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 ) 2 , b ) 4 , c ) 8 , d ) 12 , e ) 15 | c | divide(add(11, 5), const_2) | in one hour , a boat goes 11 km along the steram and 5 km against the stream . the speed of the boat in still waer ( in km / hr ) is : | sol . speed in still water = 1 / 2 ( 11 + 5 ) kmph = 8 kmph . answer c | a = 11 + 5
b = a / 2
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a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | d | add(add(add(divide(20, 2), divide(20, power(2, 2))), floor(divide(20, power(power(2, 2), 2)))), floor(divide(20, power(2, const_3)))) | if n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2 ^ k is a factor of n ? | "given : n = 20 ! n = 20 ! . the highest power k for which 2 ^ k is a factor of n can be found with the above formula : k = 20 / 2 + 20 / 4 + 20 / 8 + 20 / 16 = 10 + 5 + 2 + 1 = 18 answer : d ." | a = 20 / 2
b = 2 ** 2
c = 20 / b
d = a + c
e = 2 ** 2
f = e ** 2
g = 20 / f
h = math.floor(g)
i = d + h
j = 2 ** 3
k = 20 / j
l = math.floor(k)
m = i + l
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a ) 20 km , b ) 16 km , c ) 6 km , d ) 3 km , e ) 61 km | a | multiply(2, 10) | a dog travelled for 2 hours . he covered the first half of the distance at 10 kmph and remaining half of the distance at 5 kmph . find the distance travelled by the dog ? | let the distance travelled be x km . total time = ( x / 2 ) / 10 + ( x / 2 ) / 5 = 2 = > x / 20 + x / 10 = 2 = > ( 1 x + 2 x ) / 20 = 2 = > x = 20 km answer : a | a = 2 * 10
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a ) 455465473 , b ) 654676854 , c ) 667676753 , d ) 725117481 , e ) 764534522 | d | multiply(subtract(9999, const_4), 72519) | find the value of 72519 x 9999 = m ? | "72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d" | a = 9999 - 4
b = a * 72519
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a ) 17 , b ) 34 , c ) 54 , d ) 64 , e ) 112 | a | add(15, const_2) | 3 , 15 , x , 51 , 53 , 159161 value of x ? | 3 * 5 = 15 15 + 2 = 17 17 * 3 = 51 51 + 2 = 53 answer : a | a = 15 + 2
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a ) 5000 , b ) 10000 , c ) 15000 , d ) 2778 , e ) 1991 | c | 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 a ? | a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 300 a ' s capital = 22500 * 2 / 3 = 15000 answer : c | a = 300 * 100
b = a / 2
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a ) 20 , b ) 24 , c ) 25 , d ) 30 , e ) 32 | b | add(add(divide(100, add(const_4, const_1)), divide(subtract(100, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(100, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros does 100 ! end with ? | "according to above 100 ! has 100 / 5 + 100 / 25 = 20 + 4 = 24 trailing zeros . answer : b ." | a = 4 + 1
b = 100 / a
c = 4 + 1
d = 100 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 100 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(add(3, multiply(14, 4)), multiply(14, 4)) | how many two - digit numbers yield a remainder of 3 when divided by both 4 and 14 ? | "easier to start with numbers that are of the form 14 p + 3 - - - > 17,31 , 45,59 , 73,87 . out of these 3 ( 31 , 59,87 ) are also of the form 4 q + 3 . thus 3 is the answer . d is the correct answer ." | a = 14 * 4
b = 3 + a
c = 14 * 4
d = b - c
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a ) 15 minutes , b ) 18 minutes , c ) 12 minutes , d ) 58 minutes , e ) 10 minutes | a | add(divide(26, 2), 2) | a squirrel can climb 5 metres of a pole in one minute but slips 2 metres in the next minute . how much time will the squirrel take to climb 26 metres ? | explanation : as the squirrel climbs 5 m in one minute and slips 2 metres in the next minute , it climbs 3 metres in 2 minute to climb 26 metres , the time required is : the squirrel will climb ( 26 – 5 ) = 21 metres = 7 * 3 metres in 7 * 2 = 14 minutes also , the last 5 metres it climbs in another 1 minute hence , total time required = 14 + 1 = 15 minutes answer : a | a = 26 / 2
b = a + 2
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a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 8 : 6 , e ) 1 : 2 | d | divide(divide(multiply(const_4, 2), multiply(const_3.0, 2)), divide(multiply(2, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 4 : 2 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 12 : 8 . the ratio of economy cars to sport utility vehicles is 4 : 2 - - > e : s = 4 : 2 = 12 : 6 . thus , l : s = 8 : 6 . answer : d ." | a = 4 * 2
b = 3 * 0
c = a / b
d = 2 * 4
e = 2 * 4
f = d / e
g = c / f
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a ) 11 , b ) 22 , c ) 99 , d ) 77 , e ) 18 | b | multiply(divide(subtract(1100, 900), 900), const_100) | a cycle is bought for rs . 900 and sold for rs . 1100 , find the gain percent ? | "900 - - - - 200 100 - - - - ? = > 22 % answer : b" | a = 1100 - 900
b = a / 900
c = b * 100
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a ) 10 sec , b ) 22 sec , c ) 82 sec , d ) 20 sec , e ) 89 sec | b | divide(add(120, 320), multiply(add(42, 30), const_0_2778)) | two trains of length 120 m and 320 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 320 = 440 m . the time required = d / s = 440 / 20 = 22 sec . answer : b" | a = 120 + 320
b = 42 + 30
c = b * const_0_2778
d = a / c
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a ) s . 1020 , b ) s . 1160 , c ) s . 1190 , d ) s . 1202 , e ) s . 1256 | a | divide(multiply(subtract(const_100, 15), 1200), const_100) | a man buys a cycle for rs . 1200 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 1200 = rs . 85 / 100 x 1200 = rs . 1020 answer : a" | a = 100 - 15
b = a * 1200
c = b / 100
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a ) 6 metre , b ) 120 metre , c ) 220 metre , d ) 1200 metre , e ) 420 metre | e | divide(multiply(multiply(20, const_60), 35), const_100) | a wheel rotates 20 times every minute and moves 35 cm during each rotation . how many metres does the wheel move in one hour ? | number of times wheel moves in 1 hour = 20 * 60 = 1200 : . distance moves = ( 1200 * 35 ) cms = 42000 cms in metres = 420 metre answer : e | a = 20 * const_60
b = a * 35
c = b / 100
