options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 280 , b ) 63 , c ) 120 , d ) 160 , e ) 1260 | a | multiply(divide(multiply(5, subtract(5, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2)) | if 5 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ? | parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 5 c 2 for vertical 8 c 2 total parallelogram is 5 c 2 * 8 c 2 = 10 * 28 = 280 answer : a | a = 5 - 1
b = 5 * a
c = b / 2
d = 8 - 1
e = 8 * d
f = e / 2
g = c * f
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | divide(24, const_2) | in a group of buffaloes and ducks , the number of legs are 24 more than twice the number of heads . what is the number of buffaloes in the group ? | let no . of buffaloes be x & no . of ducks be y buffaloes have 4 legs while ducks have 2 4 x + 2 y = 2 * ( x + y ) + 24 = > 4 x + 2 y - 2 x - 2 y = 24 = > x = 12 number of buffaloes in the group = 12 answer : a | a = 24 / 2
|
a ) 6810 , b ) 7920 , c ) 9030 , d ) 10,140 , e ) 11,250 | b | multiply(330, 24) | a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 24 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s ) | the distance the sound traveled to the man is 24 * 330 = 7920 meters the answer is b . | a = 330 * 24
|
a ) 0 , b ) 1 , c ) 3 , d ) 5 , e ) 6 | b | floor(multiply(const_100, divide(57, 5000))) | what is the thousandths digit in the decimal equivalent of 57 / 5000 ? | 57 / 5000 = 57 / ( 5 * 10 ^ 3 ) = ( 57 / 5 ) * 10 ^ - 3 = 11.4 * 10 ^ - 3 = . 0114 thousandths digit = 1 answer b | a = 57 / 5000
b = 100 * a
c = math.floor(b)
|
a ) 28 % , b ) 54 % , c ) 45 % , d ) 72 % , e ) 78 % | b | multiply(divide(multiply(divide(70, const_100), divide(subtract(8, 6), 8)), add(multiply(divide(subtract(const_100, 80), const_100), divide(6, 8)), multiply(divide(70, const_100), divide(subtract(8, 6), 8)))), const_100) | 6 / 8 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded if necessary to the nearest percent , the probability that a juliet supporter chosen at random resides in capulet is | "total population = 80 ( assume ) . 6 / 8 * 80 = 60 people from montague . 2 / 8 * 80 = 20 people from capulet . 0.2 * 60 = 12 people from montague support juliet . 0.7 * 20 = 14 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 14 / ( 12 + 14 ) = ~ 54 answer : b ." | a = 70 / 100
b = 8 - 6
c = b / 8
d = a * c
e = 100 - 80
f = e / 100
g = 6 / 8
h = f * g
i = 70 / 100
j = 8 - 6
k = j / 8
l = i * k
m = h + l
n = d / m
o = n * 100
|
a ) 10 , b ) 15 , c ) 13 , d ) 18 , e ) 19 | b | divide(multiply(multiply(45, 12), 5), multiply(30, 6)) | 45 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 45 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 45 x 5 x 12 x = ( 45 x 5 x 12 ) / ( 30 x 6 ) x = 15 answer b" | a = 45 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 75 , b ) 65 , c ) 85 , d ) 95 , e ) 80 | c | add(multiply(8, 2.5), 65) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "total increase in weight = 8 × 2.5 = 20 if x is the weight of the new person , total increase in weight = x − 65 = > 20 = x - 65 = > x = 20 + 65 = 85 answer is c ." | a = 8 * 2
b = a + 65
|
a ) 10 hr , b ) 5 hr , c ) 6 hr , d ) 4 hr 30 min , e ) 5 hr 12 min | e | add(divide(60, 4), divide(200,000, 50)) | it takes 60 identical printing machines 4 hours to print 200,000 cards . how long would it take 50 of these machines to print the same total ? | "50 % of 60 = 40 % or 0.30 4 hr x 60 min = 240 min 240 min x 0.30 = 72 min or 1 hr and 12 min 1 hr and 12 min + 4 hr = 5 hr and 12 min answer is e" | a = 60 / 4
b = 200 / 0
c = a + b
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(11, 3) | in a party , there are rows arranged with 1220 and 30 chairs and there were 3 left in all the cases . when there were 11 rows . there were no chair left . if 30 chairs were added . then what will be the reminder ? | lcm of 12 , 2030 is 120 . so the number is in the form of 120 x + 3 which is divisible by 11 , by placing x = 3 , we get 363 which is divisible by 11 and the multiple of 120 when 3 is subtracted from 363 . so the no . is 363 . so when 30 is added to it and is divided by 11 we get remainder as 8 answer : a | a = 11 - 3
|
a ) 100 , b ) 99 , c ) 98 , d ) 102 , e ) 104 | e | subtract(106, 1) | for every integer n ≥ 3 , the function g ( n ) is defined as the product of all the odd integers from 1 to n , inclusive . what is the value of g ( 106 ) – g ( 103 ) ? | "g ( 106 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 * 103 * 105 g ( 103 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 * 103 g ( 106 ) - g ( 103 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 * 103 * 105 - 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 * 103 = 1 * 3 * 5 * 7 * 9 * . . . * 103 * ( 105 - 1 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 * 103 * 104 hence : e" | a = 106 - 1
|
a ) 99.9 , b ) 91.0 , c ) 88.0 , d ) 70.9 , e ) 71.2 | a | add(90, multiply(divide(11, const_100), 90)) | if x is 11 percent greater than 90 , then x = | "11 % of 90 = ( 90 * 0.11 ) = 9.9 11 % greater than 90 = 90 + 9.9 = 99.9 answer is clearly a ." | a = 11 / 100
b = a * 90
c = 90 + b
|
a ) 150 , b ) 882 , c ) 277 , d ) 261 , e ) 120 | e | multiply(240, divide(240, add(add(5, 12), 13))) | the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 240 m , its area is ? | "5 x + 12 x + 13 x = 240 = > x = 8 a = 40 , b = 96 , c = 104 s = ( 40 + 96 + 104 ) / 2 = 120 answer : e" | a = 5 + 12
b = a + 13
c = 240 / b
d = 240 * c
|
a ) 24 , b ) 26 , c ) 21 , d ) 20 , e ) 19 | c | add(22, const_1) | total 22 matches are conducted in knockout match type . how many players will be participated in that tournament ? | "21 players answer : c" | a = 22 + 1
|
a ) 50 , b ) 35 , c ) 45 , d ) 25 , e ) 60 | d | subtract(150, add(add(15, 75), divide(add(15, subtract(110, add(add(75, 25), 15))), 2))) | in an intercollegiate competition that lasted for 3 days , 150 students took part on day 1 , 110 on day 2 and 140 on day 3 . if 75 took part on day 1 and day 2 and 25 took part on day 2 and day 3 and 15 took part on all three days , how many students took part only on day 2 ? | "day 1 & 2 = 75 ; only day 1 & 2 ( 75 - 15 ) = 60 , day 2 & 3 = 25 ; only day 2 & 3 ( 25 - 15 ) = 10 , only day 2 = 110 - ( 60 + 10 + 15 ) = 25 answer : d" | a = 15 + 75
