1. A man deposited a total amount of rs. 36000 partially in scheme A and scheme B. Scheme A offers simple interest at the rate of 14% p.a while scheme B offers compound interest at the rate of 20% p.a compounded annually. If the interest earned from scheme A after 2 years is rs.2160 less than the interest earned from scheme B after 2 years, then what is the amount deposited in scheme B ?
   15000
   17000
   23000
   14000
   17500
   Option B
   Let amount deposited in scheme B = rs. x
   amount deposited in scheme A = rs.(36000 – x)
   Rate of compound interest after 2 years = 20 + 20 + 20 * 20/100 = 44%
   x * 44/100 – (36000 – x) * 14/100 * 2 = 2160
   (44x – 108000 + 28x)/100 = 2160
   72x = 1224000
   x = 17000
   amount deposited in scheme B is rs.17000

    

2. The income of A and B are in the ratio 4 : 5 respectively. If the expenditure of A and B are rs.14000 and rs.18000 and ratio of saving of A and B is 6 : 7, then what is the saving of B ?
   8000
   9200
   7000
   6800
   8200
   Option C
   Let income of A and B = 4x and 5x respectively
   4x – 14000/5x – 18000 = 6/7
   28x – 98000 = 30x – 108000
   x = 5000
   income of B = 5 * 5000 = 25000
   saving of B = 25000 – 18000 = 7000

    

3. A takes 5 days less than B to complete a work alone. A and B together can complete the work in 13 7/11 days. Find the time taken by B to complete the whole work ?
   25 days
   30 days
   45 days
   43 days
   40days
   Option B
   Let B alone can complete the whole work = x days
   A alone can complete the whole work = (x – 5) days
   1/x – 5 + 1/x = 11/150
   2x – 5/x^2 – 5x = 11/150
   11x^2 – 55x = 300x – 750
   11x^2 – 355x + 750 = 0
   11x^2 – 330x – 25x + 750 = 0
   x = 30
   B alone can complete the whole work in 30 days

    

4. Three containers having capacity in the ratio of 4 : 5 : 3 contain mixture of milk and water in the ratio of 2 : 3, 4 : 1 and 3 : 1 respectively. If all three containers poured in a big container, then find the ratio of water to milk in final mixture ?
   28 : 137
   85 : 152
   83 : 157
   134 : 49
   5 : 2
   Option C
   Let total mixture in three container are 400, 500 and 300 respectively
   total quantity of milk in big container = 400 * 2/5 + 500 * 4/5 + 300 * 3/4 = 785
   total quantity of water in big container = 400 * 3/5 + 500 * 1/5 + 300 * 1/4 = 415
   required ratio = 415 : 785 = 83 : 157
   

    

5. B and C started a partnership business with the capital of rs.(x + 4000) and rs.(x + 2000) respectively. After one year A joined the business with capital of rs.(x + 12000) while B left the business. If at the end of three years the profit sharing ratio of A , B and C is 20 : 6 : 15, then find the difference between amount invested by A and B in the business ?
   15000
   13000
   14000
   8000
   12000
   Option D
   A : B : C = (x + 12000) * 2 : (x + 4000) * 1 : (x + 2000) * 3
   = 2x + 24000 : x + 4000 : 3x + 6000
   2x + 24000/x + 4000 = 20/6
   20x + 80000 = 12x + 144000
   8x = 64000
   x = 8000
   difference = (8000 + 12000) – (8000 + 4000) = 8000

    

6. Directions : In these questions two equations is given, solve the equations and answer the questions given below.
   I. x^2 – 42x + 392 = 0
   II. y^2 – 27y + 180 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option E
   I. x^2 – 28x – 14x + 392 = 0
   x = 28 , 14
   II. y^2 – 15y – 12y + 180 = 0
   y = 15, 12

    

7. I. x^2 – 8x – 65 = 0
   II. 2y^2 – 27y + 91 = 0
   
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option E
   I. x^2 – 13x + 5x – 65 = 0
   x = 13, -5
   II. 2y^2 – 14y – 13y + 91 = 0
   2y = 14, 13
   y = 7, 6.5
   

    

8. I. x^3 = 216
   II. 5y^2 + 2y – 185 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option B
   I. x^3 = 216
   x = 6, 6
   II. 5y^2 + 15y – 13y – 37 = 0
   5y = – 15, 13
   y = -3 , 2.6

