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 ) 60 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 70 mph , e ) 66.67 mph | d | add(divide(add(multiply(90, 3), multiply(40, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 90 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles . distance covered in the next 3 hours = speed * time = 90 * 3 = 270 miles . therefore , total distance covered = 80 + 270 = 350 miles . total time taken = 2 + 3 = 5 hours . hence , average speed = total distance travelled / total time taken = 350 / 5 = 70 miles per hour . choice d" | a = 90 * 3
b = 40 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
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a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | b | divide(power(2, 2), multiply(choose(4, 2), choose(2, 1))) | if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be odd ? | "probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 1 / 3 . answer : b ." | a = 2 ** 2
b = math.comb(4, 2)
c = math.comb(2, 1)
d = b * c
e = a / d
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a ) 83 , b ) 56 , c ) 34 , d ) 12 , e ) 26 | a | add(multiply(divide(multiply(divide(subtract(const_100, 10), const_100), 80), add(subtract(const_100, 80), multiply(divide(subtract(const_100, 10), const_100), 80))), const_100), const_4) | in march mrs bil ' s earnings were 80 percent of the bil family ' s total income . in april mrs bil earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs bil ' s earnings were approximately what percent of the bil family ' s total income ? | lets say the family income is 100 in may , lee earned 80 family income is 20 in june , lee earned 20 % more than may , so it is ( 80 + 20 * 80 / 100 = 96 ) family income is same 20 in june lee ' s income percent is 96 * 100 / 116 ~ 83 ans is a | a = 100 - 10
b = a / 100
c = b * 80
d = 100 - 80
e = 100 - 10
f = e / 100
g = f * 80
h = d + g
i = c / h
j = i * 100
k = j + 4
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | floor(const_2) | how many prime numbers are between 29 / 4 and 87 / 5 ? | 29 / 4 = 7 . xxx 87 / 5 = 17 . xxx so we need to find prime numbers between 7 ( exclusive ) - 16 ( inclusive ) there are 2 prime numbers 1113 hence answer will be ( b ) 2 b | a = math.floor(2)
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a ) 1750 , b ) 1000 , c ) 2866 , d ) 2787 , e ) 1121 | a | multiply(multiply(25, 10), 5) | in digging a pond 25 m * 10 m * 5 m the volumes of the soil extracted will be ? | "25 * 10 * 5 = 1750 answer : a" | a = 25 * 10
b = a * 5
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a ) 16 kmph , b ) 19 kmph , c ) 14 kmph , d ) 11 kmph , e ) 13 kmph | a | subtract(48, divide(multiply(48, 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 48 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 = 48 / 3 = 16 kmph . answer : a" | a = 48 * 2
b = a / 3
c = 48 - b
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a ) 51 , b ) 63 , c ) 79 , d ) 66 , e ) 42 | c | add(divide(subtract(500, 100), 5), const_1) | what is the total number of integers between 100 and 500 ( exclusive ) that are divisible by 5 ? | 105 , 110 , 115 , . . . , 490,495 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 495 - 105 ) / ( 5 ) + 1 = 78 + 1 = 79 answer is c | a = 500 - 100
b = a / 5
c = b + 1
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a ) - 10 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | d | multiply(divide(subtract(12, 10), 10), const_100) | a shopkeeper sold 10 articles at the cost price of 12 articles . then find the profit % or lost % | here 10 articles selling price = 12 articles cost price so the difference = 12 - 10 = 2 % of profit = 2 * 100 / 10 = 20 % correct option is d | a = 12 - 10
b = a / 10
c = b * 100
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a ) 64,420 , b ) 78,840 , c ) 83,160 , d ) 88,970 , e ) 94,320 | b | multiply(21.9, const_3600) | venus travels around the sun at a speed of approximately 21.9 miles per second . the approximate speed is how many miles per hour ? | 21.9 * 3600 = 78,840 the answer is b . | a = 21 * 9
