1. x^2-4x+4=0
   y^2-9y+18=0
   x>y
   x< y
   X≥Y
   x≤y
   X=Y or no relation
   Option B
   1. x^2-4x+4=0
   x^2-2x-2x+4=0
   x=2,2
   2. y^2-9y+18=0
   y^2-6y-3y+18=0
   y=6,3

    

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

    

3. x^2+7x+10=0
   y^2+12y=35=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option C
   1. x^+7x+10=0
   x^2+5x+2x+10=0
   x= -5,-2
   2. y^2+12y+35=0
   y^2+7y+5y+35=0
   y= -7,-5

    

4. 25x^2=625
   y^2-10y+25=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option D
   1. 25x^2=625
   x=+5,-5
   2. y^2-10y+25=0
   y^2-5y-5y+25=0
   y=5,5

    

5. x^2-x-42=0
   2y^2-8y+8=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   1. x^2-x-42=0
   x^2-7x+6x-42=0
   x=7,-6
   2. 2y^2-8y+8=0
   2y^2-4y-4y+8=0
   2y=4,4
   y=2,2

    

6. x^4=256
   y^2-12y+36=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option B
   1. x^4=256
   x=+4,-4
   2. y^-12y+36=0
   y^2-6y-6y+36=0
   y=6,6

    

7. 2x+3y=3
   3x+5y=2
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option A
   1. 2x+3y=3……..(1)
   3x+5y=2………..(2)
   solving euations (1) and (2) we get,
   5(2x+3y=3)=10x+15y=15
   3(3x+5y=2)=9x+15y=6
   x=9
   by putting x=9, in equation(1) we get,
   18+3y=3
   y=-5

    

8. 3x^2-7x+4=0
   4y^2-13y+9=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   1. 3x^2-4x-3x+4=0
   3x=4,4
   x=4/3, 4/3
   2. 4y^2-13y+9=0
   4y^2-9y-4y+9=0
   4y=9,4
   y=9/4, 1

    

9. x^2=16
   y^2+6y+8=0
   x>y
   x< y
   X≥Y
   X≤Y
   X=Y or no relation
   Option E
   1. x^2=16
   x=4,-4
   2. y^2+6y+8=0
   y^2+4y+2y+8=0
   y= -4,-2

    

10. 5x^2-8x+3=0
    2y^2-16y+32=0
    
    x>y
    x< y
    X≥Y
    X≤Y
    X=Y or no relation
    Option B
    1. 5x^2-8x+3=0
    5x^2-5x-3x+3=0
    5x=1, 3/5
    2. 2y^2-16y+32=0
    2y^2-8y-8y+32=0
    2y=8,8
    y=4,4

     

Click here for topic wise Quantitative Aptitude Questions

1. The perimeter of an equilateral triangle is equal to the perimeter of a square whose diagonal is 9√2 cm. Find out the area of the equilateral triangle ?
   48√3
   60√3
   30√3
   36√3
   20√3
   Option D
   side of square=9√2/√2=9
   let the side of equilateral triangle be x cm.
   3x=4*9
   x=12
   area of equilateral triangle= √3/4*12*12=36√3

    

2. If ratio of height and radius of a cone is 5:6. Volume of cone is 480π cm^3, then find slant height of cone (in cm) ?
   2√61
   4√61
   8√50
   9√68
   10√52
   Option A
   let height and radius of cone be 5x and 6x.
   volume of cone =480π
   1/3*π*6x^2*5x=480π
   x^3=8
   x=2 cm
   height=10cm , radius=12 cm
   so, slant height of cone =√(r^2+h^2)
   = √(144+100)=√244=2√61

    

3. Circumference of a circle A is 2 times perimeter of a square. Area of the square is 484 cm^2. What is the area of another circle B whose radius is half the radius of the circle A ?
   530
   640
   590
   616
   none of these
   Option D
   Area of the square =484cm
   side of square =22cm.
   perimeter of square =22*4=88cm.
   ATQ,
   circumference of a circle A=88*2 =176cm.
   2πr =176
   r=176*7/44 =28
   Radius of circle B= 28/2 =14cm.
   Area of circle B=22/7*14*14 =616cm^2

    

4. If the length of a rectangular field is increased by 20% and the breadth is reduced by 20%, the area of the rectangle will be 288 m^2 . What is the area of the original rectangle ?
   400
   300
   520
   220
   none of these
   Option B
   Percentage change in area =
   +20-20-(20*20)/100 =-4%
   96%= 288
   100%=288/96*100= 300 m^2

