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 |
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a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 70 min | e | multiply(const_10, multiply(const_1, 7)) | a pipe takes a hours to fill the tank . but because of a leakage it took 7 times of its original time . find the time taken by the leakage to empty the tank | pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 420 x = 70 min answer : e | a = 1 * 7
b = 10 * a
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a ) 2.625 cm , b ) 3 cm , c ) 10.5 cm , d ) 15 cm , e ) none of these | c | divide(multiply(3, add(const_3, const_4)), const_2) | at chennai it rained as much on tuesday as on all the others days of the week combined . if the average rainfall for the whole week was 3 cm . how much did it rain on tuesday ? | total rainfall = 3 x 7 = 21 cm hence , rainfall received on tuesday = 21 / 2 ( as it rained as much on tuesday as on all the others days of the week combined ) = 10.5 cm . answer : c | a = 3 + 4
b = 3 * a
c = b / 2
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a ) 100 % , b ) 200 % , c ) 250 % , d ) 300 % , e ) 50 % | a | divide(multiply(2, const_100), 4) | a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 2 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3 he selling at 2 a rupee , his sp = 4 profit = sp - cp = 4 - 2 = 2 profit percent = 2 / 2 * 100 = 100 % answer : a" | a = 2 * 100
b = a / 4
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a ) 14 , b ) 12 , c ) 1200 , d ) 1.2 , e ) 14.11 | b | multiply(multiply(3, power(10, 2)), multiply(4, power(10, negate(2)))) | ( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = ? | 3 * 10 ^ 2 = 300 4 * 10 ^ - 2 = 0.04 ( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = 300 * 0.04 = 12.00 the answer is option b | a = 10 ** 2
b = 3 * a
c = 10 ** negate
d = 4 * c
e = b * d
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a ) 24 , b ) 26 , c ) 20 , d ) 29 , e ) 30 | c | sqrt(divide(60, add(power(3, 2), add(power(1, 2), power(2, 2))))) | the ratio of three numbers is 1 : 2 : 3 and their sum is 60 . the second number of the three numbers is ? | "1 : 2 : 3 total parts = 6 6 parts - - > 60 1 part - - - - > 10 the second number of the three numbers is = 2 * 10 = 20 answer : c" | a = 3 ** 2
b = 1 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 60 / e
g = math.sqrt(f)
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a ) 1.5 , b ) 2.5 , c ) 3.75 , d ) 5 , e ) 7.5 | b | divide(divide(multiply(add(3, 2), 3), 2), 3) | natasha climbs up a hill , and descends along the same way she went up . it takes her 3 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "lets assume distance to top as x , so the total distance travelled by natasha = 2 x total time taken = 3 + 2 = 5 hrs avg speed = total dist / total time taken = 2 x / 5 avg speed of complete journey is given as = 3 hrs 2 x / 5 = 3 x = 7.5 miles avg speed while climbing = distance / time = 7.5 / 3 = 2.5 option b" | a = 3 + 2
b = a * 3
c = b / 2
d = c / 3
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a ) 20 , b ) k = 25 , c ) 30 , d ) k = 35 , e ) 40 | d | divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100)) | the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years k in which the population would be doubled assuming there is no emigration or immigration is | "suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 / 100 ) ) ^ n n = 35 answer is d ." | a = 10 * 3
b = 100 - a
c = 39 - 4
d = c / 1000
e = d * 100
f = b / e
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['a ) 113', 'b ) 114', 'c ) 115', 'd ) 116', 'e ) 117'] | b | add(multiply(multiply(const_1, 9), 5), add(subtract(multiply(multiply(const_1, 9), 5), multiply(add(const_4, const_3), const_3)), multiply(multiply(const_1, 9), 5))) | susan made a block with small cubes of 8 cubic cm volume to make a block , 3 small cubes long , 9 small cubes wide and 5 small cubes deep . she realizes that she has used more small cubes than she really needed . she realized that she could have glued a fewer number of cubes together to lock like a block with same dimensions , if it were made hollow . what is the minimum number of cubes that she needs to make the block ? | the total volume ( in terms of number of cubes ) of the solid = 3 * 9 * 5 = 135 the total volume ( in terms of number of cubes ) of the hollow = ( 3 - 2 ) * ( 9 - 2 ) * ( 5 - 2 ) = 21 so number of cubes required = 135 - 21 = 114 answer : b | a = 1 * 9
b = a * 5
c = 1 * 9
d = c * 5
e = 4 + 3
f = e * 3
g = d - f
h = 1 * 9
i = h * 5
j = g + i
k = b + j
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a ) 32 , b ) 50 , c ) 40 , d ) 20 , e ) 10 | c | multiply(subtract(83, 63), const_2) | a pupil ' s marks were wrongly entered as 83 instead of 63 . due to that the average marks for the class got increased by half ( 1 / 2 ) . the number of pupils in the class is : | "let there be x pupils in the class . total increase in marks = x x 1 / 2 = x / 2 x / 2 = ( 83 - 63 ) x / 2 = 20 x = 40 . answer : c" | a = 83 - 63
b = a * 2
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a ) 5 % , b ) 20 % , c ) 50 % , d ) 300 % , e ) 500 % | d | multiply(divide(120, 40), const_100) | 120 is what percent of 40 ? | "40 * x = 120 - - > x = 3 - - > 3 expressed as percent is 300 % . answer : d ." | a = 120 / 40
b = a * 100
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a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 2 | d | divide(subtract(divide(9, add(6, 9)), divide(6, 9)), subtract(divide(6, 9), divide(6, add(6, 9)))) | an alloy of copper and zinc contains copper and zinc in the ratio 9 : 6 . another alloy of copper and zinc contains copper and zinc in the ratio 6 : 9 . 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 = 9 x / 15 , and fraction of zinc in alloy _ 1 = 6 x / 15 similarly , fraction of copper in alloy _ 2 = 6 y / 15 , and fraction of zinc in alloy _ 2 = 9 y / 15 . mixing them , we get copper = 9 x / 15 + 6 y / 15 ; zinc = 6 x / 15 + 9 y / 15 so , 9 x + 6 y = 6 x + 9 y - > 3 x = 3 y - > x / y = 3 / 3 = 1 / 1 so , they must be mixed in the ratio 1 : 1 answer : d" | a = 6 + 9
b = 9 / a
c = 6 / 9
d = b - c
e = 6 / 9
f = 6 + 9
g = 6 / f
h = e - g
i = d / h
