1. Find the rate of interest per annum, if the sum 4 times itself at simple interest in 12 years.
   22
   23
   25
   26
   27
   Option C
   Let the principle be x,
   SI = 4x – x = 3x
   Given,
   3x = (x * 12 * r)/100
   r = (100 / 12) * 3
   r = 25 %
   The rate of interest (r) = 25 %
   

    

2. Friend 1 and 2 are partners in a project. friend 1 contributes 5/8th of the capital for 8 months and friend 2 received 3/7th of the profit. How long friend 2’s money was used?
   15
   10
   5
   20
   25
   Option B
   The ratio of investment of F1 and 2 = (5/8) : (3/8) = 5 : 3
   The ratio of profit of F1 and 2 = (4/7) : (3/7) = 4 : 3
   Given,
   (5 * 8) / (3 * x) = (4/3)
   x = 10
   F2’s money used for 10 months
   

    

3. The ratio of age of friend 1, after 3 years and the age of friend 2, 5 years ago is 3 : 5. The sum of the present age of friend1 and friend 2 together is 58 years. Find the age of friend 1, 7 years hence?
   25
   26
   27
   28
   29
   Option A
   The ratio of age of f1, after 3 years and the age of F2, 5 years ago = 3 : 5 (3x, 5x)
   The present age of F1 and F2 = 3x – 3 and 5x + 5
   Given,
   3x – 3 + 5x + 5 = 58
   8x + 2 = 58
   8x = 56
   x = 7
   The age of F1, 7 years hence = 3x – 3 + 7 = 3x + 4 = 25 years
   

    

4. A person invested Rs. 40000 in SI at the rate of 2x % per annum for two years and the same amount is invested in CI at the same rate of interest. The interest received by him in CI is Rs. 576 more interest than the interest received by him in S.I.Find the rate of interest.
   10
   11
   12
   13
   14
   Option C
   CI – SI = 576
   40000 * [(1 + 2x/100)2 – 1] – (40000 * 2x * 2)/100 = 576
   40000 * ([(100 + 2x) /100]2 – 1) – 1600x = 576
   40000 * [(10000 + 400x + 4×2 – 10000)/10000] – 1600x = 576
   1600x + 16×2 – 1600x = 576
   16×2 = 576
   x2 = 36
   x = 6
   The rate of interest = 2x % = 12 %
   

    

5. A yacht takes two hours more to travel upstream than travel the same distance in downstream. If the distance travelled by the yacht is 120 km and the ratio of the speed of boat in still water to speed of stream is 5: 1, then what is the speed of the boat in still water?
   15
   20
   25
   30
   35
   Option C
   Distance = speed * time
   120/4x – 120/6x = 2
   (30 – 20)/x = 2
   10 = 2x
   = > x = 5
   Speed of the yacht in still water = 5 * 5 = 25 kmph
   

    

6. A plane covered a distance of 840 km in 14 hours. Ratio of the speed of train to speed of the plane is 4: 3 and the ratio of the speed of bus to train is 1: 2. In how much distance will the bus cover in 8 hours?
   300
   310
   320
   330
   340
   Option C
   Distance = Speed * Time
   Speed of plane = 840/14 = 60 kmph
   Speed of the train = 4/3 * 60 = 80 kmph
   Speed of the bus = ½ * 80 = 40 kmph
   Distance covered by in 8 hours = 40 * 8 = 320 km
   

    

7. A container contains 90 kg of the mixture of rice and wheat in the ratio of 5: 4 and 45 kg of the mixture is taken out. Then added x kg of the wheat in the remaining mixture, the ratio of the rice and wheat becomes 2: 3. Find the value of x
   17.5 kg
   18.5 kg
   19.5 kg
   15.5 kg
   16.5 kg
   Option
   rice = 5/9 * 90 = 50 kg
   Wheat = 4/9 * 36 = 40 kg
   rice in 45 kg = 5/9 * 45 = 25 kg
   Water in 45 kg = 4/9 * 45 = 20 kg
   (50 – 25)/(40 – 20 + x) = 2/3
   40 + 2x = 75
   2x = 35
   x = 17.5 kg
   

    

8. cistern A can fill a tank in 40 hours and ratio of the efficiency of cistern A to B is 3: 2. cistern B and C together can fill the tank in 30 hours. In how many hour cisterns A, B and pipe C together can fill the tank?
   17 (1/7) hours
   19 (1/7) hours
   16 (1/7) hours
   17 (1/5) hours
   13 (1/7) hours
   Option A
   The work done by A in one hour,
   A = 1/40
   The tank will be filed by B in,
   B = 3/2 * 40 = 60 hours
   Cistern B and C together can fill the tank in 30 hours,
   B + C = 1/30
   C = 1/30 – 1/60
   C = 1/60
   A + B + C = 1/40 + 1/60 + 1/60
   = (3 + 2 + 2)/120
   = 7/120
   Required time = 120/7 hours = The work done by A in one hour,
   A = 1/40
   The tank will be filed by B in,
   B = 3/2 * 40 = 60 hours
   Cistern B and pipe C together can fill the tank in 30 hours,
   B + C = 1/30
   C = 1/30 – 1/60
   C = 1/60
   A + B + C = 1/40 + 1/60 + 1/60
   = (3 + 2 + 2)/120
   = 7/120
   Required time = 120/7 hours = 17 (1/7) hours

    