    

9. I. x^2 + 17x + 60 = 0
   II. y^2 – 2y – 15 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option A
   I. x^2 + 12x + 5x + 60 = 0
   x = -12, -5
   II. y^2 – 5y + 3y – 15 = 0
   y = 5, -3

    

10. I. x^2 – 15x + 36 = 0
    II. y^2 – 19y + 60 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option E
    I. x^2 – 12x – 3x + 36 = 0
    x = 12 , 3
    II. y^2 – 15y – 4y + 60 = 0
    y = 15, 4

     

1. Ratio of number of male of company X to company Y is 8 : 5. The ratio of female to male of company X and company Y is 3 : 4 and 7 : 5 respectively. The difference between number of female of company X and company Y is 150, then find total number of male and female in company Y ?
   2700
   2200
   1800
   1500
   1200
   Option C
   Let the number of male of company X and Y = 8a and 5a
   number of male and female of company X = 4a and 3a
   number of male and female of company Y = 5a and 7a
   The number of male and female of company X = 8a and 6a
   difference = 7a – 6a = 150
   a = 150
   sum of number of male and female of company Y = 5a + 7a = 12a
   = 12 * 15 = 1800
   

    

2. A and B entered into a partnership business with initial investments of rs.4000 and rs.5000 respectively. After 6 months A and B withdrew rs.1000 and rs.2000 respectively, while C joined with them capital of 4x. If after 14 months, C received rs.4000 as profit share out of total profit of rs.12500, then find how much amount invested by C ?
   4000
   4500
   4800
   6000
   3200
   Option D
   Profit sharing ratio of A, B and C = (4000 * 6 + 3000 * 8) : (5000 * 6 + 3000 * 8) : 4x * 8
   = 48000 : 54000 : 32x
   profit of A and B = 12500 – 4000 = 8500
   32x/102000 = 4000/8500
   x = 1500
   C invested = 4 * 1500 = 6000

    

3. Rohit invested rs.100x in a scheme offering compound interest at the rate of 20% p.a and after two years he again invested rs.80x in the same scheme. He received rs. 1243.2 as total interest after 3 years. If he rs.120x is invested at 10% p.a simple interest for 4 years the find how much interest he received ?
   1128
   672
   845
   862
   920
   Option B
   Compound interest after 2 years = 100x * 120/100 * 120/100 – 100x = 44x
   now total principal = 100x + 44x + 80x = 224x
   CI received after one year = 224x * 20/100 = 44.8x
   total CI after 3 years = 88.8x
   88.8x = 1243.2
   x = 14
   If he received 120x at 10% p.a in SI for 4 years
   120 * 14 * 40/100 = 672

    

4. The ratio of milk and water in Jar A is 5 : 3 and total mixture of container A is 480 liters. Jar B contains mixture of water to milk in the ratio of 7 : 9. Total mixture of jar A and B poured into another jar C such that quantity of milk in jar C is 60% of total mixture. Find the quantity of mixture in jar B ?
   440 lt
   480 lt
   380 lt
   240 lt
   320 lt
   Option E
   Quantity of milk in jar A = 480 * 5/8 = 300
   quantity of water in jar B = 480 * 3/8 = 180
   let the quantity of milk and water in jar B = 9x and 7x
   300 + 9x = (480 + 16x) * 60/100
   1500 + 45x = 48x + 1440
   3x = 60
   x = 20
   total mixture in jar B = (9x + 7x) * 20 = 320 liters
   

    

5. A man deposited rs.8000 in a bank offering compound interest of 20% p.a compound annually for first 1.5 years and after 1.5 years he offering simple interest 50% p.a for rest of the time. If at the end of 1.5 years, he withdrew rs.3560, then how much amount received by him at the end of 3 years ?
   12400
   10500
   18000
   12250
   15240
   Option D
   Rate of compound interest after 1.5 years = 20 + 10 + 20 * 10/100 = 32%
   total amount after 1.5 years = 8000 * 132/100 = 10560
   remaining amount after withdrawal by him = 10560 – 3560 = 7000
   simple interest received for rest of time = 7000 * 1.5 * 50/100 = 5250
   total amount received = 7000 + 5250 = 12250

    

