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a ) 70 / 8 , b ) 70 / 7 , c ) 70 / 5 , d ) 70 / 3 , e ) 70 / 1 | d | divide(200, multiply(add(36, 72), const_0_2778)) | two trains of length 200 m and 300 m are 200 m apart . they start moving towards each other on parallel tracks , at speeds 36 kmph and 72 kmph . in how much time will the trains cross each other ? | "relative speed = ( 36 + 72 ) * 5 / 18 = 6 * 5 = 30 mps . the time required = d / s = ( 200 + 200 + 300 ) / 30 = 700 / 30 = 70 / 3 sec . answer : d" | a = 36 + 72
b = a * const_0_2778
c = 200 / b
|
a ) 12028 , b ) 12000 , c ) 12019 , d ) 16000 , e ) 12012 | d | divide(1200, divide(5, const_100)) | a salesman β s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1200 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 1200 more than that by the previous schema , his sales were worth ? | "[ 1200 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 16000 answer : d" | a = 5 / 100
b = 1200 / a
|
a ) 600 , b ) 700 , c ) 120 , d ) 900 , e ) none | c | multiply(add(15, 5), 6) | i chose a number and divide it by 6 . then i subtracted 15 from the result and got 5 . what was the number i chose ? | "solution : let xx be the number i chose , then x / 6 Γ’ Λ β 15 = 5 x / 6 = 20 x = 120 answer c" | a = 15 + 5
b = a * 6
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a ) 88 , b ) 108 , c ) 110 , d ) 90 , e ) 117 | e | subtract(subtract(subtract(182, 22), subtract(44, const_1)), const_4) | veena ranks 44 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | "total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 44 + 1 = 117 answer : e" | a = 182 - 22
b = 44 - 1
c = a - b
d = c - 4
|
a ) 4 : 1 , b ) 6 : 1 , c ) 1 : 3 , d ) 2 : 1 , e ) 4 : 1 | c | divide(subtract(divide(5, add(3, 6)), divide(3, 6)), subtract(divide(3, 6), divide(3, add(3, 6)))) | an alloy of copper and zinc contains copper and zinc in the ratio 5 : 4 . another alloy of copper and zinc contains copper and zinc in the ratio 3 : 6 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 5 x / 9 , and fraction of zinc in alloy _ 1 = 4 x / 9 . similarly , fraction of copper in alloy _ 2 = 3 y / 9 , and fraction of zinc in alloy _ 2 = 6 y / 9 . mixing them , we get copper = 5 x / 9 + 3 y / 9 ; zinc = 4 x / 9 + 6 y / 9 so , 5 x + 3 y = 4 x + 6 y - > x = 3 y - > x / y = 1 / 3 so , they must be mixed in the ratio 1 : 3 answer : c | a = 3 + 6
b = 5 / a
c = 3 / 6
d = b - c
e = 3 / 6
f = 3 + 6
g = 3 / f
h = e - g
i = d / h
|
a ) 287 cm 2 , b ) 887 cm 2 , c ) 364 cm 2 , d ) 368 cm 2 , e ) 668 cm 2 | c | multiply(26, 14) | find the area of a parallelogram with base 26 cm and height 14 cm ? | "area of a parallelogram = base * height = 26 * 14 = 364 cm 2 answer : c" | a = 26 * 14
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a ) 118 , b ) 677 , c ) 700 , d ) 2688 , e ) 1991 | c | divide(532, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 532 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 532 cp = 7 * 100 = > cp = 700 answer : c" | a = 24 / 100
b = 1 - a
c = 532 / b
|
a ) 40 , b ) 45 , c ) 108 , d ) 55 , e ) 60 | c | multiply(divide(divide(add(divide(120, const_2), 120), 25), const_4), divide(120, const_2)) | a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 25 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | "a - b = 120 miles b - c = 60 miles avg speed = 25 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 25 = 180 / 4 t t = 108 b - c = 108 mph answer c" | a = 120 / 2
b = a + 120
c = b / 25
d = c / 4
e = 120 / 2
f = d * e
|
a ) 4 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 44.5 meters . how many times did the ball bounce ? | "division of total diatance travelled will be 16 + 16 + 8 + 4 + 0.5 ans : 4" | a = 16 / 2
b = a / 2
|
a ) 54 , b ) 55 , c ) 59 , d ) 71 , e ) 92 | c | add(37, multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 37))) | of 70 players on a cricket team , 37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 90 thrower = 37 rest = 90 - 37 = 33 left handed = 33 / 3 = 11 right handed = 22 if all thrower are right handed then total right handed is 37 + 22 = 59 so c . 59 is the right answer" | a = 1 / 3
b = 1 - a
c = 70 - 37
d = b * c
e = 37 + d
|
a ) 675 , rs . 3375 , b ) 575 , rs . 1350 , c ) 1350 , rs . 675 , d ) 1450 , rs . 775 , e ) 1550 , rs . 875 | a | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 5))), rectangle_area(75, 55)), 5) | a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 5 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 5 = rs . 3375 answer : a" | a = 2 * 5
b = 75 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | c | multiply(divide(divide(100, 4), 100), const_2.0) | an integer n between 1 and 100 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 4 ? | "n ( n + 1 ) to be divisible by 4 either n or n + 1 must be a multiples of 4 . in each following group of numbers : { 1 , 2 , 3 , 4 } , { 5 , 6 , 7 , 8 } , . . . , { 97 , 98 , 99 , 100 } there are exactly 2 numbers out of 3 satisfying the above condition . for example in { 1 , 2 , 3 , 4 } n can be : 3 , or 4 . thus , the overall probability is 2 / 4 = 1 / 2 . answer : c ." | a = 100 / 4
b = a / 100
c = b * 2
|
a ) 23 , b ) 25 , c ) 27 , d ) 31 , e ) 35 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 and the sum of their squares is 289 . the sum of the number is ? | let the numbers be x and y . then , xy = 120 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 120 ) = 529 x + y = 529 = 23 . option a | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | c | add(add(const_2, add(const_1, const_4)), const_2) | an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 100000 people who applied for the army and 1400 were finally selected , what was the value of n ? | "since 70 % is accepted and figure will be 1400 therefore in last stage tested / interviewed no . candidates should be 2000 so from here i will work for those two steps of decrements of 100000 candidates . fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 16000 = 8000 step ( 4 ) 50 % of 8000 = 4000 step ( 5 ) 50 % of 4000 = 2000 . . . here we are done with total 5 steps and last step of accepting 70 % of 2000 = 1400 ( our target ) total 6 steps required . answer : c" | a = 1 + 4
b = 2 + a
c = b + 2
|