9. The ratio of the ages of Akil and Bheema is 4: 5 and after 4 years the ratio of the ages of Bheema and Cimoan is 3: 4. If sum of the ages of Akil, Bheema, and Cimoan is 64 years, what is the present age of Bheema?
   10
   20
   30
   40
   50
   Option B
   A + B + C = 64
   A/B = 4/5
   A = (4/5) * B
   (B + 4)/(C + 4) = ¾
   3C + 12 = 4B + 16
   3C – 4B = 4
   3C = 4B + 4
   A + B + C = 64
   (4/5) * B + B + (4B + 4)/3 = 64
   12B + 15B + 20B + 20 = 64 * 15
   47B = 940
   B = 20 years
   

    

Directions (1-5): Study the following information carefully and answer the questions that follow:

Monthly salary of three workers P, Q and R in the ratio of 4: 6: 5. S’s monthly salary is 33.33% more than the R’s monthly salary.T’s spends is half of the S’s monthly salary which is five times the monthly savings of T. P and Q saves same amount in monthly which is 50% of the amount saved by S in monthly. S saves 3/2nd of the amount saves by T in monthly. T’s monthly salary is equal to the S’s monthly spends.R’s monthly spends is two times of the amount saves by him. Q invests his monthly savings in two different plans A and B in the ratio of 2: 1. Plan A offers 10% simple interest for two years and plan B offers 20% compound interest for two years compounded annually. Total interest obtained by Q in both the plans together is Rs. 1680.

1. What is the ratio of R’s monthly spends to that of Q’s monthly spends?
   2: 1
   2: 5
   2: 9
   2: 3
   2: 7
   Option D
   Let us take Q saves Rs. 3x amount
   by question,
   (2x * 10 * 2)/100 + (x * 44)/100 = 1680
   40x/100 + 44x/100 = 1680
   84x/100 = 1680
   = > x = 1680 * (84/100) = Rs.2000
   Q saves monthly Rs.6000
   P’s monthly savings = Q’s monthly savings = 50% of S’s monthly savings
   P‘s monthly savings = Rs.6000
   Q’s monthly savings = Rs.6000
   S’s monthly savings = 6000 * (100/50) = Rs.12000
   T’s monthly savings = 2/3rd of S’s monthly savings = 12000 * 2/3 = Rs.8000
   S’s monthly salary = 5 * T’s monthly savings = 5 * 8000 = Rs.40000
   T’s monthly spends = ½ * S’s monthly salary = 40000 * ½ = Rs.20000
   S’s monthly spends = 40000 – 12000 = Rs.28000
   T’s monthly salary = S’s monthly spends = Rs.28000
   S’s monthly salary = (100 + 33.33)% of R’s monthly salary
   40000 = (100 + 100/3)% of R’s monthly salary
   40000 = 400/300 * R’s monthly salary
   40000 * (3/4) = R’s monthly salary
   R’s monthly salary = Rs.30000
   Ratio of R’s spends to savings = 2: 1
   R’s monthly spends = 30000 * (2/3) = Rs.20000
   R’s monthly savings = 30000 * (1/3) = Rs.10000
   P’s monthly salary = 30000 * (4/5) = Rs.24000
   Q’s monthly salary = 30000 * (6/5) = Rs.36000
   Q’s monthly spends = 36000 – 6000 = Rs.30000
   Q’s monthly salary = 200% of P’s monthly spends
   36000 = 2 * P’s monthly spends
   P’s monthly spends = 36000/2 = Rs.18000
   Required ratio = 20000: 30000 = 2: 3
   

    

2. P and R invested their monthly savings in a business. After 6 months, S also joined with his monthly savings in the business. Find the share of P if the total profit at the end of the year is Rs.18700?
   Rs.5390
   Rs.6700
   Rs.5800
   Rs.3400
   Rs.5100
   Option E
   Profit ratio of P, Q and R = (6000 * 12): (10000 * 12): (12000 * 6)
   = 72: 120: 72
   = 3: 5: 3
   P’s share = 18700 * (3/11) = Rs.5100
   

    

3. P and Q’s monthly salary is what percentage of the T’s monthly spends?
   100%
   300%
   200%
   350%
   50%
   Option B
   Required percentage = (24000 + 36000)/20000 * 100
   = 60000/20000 * 100
   = 300%
   

    

4. What is the difference between the R, S and T’s monthly salary together and P, Q and S’s monthly spends?
   Rs.32000
   Rs.20000
   Rs.42000
   Rs.12000
   Rs.22000
   Option E
   Required difference = (30000 + 40000 + 28000) – (18000 + 30000 + 28000)
   = Rs.22000
   

    

5. Who among the following person has maximum savings percentage?
   p
   Q
   R
   S
   T
   Option C
   P’s savings percentage = 6000/24000 * 100 = 25%
   Q’s savings percentage = 6000/36000 * 100 = 16.67%
   ’s savings percentage = 10000/30000 * 100 = 33.33%
   S’s savings percentage = 12000/40000 * 100 = 30%
   T’s savings percentage = 8000/28000 * 100 = 28.57%
   R has maximum savings percentage.
   

    

Directions (6-10): Study the following bar graph and answer the questions that follows:
The following line graph shows the total weight loss by two persons on 5 different days of a week by exercise in a certain month.