6. Directions : What will come in place of questions (?) marks in the following questions given below ?
   30% of 168 + √1024 + 42 = ?
   84.2
   94.8
   124.4
   65
   84
   Option C
   50.4 + 32 + 42 = ?
   ? = 124.4

    

7. (25^2) + (38^2) – 1220 + ? = 1000
   151
   162
   170
   175
   320
   Option A
   625 + 1444 – 1220 + ? = 1000
   ? = 1000 – 849
   ? = 151

    

8. (32 * 8 + 440 + 15 * 12) * 6 = ?
   4340
   3240
   2015
   5256
   8450
   Option D
   (256 + 440 + 180) * 6 = ?
   ? = 5256
   

    

9. 5^2 + 387 = 180/? + 376
   6
   8
   4
   5
   7
   Option D
   25 + 387 = 180/? + 376
   412 – 376 = 180/?
   ? = 5

    

10. 75 * 8 + √1156 = ?^3 + 122
    4
    32
    14
    8
    4
    Option D
    600 + 34 = ?^3 + 122
    ?^3 = 512
    ? = 8

     

Direction: What is the approximate value in the place of questions mark (?) in the following questions ?

1. 887.97÷12.06*√16.05-36=?
   268
   260
   280
   300
   240
   Option B
   888/12*4-36=?
   ?=296-36=260

    

2. ?^2+56.01-24*5.99=168.01
   16
   15
   13
   20
   24
   Option A
   168+144-56=?^2
   ?=16^2

    

3. 45.01% of 4500÷15.05+139.88-220.01=?
   60
   65
   55
   70
   85
   Option C
   2025/15+140-220=?
   135+140-220=55

    

4. (7.99)^2+(10.01)^2-(18.05)^2+(12.99)^2=?
   15
   13
   9
   12
   28
   Option C
   8^2+10^2-18^2+13^2=?
   64+100-324+169=?
   ?=9

    

5. 539.98/?=?/60.01
   200
   160
   120
   180
   200
   Option D
   ?^2=540*60
   ?^2=540*60
   ?=180

    

Direction: What will come in the of questions mark(?) in the following questions.

6. 36% of 1000+42% of 3000=?-180+1050
   750
   650
   850
   950
   550
   Option A
   360+1260+180-1050=750

    

7. 5/11 of 1331+4/9 of 270=30% of 400+30+?
   680
   575
   675
   550
   740
   Option B
   605+120-120-30=?
   ?=575

    

8. 2748+1323-2238-1235+3248=?
   3646
   3948
   3748
   3846
   2896
   Option D
   7319-3473=3846

    

9. 4515÷15+324-25% of 1200=?
   325
   425
   320
   440
   345
   Option A
   301+324-300=325

    

10. 48/3*45/80*1/9+135-40=?
    118
    105
    78
    96
    120
    Option D
    48/3*45/80*1/9+135-40=?
    1+135-40=96

     

1. In a mixture of 80 liters, the ratio of milks and water is 3:2 . How much water must be added to this mixture so that the ratio of milk and water becomes 3:4.
   24
   28
   40
   36
   32
   Option E
   Total mixture =80 liters.
   milk =80*3/5 =48
   water =80*2/5 =32
   Let, x liters of water is added.
   48/32+x =3/4
   32+x =64
   x =64-32 =32 liters.

    

2. In a school the students of three classes are in the ratio of 2:3:5 . If 20 students are increased in every class the ratio of changes is 4:5:7. What is the total number of students in the three classes before the class increased .
   200
   100
   250
   300
   400
   Option B
   Let students in three classes be
   2x, 3x and 5x respectively.
   According to the question ,
   (2x+20):(3x+20):(5x+20) =4:5:7
   2x+20/3x+20 =4/5
   12x+80 =10x+100
   2x=20
   x=10
   Total students =(2*10)+(3*10)+(5*10)
   = 20+30+50 =100

    

3. If 16 cows cost as much as 8 bullocks, 10 bullocks cost as much as 5 horses and 3 horses cost as much as 4 camels. If the cost of one camel is rs.3300. Find the cost of one cow is,
   2100
   2200
   2300
   1100
   1200
   Option D
   1 camel cost=3300
   4 camel cost=13200
   3 horses cost=13200
   1 horses cost=4400
   5 horses cost=22000
   10 bullocks cost=22000
   1 bullocks cost=2200
   8 bullocks cost =17600
   16 cows cost =17600
   1 cows cost =1100
   