a ) 10 mps , b ) 59 mps , c ) 79 mps , d ) 90 mps , e ) 19 mps | d | multiply(const_0_2778, 342) | express a speed of 342 kmph in meters per second ? | "342 * 5 / 18 = 90 mps answer : d" | a = const_0_2778 * 342
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | divide(factorial(subtract(add(const_4, 3), const_1)), multiply(factorial(3), factorial(subtract(const_4, const_1)))) | how many positive integers less than 50 are multiples of 3 but not multiples of 5 ? | "imo the answer is c ( 8 numbers ) the lcm of 3 and 5 is 15 . if x < 50 and x is divisible by 3 not by 5 - - > x is not divisible by 15 . from 1 - - > 50 , we have 3 numbers which is divisible by 15 : 15 , 30 , 45 . from 1 - - > 50 , we have ( 45 - 3 ) / 3 + 1 = 15 numbers divisible by 3 . therefore , our answer is 15 - 3 = 12 numbers . e" | a = 4 + 3
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) rs . 292.50 , b ) rs . 297.50 , c ) rs . 298 , d ) rs . 298.50 , e ) none of these | a | add(multiply(8, 30), divide(add(multiply(8, 30), multiply(9, 20)), 8)) | out of 9 persons , 8 persons spent rs . 30 each for their meals . the ninth one spent rs . 20 more than the average expenditure of all the 9 . the total money spent by all of them was | explanation : let the average expenditure be rs . x then , 9 x = [ 8 Γ 30 + ( x + 20 ) ] β 9 x = x + 260 β x = 32.50 total money spent = 9 x = rs . ( 9 x 32.5 o ) = rs 292 . 50 answer : a | a = 8 * 30
b = 8 * 30
c = 9 * 20
d = b + c
e = d / 8
f = a + e
|
a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | c | divide(multiply(36, 5), subtract(const_100, 10)) | a driver just filled the car ' s gas tank with 36 liters of gasohol , a mixture consisting of 5 % ethanol and 95 % gasoline . if the car runs best on a mixture consisting of 10 % ethanol and 90 % gasoline , how many liters of ethanol must be added into the gas tank for the car to achieve optimum performance ? | "let x be the number of liters of ethanol added to the gas tank . 0.05 ( 36 ) + x = 0.1 ( 36 + x ) 0.9 x = 3.6 - 1.8 = 1.8 x = 2 liters the answer is c ." | a = 36 * 5
b = 100 - 10
c = a / b
|
['a ) 4 : 1', 'b ) 2 : 1', 'c ) 7 : 1', 'd ) 5 : 1', 'e ) 3 : 1'] | e | divide(volume_cylinder(const_1, const_1), volume_cone(const_1, const_1)) | a cylinder and a cone have the same height and same radius of base . the ratio between the volumes of the cylinder and the cone is : | volume of the cylinder = Ο r ( power 2 ) h volume of the cone = 1 / 3 Ο r ( power 2 ) h ( Ο r ( power 2 ) h ) / ( 1 / 3 Ο r ( power 2 ) h ) = 3 / 1 = 3 : 1 answer is e . | a = volume_cylinder / (
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a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2 | b | add(subtract(4, 2), divide(const_1, add(2, 3))) | when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in f hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of f ? | rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * f = 1 = > x = 1 / f when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * 3 / 4 = 1 = > w + x = 4 / 3 = > 1 / 2 + 1 / f = 4 / 3 = > 1 / f = ( 8 - 3 ) / 6 = 5 / 6 = > f = 6 / 5 = 1 [ 1 / 5 ] answer b | a = 4 - 2
b = 2 + 3
c = 1 / b
d = a + c
|
a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 900 , e ) $ 950 | b | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,054 . how much did she invest at 5 % ? | et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1054 / 1000 ) - 1 = 0.054 = 5.4 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of 1000 invested at 5 % ) 0.05 x + 0.06 ( 1 - x ) = 0.054 ( 0.05 ) x + 0.06 - ( 0.06 ) x = 0.054 - 0.01 x = - 0.006 x = - 0.006 / - 0.01 = 6 / 10 = 60 % so x = 60 % of the 1000 which is 600 answer : b | a = 5 * 2
b = a * 100
c = 5 * 2
d = c * 100
e = 5 * 2
f = e * 5
g = f + 4
h = d + g
i = 5 * 2
j = i * 100
k = h - j
l = k * 100
m = 5 * 2
n = m * 100
o = 5 * n
p = l - o
q = b - p
|
a ) 5 % , b ) 6 % , c ) 3 % , d ) 9 % , e ) 1 % | b | divide(multiply(const_100, 192), multiply(800, 4)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 192 in 4 years ? | "192 = ( 800 * 4 * r ) / 100 r = 6 % answer : b" | a = 100 * 192
b = 800 * 4
c = a / b
|
a ) 17 , b ) 18 , c ) 22 , d ) 26 , e ) 36 | b | divide(add(subtract(multiply(16, 10), 25), 45), 10) | the average of 10 numbers is calculated as 16 . it is discovered later on that while calculating the average , the number 45 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 20 . then the average will increase by 20 / 10 = 2 . the correct average is 18 . the answer is b ." | a = 16 * 10
b = a - 25
c = b + 45
d = c / 10
|
a ) 40 , b ) 60 , c ) 120 , d ) 80 , e ) 105 | e | multiply(70, sqrt(divide(9, 4))) | two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 70 kmph , find the rate at which the train b runs . | "if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in β a β and β b β hours respectively ( i . e . a takes β a hrs β to travel from the meeting point to his destination and b takes β b hrs β to travel from the meeting point to his destination ) , then the ratio of their speeds is given by : sa / sb = β ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = β ( 4 / 9 ) = 2 / 3 this gives us that the ratio of the speed of a : speed of b as 2 : 3 . since speed of a is 80 kmph , speed of b must be 70 * ( 3 / 2 ) = 105 kmph e" | a = 9 / 4
b = math.sqrt(a)
c = 70 * b
|
a ) 2351 , b ) 2451 , c ) 2551 , d ) 2651 , e ) 2751 | b | multiply(4673, divide(7528, 5306)) | 7528 : 5306 : : 4673 : ? | "as there is a difference of 2222 . 7528 - 2222 = 5306 . so 4673 - 2222 = 2451 answer : b" | a = 7528 / 5306
b = 4673 * a
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a ) 23 % , b ) 17 % , c ) 11 % , d ) 10 % , e ) 20 % | e | subtract(const_100, add(add(add(subtract(const_100, 70), subtract(const_100, 75)), subtract(const_100, 90)), subtract(const_100, 85))) | in a urban village of india named ` ` owlna ' ' , 70 % people have refrigerator , 75 % people have television , 90 % people got computers and 85 % got air - conditionor . how many people ( minimum ) got all these luxury . | "e 10 % 100 - [ ( 100 - 85 ) + ( 100 - 90 ) + ( 100 - 75 ) + ( 100 - 70 ) ] = 100 - ( 30 + 25 + 10 + 15 ) = 100 - 80" | a = 100 - 70
b = 100 - 75
c = a + b
d = 100 - 90
e = c + d
f = 100 - 85
g = e + f
h = 100 - g
|