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a ) 0 , b ) 1 / 2 , c ) 2 / 9 , d ) 1 / 3 , e ) 1 | b | multiply(factorial(2), power(divide(1, 2), 2)) | a certain roller coaster has 2 cars , and a passenger is equally likely to ride in any 1 of the 2 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 2 times , what is the probability that the passenger will ride in each of the 2 cars ? | "if he is to ride 2 times and since he can choose any of the 2 cars each time , total number of ways is = 2 * 2 = 4 now the number of ways if he is to choose a different car each time is = 2 * 1 = 2 so the probability is = 2 / 4 = 1 / 2 answer : b" | a = math.factorial(2)
b = 1 / 2
c = b ** 2
d = a * c
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a ) 288 , 000,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 24 , 000,000 , e ) 240 , 000,000 | a | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(6, 100)), multiply(12, 100)) | if a rectangular room measures 12 meters by 6 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "a . 288 , 000,000 12 * 100 * 6 * 100 * 4 * 100 = 288 , 000,000" | a = 4 * 100
b = 1 / 10
c = a * b
d = 6 * 100
e = c * d
f = 12 * 100
g = e * f
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a ) rs . 207.40 , b ) rs . 534.60 , c ) rs . 648 , d ) rs . 655.60 , e ) rs . 685.60 | c | multiply(divide(12, const_100), multiply(divide(4455, 8.25), 10)) | a man invested rs . 4455 in rs . 10 shares quoted at rs . 8.25 . if the rate of dividend be 12 % , his annual income is : | solution numbers of shares = ( 4455 / 825 ) = 540 face value = rs . ( 540 x 10 ) = rs . 5400 annual income = rs . ( 12 / 10 x 5400 ) = rs . 648 . answer c | a = 12 / 100
b = 4455 / 8
c = b * 10
d = a * c
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a ) 150 , b ) 750 , c ) 1,250 , d ) 1,600 , e ) 2,500 | d | multiply(50, 5) | in a forest 160 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 160 tagged deers comprise 10 % of total # of deers - - > total # of deers = 160 * 10 = 1,600 . answer : d ." | a = 50 * 5
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a ) w = 26.7 , b ) w = 30.0 , c ) w = 40.0 , d ) w = 53.3 , e ) 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his average rate w in miles per hour for the entire trip ? | "veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is not 200 % of the first half . ) the overall time is 4 hours + 12 hours or 16 hours . since the definition of average rate = total distance traveled / total time of travel , mike ' s average rate = 640 / 16 or 40 miles per hour . answer choice c is correct ." | a = 640 / 2
b = a / 80
c = b * 2
d = 640 / 2
e = d / 80
f = c + e
g = 640 / 2
h = g / 80
i = f + h
j = 640 / i
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a ) 45 , b ) 40 , c ) 35 , d ) 47.61 , e ) 30 | d | divide(50, multiply(subtract(const_1, divide(10, const_100)), add(divide(20, const_100), const_1))) | the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 10 % . if in july , a mixture containing equal quantities of tea and coffee costs 50 / kg . how much did a kg of coffee cost in june ? | let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.9 x . in july the price of 1 / 2 kg ( 500 gm ) of tea and 1 / 2 kg ( 500 gm ) of coffee ( equal quantities ) = 50 1.2 x ( 1 / 2 ) + 0.9 x ( 1 / 2 ) = 50 = > x = 47.61 d | a = 10 / 100
b = 1 - a
c = 20 / 100
d = c + 1
e = b * d
f = 50 / e
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a ) 1954404 , b ) 1981709 , c ) 18362619 , d ) 2031719 , e ) none of these | a | multiply(divide(1398, 1398), const_100) | 1398 x 1398 | "1398 x 1398 = ( 1398 ) 2 = ( 1400 - 2 ) 2 = ( 1400 ) 2 + ( 2 ) 2 - ( 2 x 1400 x 2 ) = 1954404 . answer : option a" | a = 1398 / 1398
b = a * 100
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a ) 3 , b ) 2.5 , c ) 2.25 , d ) 1.5 , e ) 4 | b | divide(7.5, const_3) | a man can swim in still water at 7.5 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 7.5 s = x ds = 7.5 + x us = 7.5 - x 7.5 + x = ( 7.5 - x ) 2 7.5 + x = 15 - 2 x 3 x = 7.5 x = 2.5 answer : b" | a = 7 / 5
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a ) 1 : 1 , b ) 3 : 4 , c ) 5 : 2 , d ) 4 : 3 , e ) 7 : 9 | b | divide(add(multiply(4, 5), multiply(3, divide(add(5, 3), const_2))), add(multiply(4, 3), multiply(3, divide(add(5, 3), const_2)))) | two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 5 , respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio 4 : 3 . what is the ratio of the composition of the two basic elements in alloy x ? | "mixture a has a total of 5 + 3 = 8 parts . if in the final mixture this represents 4 parts , then the total number of parts in mixture b should be ( 8 / 4 ) * 3 = 6 . so , we should take of mixture b a quantity with 1 and 5 parts , respectively . this will give us in the final mixture ( 5 + 1 ) : ( 3 + 5 ) , which means 6 : 8 , or 3 : 4 . answer b ." | a = 4 * 5
b = 5 + 3
c = b / 2
d = 3 * c
e = a + d
f = 4 * 3
g = 5 + 3
h = g / 2
i = 3 * h
j = f + i
k = e / j
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a ) rs . 8000 , b ) rs . 12000 , c ) rs . 15000 , d ) rs . 6000 , e ) none | c | add(divide(57000, 4), multiply(multiply(3, 5), const_10)) | in a express train passengers traveling in a . c . sleeper class , first class and sleeper class are in the ratio 1 : 2 : 3 , and fare to each class in the ratio 5 : 4 : 2 . if the income from this train is rs . 57000 , the income of a . c . sleeper class is | income is divided in the ratio 1 × 5 : 2 × 4 : 3 × 2 = 5 : 8 : 6 . now 5 x + 8 x + 6 x = 19 x = 57000 . ∴ x = 3000 . therefore income from a . c . sleeper class = 30005 = 15000 . answer : c . | a = 57000 / 4
b = 3 * 5
c = b * 10
d = a + c
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a ) 8 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(90, const_10), const_10) | how many positive factors do 150 and 90 have in common ? | the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 150 and 90 is 30 number of factors of 30 = 8 answer : a | a = 90 - 10
b = a / 10