b = 75 + 25
c = b + 15
d = 110 - c
e = 15 + d
f = e / 2
g = a + f
h = 150 - g
|
a ) 113 miles , b ) 277 miles , c ) 456 miles , d ) 887 miles , e ) 767 miles | b | subtract(1200, 923) | jim drove 923 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ? | the number of miles to drive to finish his journey is given by 1200 - 923 = 277 miles correct answer b | a = 1200 - 923
|
a ) 4 , b ) 5 , c ) 36 , d ) 2 , e ) 7 | c | subtract(add(const_100, 70), add(divide(multiply(add(const_100, 70), 20), const_100), const_100)) | on increasing the price of t . v . sets by 70 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ? | "explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 × 100 ) = rs . 10000 new sale value = rs . ( 170 × 80 ) = rs . 13600 increase % = 3600 / 10000 × 100 = 36 % answer : c" | a = 100 + 70
b = 100 + 70
c = b * 20
d = c / 100
e = d + 100
f = a - e
|
a ) 4000 , b ) 2000 , c ) 3000 , d ) 5000 , e ) 1000 | e | divide(volume_cube(1), volume_cube(divide(10, const_100))) | how many cubes of 10 cm edge can be put in a cubic box of 1 m edge ? | "( 110 × 100 × 100 ) / 10 × 10 × 10 = 1000 answer is e ." | a = volume_cube / (
|
a ) 150 meter , b ) 600 meter , c ) 167 meter , d ) 719 meter , e ) 169 meter | b | multiply(divide(multiply(60, const_1000), const_3600), 36) | a train running at the speed of 60 km / hr crosses a pole in 36 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 36 = 600 meter answer : b" | a = 60 * 1000
b = a / 3600
c = b * 36
|
['a ) 8', 'b ) 7', 'c ) 6', 'd ) 5', 'e ) 2'] | a | sqrt(divide(128, const_2)) | the area of a parallelogram is 128 sq m and its altitude is twice the corresponding base . then the length of the base is ? | 2 x * x = 128 = > x = 8 answer : a | a = 128 / 2
b = math.sqrt(a)
|
a ) 40 , b ) 99 , c ) 88 , d ) 77 , e ) 66 | a | divide(add(60, 20), const_2) | the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together ? | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a" | a = 60 + 20
b = a / 2
|
a ) $ 5670 , b ) $ 18900 , c ) $ 21000 , d ) $ 13500 , e ) $ 27000 | c | multiply(divide(10, add(add(8, 9), 10)), 56700) | a , b , and c were started a business , the total profit for one year is $ 56700 . the profit sharing ratio is 8 : 9 : 10 ( according to their investment they divide their profit ) . what is the profit of c ? | a : b : c = 8 + 9 + 10 = 27 ; c ' share = 10 / 27 of the total profit . c will get = 10 / 27 * $ 56700 = $ 21000 . answer : c | a = 8 + 9
b = a + 10
c = 10 / b
d = c * 56700
|
a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 35 | c | multiply(divide(divide(divide(1029, const_3), add(const_3, const_4)), add(const_3, const_4)), const_3) | if the product of the integers from 1 to n is divisible by 1029 , what is the least possible value of n ? | 1029 = 7 x 7 x 7 x 3 n must include at least 7 , 2 * 7 , and 3 * 7 . the answer is c . | a = 1029 / 3
b = 3 + 4
c = a / b
d = 3 + 4
e = c / d
f = e * 3
|
a ) $ 1000 , b ) $ 2000 , c ) $ 3000 , d ) $ 4000 , e ) $ 5000 | b | divide(multiply(3000, 2), 3) | average families of 4 pay $ 3000 for their health insurance deductibles annually . there will be a 2 / 3 increase in the deductibles for next year . how much more will the average family pay in deductibles next year ? | a certain percentage of questions in the quant section of the gmat are just ' math questions ' - you ' ll use a formula , do some calculations and you ' ll have the answer . this is one of those types of questions . you still have to write everything down and stay organized , but the work involved is relatively straight - forward . here , we ' re told that the current annual deductible is $ 3000 and that the deductible is to be increased by 2 / 3 . we ' re asked how much more will the deductible cost a family next year . original deductible = $ 3000 increase in deductible = 2 / 3 the answer is calculated by the equation : $ 3000 × 2 / 3 = $ 2000 . b | a = 3000 * 2
b = a / 3
|
a ) 42 , b ) 70 , c ) 140 , d ) 135 , e ) 315 | d | multiply(multiply(10, 3), 3) | a certain university will select 1 of 3 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 of 3 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more than one departments . we can choose any of the 3 candidates for the math dep . , which gives as 3 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections and 8 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 2 ! * 8 ! = ( 9 * 10 ) / 2 = 90 / 2 = 45 . we are multiplying our individual selections : 3 * 45 = 135 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans d" | a = 10 * 3
b = a * 3
|
a ) 15 , b ) 10 , c ) 5 , d ) 20 , e ) 25 | c | divide(divide(6, const_2), divide(60, const_100)) | at the end of the month , a certain ocean desalination plant ’ s reservoir contained 6 million gallons of water . this amount is twice the normal level . if this amount represents 60 % of the reservoir ’ s total capacity , how many million gallons short of total capacity is the normal level ? | "the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 6 mg = 60 % of total . . total = 6 / . 6 = 10 mg . . normal level = 1 / 2 of 10 = 5 mg . . shortage of normal level = 10 - 5 = 5 mg . . c" | a = 6 / 2
b = 60 / 100
c = a / b
|
a ) 6 : 5 , b ) 9 : 25 , c ) 5 : 6 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(60, const_2))) | two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 60 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 30 / speed b since the times are same , 50 pi / speed a = 60 pi / speed b speeda / speed b = 50 pi / 60 pi = 5 / 6 correct option : c" | a = 50 / 2
b = circumface / (
|