a ) 23 % , b ) 24 % , c ) 25 % , d ) 26 % , e ) 27 % | c | multiply(divide(divide(20, const_100), subtract(const_1, divide(20, const_100))), const_100) | a merchant has selected two items to be placed on sale , one of which currently sells for 20 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | expensive item = $ 100 ; cheap item = $ 80 ; we must increase $ 80 to $ 100 , so by $ 20 , which is 25 % increase : ( 100 - 80 ) / 80 = 1 / 4 = 0.25 answer : c . | a = 20 / 100
b = 20 / 100
c = 1 - b
d = a / c
e = d * 100
|
a ) 5 : 1 , b ) 3 : 1 , c ) 5 : 8 , d ) 3 : 2 , e ) 8 : 3 | a | divide(5, divide(5, 5)) | what is the ratio between perimeters of two squares one having 5 times the diagonal then the other ? | "d = 5 d d = d a β 2 = 5 d a β 2 = d a = 5 d / β 2 a = d / β 2 = > 5 : 1 answer : a" | a = 5 / 5
b = 5 / a
|
a ) 1730 cm 2 , b ) 1728 cm 2 , c ) 1870 cm 2 , d ) 1890 cm 2 , e ) 668 cm 2 | b | multiply(48, 36) | find the area of a parallelogram with base 48 cm and height 36 cm ? | "area of a parallelogram = base * height = 48 * 36 = 1728 cm 2 answer : b" | a = 48 * 36
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a ) 3.75 days , b ) 4.75 days , c ) 3.25 days , d ) 3.15 days , e ) 2.75 days | b | inverse(add(inverse(3), inverse(9))) | a and b complete a work in 3 days . a alone can do it in 9 days . if both together can do the work in how many days ? | "1 / 3 + 1 / 9 = 4 / 9 9 / 4 = 4.75 days answer : b" | a = 1/(3)
b = 1/(9)
c = a + b
d = 1/(c)
|
a ) 1100 , b ) 1200 , c ) 1235 , d ) 1354 , e ) 1140 | e | divide(multiply(choose(10, 8), choose(10, 5)), 10) | a question paper has 2 parts , a & b , each containing 10 questions . if a student has to choose 8 from part a & 5 from part b , in how many ways can he choose the questions ? | there 10 questions in part a out of which 8 question can be chosen as = 10 c 8 . similarly , 5 questions can be chosen from 10 questions of part b as = 10 c 5 . hence , total number of ways , = 10 c 8 * 10 c 5 = [ 10 ! / ( 2 ! 8 ! ) ] * [ 10 ! / ( 5 ! * 5 ) ] = { 10 * 9 / 2 } * { 10 * 9 * 8 * 7 * 6 / ( 5 * 4 * 3 * 2 * 1 ) } = 1140 . e | a = math.comb(10, 8)
b = math.comb(10, 5)
c = a * b
d = c / 10
|
a ) 18 , b ) 29 , c ) 39 , d ) 49 , e ) 59 | b | divide(subtract(multiply(3, 15), multiply(4, 4)), subtract(4, 3)) | what must be added to each term of the ratio 4 : 15 , so that it may become equal to 3 : 4 ? | "let x be added to each term of the ratio 4 : 15 . new ratio becomes ( 4 + x ) / ( 15 + x ) , which is given to be equal to 3 : 4 . this means , ( 4 + x ) / ( 15 + x ) = 3 / 4 solving we get , x = 29 . ans b" | a = 3 * 15
b = 4 * 4
c = a - b
d = 4 - 3
e = c / d
|
a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 11 | d | add(add(divide(add(multiply(11, 10), const_1), 111), const_3), const_3) | what is the greatest prime factor of ( 11 ! Γ 10 ! β + 10 ! Γ 9 ! ) / 111 ? | ( 11 ! Γ 10 ! β + 10 ! Γ 9 ! ) / 111 in numerator take out 10 ! * 9 ! as common term 10 ! β 9 ! ( 11 Γ 10 β + 1 ) / 111 10 ! β 9 ! β 111 / 111 10 ! β 9 ! 10 ! has 7 as the greatest prime factor , so ans is 7 answer : d | a = 11 * 10
b = a + 1
c = b / 111
d = c + 3
e = d + 3
|
a ) 4000 , b ) 5000 , c ) 2000 , d ) 1000 , e ) 6000 | c | multiply(multiply(subtract(4, 3), 1000), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 1000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 1000 x = 1000 . b ' s share = rs . 2 x = rs . ( 2 x 1000 ) = rs . 2000 . answer is c ." | a = 4 - 3
b = a * 1000
c = b * 3
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a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | d | divide(factorial(11), multiply(factorial(subtract(11, const_2)), factorial(const_2))) | there are 11 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "11 c 2 = 55 the answer is d ." | a = math.factorial(11)
b = 11 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23 | b | divide(add(multiply(10, 17), subtract(56, 26)), 10) | the average of 10 numbers was calculated as 17 . it is discovered later on that while calculating the average , one number , namely 56 , was incorrectly read as 26 . what is the correct average ? | 10 * 17 - 26 + 56 = 200 200 / 10 = 20 the answer is b . | a = 10 * 17
b = 56 - 26
c = a + b
d = c / 10
|
a ) 1 , b ) 8 , c ) 5 , d ) 6 , e ) 7 | b | subtract(multiply(subtract(15, 1), add(6, 1)), multiply(15, 6)) | the average of 6 observations is 15 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 90 + x ) / 7 = 14 = > x = 8 hence , the seventh observation is 8 answer : b" | a = 15 - 1
b = 6 + 1
c = a * b
d = 15 * 6
e = c - d
|
a ) 1 / 7 , b ) 1 / 5 , c ) 3 / 14 , d ) 2 / 7 , e ) 1 / 2 | c | divide(choose(4, 4), choose(add(const_2.0, 4), const_2)) | a bag holds 4 red marbles and 4 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "probability of selecting first red marble = 4 / 8 probability of selecting second red marble without replacement = 3 / 7 final probability = 4 / 8 * 3 / 7 = 3 / 14 the correct answer is c ." | a = math.comb(4, 4)
b = 2 + 0
c = math.comb(b, 2)
d = a / c
|
a ) 1 : 2 , b ) 1 : 5 , c ) 1 : 4 , d ) 2 : 3 , e ) 1 : 8 | d | divide(divide(288, divide(288, 32)), divide(288, subtract(divide(288, 32), const_3))) | a motorcyclist goes from bombay to pune , a distance of 288 kms at an average of 32 kmph speed . another man starts from bombay by car 2 Β½ hours after the first , and reaches pune Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | "t = 288 / 32 = 9 h t = 9 - 3 = 6 time ratio = 9 : 6 = 3 : 2 speed ratio = 2 : 3 answer : d" | a = 288 / 32
b = 288 / a
c = 288 / 32
d = c - 3
e = 288 / d
f = b / e
|
a ) 872 , b ) 738 , c ) 837 , d ) 780 , e ) 83 | d | divide(subtract(multiply(1500, divide(15, const_100)), 30), divide(25, const_100)) | if 25 % of x is 30 less than 15 % of 1500 , then x is ? | "25 % of x = x / 4 ; 15 % of 1500 = 15 / 100 * 1500 = 225 given that , x / 4 = 225 - 30 = > x / 4 = 195 = > x = 780 . answer : d" | a = 15 / 100
b = 1500 * a
c = b - 30
d = 25 / 100
e = c / d
|
a ) s . 600 , b ) s . 480 , c ) s . 500 , d ) s . 450 , e ) s . 550 | a | subtract(divide(multiply(1000, const_100), add(25, const_100)), divide(multiply(divide(multiply(1000, const_100), add(25, const_100)), 25), const_100)) | by selling an article at rs . 1000 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ? | "sp = 1000 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 1000 * [ 100 / 125 ] = 800 loss = 25 % = 25 % of 800 = rs . 200 sp = cp - loss = 800 - 200 = rs . 600 answer : a" | a = 1000 * 100
