   Option D
   65 -5-4=56
   56-4-3=49
   49-3-2=44(not 42)
   44-2-1=41
   

    

9. 200.., 45.., 14.., 7.8.., 6.6
   200
   45
   14
   7.8
   6.6
   Option E
   200/5 +5=45
   45/5 +5=14
   14/5+5=7.8
   7.8/5+5=6.56 (not 6.6)
   

    

10. 231.., 222.., 230.., 223.., 234.., 224
    231
    222
    230
    223
    234
    Option E
    231-9=222
    222+8=230
    230-7=223
    223+6=229 (not 234)
    229-5=224
    

     

1. 5 years ago, the age of Mayur and 7 years hence, the age of Sumain is in the ratio of 1: 2. 5 years hence, the age of Mayur and Sumain is in the ratio of 15: 19. Find the present age of Sumain?
   31
   32
   33
   34
   35
   Option C
   5 years ago, the age of Mayur and 7 years hence, the age of Sumain is in the ratio = 1 : 2 (x, 2x)

   Present age of Mayur and Sumain = x + 5, 2x – 7

   5 years hence, the age of Mayur and Sumain is in the ratio = 15 : 19

   According to the question,

   = > (x + 5 + 5) / (2x – 7 + 5) = (15/19)

   = > (x + 10) / (2x – 2) = (15/19)

   = > 19x + 190 = 30x – 30

   = > 11x = 220

   = > x = 20

   The present age of sumain= 2x – 7 = 33 years

    

2. Three persons started a business by investing Rs. 30000, Rs. 40000 and Rs. 50000 respectively. A is a working partner and B and C is a sleeping partner in a business. For managing the business, A will get 10 % of profit and the remaining will be shared by three of them. Find the difference between the share of A and B, if the total profit at the end of the year is Rs. 92000?
   Rs. 2350
   Rs. 2340
   Rs. 2000
   Rs. 5300
   Rs. 2300
   Option E
   A get = 92000 * (10/100) = Rs. 9200

   Remaining = 92000 – 9200 = Rs. 82800

   The share of three,

   = > 30000 : 40000 : 50000

   = > 3 : 4 : 5

   12’s = 82800

   1’s = 6900

   The share of A = 9200 + (3 * 6900) = Rs. 29900

   The share of B = (4 * 6900) = Rs. 27600

   Required difference = 29900 – 27600 = Rs. 2300

    

3. Find the corresponding simple interest, If the compound interest on a certain sum for 3 years at 15 % per annum is Rs. 20835.
   
   Rs. 19000
   Rs. 18000
   Rs. 17000
   Rs. 16000
   Rs. 15000
   Option B
   Let the sum of money be x,

   x * [(1 + (15/100))^3 – 1] = 20835

   x * [(115/100)^3 – 1] = 20835

   x * [(12167/8000) – 1] = 20835

   x * (4167/8000) = 20835

   x = 20835 * (8000 / 4167) = Rs. 40000

   Corresponding SI:

   SI = (40000 * 3 * 15) / 100 = Rs. 18000

    

4. 8 boys and 6 girls can do a work in 12 days. 10 boys and 6 girls can do the same work in 10 days. How long will 12 boys and 8 girls can take to do the work?
   7 2/11 days
   8 2/13 days
   8 2/11 days
   8 2/15days
   8 7/11 days
   Option C
   (8 b + 6 g) * 12 = (10 b + 6 g) * 10

   48 b + 36 g = 50 b + 30 g

   6 g = 2 b

   3 g = 1 b

   8 b + 6 g = 24 b + 6 g = 30 g

   12 b + 8 g = 36 g + 8 g = 44 g

   girls days

   30 12

   44 ?

   (30 * 12) = (44 *x)

   x = (30 * 12) / 44 = 8 2/11 days

    

5. Two dice are thrown simultaneously. Find the Sum of probabilities of getting sum of the dice is divisible by 2 and the difference of the dice is divisible by 3?
   1/3
   2/3
   1/5
   2/5
   3/5
   Option B
   Possibilities of getting sum of the dice is divisible by 2

   = > (1, 1), (1, 3), (1,5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3,5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5,5), (6, 2), (6, 4), (6, 6) = 18

   Possibilities of getting the difference of the dice is divisible by 3

   = > (1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3) = 6

   Required difference = (18/36) + (6/36) = 2/3

    