a ) $ 500 , b ) $ 520 , c ) $ 540 , d ) $ 560 , e ) $ 580 | a | subtract(600, multiply(subtract(850, 600), divide(2, 5))) | a sum of money lent out at s . i . amounts to a total of $ 600 after 2 years and to $ 850 after a further period of 5 years . what was the initial sum of money that was invested ? | s . i for 5 years = $ 850 - $ 600 = $ 250 the s . i . is $ 50 / year s . i . for 2 years = $ 100 principal = $ 600 - $ 100 = $ 500 the answer is a . | a = 850 - 600
b = 2 / 5
c = a * b
d = 600 - c
|
a ) 227 , b ) 240 , c ) 460 , d ) 450 , e ) 455 | d | multiply(16, multiply(54, const_0_2778)) | a train passes a station platform in 46 sec and a man standing on the platform in 16 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 16 = 240 m . let the length of the platform be x m . then , ( x + 240 ) / 46 = 15 = > x = 450 m . answer : d" | a = 54 * const_0_2778
b = 16 * a
|
a ) 4 , b ) 8 , c ) 12 , d ) 7 , e ) 18 | d | divide(power(5, const_2), 4) | if the area of a square with sides of length 5 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 5 ^ 2 = l * 4 = > l = 25 / 4 = 7 answer d" | a = 5 ** 2
b = a / 4
|
a ) rs 680 , b ) rs 780 , c ) rs 880 , d ) rs 480 , e ) rs 980 | a | divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 80), const_100) | a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . metre | "area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 80 = 68000 p in rs = 68000 / 100 = rs 680 answer : a" | a = 110 * 65
b = 2 * 5
c = 110 - b
d = 2 * 5
e = 65 - d
f = c * e
g = a - f
h = g * 80
i = h / 100
|
a ) 1 / 12 , b ) 5 / 36 , c ) 1 / 6 , d ) 1 / 3 , e ) 17 / 36 | b | divide(add(const_4, const_1), multiply(6, 6)) | a certain board game is played by rolling a pair of fair 6 - sided dice and then moving one ' s piece forward the number of spaces indicated by the sum showing on the dice . a player is frozen if her opponent ' s piece comes to rest in the space already occupied by her piece . if player a is about to roll and is currently 8 spaces behind player b , what is the probability that player b will be frozen after player a rolls ? | no . of possible outcomes = 6 * 6 = 36 no . of outcomes that result a total of 8 ( as a is 8 spaces behind b ) = 5 ( ( 2,6 ) , ( 3,5 ) , ( 4,4 ) , ( 5,3 ) , ( 6,2 ) ) so , the probability = 5 / 36 ( option b ) | a = 4 + 1
b = 6 * 6
c = a / b
|
a ) 36 , b ) 38 , c ) 40 , d ) 42 , e ) 44 | b | subtract(548, multiply(85, 6)) | we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 548 . how many green hats did we buy ? | "let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 548 . 6 ( 85 - g ) + 7 g = 548 . 510 - 6 g + 7 g = 548 . g = 548 - 510 = 38 . the answer is b ." | a = 85 * 6
b = 548 - a
|
a ) 13 seconds , b ) 25 seconds , c ) 29 seconds , d ) 21 seconds , e ) 6.25 seconds | a | divide(subtract(100, 35), 5) | in a 100 m race , sam beats john by 5 seconds . on the contrary , if sam allowed john to start 35 m ahead of sam , then sam and john reach the finishing point at the same time . how long does sam take to run the 100 m race ? | "their difference is 5 second but this difference is 0 if john allows sam to start the race from 35 m ahead . that means jhon was 35 m away from finishing line when they started together . so he will cover 35 m in 5 seconds . so his speed = 35 / 5 = 7 metre / second . so time taken = 100 / 5 = 20 seconds . so sam took = 13 seconds . correct answer = a" | a = 100 - 35
b = a / 5
|
a ) 12.5 km , b ) 50 km , c ) 60 km , d ) 8.5 km , e ) 9.5 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | when a train travels at a speed of 110 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 110 km / h 50 km / h - - - - > x + 15 distance is equal so 110 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 110 x = 50 x + 750 so the would be in km and x = 12.5 answer : a" | a = 15 / const_60
b = a * 50
c = b / 50
d = const_60 * c
|
['a ) 40 h', 'b ) 44 h', 'c ) 38 h', 'd ) 20 h', 'e ) 22 h'] | b | multiply(multiply(2, 4), 5.5) | three runners running around a circular track can complete one revolution in 2 , 4 and 5.5 respectively . when will they meet at starting point ? | time at which they meet at starting point = lcm of 2 , 4 and 5.5 = 44 h answer : b | a = 2 * 4
b = a * 5
|
a ) 64 : 65 , b ) 65 : 64 , c ) 19 : 65 , d ) 65 : 19 , e ) none of these | d | divide(add(add(6, 5), 3), add(add(1, 2), 1)) | 3 bottles contain equal mixtures of spirit and water in the ratio 6 : 1 , 5 : 2 and 3 : 1 respectively . if all the solutions are mixed together , the ratio of spirit to water in the final mixture will be | given that all bottles contain equal amount of mixture say v . so in the first vessel - > water : spirit = 1 / 7 : 6 / 7 in the second bottle - > 2 / 7 : 5 / 7 in the third bottle - > l / 4 : 3 / 4 . hence , the final ratio is = ( ( 6 / 7 ) + ( 5 / 7 ) + ( 3 / 4 ) ) / ( ( 1 / 7 ) + ( 2 / 7 ) + ( 1 / 4 ) ) = 65 / 19 answer : d | a = 6 + 5
b = a + 3
c = 1 + 2
d = c + 1
e = b / d
|
a ) 10 , b ) 3 , c ) 4 , d ) 7 , e ) 5 | b | divide(multiply(12, 10), 40) | 12 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | required number of machines = 12 * 10 / 40 = 3 answer is b | a = 12 * 10
b = a / 40
|
a ) 11 , b ) 13 , c ) 14 , d ) 12 , e ) 20 | e | divide(subtract(multiply(add(const_4, const_1), 10), 10), subtract(add(const_4, const_1), const_3)) | the age of somu is one - third his father ' s . 10 years back he was one - fifth of his father ' s age . what is his persent age ? | "explanation : let somu ' s age be x and that of his father be 3 x . so , x - 10 = 3 x - 10 / 5 = x = 20 answer : option e" | a = 4 + 1
b = a * 10
c = b - 10
d = 4 + 1
e = d - 3
f = c / e
|
a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | e | multiply(divide(const_100, add(const_100, 8)), 45) | from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 8 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 8 x = 45 - - > 27 / 25 * x = 45 - - > x = 45 * 25 / 27 = 125 / 3 = ~ 42 . answer : e ." | a = 100 + 8
b = 100 / a
c = b * 45
|
a ) 57 , b ) 67 , c ) 77 , d ) 87 , e ) 96 | e | sqrt(multiply(92.16, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 92.16 , the number of the member is the group is : | "explanation : money collected = ( 92.16 x 100 ) paise = 9216 paise . ∴ number of members = √ ( 9216 ) = 96 . answer : e" | a = 92 * 16