b = 25 + 100
c = a / b
d = 1000 * 100
e = 25 + 100
f = d / e
g = f * 25
h = g / 100
i = c - h
|
a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) 160 | d | multiply(divide(const_60, 20), 1) | if the population of a certain country increases at the rate of one person every 20 seconds , by how many persons does the population increase in 1 hour ? | "answer = 3 * 60 = 180 answer is d" | a = const_60 / 20
b = a * 1
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a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 200 / 9 , e ) 11 / 4 | d | divide(12, divide(const_1, add(divide(const_1, 40), divide(const_1, 50)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 40 , and 50 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 40 + 1 / 50 = 9 / 200 therefore they take 200 / 9 ratio = 200 / 9 = d" | a = 1 / 40
b = 1 / 50
c = a + b
d = 1 / c
e = 12 / d
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a ) 67 , b ) 73 , c ) 79 , d ) 85 , e ) 91 | e | add(add(choose(9, const_1), choose(9, const_1)), choose(const_4, const_1)) | jane and thomas are among the 9 people from which a committee of 4 people is to be selected . how many different possible committees of 4 people can be selected from these 9 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 4 people from 9 is 9 c 4 = 126 . the number of committees without jane or thomas is 7 c 4 = 35 . there are 126 - 35 = 91 possible committees which include jane and / or thomas . the answer is e ." | a = math.comb(9, 1)
b = math.comb(9, 1)
c = a + b
d = math.comb(4, 1)
e = c + d
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a ) 8 , b ) 9 , c ) 11 , d ) 15 , e ) 18 | c | divide(multiply(2, 22), 4) | albert is 2 times mary β s age and 4 times as old as betty . mary is 22 years younger than albert . how old is betty ? | "a = 2 m = m + 22 m = 22 a = 44 a = 4 b , and so b = 11 the answer is c ." | a = 2 * 22
b = a / 4
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a ) 300 m . , b ) 250 m . , c ) 100 m . , d ) 200 m . , e ) 150 m . | c | multiply(150, subtract(const_2, const_1)) | a train speeds past a pole in 10 seconds and a platform 150 m long in 25 seconds . its length is : | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 150 / 25 = x / 10 x = 100 m . answer : option c" | a = 2 - 1
b = 150 * a
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a ) 2 / 69 , b ) 2 / 20 , c ) 2 / 18 , d ) 2 / 29 , e ) 2 / 15 | e | multiply(divide(1, 5), divide(2, 3)) | two brother x and y appeared for an exam . the probability of selection of x is 1 / 5 and that of b is 2 / 3 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 5 , p ( b ) = 2 / 3 . let c be the event that both are selected . p ( c ) = p ( a ) Γ£ β p ( b ) as a and b are independent events : = ( 1 / 5 ) Γ£ β ( 2 / 3 ) = 2 / 15 answer : e ) 2 / 15" | a = 1 / 5
b = 2 / 3
c = a * b
|
a ) 3,551 , b ) 3,668 , c ) 3,733 , d ) 3,822 , e ) 3,957 | b | add(lcm(lcm(25, 49), lcm(15, 21)), 7) | what is the smallest number which when increased by 7 is divisible by 25 , 49 , 15 , and 21 ? | "first factor each of 25 , 49 , 15 , and 21 into primes : 25 = 5 * 5 ; 49 = 7 * 7 ; 15 = 3 * 5 ; 21 = 3 * 7 the smallest number divisible by each of the numbers is 3 * 5 * 5 * 7 * 7 = 3,675 3,675 - 7 = 3,668 thus 3,668 + 7 is the smallest number divisible by 25 , 49 , 15 , and 21 b" | a = math.lcm(25, 49)
b = math.lcm(15, 21)
c = math.lcm(a, b)
d = c + 7
|
a ) 1.4 % , b ) 50 % , c ) 11.1 % , d ) 12.5 % , e ) 23.6 % | b | multiply(subtract(divide(15, 10), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 10 . if the price per share of stock k was $ 15 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | opening = 10 closing = 15 rise in price = 5 so , percent increase = 5 / 10 * 100 = 50 % answer : b | a = 15 / 10
b = a - 1
c = b * 100
|
a ) 1638 , b ) 1863 , c ) 1683 , d ) 1836 , e ) none of them | c | add(8, lcm(5, 6,7)) | find the least number which when divided by 5 , 6,7 , and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder . | "l . c . m . of 5,6 , 7,8 = 840 . required number is of the form 840 k + 3 least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . required number = ( 840 x 2 + 3 ) = 1683 answer is c ." | a = math.lcm(5, 6)
b = 8 + a
|
a ) 10.55 meters , b ) 11.55 meters , c ) 33.55 meters , d ) 55.55 meters , e ) none of these | d | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 150)), subtract(multiply(const_100, const_10), 100))) | a can give b 100 meters start and c 150 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "explanation : a runs 1000 meters while b runs 900 meters and c runs 850 meters . therefore , b runs 900 meters while c runs 850 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 850 ) / 900 = 944.44 meters thus , b can give c ( 1000 - 944.44 ) = 55.55 meters start answer : d" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 150
e = b * d
f = 100 * 10
g = f - 100
h = e / g
i = a - h
|
a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these | b | multiply(1.6, 2.13) | if 213 Γ 16 = 3408 , then 1.6 Γ 2.13 is equal to : | "solution 1.6 Γ 2.13 = ( 16 / 10 x 213 / 100 ) = ( 16 x 213 / 1000 ) = 3408 / 100 = 3.408 . answer b" | a = 1 * 6
|
a ) 648 , b ) 512 , c ) 252 , d ) 225 , e ) 26 | d | divide(divide(1000, 100), 3) | in how many no . between 100 and 1000 exactly one of the digits is 3 ? | "3 scenarios possible 3 xy = 1 * 9 * 9 = 81 x 3 y ( with x β β 3 or 0 ) = 8 * 1 * 9 = 72 y 3 x ( with y β β 3 or 0 ) = 8 * 1 * 9 = 72 total cases possible = 144 + 81 = 225 d is the correct answer" | a = 1000 / 100
b = a / 3
|
a ) rs . 6672 , b ) rs . 6727 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 6728 | a | divide(8340, add(const_1, divide(25, const_100))) | the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 8340 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8340 ( 100 / 125 ) = rs . 6672 . answer : a" | a = 25 / 100
b = 1 + a
c = 8340 / b
|
a ) 13 , b ) 12 , c ) 11 , d ) 15 , e ) 16 | a | divide(divide(multiply(78, const_2), const_3), const_4) | everyone shakes hands with everyone else in a room . total number of handshakes is 78 . number of persons = ? | in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 78 or n ( n - 1 ) = 156 or n = 13 answer is ( a ) | a = 78 * 2