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | 150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration of the acid in the mixture ? | 150 ml solution contains 45 ml sulphuric acid 400 ml solution contains 48 ml sulphuric acid adding 550 ml solution contains 93 ml sulphuric acid 93 / 550 ≈ 90 / 540 = 1 / 6 answer is d | a = 150 * 30
b = a / 100
c = 400 * 12
d = c / 100
e = b + d
f = 150 + 400
g = e / f
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a ) 22 , b ) 15 , c ) 77 , d ) 9 , e ) 182 | d | divide(multiply(subtract(18, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 18 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 18 = > x = 3 so deepak age is = 3 ( 3 ) = 9 answer : d" | a = 18 - 6
b = a * 3
c = b / 4
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a ) 3.33 days , b ) 8.33 days , c ) 3.99 days , d ) 7.38 days , e ) 2.88 days | d | inverse(divide(add(inverse(16), add(inverse(8), inverse(12))), const_2)) | a and b can do a piece of work in 8 days . b and c can do it in 12 days and a and c in 16 days . working together they will complete the work in how many days ? | a + b = 1 / 8 b + c = 1 / 12 c + a = 1 / 16 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ( a + b + c ) = 1 / 8 + 1 / 12 + 1 / 16 = 13 / 48 a + b + c = 13 / 96 96 / 13 = 7.38 days answer : d | a = 1/(16)
b = 1/(8)
c = 1/(12)
d = b + c
e = a + d
f = e / 2
g = 1/(f)
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a ) 8 ° , b ) 90 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | b | divide(multiply(subtract(const_100, add(add(add(add(14, 19), 10), 24), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 19 % home electronics ; 10 % food additives ; 24 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "14 % microphotonics ; 19 % home electronics ; 10 % food additives ; 24 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 14 + 19 + 10 + 24 + 8 ) = 25 % basic astrophysics . 25 % of 360 ° is 90 ° . answer : b ." | a = 14 + 19
b = a + 10
c = b + 24
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
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a ) $ 12436 , b ) $ 13344 , c ) $ 71234 , d ) $ 15736 , e ) $ 19403 | e | subtract(40000, multiply(multiply(multiply(40000, subtract(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100)))) | a will was written such that an estate would be divided among 4 beneficiaries with the remainder donated to charity . one stipulation in the will is that no one beneficiary can receive an amount within 20 % of another ' s amount . if one of the beneficiaries received $ 40000 , what is the smallest possible range between the highest and lowest amounts ( rounded to the nearest dollar ) among the 4 beneficiaries ? | 1 st - 40,000 . 2 nd - 0.8 * 40,000 = 32,000 3 rd - 0.8 * 32,000 = 25,600 4 th - 0.8 * 25,600 = ~ 20,480 range = 40,000 - 20,480 = 19520 answer : e . | a = 20 / 100
b = 1 - a
c = 40000 * b
d = 20 / 100
e = 1 - d
f = c * e
g = 20 / 100
h = 1 - g
i = f * h
j = 40000 - i
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a ) 1 / 4 , b ) 7 / 5 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3 | b | divide(add(2, 5), 5) | a box contains 2 red balls and 5 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 5 black balls be bbbbb and 2 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 5 ! x 2 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 6 slots and 4 black ( bbbb ) and 2 red ( rrr ) they can be arranged in ( 7 ! ) / ( 4 ! x 2 ! ) hence required probability = [ ( 7 ! ) / ( 5 ! x 2 ! ) ] / [ ( 6 ! ) / ( 4 ! x 2 ! ) ] = 7 / 5 b" | a = 2 + 5
b = a / 5
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a ) 24.9 , b ) 24.6 , c ) 24.2 , d ) 24.8 , e ) 24.5 | e | multiply(divide(4, subtract(9, 7)), 7) | sachin is younger than rahul by 4 years . if their ages are in the respective ratio of 7 : 9 , how old is sachin ? | "let rahul ' s age be x years . then , sachin ' s age = ( x - 7 ) years . ( x - 7 ) / x = 7 / 9 2 x = 63 = > x = 31.5 hence , sachin ' s age = ( x - 7 ) = 24.5 years . answer : e" | a = 9 - 7
b = 4 / a
c = b * 7