    

5. Side of a rhombus is 20cm and the ratio of diagonals of a rhombus is 3:4, then find the area of the rhombus .
   384
   264
   544
   288
   484
   Option A
   Side of rhombus =20cm.
   Let,
   two diagonals of a rhombus be 3x and 4x.
   side =√(3x)^2+(4x)^2
   = 9x^2+16x^2 =5x
   5x=20
   x=4
   two diagonals =(4*3*2) and (4*4*2) =24 and 32
   Area =1/2*24*32 =384 cm^2.

    

6. The side of a square is 20% more than the side of an equilateral triangle whose perimeter is 75m. Find the ratio between perimeter of square to area of triangle.
   96:125√3
   48:55√3
   32:58
   96:130√3
   32:42
   Option A
   perimeter of triangle=75
   side of triangle=75/3=25m
   side of square=25*120/100=30 m
   perimeter of square=4*30=120 m
   area of triangle=√3/4*25*25=625√3/4 m^2
   ratio=120:625√3/4=480:625√3=96:125√3
   

    

7. Radius of circular field is 40% more than length of a rectangular field, having the area 80 m^2. If circumference of a circle is 88m, then radius of circular field is what percent more than breadth of rectangular field ?
   60
   50
   120
   75
   100
   Option D
   let radius of circle =7x
   and length of rectangular field=10m
   area of rectangular field=80 m^2
   10*breadth=80
   breadth=8m
   required percentage=14-8?8*100=6/8*100=75%

    

8. Perimeter of a circle is 88cm. Find volume of a cylindrical vessel whose radius is equal to the radius of circle and height of cylindrical vessel is 20 cm.
   15000
   12320
   14220
   16780
   20400
   Option B
   Perimeter of circle =88
   r=88*7/44=14 cm
   radius of cylindrical vessel =22/7*14*14*20 = 12320 cm^2

    

9. The ratio between length and breadth of a rectangular field is 5:4 . If area of rectangular field is 320 m^2, then find the perimeter of rectangular field.
   64
   40
   100
   72
   84
   Option D
   Let length of field =5x
   breadth of field =4x
   ATQ,
   5x*4x =320
   20x^2=320
   x^2 =320/20 =16
   x=4
   Perimeter =2(5x+4x) =18x =18*4= 72

    

10. The total area of a circle and a square is equal to 1241 cm^2. The diameter of the circle is 28cm. What is the sum of the circumference of the circle and the perimeter of the square ?
    132
    188
    192
    288
    196
    Option B
    diameter of circle=28cm
    radius=14cm
    area of circle=22/7*14*14=616
    ATQ,πr^2+a^2=1241
    a^2=1241-616=625
    a=25
    circumference of the circle=2*22/7*14=88
    perimeter of square=4a=4*25=100
    sum=88+100=188

     

Click here for topic wise Quantitative Aptitude Questions

1. A train running at the speed of 90 kmph passes a pole in 10 sec, in how much time the train can cross 350m long another train B, which running opposite direction with the speed of 180 kmph.
   10 sec
   8 sec
   12 sec
   15 sec
   20 sec
   Option B
   length of the train=90*5/18*10=250 m
   speed of train=90*5/18=25 m/sec
   speed of another train=180*5/18=50 m/sec
   required time=250+350/25+50=6000/75=8 sec

    

2. A bus can cover a certain distance in 5 hours and when same bus starts running with 20% less speed then it can cover 300 km in 5 hours. Find the distance of the journey.
   375
   400
   425
   500
   575
   Option A
   let the distance of the journey=D km
   original speed of the bus=5x
   decreased speed of the bus=4x
   4x=300/5
   x=15
   distance=15*5*5=375 km
   

    

3. A train running at the speed of 72 kmph cross pole in 18 sec. If same train cross a ‘x’ meters long platform in 30 sec, then find value of ‘x’ ?
   300
   360
   420
   240
   180
   Option D
   speed of train=72*5/18=20 m/sec
   length of train=72*5/18*18=360 m
   ATQ, (360+x)/20=30
   x=600-360=240 m
   length of platform=240 km

    