b = a / 3
c = b / 4
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | c | multiply(sqrt(divide(134, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 134 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose 12 at first , but then changed to 48 . i ' m not a native speaker , so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) , so is its largest factor . please correct if i ' m wrong here . i know that if n = 12 , n ^ 2 = 144 = 2 * 72 ( satisfy the condition ) . when n = 12 , the largest factor of n is n itself , which is 12 . check : 12 is the largest positive number that must divide 12 - - > true however if n = 48 , n ^ 2 = 48 * 48 = 32 * 72 ( satisfy the condition too ) . when n = 48 , the largest factor of n is n itself , which is 48 . check : 48 is the largest positive number that must divide 48 - - > true so , i also notice that the keyword ismust , notcould . the question is , why is 48 notmust divide 48 , but instead onlycould divide 48 ? i ' m not clear right here . why is 12 must divide 12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ 2 is divisible by 72 . the least value of n for which n ^ 2 is divisible by 72 is 12 , thus nmustbe divisible by 12 ( n is in any case divisible by 12 ) . for all other values of n , for which n ^ 2 is divisible by 72 , n will still be divisible by 12 . this means that n is always divisible by 12 if n ^ 2 is divisible by 72 . now , ask yourself : if n = 24 , is n divisible by 48 ? no . so , n is not always divisible by 48 . c" | a = 134 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 70 , b ) 60 , c ) 90 , d ) 100 , e ) 110 | b | subtract(add(350, 110), 400) | in a group of 400 players some are outdoor players , some are indoor players and some plays both . 350 plays outdoors and 110 plays indoor . how many plays indoor as well as outdoor games . | consider total number of players n ( o u i ) = 400 players who plays outdoors n ( o ) = 350 players who plays indoors n ( i ) = 110 players who plays both n ( o Γ’ Λ Β© i ) = ? n ( o u i ) = n ( o ) + n ( i ) - n ( o Γ’ Λ Β© i ) 400 = 350 + 110 - n ( o Γ’ Λ Β© i ) n ( o Γ’ Λ Β© i ) = 460 - 400 n ( o Γ’ Λ Β© i ) = 60 so people read both science fiction and literacy works are 60 answer : b | a = 350 + 110
b = a - 400
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a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 90 | b | divide(20, divide(50, const_100)) | if 50 % of x equals the sum of y and 20 , then what is the value of x β 2 y ? | explanation : given , 50 % of x equals the sum of y and 20 . = > ( 50 / 100 ) Γ x = y + 20 . = > x / 2 = y + 20 . = > x = 2 y + 40 . = > x β 2 y = 40 . answer : b | a = 50 / 100
b = 20 / a
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a ) 12.5 % , b ) 17.5 % , c ) 19 % , d ) 21 % , e ) none of these | c | add(multiply(divide(4, 5), subtract(15, 10)), 15) | weights of two friends ram and shyam are in the ratio of 4 : 5 . ram β s weight increases by 10 % and the total weight of ram and shyam together becomes 82.8 kg , with an increase of 15 % . by what per cent did the weight of shyam increase ? | let the weights of ram and shyam be 4 x and 5 x . now , according to question , 4 x Γ 110 / 100 + shyam ? snewwt 4 x Γ 110100 + shyam ? s new wt = 82.8 . . . . . . . . ( i ) and 4 x + 5 x = 9 x + 115 / 100 = 82.8 . . . . . . . . ( ii ) from ( ii ) , x = 8 putting in ( i ) , we get shyam β s new wt = ( 82.8 β 35.2 ) = 47.6 % increase in shyam β s wt = ( 47.6 β 40 / 40 Γ 100 ) = 19 answer c | a = 4 / 5
b = 15 - 10
c = a * b
d = c + 15
|
a ) 1951609 , b ) 1981709 , c ) 18362619 , d ) 2031719 , e ) none of these | a | multiply(divide(1397, 1397), const_100) | 1397 x 1397 | "1397 x 1397 = ( 1397 ) 2 = ( 1400 - 3 ) 2 = ( 1400 ) 2 + ( 3 ) 2 - ( 2 x 1400 x 3 ) = 1960000 + 9 - 8400 = 1960009 - 8400 = 1951609 . answer : option a" | a = 1397 / 1397
b = a * 100
|
a ) 1 , b ) 7 , c ) 3 , d ) 5 , e ) 6 | b | multiply(2, 8) | if n divided by 9 has a remainder of 8 , what is the remainder when 2 times n is divided by 9 ? | "as per question = > n = 9 p + 8 for some integer p hence 2 n = > 18 q + 16 but again , 16 can be divided by 9 to get remainder 7 for some integer q hence b" | a = 2 * 8
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a ) 1200 , b ) 3000 , c ) 2400 , d ) 3600 , e ) 2450 | c | divide(subtract(multiply(divide(6, const_100), 2400), multiply(2400, divide(4, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | barbata invests $ 2400 in the national bank at 4 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | "let the additional invested amount for 8 % interest be x ; equation will be ; 2400 + 0.04 * 2400 + x + 0.08 x = 2400 + x + 0.06 ( 2400 + x ) 0.04 * 2400 + 0.08 x = 0.06 x + 0.06 * 2400 0.02 x = 2400 ( 0.06 - 0.04 ) x = 2400 * 0.02 / 0.02 = 2400 ans : c" | a = 6 / 100
b = a * 2400
c = 4 / 100
d = 2400 * c
e = b - d
f = 8 / 100
g = 6 / 100
h = f - g
i = e / h
|
a ) 1 / 602 , b ) 1 / 120 , c ) 1 / 650 , d ) 1 / 1024 , e ) 1 / 1000 | d | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of ( 4 ) ^ - 5 ? | "4 ^ - 5 = 1 / ( 4 ) ^ 5 = 1 / 1024 answer : d" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(12, add(2, const_1)) | if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 12 - m ) then m = ? | "2 m = 12 - m 3 m = 12 m = 4 the answer is d ." | a = 2 + 1
b = 12 / a
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a ) $ 9 , b ) $ 15 , c ) $ 18 , d ) $ 21 , e ) $ 27 | a | subtract(divide(540, divide(10, subtract(divide(const_3, const_2), const_1))), divide(540, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | p and q are the only two applicants qualified for a short - term research project that pays 540 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job . candidate p β s hourly wage is how many dollars greater than candidate q β s hourly wage ? | "let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 540 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 540 / 30 = $ 18 p ' s hourly wage is 540 / ( t - 10 ) = $ 27 which is $ 9 per hour more . the answer is a ." | a = 3 / 2
b = a - 1
c = 10 / b
d = 540 / c
e = 3 / 2
f = e - 1
g = 10 / f
h = g + 10
i = 540 / h
j = d - i
