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 ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | c | sqrt(add(power(multiply(8, divide(1, const_2.0)), 2), power(multiply(subtract(8, 2), divide(1, 6)), 6))) | 9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 6 miles per hour faster than the first runner ' s rate . how far apart , to the nearest mile , will they be after 1 / 2 hour ? | "if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 8 m / h - - > 4 m in 1 / 2 hr side 2 = 14 m / h - - > 7 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 7 ^ 2 d ^ 2 = 65 = ~ 8 answer option c" | a = 1 / 2
b = 8 * a
c = b ** 2
d = 8 - 2
e = 1 / 6
f = d * e
g = f ** 6
h = c + g
i = math.sqrt(h)
|
a ) 150 m , b ) 200 m , c ) 250 m , d ) 270 m , e ) 300 m | a | subtract(multiply(20, multiply(72, const_0_2778)), 250) | a train 250 m long running at 72 kmph crosses a platform in 20 sec . what is the length of the platform ? | "d = 72 * 5 / 18 = 20 = 400 â € “ 250 = 150 m answer : a" | a = 72 * const_0_2778
b = 20 * a
c = b - 250
|
a ) 75 , b ) 65 , c ) 45 , d ) 25 , e ) 85 | d | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 50), add(const_100, 50))), const_100) | raman ' s salary was decreased by 50 % and subsequently increased by 50 % . how much percent does he loss . | explanation : let the origianl salary = rs . 100 it will be 150 % of ( 50 % of 100 ) = ( 150 / 100 ) * ( 50 / 100 ) * 100 = 75 so new salary is 75 , it means his loss is 25 % option d | a = 100 * 100
b = 100 - 50
c = 100 + 50
d = b * c
e = a - d
f = e / 100
|
a ) 5 : 3 , b ) 1 : 4 , c ) 3 : 7 , d ) 6 : 11 , e ) 2 : 3 | a | divide(divide(const_1, const_4), divide(80, const_100)) | if 80 % of a number is equal to four - third of another number , what is the ratio of first number to the second number ? | "let 80 % of a = 4 / 3 b then 80 a / 100 = 4 b / 3 4 a / 5 = 4 b / 3 a / b = 5 / 3 a : b = 5 : 3 answer is a" | a = 1 / 4
b = 80 / 100
c = a / b
|
a ) 11 minutes , b ) 17 minutes , c ) 18 minutes , d ) 10 minutes , e ) 16 minutes | d | subtract(add(divide(const_1, 18), divide(const_1, 15)), divide(const_1, 45)) | two pipes can fill a tank in 18 minutes and 15 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 18 + 1 / 15 - 1 / 45 = ( 5 + 6 - 2 ) / 90 = 9 / 90 = 1 / 10 so , the tank becomes full in 10 minutes . answer : d :" | a = 1 / 18
b = 1 / 15
c = a + b
d = 1 / 45
e = c - d
|
a ) 983.578 , b ) 459.121 , c ) 781.189 , d ) 656.112 , e ) 457.358 | e | multiply(223.124, power(11.98, 6.001)) | 223.124 x 11.98 ã · 6.001 + 11.11 = ? | "explanation : ? = 223.124 x 11.98 ã · 6.001 + 11.11 â ‰ ˆ ( 223.124 x 12 / 6 ) + 11.11 â ‰ ˆ 446.248 + 11.11 â ‰ ˆ 457.358 answer : option e" | a = 11 ** 98
b = 223 * 124
|
a ) 2 hours , b ) 2.4 hours , c ) 3 hours , d ) 3.4 hours , e ) 4.5 hours | d | inverse(add(divide(const_1, 24), add(divide(const_1, 6), divide(const_1, 12)))) | pipe p can fill a tank in 6 hours , pipe q in 12 hours and pipe r in 24 hours . if all the pipes are open , in how many hours will the tank be filled ? | "explanation : part filled by ( p + q + r ) in 1 hour = ( 1 / 6 + 1 / 12 + 1 / 24 ) = 7 / 24 all the three pipes together will fill the tank = 24 / 7 = 3.4 hours answer d" | a = 1 / 24
b = 1 / 6
c = 1 / 12
d = b + c
e = a + d
f = 1/(e)
|
a ) $ 320 , b ) $ 160 , c ) $ 200 , d ) $ 410 , e ) $ 325 | a | divide(subtract(subtract(add(1200, 11), 20), multiply(divide(add(const_100, 15), const_100), 20)), add(add(add(const_1, divide(const_1, const_2)), const_1), divide(add(const_100, 15), const_100))) | 4 friends , peter , john , quincy , and andrew , are pooling their money to buy a $ 1200 item . peter has twice as much money as john . quincy has $ 20 more than peter . andrew has 15 % more than quincy . if they put all their money together and spend the $ 1200 , they will have $ 11 left . how much money does peter have ? | p = 2 j ; q = p + 20 ; a = 1.15 ( q ) p + j + q + r - 1200 = 11 p + 0.5 p + p + 20 + 1.15 ( p + 20 ) = 1211 p + 0.5 p + p + 20 + 1.15 p + 23 = 1211 p + 0.5 p + p + 1.15 p + 43 = 1211 p + 0.5 p + p + 1.15 p = 1168 3.65 p = 1168 p = 320 answer : a | a = 1200 + 11
b = a - 20
c = 100 + 15
d = c / 100
e = d * 20
f = b - e
g = 1 / 2
h = 1 + g
i = h + 1
j = 100 + 15
k = j / 100
l = i + k
m = f / l
|
a ) 20 , b ) 87 , c ) 276 , d ) 198 , e ) 171 | a | divide(multiply(divide(add(25, 4), 5), divide(multiply(multiply(25, 4), const_2), add(25, 4))), const_2) | a man travelled from the village to the post - office at the rate of 25 kmph and walked back at the rate of 4 kmph . if the whole journey took 5 hours 48 minutes , find the distance of the post - office from the village | explanation : average speed = ( 2 × a × b ) / ( a + b ) here a = 25 b = 4 average speed = 2 × 25 × 4 / ( 25 + 4 ) = 200 / 29 kmph . distance covered in 5 hours 48 minutes = speed × time distance = ( 200 / 29 ) × ( 29 / 5 ) = 40 kms distance covered in 5 hours 48 minutes = 40 kms distance of the post office from the village = ( 40 / 2 ) = 20 km . answer : a | a = 25 + 4
b = a / 5
c = 25 * 4
d = c * 2
e = 25 + 4
f = d / e
g = b * f
h = g / 2
|
a ) 225.623 , b ) 224.777 , c ) 233.523 , d ) 414.637 , e ) none of these | a | subtract(893.7, divide(573.07, 95.007)) | 893.7 – 573.07 – 95.007 = ? | "solution given expression = 893.7 - ( 573.07 + 95.007 ) = 893.7 - 668.077 = 225.623 . answer a" | a = 573 / 7
b = 893 - 7
|
a ) 11 km , b ) 13 km , c ) 16 km , d ) 18 km , e ) 19 km | c | add(subtract(add(9, 9), 4), const_2) | one man traveled a distance of 61 km in 9 hrs . he traveled partly on footat 4 km / hr and partly on bicycle ta 9 km / hr . the distance traveled on foot is ? | let the distance travelled on foot be x km . then , distance travelled on bicycle = ( 61 - x ) km . so , x + ( 61 - x ) = 9 4 9 9 x + 4 ( 61 - x ) = 9 x 36 5 x = 80 x = 16 km . c | a = 9 + 9
b = a - 4
c = b + 2
|
a ) 22 , b ) 77 , c ) 17.3 , d ) 99 , e ) 88 | c | multiply(divide(subtract(64500, add(42000, 13000)), add(42000, 13000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 64500 . what is his profit percent ? | "total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 64500 profit ( % ) = ( 64500 - 55000 ) / 55000 * 100 = 17.3 % answer : c" | a = 42000 + 13000
b = 64500 - a
c = 42000 + 13000
d = b / c
e = d * 100
|
a ) 9 , b ) 8 , c ) 7 , d ) 5 , e ) 6 | b | multiply(multiply(5, 5), divide(3, 5)) | in the coordinate plane , points ( x , 3 ) and ( 5 , y ) are on line k . if line k passes through the origin and has slope 3 / 5 , then x + y = | "line k passes through the origin and has slope 3 / 5 means that its equation is y = 3 / 5 * x . thus : ( x , 3 ) = ( 3 , 5 ) and ( 5 , y ) = ( 3 , 5 ) - - > x + y = 3 + 5 = 8 . answer : b" | a = 5 * 5