    

4. The ratio of the earning of A and B is 3:5. If the earning of a increased by one-forth and the earning of B decreased by one-forth, then find the new ratio of their earning ?
   2:3
   1:1
   3:4
   5:2
   2:5
   Option B
   Let earning of A and B be 3x and 5x.
   New ratio of their earning = 3x*5/4:5x*3/4 =1:1

    

5. A some of money is to be distributed among A,B,C,D in the proportion of 3:5:2:4 . If D gets 3000 more than C, then find total money which is distribute among all.
   30000
   21000
   25000
   28000
   24000
   Option B
   Let the money is distributed among A,B,C and D be 3x,5x,2x and 4x respectively.
   4x-2x =3000
   2x =3000
   x =1500
   total amount of money distributed =3x+5x+2x+4x =14x
   = 14*1500 =21000

    

6. The ratio of the number of boys and girls in a college is 7:9. If the percentage increase in the number of boys and girls be 20% and 30% respectively. The total number of boys and girls in college before increase is 1440, then find ratio of boys and girls in the college after increased.
   28:39
   38:43
   25:28
   35:28
   30:37
   Option A
   ratio of boys and girls=7:9
   total boys and girls=1440
   boys=1440*7/16=630
   girls=1440*9/16=810
   the number of boys after increased=630*120/100=756
   the number of girls after increased=810*130/100=1053
   ratio of boys and girls after increased=756:1053=28:39

    

7. Salaries of Ram and Rohit are in the ratio of 4:5. If the salary of each is increased by rs.3000, the new ratio becomes 10:11. What is initial salary of Rohit.
   2700
   3000
   4200
   2500
   3500
   Option D
   let salaries of Ram and Rohit be 4x and 5x
   (4x+3000)/(5x+3000)=10/11
   44x+33000=50x+30000
   6x=3000
   x=500
   initial salary of Rohit=5x=5*500=2500
   

    

8. A contains a mixture of 56 liters of wine and water in the ratio of 3:4 . If 28 liter of mixture is taken and 8 liter of wine is added , then find the ratio of new mixture .
   3:2
   8:3
   5:5
   5:4
   3:4
   Option D
   Ratio of wine and water =3:4
   wine =56*3/7 =24
   water =56*4/7 =32
   28 liter of mixture taken out wine = 24-(28*3/7) =24-12 =12
   water=32-(24*4/7) =32-16 =16
   8 liters of wined is added then ,
   new ratio of wine and water =(12+8):16 =20:16 =5:4

    

9. 48% of A is equal to 64% of B, then find the ratio of A and B ?
   3:2
   3:4
   7:3
   5:4
   4:3
   Option E
   48% of A = 64% of B
   A/B=64/48=4:3

    

10. The ratio of three numbers is 2:4:5 and sum of their squares is 1125. Find the sum of the numbers.
    60
    45
    50
    55
    65
    Option D
    ratio of three numbers=2:4:5
    let the numbers be 2x, 4x and 5x
    sum of their squares=1125
    4x^2+16x^2+25x^2=1125
    45x^2=1125
    x^2=25
    x=5
    the sum of the numbers=2x+4x+5x=11x=11*5=55

     

1. x^2+14x+24 =0
   y^2+18x+65 =0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   I. x^2+12x+2x+24 =0
   x= -12, -2
   II. x^2+13x+5x+65 =0
   x= -13, -5

    

2. I. 2x^2+23x++65 =0
   II. 3x^2+25x+50=0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option D
   I. 2x^2+13x+10x+65=0
   2x=-13, -10
   x=-6.5, -5
   II. 3x^2+15x+10x+50 =0
   3x= -15,-10
   x=-5, -3.33

    

3. I. x^2-18x+72=0
   II. y^2-8y+15 =0
   
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option A
   I. x^2-12x-6x+72=0
   x =12, 6
   II. y^2-5y-3y+15 =0
   y=5, 3

    

4. I. x^3= 729
   II. y =√81
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   I. x^3=729
   x=9
   II. y=9

    

5. I. x^2-11x+18 =0
   II. y^2-23y+132 =0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option B
   I. x^2-9x-2x+18 =0
   x=9, 2
   II. y^2-12y-11y+132 =0
   y^2 =12, 11

    