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a ) 160 , b ) 120 , c ) 100 , d ) 150 , e ) 170 | c | multiply(50, const_2) | the average of the marks of 12 students in a class is 50 . if the marks of each student are doubled , find the new average ? | "sum of the marks for the 12 students = 12 * 50 = 600 . the marks of each student are doubled , the sum also will be doubled . the new sum = 600 * 2 = 1200 . so , the new average = 1200 / 12 = 100 . answer : c" | a = 50 * 2
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a ) 10 % , b ) 20 % , c ) 30 % , d ) 50 % , e ) 65 % | b | subtract(add(add(75, 25), 20), const_100) | if 75 percent of a class answered the first question on a certain test correctly , 25 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "{ total } = { first } + { second } - { both } + { neither } 100 % = 75 % + 25 % - { both } + 20 % - - > { both } = 20 % . answer : b ." | a = 75 + 25
b = a + 20
c = b - 100
|
a ) 13 , b ) 24 , c ) 18 , d ) 19 , e ) 12 | b | divide(multiply(8, 6), subtract(8, 6)) | pipe a can fill a tank in 6 hours . due to a leak at the bottom , it takes 8 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 6 - 1 / x = 1 / 8 = > 1 / x = 1 / 6 - 1 / 8 = ( 4 - 3 ) / 24 = 1 / 24 = > x = 24 . answer : b" | a = 8 * 6
b = 8 - 6
c = a / b
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a ) 11 / 122 , b ) 11 / 120 , c ) 11 / 121 , d ) 11 / 140 , e ) 11 / 111 | b | divide(divide(55, 5), 120) | if the sum of two numbers is 55 and the h . c . f . and l . c . m . of these numbers are 5 and 120 respectively , then the sum of the reciprocals of the numbers is equal to : | explanation : let the numbers be a and b . then , a + b = 55 and ab = 5 x 120 = 600 . the required sum answer : b ) 11 / 120 | a = 55 / 5
b = a / 120
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a ) 62 % , b ) 57 % , c ) 52 % , d ) 85 % , e ) 42 % | d | subtract(const_100, add(multiply(60, divide(100, const_100)), multiply(divide(60, const_100), 100))) | in a factory , there are 60 % technicians and 100 % non - technicians . if the 60 % of the technicians and 100 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ? | "total = 160 t = 60 nt = 100 60 * ( 100 / 100 ) = 37.5 100 * ( 60 / 100 ) = 37.5 37.5 + 37.5 = 75 = > 160 - 75 = 85 % answer : d" | a = 100 / 100
b = 60 * a
c = 60 / 100
d = c * 100
e = b + d
f = 100 - e
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a ) 2177 , b ) 2876 , c ) 4500 , d ) 5700 , e ) 6711 | d | subtract(multiply(8000, const_4), subtract(multiply(8200, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8200 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8200 = 32800 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 800 salary of may is rs . 6500 salary of january = rs . 5700 . answer : d" | a = 8000 * 4
b = 8200 * 4
c = b - 6500
d = a - c
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a ) 11 years , b ) 17 years , c ) 21 years , d ) 14 years , e ) 12 years | c | subtract(multiply(15, 15), add(multiply(5, 12), multiply(9, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 12 years and that of the other 9 students is 16 years . the age of the 15 th student is ? | "age of the 15 th student = [ 15 * 15 - ( 12 * 5 + 16 * 9 ) ] = ( 225 - 204 ) = 21 years . answer : c" | a = 15 * 15
b = 5 * 12
c = 9 * 16
d = b + c
e = a - d
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a ) 8 m , b ) 10 m , c ) 12 m , d ) 20 m , e ) 17 m | d | divide(sqrt(divide(800, divide(const_1, const_2))), const_2) | the width of a rectangular hall is Β½ of its length . if the area of the hall is 800 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 800 = x * 1 x / 2 x Β² = 1600 x = 40 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 40 / 2 = 20 m answer : d" | a = 1 / 2
b = 800 / a
c = math.sqrt(b)
d = c / 2
|
a ) . 2 , b ) 0.002 , c ) . 05 , d ) . 02 , e ) none of these | d | divide(20, const_1000) | what decimal fraction is 20 ml of a litre ? | "answer required fraction = 20 / 1000 = 2 / 100 = . 02 correct option : d" | a = 20 / 1000
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a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 14 | a | divide(divide(divide(divide(210, const_2), const_3), const_4), divide(const_10, const_2)) | if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 210 , what is the least possible value of n ? | "210 = 2 * 3 * 5 * 7 , so n must be at least 7 . the answer is a ." | a = 210 / 2
b = a / 3
c = b / 4
d = 10 / 2
e = c / d
|
a ) 30 km , b ) 60 km , c ) 80 km , d ) 20 km , e ) 40 km | e | subtract(divide(multiply(const_100, 40), 40), multiply(divide(40, 2), const_3)) | a man covers a distance on scooter . had he moved 3 kmph faster he would have taken 40 min less . if he had moved 2 kmph slower , he would have taken 40 min more . the distance is . | e 40 km let distance = x m usual rate = y kmph x / y β x / y + 3 = 40 / 60 hr 2 y ( y + 3 ) = 9 x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( 1 ) x / y - 2 β x / y = 40 / 60 hr y ( y - 2 ) = 3 x - - - - - - - - - - - ( 2 ) divide 1 & 2 equations by solving we get x = 40 km | a = 100 * 40
b = a / 40
c = 40 / 2
d = c * 3
e = b - d
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a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | divide(multiply(add(divide(50, const_100), const_1), 25), add(add(divide(50, const_100), const_1), const_1)) | p is 50 % more efficient than q . p can complete a work in 25 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 25 let work done by q in 1 day = q q Γ ( 150 / 100 ) = 1 / 25 q = 100 / ( 25 Γ 150 ) = 10 / ( 25 Γ 15 ) the work done by p and q in 1 day = 1 / 25 + 10 / ( 25 Γ 15 ) = 25 / ( 25 Γ 15 ) = 1 / 15 p and q together can do the work in 15 days . the answer is c ." | a = 50 / 100
b = a + 1
c = b * 25
d = 50 / 100
e = d + 1
f = e + 1
g = c / f
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a ) 65 , b ) 85 , c ) 48 , d ) 12 , e ) 25 | d | inverse(subtract(inverse(3), subtract(inverse(2), inverse(4)))) | a can do a piece of work in 4 hours . a and c together can do it in just 2 hours , while b and c together need 3 hours to finish the same work . b alone can complete the work in - - - hours . | explanation : work done by a in 1 hour = 1 / 4 work done by b and c in 1 hour = 1 / 3 work done by a and c in 1 hour = 1 / 2 work done by a , b and c in 1 hour = 1 / 4 + 1 / 3 = 7 / 12 work done by b in 1 hour = 7 / 12 β 1 / 2 = 1 / 12 = > b alone can complete the work in 12 hours answer : d | a = 1/(3)