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a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 5 | e | subtract(const_1, divide(12, multiply(12, const_3))) | a searchlight on top of the watchtower makes 4 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 12 seconds ? | "the searchlight completes one revolution every 15 seconds . the probability that the man ' s area will be lit up is 12 / 15 = 4 / 5 . the probability that he will stay in the dark is 1 - 4 / 5 = 1 / 5 the answer is e ." | a = 12 * 3
b = 12 / a
c = 1 - b
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['a ) 4', 'b ) 9', 'c ) 3', 'd ) 5', 'e ) 7'] | d | divide(divide(divide(divide(divide(divide(2880, const_4), const_3), const_4), const_3), const_2), const_2) | which is the smallest no which divides 2880 and gives a perfect square ? | 2880 = 20 * 144 ( because w . k . t 144 is perfect square ) = 2 * 10 * 144 = 2 * 2 * 5 * 144 = 4 * 5 * 144 ( here 4 and 144 are perfect square ) so the smallest number is 5 answer : d | a = 2880 / 4
b = a / 3
c = b / 4
d = c / 3
e = d / 2
f = e / 2
|
a ) 10.09 , b ) 1.06 , c ) 10.06 , d ) 201.8 , e ) 100.9 | d | divide(4.036, 0.02) | 4.036 divided by 0.02 gives : | "= 4.036 / 0.02 = 403.6 / 2 = 201.8 answer is d ." | a = 4 / 36
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a ) 2.5 mile , b ) 1.9 mile , c ) 1.7 mile , d ) 1.5 mile , e ) 1.18 mile | e | multiply(divide(subtract(5, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | stacy and heather are 5 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ? . | "ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 5 - 2.4 = 2.6 time taken = distance / speed = 2.6 / 11 = 0.23 hrs heather will cover = 5 * 0.23 = 1.18 miles . answer e" | a = 1 + 5
b = a / const_60
c = 24 * b
d = 5 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
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a ) 30 , b ) 887 , c ) 167 , d ) 197 , e ) 161 | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(180, divide(20, divide(15, const_3)))) | a train crosses a platform of 180 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be β x β x + 180 / 15 = x + 250 / 20 20 x + 3600 = 15 x + 3750 5 x = 150 x = 30 m answer : a" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 180 * e
g = c - f
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a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 16 , d ) 2 ^ 35 , e ) 2 ^ 8 | e | divide(multiply(2, add(2, 2)), 2) | 2 + 2 + 2 Β² + 2 Β³ . . . + 2 ^ 7 | "2 + 2 = 2 ^ 2 2 ^ 2 + 2 ^ 2 = ( 2 ^ 2 ) * ( 1 + 1 ) = 2 ^ 3 2 ^ 3 + 2 ^ 3 = ( 2 ^ 3 ) * ( 1 + 1 ) = 2 ^ 4 so you can notice the pattern . . . in the end you will have 2 ^ 7 + 2 ^ 7 , which will give you 2 ^ 8 answer e" | a = 2 + 2
b = 2 * a
c = b / 2
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a ) rs . 31 , b ) rs . 32.10 , c ) rs . 40.40 , d ) rs . 64.10 , e ) none | d | subtract(subtract(multiply(power(add(divide(10, const_100), const_1), const_4), 1000), 1000), divide(multiply(4, multiply(10, 1000)), const_100)) | what will be the difference between simple interest and compound interest @ 10 % per annum on a sum of rs . 1000 after 4 years ? | solution s . i . = rs ( 1000 x 10 x 4 / 100 ) = rs . 400 . c . i . = rs . [ 1000 x ( 1 x 10 / 100 ) 4 - 1000 ] = rs . 464.10 β΄ difference = rs . ( 464.10 - 400 ) = rs . 64.10 answer d | a = 10 / 100
b = a + 1
c = b ** 4
d = c * 1000
e = d - 1000
f = 10 * 1000
g = 4 * f
h = g / 100
i = e - h
|
a ) 6 km , b ) 7 km , c ) 8 km , d ) 9 km , e ) none of these | a | divide(multiply(3, 4), const_2) | two men start together to walk a certain distance , one at 4 kmph and another at 3 kmph . the former arrives half an hour before the latter . find the distance . | let the distance be x km . then , x / 3 - x / 4 = 1 / 2 ( 4 x - 3 x ) / 12 = 1 / 2 x = 6 km answer : a | a = 3 * 4
b = a / 2
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a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 100 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 25) | cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 25 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 25 shows a cube of a number could be right . here 64 + 25 = 89 72 + 25 = 97 86 + 25 = 111 98 + 25 = 123 100 + 25 = 125 - - - ( 5 * 5 * 5 ) so we have 100 as the answer ! e" | a = 1 + 3
b = 1 + a
c = b ** 3
d = c - 25
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a ) 12 , b ) 32 , c ) 5 , d ) 28 , e ) 35 | d | subtract(35, subtract(26, 19)) | there are 35 kids in a play school . 26 kids play with white color balls and some kids play with yellow color balls and 19 kids play with both color balls . how many kids use yellow color balls ? | explanation : n ( aub ) = n ( a ) + n ( b ) - n ( anb ) . . . . ( 1 ) n ( a ) = number of kids playing with white color balls = 26 n ( aub ) = total number of kids in the play school = 35 n ( anb ) = number of kids playing with both white and yellow color balls = 19 we have to find , n ( b ) = number of kids playing with yellow color balls from ( 1 ) , 35 = 26 + n ( b ) - 19 n ( b ) = 35 + 19 - 26 = 28 hence , answer is 28 kids play with yellow color balls . answer : d | a = 26 - 19
b = 35 - a
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a ) 70 , b ) 245 , c ) 150 , d ) 35 , e ) 350 | c | divide(21, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 21 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 21 , so 0.14 * { total } = 21 - - > { total } = 150 . answer : c ." | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 21 / d
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a ) 5 , b ) 6 , c ) 4.5 , d ) 8 , e ) 9 | c | divide(110, multiply(add(80, 8), const_0_2778)) | a train 110 m long is running with a speed of 80 km / hr . in what time will it pass a man who is running at 8 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 80 + 8 = 88 km / hr . = 88 * 5 / 18 = 220 / 9 m / sec . time taken to pass the men = 110 * 9 / 220 = 4.5 sec . answer : option c" | a = 80 + 8
b = a * const_0_2778
c = 110 / b
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a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 / 4 | c | divide(2, const_10) | tickets numbered 1 to 10 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of 2 ? | "here , s = { 1 , 2,3 , . . . . . . , 9,10 } e = event getting a multiple of 3 = { 2 , 4,6 , 8,10 } probability = 5 / 10 = 1 / 2 correct option is c" | a = 2 / 10
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a ) 2300 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(106, 6), add(divide(subtract(50, 6), const_2), const_1)) | set a contains all the even numbers between 6 and 50 inclusive . set b contains all the even numbers between 106 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "set a contains 6 , 8 , . . . 50 set b contains 102 , 104 , 106 . . . 150 number of terms in each set = 23 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 23 = 2300 answer a" | a = 106 - 6