4. Train P having length of 200m crosses a platform twice of its length in 30 sec. Another train Q having speed of 180 kmph crosses a pole in 10 sec. Find time taken by both trains to cross each other when running in opposite direction.
   12 sec
   10 sec
   18 sec
   22 sec
   25 sec
   Option B
   speed of train P=200+400/30=600/30=20 m/sec
   length of train Q=180*5/18=50 m/sec
   required time=(500+200)/20+50=10 sec

    

5. Distance between A and B is 880 km and speed of car is 20% more than speed of a bike. If car starts from A towards B and at the same time bike starts from B towards A and they meet after 4 hours, then find total distance travelled by car.
   400
   480
   540
   600
   360
   Option B
   let speed of bike=5x
   speed of car=6x
   ATQ, 880/6x+5x=4
   880/11x=20 sec
   880/11x=4
   x=20 km
   speed of car=6*20=120 kmph
   distance travelled by car=120*4=480 km

    

6. ‘L’ meters of long train cross a 480 m long tunnel in 40 sec running at the speed of 108 kmph. If train cross a man running in opposite direction in 8 sec, find speed of man (in km/hour).
   108
   210
   110
   280
   300
   Option A
   speed of train=108*5/18=30 sec
   ATQ, L+480/40=30 sec
   L=1200-480=720
   let, the speed of man=’x’ sec
   720/30+x=12
   x=30m/sec
   sped of main =30*18/5=108 km/hr

    

7. A train can cross a tunnel of length 360m in 15sec, while a pole in 3 sec. In what time the train can cross a platform of length of 210 m.
   10 sec
   8 sec
   12 sec
   16 sec
   20 sec
   Option A
   let length of train be x meters
   (x+360)/15=x/3
   15x=3x+1080
   x=90 meters
   speed of train=90+360/15=450/15=30 m/sec
   required time=90+210/30=10 sec

    

8. Ritu goes to office at a speed of 7 kmph and returns to the home at a speed of 4 kmph. If he takes 5 hours 30 minutes in all, what is the distance between the office and his home ?
   24
   15
   18
   20
   14
   Option E
   let distance between office and Ritu’s home=D km
   D/7+D/4=11/2
   11D/28=11/2
   D=14 km

    

9. A train running at 11/15 of his own speed reached a place in 60 hours. How much time could be saved if the train would run at its own speed ?
   18
   16
   26
   18
   20
   Option B
   let the original speed of train=15x
   current speed of train=11x
   original time required=60*11x/15x=44
   time saved=60-44=16 hours

    

10. In covering a certain distance, the speed of A and B are in the ratio of 3:4. A takes 6 hours more than B to reach the destination. What is the time taken by A to reach the destination ?
    20
    32
    30
    24
    36
    Option D
    speed of A and B =3:4
    time ratio of A and B =4:3
    let time taken by A=4x
    time taken by=3x
    4x-3x=6
    x=6
    time take by A to reach the destination=4*6=24 hours

     

Click here for topic wise Quantitative Aptitude Questions

1. In how many different ways the letter of ‘MORBID’ can be arranged ?
   780
   720
   620
   680
   none of these
   Option B
   required no. of ways=6!=720

    

2. When three coins are tossed simultaneously, then find the probability of getting at least one head.
   7/8
   1/8
   3/10
   6/15
   5/8
   Option A
   required no. of outcome= 3C1+ 3C2+ 3C3
   =3+3+1=7
   total no. of outcome=2^3=8
   probability=7/8

    

3. A bag contains 3 red balls, 4 black balls and 7 green balls. 3 balls are drawn randomly, then find the probability of getting all balls are of green colours.
   5/52
   9/52
   11/52
   15/52
   17/52
   Option A
   required no. of outcome= 7C3
   total no. of outcome= 14C3
   probability=(7C3)/ 14C3
   =7*6*5/14*13*12=5/52

    

4. When two dice are thrown simultaneously, then find the probability that sum of the digits on both the faces is 8.
   9/37
   7/36
   3/37
   5/36
   11/36
   Option D
   total no. of outcome=6^2=36
   required no. of outcome=(3,5), (5,3), (6,2), (2,6), (4,4)=5
   probability=5/36

    

5. When two dice are thrown simultaneously, then find the probability that sum of the digits is a multiple of 4.
   1/4
   3/10
   9/10
   13/36
   17/36
   Option A
   total no. of outcome=6^2=36
   required no. of outcome=(4,8,12)
   4=(2,2), (3,1), (1,3)
   8=(3,5), (5,3), (6,2), (2,6), (4,4)
   12=(6,6)
   probability=(3+5+1)/36
   =9/36=1/4