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a ) 120 , b ) 80 , c ) 60 , d ) 136 , e ) 150 | b | divide(400, multiply(subtract(36, 18), const_0_2778)) | mona and sona go around a circular track of length 400 m on a bike at speeds of 18 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | time taken to meet for the first time at the starting point = lcm { length of the track / speed of mona , length of the track / speed of sona } = lcm { 400 / ( 18 * 5 / 18 ) , 400 / ( 36 * 5 / 18 ) } = lcm ( 80 , 40 ) = 80 sec . answer : b | a = 36 - 18
b = a * const_0_2778
c = 400 / b
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a ) 1,000 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,540 | a | divide(540, 540) | the rate of interest on a sum of money is 11 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | "explanation : i 1 = ( p x 3 x 11 ) / 100 = p / 3 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 p / 3 + 4 p / 25 + p / 20 = 540 27 p / 50 = 540 p = 1000 answer : option a" | a = 540 / 540
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['a ) 6 cm', 'b ) 8.25 cm', 'c ) 11.25 cm', 'd ) 15.12 cm', 'e ) 3.33 cm'] | e | divide(volume_cube(10), multiply(20, 15)) | a cube of edge 10 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ? | increase in volume = volume of the cube = 10 * 10 * 10 cm ^ 3 rise in water level = volume / area = 10 * 10 * 10 / 20 * 15 = 3.33 cm answer is e | a = volume_cube / (
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a ) 25 , b ) 27 , c ) 29 , d ) 28 , e ) 31 | d | multiply(7, divide(multiply(add(7, 12), subtract(12, multiply(divide(5, add(7, 5)), 12))), subtract(multiply(12, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 12 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | "ci * vi = cf * vf ( 7 / 12 ) * ( v 1 - 12 ) = ( 7 / 16 ) * v 1 ( v 1 - 12 ) / v 1 = 3 / 4 12 accounts for the difference of 1 on ratio scale so initial volume = v 1 = 4 * 12 = 48 litres . 7 / 12 of the initial mixture was liquid a so liquid a was ( 7 / 12 ) * 48 = 28 litres . answer : d" | a = 7 + 12
b = 7 + 5
c = 5 / b
d = c * 12
e = 12 - d
f = a * e
g = 12 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
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a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 40 % | e | multiply(divide(subtract(200, 120), 200), const_100) | a bag marked at $ 200 is sold for $ 120 . the rate of discount is ? | "rate of discount = 80 / 200 * 100 = 40 % answer is e" | a = 200 - 120
b = a / 200
c = b * 100
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a ) rs 1000 , b ) rs 1515 , c ) rs 800 , d ) rs 900 , e ) none of these | c | divide(subtract(1120, multiply(1120, divide(12, const_100))), add(divide(25, const_100), const_1)) | the sale price of a trolley bag including the sale tax is rs . 1120 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is : | "explanation : 112 % of s . p . = 1120 s . p . = rs . ( 1120 x 100 / 112 ) = rs . 1000 . c . p . = rs ( 100 / 125 x 1000 ) = rs 800 . answer : c" | a = 12 / 100
b = 1120 * a
c = 1120 - b
d = 25 / 100
e = d + 1
f = c / e
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a ) 65 km / hr , b ) 17 km / hr , c ) 55 km / hr , d ) 50 km / hr , e ) 15 km / hr | c | divide(divide(subtract(125, multiply(multiply(10, const_0_2778), 10)), 10), const_0_2778) | a train 125 m long passes a man , running at 10 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 - 10 ) km / hr . x - 10 = 45 = > x = 55 km / hr . answer : c" | a = 10 * const_0_2778
b = a * 10
c = 125 - b
d = c / 10
e = d / const_0_2778
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a ) 3 . , b ) 7 . , c ) 10 . , d ) 12 . , e ) 15 . | c | multiply(5, 5) | the distance between west - town to east - town is 14 kilometers . two birds start flying simultaneously towards one another , the first leaving from west - town at a speed of 5 kilometers per minute and the second bird , leaving from east - town , at a speed of 2 kilometers per minute . what will be the distance , in kilometers , between the meeting point and west - town ? | "time taken by the birds to meet = 14 / ( 5 + 2 ) = 2 mins distance traveled by the bird traveling from west - town = 5 * 2 = 10 answer : c" | a = 5 * 5
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a ) 32 , b ) 1136 , c ) 40 , d ) 1344 , e ) 104 | d | subtract(multiply(add(subtract(divide(220, const_10), const_2), const_10), add(divide(220, const_10), subtract(divide(220, const_10), const_2))), 220) | a rectangular room has the rectangular shaped rug shown as above figure such that the rug ’ s area is 220 square feet and its length is 12 feet longer than its width . if the uniform width between the rug and room is 12 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | "rug ' s area = 220 which is ( x ) x ( 12 + x ) = 220 so x = 10 rug maintains a uniform distance of 12 feet so room has dimension 10 + 24 and 22 + 24 i . e . 34 and 46 area of room 34 x 46 = 1564 area covered is 120 so uncovered area is 1564 - 220 = 1344 ( answer d )" | a = 220 / 10
b = a - 2
c = b + 10
d = 220 / 10
e = 220 / 10
f = e - 2
g = d + f
h = c * g
i = h - 220
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a ) $ 2.45 , b ) $ 2.54 , c ) $ 2.35 , d ) $ 2.48 , e ) $ 2.49 | a | subtract(10.10, add(3.25, add(2.20, 2.20))) | little john had $ 10.10 . he spent $ 3.25 on sweets and gave to his two friends $ 2.20 each . how much money was left ? | "john spent and gave to his two friends a total of 3.25 + 2.20 + 2.20 = $ 7.65 money left 10.10 - 7.65 = $ 2.45 answer : a" | a = 2 + 20
b = 3 + 25
c = 10 - 10
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a ) 22500 , b ) 15500 , c ) 18500 , d ) 25000 , e ) none of them | c | subtract(37, multiply(multiply(181, 37), 319)) | evaluate : 37 x 181 + 37 x 319 | "37 x 181 + 37 x 319 = 37 x ( 181 + 319 ) = 37 x 500 = 18500 . answer is c ." | a = 181 * 37
b = a * 319
c = 37 - b
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a ) 2 , b ) 4 , c ) 16 , d ) 42 , e ) 40 | d | divide(subtract(power(22, const_2), power(20, const_2)), const_2) | the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 22 - inch screen would be how many square inches greater than the area of a square 20 - inch screen ? | "pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 22 ^ 2 / 2 - 20 ^ 2 / 2 = 84 / 2 = 42 ans d ." | a = 22 ** 2