6. I. 4x^2-12x+5 =0
   II. 2y^2+3y-5=0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   I. 4x^2-10x-2x+5 =0
   4x=10, 2
   x= 2.5, .5
   II. 2y^2+5y-2y-5 =0
   2y= -5, 2
   y= -2.5, 1

    

7. I. x^2-3x+28 =0
   II. y^2+3y-18=0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   I. x^2-7x+4x+28=0
   x=7, -4
   II. y^2+6y-3y-18=0
   y=-6, 3

    

8. I. x^4=256
   II. y^3=64
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option D
   I. x^4=256
   x=4, -4
   II. y^3=64
   y=4, 4

    

9. I. x^2-35x+250=0
   Ii. y^2-14y+40=0
   X>Y
   X
   X≥Y
   X≤Y
   X=Y or no relation
   Option C
   I. x^2-25x-10x+250=0
   x=25,10
   II. y^2-10y-4y+40=0
   y=10,4

    

10. I. x+5=√100
    II. y+8=√144
    X>Y
    X
    X≥Y
    X≤Y
    X=Y or no relation
    Option A
    I. x+5=√100
    x+5=10
    x=5
    II. y+8=√144
    y+8=12
    y=4

     

1. 32, 80, 200, 500, 1250 ?
   3125
   4125
   2500
   3750
   5125
   Option A
   The series is (32*2.5), (80*2.5), (200*2.5)…..
   

    

2. 12, 29, 63, 131, 267, ?
   541
   538
   429
   539
   342
   Option D
   The series is (*2+5), (*2+5)…

    

3. 48, 51, 56, 63, ?, 87
   72
   74
   75
   78
   76
   Option B
   The series (+3), (+5), (+7), (+11), (+13) the addition of prime numbers.

    

4. 1248, 1184, 1135, 1099, 1074, ?
   1158
   538
   1058
   1220
   625
   Option C
   The series (-8^2), (-7^2), (-6^2)…

    

5. 6420, 6435, 6465, 6510, ?, 6645
   6420
   6500
   6530
   6570
   6424
   Option D
   The series is (+15), (+30), (+45)…

    

6. 8, 8, 12, 28, 88, 932
   28
   88
   932
   12
   8
   Option B
   The series is (*.5+4), (*1+4), (*2+4), (*4+4), (*8+4)

    

7. 320, 330, 310, 340, 300, 360
   300
   360
   340
   320
   310
   Option B
   The series is (+10), (-20), (+30), (-40), (+50)

    

8. 32, 33, 41, 68, 114, 257
   257
   68
   41
   114
   33
   Option D
   The series is (+1^3), (+2^3), (+3^3), (+4^3), (+5^3 )

    

9. 18, 36, 108, 432, 2160, 12360
   12360
   2160
   432
   108
   36
   Option A
   The series is (*2), (*3), (*4)…

    

10. 45, 49, 41, 45, 37, 42
    45
    49
    41
    42
    37
    Option D
    The series is (+4), (-8), (+4), (-8), (+4)

     

1. 960, 839, 758, 709, 684, ?
   685
   550
   485
   675
   745
   Option D
   the series is, (-11^2), (-9^2), (-7^2)…

    

2. 340, 356, 382, 418, 464, ?
   520
   480
   540
   516
   496
   Option A
   the series is, +16, +26, +36, +46, +56

    

3. 15, 41, 78, 128, 193, ?
   278
   275
   385
   345
   264
   Option B
   the series is, +(5^2+1), +(6^2+1), +(7^2+1)…

    

4. 335, 351, 383, 431, ?, 575
   487
   535
   530
   485
   495
   Option E
   the series is, +16, +32, +48, +64, +80

    

5. 13, 15, 21, 33, 53, ?
   83
   96
   95
   85
   80
   Option A
   the series is, +(1*2), +(2*3), +(3*4), +(4*5), +(5*6)

    

6. 8, 18, 57, 232, 1165, 6994
   1165
   232
   57
   6994
   8
   Option D
   the series is, (*2+2), (*3+3), (*4+4)…

    

7. 14, 20, 32, 44, 58, 74
   14
   20
   32
   74
   44
   Option B
   the series is, +8, +10, +12, +14, +16

    

8. 244, 202, 172, 152, 140, 132
   132
   140
   244
   172
   152
   Option A
   the series is, -(6^2+6), -(5^2+5), -(4^2+4)…