b = math.sqrt(a)
|
a ) 50 km , b ) 76 km , c ) 20 km , d ) 16 km , e ) 97 km | c | multiply(5, divide(20, subtract(10, 5))) | if a person walks at 10 km / hr instead of 5 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 5 = ( x + 20 ) / 10 5 x - 100 = > x = 20 km . answer : c" | a = 10 - 5
b = 20 / a
c = 5 * b
|
a ) 100 , b ) 110 , c ) 120 , d ) 175 , e ) 150 | d | divide(multiply(700, subtract(const_100, 75)), const_100) | in a public show 75 % of the seats were filled . if there were 700 seats in the hall , how many seats were vacant ? | "75 % of 600 = 75 / 100 × 700 = 450 therefore , the number of vacant seats = 700 - 450 = 175 . answer : d" | a = 100 - 75
b = 700 * a
c = b / 100
|
a ) 724533811 , b ) 353654655 , c ) 825107481 , d ) 725117481 , e ) 477899932 | c | multiply(subtract(9999, const_4), 82519) | find the value of m 82519 x 9999 = m ? | "82519 x 9999 = 82519 x ( 10000 - 1 ) = 82519 x 10000 - 82519 x 1 = 825190000 - 82519 = 825107481 c" | a = 9999 - 4
b = a * 82519
|
a ) 4 / 9 , b ) 5 / 9 , c ) 17 / 27 , d ) 23 / 27 , e ) 65 / 81 | e | subtract(const_1, power(divide(const_1, 3), 4)) | on a ranch , a rancher can place a loop of rope , called a lasso , once in every 3 throws around a cow ’ s neck . what is the probability that the rancher will be able to place a lasso around a cow ’ s neck at least once in 4 attempts ? | "p ( missing all 3 ) = ( 2 / 3 ) ^ 4 = 16 / 81 p ( success on at least one attempt ) = 1 - 16 / 81 = 65 / 81 the answer is e ." | a = 1 / 3
b = a ** 4
c = 1 - b
|
a ) 35 , b ) 40 , c ) 44 , d ) 45 , e ) 56 | c | add(multiply(2, power(4, 2)), multiply(3, 4)) | if one root of the equation 2 x ^ 2 + 3 x – k = 0 is 4 , what is the value of k ? | "we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 4 ^ 2 + 3 * 4 - k = 0 k = 32 + 12 = 44 the answer is c" | a = 4 ** 2
b = 2 * a
c = 3 * 4
d = b + c
|
a ) $ 475 , b ) $ 500 , c ) $ 525 , d ) $ 550 , e ) $ 575 | d | add(multiply(100, 4), multiply(add(multiply(2, 60), multiply(2, add(divide(multiply(60, 50), 100), 60))), divide(const_1, const_2))) | 4 different airplanes owned by a rock band will be refueling today at the same airport . there are 2 larger planes carrying people and 2 smaller planes carrying equipment . all tanks are empty and will need to be filled completely . the 2 smaller tanks hold 60 liters each and the larger tanks are 50 % bigger . fuel is . 50 cents a liter and there is a charge of $ 100 service for each plane . how much will it cost to fill all 4 planes ? | lots of calculations . 100 * 4 + 2 * 60 * . 50 + 60 * ( 3 / 2 ) * 2 * . 50 answer = $ 550 the correct option is d | a = 100 * 4
b = 2 * 60
c = 60 * 50
d = c / 100
e = d + 60
f = 2 * e
g = b + f
h = 1 / 2
i = g * h
j = a + i
|
a ) 3.6 , b ) 4.8 , c ) 3.5 , d ) 2.3 , e ) 4.4 | b | multiply(divide(40, const_100), 12) | how many litres of pure acid are there in 12 litres of a 40 % solution | explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 12 * 40 / 100 = 4.8 answer : option b | a = 40 / 100
b = a * 12
|
a ) 45 kmph , b ) 50 kmph , c ) 55 kmph , d ) 60 kmph , e ) 70 kmph | c | subtract(divide(100, multiply(6, const_0_2778)), 5) | a train 100 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | "explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 / 18 m / sec . therefore 100 / ( ( x + 5 ) × 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 1800 < = > x = 55 speed of the train is 55 kmph . answer : option c" | a = 6 * const_0_2778
b = 100 / a
c = b - 5
|
a ) - 2 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | a | divide(add(divide(subtract(4, 8), 2), 4), divide(add(2, 6), 2)) | line m lies in the xy - plane . the y - intercept of line m is - 4 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 8 ) . what is the slope of line m ? | "ans : a solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 2 ) as we can see that the y axis of intercept point is ( 0 , - 4 ) means line m is parallel to x axis slope m = - 2 ans : a" | a = 4 - 8
b = a / 2
c = b + 4
d = 2 + 6
e = d / 2
f = c / e
|
a ) 2 hrs , b ) 4 hrs , c ) 5 hrs , d ) 6 hrs , e ) 7 hrs | c | divide(const_1, add(divide(const_1, 6), divide(const_1, 4))) | two pipes a and b can fill a tank in 6 hours and 4 hours respectively . if they are opened on alternate hours and if pipe a is opened first , in how many hours , the tank shall be full ? | "a ' s work in 1 hour = 1 / 6 b ' s work in 1 hour = 1 / 4 a + b 2 hour ' s work when opened alternately = 1 / 6 + 1 / 4 = 5 / 12 a + b 4 hour ' s work when opened alternately = 10 / 12 = 5 / 6 remaining part = 1 - 5 / 6 = 1 / 6 it is a ' s turn and 1 / 6 part is filled by a in 1 hour . total time taken to fill the tank = 4 + 1 = 5 hrs answer is c" | a = 1 / 6
b = 1 / 4
c = a + b
d = 1 / c
|
a ) 48.5 , b ) 62.5 , c ) 73.5 , d ) 87.5 , e ) 96.5 | b | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 45), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 45 seconds . what is the length of each train ( in meters ) ? | the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 45 seconds , the relative difference in distance traveled is 45 * 25 / 9 = 125 meters this distance is twice the length of each train . the length of each train is 125 / 2 = 62.5 meters the answer is b . | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 45
e = d / 2
|
a ) 32 , b ) 26 , c ) 36 , d ) 28 , e ) 24 | c | subtract(divide(20, divide(4, 9)), 20) | a certain lab experiments with white and brown mice only . in one experiment , 4 / 9 of the mice are brown . if there are 20 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of brown mice = 4 / 9 m number of white mice = 5 / 9 m = 20 = > m = 36 answer c" | a = 4 / 9
b = 20 / a
c = b - 20
|
a ) 223 , b ) 233 , c ) 240 , d ) 243 , e ) 253 | e | divide(divide(factorial(add(21, const_2)), factorial(21)), const_2) | if a + b + c = 21 , what is the total number of non - negative integral solutions ? | think of it is a problem of diving 21 items to three people a , b & c which may or may not receive an item . so simply use combination here as ( n - r + 1 ) c ( r - 1 ) i . e . ( 21 + 3 - 1 ) c ( 3 - 1 ) = > 23 c 2 = = > 253 solutions . answer : e | a = 21 + 2