b = 20 ** 2
c = a - b
d = c / 2
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a ) 48 , b ) 130 , c ) 66 , d ) 68 , e ) 84 | b | subtract(subtract(subtract(600, 1), const_4), const_1) | how many positive integers less than 600 can be formed using the numbers 1 , 2 , 3 , 5 and 6 for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 5 6 6 number of possibilities = 5 * 5 * 5 = 125 . then , just add up the number of 1 digits numbers = 5 , so total is 125 + 5 = 130 . answer : b" | a = 600 - 1
b = a - 4
c = b - 1
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a ) 36 , b ) 24 , c ) 15 , d ) 6 , e ) 5 | c | add(power(divide(subtract(5, sqrt(subtract(power(5, 2), multiply(const_4, 5)))), 2), 2), power(divide(add(5, sqrt(subtract(power(5, 2), multiply(const_4, 5)))), 2), 2)) | if a and b are the roots of the equation x 2 - 5 x + 5 = 0 , then the value of a 2 + b 2 is : | "sol . ( b ) the sum of roots = a + b = 5 product of roots = ab = 5 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 25 - 10 = 15 answer c" | a = 5 ** 2
b = 4 * 5
c = a - b
d = math.sqrt(c)
e = 5 - d
f = e / 2
g = f ** 2
h = 5 ** 2
i = 4 * 5
j = h - i
k = math.sqrt(j)
l = 5 + k
m = l / 2
n = m ** 2
o = g + n
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a ) 2263 minutes , b ) 24 minutes , c ) 32 minutes , d ) 36 minutes , e ) none of these | c | multiply(divide(88, add(multiply(90, divide(5, 6)), 90)), const_60) | two cars a and b are running towards each other from different places 88 km apart . if the ratio of the speeds of the cars a and b is 5 : 6 and the speed of the car b is 90 km per hour then after how long will the two meet each other ? | speed of the car a = 5 ⁄ 6 × 90 = 75 km / hr \ reqd time = 88 / 90 + 75 × 60 = 32 minutes answer c | a = 5 / 6
b = 90 * a
c = b + 90
d = 88 / c
e = d * const_60
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a ) 47.5 , b ) 46.5 , c ) 45.5 , d ) 48.5 , e ) 44.5 | a | subtract(const_60, divide(multiply(divide(const_60, const_2), multiply(subtract(3.25, const_3), const_100)), const_60)) | find the angle between the hour hand and the minute hand of a clock when 3.25 . | angle = 30 ( hours ) - ( 11 / 2 ) ( minutes ) = 30 * 3 - ( 5.5 * 25 ) ( hour = 3 and minutes = 25 ) = 90 - 137.5 = 47.5 answer : a | a = const_60 / 2
b = 3 - 25
c = b * 100
d = a * c
e = d / const_60
f = const_60 - e
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a ) 123 , b ) 213 , c ) 440 , d ) 550 , e ) 540 | c | add(400, 40) | a chocolate manufacturer produces 400 units of a certain model each month at a cost to the manufacturer of £ 40 per unit and all of the produced units are sold each month . what is the minimum selling price per unit that will ensure that the monthly profit ( revenue from sales minus production costs ) on the sales of these units will be at least £ 40000 ? | 400 ( x - 40 ) ≥ 40000 x - 40 ≥ 400 x ≥ 440 answer : option c | a = 400 + 40
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a ) 6 , 030,053 , b ) 6 , 030,054 , c ) 6 , 030,055 , d ) 6 , 042,091 , e ) 6 , 030,057 | d | multiply(2, const_3) | 1,000 ^ 2 + 1,001 ^ 2 + 1,002 ^ 2 + 1,003 ^ 2 + 1,004 ^ 2 + 1,005 ^ 2 + 1,006 ^ 2 = | "interesting problem . i think key is to notice that all the given answer choices differs in last two digits . therefore , our entire focus should be to figure out how the given terms contribute to last two digits of total . 1000 ^ 2 - > 00 1001 ^ 1 - > 01 . . . 1006 ^ 2 - > 36 total - > * 91 answer d ." | a = 2 * 3
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a ) 67000 , b ) 70000 , c ) 76500 , d ) 77200 , e ) 85200 | b | multiply(112, power(add(const_4, const_1), const_4)) | ( 112 x 54 ) = ? | explanation : ( 112 x 54 ) = 112 x ( 10 / 2 ) 4 = 112 x 10 4 / 2 4 = 1120000 / 16 = 70000 answer is b | a = 4 + 1
b = a ** 4
c = 112 * b
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a ) 3312 , b ) 2888 , c ) 4312 , d ) 2881 , e ) 1221 | c | subtract(circle_area(divide(352, multiply(const_2, const_pi))), circle_area(divide(264, multiply(const_2, const_pi)))) | the circumferences of two circles are 264 meters and 352 meters . find the difference between the areas of the larger and the smaller circles . | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 264 and 2 ∏ l = 352 s = 264 / 2 ∏ and l = 352 / 2 ∏ difference between the areas = ∏ l 2 - ∏ s 2 = ∏ { 1762 / ∏ 2 - 1322 / ∏ 2 } = 1762 / ∏ - 1322 / ∏ = ( 176 - 132 ) ( 176 + 132 ) / ∏ = ( 44 ) ( 308 ) / ( 22 / 7 ) = ( 2 ) ( 308 ) ( 7 ) = 4312 sq m answer : c" | a = 2 * math.pi
b = 352 / a
c = circle_area - (
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a ) 25.8 , b ) 19.5 , c ) 18.0 , d ) 12.2 , e ) 29.8 | b | divide(multiply(multiply(const_2, 15), 28), add(15, 28)) | find avrg speed if a man travels at speed of 15 km / hr up and 28 km / hr dawn at an altitude of 230 m . | "avg speed = 2 * x * y / ( x + y ) = 2 * 15 * 28 / ( 15 + 28 ) = 19.5 answer : b" | a = 2 * 15
b = a * 28
c = 15 + 28
d = b / c
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a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 20 % | e | multiply(divide(subtract(21168, 17640), 17640), const_100) | an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 21168 / - in 3 years at the same rate of interest . find the rate percentage ? | "explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 21168 / - - 17640 / - = rs . 3528 / - rate of interest = ( 3528 / 17640 ) × ( 100 / 1 ) = > 20 % answer : option e" | a = 21168 - 17640
b = a / 17640
c = b * 100
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a ) 0 , b ) 15 , c ) 20 , d ) 25 , e ) 35 | e | multiply(add(const_3, const_4), add(const_2, const_3)) | in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 25 , what is the greatest possible value of b x a ? | i also was confused when i was looking forabove number : d as far as i understood , 25 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 25 x 1 ( a = 2 , b = 5 respectively we have = 10 ) . but we do n ' t have this number in answer choices , move on . ( avoid even multiples of 2,4 , 6,8 etc ( we will have 0 in units thus making our result 0 ) 25 x 3 = 75 ( a = 7 b = 5 respectively . hey ! that ' s 35 . this is the greatest possible value of b x a imo e . | a = 3 + 4