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a ) − 5 , b ) − 9 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(6, const_10), divide(6, divide(6, const_10))) | if a ( a + 5 ) = 6 and b ( b + 5 ) = 6 , where a ≠ b , then a + b = | "i . e . if a = 1 then b = - 6 or if a = - 6 then b = 1 but in each case a + b = - 6 + 1 = - 5 answer : option a" | a = 6 / 10
b = 6 / 10
c = 6 / b
d = a + c
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a ) 28 , b ) 30 , c ) 40 , d ) 53 , e ) 54 | b | add(multiply(divide(36, multiply(2, 3)), 3), multiply(divide(36, multiply(2, 3)), 2)) | the l . c . m . of two numbers is 36 . the numbers are in the ratio 2 : 3 . then sum of the number is : | "let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 36 or x = 6 . the numbers are 12 and 18 . hence , required sum = ( 12 + 18 ) = 30 . answer : option b" | a = 2 * 3
b = 36 / a
c = b * 3
d = 2 * 3
e = 36 / d
f = e * 2
g = c + f
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a ) 5 , b ) 7 , c ) 6 , d ) 8 , e ) 9 | c | divide(multiply(6, 3), subtract(6, 3)) | a man can do a piece of work in 6 days , but with the help of his son , he can do it in 3 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 3 ) - ( 1 / 6 ) = 1 / 6 the son alone can do the work in 6 days answer is c" | a = 6 * 3
b = 6 - 3
c = a / b
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a ) 160 , b ) 240 , c ) 385 , d ) 440 , e ) 560 | c | add(70, multiply(divide(multiply(70, 3), const_2), 3)) | a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 70 square feet of the wall , and no tiles overlap , what is the area of the entire wall ? | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 70 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 70 = 315 . total area = 70 + 315 = 385 answer : c ." | a = 70 * 3
b = a / 2
c = b * 3
d = 70 + c
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | add(divide(subtract(57, multiply(3, const_3)), add(add(1, 2), 3)), 3) | the sum of the present ages of a , b , c is 57 years . 3 years ago their ages were in the ratio 1 : 2 : 3 . what is the present age of a ? | three years ago : a : b : c = 1 : 2 : 3 let a = 1 x , b = 2 x , and c = 3 x . today : ( x + 3 ) + ( 2 x + 3 ) + ( 3 x + 3 ) = 57 x = 8 so the present age of a is x + 3 = 11 the answer is d . | a = 3 * 3
b = 57 - a
c = 1 + 2
d = c + 3
e = b / d
f = e + 3
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a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 264 | e | subtract(multiply(add(multiply(12, add(const_3, const_2)), const_2), 12), multiply(30, const_12)) | a parking garage rents parking spaces for $ 12 per week or $ 30 per month . how much does a person save in a year by renting by the month rather than by the week ? | "12 $ per week ! an year has 52 weeks . annual charges per year = 52 * 12 = 624 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 30 = 360 $ 624 - 360 = 264 ans e" | a = 3 + 2
b = 12 * a
c = b + 2
d = c * 12
e = 30 * 12
f = d - e
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a ) 706 , b ) 786 , c ) 906 , d ) 896 , e ) 696 | c | multiply(subtract(753, divide(subtract(753, 455), const_2)), add(const_1, divide(50, const_100))) | the profit earned by selling an article for rs . 753 is equal to the loss incurred when the same article is sold for rs . 455 . what should be the sale price for making 50 % profit ? | "let c . p = rs . x then as given , ( 753 - x ) = ( x - 455 ) 2 x = 1208 x = 604 there fore s . p = 150 % of 604 = > rs . = rs . 906 answer : c" | a = 753 - 455
b = a / 2
c = 753 - b
d = 50 / 100
e = 1 + d
f = c * e
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['a ) 1409', 'b ) 1402', 'c ) 1221', 'd ) 1406', 'e ) 8828'] | d | divide(add(sqrt(subtract(const_1, multiply(multiply(add(add(const_1000, multiply(const_4, const_100)), multiply(const_2, const_3)), subtract(const_1, const_2)), const_4))), subtract(const_1, const_2)), const_2) | the sum of a number and its square is 1406 . what is the number ? | explanation : by option verification 37 + 37 × 37 = 1406 answer : d | a = 4 * 100
b = 1000 + a
c = 2 * 3
d = b + c
e = 1 - 2
f = d * e
g = f * 4
h = 1 - g