6. Find the difference between the average number of weights loss by P to that of Q on all the given days together.
   15
   16
   17
   18
   19
   Option D
   The average number of weights loss by P on all the given days together
   = > (120 + 80 + 180 + 150 + 220) / 5 = 150
   The average number of weights loss by Q on all the given days together
   = > (170 + 120 + 90 + 200 + 260) / 5 = 168
   Required difference = 168 – 150 = 18
   

    

7. Find the difference between total weights loss by P on Monday and Friday together to that of total weights loss by Q on Tuesday and Thursday together.
   20
   30
   24
   17
   19
   Option A
   The total weights loss by P on Monday and Friday together
   = >120 + 220 = 340
   The total weights loss by Q on Tuesday and Thursday together
   = >120 + 200 = 320
   Required difference = 340 – 320 = 20
   

    

8. If the average number of weights loss by P and Q from Monday to Saturday is 160 and 168 respectively, then find the total number of weights loss by P and Q on Saturday.
   234
   378
   267
   438
   213
   Option B
   The number of weights loss by P on Saturday
   = > (160 * 6) – (120 + 80 + 180 + 150 + 220)
   = >960 – 750 = 210
   The number of weights loss by Q on Saturday
   = > (168 * 6) – (170 + 120 + 90 + 200 + 260)
   = > 1008 – 840 = 168
   The total number of weights loss by P and Q on Saturday
   = >210 + 168 = 378
   

    

9. Total weights loss by both of them together on Tuesday is approximately what percentage of total weights loss by both of them together on Thursday?
   37
   47
   57
   67
   68
   Option C
   Total weights loss by both of them together on Tuesday
   = > 80 + 120 = 200
   Total weights loss by both of them together on Thursday
   = > 150 + 200 = 350
   Required % = (200 / 350) * 100 = 57.14 % = 57 %
   

    

10. The average marks of English of a class of 40 students is 36. If the marks of three students were told wrong as 48, 52 and 56 of the actual marks 58, 66 and 72 respectively, then what would be the correct average?
    33
    34
    35
    36
    37
    Option E
    The average marks of English of a class of 40 students = 36
    Total marks = 40 * 36 = 1440
    Correct average = [1440 – (48 + 52 + 56) + (58 + 66 + 72)] / 40
    = > [1440 – 156 + 196] / 40
    = > 1480 / 40 = 37
    

     

Directions(1-5):
There are total x tourists in a area. Each tourists visit one or more places among three places viz. Mahabalipuram, Goa and Agra. 20% of the total tourists visit only Mahabalipuram, which is equal to the tourists visit only Goa. 60% of the total tourists who visit only Mahabalipuram is the tourists who visit Mahabalipuram and Agra but not Goa. The tourists who visit Goa and Agra but not Mahabalipuram is 25% of the total tourists who visit Only Goa and Only Mahabalipuram together. 5% of the total tourists visit all the three places and the ratio of the number of tourists visit only Agra to that of the tourists who visit Goa and Mahabalipuram but not Agra is 5: 6. The difference between the tourists visit only Agra and the tourists visit Mahabalipuram and Goa but not Agra is 36.

1. Total tourists who visit only Agra is what percentage of the total tourists who visit only Mahabalipuram?
   
   70
   75
   80
   85
   90
   Option B
   Total tourists in a area is x
   Tourists visit only Mahabalipuram = 20%
   Tourists visit only Goa = 20%
   Tourists visit Mahabalipuram and Agra but not Goa = 60% of 20
   = 3/5 * 20 = 12%
   Total tourists visit only Goa and only Mahabalipuram = 20% + 20% = 40%
   Tourists visit Goa and Agra but not Mahabalipuram = 25% of 40
   = ¼ * 40 = 10%
   Tourists visit all the three places = 5%
   Remaining tourists = 100% – (20% + 20% + 12% + 10% + 5%)
   = 100% – 67% = 33%
   Tourists visit only Agra = 33% * 5/11 = 15%
   Tourists visit Mahabalipuram and Goa but not Agra = 33% – 15% = 18%
   Difference between the tourists visit only Agra and the tourists visit Mahabalipuram and Goa but not Agra = 18% – 15% = 3%
   3% of total tourists = 36
   Total tourists = 36/3 * 100 = 1200

   Required percentage = 180/240 * 100
   = ¾ * 100 = 75%
   

    

2. Find the total tourists who visit at most one place
   550
   770
   660
   440
   330
   Option C
   Required total = (20% + 20% + 15%) * 1200
   = 55/100 * 1200
   = 660
   

    

3. What is the ratio of the total tourists who visit Mahabalipuram and Agra but not Goa to the total tourists who visit Mahabalipuram and Goa but not Agra?
   2: 3
   2: 7
   2: 5
   2: 9
   5: 3
   Option A
   Required ratio = 144: 216
   = 2: 3
   

    

4. Find the total tourists who visit Agra
   500
   501
   502
   503
   504
   Option E
   Required total = (15% + 5% + 12% + 10%) * 1200
   = 42/100 * 1200
   = 504
   

    

5. Find the total tourists who visit at least two places
   520
   530
   540
   550
   560
   Option C
   Required total = (18% + 12% + 10% + 5%) of 1200
   = 45/100 * 1200
   = 540
   

    

Directions (6-10):
Four friends (F1, F2, F3 and F4) are started a shop together by sharing the amount of Rs. 20000, Rs. 30000, Rs. 50000 and Rs. 60000 respectively. After 4 months, F1 and F3 increased the initial sharing 20 % and 40 % respectively. At the end of the year, they earned a total profit of Rs. 171600.
The friend A and B borrows F1 and F3’s share. They are giving 8 % simple interest for 4 years.
The friend C borrows F2’s total amount for 2 years at the rate of 6 % compound interest.