    

9. 35, 44, 28, 53, 17, 65
   35
   65
   17
   53
   28
   Option B
   the series is, +3^2, -4^2, +5^2, -6^2, +7^2

    

10. 36, 23, 28, 47, 99, 252
    99
    47
    28
    252
    36
    Option D
    the series is, (*.5+5), (*1+5), (*1.5+5), (*2+5), (*2.5+5)

     

1. Direction: What will come in place of question(?) marks in the following questions ?
   13540+12981+1519-? =25000
   4260
   3140
   3340
   3040
   none of these
   Option D
   13540+12981+1519-25000 =3040

    

2. (1696÷4-?)10=1200
   404
   304
   340
   420
   520
   Option B
   (424-x)10 =1200
   4240-10x =1200
   x =304

    

3. 19*12*8+94 =?
   1918
   2018
   2200
   1900
   1920
   Option A
   1824+94=1918

    

4. 120% of 55+240% of 25+? =140
   16
   14
   20
   22
   24
   Option B
   66+60+? =140
   ?=14

    

5. (196/27*4)*(8/20)*(180/9)*(81/7)= ?
   178
   210
   180
   168
   none of these
   Option D
   7*8*3 =168

    

6. Direction: What will come in place of ‘x’ in the following questions given below ?
   20% of 25% of 40% of 1200 =x+8
   22
   16
   12
   18
   15
   Option B
   1/5*1/4*2/5*1200 =x+8
   x= 24-8 =16

    

7. 52^2+62^2+75^2-85^2 =x+48
   5000
   6000
   5200
   4200
   none of these
   Option A
   2704+3844+5625-7225-48 =x
   12273-7225-48 =x
   x =5000

    

8. √3025+√1600-81+x=79
   55
   64
   60
   65
   70
   Option D
   55+40-81+x =79
   x= 65

    

9. 520/13+980-1020/17+20% of 55 =x+88
   883
   880
   920
   780
   783
   Option A
   40+980-60+11 =x+88
   x= 883

    

10. 168÷21*8+218+x =525
    345
    445
    355
    245
    280
    Option D
    64+218+x =527
    x= 527-282
    x= 245

     

1. The ratio of amount of A and B is 4 : 3 and the ratio of amount of B and C is 4 : 5 . If the total number of amount A , B and C have rs. 3870, then find the how much amount B have ?
   1200
   2340
   1080
   880
   960
   Option C
   Ratio of amount of A and B = 4 : 3
   ratio of amount of B and C = 4 : 5
   By multiplying ‘4’ in ratio of A and B
   A : B = (4 : 3) 4 = 16 : 12
   By multiplying ‘3’ in ratio of B and C
   B : C = (4 : 5) 3 = 12 : 15
   A : B : C = 16 : 12 : 15
   let amount of A , B and C have = 16x , 12x and 15x
   total = 43x
   43x = 3870
   x = 90
   12x = 1080
   B have = 1080

    

2. Rabin has three types of coins i.e 10 paisa , 20 paisa and 50 paisa and total value of this coins is rs. 78. The number of coins is in the ratio of 6 : 4 : 5. Find the number 10 paisa coins.
   180
   300
   120
   140
   420
   Option C
   Let the number coins of 10 paisa , 20 paisa and 50 paisa is 6x, 4x and 5x respectively.
   (10 * 6x) + (20 * 4x) + (50 * 5x) = rs. 78
   60x + 80x + 250x = 78
   390x = 7800
   x = 20
   The number 10 paisa coins = 6 * 20 = 120

    

3. A sum of amount of rs. 3500 is divided between three friends. The ratio of amount given to P and Q is 2 : 3 and Q and R is 3 : 5. What is difference the amount given to P and R ?
   1240
   940
   480
   1050
   1280
   Option D
   Let the amount given to P and Q = 2x and 3x
   Q and R = 3x and 5x
   total amount = 3500
   2x + 3x + 5x = 3500
   10x = 3500
   x = 350
   difference = 5x – 2x = 3x = 3 * 350 = 1050

    