b = math.factorial(a)
c = math.factorial(21)
d = b / c
e = d / 2
|
a ) 30 , b ) 32 , c ) 36 , d ) 40 , e ) 42 | b | divide(60, add(divide(30, 48), divide(subtract(60, 30), 24))) | a driver goes on a trip of 60 kilometers , the first 30 kilometers at 48 kilometers per hour and the remaining distance at 24 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | "the time for the first part of the trip was 30 / 48 = 5 / 8 hours . the time for the second part of the trip was 30 / 24 = 5 / 4 hours . the total time fro the trip was 5 / 8 + 5 / 4 = 15 / 8 hours . the average speed for the trip was 60 / ( 15 / 8 ) = 32 kph the answer is b ." | a = 30 / 48
b = 60 - 30
c = b / 24
d = a + c
e = 60 / d
|
a ) 480 . , b ) 500 . , c ) 510 . , d ) 600 . , e ) 800 . | a | multiply(add(divide(120, subtract(160, 120)), const_1), 120) | a train travelled from point a to point b at 160 km / h . on its way back the train travelled at 120 km / h and therefore the trip back lasted one hour longer . what is the distance ( in km ) between a and b ? | distance = speed * time d 1 = s 1 t 1 d 2 = s 2 t 2 the distance from point a to point b is the same for each trip so , d 1 = d 2 and t 2 = t 1 + 1 thus , s 1 t 1 = s 2 ( t 1 + 1 ) 160 t 1 = 120 ( t 1 + 1 ) t 1 = 3 160 * 3 = 480 answer : a | a = 160 - 120
b = 120 / a
c = b + 1
d = c * 120
|
a ) 140 / 217 , b ) 156 / 185 , c ) 217 / 140 , d ) 263 / 149 , e ) 241 / 163 | c | subtract(divide(subtract(15, const_1), add(6, const_1)), divide(add(8, const_1), subtract(21, const_1))) | if a is an integer greater than 8 but less than 15 and b is an integer greater than 6 but less than 21 , what is the range of a / b ? | the way to approach this problem is 8 < a < 15 and 6 < b < 21 minimum possible value of a is 9 and maximum is 14 minimum possible value of b is 7 and maximum is 20 range = max a / min b - min a / max b ( highest - lowest ) 14 / 7 - 9 / 20 = 217 / 140 hence c | a = 15 - 1
b = 6 + 1
c = a / b
d = 8 + 1
e = 21 - 1
f = d / e
g = c - f
|
a ) 450 , b ) 375 , c ) 500 , d ) 425 , e ) 400 | d | subtract(divide(add(25, 225), const_2), divide(add(50, 1050), const_2)) | the average ( arithmetic mean ) of the integers from 25 to 225 , inclusive , is how much lower than the average of the integers from 50 to 1050 inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 50 to 1050 , inclusive = ( 50 + 1050 ) / 2 = 550 average of numbers between 25 to 225 , inclusive = ( 25 + 225 ) / 2 = 125 difference = 550 - 125 = 425 answer is d" | a = 25 + 225
b = a / 2
c = 50 + 1050
d = c / 2
e = b - d
|
a ) 2.25 hours , b ) 8.75 hours , c ) 12 hours , d ) 14.25 hours , e ) 9.6 hours | e | divide(multiply(6, 8), 5) | 8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 5 such machines to perform the same job ? | "let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 5 machines 5 units of work ( 5 machines ) - - - > 1 hour for 48 units of work 5 * 9.6 - - - > 1 * 9.6 hours e 9.6 hours" | a = 6 * 8
b = a / 5
|
['a ) 37.2 %', 'b ) 37.5 %', 'c ) 30.2 %', 'd ) 37.7 %', 'e ) 33.2 %'] | b | subtract(divide(multiply(add(const_100, 10), add(const_100, 25)), const_100), const_100) | the length and breadth of a rectangle is increased by 10 % and 25 % respectively . what is the increase in the area ?. a . 27.5 % | explanation : 100 * 100 = 10000 110 * 125 = 13750 - - - - - - - - - - - 3750 10000 - - - - - - 3750 100 - - - - - - - ? = > 37.5 % answer : option b | a = 100 + 10
b = 100 + 25
c = a * b
d = c / 100
e = d - 100
|
a ) 20 , b ) 30 , c ) 38 , d ) 48 , e ) none of these | b | multiply(divide(98, add(add(2, 3), divide(multiply(8, 3), 5))), 3) | the sum of 3 numbers is 98 . if the ratio of the first to the second is 2 : 3 and that of the second to the third is 5 : 8 , then the second number is : | a : b = 2 : 3 = 2 × 5 : 3 × 5 = 10 : 15 and b : c = 5 : 8 = 5 × 3 : 8 × 3 = 15 : 24 therefore , a : b : c = 10 : 15 : 24 ∴ a : b : c = 10 : 15 : 24 let the number be 10 x , 15 x and 24 x . then , 10 x + 15 x + 24 x = 98 or 49 x = 98 or x = 2 ⇒ second number = 15 x = 15 × 2 = 30 answer b | a = 2 + 3
b = 8 * 3
c = b / 5
d = a + c
e = 98 / d
f = e * 3
|
a ) 3 m , b ) 4 m , c ) 7 m , d ) 9 m , e ) 10 m | c | multiply(divide(56, multiply(20, 42)), multiply(35, 3)) | if 20 men can build a water fountain 56 metres long in 42 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 42 : 3 : : 56 : x therefore ( 20 x 42 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 840 = 7 hence , the required length is 7 m . answer : c" | a = 20 * 42
b = 56 / a
c = 35 * 3
d = b * c
|
a ) 1234 , b ) 1345 , c ) 1456 , d ) 1797 , e ) 1635 | d | multiply(divide(subtract(1500, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1500 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1500 ) . x + 1500 = 6 x + 15 5 x = 1485 x = 297 large number = 297 + 1500 = 1797 d" | a = 1500 - 15
b = 6 - 1
c = a / b
d = c * 6
|
a ) a ) 4 , b ) b ) 8 , c ) c ) 9 , d ) d ) 10 , e ) e ) 911 | c | sqrt(27) | from below option 27 is divisible by which one ? | "27 / 9 = 3 c" | a = math.sqrt(27)
|
a ) 80 , b ) 70 , c ) 56 , d ) 16 , e ) 40 | e | multiply(divide(80, add(add(2, 3), 5)), 5) | the amounts of time that three secretaries worked on a special project are in the ratio of 2 to 3 to 5 . if they worked a combined total of 80 hours , how many hours did the secretary who worked the longest spend on the project ? | "10 x = 80 = > x = 8 therefore the secretary who worked the longest spent 8 x 5 = 40 hours on the project option ( e )" | a = 2 + 3
b = a + 5
c = 80 / b
d = c * 5
|