    

6. If 2 cards are picked randomly from a pack of 52 cards, then find the probability of getting both queen cards.
   5/221
   1/221
   9/221
   37/221
   38/221
   Option B
   required probability= (4C2)/(52C2)
   =1/221

    

7. How many 4 digit numbers can be formal by using (0,9), if repetition is not allowed such that unit digit of the number will always be 3 ?
   628
   220
   548
   346
   448
   Option E
   required number of 4 digit number=8*8*7=448

    

8. Find the probability of selecting 2 green balls out of 4 green and 3 red balls.
   5/7
   6/7
   4/7
   2/7
   1/7
   Option D
   required probability=(4C2)/(7C2)
   4*3/7*6=2/7

    

9. A basket contains 4 blue balls, 3 red balls, 4 green balls and 5 green balls. If 4 balls are picked at randomly, then find the probability that 2 are blue and 2 are yellow ?
   4/91
   6/91
   3/91
   9/96
   11/96
   Option C
   required probability=(4C2)*(5C2)/(16C4)
   = (4*3/2*1)*(5*4/2*1)/(16*15*14*13)/(4*3*2*1)
   =3/91
   

    

10. A bag contains 4 yellow, 5 blue and 6 green balls. If 3 balls are drawn at randomly, then find the probability that all ball is not yellow ?
    452/455
    454/455
    6/355
    451/455
    2/455
    Option D
    required probability=(15C3)-(4C3)/15C3
    =1-(4*3*2)/(15*14*13)=451/455

     

Click here for topic wise Quantitative Aptitude Questions

1. A scooter and a bike were sold for rs. 9100 each. The scooter was sold at a loss of 30% and the bike at a gain of 30%. The entire transactions has resulted in how much amount loss?
   1800
   1400
   1100
   1600
   1700
   Option A
   Total SP on scooter and bike =9100+9100
   According to question,
   Total SP on both =9100/70*100+
   9100/130*100 =13000+7000=20,000
   The result in whole transaction is = 20,000-18,200=1800 loss.

    

2. The selling price of a book decreased by〖33〗_3^1 but the revenue generated by the sale of DVDs remained same. What is the percentage increase in the number of DVDs sold ?
   40%
   50%
   30%
   20%
   10%
   Option B
   Let the old SP and revised SP =3x and 2x
   Price * no. items =Revenue
   ATQ, revenue in both case is equal
   Then , Ratio of no. of times = 2.3
   revenue = 3x*2 =2x*3
   increased in no. of times =1/2*100 =50%

    

3. The cost price 20 articles is the same as the selling price of x articles . If the profit is 25%, then the value of x is,
   14 articles
   16 articles
   17 articles
   18 articles
   19 articles
   Option B
   CP of 20 articles = SP of x articles
   CP/SP= x/20
   profit= 20-x
   profit%= 20x-x/x*100
   25=20-x/x*100
   x=16 articles

    

4. A loss of 19% is converted to profit of 17%, when the selling price of articles is increased by rs. 360. Find the cost price of article.
   2000
   1000
   3000
   4000
   5000
   Option B
   Let CP of articel = 100
   Difference in % increased in profit and loss =+17-(-19) = 17+19 =36%
   36%=360
   CP=360/36*100= 1000

    

5. A person earns a profit of 20% by selling the pen after allowing a discount of 15% . If marked price of the pen is 600. Then find the cost price of the pen.
   324
   425
   556
   226
   110
   Option B
   CP= 600*85/100*100/120=rs 425

    

6. A man buys a rice packet for rs.3600. He sells 1/3rd at a loss of 20% and 2/5th at a loss of 25%. At what price must he sell the remaining rice so as make an overall profit of 10%?
   1400
   1200
   1800
   2000
   none of these
   Option B
   Total CP= rs.3600
   overall profit=10%
   total SP=3600*110/100=3960
   ATQ, in 1st case CP=1/3rd of 3600=1200
   loss=20%
   SP=960
   in 2nd case CP=2/5th 3600=1440
   profit=25%
   SP=1440*125/100=1800
   total SP in both cases=1800+960=2760
   remaining SP=3960-2760=1200

    