b = 1/(2)
c = 1/(4)
d = b - c
e = a - d
f = 1/(e)
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a ) 18.97 miles , b ) 20 miles , c ) 22 miles , d ) 23 miles , e ) 25 miles | a | sqrt(add(power(add(6, 12), const_2), power(4, const_2))) | beginning in town a , biker bob rides his bike 12 miles west , 6 miles north , 4 miles east , and then 12 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 12 + 6 = 18 m other being the base ie distance traveled west - distance traveled eat = 12 - 6 = 6 m now this third side or the distance between town a and town b = 18 ^ 2 + 6 ^ 2 = sq root 360 = 18.97 mile answer : a" | a = 6 + 12
b = a ** 2
c = 4 ** 2
d = b + c
e = math.sqrt(d)
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a ) a ) 8 , b ) b ) 25 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | a | subtract(multiply(16, 60.5), multiply(subtract(16, const_1), 64.0)) | the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 60.5 . when one student dropped the class , the average of the remaining scores increased to 64.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 60.50 = 968 total score of 15 students is 15 * 64 = 960 so , the score of the person who left is 8 ( 968 - 960 ) answer will be ( a" | a = 16 * 60
b = 16 - 1
c = b * 64
d = a - c
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a ) 38000 , b ) 47000 , c ) 48000 , d ) 51000 , e ) 56000 | e | subtract(57000, multiply(const_100, const_10)) | median w of price of the car is 52000 . which could be the median of the other two . 35000 , 44000 , x , y , 57000 . | median of the car is 52000 therefore 52000 is the middle number . assuming y > = x , x will have to be the median value i . e . 52000 . therefore the remaining value would have to be between 52000 and 57000 inclusive . the only answer choice that fits is w = 56000 . e | a = 100 * 10
b = 57000 - a
|
a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 120 | b | multiply(divide(add(100, const_10), subtract(add(const_1, divide(divide(multiply(const_4, const_10), const_100), divide(multiply(add(const_2, const_3), add(const_2, const_3)), const_100))), divide(multiply(const_4, const_10), const_100))), divide(divide(multiply(const_4, const_10), const_100), divide(multiply(add(const_2, const_3), add(const_2, const_3)), const_100))) | each man in a gang of 100 investors has investments ineither equities or securities or both . exactly 2525 of the investors in equities haveinvestments in securities , exactly 4040 of the investorsin securities have investments inequities . how many have investmentsin equities ? | the investors can be categorized into three groups : ( 1 ) those who have investments in equities only . ( 2 ) those who have investments in securities only . ( 3 ) those who have investments in both equities and securities . let xx , yy , and zz denote the number of people in the respective categories . since the total number of investors is 110 , we have x + y + z = 110 x + y + z = 110 - - - - - - - - - - - - - ( 1 ) also , the number of people with investments in equities is x + zx + z and the number of people with investments in securities is y + zy + z . since exactly 25 % 25 % of the investors in equities have investments in securities , we have the equation 25100 Γ ( x + z ) = z 25100 Γ ( x + z ) = z 25100 Γ x = 25100 Γ z 25100 Γ x = 25100 Γ z x = 3 zx = 3 z - - - - - - - - - - - - - - - - - - - ( 2 ) since exactly 40 % 40 % of the investors in securities have investments in equities , we have the equation 40100 Γ ( y + z ) = z 40100 Γ ( y + z ) = z ( y + z ) = 5 z 2 ( y + z ) = 5 z 2 y = 3 z 2 y = 3 z 2 substituting equations ( 2 ) and ( 3 ) into equation ( 1 ) yields 3 z + 3 z 2 + z = 1103 z + 3 z 2 + z = 110 11 z 2 = 11011 z 2 = 110 z = 110 Γ 211 = 20 z = 110 Γ 211 = 20 hence , the number of people with investments in equities is : x + z = 3 z + z = 3 Γ 20 + 20 = 60 + 20 x + z = 3 z + z = 3 Γ 20 + 20 = 60 + 20 = 80 b | a = 100 + 10
b = 4 * 10
c = b / 100
d = 2 + 3
e = 2 + 3
f = d * e
g = f / 100
h = c / g
i = 1 + h
j = 4 * 10
k = j / 100
l = i - k
m = a / l
n = 4 * 10
o = n / 100
p = 2 + 3
q = 2 + 3
r = p * q
s = r / 100
t = o / s
u = m * t
|
a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 4,000 | e | divide(multiply(9.6, power(10, 11)), multiply(240, power(10, add(const_4, const_2)))) | a certain country had a total annual expenditure of $ 9.6 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ? | total expenditure / population = per capita expenditure hence , ( 9,6 x 10 ^ 11 ) / 240 000 000 = ( 9,6 x 10 ^ 11 ) / ( 2,4 x 10 ^ 8 ) = 4 x 10 ^ ( 11 - 8 ) = 4 x 10 ^ 3 = 4000 . answer is e . | a = 10 ** 11
b = 9 * 6
c = 4 + 2
d = 10 ** c
e = 240 * d
f = b / e
|
a ) 0.125 % , b ) 0.24 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 % | b | multiply(6, divide(4, const_100)) | in the manufacture of a certain product , 6 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | "percent of defective produced = 6 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 6 / 100 ) * 100 % = ( 24 / 10000 ) * 100 % = ( 24 / 100 ) % = . 24 % answer b" | a = 4 / 100
b = 6 * a
|
a ) 399 , b ) 272 , c ) 732.6 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.10), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.10 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 Γ’ β¬ β 6 = 666 * 1.1 = 732.6 answer : c" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 1
f = e * 3
|
a ) 11 am . , b ) 12 p . m . , c ) 3 pm . , d ) 2 p . m . , e ) 1 p . m . | e | add(divide(add(2, 9), add(8, 9)), 1000) | the distance between 2 cities a and b is 1000 km . a train starts from a at 8 a . m . and travels towards b at 100 km / hr . another starts from b at 9 a . m . and travels towards a at 150 km / hr . at what time do they meet ? | "suppose they meet x hrs after 8 a . m . distance moved by first in x hrs + distance moved by second in ( x - 1 ) hrs = 1000 100 x + 150 ( x - 1 ) = 1000 x = 4.60 = 5 hrs they meet at 8 + 5 = 1 p . m . answer is e" | a = 2 + 9
b = 8 + 9
c = a / b
d = c + 1000
|