a ) 56 , b ) 70 , c ) 78 , d ) 80 , e ) 85 | c | multiply(divide(add(multiply(subtract(1, divide(20, const_100)), 3), multiply(subtract(1, divide(25, const_100)), 2)), add(3, 2)), const_100) | the ratio of number of boys and girls in a class is 3 : 2 . in the 1 st semester exam 20 % of boys and 25 % of girls get more than or equal to 90 % marks . what percentage of students get less than 90 % marks ? | let boys = 3 x and girls = 2 x . number of those who get less than 90 % mark = ( 80 % of 3 x ) + ( 75 % of 2 x ) = ( 80 / 100 ) * 3 x + ( 75 / 100 * 2 x ) = 39 x / 10 required percentage = ( 39 x / 10 * 1 / 5 x * 100 ) % = 78 % . answer : c | a = 20 / 100
b = 1 - a
c = b * 3
d = 25 / 100
e = 1 - d
f = e * 2
g = c + f
h = 3 + 2
i = g / h
j = i * 100
|
a ) − 100 , b ) 0.015 , c ) 0.25 , d ) 4 , e ) 8 | b | divide(divide(2, 2), const_1000) | if x = 4 and y = − 2 , what is the value of ( x − 2 y ) ^ y ? | "quickly we can spot that answer is neither integer nor negative . eliminate a , de by inversing and squaring 0.015 answer : b" | a = 2 / 2
b = a / 1000
|
a ) 20 / 323 , b ) 40 / 323 , c ) 60 / 273 , d ) 50 / 373 , e ) 40 / 373 | b | divide(multiply(choose(6, 1), choose(5, 2)), choose(add(add(5, 6), 8), 4)) | an urn contains 5 red , 6 blue and 8 green balls . 4 balls are randomly selected from the urn , find the probability that the drawn ball are 1 blue and 2 red and 1 green balls ? | "sample space = no . of ways 4 balls were drawn from urn = 19 c 4 = 3876 no . ways 1 blue and 2 red and 1 green balls were drawn from bag = 6 c 1 * 5 c 2 * 8 c 1 = 480 probability = 480 / 3876 = 40 / 323 ans - b" | a = math.comb(6, 1)
b = math.comb(5, 2)
c = a * b
d = 5 + 6
e = d + 8
f = math.comb(e, 4)
g = c / f
|
a ) 48 kmph , b ) 50 kmph , c ) 52 kmph , d ) 56 kmph , e ) 60 kmph | d | divide(504, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 504 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | time = 6 distence = 504 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 504 / 9 = 56 kmph d | a = 6 * 3
b = a / 2
c = 504 / b
|
a ) 2.3 % , b ) 3.4 % , c ) 4.5 % , d ) 5.6 % , e ) 6.7 % | b | multiply(divide(multiply(divide(1, add(const_3, const_4)), inverse(add(const_2, const_3))), add(multiply(divide(subtract(15, 1), 15), divide(add(const_3, const_3), add(const_3, const_4))), multiply(divide(1, add(const_3, const_4)), inverse(add(const_2, const_3))))), const_100) | one - seventh of the light switches produced by a certain factory are defective . 4 - fifths of the defective switches are rejected and 1 / 15 of the non defective switches are rejected by mistake . if all the switches not rejected are sold , what percent of the switches sold by the factory are defective ? | 1 / 7 of the switches are defective . the defective switches that are not rejected are 1 / 5 * 1 / 7 = 1 / 35 = 3 / 105 of all switches . the non defective switches that are sold are 6 / 7 * 14 / 15 = 84 / 105 of all switches . the percent of switches sold that are defective is 3 / 87 which is about 3.4 % . the answer is b . | a = 3 + 4
b = 1 / a
c = 2 + 3
d = 1/(c)
e = b * d
f = 15 - 1
g = f / 15
h = 3 + 3
i = 3 + 4
j = h / i
k = g * j
l = 3 + 4
m = 1 / l
n = 2 + 3
o = 1/(n)
p = m * o
q = k + p
r = e / q
s = r * 100
|
a ) a ) 16 , b ) b ) 18 , c ) c ) 19 , d ) d ) 22 , e ) e ) 24 | e | add(23, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 23 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 23 + 36 – 26 = 240 = > 240 / 10 = 24 e )" | a = 36 - 26
b = a / 10
c = 23 + b
|
a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 63 % | b | floor(multiply(subtract(divide(9, 63), divide(7, 80)), const_100)) | a survey was sent to 80 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 80 ) = x / 100 x = 9 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 9 ) % = 5 % answer is b" | a = 9 / 63
b = 7 / 80
c = a - b
d = c * 100
e = math.floor(d)
|
a ) 1200 , b ) 120 , c ) 360 , d ) 200 , e ) none of these | d | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(42, add(32, 4))), 4) | average age of students of an adult school is 42 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | "explanation : let the original no . of students be x . according to situation , 42 x + 120 * 32 = ( x + 120 ) 36 ⇒ x = 80 so , required no . of students after joining the new students = x + 120 = 200 . answer : d" | a = 32 + 4
b = a * 120
c = 120 * 32
d = b - c
e = 32 + 4
f = 42 - e
g = d / f
h = g * 4
|
a ) 183 , b ) 185 , c ) 182 , d ) 184 , e ) 181 | b | add(add(add(50, 53), 46), 46) | you have been given a physical balance and 7 weights of 45 , 50 , 53 , 46 , 52 , 46 and 80 kgs . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 188 kgs . | "80 + 52 + 53 = 185 answer : b" | a = 50 + 53
b = a + 46
c = b + 46
|
a ) 580 seconds , b ) 680 seconds , c ) 500 seconds , d ) 540 seconds , e ) 560 seconds | a | multiply(const_4, subtract(146, const_1)) | a person has to make 146 pieces of long bar . he take $ seconds to cut piece . what is total time taken by him to make 146 pieces ? | if it is 4 seconds , as u have mistaken with shift key , then . . . . we ' ll have to make 145 cuts for 146 pieces . so 145 * 5 = 580 seconds answer : a | a = 146 - 1
b = 4 * a
|
a ) 15 , b ) 18 , c ) 20 , d ) 24 , e ) 36 | e | divide(multiply(divide(multiply(10, 45), subtract(60, 45)), 60), add(20, divide(multiply(10, 45), subtract(60, 45)))) | a tank can supply water to a village for 60 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 45 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | losing 10 liters per day results in a loss of 450 liters in 45 days . so , those 450 liters were for 15 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 60 = 1800 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last 1800 / 50 = 36 days . the answer is e . | a = 10 * 45
b = 60 - 45
c = a / b
d = c * 60
e = 10 * 45
f = 60 - 45
g = e / f
h = 20 + g
i = d / h
|
a ) 18 , b ) 750 , c ) 24 , d ) 20 , e ) none of these | d | divide(circle_area(divide(32, multiply(2, const_pi))), 4) | how many plants will be there in a circular bed whose outer edge measure 32 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 32 cm area of circular bed = ( 32 ) 2 â „ 4 ï € space for each plant = 4 cm 2 â ˆ ´ required number of plants = ( 32 ) 2 â „ 4 ï € ã · 4 = 20.36 = 20 ( approx ) answer d" | a = 2 * math.pi