7. A shopkeeper marked MRP of a product after giving 10% discount, he earns 20% profit. What will be his profit or loss percent if he allows a discount of 15%?
   40/3
   25/3
   20/3
   10/3
   none of these
   Option A
   Let MP be rs.100x
   SP=rs.90x
   CP=90x*100/120=75x
   new SP=100-15=85x
   profit%=10/75*100=40/3%

    

8. Sriman sold a table for rs.950. If he had sold it for rs.50 more, the selling price would be 25% more than cost price, what is the profit percent when article is sold for rs.200 more than original selling price?
   40
   38
   42.75
   43.75
   50
   Option D
   

    

9. A book vendor sold a book at a loss of 10%. Had he sold it for rs.108 more, he would have earned a profit of 10%. Find the CP of book.
   540
   630
   600
   580
   none of these
   Option A
   Let CP of book=100x
   loss=10%
   SP=90x
   when,profit 10%, SP=110x
   difference in SP=110x-90x=20x
   CP=108/20x*100x=rs.540
   

    

10. One trader calculates the percentage of profit of a ticket on the buying price and another calculates on the selling price. When their selling price are the same, then difference of their actual profit is rs.80 and both claim to have made 20%profit. What is the selling price of each.
    2000
    3200
    3400
    2800
    2400
    Option E
    Let CP of ticket in both cases=100x
    profit=20%
    in 1st case CP=100x
    when profit 20% on CP, profit=20x
    SP=120x
    in 2nd case, CP=100x
    SP=120x
    when profit 20% on SP, profit=24x
    difference of profit in both cases,4x=80
    SP of ticket=80/4x*120x=2400
    

     

Click here for topic wise Quantitative Aptitude Questions

1. 4 2 3 7.5 ? 118.125
   26.25
   38.75
   42.25
   28.75
   none of these
   Option A
   Series Pattern Given Series
   4 4
   4 × 0.5 = 2 2
   2 × 1.5 = 3 3
   3 × 2.5 = 7.5 7.5
   7.5 × 3.5 = 26.25 26.25 ✓
   26.25 × 4.5 = 118.125 118.125

   Hence, option (A) is correct.

    

2. 256 252 279 263 388 ?
   341
   352
   323
   336
   none of these
   Option B

   Series Pattern Given Series
   256 256
   256 – 22 = 252 252
   252 + 33 = 279 279
   279 – 42 = 263 263
   263 + 53 = 388 388
   388 – 62 = 352 352 ✓

   Hence, option (B) is correct.

    

3. 
   188 269 333 382 418 ?
   181
   126
   443
   315
   none of these
   Option C
   Series Pattern Given Series
   188 188
   188 + (9)2 = 269 269
   269 + (8)2 = 333 333
   333 + (7)2 = 382 382
   382 + (6)2 = 418 418
   418 + (5)2 = 443 443 ✓

   Hence, option (C) is correct.

    

4. 420, 462, 506, 552, 600, ?
   725
   800
   350
   650
   675
   Option D
   20*21=420
   21*22=462
   22*23=506
   23*24=552
   24*25=600
   25*26=650
   Hence, option (D) is correct.

    

5. 40 44 97 307 1253 ?
   6301
   6201
   6401
   6403
   6501
   Option A
   Correct Option: A
   The series is × 1 + 2*2, × 2 + 3*3, × 3 + 4*4, × 4 + 5*5, × 5 + 6*6, × 6 + 7*7
   

    

6. 129 163 202 246 ? 349
   265
   285
   295
   275
   None of these
   Option C
   Correct Option: C
   The series is + (5 × 7) – 1, + (5 × 8) – 1, + (5 × 9) – 1, + (5 × 10) – 1, + (5 × 11) – 1, ….

    

7. 3 13 39 89 ? 293
   172
   171
   174
   181
   183
   Option B
   Series Pattern Given Series
   3 3
   3 + (32 + 1) = 13 13
   13 + (52 + 1) = 39 39
   39 + (72 + 1) = 89 89
   89 + (92 + 1) = 171 171 ✓
   171 + (112 + 1) = 293 293
   Hence, option B is correct.
   

    

8. 3, 2, 3, 6, ?, 37.5, 115.5
   11
   14
   16
   18
   none of these
   Option B
   Correct Option: B
   The series is ×0.5+0.5, ×1+1, ×1.5+1.5, *2+2, *2.5+2.5….