6. Find the sum of the total amount, friend A and B have to pay for the friend F1 and F3?
   
   Rs. 110600
   Rs. 11082
   Rs. 1106
   Rs. 10682
   Rs. 110682
   Option
   [20000 * 4 + 20000 * (120/100) * 8] : [30000 * 12] : [50000 * 4 + 50000 * (140/100) * 8] : [60000 * 12]
   = > 272000 : 360000 : 760000 : 720000
   = > 34 : 45 : 95 : 90
   Total profit = 264’s = Rs. 171600
   1’s = 650
   share of F1 = Rs. 22100
   share of F2= Rs. 29250
   share of F3= Rs. 61750
   share of F4= Rs. 58500
   The friend A borrows Rs. 22100 from the friend F1. So, the principle amount will be Rs. 22100.
   SI for friend A = PNR/100
   = > (22100 * 4 * 8) / 100 = Rs. 7072
   Total amount, the friend F1 has received = 22100 + 7072 = Rs. 29172
   The friend B borrows Rs. 61750 from the friend F3. So, the principle amount will be Rs. 61750.
   SI for friend B = (61750 * 4 * 8)/100 = Rs. 19760
   Total amount, the friend F3 has received = 61750 + 19760 = Rs. 81510
   Required sum = 29172 + 81510 = Rs. 110682

    

7. Find the average share of F1 and F4 together at the end of the year of shop?
   Rs. 60300
   Rs. 40700
   Rs. 40560
   Rs. 40590
   Rs. 40300
   Option E
   The average share of F1 and F4 together at the end of the year of shop
   = > (22100 + 58500) / 2
   = > 80600 / 2 = Rs. 40300
   

    

8. Find the interest amount, the friend A have to paid for the friend F1?
   Rs. 7072
   Rs. 6072
   Rs. 7067
   Rs. 7000
   Rs. 7052
   Option A
   The friend A borrows Rs. 22100 from the friend F1. So, the principle amount will be Rs. 22100.
   SI = PNR/100
   SI = (22100 * 4 * 8)/100 = Rs. 7072
   

    

9. The profit of friend F2 is approximately what percentage of the total amount (Initial sharing + Profit), the friend F4 received at the end of the year of shop?
   25 %
   20%
   12 %
   10%
   23%
   Option A
   The profit of friend F2 = Rs. 29250
   The total amount (Initial sharing + Profit), the friend F4 received at the end of the year of shop
   = > 60000 + 58500 = Rs. 118500
   Required % = (29250 / 118500) * 100 = 24.68 % = 25 %
   

    

10. Find the compound interest, friend C has to pay for the friend F2?
    Rs. 723.30
    Rs. 7323.30
    Rs. 7300.30
    Rs. 7823.30
    Rs. 8323.30
    Option B
    The friend C borrows Rs. 59250 from friend F2. So, the principle amount will be Rs. 59250.
    Total amount = 59250 * (106/100) * (106/100) = Rs. 66573.30
    CI = 66573.30 – 59250 = Rs. 7323.30
    

     

1. Shyam,Seenu,Rio,Tippu,and Neetu are five friends .The sums of the weights of each group of four are 130,140,113,131 and 126 .What is the positive difference of the weights of the heaviest and weakest among them?
   38
   27
   43
   12
   45
   Option B
   Let the sum of weight of five friends be W.Let the weight of friends be x,y, z,t,s.
   S= x+y+z+t+s
   So S-x=130
   S-y=140
   S-z=113
   S-t=131
   S-s=126
   Add,5S-S=640
   ie) 4S=640 implies S=160.
   For heaviest, take S-y=140160-y=140y=20
   Weakest ,Take S-z=113160-z=113z=47
   Therefore, difference =47-20=27.
   

    

2. 15 pairs of competitors were participated in a competition . Each one of the pair gave a pack of gifts to everyone except their pair and no exchange of gifts took place between the pairs. How many gifts were exchanged among them?
   230
   450
   420
   340
   320
   Option C
   Total no of competitors = 15 pairs= 30 members .
   In 15 pairs ,One of the pair exchange gifts to others except their own pair.
   1 member exchange 28 gifts. ie) to 14 pair. (except themselves and their pair)
   15 members exchange 15×28=420 gifts.
   

    

3. Anu deposits certain amount on a bank and got simple interest Rs.280 for two years . And the compound interest of same amount be Rs. 291.2.If the rate of interest is same in two cases , then what is the principal amount Anu deposits?
   Rs. 1750.
   Rs. 2750.
   Rs. 1050.
   Rs. 1700
   Rs. 3750.
   Option A
   Let Simple interest be x for first year and it will be remained in second year also.
   Since rate of interest same, Compound interest of first year also x.
   But for Second year compound interest =x + a ( a is the inerest of x).
   Simple interest for 2 years = 2x = 280 and compound interest = 2x+a = 291.2
   280+ a= 291.2
   a = 11.2
   Also x=140

   Since‘a’ is the interest of x,
   Rate of interest r = a/x x 100
    r = 11.2/140 x 100 r= 8
   Rate of interest of 2x is 16 %.
   16% of principal P
   = Rs.280
    P= 280/ 16 x 100 = Rs. 1750.
   

    

4. A train travels at 20 kmph, and reaches its destination at 3.pm.If it travels at 30kmph it reaches at 11.am. At what speed it should be travel to reach at 1.pm.?
   22
   23
   24
   25
   26
   Option C
   Let the time of a train to reach 1pm be t. ie) t= 1.pm
   t+2 = 11.am
   t+4 = 3.pm
   t, t+2,t+4 are in arithmetic progression
   if time is in A.P then corresponding speed will be in HP.
   20, x, 30 in HP. ( x = speed of a train to reach at 1.pm)
   Then Harmonic mean of 20 and 30 x= 2 x 20 x 30/ 20 +30
   x= 24
   The speed of the train is 24kmph to reach 1.pm.
   

    

5. In school hostel rice is stocked for 10 people for 40 days. After 15 days 5 students went for vacation .How many number of days for which the remaining food last?
   33
   16
   34
   89
   80
   Option E
   Let 15 days after 5 people had food be x days.
   Less people  more food (indirect proportion)
   ie) 5:10:: 40:x
   5 X x = 10 x 40
   x = 80.
   Therefore rice remains 80 days for 5 people.
   