i = math.sqrt(h)
j = 1 - 2
k = i + j
l = k / 2
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(const_3.0, 1) | 1 , 1 , 2 , 1 , 2 , 3 , 1 , 2 , 3 , 4 , 1 , 2 , __ ? | "1,1 , 2,1 , 2,3 , 1,2 , 3,4 , 1,2 , __ ? looking at the series 1 12 123 1234 12 ( 3 ) 45 so the answer is 3 . answer : c" | a = 3 * 0
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a ) 1240 , b ) 1120 , c ) 1190 , d ) 1806 , e ) none of these | d | add(756, divide(multiply(756, const_100), multiply(12, 6))) | the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 756 . what is the banker ' s discount ? | "explanation : td = ( bg × 100 ) / tr = ( 756 × 100 ) / ( 6 × 12 ) = rs . 1050 bg = bd – td = > 756 = bd - 1050 = > bd = 1050 + 756 = 1806 answer : option d" | a = 756 * 100
b = 12 * 6
c = a / b
d = 756 + c
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a ) a ) 140 , b ) b ) 420 , c ) c ) 180 , d ) d ) 200 , e ) e ) 220 | b | divide(multiply(divide(35, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 35 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 35 then x = 35 * 30 = 1050 40 % of 1050 = 420 answer : option b" | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 35 / f
h = g * 40
i = h / 100
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a ) 12.5 m , b ) 10.5 m , c ) 14 , d ) 12 , e ) 10 | a | multiply(28.75, divide(17.5, 40.25)) | a flagstaff 17.5 m high casts a shadow of length 40.25 m . what will be the height of a building , which casts a shadow of length 28.75 m under similar conditions ? | explanation : let the required height of the building be x meter more shadow length , more height ( direct proportion ) hence we can write as ( shadow length ) 40.25 : 28.75 : : 17.5 : x ⇒ 40.25 × x = 28.75 × 17.5 ⇒ x = 28.75 × 17.5 / 40.25 = 2875 × 175 / 40250 = 2875 × 7 / 1610 = 2875 / 230 = 575 / 46 = 12.5 answer : option a | a = 17 / 5
b = 28 * 75
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | divide(add(sqrt(add(multiply(multiply(153, const_2), const_4), const_1)), const_1), const_2) | in a party every person shakes hands with every other person . if there were a total of 153 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 153 n ( n - 1 ) / 2 = 153 â ˆ ´ n = 18 answer : option d" | a = 153 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
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a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | a | divide(subtract(add(5, 1), 5), 2) | in the xy - coordinate system , if ( m , n ) and ( m + 1 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 1 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 4 - - - > 5 = 2 k + 4 - - - > k = 1 / 2 . the answer is , therefore , ( a ) ." | a = 5 + 1
b = a - 5
c = b / 2
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a ) 1125 , b ) 1225 , c ) 1325 , d ) 1145 , e ) 2125 | a | subtract(divide(multiply(multiply(25000, 11.5), 3), const_100), divide(multiply(multiply(25000, 10), 3), const_100)) | if a lends rs . 25000 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 25000 * 1.5 * 3 ) / 100 = > 1125 answer : a" | a = 25000 * 11
b = a * 3
c = b / 100
d = 25000 * 10
e = d * 3
f = e / 100
g = c - f
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a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 55 % | b | multiply(divide(50, 125), const_100) | 50 is what percent of 125 ? | 50 / 125 × 100 = 50 % answer : b | a = 50 / 125
b = a * 100
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a ) 1 / 5 , b ) 5 / 15 , c ) 6 / 15 , d ) 3 / 10 , e ) 4 / 15 | c | add(divide(multiply(subtract(const_1, divide(40, multiply(10, 10))), 150), 450), divide(add(multiply(divide(10, multiply(10, 10)), 300), multiply(divide(40, multiply(10, 10)), 150)), 450)) | in a certain corporation , there are 300 male employees and 150 female employees . it is known that 10 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.1 * 300 / 450 = 30 / 450 = 1 / 15 the sum of the probabilities is 6 / 15 = 2 / 5 the answer is c . | a = 10 * 10
b = 40 / a
c = 1 - b
d = c * 150
e = d / 450
f = 10 * 10
g = 10 / f
h = g * 300
i = 10 * 10
j = 40 / i
k = j * 150
l = h + k