4. Rohit gave half of his monthly salary to his parents and one-third of remaining salary to his wife and remaining his salary gave to his son and daughter in the ratio of 13 : 12. If his monthly salary is rs.45000, then find how much amount he gave to his son ?
   4500
   3450
   6300
   5200
   7800
   Option E
   Rohit gave to his parents = 45000/2 = 22500
   remaining salary = 45000 – 22500 = 22500
   he gave to his wife = 22500 * 3 = 7500
   remaining salary = 22500 – 7500 = 15000
   he gave to his son = 15000 * 13/25 = 7800
   

    

5. Two numbers are in the ratio of 4 : 3. When we added 60 in each number the ratio becomes 30 : 23. Find the sum of both initial numbers.
   1520
   2000
   1470
   3800
   2540
   Option C
   Let two numbers is 4x and 3x
   When we added 60
   4x + 60 / 3x + 60 = 30/23
   92x + 1380 = 90x + 1800
   2x = 420
   x = 210
   total sum = 4x + 3x = 7x = 7 * 210 = 1470

    

6. The ratio of income of Rahul and Raju in the ratio of 6 : 5 and they saved rs. 1500 each, Their expenditure ratio 9 : 7 , then find what is the sum of their incomes ?
   14000
   18000
   35000
   20000
   11000
   Option E
   Let income of Rahul and Raju = 6x and 5x
   they saved = 1500 each
   6x – 1500 / 5x – 1500 = 9/7
   42x – 10500 = 45x – 13500
   3x = 3000
   x = 1000
   sum = 6x + 5x = 11 * 1000 = 11000

    

7. There are total of 300 employees in a company. The ratio of male employees to female employees are 8 : 7. How many female employees have to be recruited so that the ratio
   becomes male and female is equal ?
   40
   80
   25
   62
   20
   Option E
   Number of male = 300 * 8/15 = 160
   number of female = 300 * 7/15 = 140
   let ‘x’ number of female we added
   160 = 140 + x
   x = 20

    

8. If the numerator is increased by 20 % and the denominator is decreased by 20% so that fraction becomes 2 : 3. What is the original fraction ?
   2/5
   5/9
   4/9
   2/3
   4/11
   Option C
   Let the numerator and denominator be x and y respectively
   (x * 120/100) / (y * 80/100) = 2/3
   (6x/5) /(4y/5) = 2/3
   6x / 4y = 2/3
   18x = 8y
   x/y = 8 /18 = 4/9

    

9. Total number of workers worked in a company is 50 and the ratio of male workers to female workers is 3 : 2. The ratio of amount of wages received by a male and a female worker is 7 : 6. If the wages received by each female workers is rs.120, then how much total wages received by all male workers ?
   3800
   3200
   4500
   4200
   6500
   Option D
   Total workers = 50
   male workers = 50 * 3/5 = 30
   female workers = 50 * 2/5 = 20
   wages received by a female workers = 120
   wages received by a male workers = 120/6 *7 = 140
   total wages received by all male workers = 140 * 30 = 4200
   

    

10. Three friends A , B and C got a total prize money is rs.2149. If B got 15% more A and C got 20% less than B, then find the how much amount of money C get ?
    520
    644
    440
    634
    460
    Option B
    Total prize money = 2149
    let A got = 100x
    B got = 100x * 115/100 = 115x
    C got = 115x * 80/100 = 92x
    total = 100x + 115x + 92x = 307x
    amount of money C got = 2149/307x * 92x = 644

     

1. A man purchase is a bike for rs.75400 and spends rs. 2600 for his its maintenance. After some days he sells the bike at rs. 89700. What is his profit percentage ?
   10%
   18%
   11%
   15%
   8%
   Option D
   CP of bike = 75400
   maintenance cost = 2600
   total cost price of Bike = 75400 + 2600 = 78000
   profit = 89700 – 78000 = 11700
   profit percentage = 11700/78000 * 100 = 15%
   

    

2. Ayush made a loss of 10% on selling a cycle. If he would have made a profit of 40%. In what price he would sell so that he can get 22% profit ?
   1520
   1840
   2480
   1830
   2240
   Option D
   Let CP of cycle = 100x
   loss = 10%
   SP of cycle = 100x * 90/100 = 90x
   If he sold 750 more, he would have 40% profit
   SP = 100x * 140/100 = 140x
   90x + 750 = 140x
   50x = 750
   x = 15
   CP = 100x = 1500
   SP = 1500 * 122/100 = 1830

    