b = 32 / a
c = circle_area / (
|
a ) 75 kg , b ) 55 kg , c ) 80 kg , d ) 85 kg , e ) 25 kg | c | add(multiply(2.5, 8), 60) | he average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 60 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 60 + 20 ) kg = 80 kg . answer : c" | a = 2 * 5
b = a + 60
|
a ) a ) 30000 , b ) b ) 34000 , c ) c ) 34098 , d ) d ) 33007 , e ) e ) 44098 | a | subtract(65000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 65000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 65000 share of b = 6 / 13 ( 65000 ) = rs . 30000 answer : a" | a = const_60 * 100
b = 65000 - a
|
a ) 10 % decrease , b ) 19 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 50 % decrease | b | subtract(const_100, multiply(power(divide(10, const_100), const_2), const_100)) | if the radius of a circle is decreased 10 % , what happens to the area ? | "area of square = pi * radius ^ 2 new radius = 0.9 * old radius so new area = ( 0.9 ) ^ 2 old area = > 0.81 of old area = > 81 % old area ans : b" | a = 10 / 100
b = a ** 2
c = b * 100
d = 100 - c
|
a ) 1200 , b ) 180 , c ) 360 , d ) 240 , e ) none of these | b | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(44, add(32, 4))), 4) | average age of students of an adult school is 44 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | "explanation : let the original no . of students be x . according to situation , 44 x + 120 * 32 = ( x + 120 ) 36 ⇒ x = 60 so , required no . of students after joining the new students = x + 120 = 180 answer : b" | a = 32 + 4
b = a * 120
c = 120 * 32
d = b - c
e = 32 + 4
f = 44 - e
g = d / f
h = g * 4
|
a ) 36 , b ) 39 , c ) 42 , d ) 45 , e ) none of the above | e | add(subtract(87, multiply(17, 2)), 2) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his average by 2 . find his average after 17 th inning . | "let the average after 17 th inning = x . then , average after 16 th inning = ( x – 2 ) . ∴ 16 ( x – 2 ) + 87 = 17 x or x = ( 87 – 32 ) = 55 . answer e" | a = 17 * 2
b = 87 - a
c = b + 2
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | divide(subtract(multiply(24, 5), 25), subtract(5, divide(1, 4))) | 24 oz of juice p and 25 oz of juice v are mixed to make smothies a and y . the ratio of p to v in smothie a is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie a ? | the ratio of p to v in smothie x is 4 is to 1 and that in y is 1 is to 5 . p 1 + p 2 = 24 v 1 + v 2 = 25 p 1 = 4 v 1 p 2 = v 2 / 5 4 v 1 + v 2 / 5 = 24 v 1 + v 2 = 25 4 v 2 - v 2 / 5 = 76 19 v 2 / 5 = 76 = > v 2 = 20 = > v 1 = 5 = > p 1 = 20 answer - d | a = 24 * 5
b = a - 25
c = 1 / 4
d = 5 - c
e = b / d
|
a ) 80 , b ) 70 , c ) 56 , d ) 50 , e ) 105 | d | add(100, divide(subtract(60, 70), subtract(divide(70, 100), divide(50, 100)))) | during a certain season , a team won 60 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ? | "we are first given that a team won 60 percent of its first 100 games . this means the team won 0.6 x 100 = 60 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 equals the number of remaining games in the season . thus we can say : 0.5 ( t – 100 ) = number of wins for remaining games 0.5 t – 50 = number of wins for remaining games lastly , we are given that team won 70 percent of all games played in the season . that is , they won 0.7 t games in the entire season . with this we can set up the equation : number of first 100 games won + number of games won for remaining games = total number of games won in the entire season 60 + 0.5 t – 50 = 0.7 t 10 = 0.2 t 100 = 2 t 50 = t answer is d ." | a = 60 - 70
b = 70 / 100
c = 50 / 100
d = b - c
e = a / d
f = 100 + e
|
a ) 400 , b ) 625 , c ) 1250 , d ) 2500 , e ) 500 | e | divide(20, divide(2, 20)) | in a certain pond , 20 fish were caught , tagged , and returned to the pond . a few days later , 20 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "if x is total number of fish in the pond : 4 = 20 / x * 100 = > x = 500 so answer is e" | a = 2 / 20
b = 20 / a
|
a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | d | add(40, divide(subtract(976, multiply(18, 40)), divide(multiply(18, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 18 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 18 = 720 excess = 976 - 720 = 252 for extra hours = . 75 ( 18 ) = 13.5 + 18 = 31.5 number of extra hrs = 252 / 31.5 = 8 total hrs = 40 + 8 = 48 answer d 48" | a = 18 * 40
b = 976 - a
c = 100 + 75
d = 18 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 200 , b ) 768 , c ) 276 , d ) 750 , e ) 279 | d | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 75)), 1000), multiply(divide(const_100, add(const_100, 75)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 75)), 1000), multiply(divide(const_100, add(const_100, 75)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 75 % profit on the first article and 75 % loss on second article . find the net profit or loss percent ? | "profit on first article = 75 % of 1000 = 750 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : d" | a = 100 - 75
b = 100 / a
c = b * 1000
d = 100 + 75
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 75
l = 100 / k
m = l * 1000
n = 100 + 75
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 13 , b ) 33 , c ) 50 , d ) 51 , e ) 52 | c | multiply(multiply(50, divide(1, add(1, 2))), add(1, 2)) | a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 50 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ? | "the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . it means there is no effect of change in volume of water to the costing of the solution so a solution of 1 liter of spirit and 2 liters of water will cost 50 cents answer = 50 answer : c" | a = 1 + 2
b = 1 / a
c = 50 * b
d = 1 + 2
e = c * d
|
a ) 50 , b ) 55 , c ) 60 , d ) 65 , e ) 70 | b | subtract(90, subtract(subtract(180, 110), 35)) | in a company of 180 employees , 110 are females . a total of 90 employees have advanced degrees and the rest have a college degree only . if 35 employees are males with a college degree only , how many employees are females with advanced degrees ? | "the number of males is 180 - 110 = 70 . the number of males with advanced degrees is 70 - 35 = 35 . the number of females with advanced degrees is 90 - 35 = 55 . the answer is b ." | a = 180 - 110