b = 50 - 6
c = b / 2
d = c + 1
e = a * d
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a ) 3.2 days , b ) 8 days , c ) 10 days , d ) 12 days , e ) 14 days | a | inverse(add(inverse(9), inverse(5))) | david completes a piece of work in 5 days , john completes the same work in 9 days . if both of them work together , then the number of days required to complete the work is ? | ifa can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 5 Γ 9 / 14 = 3.2 days answer : a | a = 1/(9)
b = 1/(5)
c = a + b
d = 1/(c)
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a ) 192 , b ) 200 , c ) 196 , d ) 198 , e ) none | a | multiply(3, power(8, const_2)) | the citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8 . a certain street in nigiet contains 1000 buildings numbered 1 to 1000 in base 8 . how many 3 s are used in numbering these buildings ? express your answer in base 10 | "explanation : number 3 in 1 - 100 comes 20 times . 101 - 200 - > 20 times 201 - 300 - > 20 times 301 - 400 - > 120 times 401 - 500 - > 20 times 501 - 600 - > 20 times 601 - 700 - > 20 times 701 - 800 - > 20 times 801 - 900 - > 20 times 901 - 1000 - > 20 times total times : 300 times ( any no . from 1 to 9 except 0 will occur 300 times ) 8 based planet . . then 3 * 8 ( 2 ) + 0 * 8 ( 1 ) + 0 * 8 ( 0 ) = 192 answer : a" | a = 8 ** 2
b = 3 * a
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a ) 0.8 , b ) 0.7 , c ) 0.6 , d ) 0.5 , e ) 0.4 | d | divide(divide(multiply(30, 60), 60), const_2) | mixture a is 30 percent alcohol , and mixture b is 60 percent alcohol . if the two are poured together to create a 4 - gallon mixture that contains 40 percent alcohol , approximately how many gallons of mixture a are in the mixture ? | "( 60 - 50 ) / ( 50 - 30 ) = qa / qb 10 / 20 = qa / qb 1 / 2 = qa / qb qa = ( 1 / 6 ) * 4 = 4 / 6 = 1 / 2 = 0.50 answer : d" | a = 30 * 60
b = a / 60
c = b / 2
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a ) 15 days , b ) 12 days , c ) 16 days , d ) 18 days , e ) 20 days | c | divide(multiply(multiply(12, 10), 8), multiply(4, 15)) | 12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 4 days , working 15 hours a day , the number of men required ? | that is , 1 work done = 12 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 6 ? ( i . e . no . of men required ) = 12 Γ 8 Γ 10 / 15 Γ 6 = 16 days c ) | a = 12 * 10
b = a * 8
c = 4 * 15
d = b / c
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a ) 5 , b ) 8 , c ) 10 , d ) 15 , e ) 2 | e | max(multiply(subtract(add(55, 4), const_1), subtract(divide(4, 35), divide(4, 55))), const_4) | due to construction , the speed limit along an 4 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "4 / 35 - 4 / 55 = 4 / 5 * ( 11 - 7 ) / 77 = 4 / 5 * 4 / 77 * 60 min = 4 * 12 * 4 / 77 = 192 / 77 ~ 2.4 answer - e" | a = 55 + 4
b = a - 1
c = 4 / 35
d = 4 / 55
e = c - d
f = b * e
g = max(f)
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a ) 12 , b ) 15 , c ) 18 , d ) 24 , e ) 36 | c | divide(divide(162, const_2), const_2) | if k ^ 3 is divisible by 162 , what is the least possible value of integer k ? | "k ^ 3 = 162 * x = 2 * 3 ^ 4 * x the factors of k must at minimum include 2 * 3 ^ 2 = 18 . the answer is c ." | a = 162 / 2
b = a / 2
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a ) 1 , b ) 0 , c ) - 1 , d ) - 2 , e ) - 4 | a | subtract(multiply(13, divide(subtract(multiply(2, 4), 3), subtract(multiply(8, 3), multiply(2, 11)))), add(multiply(11, divide(subtract(multiply(2, 4), 3), subtract(multiply(8, 3), multiply(2, 11)))), 4)) | when positive integer x is divided by 11 , the quotient is y and the remainder is 4 . when 2 x is divided by 8 , the quotient is 3 y and the remainder is 3 . what is the value of 13 y β x ? | "( 1 ) x = 11 y + 4 ( 2 ) 2 x = 24 y + 3 let ' s subtract equation ( 1 ) from equation ( 2 ) . 13 y - 1 = x 13 y - x = 1 the answer is a ." | a = 2 * 4
b = a - 3
c = 8 * 3
d = 2 * 11
e = c - d
f = b / e
g = 13 * f
h = 2 * 4
i = h - 3
j = 8 * 3
k = 2 * 11
l = j - k
m = i / l
n = 11 * m
o = n + 4
p = g - o
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a ) 3 , b ) 4 , c ) 6 , d ) 9 / 2 , e ) 12 | d | divide(const_1, multiply(12, add(divide(const_1, 36), divide(const_1, const_2.0)))) | a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 12 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 12 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 12 lets say x machines are used to attain the desired rate , thus x / 36 + x / 12 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + 3 x ) / 36 = 1 / 2 4 x / 36 = 1 / 2 x = 9 / 2 . qa = 9 / 2 ( answer d )" | a = 1 / 36
b = 1 / 2
c = a + b
d = 12 * c
e = 1 / d
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a ) 24 days , b ) 60 days , c ) 86 days , d ) 45 days , e ) 17 days | b | divide(10, subtract(const_1, divide(add(10, 10), 20))) | a can do a piece of work in 20 days ; b can do the same in 30 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 20 + 10 / 30 + 10 / x = 1 x = 60 days answer : b" | a = 10 + 10
b = a / 20
c = 1 - b
d = 10 / c
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a ) 3.5 , b ) 3.72 , c ) 4 , d ) 4.25 , e ) 4.5 | b | divide(350, add(divide(multiply(3, 800), const_100), divide(multiply(1400, 5), const_100))) | a money lender lent rs . 800 at 3 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years . | ( 800 xtx 3 / 100 ) + ( 1400 xtx 5 / 100 ) = 350 Γ’ β β t = 3.72 answer b | a = 3 * 800
b = a / 100
c = 1400 * 5
d = c / 100
e = b + d
f = 350 / e
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a ) 5 : 00 , b ) 4 : 00 , c ) 4 : 10 , d ) 5 : 20 , e ) 6 : 00 | b | subtract(add(4, 12), multiply(subtract(add(4, 12), 10), 2)) | two watches are set for an alarm at 10 0 ' clock in the night . one watch gains 2 minutes per hour . when the faster watch shows 4 : 12 in the morning when the alarm rings what does the correct watch show ? | the faster watch gains 2 minutes per hour the two watches show 10 : 00 at 10 0 ' clock at 11 0 ' clock - the correct watch is at 11 : 00 the faster watch is at 11 : 02 ( 1 hour + gains 2 minutes per hour ) at 12 midnight - the correct watch is at 12 : 00 the faster watch is at 12 : 04 ( 2 hours + gains 4 minutes per 2 hours ) in 6 hours the faster watch gains 6 * 2 = 12 minutes and is at 4 : 12 and the correct watch is at 4 : 00 answer is b | a = 4 + 12