a ) 8 , b ) 7 , c ) 42 , d ) 25 , e ) 63 | d | subtract(negate(53), multiply(subtract(221, 109), divide(subtract(221, 109), subtract(445, 221)))) | 445 , 221 , 109 , 53 , . . . | "explanation : to obtain next number , subtract 3 from the previous number and divide the result by 2 445 ( 445 - 3 ) / 2 = 221 ( 221 - 3 ) / 2 = 109 ( 109 - 3 ) / 2 = 53 ( 53 - 3 ) / 2 = 25 answer : d" | a = negate - (
|
a ) 11 : 6 , b ) 52 : 35 , c ) 13 : 7 , d ) 15 : 6 , e ) 13 : 6 | b | divide(multiply(65, 8), multiply(70, 5)) | car a runs at the speed of 65 km / hr & reaches its destination in 8 hr . car b runs at the speed of 70 km / h & reaches its destination in 5 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 65 Γ£ β 8 = 520 km distance travelled by car b = 70 Γ£ β 5 = 350 km ratio = 520 / 350 = 52 : 35 b" | a = 65 * 8
b = 70 * 5
c = a / b
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(568219, multiply(floor(divide(568219, 5)), 5)) | find the least number must be subtracted from 568219 so that remaining no . is divisible by 5 ? | "on dividing 568219 by 5 we get the remainder 4 , so 4 should be subtracted e" | a = 568219 / 5
b = math.floor(a)
c = b * 5
d = 568219 - c
|
a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 60 % profit , e ) 75 % profit | d | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 40)), divide(multiply(const_100, 25), 40))) | if the cost price of 40 articles is equal to the selling price of 25 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 40 articles = 40 * 1 = $ 40 the selling price of 25 articles = cost price of 40 articles = $ 40 . now , we know the selling price of 25 articles . let us find the cost price of 25 articles . cost price of 25 articles = 25 * 1 = $ 25 . therefore , profit made on sale of 25 articles = selling price of 25 articles - cost price of 25 articles = 40 - 25 = $ 15 . as the profit is in the positive , the merchant has made a profit of $ 15 . therefore , % loss = loss / cp * 100 % loss = 15 / 25 * 100 = 60 % profit . d" | a = 100 * 25
b = a / 40
c = 100 - b
d = 100 * 25
e = d / 40
f = c / e
g = 100 * f
|
a ) 7 , b ) 8 , c ) 9 , d ) 4 , e ) 5 | a | add(reminder(gcd(subtract(272738, 13), subtract(232342, 17)), const_10), const_2) | the numbers 272738 and 232342 , when divided by n , a two digit number , leave a remainder of 13 and 17 respectively . find the sum of the digits of n ? | from the given information , ( 272738 - 13 , 232342 - 17 ) are exactly divisible by that two digit number . we have to find the hcf of the given numbers 272725 , 232325 . hcf = 25 . so sum of the digits = 7 . answer : a | a = 272738 - 13
b = 232342 - 17
c = math.gcd(a, b)
d = reminder + (
|
a ) 287 cm 2 , b ) 887 cm 2 , c ) 384 cm 2 , d ) 268 cm 2 , e ) 648 cm 2 | e | multiply(36, 18) | find the area of a parallelogram with base 36 cm and height 18 cm ? | "area of a parallelogram = base * height = 36 * 18 = 648 cm 2 answer : e" | a = 36 * 18
|
a ) a ) 51 , b ) b ) 63 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36 | b | divide(multiply(42, 3), subtract(3, const_1)) | if the number is divided by 3 , it reduced by 42 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 42 = > 2 x / 3 = 42 = > x = 63 answer : option b" | a = 42 * 3
b = 3 - 1
c = a / b
|
a ) 82,650 , b ) 64,850 , c ) 64,749 , d ) 49,700 , e ) 56,720 | a | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 6), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 6 ? | "find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 6 seeing the options after dividing an finding the reminder of 2 my answer was a" | a = 1000 - 3
b = 3 * 3
c = b * 10
d = 3 * 3
e = c + d
f = e + 3
g = a - f
h = g / 6
i = h + 1
j = i / 2
k = 1000 - 3
l = 3 * 3
m = l * 10
n = 3 * 3
o = m + n
p = o + 3
q = k + p
r = j * q
|
a ) 6400 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3200 | e | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(50, 11.25), 6)) | how many bricks , each measuring 50 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 50 x 11.25 x 6 ) = 3200 answer : e" | a = 8 * 100
b = 6 * 100
c = a * b
d = c * 22
e = 50 * 11
f = e * 6
g = d / f
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind yoe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of yoe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . yoe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena can cover 7.5 miles in 30 mins teena will cover 15 miles in 60 mins so answer 30 + 60 = 90 mins . d" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 24 , b ) 30 , c ) 48 , d ) 72 , e ) 132 | b | floor(multiply(divide(82, subtract(11, 3)), 3)) | at a certain zoo , the ratio of lions to penguins is 3 to 11 . if there are 82 more penguins than lions at the zoo , how many lions are there ? | 3 / 11 = x / x + 82 3 * 82 = 8 x x = 30 b . 30 | a = 11 - 3
b = 82 / a
c = b * 3
d = math.floor(c)
|
a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), const_3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed q of the train . | let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y / x * 1 / 3 = 7 / 12 y / x = 7 / 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) / ( 3 x / 4 ) - ( y - 24 ) / x = 25 / 60 4 ( y - 24 ) / 3 x - ( y - 24 ) / x = 5 / 12 ( y - 24 ) / x ( 4 / 3 - 1 ) = 5 / 12 ( y - 24 ) / x * 1 / 3 = 5 / 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = q ans = c | a = 24 * 4
b = a / 3
c = b - 24
d = 35 - 25
e = d + 50
f = 35 - 25
g = e / f
h = c * g
|
a ) 5200 , b ) 3050 , c ) 4400 , d ) 4500 , e ) none of these | a | divide(117, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100))) | if 15 % of 30 % of 50 % of a number is 117 , then what is the number ? | "let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 117 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 117 = > a = 10 * 20 * 10 * 2 = 5200 . answer : a" | a = 50 / 100
b = 30 / 100
c = a * b
d = 15 / 100
e = c * d
f = 117 / e
|
a ) 100 , b ) 90 , c ) 80 , d ) 30 , e ) 50 | c | multiply(multiply(divide(100, 20), const_2), 8) | machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 20 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 8 minutes ? | "machine b produces 100 part in 20 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 20 / 2 = 10 minutes . now , machine a produces 100 parts in 10 minutes which is 100 / 10 = 10 parts / minute . 10 parts x a total of 8 minutes = 80 c" | a = 100 / 20
b = a * 2
c = b * 8
|
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