3. Marked price of an article is 45% above its cost price and by giving 20% discount on marked price he earned rs.128 as profit. What is profit percent incurred when he sells same article rs.72 more than its selling price.
   22%
   28%
   34%
   25%
   12%
   Option D
   Let CP of article = 100x
   MP of article = 100x * 145/100 = 145x
   SP of article = 145x * 4/5 = 116x
   profit = rs.128
   116 x – 100x = 16x
   16x = 128
   x = 8
   CP = 800
   SP = 928
   When he sells rs.72 more than its SP,
   SP = 928 + 72 = 1000
   Profit % = 1000 – 800/800 * 100 = 25%

    

4. Rohit purchased a mobile rs.4500 and sold it at 20% loss to Reema. Which amount he got from Reema, he purchased a TV and sold it at a profit of 25%. How much amount of overall profit or loss made by him in whole transactions ?
   180
   470
   200
   350
   No profit no loss
   Option E
   CP of mobile = 4500
   he sold at 20% loss,
   SP of mobile = 4500 * 4/5 = 3600
   CP of TV = 3600
   he sold at 25% profit,
   SP = 3600 * 5/4 = 4500
   there is no profit no loss

    

5. A shopkeeper marked up the price of an article 60% above the cost price. If he gives a successive discount of 15% and 10% on marked price, so that selling price of the article is rs.24480. Find the cost price.
   2200
   2000
   1240
   1350
   1430
   Option B
   Let CP of article = 100x
   MP of article = 160x
   SP of article = 160x * 85/100 * 90/100 = 122.4x
   122.4x = 2448
   x = 2448/122.4 = 20
   CP = 100x = 2000

    

6. A man buys 40 mangoes at rs. 5 each and 60 oranges at rs.4 each. 10% of the mangoes and oranges are rotted. The man sells the remaining mangoes at rs. 7 each and oranges at rs 3 each. Find the overall gain or loss percentage ?
   8% profit
   10% loss
   22% loss
   65/11% loss
   14%
   Option D
   Total CP of 40 mangoes = 40 * 5 = 200
   Total CP of 60 mangoes = 60 * 4 = 240
   10% of mangoes and oranges are rotted
   remaining mangoes = 40 * 90/100 = 36
   and oranges = 60 * 90/100 = 54
   total SP of oranges = 54 * 3 = 162
   total SP = 252 + 162 = 414
   loss % = 440 – 414 /440 * 100 = 65/11%

    

7. A shopkeeper sold a product at 20% discount and profit earned on that product is 16.67%. If the marked price of that product is rs.875, then what is the cost price of that product ?
   820
   340
   680
   600
   485
   Option D
   MP of the product = 875
   SP of product = 875 * 4/5 = 700
   profit = 16.67% = 1/6
   let CP = 6x
   profit = x
   SP = 7x
   7x = 700
   x = 100
   CP = 6x = 600

    

8. A dishonest shopkeeper sold rice at cost price. The 1kg weight of rice that he uses actually weight of 780 gm. How much percentage profit does he earned on the sale of 1kg price ?
   200/6%
   120/7%
   18%
   1100/39%
   204/5%
   Option D
   Actual weight = 1kg
   false weight = 780 gm
   profit = 100 – 780 = 220
   profit % = 220/780 * 100 = 1100/39%
   

    

9. Sekhar sells two products A and B have equal cost price. Product A is sold at 30% profit and product b is sold at an amount of 15% less than the selling price of product A. If the total profit is rs.162, then what is cost price of B ?
   480
   400
   525
   340
   260
   Option B
   Let CP of the product A = x
   SP of product A = x * 130/100 = 13x/10
   SP of product B = 13x /10 * 17/20 = 221x/200
   profit of product A = 13x/10 – x = 3x /10
   profit of product B = 221x/200 = 21x/200
   3x/10 + 21x/200 = 162
   60x + 21x / 200 = 162
   x = 400

    

10. An article is sold at a profit of 15%. If it had been sold at a profit of 22%, it would have fetched rs. 340 more. What is the cost price of article ?
    2500
    3240
    1450
    2000
    4420
    Option D
    Let CP of the article = 100x
    SP of the article = 100x * 115/100
    when it sold at 22% profit
    SP = 100 * 122/100 = 122x
    122x – 115x = 7x
    7x = 340
    x = 20
    CP of the product = 100x = 100 * 20 = 2000