6. In a bag of 4 red balls and 3 blue balls, 3 balls are to be selected. In how many different ways can they be selected such that at least one red ball should be there?
   32
   34
   36
   38
   40
   Option B
   = > (4C1 and 3C2) or (4C2 and 3C1) or 4C3

   = > (4 * 3) + (6 * 3) + 4

   = >12 + 18 + 4 = 34

    

7. The first class of Dal priced at Rs. 52/kg is mixed with another variety of Dal, which is in the ratio of 3 : 4. The mixed variety of Dal is priced at Rs. 60/kg. Find the price of second variety of Dal?
   60
   62
   64
   66
   68
   Option D
   (x – 60) / 8 = 3/4

   x – 60 = 6

   x = 66

   The price of second variety of Dal = Rs. 66/Kg

    

8. Tap 1 and Tap 2 fill the tank in 30 hours and 20 hours respectively. If Tap 1 and 2 together opened simultaneously and after 10 hours Tap 2 closed, in how many hour will Tap 2 take to fill the remaining tank?
   5
   10
   15
   12
   6
   Option A
   LCM of (30, 20) = 60

   Tap 1 done the work in a hour = 2 units

   Tap2 done the work in a hour = 3 units

   1 + 2 together fill the tank in 10 hours = 5 * 10 = 50 units

   Remaining units of work = 60 – 50 = 10 units

   A lone fill the remaining tank in = 10/2 = 5 hours

    

9. An average wage of all the labours in a company is Rs.10000. The average wage of 10 mechanics is Rs.14000 and the average wage of the rest is Rs.5000. What is the total number of labours in the company?
   12
   16
   18
   20
   22
   Option C
   Let the total number of employees be x.

   10000x = 14000 * 10 + 5000 * (x – 10)

   => 10000x = 140000 + 5000x – 50000

   => x = 90/5 = 18

    

10. What would have been the percentage of profit earned by the seller if no discount was offered, and if sold by offering a discount of 3% and earned a profit of 16.40%.
    15
    20
    25
    30
    35
    Option B
    Let the C.P be Rs.100

    Then, S.P = Rs.116.40

    Let the marked price be Rs. x

    Then, 97x/100 = 116.40

    => x = 120

    Hence, S.P = Rs.120, C.P = Rs.100

    So, Profit % = 20%

     

1. 3239.8765 ÷ 29.56 * 14.67 =? * 16.324
   123
   101
   110
   145
   134
   Option B
   ? * 16 = (3240/30) * 15

   ? * 16 = 3240/2

   ? = 101(approx)

    

2. 60.67% of 985.0098 – 150.9046 =? – 62.67% of 698.008
   881
   882
   883
   884
   885
   Option A
   61% of 985 – 151.046 =? – 63% of 698

   60/100 * 985 – 151 =? – 63/100 * 700

   591 – 151 =? – 441

   ? = 881

    

3. (448.068/32) ÷ (18.923/778.69) * (227.54/76) =?
   56
   57
   58
   59
   60
   Option C
   ? = (448/32) * (779/19) * (228/76)

   ? = 14 + 41 + 3

   ? = 58

    

4. 1235.0067 + 656.9009 + 90.1763 + 452.987 + 6753.099 =?
   9009
   9809
   9188
   9006
   9999
   Option C
   1235 + 657 + 90 + 453 + 6753 =?

   9188 =?

    

5. √7395.99 * √11024.999 + √624.9999 =? – √728.9999
   9082
   9077
   9045
   9023
   9000
   Option A
   √7396 * √11025 + √625 =? – √729

   86 * 105 + 25 =? – 27

   9030 + 25 + 27 =?

   9082 =?

    

6. √16383.56 ÷ 3.99 × 15.89 = ? – 18311.88 ÷ 11.99 + 141.89
   1234
   1879
   1842
   1456
   1786
   Option C
   √16384 ÷ 4 × 16 = x – 18312 ÷ 12 + 142
   (128/4) * 16 = x – (18312 / 12) + 196
   512 = x – 1526 + 196
   x = 512 + 1526 – 196 = 1842
   

    

7. (6.99/8.99) of 4067.99 + 76.99 % of 1199.99 = (?)2 – (127.89 ÷ 15.99)
   63
   64
   65
   66
   67
   Option B
   (7/9) * 4068 + (77/100) * 1200 = x2 – (128/16)
   3164 + 924 + 8 = x2
   x2 = 4096
   x = 64
   

    