    

6. How many multiples of 2 are there in 10 to 80?
   11
   36
   13
   14
   15
   Option B
   Multiples of 2 =>last digits must be 0, 2, 4, 6, 8
   And first digit 1,2 ,3,4,5,6,7,
   So there are 7 choices for the first number and 5 choices for the second number.
   So there are 5x 7 =3 5 + 1= 36 (include 80) multiples of 2 in between 10 and 80.
   

    

7. In a shop two partners invested in the ratio 10:20 and they got the profits in the ratio 5:4.If the first partner invest the money for 10 months the how many months the second partner invested?
   17
   15
   18
   16
   10
   Option D
   Investment ratio: 10:20
   Profit ratio :5:4
   First partner invested 10 months.
   Second parner invested x (take) months.
   So 10 x 20 / 10 x x = 5/4
   20/x = 5/4
   x = 16 months

    

8. Two glasses of equal capacity contains sand and color powder.The ratio of the mixing is 2:3 and 5:4 respectively. These mixture are transferred into a big bowl.Then what is the ratio of sand and color powder in the bowl?
   9: 4
   9: 3
   5: 2
   9: 2
   7: 7
   Option E
   Ratio= 2 + 5 : 3+4 = 7: 7

    

9. Two triangles having the ratio s of area 5:3. Height ratio 3:2.Find the ratio of the base.
   5:8
   5:4
   11:3
   5:7
   5:3
   Option E
   Let the base be x and y of two triangles.
   Heights are 3h and 2h.
   (½ X x X 3h) /( ½ X y X 2h) = 5/3
   x/ y = 5/ 3 x 2/ 3 = 10/6 =>x:y= 10: 6=5:3
   

    

10. If (64 ^2) – (36^ 2) = 20 x X .Find the value of X?
    100.
    140.
    105.
    108.
    101
    Option B
    20 x X = (64 2) – (36 2)
    = ( 64 + 36) (64 -36) = 100 x 28 =2800
    X= 2800/20 =140
    

     

1. (3564.09 ÷ 12.32) ÷ 2.99 × 35.89 + 32.008 % of 498.90 – 36.87 % of 1300.113 = ?
   3250
   3460
   3490
   3560
   3300
   Option A
   x = (3564/12) * (1/3) * 36 + (32/100) * 500 – (37/100) * 1300
   x = 3564 + 160 – 481
   x = 3243 = 3250

    

2. 54.99 % of 1218.9 + (5/12.78) of 233.008 – 17.77 % of 1398.90 = ?
   540
   560
   530
   550
   510
   Option E
   (55/100) * 1220 + (5/13) * 234 – (18/100) * 1400 = x
   x = 671 + 90 – 252
   x = 509 = 510
   

    

3. (11.99/36.99) ÷ (51.99/18.99) × 102.99 – 738.57 + 48.99 % of 1598.99 = ?
   65
   23
   45
   60
   45
   Option A
   (12/38) * (19/52) × 104 – 739 + (49/100) * 1600 = x
   x = 20 – 739 + 784
   x = 65
   

    

4. 13.65 × 6.105 + (78.401)2 ÷ 12.691 = ? – 47.66 % of 1498.67
   2350
   1260
   3400
   3800
   1400
   Option B
   (14 * 6) + (78 * 78)/13 = x – (47/100) * 1500
   84 + 468 = x – 705
   x = 84 + 468 + 705
   x = 1257 = 1260
   

    

5. 800 ÷ 40 = ? + 366 – (157 ÷ 3 × 14)
   390
   410
   540
   320
   450
   Option A
   (800 / 40) = x + 366 – (157 * 14) / 3
   20 = x + 366 – 733
   x = 20 – 366 + 733 = 387
   x = 390
   

    

6. 55.55 % of 4249.89 + 35.67 % of 749.89 + 22.56 % of ? = 2787.88
   400
   500
   550
   600
   650
   Option D
   2380 + 270 + (23/100) * x = 2788
   (23/100) * x = 2788 – 2380 – 270
   (23/100) * x = 138
   x = 138 * (100/23)
   x = 600
   

    

7. 2716.66 ÷ (21.66 % of 649.99) = 34.77 × 5 3/7 of (? ÷ 16.55)
   4
   2
   1
   1.7
   3
   Option D
   2717 ÷ [(22/100) * 650] = 35 × 5 3/7 of (x ÷ 17)
   2717 ÷ 143 = 35 * (38/7) * (x/17)
   19 = 35 * (38/7) * (x/17)
   x = 17/10 = 1.7
   

    

8. 6 5/8 % of 9599.99 + (1424.66 ÷ 24.88 ÷ 2.99] = 126.88 + ?
   528
   456
   768
   567
   757
   Option A
   (53/800) * 9600 + ((1425/25) / 3) = 127 + x
   636 + 19 – 127 = x
   x = 528
   

    

9. ∛85183.55 – √960.99 × ? + (2/5) of 375.12 – 48.43 = 22.22
   1
   2
   4
   5
   6
   Option C
   44 – (31 * x) + 150 – 48 = 22
   44 + 150 – 48 – 22 = (31 * x)
   124 = (31 * x)
   x = 124/31 = 4
   

    

10. (3/5) of (10/17) of 424.55 – ? + 47.67 % of 649.95 = 57.9
    646
    467
    635
    456
    404
    Option E
    (3/5) * (10/17) * 425 – x + (48/100) * 650 = 58
    150 – x + 312 = 58
    404 = x
    

     

The pie chart given below shows the distribution of the number of men in percentage or in absolute value or in degree in six different cities in china affected by corona. 42.857% of the number of women in city U is 300.