b = 2 + 3
c = a * b
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a ) 60 kmph , b ) 76 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph | a | divide(divide(add(100, 300), const_1000), divide(24, const_3600)) | a train 100 meters long completely crosses a 300 meters long bridge in 24 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 24 = 400 / 24 * 18 / 5 = 60 answer : a" | a = 100 + 300
b = a / 1000
c = 24 / 3600
d = b / c
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a ) 0.09 , b ) 0.08 , c ) 0.1 , d ) 0.07 , e ) none of them | a | divide(subtract(power(0.15, 3), power(0.06, 3)), add(add(power(0.15, 2), 0.009), power(0.06, 2))) | ( 0.15 ) ( power 3 ) - ( 0.06 ) ( power 3 ) / ( 0.15 ) ( power 2 ) + 0.009 + ( 0.06 ) ( power 2 ) is : | "given expression = ( 0.15 ) ( power 3 ) - ( 0.06 ) ( power 3 ) / ( 0.15 ) ( power 2 ) + ( 0.15 x 0.06 ) + ( 0.06 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.15 - 0.06 ) = 0.09 answer is a ." | a = 0 ** 15
b = 0 ** 6
c = a - b
d = 0 ** 15
e = d + 0
f = 0 ** 6
g = e + f
h = c / g
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(12, 6) | if the remainder is 8 when positive integer n is divided by 12 , what is the remainder when n is divided by 6 ? | "assume x is quotient here , n = 12 x + 8 - - - - - - - - - - ( 1 ) and n = 6 x + ? we can also write equation ( 1 ) as : n = ( 12 x + 6 ) + 2 . ie 6 ( 2 x + 1 ) + 2 ie the first term is perfectly divisible by 6 . so , the remainder left is 2 . so , answer ( c ) is right choice ." | a = 12 / 6
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a ) 18 days , b ) 36 days , c ) 42 days , d ) 48 days , e ) 44 days | b | divide(multiply(9, 80), 20) | if 9 men do a work in 80 days , in how many days will 20 men do it ? | "9 * 80 = 20 * x x = 36 days answer : b" | a = 9 * 80
b = a / 20
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a ) 300 % , b ) 400 % , c ) 500 % , d ) 600 % , e ) 750 % | c | multiply(const_100, divide(multiply(surface_cube(1), surface_cube(6)), surface_cube(6))) | if a 6 cm cube is cut into 1 cm cubes , then what is the percentage increase in the surface area of the resulting cubes ? | "the area a of the large cube is 6 * 6 * 6 = 216 square cm . the area of the 216 small cubes is 216 * 6 = 6 a , an increase of 500 % . the answer is c ." | a = surface_cube * (
b = a / surface_cube
c = 100 * b
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a ) 17.2 , b ) 18.2 , c ) 19.2 , d ) 15.2 , e ) 16.2 | a | divide(29.94, 1.45) | if 2994 ÷ 14.5 = 172 , then 29.94 ÷ 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 172 in the place of 2994 / 14.5 ] = 172 / 10 = 17.2 answer is a ." | a = 29 / 94
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a ) 5 % , b ) 6 % , c ) 12.5 % , d ) 95 % , e ) 1 % | c | divide(multiply(const_100, 500), multiply(2000, 2)) | what is the rate percent when the simple interest on rs . 2000 amount to rs . 500 in 2 years ? | "interest for 2 yrs = 500 interest for 1 yr = 250 interest rate = 250 / 2000 x 100 = 12.5 % answer : c" | a = 100 * 500
b = 2000 * 2
c = a / b
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a ) 1000 , b ) 1250 , c ) 1300 , d ) 1600 , e ) 1750 | e | divide(add(divide(multiply(400, const_100), 30), 400), divide(80, const_100)) | a small pool filled only with water will require an additional 400 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 400 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "since pumping in additional 400 gallons of water will increase the amount of water in the pool by 30 % , then initially the pool is filled with 1,000 gallons of water . so , we have that 1,000 + 400 = 0.8 * { total } - - > { total } = 1,750 . answer : e ." | a = 400 * 100
b = a / 30
c = b + 400
d = 80 / 100
e = c / d
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a ) 15.6 seconds , b ) 24.6 seconds , c ) 28.6 seconds , d ) 13.6 seconds , e ) 35.6 seconds | d | divide(add(90, 80), multiply(20, const_0_2778)) | a train 90 m long is running at 20 kmph . in how much time will it pass a platform 80 m long ? | distance travelled = 90 + 80 m = 170 m speed = 20 * 5 / 8 = 25 / 2 m time = 170 * 2 / 25 = 13.6 seconds answer : d . | a = 90 + 80
b = 20 * const_0_2778
c = a / b
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 9 , e ) 1 / 3 | d | divide(multiply(1, power(2, const_2)), power(6, const_2)) | x varies inversely as square of y . given that y = 2 for x = 1 . the value of x for y = 6 will be equal to : | "given x = k / y 2 , where k is a constant . now , y = 2 and x = 1 gives k = 4 . x = 4 / y 2 = > x = 4 / 62 , when y = 6 = > x = 4 / 36 = 1 / 9 . answer : d" | a = 2 ** 2
b = 1 * a
c = 6 ** 2
d = b / c
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a ) 1 / 5 , b ) 1 / 4 , c ) 17 / 20 , d ) 3 / 5 , e ) it can not be determined from the given information . | c | subtract(divide(5, 4), divide(2, 4)) | 5 / 4 of all married couples have more than one child . 2 / 5 of all married couples have more than 3 children . what fraction of all married couples have 2 or 3 children ? | "plug in simple numbers . take 100 couples for example . 5 / 4 of 100 couples have more than one child = 125 couples . 2 / 5 of 100 couples have more than 3 kids = 40 couples . this implies that 40 couples are a subset of 125 couples we need to find couples that have 2 or 3 kids , so essentially , it is 125 - 40 = 85 . fraction will be 85 / 100 = 17 / 20 . option c" | a = 5 / 4
b = 2 / 4
c = a - b
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a ) 4.25 , b ) 5.9 , c ) 6.25 , d ) 7 , e ) 7.5 | b | divide(subtract(282, multiply(10, 4.6)), 40) | in the first 10 overs of a cricket game , the run rate was only 4.6 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 4.6 x 10 ) = 236 236 / 40 = 5.9 b )" | a = 10 * 4
b = 282 - a
c = b / 40
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a ) 3 , b ) 9 , c ) 3 , d ) 4 , e ) 5.04 | e | divide(multiply(210, 72), 3000) | calculate the dividend from moses ' s stock , if he invested rs . 3000 at 72 to obtain an income of rs . 210 . | by investing rs . 3000 , income = rs . 210 by investing rs . 72 , income = 210 ã — 72 / 3000 = 5.04 ie , dividend = 5.04 % answer is e . | a = 210 * 72