    

9. 567 126 36 14.4 9.6 ?
   16.5
   17.9
   18.1
   12.6
   none of these
   Option E
   Correct Option: E
   Series Pattern Series
   567 567
   567 ÷ 4.5 126
   126 ÷ 3.5 36
   36 ÷ 2.5 14.4
   14.4 ÷ 1.5 9.6
   9.6 ÷ .5 19.2 ✓

    

10. 441 428 407 370 ? 168
    288
    291
    301
    311
    245
    Option C
    Correct Option: C
    Series I : 441 428 407 370 ? 168
    Sereis II : –13 –21 –37 ? ?
    Series III : 8 16 32 ?
    Series IV ×2 ×2 ×2
    Clearly, the pattern in series IV is multiplication of 2.

    So, the missing term in series III = 32 × 2 = 64

    ∴ missing term in series II = – 37 – 32 = – 69, – 69 – 64 = – 133
    ∴ missing term in series I = 370 – 69 = 301

     

1. 1366 + 336 – 666 – 33 + 333 = ?
   1336
   1136
   1236
   1363
   none of these
   Option A
   Correct Option: A
   1366 + 336 – 666 – 33 + 333 = ?

   ⇒ 1336 = ?

   Hence, option A is correct

    

2. ? % of (19154 – 15644) = 81 + 6*6
   5
   3
   10/3
   20/3
   6
   Option C
   ?% of 3510 = 117

   or, (? × 3510 / 117) =100

   or, ? = 100 / 30 = 10 / 3

   Hence, option C is correct.

    

3. 
   √(1686 – 317) + √(3440 – 524) = ?
   92
   95
   91
   99
   none of these
   Option C
   √(1686 – 317) + √(3440 – 524) = ?

   √1369 + √2916 = ?

   ? = 37 + 54 = 91

   Hence, option C is correct

    

4. 64% of √409600 ÷ 1.6 = ? × 2.56
   64
   81
   10
   100
   110
   Option D
   64% of √409600 ÷ 1.6 = ? × 2.56

   ? × 2.56 = 64% of 640 ÷ 1.6

   ? × 2.56 = (64 × 640 / 100) ÷ 1.6 = (64 × 640) / 100 × 1.6) = 256

   ∴ ? = (256/2.56) = 100

   Hence, option D is correct.

    

5. 
   10*0.5 × √? = 735 ÷ 3
   7
   49
   2401
   97
   68
   Option C
   10*0.5 × √? = 735 ÷ 3

   5 × √? = 245

   √? = 49
   ? = 2401
   Hence, option C is correct
   

    

6. 534.255 – 786.55 + 678.45 + 45.324 – 23.122 = ?
   448.348
   438.438
   443.838
   438.448
   none of these
   Option A
   534.255 – 786.55 + 678.45 + 45.324 – 23.122 = 448.348

   Hence, option (A) is correct.

    

7. 221 ÷ 23.6 × 94.4 ÷ 169 ÷ 17 = ? ÷ 91
   10
   28
   14
   18
   17
   Option B
   221 ÷ 23.6 × 94.4 ÷ 169 ÷ 17 = ? ÷ 91

   (221/23.6) × {94.4/ (169 × 17)} × 91 = ?

   (4/13) × 91 = ?

   ? = 28

   Hence, option B is correct.

    

8. 2 + [ 2 – { 2 – 2 ( 2 + 2 ) } ] =?
   10
   -10
   6
   -6
   none of these
   Option A
   Given expression = 2 + [2 – {2 – 2 x 4 }]

   ⇒ 2 + [ 2 – {2 – 8}] ⇒ 2 + [ 2 – (–6)]

   ⇒ 2 + [2 + 6] ⇒ 2 + 8

   ⇒ 10.

   Hence, option A is correct.

    

9. 
   100 × 10 – 100 – 2000 ÷ 100 = ?
   290
   980
   880
   780
   740
   Option C
   Given expression = 100 x 10 – 100 – 20

   ⇒ 1000 – 120

   ⇒ 880

   Hence, option C is correct.