1. Find the total number of affected person in city S
   
   200
   220
   210
   190
   205
   Option B
   Percentage of mens in city P = 20%
   Percentage of mens in city Q = 90˚ * 100/360 = 25%
   Percentage of mens in city R = 36˚ * 100/360 = 10%
   Percentage of mens in city S = 5%
   Percentage of mens in city T = 54˚ * 100/360 = 15%
   Percentage of mens in city U = 100% – (20% + 25% + 10% + 5% + 15%)
   = 100% – 75% = 25%
   Number of women affected in city U = 300/42.857 * 100
   = 300 * 7/3 = 700
   Number of men affected in city U = 700 *(100/140) = 500
   Total number of men affected in all the cities together = 500 * (100/25)
   = 2000
   Total number of persons in city S = 100 + 120 = 220

    

2. If x number of women affected in city Q in the month jan and the number of women affected in city Q in the month jan is 40% of the number of persons in city U in the month Dec. What is the value of x?
   80
   85
   90
   95
   100
   Option A
   Number of women affected in city Q in the year 2018
   = 1200 * 2/5 = 480
   Number of women affected in city Q in the month Jan = 480 – 400 = 80
   

    

3. What is the ratio of the total men affected in city R and T together to the difference between the women affected in city U and P?
   5:2
   3:4
   1:5
   6:5
   11:3
   Option A
   Required ratio = (200 + 300): (700 – 500)
   = 500: 200 = 5: 2
   

    

4. Number of women affected in city S is what percentage of the average number of men affected in city Q, R and U together?
   10
   20
   30
   40
   50
   Option C
   Required percentage = 120/[(500 + 200 + 500)/3] * 100
   = 120/400 * 100 = 30%
   

    

5. If the ratio of the number of persons eats Veg to Non-veg in city T is 2: 1 and the number of women in city T eat Veg is 33.33% more than the number of women in city S. What is the ratio of the number of women to men eat Non – veg?
   1:3
   4:5
   3:7
   5:8
   5: 4
   Option E
   Total persons in city T = 300 + 240 = 540
   Number of persons like Veg = 540 * 2/3 = 360
   Number of persons like Non – veg = 540 * 1/3 = 180
   Number of women persons like veg = 120 * (100 + 33.33)/100
   = 120 * (4/3) = 160
   Number of women persons like Non – veg = 240 – 160 = 80
   Number of men persons like Non – veg = 180 – 80 = 100
   Required ratio = 100: 80 = 5: 4
   

    

Study the following information carefully and answer the questions given below?



6. The number of girls in chemistry department is approximately what percent of the total number of students in that department?
   23
   45
   54
   67
   52
   Option E
   Required percentage = (250 – 120)/250 * 100 = 52%

    

7. What is the ratio of the number of boys to girls in COMPUTER and ENGLISH department together?
   4:1
   5:1
   2:1
   3:1
   6:1
   Option C
   Number of girls in COMPUTER = 400 – 280 = 120
   Number of girls in ENGLISH = 500 – 320 = 180
   Required ratio = (280 + 320):(120 + 180)
   = 600:300
   = 2:1
   

    

8. What is the average number of girls in all the departments together?
   123
   156
   177
   189
   162
   Option E
   Girls in COMPUTER = 400 – 280 = 120
   Girls in ENGLISH = 500 – 320 = 180
   Girls in BIOLOGY = 350 – 200 = 150
   Girls in Maths = 450 – 220 = 230
   Girls in Chemistry = 250 – 120 = 130
   Required Average = (120 + 180 + 150 + 230 + 130)/5
   = 162
   

    

9. What is the difference between the number of boys and girls in BIOLOGY department?
   50
   20
   54
   34
   16
   Option A
   Number of girls in BIOLOGY = 350 – 200 = 150
   Required difference = 200 – 150 = 50
   

    

10. What is the difference between the average number of boys and girls in all the departments together?
    66
    34
    51
    67
    87
    Option A
    Girls in COMPUTER = 400 – 280 = 120
    Girls in ENGLISH = 500 – 320 = 180
    Girls in BIOLOGY = 350 – 200 = 150
    Girls in Maths = 450 – 220 = 230
    Girls in Chemistry = 250 – 120 = 130
    Average of girls = (120 + 180 + 150 + 230 + 130)/5
    = 162
    Average of boys = (280 + 320 + 200 + 220 + 120)/5 = 228
    Difference = 228 – 162 = 66

     

1. What is the sum if the difference between the simple interest and Compound interest on that sum for 2 years at 8 % is Rs. 1600.
   270000
   210000
   2500
   25000
   250000
   Option E
   Diff = Sum*(r/100)2
   1600 = Sum*(8/100)2
   1600*100*100/64 = Sum
   Sum = Rs. 250000
   

    

2. 6 years ago, the ratio of ages of Me and my husband is 2: 3. After 6 years, the ratio of ages of Me and my husband is 3: 4. Find the sum of our present ages?
   40
   89
   69
   72
   90
   Option D
   6 years ago, the ratio of ages of ours = 2: 3 (2x, 3x)
   Present ages of ours = 2x + 6, 3x + 6
   After 6 years, the ratio of ages ours = 3: 4
   According to the question,
   (2x + 12)/(3x + 12) = (3/4)
   8x + 48 = 9x + 36
   x = 12
   Sum of the present ages ours = 2x + 3x + 12 = 72 years
   

    

3. Cistern A opened to fill the tank and after 4 hours is closed and then Cistern B opened and fills the remaining tank in 10 hours. If Cistern B alone fill 25% of the tank in 5 hours, in how many hours Cistern A and B together fill the tank completely?
   5(1/7)
   5(5/9)
   5(6/7)
   4(5/7)
   5(5/7)
   Option E
   B = 5 * 4/1 = 20 hours
   4/x + 10/20 = 1
   4/x = ½
   1/x = 1/8
   A + B = 1/20 + 1/8
   = 7/40
   Time = 40/7 hours = 5(5/7) hours
   