b = 3 / 5
c = a * b
|
a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 14.28 % | e | divide(const_100, multiply(multiply(divide(10, const_100), divide(30, const_100)), const_100)) | on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 30 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "0.1 m = 0.70 e = > e / m = 1 / 7 * 100 = 14.28 % so answer is e . m - # of motorists e - # of motorists exceeding speed" | a = 10 / 100
b = 30 / 100
c = a * b
d = c * 100
e = 100 / d
|
a ) 100 , b ) 200 , c ) 350 , d ) 150 , e ) 140 | d | divide(multiply(choose(6, 4), choose(6, 3)), 6) | a question paper has 2 parts , a & b , each containing 6 questions . if a student has to choose 4 from part a & 3 from part b , in how many ways can he choose the questions ? | "there 6 questions in part a out of which 4 question can be chosen as = 6 c 4 . similarly , 3 questions can be chosen from 6 questions of part b as = 6 c 3 . hence , total number of ways , = 6 c 4 * 6 c 3 = [ 6 ! / ( 2 ! 4 ! ) ] * [ 6 ! / ( 3 ! * 3 ! ) ] = { 15 } * { 6 * 5 * 4 * 3 / ( 3 * 2 * 1 ) } = 300 . d" | a = math.comb(6, 4)
b = math.comb(6, 3)
c = a * b
d = c / 6
|
a ) 3 : 4 , b ) 3 : 5 , c ) 11 : 8 , d ) 16 : 15 , e ) none | c | divide(multiply(132, const_2), multiply(80, const_3)) | a man invests some money partly in 12 % stock at 132 and partly in 10 % stock at 80 . to obtain equal dividends from both , he must invest the money in the ratio : | "solution for an income of rs . 1 in 12 % stock at 132 , investment = rs . ( 132 / 12 ) = rs . 11 . for an income of rs . 1 in 10 % stock at 80 , investment = rs . ( 80 / 10 ) = rs . 8 . ∴ ratio of investments = 11 : 8 = 11 : 8 answer c" | a = 132 * 2
b = 80 * 3
c = a / b
|
a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 2 | b | divide(subtract(subtract(subtract(150, multiply(150, divide(3, 7))), multiply(subtract(150, multiply(150, divide(3, 7))), divide(3, 7))), 3), subtract(subtract(subtract(150, multiply(150, divide(3, 7))), multiply(subtract(150, multiply(150, divide(3, 7))), divide(3, 7))), 3)) | george baked a total of 150 pizzas for 7 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ? | "3 / 5 of the 150 pizzas cooked on saturday = 90 pizzas 3 / 5 of the remaining pizzas on sunday = 36 pizzas we ' re left with ( 150 - 90 - 36 ) = 24 pizzas for the remaining 5 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w th f 8 7 6 2 1 = 24 w = 6 b" | a = 3 / 7
b = 150 * a
c = 150 - b
d = 3 / 7
e = 150 * d
f = 150 - e
g = 3 / 7
h = f * g
i = c - h
j = i - 3
k = 3 / 7
l = 150 * k
m = 150 - l
n = 3 / 7
o = 150 * n
p = 150 - o
q = 3 / 7
r = p * q
s = m - r
t = s - 3
u = j / t
|
a ) 5 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 15 sec | c | divide(200, multiply(subtract(68, 8), const_0_2778)) | a train 200 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "answer : c . speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 200 m at 50 / 3 m / sec = 200 * 3 / 50 sec = 12 sec" | a = 68 - 8
b = a * const_0_2778
c = 200 / b
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 65 | e | multiply(40, const_1) | at veridux corporation , there are 200 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ? | "well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 200 â € “ 90 = 110 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 200 â € “ 40 = 160 must be associates . now , in the â € œ associate â € row , 135 + e = 160 , which means e = 25 â € ” the other 25 associates must be female . now , to find b , which is what the question is asking , we need only look at the sum in the â € œ female â € column : b + 25 = 90 , which means b = 65 . there are fifteen female managers in this company . thus , the answer = ( e ) ." | a = 40 * 1
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a ) 195 / 63 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | e | add(divide(sqrt(1.21), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | find the value of ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) 11 / 9 + 12 / 7 = > ( 77 + 108 ) / 63 = > 185 / 63 answer is e ." | a = math.sqrt(1)
b = math.sqrt(0)
c = a / b
d = math.sqrt(1)
e = math.sqrt(0)
f = d / e
g = c + f
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | d | add(reminder(multiply(reminder(49, const_4), 93), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 49 ) is : | "the units digit of powers of 3 , cycles in a group of 4 : { 3 , 9 , 7 , 1 } 49 has the form 4 k + 1 , so the units digit of 93 ^ 49 is 3 . the units digit of powers of 5 is always 5 . 3 + 5 = 8 , so the units digit is 8 . the answer is d ." | a = reminder * (
b = reminder + (
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a ) 25 % loss , b ) 33.33 % profit , c ) 20 % loss , d ) 20 % profit , e ) 5 % profit | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 15), 20)), divide(multiply(const_100, 15), 20))) | if the cost price of 20 articles is equal to the selling price of 15 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 15 articles = cost price of 20 articles = $ 15 . now , we know the selling price of 15 articles . let us find the cost price of 15 articles . cost price of 15 articles = 15 * 1 = $ 15 . therefore , profit made on sale of 15 articles = selling price of 15 articles - cost price of 15 articles = 20 - 15 = $ 5 . as the profit is in the positive , the merchant has made a profit of $ 5 . therefore , % loss = loss / cp * 100 % loss = 5 / 15 * 100 = 33.33 % profit . b" | a = 100 * 15
b = a / 20
c = 100 - b
d = 100 * 15
e = d / 20
f = c / e
g = 100 * f
|
a ) 20 , b ) 31 , c ) 42 , d ) 43 , e ) 64 | d | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 255 are multiple of 4 but not multiples of 6 ? | from here , it might be easier to go up in bounds of 60 , so we know that 61 - 120 gives 10 more numbers . 121 - 180 and 181 - 240 as well . this brings us up to 240 with 40 numbers . a cursory glance at the answer choices should confirm that it must be 42 , as all the other choices are very far away . the numbers 244 and 248 will come and complete the list that ’ s ( naughty or nice ) under 250 . answer choice d is correct here . | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 3 / 2 , b ) sqrt ( 15 ) , c ) sqrt ( 6 ) / 2 , d ) 15 , e ) 18 | b | divide(subtract(10, divide(add(multiply(subtract(multiply(20, 2), 2), 2), 10), add(multiply(multiply(2, 2), 2), const_1))), 2) | if 2 x + y = 10 , 2 y + z = 20 and 2 x + z = 30 . where x , y , and z are real number . what is the value of 2 x ? | "explanation : let , 2 x = a , 2 y = b and 2 z = c then from the question : 2 x + y = 2 x * 2 y = ab = 10 similarly , bc = 20 , ac = 30 ab / bc = 1 / 2 = > a / c = 1 / 2 and ac = 30 ( a / c ) * ac = 1 / 2 * 30 = 15 a = sqrt ( 15 ) answer : b" | a = 20 * 2