    

10. 85% of 620 + x% of 480 = 70% of 890
    10%
    20%
    33%
    45%
    64%
    Option B
    85% of 620 + x% of 480 = 70% of 890

    527 + x × 480 ÷ 100 = 623

    x × 4.8 = 623 – 527

    x × 4.8 = 96

    x = 20%

    Hence, option B is correct

     

1. If 45% of P is equal to 60% of Q, then find 9P is what percent of 12Q ?
   100%
   50%
   80%
   90%
   120%
   Option A
   solution: 45*P=60*Q
   P/Q=60/45=4/3
   percentage=(9*4/12*3)*100=100%

    

2. Two numbers are respectively 20% and 50% more than the 3rd number. What percent is the 1st number of the 2nd number ?
   90%
   85%
   80%
   60%
   100%
   Option C
   solution: let the 3rd number =100
   1st number=120
   2nd number=150
   % required = 120/150*100=80%

    

3. A man spends 50% of his monthly salary in food and rent, 20% of remaining spends education of children and 25% of the remaining in entertainment. Find his yearly salary if he s left with rs.4200 in end of monthly as savings.
   144000
   168000
   280000
   180000
   240000
   Option B
   solution: let his monthly salary =x
   x*50/100*80/100*75/100=4200
   x=14000
   his yearly salary=12*14000=168000

    

4. The population of a village is increased by 20% in 2019 and decreased by 25% in 2020. Find the population of village in the end of 2020, if the population of village in 2018 is 9600.
   7500
   8500
   9000
   8640
   none of these
   Option D
   solution: the population of village in 2020=9600*120/100*75/100=8640

    

5. The length of a rectangle is 20cm and breadth is 5cm. The length decreases by 10% and the breadth deacreses by 20%, what is % change in area.
   -28
   -30
   -32
   -40
   none of these
   Option A
   solution:Now percentage change in area=[-x-y+xy/100]%
   =-10-20+10*20/100]%
   =-28%

    

6. 720 sweets were distributed equally amongst children in such a way that number of sweets received by each child is 20% of the total number of children. How much sweets did each child received ?
   18
   15
   12
   24
   30
   Option C
   solution: let the total number of children be x
   x*20/100*x=3600
   x=60
   number of sweets received by each child=720/60=12

    

7. Minu scored 35% marks and failed by 60 marks. If she scored 55% marks and she passes by 40 marks. Find by how much marks would she fail or pass if she scored 49% marks ?
   10 marks , fail
   10 marks ,pass
   12 marks, pass
   20 marks, fail
   none of these
   Option B
   solution: 35%+60=55%-40
   20%=100
   pass mark=(100/20*35)+60
   =175+60=235
   49% marks =245
   she will pass by (245-235)=10 marks
   

    

8. An amount of rs.44000 is divided among P, Q & R. The share of R is 10% more than share of P, while share of P is 20% more than share of Q, then find the sum of share of P & R ?
   37500
   35000
   30000
   31500
   32500
   Option D
   solution: let share of Q=100
   share of P=100*120/100=120
   share of R=120*110/100=132
   Ratio f P:Q:R= 120:100:132
   =30:25:33
   ATQ, 88x=44000
   x=500
   required= (30x+33x)=63x
   =63*500=31500

    

9. A fraction is such that it we increase the numerator by 20% and decrease the denominator by 10%, we get 5/9 as a fraction, then find the original fraction.
   6/13
   4/9
   9/12
   5/12
   none of these
   Option D
   solution: (x*120/100)/(y*90/100)=5/9
   x/y=5/12
   

    

10. If Rahul’s salary is 30% more than that of Ram’s salary, then how much percent is salary of Ram less than that of Rahul ?
    18.67%
    22.87%
    23.07%
    24%
    25%
    Option C
    solutions: let Ram’s salary =100
    Rahul’s salary=130
    required %= 30/130*100
    =23.07%
    

     

Click here for topic wise Quantitative Aptitude Questions

1. 3.20% of 500 × 2.40 % of x = 288.Find the value of x?
   750
   800
   820
   715
   None of these
   Option A
   =16 * 2.40% of x = 288
   x = 750

    

2. (–251 × 21.00 × (– 12)) ÷ x = 158.130.Find the value of x?
   350
   400
   450
   375
   None of these
   Option B
   63252 / x = 158.130
   x= 400

    

3. [(𝟏𝟑𝟎) * 𝟐 ÷ 𝟐𝟓.0 × 𝟏𝟓.0] ÷ 𝟑𝟎 = x.Find the value of x?
   428
   335
   338
   425
   None of these
   Option C
   

    

4. √√4900.0 + √5476.0= x.Find the value of x?
   14
   16
   18
   24
   12
   Option E
   √70 + 74 = √144 = 12