    

4. A box contains x+4 pink chocolates, 6 white and 8 brown colour chocolates; if two chocolates are taken random and the probability of getting both are white colour chocolates is 5/92, then find the difference between the no. of pink colour chocolates and the no. of brown colour chocolates.
   2
   3
   4
   5
   6
   Option A
   Given,
   6c2/(x+18)c2=5/92
   X2+35x-246=0
   we get x=6
   Required difference = 10-8=2
   

    

5. Average ages of 5 men M1, M2, M3, M4 and M5 is 38 years. Ratio of the ages of M1 to M3 is 3:4 and after 5 years the ratio of the ages of M2 to M4 becomes 2: 3. If the difference between the ages of M1 and M2 is 2 years, then what is the present age of M5?
   13
   45
   66
   67
   We cannot find the answer.
   Option E
   M1 + M2 + M3 + M4 + M5 = 190 years
   M1/M3 = 3/4
   M2 + 5/M4 + 5 = 2/3
   2M4 + 10 = 3M2 + 15
   2M4 – 3M2 = 5
   We cannot find the answer.
   

    

6. If the certain number of candidates in the police selection and the average weight is x kg. If after one month 5 candidates is reduced their weight by 4 kg, then the average weight is reduced by 2 kg. What is the total number of candidates in the selection?
   50
   40
   30
   20
   10
   Option E
   Number of candidates = y
   Total weight = xy
   Total weight after reduced the weight = xy – 5 * 4 = xy – 20
   (x – 2) * y = xy – 20
   xy – 2y = xy – 20
   2y = 20
   y = 10
   

    

7. Kailash is 8 years older than Akash and Akash is 12 years younger than Prakash. If the sum of the ages of Kailash, Akash and Prakash is 80 years, then what is Akash’s age after 10 years?
   50
   40
   30
   20
   10
   Option C
   K– A = 8
   P–A = 12
   K + A + P = 80
   A + 8 + A + 12 + B = 80
   3A = 60
   A = 20 years
   After 10 years Akash’s age = 20 + 10 = 30 years
   

    

8. Man A spends 60% of his salary on education fee. He spends 30% of remaining on food and 50% of the remaining on shopping. If he left with him Rs.2800, then what is the amount he spends on education fee?
   12000
   15000
   12500
   12340
   32000
   Option A
   Education fee = 60/100 * x
   Remaining = 40/100 * x
   Food = 40/100 * x * 30/100 = 12x/100
   Remaining = 40x/100 – 12x/100 = 28x/100
   Shopping = 28x/100 * 50/100 = 14x/100
   Remaining = 14x/100
   14x/100 = 2800
   x = 20000
   Education fee = (20000/100) * 60 = 12000
   

    

9. Ratio of the length to breadth of the rectangular board is 2: 1. If Rs.1440 is required to paint the board at the rate of Rs.5 per square meter, then what is the difference between the length and breadth of the rectangular board?
   12
   13
   14
   15
   16
   Option A
   Area of the board = l * b
   2x * x = 1440/5
   x = 12 m
   Length = 2 * 12 = 24 m
   Breadth = 12 cm
   Difference = 24 – 12 = 12 cm
   

    

10. The ratio of the income of P and Q is 3: 2 and the ratio of their spends is 6: 5. If the ratio of the income to savings of P is 3: 1, then what is the ratio of the savings of P to Q?
    3: 5
    3: 7
    3: 2
    3: 1
    3: 4
    Option D
    Income of P = 3x
    Income of Q = 2x
    Spends of A = 6y
    Spends of B = 5y
    P’s savings = 3x – 6y
    P’s savings = 1/3 * 3x = x
    3x – 6y = x
    2x = 6y
    x = 3y

    Q’s savings = 2x – 5y = 2x – 5 * (x/3)
    = 6x – 5x/3 = x/3
    Required ratio = x: x/3 = 3: 1
    

     

Find the wrong term:-

1. 289.., 292.., 297.., 306.., 322
   289
   292
   297
   306
   322
   Option E
   289 + (21 + 1) = 292
   292 + (22 + 1) = 297
   297 + (23 + 1) = 306
   306 + (24 + 1) = 323
   

    

2. 21.., 10.., 11.., 16.5.., 33..,
   21
   10
   11
   16.5
   32
   Option E
   21 * 0.5 – 0.5 = 10
   10 * 1 + 1 = 11
   11 * 1.5 – 1.5 = 16.5
   16.5 * 2 + 2 = 32
   

    

3. 225.., 223…, 111…, 36…, 8
   225
   223
   111
   36
   8
   Option B
   (225 – 1) / 1 = 224,
   (224 – 2) / 2 = 111,
   (111 – 3) / 3 = 36,
   (36 – 4) / 4 = 8
   

    

4. 1…, 2…, 8…, 32…, 148…,
   1
   2
   8
   32
   148
   Option D
   1 * 1 + 12 = 2
   2 * 2 + 22 = 8
   8 * 3 + 32 = 33
   33 * 4 + 42 = 148
   

    

5. 
   27…, 55…, 114…, 227…, 458
   27
   55
   114
   227
   458
   Option C
   27 * 2 + 1 = 55
   55 * 2 + 2 = 112
   112 * 2 + 3 = 227
   227 * 2 + 4 = 458
   

    

Find the missing numbers:-

6. 1024.., 64.., 512.., 128.., 256.., ?
   234
   256
   276
   244
   250
   Option B
   1024 ÷ 16 = 64
   64 * 8 = 512
   512 ÷ 4 = 128
   128 * 2 = 256
   256 ÷ 1 = 256
   

    