b = a - 2
c = b * 2
d = c + 10
e = 2 * 2
f = e * 2
g = f + 1
h = d / g
i = 10 - h
j = i / 2
|
a ) 50 % , b ) 57 % , c ) 52 % , d ) 22 % , e ) 42 % | a | subtract(const_100, add(multiply(50, divide(50, const_100)), multiply(divide(50, const_100), 50))) | in a factory , there are 50 % technicians and 50 % non - technicians . if the 50 % of the technicians and 50 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ? | "total = 100 t = 50 nt = 50 50 * ( 50 / 100 ) = 25 50 * ( 50 / 100 ) = 25 25 + 25 = 50 = > 100 - 50 = 50 % answer : a" | a = 50 / 100
b = 50 * a
c = 50 / 100
d = c * 50
e = b + d
f = 100 - e
|
a ) 0 , b ) 2 , c ) 6 , d ) 8 , e ) 12 | a | subtract(15, 15) | 3 interviewers , x , y , and z are interviewing 40 applicants . only with 3 interviewers ' admission can an applicant be admitted . if interviewer x admitted 15 applicants , y admitted 17 applicants , and z admitted 20 applicants , at least how many applicants get the admission ? | if x admitted 15 are overlapping with y admission of 17 but z does not overlap with anybody . then no student will get nod from all the 3 . hence 0 student will get admission . answer : a | a = 15 - 15
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a ) 60.40 mph , b ) 60.20 mph , c ) 55.32 mph , d ) 65.50 mph , e ) 70.25 mph | a | divide(50, add(divide(45, subtract(50, 15)), divide(15, 20))) | tom traveled the entire 50 miles trip . if he did the first 15 miles of at a constant rate 20 miles per hour and the remaining trip of at a constant rate 45 miles per hour , what is the his average speed , in miles per hour ? | "avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 15 + 35 ) / ( ( 15 / 20 ) + ( 35 / 45 ) ) = 50 * 180 / 149 = 60.40 mph a" | a = 50 - 15
b = 45 / a
c = 15 / 20
d = b + c
e = 50 / d
|
a ) 20 , b ) 34 , c ) 58 , d ) 40 , e ) 46 | c | divide(subtract(multiply(const_3, 60), 60), const_3) | shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 60 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 80 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 58 hence answer is c" | a = 3 * 60
b = a - 60
c = b / 3
|
a ) 8.6 , b ) 9.8 , c ) 10.4 , d ) 11.2 , e ) 12.6 | c | add(divide(multiply(divide(1, 5), subtract(17, 8)), divide(20, const_100)), multiply(subtract(15, 8), divide(1, 5))) | a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 / 5 - inch increase in the length of the shoe . if the largest size of this shoe is 20 % longer than the smallest size , how long , in inches , is the shoe in size 15 ? | "let x be the length of the size 8 shoe . then 0.2 x = 9 / 5 x = 9 inches the size 15 shoe has a length of 9 + 7 / 5 = 10.4 inches the answer is c ." | a = 1 / 5
b = 17 - 8
c = a * b
d = 20 / 100
e = c / d
f = 15 - 8
g = 1 / 5
h = f * g
i = e + h
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | c | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 60)), divide(multiply(const_100, 50), 60))) | if the cost price of 60 articles is equal to the selling price of 50 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 50 y = 60 x y = 1.2 x the answer is c ." | a = 100 * 50
b = a / 60
c = 100 - b
d = 100 * 50
e = d / 60
f = c / e
g = 100 * f
|
['a ) 1 : 4', 'b ) 1 : 16', 'c ) 4 : 1', 'd ) 16 : 1', 'e ) none'] | d | divide(surface_sphere(40), surface_sphere(10)) | spheres a and b have their radil 40 cm and 10 cm respectively . the ratio of the surface area of a to the surface area of b is | solution let the radil of a and b be r and r respectively . required ratio = 4 π r 2 / 4 π r 2 ‹ = › r 2 / r 2 ‹ = › ( r / r ) 2 ‹ = › ( 40 / 10 ) 2 ‹ = › 16 : 1 . answer d | a = surface_sphere / (
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a ) 1.5 , b ) 1.8 , c ) 2.1 , d ) 2.4 , e ) 2.7 | b | subtract(6, multiply(divide(multiply(6, 35), const_100), const_2)) | a 6 - liter solution is 35 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | "let x be the amount of pure alcohol required . 0.35 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 2.1 x = 1.8 liters the answer is b ." | a = 6 * 35
b = a / 100
c = b * 2
d = 6 - c
|
a ) 2500 , b ) 2516 , c ) 1600 , d ) 36 , e ) 2116 | e | multiply(divide(46, 46), const_100) | 46 x 46 = ? | "46 x 46 = ( 46 ) 2 = ( 50 - 4 ) 2 = ( 50 ) 2 + ( 4 ) 2 - ( 2 x 50 x 4 ) = 2500 + 16 - 400 = 2516 - 400 = 2116 e )" | a = 46 / 46
b = a * 100
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a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) none of these | a | divide(multiply(130, 250), multiply(1000, 65)) | if a * b * c = 130 , b * c * d = 65 , c * d * e = 1000 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a ∗ b ∗ c / b ∗ c ∗ d = 130 / 65 = > a / d = 2 d ∗ e ∗ f / c ∗ d ∗ e = 250 / 1000 = > f / c = 14 a / d * f / c = 2 * 1 / 4 = 1 / 2 answer : a" | a = 130 * 250
b = 1000 * 65
c = a / b
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a ) 99111 , b ) 99946 , c ) 99869 , d ) 99792 , e ) 99654 | b | multiply(add(const_100, const_2), 77) | calculate the largest 5 digit number which is exactly divisible by 77 ? | "largest 4 digit number is 99999 after doing 99999 ÷ 77 we get remainder 53 hence largest 5 digit number exactly divisible by 77 = 99999 - 53 = 99946 b" | a = 100 + 2
b = a * 77
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a ) 12 , b ) 15 , c ) 20 , d ) 21 , e ) 24 | e | multiply(divide(subtract(const_1, multiply(add(inverse(4), inverse(20)), 3)), inverse(4)), const_60) | working at a constant rate , p can finish a job in 4 hours . q , also working at a constant rate , can finish the same job in 20 hours . if they work together for 3 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | "each hour they complete 1 / 4 + 1 / 20 = 3 / 10 of the job . in 3 hours , they complete 3 ( 3 / 10 ) = 9 / 10 of the job . the time for p to finish is ( 1 / 10 ) / ( 1 / 4 ) = ( 2 / 5 ) hour = 24 minutes the answer is e ." | a = 1/(4)
b = 1/(20)
c = a + b
d = c * 3
e = 1 - d
f = 1/(4)
g = e / f
h = g * const_60
|
a ) 10 am , b ) 12 am , c ) 10.30 am , d ) 12.30 pm , e ) 2 pm | e | add(divide(add(290, 25), add(20, 25)), 7) | two stations p and q are 290 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ? | "assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 290 = > 20 x + 25 ( x - 1 ) = 290 = > 45 x = 315 = > x = 7 means , they meet after 7 hours after 7 am , ie , they meet at 2 pm answer is e ." | a = 290 + 25