b = a / 3000
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a ) 315 , b ) 325 , c ) 335 , d ) 345 , e ) 305 | a | multiply(9, divide(multiply(10, 7), subtract(9, 7))) | perimeter of the back wheel = 9 feet , front wheel = 7 feet on a certain distance , the front wheel gets 10 revolutions more than the back wheel . what is the distance ? | for 1 revolution : front wheel goes 7 ft . and back wheel goes 9 feet . let distance is x feet . x / 7 = ( x / 9 ) + 10 or , 2 x / 63 = 10 or , x = ( 10 * 63 ) / 2 therefore , x = 315 answer : a | a = 10 * 7
b = 9 - 7
c = a / b
d = 9 * c
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a ) 56 , b ) 57 , c ) 58 , d ) 59 , e ) 60 | a | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 9,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . a" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | b | divide(110, add(const_3, const_4)) | a certain airline ' s fleet consisted of 110 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ? | let x be the number of years . 4 x > 110 - 3 x 7 x > 110 x > 15 + 5 / 7 the answer is b . | a = 3 + 4
b = 110 / a
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 80 are multiples of 4 but not multiples of 6 ? | "the lcm of 4 and 6 is 12 . if x < 80 and x is divisible by 4 not by 6 - - > x is not divisible by 12 . from 1 - - > 80 , we have 6 numbers which is divisible by 12 : 12 , 24 , 36 , 48 , 60 , 72 . from 1 - - > 80 , we have ( 72 - 4 ) / 4 + 1 = 18 numbers divisible by 4 . therefore , our answer is 18 - 6 = 12 numbers . e" | 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
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a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 350 , e ) rs . 370 | b | divide(multiply(550, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 550 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 550 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 550 ⇒ y [ 12 / 10 + 1 ] = 550 ⇒ 22 y / 10 = 550 ⇒ 22 y = 5500 ⇒ y = 5500 / 22 = 500 / 2 = rs . 250 b )" | a = 1 + 4
b = a * 2
c = 550 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
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a ) 450 , b ) 550 , c ) 650 , d ) 750 , e ) 850 | d | multiply(subtract(62, 32), add(divide(subtract(80, 32), const_2), const_1)) | set a contains all the even numbers between 32 and 80 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 30 more than the corresponding term in set a . the difference of the sums = 25 * 30 = 750 . the answer is d ." | a = 62 - 32
b = 80 - 32
c = b / 2
d = c + 1
e = a * d
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a ) $ 2,250 , b ) $ 2,900 , c ) $ 2,600 , d ) $ 3,250 , e ) $ 2,500 | c | floor(divide(subtract(subtract(multiply(500, 9.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000)) | company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 9.00 each , what was company c ’ s gross profit ? | "cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 9 = $ 4500 profit : 4500 - 1900 = $ 2600 option c is correct" | a = 500 * 9
b = 5 * 0
c = a - b
d = 500 - 100
e = d * 3
f = c - e
g = f / 1000
h = math.floor(g)
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a ) 30000 , b ) 50000 , c ) 40000 , d ) 20000 , e ) 60000 | a | multiply(divide(12000, const_2.0), 5) | p and q invested in a business . the profit earned was divided in the ratio 3 : 5 . if p invested rs 12000 , the amount invested by q is | "let the amount invested by q = q 12000 : q = 3 : 5 ⇒ 12000 × 5 = 3 q ⇒ q = ( 12000 × 5 ) / 3 = 30000 answer is a" | a = 12000 / 2
b = a * 5
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a ) 125 , b ) 280 , c ) 384 , d ) 400 , e ) 576 | e | divide(multiply(multiply(300, 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 300 more than the correct answer . find the number . | explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 300 25 * x / 48 = 300 x = 576 answer e | a = 300 * 16
b = a * 6
c = 5 * 16
d = 5 * 6
e = c - d
f = b / e
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a ) 10 % , b ) 33.33 % , c ) 40 % , d ) 50 % , e ) 66.66 % | b | divide(subtract(30, 25), subtract(divide(40, const_100), divide(25, const_100))) | seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 30 percent ryegrass , what percent of the weight of this mixture is x ? | "qnty . ( cheaper ) / qty . ( dearer ) = ( dearer - mean ) / ( mean - cheaper ) thus qy / qx = ( 40 - 30 ) / ( 30 - 25 ) = 2 therefore qy = 2 qx and qy + qx = 100 % ( total ) . hence qx = 33.33 % answer : b" | a = 30 - 25
b = 40 / 100
c = 25 / 100
d = b - c
e = a / d
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a ) 21 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | c | add(divide(subtract(50, 1), 2), const_1) | how many multiples of 2 are there between 1 and 50 , exclusive ? | "24 multiples of 2 between 1 and 50 exclusive . from 2 * 1 upto 2 * 24 , ( 1,2 , 3,4 , . . . , 24 ) . hence , 24 multiples ! correct option is c" | a = 50 - 1
b = a / 2
c = b + 1
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a ) 2.5 % , b ) 17.67 % , c ) 28.3 % , d ) 45.2 % , e ) 73.6 % | c | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 3))), subtract(1, divide(1, 3))), const_100) | a library branch originally contained 18365 volumes , 30 % of which were fiction novels . 1 / 3 of the volumes were transferred to another location and 1 / 3 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ? | "fiction novels = 5,509 transferred to another location = 6,121 transferred fiction novels = 2,040 non transferred fiction novels = 3,469 percent of the remaining collection was fiction novels = 3469 / ( 18365 - 6121 ) * 100 = > 28.332 . . . % hence answer will be ( c )" | a = 30 / 100
b = 1 / 3
c = 1 - b
d = a * c
e = 1 / 3
f = 1 - e
g = d / f
h = g * 100