b = 4 + 12
c = b - 10
d = c * 2
e = a - d
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a ) 9 , b ) 6 , c ) 4 , d ) 3 , e ) none of these | d | subtract(divide(5, subtract(3, multiply(subtract(inverse(divide(5, 7)), const_1), 5))), multiply(subtract(inverse(divide(5, 7)), const_1), divide(5, subtract(3, multiply(subtract(inverse(divide(5, 7)), const_1), 5))))) | if 3 x β 5 y = 5 and x / x + y = 5 / 7 , then what is the value of x β y ? | 3 x β 5 y = 5 . . . ( i ) and xx + y = 57 xx + y = 57 β 7 x = 5 x + 5 y β 2 x = 5 y . . . ( ii ) from ( i ) and ( ii ) , x = 5 and y = 2 β΄ x β y = 3 answer d | a = 5 / 7
b = 1/(a)
c = b - 1
d = c * 5
e = 3 - d
f = 5 / e
g = 5 / 7
h = 1/(g)
i = h - 1
j = 5 / 7
k = 1/(j)
l = k - 1
m = l * 5
n = 3 - m
o = 5 / n
p = i * o
q = f - p
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a ) 3750 , b ) 7799 , c ) 2500 , d ) 2772 , e ) 2912 | a | divide(multiply(300, const_100), 2) | a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 300 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 300 b capital = 22500 * 1 / 6 = 3750 . answer : a" | a = 300 * 100
b = a / 2
|
a ) 12 / 91 , b ) 3 / 15 , c ) 12 / 15 , d ) 33 / 91 , e ) 3 / 91 | d | divide(choose(subtract(15, 3), 4), choose(15, 4)) | john borrowed 3 soccer boots from jake , and forgot them on the field , if peter came across a total of 15 boots on the field jake ' s boots inclusive , and he took 4 boots at random , what is the probability that jake ' s boots were not amongst the 4 taken ? . | since jake owns 3 of the boots , the subset from which the 4 boots should be chosen are the 12 boots not owned by jake from the universe of 15 . the first boot can be one of the 12 from the 15 with probability 12 / 15 . the second boot can be one of the 11 from the 14 remaining with probability 11 / 14 . the third boot can be one of the 10 from the 13 remaining with probability 10 / 13 . the fourth boot can be one of the 9 from the 12 remaining with probability 9 / 12 . the total probability will be 12 / 15 Γ 11 / 14 Γ 10 / 13 Γ 9 / 12 . on cancellation , this comes to 33 / 91 . thus , the answer is d - 33 / 91 | a = 15 - 3
b = math.comb(a, 4)
c = math.comb(15, 4)
d = b / c
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a ) 400 , b ) 500 , c ) 600 , d ) 700 , e ) 800 | a | divide(add(80, 40), divide(30, const_100)) | a student needs 30 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 40 marks , find the maximum marks set for the test . | 30 % = 120 marks 1 % = 4 marks 100 % = 400 marks the answer is a . | a = 80 + 40
b = 30 / 100
c = a / b
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a ) 5 , b ) 10 , c ) 8 , d ) 2 , e ) 4 | e | add(1, 5) | set a consists of the integers from 5 to 10 , inclusive , while set b consists of the integers from 1 to 8 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 5,6 , 7 , 8 , 9 , 10 } b = { 1,2 , 3,4 , 5,6 , 7,8 } common elements = { 5,6 , 7 , 8 } = 4 elements answer : option e ." | a = 1 + 5
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a ) 75 kg , b ) 95 kg , c ) 45 kg , d ) 85 kg , e ) 25 kg | b | add(multiply(2.5, 8), 75) | he average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 75 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 = ( 75 + 20 ) kg = 95 kg . answer : b" | a = 2 * 5
b = a + 75
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a ) $ 94.03 , b ) $ 96.75 , c ) $ 98.75 , d ) $ 102.07 , e ) $ 105.3 | c | add(add(multiply(const_2, 32.25), multiply(const_1, 20.40)), multiply(const_1, 13.85)) | a garden center sells a certain grass seed in 5 - pound bags at $ 13.85 per bag , 10 - pound bags at $ 20.40 per bag , and 25 - pound bags $ 32.25 per bag . if a customer is to buy at least 65 pounds of the grass seed , but no more than 80 pounds , what is the least possible cost of the grass seed that the customer will buy ? | "there can be 2 cases 1 ) 25 + 25 + 10 + 5 = $ 98.75 or 2 ) 25 + 25 + 25 = $ 96.75 c" | a = 2 * 32
b = 1 * 20
c = a + b
d = 1 * 13
e = c + d
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a ) 4500 , b ) 7500 , c ) 5000 , d ) 8000 , e ) none of these | d | divide(20, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 20 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the difference is 20 , the sum = 400 Γ 20 = 8000 answer d" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 5 / 100
e = 2 * d
f = 1 + e
g = c - f
h = 20 / g
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a ) rs . 12 , b ) rs . 16 , c ) rs . 14 , d ) rs . 10 , e ) rs . 18 | c | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 8400), const_100) | if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of rs . 8400 ? | "required difference = [ 3 Β½ % of rs . 8400 ] β [ 3 1 / 3 % of rs . 8400 ] = [ ( 7 / 20 - ( 10 / 3 ) ] % of rs . 8400 = 1 / 6 % of rs . 8400 = rs . [ ( 1 / 6 ) 8 ( 1 / 100 ) * 8400 ] = rs . 14 . answer is c ." | a = 1 / 2
b = a + 3
c = 1 / 3
d = c + 3
e = b - d
f = e * 8400
g = f / 100
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a ) none , b ) five , c ) six , d ) seven , e ) eight | c | add(add(2, 2), 2) | how many two - digit whole numbers yield a remainder of 2 when divided by 12 and also yield a remainder of 2 when divided by 6 ? | "the possible number n can be written as follow : n = multiple of lcm ( 12,6 ) + 1 st such number n = 12 x + 2 possible values = 2 , 26,38 , 50,62 , 86,98 answer : 6 such 2 digit number . c ." | a = 2 + 2
b = a + 2
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | e | divide(add(multiply(multiply(divide(3, 2), 2), 7), 8), subtract(multiply(divide(3, 2), 2), 2)) | find ( 7 x + 8 y ) / ( x - 2 y ) if x / 2 y = 3 / 2 ? | x / 2 y = 3 / 2 = > x = 6 y / 2 = 3 y = > ( 7 x + 8 y ) / ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 8 y ) / ( 3 y - 2 y ) = > 29 y / y = 29 answer : e | a = 3 / 2
b = a * 2
c = b * 7
d = c + 8
e = 3 / 2
f = e * 2
g = f - 2
h = d / g
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a ) 100 , b ) 120 , c ) 240 , d ) 150 , e ) 200 | c | divide(multiply(multiply(const_60, 2), 10), 5) | kramer can pack 10 boxes of cigarettes per minute . if there are 5 boxes of cigarettes in one case , how many cases can kramer pack in 2 hours ? | kramer can pack 10 boxes per min . therefore in 2 hrs ( 120 mins ) he can pack 120 * 10 = 1200 boxes . if there are 5 boxes in one case then then the total number of cases packed by kramer = 1200 / 5 = 240 answer is c | a = const_60 * 2