8. (8.23 × 8.007)^5 ÷ (511.8 ÷ 7.99)^2 × (7.56 × 64)^3 = (8.008)^(? + 4.5)
   10
   15
   20
   25
   12
   Option A
   (8^2)^5 ÷ (512/^8)^2 × (8 × 8^2)^3 = (8)^x + 5
   8^10 ÷ (64)^2 × (8^3)^3 = (8)^x + 5
   8^10 ÷ 8^4 × 8^9 = (8)^x + 5
   8^10 – 4 + 9 = 8^x + 5
   8^15 = 8^x + 5
   15 = x + 5
   x = 10
   

    

9. ? % of (74.99 × 3.89 – 149.99) = 419.88 + 74.5^2
   1020
   2040
   4030
   6070
   3050
   Option C
   x % of (75 × 4 – 150) = 420 + 75^2
   (x/100) * (300 – 150) = 420 + 5625
   (x/100) * 150 = 6045
   3x/2 = 6045
   x = 6045 * (2/3) = 4030
   

    

10. (6.99/84.23) of (11.99/156) of 4679.99 = 45.0001 – ?
    10
    15
    20
    25
    30
    Option B
    (7/84) * (12/156) * 4680 = 45 – x
    30 = 45 – x
    x = 45 – 30 = 15
    

     

1. I) m² – 3m – 28 = 0
   II) n² – n – 72 = 0
   m
   m<=n
   m>n
   m>=n
   Can’t be determined
   Option E
   I) m² – 3m – 28 = 0
   (m + 4) (m – 7) = 0
   M = -4, 7
   II) n² – n – 72 = 0
   (n + 8) (n – 9) = 0
   N = -8, 9
   Can’t be determined
   

    

2. I) 3m² + 19m + 28 = 0
   II) 2n² + 13n + 21 = 0
   
   m
   m<=n
   m>n
   m>=n
   Can’t be determined
   Option E
   I) 3m² + 19m + 28 = 0
   3m² + 12m + 7m + 28 = 0
   3m (m + 4) + 7 (m + 4) = 0
   (3m + 7) (m + 4) = 0
   m = -7/3, -4 = -2.33, – 4
   II) 2n² + 13n + 21 = 0
   2n² + 6n + 7n + 21 = 0
   2n (n + 3) + 7 (n + 3) = 0
   (2n + 7) (n + 3) = 0
   n = -7/2, -3 = -3.5, -3
   I) 3m² + 19m + 28 = 0
   3m² + 12m + 7m + 28 = 0
   3m (m + 4) + 7 (m + 4) = 0
   (3m + 7) (m + 4) = 0
   m = -7/3, -4 = -2.33, – 4
   II) 2n² + 13n + 21 = 0
   2n² + 6n + 7n + 21 = 0
   2n (n + 3) + 7 (n + 3) = 0
   (2n + 7) (n + 3) = 0
   n = -7/2, -3 = -3.5, -3
   Can’t be determined
   

    

3. I) 2m – 3n = -6
   II) 3m + 4n = 25
   
   m < n
   m<=n
   m>n
   m>=n
   can’t be determined
   Option A
   2m-3n = -6 –> (1)
   3m + 4n = 25 –> (2)
   Bn solving the equation (1) and (2),
   m = 3, n = 4
   m < n
   

    

4. I) 12m²– 37m + 21 = 0
   II) 15n² + 54n + 27 = 0
   
   m < n
   m<=n
   m > n
   m>=n
   can’t be determined
   Option C
   I) 12m² – 37m + 21 = 0
   12m² – 28m – 9m + 21 = 0
   4m (3m – 7) – 3 (3m – 7) = 0
   (4m – 3) (3m – 7) = 0
   m = ¾, 7/3 = 0.75, 2.33
   II) 15n² + 54n + 27 = 0
   15n² + 45n+ 9n + 27 = 0
   15n (n + 3) + 9 (n + 3) = 0
   (15n + 9) (n + 3) = 0
   n = -9/15, -3 = -3/5, -3
   

    

5. I) m² + 3√7 m – 70 = 0
   II) n² + 2√3 n – 105 = 0
   
   m < n
   m<=n
   m>n
   m>=n
   Can’t be determined
   Option E
   I) m² + 3√7 m – 70 = 0
   m² + 5√7 m – 2√7 m – 70 = 0
   (m + 5√7) (m – 2√7) = 0
   m = 2√7, – 5√7
   II) n² + 2√3 n – 105 = 0
   n² + 7√3 n – 5√3 n – 105 = 0
   (n + 7√3) (n – 5√3) = 0
   n = 5√3, – 7√3
   Can’t be determined
   