b = 20 + 25
c = a / b
d = c + 7
|
a ) 430 , b ) 450 , c ) 460 , d ) 480 , e ) 512 | e | multiply(multiply(multiply(divide(const_1, const_2), 16), 8), 8) | free notebooks were distributed in a class among all the students . each student got notebooks which were 1 / 8 th of the number of student . if number of student been half , then each child would have received 16 notebooks in total . find the total number of books distributed among students ? | let suppose total number of students in class are x . then from the question we can conclude it that , x ∗ 18 x = x 2 ∗ 16 = > x = 64 total notebooks , = 18 x 2 = ( 18 ∗ 64 ∗ 64 ) = 512 e | a = 1 / 2
b = a * 16
c = b * 8
d = c * 8
|
a ) 100 , b ) 90 , c ) 30 , d ) 71 , e ) 80 | b | subtract(multiply(const_100, add(const_10, multiply(const_3, const_2))), 990) | what is the least number to be subtracted from 990 to make it a perfect square ? | "the numbers less than 990 and are squares of certain number is 900 . the least number that should be subtracted from 990 to make it perfect square = 990 - 900 = 90 . answer : b" | a = 3 * 2
b = 10 + a
c = 100 * b
d = c - 990
|
a ) 1 / 49 , b ) 4 / 7 , c ) 3 / 7 , d ) 1 / 3 , e ) 40 / 49 | d | subtract(const_1, sqrt(divide(16, 36))) | jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 16 / 36 , what is the probability that the next one she draws will be pink ? | "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 16 / 36 = ( 2 / 3 ) * ( 2 / 3 ) therefore probability that the next one she draws is pink = 1 / 3 option ( d )" | a = 16 / 36
b = math.sqrt(a)
c = 1 - b
|
a ) 1 , b ) 13 , c ) 26 , d ) 52 , e ) 1014 | c | multiply(add(subtract(13, const_10), const_2), add(subtract(13, const_10), const_2)) | if x is a sum of all even integers on the interval 13 . . . 65 and y is their number , what is the gcd ( x , y ) ? | "x = 14 + 16 + . . . + 64 = ( largest + smallest ) / 2 * ( # of terms ) = ( 14 + 64 ) / 2 * 26 = 39 * 26 . gcd of 26 and 39 * 26 is 26 . answer : c ." | a = 13 - 10
b = a + 2
c = 13 - 10
d = c + 2
e = b * d
|
a ) 210 , b ) 1365 , c ) 3003 , d ) 5005 , e ) 3030 | c | choose(15, 5) | a child want to buy 5 new toys . if there are 15 toys that satisfy the child ' s criteria , in how many different ways can the child select the 5 new toys if the order of selection does n ' t matter . | the number of ways of choosing 5 toys out of 15 are counted by = 15 c 5 ways = 3003 answer : option c | a = math.comb(15, 5)
|
a ) 105 , b ) 106 , c ) 107 , d ) 108 , e ) 109 | d | divide(4800, add(multiply(60, const_0_2778), multiply(100, const_0_2778))) | two boys started running simultaneously around a circular track of length 4800 m from the same point at speeds of 60 km / hr and 100 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? | time taken to meet for the first time anywhere on the track = length of the track / relative speed = 4800 / ( 60 + 100 ) 5 / 18 = 4800 * 18 / 160 * 5 = 108 seconds . answer : d | a = 60 * const_0_2778
b = 100 * const_0_2778
c = a + b
d = 4800 / c
|
a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 17 | b | subtract(30, divide(add(multiply(30, 7.50), 425), add(25, 7.50))) | a contractor is engaged for 30 days on the condition that he receives rs . 25 for each dayhe works & is fined rs . 7.50 for each day is absent . he gets rs . 425 in all . for how many dayswas he absent ? | "30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = 10 b" | a = 30 * 7
b = a + 425
c = 25 + 7
d = b / c
e = 30 - d
|
a ) $ 160,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 540,000 | a | divide(const_3600, const_10) | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 28 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 28 years , the investment will get doubled 28 / 14 = 2 times . so the amount invested will get doubled thrice . so , 40000 * 2 = 80000 80000 * 2 = 160000 hence , the answer is a ." | a = 3600 / 10
|
a ) 4.5 sec , b ) 4.8 sec , c ) 4.2 sec , d ) 5.0 sec , e ) 5.2 sec | a | divide(add(100, 100), multiply(add(80, 80), const_0_2778)) | in what time will two trains cross each other completely , which are running on the same parallel lines in opposite directions , each 100 m in length both trains running with a speed of 80 kmph ? | d = 100 m + 100 m = 200 m * 1 / 1000 = 0.2 kms rs = 80 + 80 = 160 kmph t = ( 0.2 / 160 ) * 3600 = 4.5 sec answer : a | a = 100 + 100
b = 80 + 80
c = b * const_0_2778
d = a / c
|
a ) 960 , b ) 1005 , c ) 1,200 , d ) 920 , e ) none of these | b | multiply(divide(add(892, 448), const_2), add(const_1, divide(50, const_100))) | the profit earned by selling an article for 892 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ? | let the profit or loss be x and 892 – x = 448 + x or , x = 444 ⁄ 2 = 222 \ cost price of the article = 892 – x = 448 + x = 670 \ sp of the article = 670 × 150 ⁄ 100 = 1005 answer b | a = 892 + 448
b = a / 2
c = 50 / 100
d = 1 + c
e = b * d
|
a ) s . 7000 , b ) s . 9000 , c ) s . 14000 , d ) s . 12000 , e ) s . 27000 | d | divide(720, divide(multiply(subtract(15, 12), const_2), const_100)) | a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 720 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 720 = > 30 x / 100 - 24 x / 100 = 720 = > 6 x / 100 = 720 = > x = 12000 . answer : d" | a = 15 - 12
b = a * 2
c = b / 100
d = 720 / c
|
a ) 45 % , b ) 13 % , c ) 11 % , d ) 12 % , e ) 15 % | a | multiply(subtract(divide(18, const_100), divide(subtract(4, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 4 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ? | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 4 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 4 or 45 * 18 - 45 * x + 36 * x = 400 or - 9 x = 400 - 45 * 18 or x = 410 / 9 = 45 % = 45 % a" | a = 18 / 100
b = 18 / 100
c = 18 * 2
d = b * c
e = 4 - d
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
|
['a ) 320', 'b ) 340', 'c ) 300', 'd ) 330', 'e ) 360'] | b | multiply(multiply(4, 5), 17) | in a building there are 5 rooms . each having a equal area . the length of the room is 4 m and breadht is 5 m . the height of the rooms are 2 m . if 17 bricks are needed to make a square meter then how many bricks are needed to make the floor of a particular room ? | explanation : area of the floor = 4 m x 5 m = 20 m 2 17 bricks are needed to make 1 m 2 area . so to make 20 m 2 no of bricks needed = 17 x 20 = 340 answer : b | a = 4 * 5
b = a * 17
|