    

5. (6.50 % of 375) – (0.850 % of 230) =x.Find the value of x?
   22.42
   21.00
   23.50
   24.00
   25.55
   Option A
   

    

6. 17% oof 400 + 23% of 900 + 17% of 500 = x.Find the value of x?
   360
   370
   385
   390
   410
   Option A
   = 17 * 4 + 23 * 9 + 17 * 5
   =68 +207 + 85

    

7. 333 / 300.0 + 7777 / 7000 = 6600.0 / 4400.0 + x.Find the value of x?
   0.710
   0.721
   0.715
   0.814
   0.725
   Option B
   1.11 + 1.111 – 1.50

    

8. 2352 + 1545 + 5154 – 5415 = x.Find the value of x?
   3854
   3636
   3745
   3650
   None of these
   Option B
   =9051 – 5415 = 3636

    

9. 2197.0 / 13 + 242.0 / 11 + 1024.0 / 8 = x.Find the value of x?
   325
   319
   330
   345
   None of these
   Option B
   = 169 + 22 +128

    

10. ( 17% * 4600) / (23% of 8500 ) =x.Find the value of x?
    4/7
    3/7
    2/5
    1/5
    1/3
    Option C
    

     

Pie-chart given below shows number of persons come to watch films in five different malls. Study the data and answer the following questions:(Total=4400)

1. If ratio of boys and girls come in Sahu theatre to watch movies is 4 : 7. Then find the difference between number of boys to number of girls come in Sahu theatre to watch movie?
   366
   355
   480
   336
   None of these
   Option D
   Total person to Sahu=28/100 * 4400 =1232
   Therefore ,Required value=(7-4)/11 * 1232 = 336

    

2. PVR theatre shows 3 films.25% of persons watch Partner, 100/3% of person watch pitamber and remaining watch aryavrat.Find the average number of persons who watch Partner and Aryavrat?
   366
   380
   452
   441
   352
   Option E
   Total person to PVR=1056
   Value=1/200 * 200 / 3 * 1056 = 352

    

3. 50/3% of the person came in Phoenix bought pasta,200/3% of the person bought coldrink and remainng persons bought both the things.Find the total number of persons who bought almost one thing?
   660
   650
   710
   800
   None of these
   Option A
   Total to Phoenix=792
   Value=(50/3 + 200/3) * 792/100
   =660

    

4. Find the difference between the person who visit PVR and Phoenix theatre together to the person who visit BigE & Wave theatre together?
   505
   515
   530
   528
   None of these
   Option D
   Difference=12/100 * 4400

    

5. The person came to Phoenix is what percent of the person who come to PVR?
   78%
   82%
   75%
   64%
   60%
   Option C
   
   =18/24 * 100

    

The following table given below show the number of mobiles sold by five shopkeepers and ratio of Apple mobile sold to Samsung mobile sold. Study the data and answer the following questions.



6. Apple mobile sold by A is what percent less than Apple mobile sold by B?
   36%
   38%
   42%
   44%
   52%
   Option A
   Apple mobile sold by A=2376 * 6/11 =1296
   Apple mobile sold by B=3150 * 9/14=2025
   Perentage=2025-1296 / 1296 *100 = 36%

    

7. Find the average number of Samsung mobile sold by B, C and D all together?
   1675
   1805
   1755
   1860
   None of these
   Option A
   

    

8. Find the ratio of Apple mobile sold by ‘C’ to Apple mobile sold by ‘E’ ?
   15:22
   63:65
   65:56
   67:71
   None of these
   Option C
   =1300 / 1120

    

9. Samsung mobile sold by ‘D’ and ‘E’ together is how much more than Samsung mobile sold by ‘B’ and ‘C’ together?
   3455
   3500
   3650
   3755
   None of these
   Option A
   Samsung mobile by D and E together=5280 * 13/22 + 3360 * 2/3= 5360
   Samsung mobile by B and C together=1905
   Difference=5360 – 1905 = 3455

    

10. Samsung mobile sold by C is what percent of the Apple mobile sold by B?
    61 1/2%
    63 1 /5%
    72 1/4%
    67 1/2%
    69 1/3%
    Option E
    Samsung mobile sold by C=2080 * 3/8=780
    Samsung mobile sold by B=3150 * 5/14=1125
    
    Percentage=780/1125 * 100