7. 578.., 589.., 611…, 655.., ?.., 919
   657
   743
   890
   790
   888
   Option B
   578.., 589.., 611…, 655.., 743.., 919
   11 22 44 88 176
   11 22 44 88

    

8. 13…, 156…, 297.., 436.., 573.., 708.., ?
   870
   841
   887
   860
   790
   Option B
   13 + 143 = 156
   156 + 141 = 297
   297 + 139 = 436
   436 + 137 = 573
   573 + 135 = 708
   708 + 133 = 841
   

    

9. 9360.., 9360.., 4680.., 1560.., 390.., 78.., ?
   12
   13
   14
   15
   16
   Option B
   9360 / 1 = 9360
   9360 / 2 = 4680
   4680 / 3 = 1560
   1560 / 4 = 390
   390 / 5 = 78
   78 / 6 = 13
   

    

10. 23…, 34.5…, 69…, 172.5…, 517.5…, 1811.25…, ?
    3456
    7245
    3467
    7690
    1128
    Option B
    23 * 1.5 = 34.5
    34.5 * 2 = 69
    69 * 2.5 = 172.5
    172.5 * 3 = 517.5
    517.5 * 3.5 = 1811.25
    1811.25 * 4 = 7245
    

     

What will come in the missing place?

1. 11.., 43.., 170.., 677.., 2704.., 10811.., ?
   245667
   46367
   56577
   34478
   43238
   Option E
   11*4-1=43
   43*4-2=170
   170*4-3=677
   677*4-4=2704
   2704*4-5=10811
   10811*4-6=43238
   

    

2. 599.., 722.., 968.., 1337.., ?
   1829
   3352
   1335
   1343
   1241
   Option A
   599 + (1*123) = 722
   722 + (2*123) = 968
   968 + (3*123) = 1337
   1337 + (4*123) = 1829

    

3. 23.., 34.5.., 69.., 172.5.., ?
   564
   517.5
   546
   347
   643
   Option B
   23*1.5=34.5
   34.5*2=69
   69*2.5=172.5
   172.5*3=517.5
   

    

4. 19…, 23…, 32…, 48…, ..?
   70
   78
   73
   75
   77
   Option C
   19+22=23
   23+32=32
   32+42=48
   48+52=73
   

    

5. 2209.., 1849.., 1681.., 1369..,?
   879
   961
   876
   865
   990
   Option B
   2209=47*47
   1849=43*43
   1681=41*41
   1369=37*37
   961=31*31
   

    

Find the wrong term in the series:-


6. 534.., 266.., 132.., 64.., 31.5.., 14.75..
   534
   266
   132
   64
   31.5
   Option D
   The pattern is, ÷ 2 – 1
   

    

7. 289.., 292.., 297.., 306.., 322.., 356…
   289
   292
   297
   306
   322
   Option E
   The pattern is, +(21 + 1), +(22 + 1), +(23 + 1), +(24 + 1), +(25 + 1),…

    

8. 1924.., 1790.., 1610.., 1370.., 1064.., 684
   1924
   1790
   1610
   1370
   1064
   Option B
   The pattern is, -(112 + 11), -(132 + 13), -(152 + 15), -(172+17), -(192 + 19),..

    

9. 172.., 200.., 263.., 388.., 604.., 947.., 1459
   172
   200
   263
   388
   604
   Option B
   The pattern is, +3^3, +4^3, +5^3, +6^3, +7^3, +8^3,..

    

10. 9.., 16.., 45.., 175.., 875.., 5244
    9
    16
    45
    175
    875
    Option D
    The pattern is, *2 – 2, *3 – 3, *4 – 4, *5 – 5, *6 – 6,..

     

1. 40% of 80 + √3969 * 5 – 12 * 7 = ?
   263
   354
   454
   122
   223
   Option A
   32 + 315 – 84 = ?
   263 = ?
   

    

2. 1900 ÷ √1444 – 120% of 35 + 5 * 104 = ?
   345
   456
   353
   528
   346
   Option D
   50 – 42 + 520 = ?
   528 = ?
   

    

3. 5/6 * 102 + 5/11 * 187 + ? = 50 * 3
   -32
   -12
   -27
   -11
   – 20
   Option E
   85 + 85 + ? = 150
   ? = – 20
   

    

4. 175 + ? * 43 = 195% of 200
   5
   6
   7
   8
   9
   Option A
   175 + ? * 43 = 390
   ? = 5
   

    

5. x * 43 = 1/6 * 3600 + 15 * 3
   13
   14
   15
   16
   17
   Option C
   x = 645/43 = 15

    

6. 17.99 * 10.08 + 12.54 * 24.23 – 48.3 * 8.1 =? + √256
   80
   60
   70
   90
   85
   Option A
   180 + 300 – 384 =? + 16
   ?=80
   

    

7. 80.009% of 420.45 + (64.34 ÷ 136.23) * (17.11 ÷ √16) + ? = 15 * 25
   35
   36
   37
   38
   39
   Option C
   336 + 2 + ? = 375
   ? = 37
   

    

8. 720.12 ÷ 8.43 + 25% of 680.33 + 15% of 60.01 + 30 % of 40.44 = ?
   123
   281
   332
   235
   443
   Option B
   90 + 170 + 9 + 12 = ?
   ? = 281
   

    

9. 3200.009 ÷ 40.112 + 120% of 40.34 + ? = √729 * 8.123
   22
   33
   66
   77
   88
   Option E
   80 + 48 + ? = 216
   ? = 88
   

    

10. 144.5 % of 79.9 – √675.6 + 39.9 * 11.8 – 380.9=?
    189
    112
    156
    178
    165
    Option A
    116 – 26 + 480 – 381=?
    ? = 189