a ) 278,889 , b ) 308,889 , c ) 338,889 , d ) 368,889 , e ) 398,889 | c | add(add(add(subtract(const_10, 1), multiply(subtract(const_100, const_10), const_2)), multiply(subtract(const_100, const_10), const_3)), multiply(subtract(const_100, const_10), const_4)) | meena wrote all the numbers from 1 to 69,999 inclusive . how many digits did she write in total ? | "1 - 9 = > 1 * 9 digits 10 - 99 = > 2 * 90 = 180 ( numbers between 10 - 99 is 90 where each has 2 digits ) 100 - 999 = > 3 * 900 = 2700 1000 - 9999 = > 4 * 9000 = 36,000 10000 - 69999 = > 5 * 60,000 = 300,000 the answer is 338,889 the answer is c ." | a = 10 - 1
b = 100 - 10
c = b * 2
d = a + c
e = 100 - 10
f = e * 3
g = d + f
h = 100 - 10
i = h * 4
j = g + i
|
a ) a ) 1120 , b ) b ) 1065 , c ) c ) 1070 , d ) d ) 1075 , e ) e ) 1080 | a | add(multiply(9, 70), multiply(9, 55)) | harkamal purchased 9 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 9 kg grapes = 70 × 9 = 630 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 630 + 490 = 1120 a" | a = 9 * 70
b = 9 * 55
c = a + b
|
a ) 104 , b ) 60 , c ) 52 , d ) 50 , e ) 142 | e | add(subtract(100, multiply(power(2, 2), 2)), multiply(2, power(5, 2))) | if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 100 , what is the value of f ( 5 ) ? | "f ( x ) = 2 x ^ 2 + y f ( 2 ) = 100 = > 2 * ( 2 ) ^ 2 + y = 100 = > 8 + y = 100 = > y = 92 f ( 5 ) = 2 * ( 5 ) ^ 2 + 92 = 142 answer e" | a = 2 ** 2
b = a * 2
c = 100 - b
d = 5 ** 2
e = 2 * d
f = c + e
|
a ) 1672 , b ) 1640 , c ) 1677 , d ) 1698 , e ) 1679 | a | subtract(2717, divide(multiply(multiply(3, 5), 2717), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2717 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2717 - x ) * 3 * 5 ) / 100 24 x / 100 = 40755 / 100 - 15 x / 100 39 x = 40755 = > x = 1045 second sum = 2717 â € “ 1045 = 1672 answer : a" | a = 3 * 5
b = a * 2717
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2717 - f
|
a ) 110 , b ) 330 , c ) 572 , d ) 430 , e ) 880 | c | multiply(divide(880, const_100), subtract(const_100, 35)) | if 35 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "students enrolled in biology are 35 % and therefore not enrolled are 65 % . so of 880 is 880 * . 65 = 572 answer is c 572" | a = 880 / 100
b = 100 - 35
c = a * b
|
a ) 16 , b ) 24 , c ) 32 , d ) 40 , e ) 48 | a | subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 28), subtract(const_100, 75)), 800), const_100), divide(multiply(subtract(const_100, 75), 800), const_100)) | in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 28 percent are juniors and 75 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ? | "200 are sophomores . the number of freshmen is 600 - 160 - 0.28 ( 800 ) = 216 the answer is a ." | a = 160 / 800
b = a * 100
c = 100 - b
d = c - 28
e = 100 - 75
f = d - e
g = f * 800
h = g / 100
i = 100 - 75
j = i * 800
k = j / 100
l = h - k
|
a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 300 , e ) none | d | subtract(subtract(multiply(divide(1700, const_10), const_2), const_12), const_12) | a sum of rs . 1700 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is : | "explanation let c ’ s share = rs . x then , b ’ s share = rs . x / 4 , a ’ s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1700 = > 17 x / 12 = 1700 = > 1700 x 12 / 17 = rs . 1200 hence , b ’ s share = rs . ( 1200 / 4 ) = rs . 300 . answer d" | a = 1700 / 10
b = a * 2
c = b - 12
d = c - 12
|
a ) 10200 , b ) 10000 , c ) 9500 , d ) 9000 , e ) none | b | add(multiply(divide(subtract(10000, multiply(divide(5, const_100), 10000)), const_2), const_2), multiply(multiply(divide(5, const_100), 10000), const_1)) | on a certain island , 5 % of the 10000 inhabitants are one - legged and half of the others go barefooted . what is the least number of shoes needed in the island ? | explanation : = > one - legged = 5 % of 10000 = 500 = > remaining = 10000 - 500 = 9500 = > barefooted = 9500 / 2 = 4750 = > remaining people = 9500 - 4750 = 4750 = > hence required number of shoes = 4750 * 2 + 500 * 1 = 10000 hence ( b ) is the correct answer . answer : b | a = 5 / 100
b = a * 10000
c = 10000 - b
d = c / 2
e = d * 2
f = 5 / 100
g = f * 10000
h = g * 1
i = e + h
|
a ) 100 rs , b ) 150 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | d | divide(subtract(divide(multiply(multiply(9000, 6), 2), const_100), divide(multiply(multiply(9000, 4), 2), const_100)), 2) | a person borrows rs . 9000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 9000 * 6 * 2 ) / 100 ] - [ ( 9000 * 4 * 2 ) / 100 ] 1080 - 720 = 360 gain in 1 year = ( 360 / 2 ) = 180 rs answer : d" | a = 9000 * 6
b = a * 2
c = b / 100
d = 9000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 37.5 , b ) 75 , c ) 105 , d ) 150 , e ) 175 | c | divide(subtract(multiply(14, const_2), 7), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 7 gallons of water and 14 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 14 gallons salt is added and 7 gallons of water is added . so salt now becomes - > ( 0.2 x + 14 ) and water - - > ( 0.6 x + 7 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 14 ) = ( 0.6 x + 7 ) / 2 = > 0.2 x = 21 solving , x = 105 answer : c" | a = 14 * 2
b = a - 7
c = 20 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = 1 / 4
h = f * g
i = d - h
j = 20 / 100
k = 2 * j
l = i - k
m = b / l
|
a ) 25 kmph , b ) 60 kmph , c ) 100 kmph , d ) 120 kmph , e ) 180 kmph | b | divide(540, divide(multiply(12, 3), 4)) | a car takes 12 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 4 th of the previous time ? | "time = 12 distence = 540 3 / 4 of 12 hours = 12 * 3 / 4 = 9 hours required speed = 540 / 9 = 60 kmph b )" | a = 12 * 3
b = a / 4
c = 540 / b
|
a ) 140 , b ) 94 , c ) 104 , d ) 70 , e ) 80 | d | divide(multiply(50, 140), const_100) | a soccer team played 140 games and won 50 percent of them . how many games did it win ? | "50 % of 140 = x 0.50 * 140 = x 70 = x answer : d" | a = 50 * 140
b = a / 100
|
a ) 100 , b ) 610 , c ) 729 , d ) 900 , e ) 900000 | e | multiply(multiply(multiply(multiply(11, const_10), const_10), const_10), const_10) | how many 11 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 12345654321 . | "take the task of building palindromes and break it intostages . stage 1 : select the 11 th digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the 10 th 9 th , 8 th , 7 th , 6 th digit we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 2 in 10 ways important : at this point , the remaining digits are alreadylocked in . stage 3 : select the , 5 th , 4 th , 3 rd , 2 nd , 1 st digit so , we can complete this stage in 1 way . by thefundamental counting principle ( fcp ) , we can complete all 5 stages ( and thus build a 11 - digit palindrome ) in ( 9 ) ( 10 ) ( 10 ) ( 10 ) ( 10 ) ( 10 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ways ( = 900000 ways ) answer : e" | a = 11 * 10