    

6. I) 14x² – 5√15 x – 90 = 0

   II) 6y² + √21 y – 21 = 0

   x < y
   x<=y
   x>y
   x>=y
   relationship between x and y cannot be determined
   Option E
   I) 14x²-5√15 x-90=0

   14x²-12√15 x+7√15 x – 90 = 0

   2x(7x – 6√15)+ √15(7x – 6√15) = 0

   (2x + √15)(7x – 6√15) = 0

   x = -√15/2, (6√15)/7

   II) 6y²+√21 y-21=0

   6y²+3√21 y-2√21 y -21=0

   3y(2y+√21)- √21(2y+√21)=0

   (3y- √21)(2y+√21)=0

   y =√21/3 ,-√21/2

   Hence, relationship between x and y cannot be determined

    

7. I) 3×2– 13√2x + 24 = 0

   II) y2– 4√2y + 6 = 0

   x < y
   x<=y
   x>y
   x>=y
   relationship between x and y cannot be determined
   Option E
   I) 3×2– 13√2x + 24 = 0

   3x² – 9√2x – 4√2x + 24 = 0

   3x(x – 3√2) – 4√2 (x – 3√2) = 0

   (3x – 4√2)(x – 3√2) = 0

   x = 4√2/3, 3√2

   II)y2– 4√2y + 6 = 0

   y2 – √2y – 3√2y + 6 = 0

   y(y – √2) – 3√2 (y – √2) = 0

   (y – √2) (y – 3√2) = 0

   y = √2, 3√2

   Hence, relationship between x and y cannot be determined

    

8. I) 3×2– (6 + √5)x + 2√5 = 0

   II) 8y2– (16 + 3√5)y + 6√5 = 0

   x < y
   x<=y
   x>y
   x>=y
   relationship between x and y cannot be determined
   Option E
   I) 3×2– (6 + √5)x + 2√5 = 0

   3x² – 6x – √5x + 2√5 = 0

   3x (x – 2) – √5 (x – 2) = 0

   (3x – √5) (x – 2) = 0

   x = √5/3,2

   II) 8y2– (16 + 3√5)y + 6√5 = 0

   8y2 – 16y – 3√5y + 6√5 = 0

   8y (y – 2) – 3√5 (y – 2) = 0

   (8y – 3√5) (y – 2) = 0

   y = (3√5)/8, 2

   Hence, relationship between x and y cannot be determined

    

9. I) 18x² – 63x + 40 = 0

   II) 12y² + 47y + 45 = 0

   x < y
   x<=y
   x>y
   x>=y
   relationship between x and y cannot be determined
   Option C
   I) 18x² – 63x + 40 = 0

   18x²-15x-48x+40=0

   3x(6x-5)-8(6x-5)=0

   (3x-8)(6x-5)=0

   x=8/3,5/6

   II) 12y²+47y+45=0

   12y²+27y+20y+45=0

   3y(4y+9)+5(4y+9)=0

   (3y+5)(4y+9)=0

   Y =-5/3,-9/4

   Hence, x > y

    

10. I) 20x²-119x+176=0

    II) 45y²+200y+155=0

    x < y
    x<=y
    x>y
    x>=y
    relationship between x and y cannot be determined
    Option C
    I) 20x²-119x+176=0

    20x²-64x-55x+176=0

    4x(5x-16)-11(5x-16)=0

    (4x-11)(5x-16)=0

    x=11/4,16/5

    II)

    45y²+200y+155=0

    45y²+45y+155y+155=0

    45y(y+1)+155(y+1)=0

    (45y+155)(y+1)=0

    y=-155/45,-1

    Hence, x > y

     

1. In a fraction the numerator is increased by 80% and the denominator of the fraction is increased by 120%, then it becomes 5/11. Find the original fraction?
   1/8
   2/5
   5/9
   2/11
   4/7
   Option C
   The original fraction be (x/y)
   [X*(180/100)]/[Y*(220/100)] = 5/11
   9x/11y = 5/11
   x/y = 5/9
   

    

2. Akil, Brahma and kannan together have Rs. 33250. If 3/7th of Akil’s amount is equal to 2/3rd of Brahma’s amount and 1/4th of Brahma’s amount is equal to 5/8th of kannan’s amount, then how much amount does Akil have?
   Rs. 17000
   Rs. 47500
   Rs. 17500
   Rs. 27500
   Rs. 15600
   Option C
   A kil+ Brahma + kannan = Rs. 33250
   (3/7)* A = (2/3)* B
   (A/B) = (2/3) * (7/3) = 14/9
   (1/4)* B = (5/8)* k
   (B/k) = (5/8) * (4/1) = 5/2
   The ratio of A, B and k = 70: 45: 18
   133’s = 33250
   1’s = 250
   70’s = Rs. 17500
   