b = a * 10
c = b / 5
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a ) 21 , b ) 29 , c ) 28 , d ) 27 , e ) 26 | e | add(add(add(const_2.0, 11), 11), 11) | what number should replace the question mark ? 11 , 15 , 16 , 12 , 21 , 9 , - - - ? | "answer : e 11 , 15 , 16 , 12 , 21 , 9 , - - - ? there are two alternate sequences : + 5 and - 3 ." | a = 2 + 0
b = a + 11
c = b + 11
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a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12 | b | subtract(subtract(subtract(12, const_1), 4), const_1) | list d consists of 12 consecutive integers . if - 4 is the least integer in list d , what is the range of positive integers in list d ? | since - 4 is the least integer in list d , then 7 is the largest integer in that list . thus the range of the positive integers in the list is 7 - 1 = 6 . answer : b . | a = 12 - 1
b = a - 4
c = b - 1
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a ) 40 , b ) 50 , c ) 60 , d ) 79 , e ) 80 | c | divide(divide(subtract(125, multiply(multiply(15, const_0_2778), 15)), 15), const_0_2778) | a train 125 m long passes a man , running at 15 km / hr in the same direction in which the train is going , in 10 sec . the speed of the train is ? | "speed of the train relative to man = 125 / 10 = 25 / 2 m / sec . = 25 / 2 * 18 / 5 = 45 km / hr let the speed of the train be x km / hr . then , relative speed = ( x - 15 ) km / hr . x - 15 = 45 = > x = 60 km / hr . answer : option c" | a = 15 * const_0_2778
b = a * 15
c = 125 - b
d = c / 15
e = d / const_0_2778
|
a ) 1 : 12 , b ) 81 : 13 , c ) 46 : 51 , d ) 86 : 16 , e ) 56 : 13 | c | power(divide(729, 1000), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 729 : 1000 . what is the ratio of their total surface areas ? | "ratio of the sides = Β³ β 729 : Β³ β 1000 = 9 : 10 ratio of surface areas = 92 : 102 = 46 : 51 answer : c" | a = 729 / 1000
b = 1 / 3
c = a ** b
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a ) 72 kg , b ) 77.5 kg , c ) 135.5 kg , d ) data inadequate , e ) none of these | c | add(multiply(9, 5.5), 86) | the average weight of 9 person ' s increases by 5.5 kg when a new person comes in place of one of them weighing 86 kg . what might be the weight of the new person ? | "total weight increased = ( 9 x 5.5 ) kg = 49.5 kg . weight of new person = ( 86 + 49.5 ) kg = 135.5 kg . answer : option c" | a = 9 * 5
b = a + 86
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a ) 96 , b ) 280 , c ) 384 , d ) 400 , e ) 500 | a | divide(multiply(multiply(50, 16), 6), subtract(multiply(5, 16), multiply(5, 6))) | in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 50 more than the correct answer . find the number . | "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 50 25 * x / 48 = 50 x = 96 answer a" | a = 50 * 16
b = a * 6
c = 5 * 16
d = 5 * 6
e = c - d
f = b / e
|
a ) 25 % , b ) 9.09 % , c ) 40 % , d ) 75 % , e ) none of these | b | multiply(add(const_1, divide(divide(multiply(150, 40), add(150, divide(multiply(multiply(const_3, const_10), 150), const_100))), 40)), const_10) | a rectangle having length 150 cm and width 40 cm . if the length of the rectangle is increased by fifteen percent then how much percent the breadth should be decreased so as to maintain the same area . | "explanation : solution : ( 15 / ( 150 + 15 ) * 100 ) % = 9.09 % answer : b" | a = 150 * 40
b = 3 * 10
c = b * 150
d = c / 100
e = 150 + d
f = a / e
g = f / 40
h = 1 + g
i = h * 10
|
a ) 35 , b ) 56 , c ) 21 , d ) 87 , e ) 24 | c | divide(multiply(42, 6), 12) | two numbers n and 12 have lcm = 42 and gcf = 6 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 12 * n = 42 * 6 n = 42 * 6 / 12 = 21 correct answer c" | a = 42 * 6
b = a / 12
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a ) rs . 40000 , b ) rs . 36000 , c ) rs . 50000 , d ) rs . 45000 , e ) none of these | a | divide(9000, divide(subtract(const_100, add(add(45, 25), 7.5)), const_100)) | ajay spends 45 % of his monthly income on household items , 25 % of his monthly income on buying cloths , 7.5 % of his monthly income on medicines and saves the remaining amount which is rs . 9000 . find his monthly income . | "let the monthly income of ajay be rs . x savings of ajay = x - ( 45 + 25 + 7.5 ) / 100 * x = 22.5 / 100 x 22.5 / 100 x = 9000 x = 40000 . answer : a" | a = 45 + 25
b = a + 7
c = 100 - b
d = c / 100
e = 9000 / d
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a ) 0.44202 , b ) 0.44204 , c ) 0.442022 , d ) 0.442025 , e ) none | d | divide(add(multiply(0.44204, 0.44203), multiply(0.44204, 0.44202)), 0.44204) | the average of numbers 0.44204 , 0.44203 , 0.44202 and 0.44201 is ? | "answer average = ( 0.44204 + 0.44203 + 0.44202 + 0.44201 ) / 4 = 1.7681 / 4 = 0.442025 correct option : d" | a = 0 * 44204
b = 0 * 44204
c = a + b
d = c / 0
|
a ) 135 , b ) 125 , c ) 120 , d ) 110 , e ) 140 | a | divide(multiply(subtract(19, const_1), subtract(subtract(19, const_1), const_3)), const_2) | how many diagonals does a polygon with 19 sides have , if one of its vertices does not connect to any diagonal ? | "if i calculate it using the formulae , # diagonals = n ( n - 3 ) / 2 each vertex sends of n - 3 diagonals n = 19 - 1 then 18 * ( 18 - 3 ) / 2 = 135 correct option : a" | a = 19 - 1
b = 19 - 1
c = b - 3
d = a * c
e = d / 2
|
a ) 22.5 years , b ) 24.5 years , c ) 12.5 years , d ) 28.5 years , e ) 29.5 years | b | multiply(divide(7, subtract(9, 7)), 7) | sachin is younger than rahul by 7 years . if their ages are in the respective ratio of 7 : 9 , how old is sachin ? | "let rahul ' s age be x years . then , sachin ' s age = ( x - 7 ) years . therefore ( x - 7 ) / x = 7 / 9 9 x - 63 = 7 x 2 x = 63 x = 31.5 hence , sachin ' s age = ( x - 7 ) = 24.5 years . answer : b" | a = 9 - 7
b = 7 / a
c = b * 7
|
a ) 288 , b ) 279 , c ) 277 , d ) 272 , e ) 180 | e | multiply(divide(multiply(72, const_1000), const_3600), 9) | a train running at the speed of 72 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 72 * 5 / 18 = 20 m / sec length of the train = speed * time = 20 * 9 = 180 m answer : e" | a = 72 * 1000
b = a / 3600
c = b * 9
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a ) 45 , b ) 38 , c ) 44 , d ) 90 , e ) 96 | d | add(multiply(subtract(12, const_1), 8), 2) | find the 12 th term of an arithmetic progression whose first term is 2 and the common difference is 8 . | "n th term of a . p = a + ( n - 1 ) * d = 2 + ( 12 - 1 ) * 8 , = 2 + 88 = 90 . answer : d" | a = 12 - 1