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a ) 22,550 , b ) 32,550 , c ) 52,550 , d ) 62,550 , e ) 12,550 | e | divide(add(multiply(add(const_4, const_1), const_100), 20), sqrt(const_100)) | a person spent rs . 5,040 from his salary on food and 5,000 on house rent . after that he was left with 20 % of his monthly salary . what is his monthly salary ? | total money spent on food and house rent = 5,040 + 5,000 = 10,040 which is 100 - 20 = 80 % of his monthly salary ∴ his salary = 10040 x 100 / 80 = 12550 answer : e | a = 4 + 1
b = a * 100
c = b + 20
d = math.sqrt(100)
e = c / d
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a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 4.00 . , e ) 6.25 . | d | divide(divide(5376, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 5376 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 5376 / ( 28 * 24 * 2 ) = 4.00 answer d" | a = 5376 / 28
b = 28 - 4
c = b * 2
d = a / c
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a ) 44 % , b ) 120 % , c ) 96 % , d ) 40 % , e ) none of the above | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(40, const_100)), const_2)), const_100) | if the radius of a circle is increased by 40 % then the area is increased by : | "initially a = pi * r 2 now r 2 = 140 / 100 r then area = pi * ( 140 r / 100 ) 2 area = 196 r / 100 that is area increases from 100 to 196 = increase in area = 96 % answer : c" | a = 100 / 100
b = 100 / 100
c = 40 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
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a ) 15000 , b ) 25000 , c ) 45000 , d ) 32000 , e ) 35000 | e | divide(subtract(15000, 8000), subtract(0.7, 0.5)) | if a cost college rs 0.70 a companyto produce a program for yhe home coming football game . if rs 15000 was received for addvertisement in the pgm , how many copies rs 0.50 a copy must be sold to make a profit of rs 8000 | let no . of pages be x while selling with 0.70 paise / page total cost = 0.70 * x while selling with 0.50 paise / page toatal cost = 0.50 * x now according to question - ( 0.70 x - 0.50 x ) + 8000 = 15000 therefore , x = 35000 copies must be sold answer : e | a = 15000 - 8000
b = 0 - 7
c = a / b
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a ) 1.44 % , b ) 1.74 % , c ) 1.84 % , d ) 1.96 % , e ) 1.24 % | d | divide(multiply(14, 14), const_100) | if a trader sold two cars each at rs . 325475 and gains 14 % on the first and loses 14 % on the second , then his profit or loss percent on the whole is ? | "sp of each car is rs . 325475 , he gains 14 % on first car and losses 14 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 14 ) ( 14 ) / 100 % = 1.96 % answer : d" | a = 14 * 14
b = a / 100
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a ) 3 , b ) 5 , c ) 26 , d ) 12 , e ) 6 | e | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 28)) | if the wheel is 28 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 28 * x = 1056 = > x = 6 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 28
h = 1056 / g
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a ) 482 , b ) 725 , c ) 992 , d ) 837 , e ) 723 | e | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 30000), multiply(power(add(divide(20, const_100), const_1), const_2), 30000)) | how much more would rs . 30000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | 30000 ( 11 / 10 ) 4 - 30000 ( 6 / 5 ) 2 = 723 answer : e | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 4
e = d * 30000
f = 20 / 100
g = f + 1
h = g ** 2
i = h * 30000
j = e - i
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a ) 2 % , b ) 6 % , c ) 4 % , d ) 28 % , e ) 63 % | c | floor(multiply(subtract(divide(9, 63), divide(7, 70)), const_100)) | a survey was sent to 70 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | case 1 : ( 7 / 70 ) = x / 100 x = 10 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 10 ) % = 4 % answer is c | a = 9 / 63
b = 7 / 70
c = a - b
d = c * 100
e = math.floor(d)
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a ) 24 , b ) 22 , c ) 27 , d ) 25 , e ) 28 | c | add(add(add(3, 3), 3), 3) | what number should replace the question mark ? 3 , 30 , 11 , 24 , 19 , 18 , - - - ? | "answer : c 3 , 30 , 11 , 24 , 19 , 18 , 27 ? there are two alternate sequences : + 8 and - 6 ." | a = 3 + 3
b = a + 3
c = b + 3
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(25, 20) | a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 104 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 105 / 25 = 4 . . . . so the max is 4 105 / 20 = 5 . . . so the max is 5 105 - 100 = 5 answer : e | a = 25 - 20
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a ) 18 sec , b ) 15 sec , c ) 21 sec , d ) 16 sec , e ) 20 sec | a | divide(add(200, 250), multiply(90, const_0_2778)) | how many seconds will a train 250 meters long take to cross a platform 200 meters long if the speed of the train is 90 kmph ? | "d = 250 + 200 = 450 s = 90 * 5 / 18 = 25 mps t = 450 / 25 = 18 sec a ) 18 sec" | a = 200 + 250
b = 90 * const_0_2778
c = a / b
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a ) 5 , b ) 5 / 2 , c ) - 1 / 6 , d ) - 1 / 5 , e ) - 5 | b | divide(add(7, 3), subtract(7, 3)) | if x / y = 7 / 3 , then ( x + y ) / ( x - y ) = ? | "any x and y satisfying x / y = 7 / 3 should give the same value for ( x + y ) / ( x - y ) . say x = 7 and y = 3 , then ( x + y ) / ( x - y ) = ( 7 + 3 ) / ( 7 - 3 ) = 5 / 2 . answer : b ." | a = 7 + 3
b = 7 - 3
c = a / b
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a ) 250 , b ) 600 , c ) 675 , d ) 700 , e ) 750 | b | multiply(add(divide(100, 10), divide(150, 15)), 30) | working at their respective constant rates , machine a makes 100 copies in 10 minutes and machine b makes 150 copies in 15 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ? | "machine a can produce 100 * 30 / 10 = 300 copies and , machine b can produce 150 * 30 / 15 = 300 copies total producing 600 copies . b is the answer" | a = 100 / 10
b = 150 / 15
c = a + b
d = c * 30
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a ) rs . 1360 , b ) rs . 1000 , c ) rs . 360 , d ) rs . 640 , e ) none of these | a | add(360, divide(multiply(360, const_100), multiply(12, 3))) | the banker ' s gain on a sum due 3 years hence at 12 % per annum is rs . 360 . the banker ' s discount is : | "explanation : bg = rs . 360 t = 3 years r = 12 % td = ( bg × 100 ) / tr = ( 360 × 100 ) / ( 3 × 12 ) = rs . 1000 bg = bd – td = > bd = bg + td = 360 + 1000 = rs . 1360 answer : option a" | a = 360 * 100