b = a * 10
c = b * 10
d = c * 10
|
a ) 246 , b ) 248 , c ) 892 , d ) 15,128 , e ) 20,256 | c | multiply(multiply(222, subtract(224, 222)), subtract(224, 222)) | the telephone company wants to add an area code composed of 2 letters to every phone number . in order to do so , the company chose a special sign language containing 224 different signs . if the company used 222 of the signs fully and two remained unused , how many additional area codes can be created if the company uses all 224 signs ? | "# of 2 - letter codes possible from 224 different signs = 224 * 224 . # of 2 - letter codes possible from 222 different signs = 222 * 222 . the difference = 224 ^ 2 - 222 ^ 2 = ( 224 - 222 ) ( 224 + 222 ) = 892 . answer : c ." | a = 224 - 222
b = 222 * a
c = 224 - 222
d = b * c
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | multiply(divide(subtract(subtract(42, const_1), multiply(subtract(15, const_1), const_2)), subtract(multiply(subtract(15, const_1), const_4), multiply(subtract(15, const_1), const_2))), subtract(15, const_1)) | a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and captain ) to a nearby port . altogether , these passengers have 15 heads and 42 legs . how many cats does the ship host ? | "sa ' s + co + ca + cats = 15 . sa ' s + 1 + 1 + cats = 15 or sa ' s + cats = 13 . sa ' s ( 2 ) + 2 + 2 + cats * 4 = 42 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 13 - cats + cat * 2 = 19 then cats = 6 b" | a = 42 - 1
b = 15 - 1
c = b * 2
d = a - c
e = 15 - 1
f = e * 4
g = 15 - 1
h = g * 2
i = f - h
j = d / i
k = 15 - 1
l = j * k
|
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 4 | e | divide(add(multiply(5, 3), add(multiply(1, 1), multiply(4, const_2))), 6) | in 6 successive games , a baseball team scored 1 runs once , 4 runs twice , and 5 runs 3 times . what was the average ( arithmetic mean ) number of runs the team scored per game over the 6 - game period ? | 1 runs once ( one game ) 4 runs twice ( two game ) 5 runs three times ( three games ) so we have ( 1 * 1 + 4 * 2 + 5 * 3 ) / 6 = 24 / 6 = 4 . ans : e . | a = 5 * 3
b = 1 * 1
c = 4 * 2
d = b + c
e = a + d
f = e / 6
|
a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 7.2 hrs , e ) 30 hrs | e | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 6))) | a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 6 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 5 - 1 / 6 = 1 / 30 therefore the cistern will be filled in 30 hours . answer : e" | a = 1 / 5
b = 1 / 6
c = a - b
d = 1 / c
|
a ) 6 miles , b ) 8,4 miles , c ) 9 miles , d ) 9,6 miles , e ) 15 miles | e | multiply(divide(23, add(3, 2)), 2) | one hour before john started walking from p to q , a distance of 23 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 23 - 3 = 20 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 20 miles so it will take them 20 / 5 = 4 hours to meet . in 4 hrs , ann would have covered 4 hrs * 3 miles per hour = 12 miles . adding this to the 3 miles she covered before john , ann covered a total of 3 + 12 = 15 miles . answer ( e ) | a = 3 + 2
b = 23 / a
c = b * 2
|
a ) 524.48 , b ) 344.28 , c ) 574.36 , d ) 585.64 , e ) 595.46 | b | multiply(128.5, power(add(const_4, const_1), const_4)) | ( 128.5 x 60 ) + ( 13.8 x 65 ) = ? x 25 | "explanation : ? = ( 128.5 x 60 ) + ( 13.8 x 65 ) / 25 = 7710 + 897 / 25 = 344.28 answer : option b" | a = 4 + 1
b = a ** 4
c = 128 * 5
|
a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | multiply(divide(12, const_4.0), divide(12, 5)) | at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 11 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 1 to 5 . how many cooks does the restaurant have ? | "originally there were 3 k cooks and 11 k waiters . the new ratio is 1 : 5 which equals 3 : 15 . 15 k = 11 k + 12 k = 3 there are 9 cooks . the answer is c ." | a = 12 / 4
b = 12 / 5
c = a * b
|
a ) $ 5.5 , b ) $ 15.5 , c ) $ 23.5 , d ) $ 100 , e ) $ 105 | c | subtract(subtract(multiply(600, power(add(const_1, divide(10, const_100)), 2)), 600), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 1,000), 1,000)) | on a certain day , tim invested $ 600 at 10 percent annual interest , compounded annually , and lana invested 1,000 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ s investment in the first 2 years ? | "compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 600 ( 1.10 ) ^ 2 = 600 ( 1.21 ) = 726 and lana would have 1,000 ( 1.05 ) ^ 2 = 1,000 ( 1.1025 ) = 1102.5 thus , tim earned 126 dollars , while lana earned 102.5 dollars the difference is $ 23.5 and the answer is c ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 600 * c
e = d - 600
f = 5 / 100
g = 1 + f
h = g ** 2
i = h * 1
j = i - 1
k = e - j
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | subtract(10, const_3) | john has 10 pairs of matched socks . if he loses 9 individual socks , what is the greatest number of pairs of matched socks he can have left ? | "because we have to maximize the pair of matched socks , we will remove 4 pairs ( 8 socks ) out of 10 pairs 1 sock from the 5 th pair . thus the no of matching socks pair remaining = 10 - 5 = 5 answer c" | a = 10 - 3
|
a ) 9 % , b ) 10 % , c ) 105 / 8 % , d ) 11 % , e ) 12 % | a | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(8.5, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | a $ 500 investment and a $ 1,500 investment have a combined yearly return of 8.5 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1,500 investment have ? | "the equation we can form the question : return on total investment = sum of individual investments ( 500 + 1500 ) ( 8.5 ) = ( 500 ∗ 7 ) + ( 1500 x ) , where x is the return on investment of 1500 . solving the equation , we get x = 9 % ( option a ) answer : a" | a = 3 * 500
b = 500 + a
c = 8 / 5
d = b * c
e = 7 / 100
f = 500 * e
g = d - f
h = 3 * 500
i = g / h
j = i * 100
|
a ) 18 % , b ) 13 % , c ) 12 % , d ) 32 % , e ) 19 % | e | multiply(subtract(divide(divide(multiply(subtract(const_100, 30), add(const_100, 70)), const_100), const_100), const_1), const_100) | a trader bought a car at 30 % discount on its original price . he sold it at a 70 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 70 s = 70 * ( 170 / 100 ) = 119 100 - 119 = 19 % answer : e" | a = 100 - 30
b = 100 + 70