b = a * 8
c = b + 2
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(20, 12) | a set consists of 20 numbers , all are even or multiple of 5 . if 9 numbers are even and 12 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 20 = 9 + 12 - { both } + 0 ; { both } = 1 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : b ." | a = 20 - 12
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a ) 20 minutes , b ) 110 minutes , c ) 120 minutes , d ) 140 minutes , e ) 160 minutes | c | divide(subtract(12.00, 6.00), subtract(0.25, 0.20)) | united telephone charges a base rate of $ 6.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ? | "lets take number of minutesx . given that , 6 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 120 minutes ans c" | a = 12 - 0
b = 0 - 25
c = a / b
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a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 14 , e ) e ) 9 | a | sqrt(divide(2600, add(power(4, 3), add(power(1, const_2), power(3, 3))))) | the ratio of three numbers is 1 : 3 : 4 and the sum of their squares is 2600 . the sum of the numbers is ? | "let the numbers be x , 3 x , 4 x then , x ^ 2 + 9 x ^ 2 + 16 x ^ 2 = 2600 26 x ^ 2 = 2600 x ^ 2 = 100 x = 10 answer is a" | a = 4 ** 3
b = 1 ** 2
c = 3 ** 3
d = b + c
e = a + d
f = 2600 / e
g = math.sqrt(f)
|
a ) rs . 7500 , b ) rs . 7200 , c ) rs . 6000 , d ) rs . 5750 , e ) rs . 6750 | c | multiply(multiply(const_1, const_12), divide(18000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18000 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18000 = 6000 answer : c" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 18000 / h
j = a * i
|
a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) 365 | b | divide(add(1365, 15), subtract(6, const_1)) | the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and 15 as remainder . what is the smaller number ? | "let the smallest number be x . then larger number = ( x + 1365 ) x + 1365 = 6 x + 15 = 5 x = 1350 x = 270 smaller number = 270 . answer : b" | a = 1365 + 15
b = 6 - 1
c = a / b
|
a ) 4 seconds , b ) 25 seconds , c ) 19 seconds , d ) 21 seconds , e ) 6.25 seconds | c | divide(subtract(100, 24), 4) | in a 100 m race , sam beats john by 4 seconds . on the contrary , if sam allowed john to start 24 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 4 second but this difference is 0 if john allows sam to start the race from 24 m ahead . that means jhon was 24 m away from finishing line when they started together . so he will cover 24 m in 4 seconds . so his speed = 24 / 4 = 6 metre / second . so time taken = 100 / 4 = 25 seconds . so sam took = 19 seconds . correct answer = c" | a = 100 - 24
b = a / 4
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a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | c | subtract(add(14, 20), divide(14, divide(70, const_100))) | the contents of a certain box consist of 14 apples and 20 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ? | the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 20 oranges , so to get to 6 oranges , we should remove 20 - 6 = 14 oranges . answer c | a = 14 + 20
b = 70 / 100
c = 14 / b
d = a - c
|
a ) 25 , b ) 60 , c ) 70 , d ) 80 , e ) 20 | a | divide(divide(multiply(500, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 500 , what is the reduced price for kg ? | "a 500 * ( 25 / 100 ) = 125 - - - - 5 ? - - - - 1 = > rs . 25" | a = 500 * 25
b = a / 100
c = b / 5
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a ) 225 , b ) 250 , c ) 275 , d ) 300 , e ) 325 | a | multiply(divide(subtract(subtract(power(multiply(8, divide(add(const_100, 25), const_100)), const_2), power(multiply(6, divide(subtract(const_100, 50), const_100)), const_2)), subtract(power(8, const_2), power(6, const_2))), subtract(power(8, const_2), power(6, const_2))), const_100) | there are two concentric circles with radii 8 and 6 . if the radius of the outer circle is increased by 25 % and the radius of the inner circle decreased by 50 % , by what percent does the area between the circles increase ? | "the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 64 pi . the area of the small circle is 36 pi . the area a 1 between the circles is 28 pi . when the big circle ' s radius increases , the new area is 100 pi . when the small circle ' s radius decreases , the new area is 9 pi . the area a 2 between the circles is 91 pi . the ratio of a 2 / a 1 is 91 / 28 = 3.25 which is an increase of 225 % . the answer is a ." | a = 100 + 25
b = a / 100
c = 8 * b
d = c ** 2
e = 100 - 50
f = e / 100
g = 6 * f
h = g ** 2
i = d - h
j = 8 ** 2
k = 6 ** 2
l = j - k
m = i - l
n = 8 ** 2
o = 6 ** 2
p = n - o
q = m / p
r = q * 100
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(100, multiply(add(15, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 15 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 100 meters ? | the speed of the boat downstream = 15 + 3 = 18 km / hr 18 km / hr * 5 / 18 = 5 m / s the time taken to cover 100 meters = 100 / 5 = 20 seconds . the answer is d . | a = 15 + 3
b = a * const_0_2778
c = 100 / b
|
a ) $ 1166 , b ) $ 1168 , c ) $ 1166.6 , d ) $ 1167 , e ) none of the above | d | divide(3500, 3) | 3 people decided to put a down payment on a car for 3500 dollars . how much money do they each have to pay , given that that they have to pay the same amount , and they can not use any bills below $ 1 ? | to keep the amount exact when dividing for 3 , we have to work backwards through division . dividing 3500 by 3 will give us 1166.66 then by subtracting . 66 , we get 1166 , and we multiply it by 3 to get 3498 . to get enough for the down payment , we add 3 to 3498 . dividing the result by 3 gives us our answer . correct option : d | a = 3500 / 3
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a ) 140 , b ) 120 , c ) 130 , d ) 150 , e ) 160 | a | multiply(2, multiply(add(2, divide(30, const_60)), divide(multiply(add(2, divide(20, const_60)), 2), subtract(add(2, divide(30, const_60)), add(2, divide(20, const_60)))))) | a biker covered half the distance between two towns in 2 hr 30 min . after that he increased his speed by 2 km / hr . he covered the second half of the distance in 2 hr 20 min . find the distance between the two towns and the initial speed of the biker . | let x km / hr be the initial speed of the biker , then his speed during the second part of the trip is x + 2 km / hr . half the distance between two cities equals 23060 β