b = 12 * 3
c = a / b
d = 360 + c
|
a ) 190 , b ) 120 , c ) 150 , d ) 199 , e ) 110 | a | multiply(add(19, const_1), divide(19, const_2)) | calculate the sum of first 19 natural numbers . | solution we know that ( 1 + 2 + 3 + . . . . . + 19 ) = n ( n + 1 ) / 2 therefore ( 1 + 2 + 3 + . . . . + 19 ) = ( 19 × 20 / 2 ) = 190 . answer a | a = 19 + 1
b = 19 / 2
c = a * b
|
a ) 15 days , b ) 88 days , c ) 21 days , d ) 11 days , e ) 13 days | a | multiply(divide(multiply(5, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 5 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 5 x = 1 / 15 = > 15 days answer : a" | a = 2 + 1
b = 5 * a
c = b / 2
d = c * 2
|
a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 2500 , e ) $ 12000 | c | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(5, const_100)), const_4)), multiply(const_100, const_100)) | carl is facing very difficult financial times and can only pay the interest on a $ 10,000 loan he has taken . the bank charges him a quarterly compound rate of 5 % . what is the approximate interest he pays annually ? | since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 5 % quarterly rate . so the bank charges 5 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 5 % of 10,000 = $ 2000 as our answer . but since , the interest is compounded , it will be a bit more than $ 2000 . option ( c ) looks correct . answer : c | a = 100 * 100
b = 5 / 100
c = 1 + b
d = c ** 4
e = a * d
f = 100 * 100
g = e - f
|
a ) 3.5 % , b ) 2.4 % , c ) 3 % , d ) 5 % , e ) 0.7 % | e | subtract(subtract(6, 5), divide(multiply(6, 5), const_100)) | in measuring the sides of a rectangle , one side is taken 6 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements . | "say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . now , the area obtained with wrong measurements would be 106 * 95 = 10,070 , which is 0.7 % greater than the actual area . answer : e ." | a = 6 - 5
b = 6 * 5
c = b / 100
d = a - c
|
a ) 20 , b ) 23 , c ) 27 , d ) 150 , e ) 255 | b | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 and the sum of their squares is 289 . the sum of the number is | "explanation : we know ( x + y ) 2 = x 2 + y 2 + 2 xy = > ( x + y ) 2 = 289 + 2 ( 120 ) = > ( x + y ) = 529 − − − √ = 23 option b" | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
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a ) 2 , b ) 9 , c ) 6 , d ) 21 , e ) 30 | c | divide(subtract(0.6, multiply(0.06, 8)), subtract(0.08, 0.06)) | a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 8 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company charges the same amount under plan a and under plan b ? | let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 8 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 8 ) = 0.08 t - - > 60 + 6 ( t - 8 ) = 8 t - - > t = 6 . answer : c . | a = 0 * 6
b = 0 - 6
c = 0 - 8
d = b / c
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a ) a ) 145 , b ) b ) 148 , c ) c ) 165 , d ) d ) 153 , e ) e ) 158 | c | add(multiply(18, 9), 3) | what is the dividend . divisor 18 , the quotient is 9 and the remainder is 3 | "d = d * q + r d = 18 * 9 + 3 d = 162 + 3 d = 165 answer : c" | a = 18 * 9
b = a + 3
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a ) 1.8 m , b ) 6 m , c ) 7 m , d ) 9 m , e ) 12 m | b | divide(add(7, 5), const_2) | the r train leaves station a moving at a constant speed , and passes by stations b and c , in this order . it takes the r train 7 hours to reach station b , and 5 additional hours to reach station c . the distance between stations a and b is m kilometers longer than the distance between stations b and c . what is the distance between stations a and c in terms of m ? | the reason it is failing for you is that you chose incorrect numbers . if the question says r it took 7 hrs to reach from a to b and 5 hrs to reach from b to c at a constant speed . it shows that distance ab and bc should be in ratio of 7 / 5 . if you take such numbers you can solve problem . ab = 7 , bc = 5 therefore ab - bc = 2 but from question , ab - bc = m = > m = 2 now total distance = ab + bc = 12 substitute 12 to get answer in terms of m total distance = 12 = 6 m ans b | a = 7 + 5
b = a / 2
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a ) 72 % , b ) 70 % , c ) 52 % , d ) 50 % , e ) 45 % | e | subtract(50, multiply(divide(50, const_100), 10)) | a shirt goes on sale for 50 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 50 then it is marked down by 10 % = 50 - 5 = 45 . hence it is 45 % od the original price . hence answer is e ." | a = 50 / 100
b = a * 10
c = 50 - b
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a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | a | multiply(2, 8) | each child has 2 pencils and 13 skittles . if there are 8 children , how many pencils are there in total ? | 2 * 8 = 16 . answer is a . | a = 2 * 8
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a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec | c | divide(56, multiply(39, const_0_2778)) | in what time will a railway train 56 m long moving at the rate of 39 kmph pass a telegraph post on its way ? | "t = 56 / 39 * 18 / 5 = 5 sec answer : c" | a = 39 * const_0_2778
b = 56 / a
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a ) 31.5 , b ) 24.5 , c ) 24.3 , d ) 24.9 , e ) 24.1 | a | multiply(divide(9, subtract(9, 7)), 7) | sachin is younger than rahul by 9 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 9 , so ( x - 9 ) / x = 7 / 9 = > 9 x - 81 = 7 x = > 2 x = 81 = > x = 40.5 so sachin age is 40.5 - 9 = 31.5 answer : a" | a = 9 - 7
b = 9 / a
c = b * 7
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a ) 35 , b ) 45 , c ) 25 , d ) 63 , e ) 55 | d | multiply(245, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 245 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 245 × 9 / 35 = 7 × 9 = 63 answer is d" | a = 15 * 3
b = 10 * 7
c = 12 * 5
d = b + c
e = 15 * 3
f = d + e
g = a / f
h = 245 * g
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