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) 7 / 9 , b ) 20 / 9 , c ) 7 / 20 , d ) 9 / 20 , e ) 11 / 20 | b | divide(multiply(5, 4), add(4, 5)) | in an electric circuit , two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is 4 ohms and y is 5 ohms ? | "1 / r = 1 / x + 1 / y 1 / r = 1 / 4 + 1 / 5 = 9 / 20 r = 20 / 9 the answer is b ." | a = 5 * 4
b = 4 + 5
c = a / b
|
a ) 23.56 , b ) 23.59 , c ) 23.55 , d ) 18.86 , e ) 23.57 | d | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(14, 10)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 14 cm and breadth 10 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 14 + 10 ) = 48 cm parameter of the square = 48 cm i . e . 4 a = 48 a = 12 diameter of the semicircle = 12 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 12 ) = 1 / 2 ( 22 / 7 ) ( 12 ) = 264 / 14 = 18.86 cm to two decimal places answer : d" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | d | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(6, 100)) | a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 6 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 6 / 100 x 12000 ) = rs . 720 . answer d" | a = 10 * 1000
b = 4 * 1000
c = a + b
d = 4 * 100
e = c + d
f = 20 / 100
g = 100 * f
h = 100 + g
i = e / h
j = 100 * i
k = 6 / 100
l = j * k
|
a ) 21 , b ) 21.5 , c ) 22 , d ) 22.5 , e ) 23 | e | multiply(divide(43, add(add(divide(15, const_100), const_1), const_1)), add(divide(15, const_100), const_1)) | two friends plan to walk along a 43 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 15 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ? | "if q complete x kilometers , then p completes 1.15 x kilometers . x + 1.15 x = 43 2.15 x = 43 x = 43 / 2.15 = 20 then p will have have walked 1.15 * 20 = 23 km . the answer is e ." | a = 15 / 100
b = a + 1
c = b + 1
d = 43 / c
e = 15 / 100
f = e + 1
g = d * f
|
a ) s . 2890 , b ) s . 2330 , c ) s . 1190 , d ) s . 3468 , e ) s . 2680 | d | multiply(divide(multiply(divide(588, 7), 17), 7), 17) | the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 588 , krishan has ? | "ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 588 : n & there n = 289 x 588 / 49 = rs . 3468 answer : d" | a = 588 / 7
b = a * 17
c = b / 7
d = c * 17
|
a ) 1 km , b ) 500 mts , c ) 600 mts , d ) 2 km , e ) 1200 mts | e | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 4), divide(const_1, 8)), divide(const_1, 12))), const_3), const_1000) | a person travels equal distances with speeds of 4 km / hr , 8 km / hr , 12 km / hr . and takes a total time of 11 minutes . find the total distance ? | let the each distance be x km total distance = 3 x then total time , ( x / 4 ) + ( x / 8 ) + ( x / 12 ) = 11 / 60 x = 0.4 total distance = 3 * 0.4 = 1.2 km = 1200 meters correct option is e | a = 11 / const_60
b = 1 / 4
c = 1 / 8
d = b + c
e = 1 / 12
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 1 hours , b ) 2 hours , c ) 3 hours , d ) 4 hours , e ) 5 hours | b | divide(multiply(40, divide(30, const_60)), subtract(50, 40)) | a man drives at a speed of 40 miles / hr . his wife left 30 mins . late with 50 miles / hr speed . when will they meet ? | in 30 minutes the man would have droven 20 miles , when his wife starts . now both are driving in the same direction . so speed of wife related to man is 10 ( 50 - 40 ) miles / hr . so wife will cover 20 miles in 2 hours to meet the man . answer : b | a = 30 / const_60
b = 40 * a
c = 50 - 40
d = b / c
|
a ) 4 , b ) 8 , c ) 5 , d ) 12 , e ) 7 | d | multiply(8, divide(6, 8)) | a and b together can complete work in 8 days . a alone starts working and leaves it after working for 6 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ? | "explanation : ( a + b ) one day work = 1 / 8 now a does half of the work in 6 day so a can complete the whole work in 12 days a ’ s one day work = 1 / 12 b ’ s one day work = 1 / 8 - 1 / 12 = 1 / 24 b alone can complete the work in 24 days so half of the work in 12 days answer : option d" | a = 6 / 8
b = 8 * a
|
a ) 70 , b ) 450 , c ) 150 , d ) 35 , e ) 350 | b | divide(63, 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 63 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 63 , so 0.14 * { total } = 63 - - > { total } = 450 . answer : b . | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 63 / d
|
a ) 15 , b ) 19 , c ) 14 , d ) 18 , e ) 26 | e | subtract(78, divide(multiply(78, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 78 kmph , find the speed of the stream ? | the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 78 / 3 = 26 kmph . answer : e | a = 78 * 2
b = a / 3
c = 78 - b
|
a ) 2 , b ) 13 , c ) 24 , d ) 36 , e ) 42 | d | multiply(divide(subtract(960, divide(480, const_2)), 40), const_2) | angelina walked 960 meters from her home to the grocery at a constant speed . she then walked 480 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | "let the speed be x . . . so time taken from home to grocery = 960 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 960 / x - 240 / x = 40 . . 720 / x = 40 . . x = 18 m / secs . . so grocery to gym = 2 * 18 = 36 m / s . . . d" | a = 480 / 2
b = 960 - a
c = b / 40
d = c * 2
|
a ) $ 3,750 , b ) $ 5,600 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 31,500 | e | multiply(3500, power(const_3, divide(28, divide(112, 8)))) | money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 3500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 3500 triples to $ 10500 , in 28 years , it will again triple to $ 10500 * 3 = $ 31,500 answer e" | a = 112 / 8
b = 28 / a
c = 3 ** b
d = 3500 * c
|
a ) 28 , 42,49 , b ) 24 , 36,42 , c ) 20 , 30,35 , d ) 16 , 24,28 , e ) 12 , 18,21 | d | multiply(6, divide(add(multiply(8, 3), 44), add(add(4, 6), 7))) | the ratio of the ages of 3 people is 4 : 6 : 7 . 8 years ago , the sum of their ages was 44 . find their present ages . | let the present ages be 4 x , 6 x , 7 x . ( 4 x - 8 ) + ( 6 x - 8 ) + ( 7 x - 8 ) = 44 17 x = 68 x = 4 their present ages are 16 , 24 , and 28 . the answer is d . | a = 8 * 3
b = a + 44
c = 4 + 6
d = c + 7
e = b / d
f = 6 * e
|
a ) 22 sec , b ) 77 sec , c ) 25 sec , d ) 18 sec , e ) 12.5 sec | e | divide(add(150, 100), multiply(72, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 72 kmph ? | "d = 100 + 150 = 250 s = 72 * 5 / 18 = 20 mps t = 250 / 20 = 12.5 sec answer : e" | a = 150 + 100
b = 72 * const_0_2778
c = a / b
|
a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | b | add(multiply(35.50, divide(15, const_100)), 35.50) | a meal cost $ 35.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be : | "10 % ( 35.5 ) = 3.55 15 % ( 35.5 ) = 5.325 total amount could have been 35.5 + 3.55 and 35.5 + 5.325 = > could have been between 39.05 and 40.625 = > approximately between 39 and 41 answer is b ." | a = 15 / 100