    

3. Person 1, 2 and 3 started a business by investing an amount of Rs. 45000, Rs. 65000 and Rs. 70000 respectively. After 3 months, 1 invested Rs. 15000 more and 2 withdraw Rs. 5000. And after 4 months, 3 withdraw Rs. 15000. If the total profit at the end of the year is Rs. 114405, then find the share of 1?
   Rs. 35500
   Rs. 37505
   Rs. 25505
   Rs. 30005
   Rs. 35505
   Option
   The share of 1, 2 and 3,
   = > [45000*3 + 60000*9]: [65000*3 + 60000*9]: [70000*7 + 55000*5]
   = > 675000: 735000: 765000
   = > 45: 49: 51
   Total profit = Rs. 114405
   145’s = 114405
   1’s = 789
   The share of 1 = 789*45 = Rs. 35505
   

    

4. A 90 litres of mixture of acid and water contains acid and water in the ratio of 3: 2. How much water should be added to the mixture to get 45 % acid on it?
   30
   45
   67
   23
   12
   Option A
   Total mixture of acid and water = 90 litres
   5’s = 90
   1’s = 18
   Total acid = 54 litres, water = 36 litres
   According to the question,
   54/(36 + x) = (45/55)
   54/(36 + x) = 9/11
   66 = 36 + x
   X = 30
   30 litres of water should be added to the mixture.
   

    

5. A yacht can sail 55 km downstream in 66 min. The ratio of the speed of the yacht in still water to the speed of the stream is 4: 1. How much time will the yacht take to cover 72 km upstream?
   1 hr 24 m
   2hr 24 m
   5 hr 20 m
   6 hr 32 m
   1 hr 15m
   Option B
   Speed of downstream = D/T = 55/(66/60) = 55*(60/66) = 50 km/hr
   The ratio of the speed of the yacht in still water to the speed of the stream
   = > 4: 1 (4x, x)
   5x = 50
   X = 10
   Speed of upstream = 4x – x = 3x = 30 km/hr
   Distance = 752 km
   Time = D/S = 72/30 = 2 2/5 hr = 2 hours 24 mins
   

    

6. 79.89% of 1400 + 17.05% of 399.99 – 20% of 520 =? * 2
   430
   220
   541
   123
   450
   Option C
   80% of 1400 + 16.5% of 400 – 20% of 520 =? * 2
   80/100 * 1400 + 16.5/100 * 400 – 20/100 * 520 =? * 2
   1120 + 66 -104 =? * 2
   ? = 541
   

    

7. 98.902 * √1681 + 22.87 * √5184 – 48 * 69 =? * 89
   27
   23
   22
   24
   25
   Option A
   99 * √1681 + 23 * √5184 – 48 * 69 =? * 89
   4059 + 1656 – 3312 =? * 89
   ? = 27
   

    

8. 34.56 % of 119.9 + 12.5 % of 231.5 + 9 * 29 =? + 132
   200
   150
   250
   300
   350
   Option A
   35 % of 120 + 12.5 % of 232 + 9 * 29 =? + 132
   42 + 29 + 261 =? + 132
   ? = 200
   

    

9. (√1023.5 + √3720.5) * √15.89 +? = 482
   120
   130
   110
   100
   90
   Option C
   (√1024 + √3721) * √16 +? = 482
   (32+ 61) *4 +? =482
   372 + ? = 482
   ? = 482 – 372
   ? = 110
   

    

10. 19.99 % of 64.67 + 111.5 * √81 -13 * 8 =? + 27 * 15
    340
    230
    123
    540
    512
    Option E
    20 % of 65 + 112 * √81 -13 * 8 =? + 27 * 15
    13 + 1008 – 104 =? + 405
    ? = 512
    

     

The following bar graph shows the prices (In thousands) of different types of Air cooler in different months.