x km and 22060 β
( x + 2 ) km . from the equation : 23060 β
x = 22060 β
( x + 2 ) we get x = 28 km / hr . the intial speed of the biker is 28 km / h . half the distance between the two towns is 2 h 30 min Γ 28 = 2.5 Γ 28 = 70 . so the distance is 2 Γ 70 = 140 km . answer is a . | a = 30 / const_60
b = 2 + a
c = 20 / const_60
d = 2 + c
e = d * 2
f = 30 / const_60
g = 2 + f
h = 20 / const_60
i = 2 + h
j = g - i
k = e / j
l = b * k
m = 2 * l
|
['a ) 600', 'b ) 900', 'c ) 800', 'd ) 500', 'e ) 400'] | c | divide(power(40, const_2), const_2) | the diagonal of a square is 40 m . the area of the square is : | arear = 1 / 2 Γ ( diagonal ) ( power ) 2 = ( 1 / 2 Γ 40 Γ 40 ) m ( power ) 2 = 1600 / 2 = 800 m ( power ) 2 answer is c . | a = 40 ** 2
b = a / 2
|
a ) 743 , b ) 154 , c ) 187 , d ) 741 , e ) 785 | c | divide(subtract(16698, 14), 89) | on dividing 16698 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 16698 - 14 ) / 89 = 187 answer ( c )" | a = 16698 - 14
b = a / 89
|
a ) 24 / 78 , b ) 24 / 709 , c ) 24 / 76 , d ) 24 / 71 , e ) 24 / 72 | a | divide(add(add(3, 6), add(const_10, add(3, 2))), multiply(add(const_10, const_3), 6)) | a box contains 4 black , 3 red and 6 green marbles . 2 marbles are drawn from the box at random . what is the probability that both the marbles are of the same color ? | explanation : total marbles in a box = 4 black + 3 red + 6 green marbles = 13 marbles 2 marbles are drawn from 13 marbles at random . therefore , n ( s ) = 13 c 2 = 78 ways let a be the event that 2 marbles drawn at random are of the same color . number of cases favorable to the event a is n ( a ) = 4 c 2 + 3 c 2 + 6 c 2 = 6 + 3 + 15 = 24 therefore , by definition of probability of event a , p ( a ) = n ( a ) / n ( s ) = 24 / 78 answer : a | a = 3 + 6
b = 3 + 2
c = 10 + b
d = a + c
e = 10 + 3
f = e * 6
g = d / f
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a ) 5000 , b ) 5030 , c ) 5010 , d ) 5040 , e ) 5020 | d | divide(80, subtract(divide(const_1, 14), divide(const_1, 18))) | when an amount was distributed among 14 boys , each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys . what was the amount ? | let the total amount be rs . x the , x / 14 - x / 18 = 80 < = > 2 x / 126 = 80 < = > x / 63 = 63 x 80 = 5040 . hence the total amount is 5040 answer is d . | a = 1 / 14
b = 1 / 18
c = a - b
d = 80 / c
|
a ) 18 , b ) 16 , c ) 19 , d ) 21 , e ) 15 | b | sqrt(add(156, multiply(50, const_2))) | sum of the squares of 3 no . is 156 and the sum of their products taken two at a time is 50 . find the sum ? | ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 156 + 2 * 50 a + b + c = Γ’ Λ Ε‘ 256 = 16 answer b | a = 50 * 2
b = 156 + a
c = math.sqrt(b)
|
a ) 55 / 3 , b ) 55 / 7 , c ) 55 / 6 , d ) 55 / 8 , e ) 55 / 9 | a | divide(150, multiply(add(36, 72), const_0_2778)) | two trains of length 150 m and 250 m are 150 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 = ( 150 + 150 + 250 ) / 30 = 550 / 30 = 55 / 3 sec . answer : a" | a = 36 + 72
b = a * const_0_2778
c = 150 / b
|
a ) 15.502 % , b ) 25.7014 % , c ) 12.502 % , d ) 21.7014 % , e ) none | d | divide(multiply(multiply(divide(500, 3840), const_100), const_100), 60) | farm tax is levied on the 60 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . willam . mr . willam paid only $ 500 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "only trick n this question is to ignore 60 % information as farm tax is levied uniformly in the village and that includes mr william ' s land . what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 500 / 3840 ) x 100 = 13.02083 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 13.02083 % x ( 100 / 60 ) = 21.7014 % and the answer is not there in the options . the correct answer is ( d ) ." | a = 500 / 3840
b = a * 100
c = b * 100
d = c / 60
|
a ) rs . 400 , b ) rs . 450 , c ) rs . 500 , d ) rs . 550 , e ) rs . 650 | c | multiply(divide(subtract(10000, 6000), 8000), 1000) | a , b and c started a business with capitals of rs . 6000 , rs . 8000 , and rs . 10000 respectively . at the end of the year , the profit share of b is rs 1000 . the difference between the profit shares of a and c is | ratio of investments of a , b and c is 6000 : 8000 : 10000 = 3 : 4 : 5 . and also given that , profit share of β b β is rs . 1000 . now required difference is 5 β 3 = 2 parts . therefore , required difference = 2 / 4 ( 1000 ) = rs . 500 . answer : c | a = 10000 - 6000
b = a / 8000
c = b * 1000
|
a ) 7 / 20 , b ) 9 / 25 , c ) 11 / 30 , d ) 13 / 35 , e ) 17 / 40 | c | subtract(add(divide(const_3.0, 5), subtract(const_1, divide(4, 5))), divide(5, 6)) | the probability that a computer company will get a computer hardware contract is 4 / 5 and the probability that it will not get a software contract is 3 / 5 . if the probability of getting at least one contract is 5 / 6 , what is the probability that it will get both the contracts ? | "let , a β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ event of getting both hardware and software contract . p ( a ) = 4 / 5 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 3 / 5 ) = 2 / 5 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ) . = > 5 / 6 = ( 4 / 5 ) + ( 2 / 5 ) - p ( ab ) . = > p ( ab ) = 11 / 30 . hence , the required probability is 11 / 30 . the answer is c ." | a = 3 / 0
b = 4 / 5
c = 1 - b
d = a + c
e = 5 / 6
f = d - e
|
['a ) 180 sq . units', 'b ) 11 sq . units', 'c ) 110 sq . units', 'd ) 140 sq . units', 'e ) 220 sq . units'] | e | multiply(multiply(sqrt(16), add(const_2, const_3)), 11) | the length of a rectangle is five times of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 16 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 11 units ? | given that the area of the square = 16 sq . units = > side of square = Γ’ Λ Ε‘ 16 = 4 units the radius of the circle = side of the square = 4 units length of the rectangle = 5 * 4 = 20 units given that breadth = 11 units area of the rectangle = lb = 20 * 11 = 220 sq . units answer : e | a = math.sqrt(16)
b = 2 + 3
c = a * b
d = c * 11
|
a ) 24 , b ) 23 , c ) 18 , d ) 21 , e ) 22 | a | add(multiply(subtract(divide(subtract(30, const_2), const_4), const_2), const_4), const_3) | second saturday and every sunday is a holiday . how many working days will be there in a month of 30 days beginning on a saturday ? | explanation : mentioned month begins on a saturday and has 30 days sundays = 2 nd , 9 th , 16 th , 23 rd , 30 th = > total sundays = 5 every second saturday is holiday . 1 second saturday in every month total days in the month = 30 total working days = 30 - ( 5 + 1 ) = 24 . answer : option a | a = 30 - 2