b = a - 35
c = 90 - b
|
a ) 57 , b ) 56 , c ) 55 , d ) 54 , e ) 53 | b | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 8,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . b" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 400 , b ) 300 , c ) 200 , d ) 100 , e ) 150 | d | divide(500, add(divide(1000, 500), divide(1500, 500))) | c and d started a business by investing rs . 1000 / - and rs . 1500 / - respectively . find the d ’ s share out of a total profit of rs . 500 : | c = rs . 1000 / - d = rs . 1500 / - c share 2 parts & d share 3 parts total 5 parts - - - - - > rs . 500 / - - - - - > 1 part - - - - - - - > rs . 100 / - d share = 3 parts - - - - - > rs . 300 / - d | a = 1000 / 500
b = 1500 / 500
c = a + b
d = 500 / c
|
a ) 32 , b ) 56 , c ) 29 , d ) 27 , e ) 21 | b | divide(divide(add(400, 300), const_1000), divide(45, const_3600)) | a train 400 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 400 + 300 ) / 45 = 700 / 45 * 18 / 5 = 56 answer : b" | a = 400 + 300
b = a / 1000
c = 45 / 3600
d = b / c
|
a ) 63 , b ) 64 , c ) 65 , d ) 66 , e ) 67 | a | multiply(reminder(70, subtract(const_10, const_1)), subtract(const_10, const_1)) | q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 70 , what is the greatest possible value of q minus r ? | "a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 70 . the greatest multiple of 9 which is less than 70 is 63 . the answer is a ." | a = 10 - 1
b = reminder * (
|
a ) 5 / 3 , b ) 2 / 7 , c ) 7 / 3 , d ) 1 / 7 , e ) 1 / 4 | c | subtract(multiply(divide(4, 3), 4), 3) | at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 7 liters of forest green to change it to verdant green ? | 7 liter of forset green have 4 liter of blue and 3 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ¾ so the equation is blue / yellow = 4 / ( 3 + x ) = ¾ 9 + 3 x = 16 = > x = 7 / 3 answer : c | a = 4 / 3
b = a * 4
c = b - 3
|
a ) 30 , b ) 33 , c ) 36 , d ) 39 , e ) 42 | b | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive multiples of 3 is 90 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 90 9 x = 81 x = 9 largest number = 3 x + 6 = 33 answer : b" | a = 3 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 3 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 3 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 3 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) $ 150 , b ) $ 220 , c ) $ 225 , d ) $ 182 , e ) $ 189 | c | subtract(subtract(multiply(3600, power(add(divide(25, const_100), const_1), 2)), 3600), multiply(multiply(3600, divide(25, const_100)), 2)) | what will be the difference between simple and compound interest @ 25 % per annum on a sum of $ 3600 after 2 years ? | "s . i . = 3600 * 25 * 2 / 100 = $ 1800 c . i . = 3600 * ( 1 + 25 / 100 ) ^ 2 - 3600 = $ 2025 difference = 2025 - 1800 = $ 225 answer is c" | a = 25 / 100
b = a + 1
c = b ** 2
d = 3600 * c
e = d - 3600
f = 25 / 100
g = 3600 * f
h = g * 2
i = e - h
|
a ) 1 , b ) 2 , c ) 3 , d ) 8 , e ) 11 | c | multiply(3, divide(8, 8)) | if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 8 people to do 8 times of that particular work ? | "3 people can do the work one time in one day . 1 person can do 1 / 3 of the work in one day . 8 people can do 8 / 3 of the work in one day . 8 people can do 8 times the work in 3 days . the answer is c ." | a = 8 / 8
b = 3 * a
|
a ) 98 days , b ) 21 days , c ) 48 days , d ) 18 days , e ) 19 days | c | multiply(32, divide(const_3, const_2)) | a is half good a work man as b and together they finish a job in 32 days . in how many days working alone b finish the job ? | "wc = 1 : 2 2 x + x = 1 / 32 = > x = 1 / 96 2 x = 1 / 96 = > 48 days answer : c" | a = 3 / 2
b = 32 * a
|
a ) 288 , b ) 2302 , c ) 2304 , d ) 8640 , e ) 964 | c | multiply(divide(const_4, 3), power(3, 3)) | the measurement of a rectangular box with lid is 25 cmx 24 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 24 , r = 12 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 12 * 12 * 12 = 2304 π cm 3 answer : c" | a = 4 / 3
b = 3 ** 3
c = a * b
|
['a ) 1 %', 'b ) 10 %', 'c ) 100 %', 'd ) 500 %', 'e ) 1000 %'] | a | divide(multiply(10, 10), const_100) | if the sides of a square are multiplied by 10 , the area of the original square is how many times as large as the area of the resultant square ? | let x be the original length of one side . then the original area is x ^ 2 . the new square has sides of length 10 x , so the area is 100 x ^ 2 . the area of the original square is 1 / 100 = 1 % times the area of the new square . the answer is a . | a = 10 * 10
b = a / 100
|
a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(2, multiply(const_3, const_2)) | machine t takes 2 more hours than machine b to make 20 widgets . if working together , the machines can make 25 widgets in 3 hours , how long will it take machine t to make 40 widgets ? | i approached this one by plugging in numbers . . . started with c . if 40 are made in 8 hours , then 20 are made in 4 hours . so time of t is 4 , and time of b is 2 . rate together : 20 / 4 + 20 / 2 = 5 + 10 = 15 . so in 1 hour , together make 15 widgets . in 3 hours = 45 . way too much . we can eliminate right away c , b , and a - because b and t reduces the time - the total # of widgets made will be even higher . now between d and e - > try only one . . if it does n ' t work , then the other one is the answer . i picked e : 12 h to make 40 widgets , and 6 hours to make 20 . this is the time of t . time of b = 4 hours . 20 / 6 + 20 / 4 = 10 / 3 + 20 / 4 find lcm of 3 and 4 = 12 . multiply first by 4 , and second by 3 : 40 + 60 / 12 = 100 / 12 divide by 4 : 25 / 3 so this is the rate given . e is the correct answer | a = 3 * 2
b = 2 * a
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.