1. Find the difference between the price of Haier to that of Axue in all the given months together?
   1200
   1300
   1400
   1500
   1600
   Option D
   The price of Haier in all the given months together
   = > 9000 + 10000 + 7500 + 7000 = 33500
   The price of Axue in all the given months together
   = > 8000 + 9000 + 9500 + 7500 = 32000
   Required difference = 33500 – 32000 = 1500
   

    

2. Find the ratio between the total price of all the given Air cooler in the Month Jan to that in the Month Apr?
   98 : 91
   98 : 93
   98 : 95
   98 : 97
   98 : 99
   Option A
   Total price of all the given Air cooler in the Month Jan
   = > 9000 + 6000 + 7500 + 8000 + 8500 + 10000 = 49000
   Total price of all the given Air cooler in the Month Apr
   = > 7000 + 5500 + 9000 + 7500 + 8000 + 8500 = 45500
   Required ratio = 49000 : 45500 = 98 : 91
   

    

3. Find the total price of all the given Air cooler in the Month Mar?
   50000
   20000
   10000
   25000
   15000
   Option A
   The total price of all the given Air cooler in the Month Mar
   = > 7500 + 6500 + 8500 + 9500 + 9000 + 9000
   = > 50000
   

    

4. The price of LG Air cooler in all the given months together is approximately what percentage less than the price of Samsung Air cooler in all the given months together?
   13
   17
   23
   27
   15
   Option D
   The price of LG Air cooler in all the given months together
   = > 6000 + 7500 + 6500 + 5500 = 25500
   The price of Samsung Air cooler in all the given months together
   = > 8500 + 9500 + 9000 + 8000 = 35000
   Required % = [(35000 – 25500)/35000]*100 = 27 %
   

    

5. The price of all the given Air cooler in the Month Feb is approximately what percentage of price of Bajaj in all the given months together?
   125
   156
   167
   123
   145
   Option C
   The price of all the given Air cooler in the Month Feb
   = > 10000 + 7500 + 8000 + 9000 + 9500 + 11000 = 55000
   The price of Bajaj in all the given months together
   = > 7500 + 8000 + 8500 + 9000 = 33000
   Required % = (55000/33000)*100 = 166.66 % = 167 %

    

Following table shows the total number of sarees (In lakhs) produced by 6 different cities and also shows the ratio of non defected and defected sarees among them.




6. Find the total number of non defect percentage in Chennai?
   10
   20
   30
   40
   50
   Option D
   Total sarees produced = 145000
   Total number of non defecting sarees = 145000*(2/5) = 58000
   Required % = (58000/145000)*100 = 40 %
   (Or)
   The ratio of non defected to that of defected sarees = 2 : 3 (2x, 3x)
   Required % = (2x/5x)*100 = 40 %
   

    

7. Find the total number of defected sarees in Bangalore, Delhi and Jaipur together?
   300000
   310000
   340000
   240000
   140000
   Option C
   The total number of defected sarees in Bangalore, Delhi and Jaipur together
   = > 180000*(5/9) + 165000*(7/11) + 225000*(3/5)
   = > 100000 + 105000 + 135000 = 340000
   

    

8. Find the difference between the total number of non defected sarees from Mumbai and Thiruvananthapuram together to that of total number of sarees produced from Chennai?
   12000
   18000
   32000
   35000
   14000
   Option D
   The total number of non defected sarees from Mumbai and Thiruvananthapuram together
   = > 240000 *(5/12) + 260000*(4/13)
   = > 100000 + 80000 = 180000
   The total number of sarees produced from Chennai = 145000
   Required difference = 180000 – 145000 = 35000
   

    

9. If the ratio of red to that of yellow sarees produced for the export from Chennai is 3 : 2, then find the ratio between the total number of red sarees produced for the export from Chennai to that of total number of sarees produced for the export from Bangalore?
   29 : 60
   2 : 60
   29 : 61
   19 : 60
   29 : 17
   Option A
   The ratio of male to that of yellow sarees produced from Chennai = 3 : 2 (3x, 2x)
   The total number of red sarees produced from Chennai
   = > 145000*(3/5) = 87000
   The total number of sarees produced from Bangalore
   = > 180000
   Required ratio = 87000 : 180000 = 29 : 60
   

    

10. Total number of sarees produced for the export from Delhi and Mumbai together is approximately what percentage of total number of sarees produced for export from Thiruvananthapuram and Jaipur together?
    45
    55
    65
    75
    85
    Option E
    Total number of sarees produced from Delhi and Mumbai together
    = > 1.65 + 2.4 = 4.05 lakhs = 405000
    Total number of sarees produced from Thiruvananthapuram and Jaipur together
    = > 2.6 + 2.25 = 4.85 lakhs = 485000
    Required % = (405000/485000)*100 = 83.5 % = 85 %