b = 35 * 50
c = b + 35
|
a ) s . 3988 , b ) s . 3900 , c ) s . 3228 , d ) s . 4200 , e ) s . 3928 | d | multiply(multiply(subtract(add(90, 60), 10), 10), 3) | a rectangular lawn of dimensions 90 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "area = ( l + b – d ) d ( 90 + 60 – 10 ) 10 = > 1400 m 2 1400 * 3 = rs . 4200 answer : d" | a = 90 + 60
b = a - 10
c = b * 10
d = c * 3
|
a ) 84 , b ) 63.2 , c ) 104 , d ) 50.8 , e ) 15.8 | b | divide(multiply(40, 158), const_100) | a soccer team played 158 games and won 40 percent of them . how many games did it win ? | 40 % of 158 = x 0.40 * 158 = x 63.2 = x answer : b | a = 40 * 158
b = a / 100
|
a ) 1000 , b ) 1500 , c ) 1240 , d ) 1630 , e ) 1400 | e | add(multiply(divide(700, 70), 70), multiply(divide(700, 70), 70)) | the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 70 kmph its consumption is 1 litre per hour . if each litre costs $ 70 and other expenses per hous is $ 70 , then what would be the minimum expenditure required to cover a distance of 700 km ? | "70 kmph consumption is 1 lt / hr so 700 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 70 so 10 lt costs $ 700 extra expenses for 1 hr - $ 70 10 hrs - $ 700 total expense - $ 700 + $ 700 = $ 1400 answer : e" | a = 700 / 70
b = a * 70
c = 700 / 70
d = c * 70
e = b + d
|
a ) 71.5 , b ) 72 , c ) 72.5 , d ) 73 , e ) 76.6 | e | divide(1, divide(add(multiply(const_3600, divide(1, 80)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 80 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 80 kilometers per hour is 1 / 80 hours = 3,600 / 80 seconds = 45 seconds ; time to cover this distance at regular speed is 45 + 2 = 47 seconds = 47 / 3,600 hours = 5 / 383 hours ; so , we get that to cover 1 kilometer 1 / ( 5 / 383 ) hours is needed - - > regular speed 76.6 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : e" | a = 1 / 80
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
|
a ) 16.67 % , b ) 23 % , c ) 18.3 % , d ) 19.75 % , e ) 21.23 % | a | multiply(divide(subtract(add(15, 3), add(multiply(divide(subtract(const_100, 20), const_100), 15), 3)), add(15, 3)), const_100) | 15 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 15 litres of mix . = 20 % of 15 litres = ( 20 * 15 / 100 ) = 3 litres water in it = 15 - 3 = 12 litres new quantity of mix . = 15 + 3 = 18 litres quantity of alcohol in it = 3 litres percentage of alcohol in new mix . = 3 * 100 / 18 = 50 / 3 = 16.67 % answer is a" | a = 15 + 3
b = 100 - 20
c = b / 100
d = c * 15
e = d + 3
f = a - e
g = 15 + 3
h = f / g
i = h * 100
|
a ) 168 , b ) 72 , c ) 56 , d ) 170 , e ) 60 | b | multiply(divide(subtract(const_100, 70), const_100), multiply(divide(subtract(const_100, 60), const_100), 600)) | total number of people attending a party is 600 . 70 % of the men are wearing black shirt and rest are wearing red . if 60 % of the total people are women then how many men are wearing red shirts ? | number of woman in the party = 0.60 * 600 = 360 number of men in the party = 600 - 360 = 240 % of men wearing red shirts = 100 - 70 = 30 % number of men wearing red shirt = 0.30 * 240 = 72 . b is the correct answer | a = 100 - 70
b = a / 100
c = 100 - 60
d = c / 100
e = d * 600
f = b * e
|
a ) $ 25.2 , b ) $ 22 . , c ) $ 13 . , d ) $ 14.5 . , e ) $ 15.5 | a | add(add(multiply(4, 4), multiply(add(subtract(7, 4), divide(60, const_100)), 2)), divide(60, const_100)) | the mall charges 60 cents for the first hour of parking and $ 4 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 2 per hour . if edward parked his car in the mall for 7 hours and 30 minutes , how much is he going to pay ? | "0.60 + 4 * 4 + 2 * 4 + 0.60 = $ 25.2 answer : $ 25.2" | a = 4 * 4
b = 7 - 4
c = 60 / 100
d = b + c
e = d * 2
f = a + e
g = 60 / 100
h = f + g
|
a ) $ 2.50 , b ) $ 5.00 , c ) $ 5.50 , d ) $ 7.50 , e ) $ 15.00 | b | divide(subtract(650, multiply(22, divide(subtract(multiply(650, const_2), 760), subtract(multiply(22, const_2), 8)))), 16) | suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 650 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 760 . how much more do pairs of boots cost than pairs of shoes at suzie ’ s discount footwear ? | "let x be pair of shoes and y be pair of boots . 22 x + 16 y = 650 . . . eq 1 8 x + 32 y = 760 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1300 8 x = 760 . 36 x = 540 = > x = 15 . sub x in eq 2 . . . . we get 120 + 32 y = 760 . . . then we get 32 y = 640 then y = 20 difference between x and y is 5 . answer : option b is correct answer ." | a = 650 * 2
b = a - 760
c = 22 * 2
d = c - 8
e = b / d
f = 22 * e
g = 650 - f
h = g / 16
|
a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | e | multiply(17, 17) | ( 17 ) 5.5 x ( 17 ) ? = 178 | "solution let ( 17 ) 5.5 * ( 17 ) x = 178 . then , ( 17 ) 5.5 + x = ( 17 ) 8 . ∴ 5.5 + x = 8 ⇔ x = ( 8 - 5.5 ) ⇔ x = 2.5 answer e" | a = 17 * 17
|
a ) 627 , b ) 528 soldiers , c ) 626 , d ) 657 , e ) 673 | b | subtract(divide(divide(multiply(multiply(1200, 3), 30), 25), 2.5), 1200) | in a fort , there are 1200 soldiers . if each soldier consumes 3 kg per day , the provisions available in the fort will last for 30 days . if some more soldiers join , the provisions available will last for 25 days given each soldier consumes 2.5 kg per day . find the number of soldiers joining the fort in that case . | assume x soldiers join the fort . 1200 soldiers have provision for 1200 ( days for which provisions last them ) ( rate of consumption of each soldier ) = ( 1200 ) ( 30 ) ( 3 ) kg . also provisions available for ( 1200 + x ) soldiers is ( 1200 + x ) ( 25 ) ( 2.5 ) k as the same provisions are available = > ( 1200 ) ( 30 ) ( 3 ) = ( 1200 + x ) ( 25 ) ( 2.5 ) x = [ ( 1200 ) ( 30 ) ( 3 ) ] / ( 25 ) ( 2.5 ) - 1200 = > x = 528 . answer : b | a = 1200 * 3
b = a * 30
c = b / 25
d = c / 2
e = d - 1200
|
a ) 3 , b ) 4 , c ) 5 , d ) 2 , e ) 7 | d | subtract(4312, subtract(4312, 4312)) | what least value should be replaced by * in 483 * 4312 so the number become divisible by 6 | "explanation : trick : number is divisible by 6 , if sum of all digits is divisible by 3 and 2 , last digit is 2 so it is dividable by 2 so ( 4 + 8 + 3 + * + 4 + 3 + 1 + 2 ) = 25 + * should be divisible by 6 , 15 + will be divisible by , so that least number is 2 . answer : option d" | a = 4312 - 4312
b = 4312 - a
|
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