b = a / 4
c = b - 2
d = c * 4
e = d + 3
|
a ) 70 , b ) 60 , c ) 10 , d ) 75 , e ) 85 | b | divide(subtract(multiply(divide(80, const_100), 90), 30), divide(70, const_100)) | if 80 % of 90 is greater than 70 % of a number by 30 , what is the number ? | "explanation : 80 / 100 * 90 - 70 / 100 * x = 30 72 - 70 / 100 * x = 30 72 - 30 = 70 / 100 * x 42 = 70 / 100 * x 42 * 100 / 70 = x 60 = x answer : option b" | a = 80 / 100
b = a * 90
c = b - 30
d = 70 / 100
e = c / d
|
a ) 8 , b ) 6 , c ) 24 , d ) 32 , e ) 48 | b | sqrt(subtract(power(divide(40, 4), const_2), power(12, const_2))) | a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is 40 inches . find b . | "from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 40 ( 2 ^ 3 * 5 ) , sob = 2 ^ 3 or 2 ^ 3 * 5 ( 8 or 40 ) . combining 2 premises shows the answer is b ( 6 ) ." | a = 40 / 4
b = a ** 2
c = 12 ** 2
d = b - c
e = math.sqrt(d)
|
a ) 1345 , b ) 1197 , c ) 1540 , d ) 1600 , e ) 1635 | b | multiply(divide(subtract(1000, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1000 . 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 + 1365 ) . x + 1000 = 6 x + 15 5 x = 985 x = 197 large number = 197 + 1365 = 1197 b" | a = 1000 - 15
b = 6 - 1
c = a / b
d = c * 6
|
['a ) 7744', 'b ) 8844', 'c ) 5544', 'd ) 4444', 'e ) 3444'] | a | power(divide(circumface(56), const_4), const_2) | a wire can be bent in the form of a circle of radius 56 cm . if it is bent in the form of a square , then its area will be ? | length of wire = 2 pier = 2 * ( 22 / 7 ) * 56 = 352 cm side of the square = 352 / 4 = 88 cm area of the square = 88 * 88 = 7744 sq cm answer : a | a = circumface / (
b = a ** 4
|
['a ) 17 m', 'b ) 45 m', 'c ) 54 m', 'd ) 77 m', 'e ) 34 m'] | b | subtract(sqrt(add(power(multiply(4, sqrt(divide(divide(600, divide(50, const_100)), multiply(4, 3)))), 2), power(multiply(3, sqrt(divide(divide(600, divide(50, const_100)), multiply(4, 3)))), 2))), add(const_2, const_3)) | a rectangular courtyard , the sides of which are in the ratio of 4 : 3 , cost rs . 600 for paving at 50 p per m 2 ; find the length of the diagonal of the courtyard ? | 1 m 2 - - - - 1 / 2 ? - - - - - 600 = > 1200 m 2 4 x * 3 x = 1200 = > x = 10 answer : b | a = 50 / 100
b = 600 / a
c = 4 * 3
d = b / c
e = math.sqrt(d)
f = 4 * e
g = f ** 2
h = 50 / 100
i = 600 / h
j = 4 * 3
k = i / j
l = math.sqrt(k)
m = 3 * l
n = m ** 2
o = g + n
p = math.sqrt(o)
q = 2 + 3
r = p - q
|
a ) 20 % , b ) 18 % , c ) 24 % , d ) 64 % , e ) none of these | c | multiply(subtract(const_1, sqrt(divide(subtract(const_100, 42), const_100))), const_100) | if the area of a circle decreases by 42 % , then the radius of a circle decreases by | "if area of a circle decreased by x % then the radius of a circle decreases by ( 100 β 10 β 100 β x ) % = ( 100 β 10 β 100 β 42 ) % = ( 100 β 10 β 58 ) % = 100 - 76 = 24 % answer c" | a = 100 - 42
b = a / 100
c = math.sqrt(b)
d = 1 - c
e = d * 100
|
a ) 19 , b ) 29 , c ) 35 , d ) 49 , e ) 59 | c | add(subtract(83, multiply(17, 3)), 3) | a batsman makes a score of 83 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 83 = 17 x therefore x = 35 answer : c" | a = 17 * 3
b = 83 - a
c = b + 3
|
a ) 10 , b ) - 2 , c ) 11 , d ) 13 , e ) 22 | a | divide(add(20, 20), add(3, const_1)) | when x is multiplied by 3 , the result is 20 more than the result of subtracting x from 20 . what is the value of x ? | "the equation that can be formed is : 3 x - 20 = 20 - x or , 4 x = 40 or , x = 10 . answer : a" | a = 20 + 20
b = 3 + 1
c = a / b
|
a ) 2.91 , b ) 2.911 , c ) 2.938 , d ) 2.986 , e ) 2.999 | b | add(add(2.75, divide(003, const_1000)), divide(158, const_1000)) | solution for 2.75 + . 003 + . 158 | "2.75 + . 003 + . 158 = 0 0 = 0 - 2.75 - 0.003 - 0.158 0 = - 2.911 answer : b" | a = 3 / 1000
b = 2 + 75
c = 158 / 1000
d = b + c
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 70), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 70 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent ? | "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 80 % b" | a = 100 - 70
b = a - 10
c = 10 / b
d = c * 100
|
a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | b | divide(multiply(multiply(multiply(multiply(8, const_3), multiply(9, const_2)), const_100), 8), multiply(12000, 9)) | ravi and sunil are partners in a business . ravi invests rs . 9,000 for 8 months and sunil invested rs . 12000 for 9 months then after one year ratio of their profits will be | "= ( 9000 * 8 ) : ( 15000 * 8 ) = 72000 : 108000 = 2 : 3 answer : b" | a = 8 * 3
b = 9 * 2
c = a * b
d = c * 100
e = d * 8
f = 12000 * 9
g = e / f
|
a ) 22 , b ) 33 , c ) 55 , d ) 42 , e ) 88 | d | subtract(divide(238, add(divide(10, const_100), divide(3, 4))), 238) | a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 238 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 238 = > . 1 x + 3 / 4 x = 238 = > x = 280 280 - 238 = pure gala = 42 . answer d" | a = 10 / 100
b = 3 / 4
c = a + b
d = 238 / c
e = d - 238
|
a ) 50 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | a | add(multiply(divide(12, 48), const_100), 25) | the purchase price of an article is $ 48 . in order to include 25 % of cost for overhead and to provide $ 12 of net profit , the markup should be | cost price of article = 48 $ % of overhead cost = 25 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 25 = 50 % answer a | a = 12 / 48
b = a * 100
c = b + 25
|
a ) 100 m , b ) 255 m , c ) 375 m , d ) can not be determined , e ) none of these | c | multiply(divide(90, const_3_6), 15) | a car is running at a speed of 90 kmph . what distance will it cover in 15 second ? | "explanation : given : speed = 108 kmph = ( 90 x ( 5 / 18 ) ) m / sec = 25 m / sec distance covered in 15 second = ( 25 x 15 ) m = 375 m . answer : c" | a = 90 / const_3_6
b = a * 15
|
a ) 6 , b ) 4 , c ) 3 , d ) 5 , e ) 2 | c | subtract(9, reminder(51234, 9)) | what should be the least number to be added to the 51234 number to make it divisible by 9 ? | "the least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits . sum of digits = 15 . nearest multiple of 9 greater than sum of digits = 18 . hence 3 has to be added . answer : c" | a = 9 - reminder
|
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