m = l / 450
n = e + m
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a ) 6 , b ) 12 , c ) 14 , d ) 18 , e ) 20 | c | divide(112, gcd(72, 112)) | the least number by which 72 must be multiplied in order to produce a multiple of 112 , is | "solution required numbers is divisible by 72 as well as by 112 , if it is divisible by their lcm , which is 1008 . now , 1008 when divided by 72 , gives quotient = 14 . required number = 14 . answer c" | a = math.gcd(72, 112)
b = 112 / a
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a ) 140 , b ) 143 , c ) 144 , d ) 145 , e ) 146 | d | multiply(divide(58, const_2), 5) | find the number , difference between number and its 3 / 5 is 58 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 58 , = > ( 2 / 5 ) x = 58 = > 2 x = 58 * 5 , = > x = 145 answer : option d" | a = 58 / 2
b = a * 5
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a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | c | subtract(multiply(multiply(multiply(891, 781), 912), 463), subtract(multiply(multiply(multiply(891, 781), 912), 463), add(const_4, const_4))) | the unit digit in the product ( 891 * 781 * 912 * 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 1 * 1 * 2 * 3 ) = 6 answer : c" | a = 891 * 781
b = a * 912
c = b * 463
d = 891 * 781
e = d * 912
f = e * 463
g = 4 + 4
h = f - g
i = c - h
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a ) 25 , b ) 13 , c ) 17 , d ) 24 , e ) 19 | d | multiply(add(divide(2, 3), 3), 8) | find the value of 3 + 2 . ( 8 - 3 ) | "3 + 2 . ( 8 - 3 ) = 3 + 2 ( 5 ) , = 3 + 2 * 5 , = 3 + 10 , = 13 , correct answer : ( d )" | a = 2 / 3
b = a + 3
c = b * 8
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a ) 2 : 3 , b ) 1 : 1 , c ) 3 : 4 , d ) 4 : 3 , e ) 5 : 5 | d | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(4, const_100)), subtract(divide(5, const_100), divide(multiply(divide(6, const_100), const_2), const_3))) | two numbers a and b are such that the sum of 5 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "5 / 100 a + 4 / 100 b = 2 / 3 ( 6 / 100 a + 8 / 100 b ) = > 1 / 20 a + 1 / 25 b = 1 / 25 a + 4 / 75 b = > ( 1 / 20 - 1 / 25 ) b = 1 / 25 a + 4 / 75 b ( 1 / 20 - 1 / 25 ) a = ( 4 / 75 - 1 / 25 ) b 1 / 100 a = 1 / 75 b a / b = 100 / 75 = 4 / 3 req ratio = 4 : 3 answer d" | a = 8 / 100
b = a * 4
c = b / 3
d = 4 / 100
e = c - d
f = 5 / 100
g = 6 / 100
h = g * 2
i = h / 3
j = f - i
k = e / j
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a ) 300 , b ) 360 , c ) 400 , d ) 500 , e ) 575 | b | multiply(subtract(divide(200, subtract(1, divide(const_3, 5))), 200), add(1, divide(1, 5))) | when jessica withdrew $ 200 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 5 of the remaining balance , what will be the final balance in her bank account ? | "as per the question 200 = 2 a / 5 thus - a which is the total amount = 500 the amount thus left = 300 she then deposited 1 / 5 of 300 = 60 total amount in her account = 360 answer b" | a = 3 / 5
b = 1 - a
c = 200 / b
d = c - 200
e = 1 / 5
f = 1 + e
g = d * f
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a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | c | floor(divide(multiply(32, 9), 7)) | in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 32 . in the company , what is the maximum number of non - managers in a department that has 9 managers ? | "9 / 7 * 32 = 41.1 the answer is c ." | a = 32 * 9
b = a / 7
c = math.floor(b)
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | subtract(subtract(22, 5), 5) | a man ' s speed with the current is 22 km / hr and the speed of the current is 5 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 22 - 5 ) km / hr = 17 km / hr . man ' s rate against the current = ( 17 - 5 ) km / hr = 12 km / hr . answer : d" | a = 22 - 5
b = a - 5
|
a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(divide(40, 20), multiply(2, 3)) | machine z takes 2 more hours than machine b to make 20 widgets . if working together , the machines can make 25 widgets in 3 hours , how long will it take machine z to make 40 widgets ? | i approached this one by plugging in numbers . . . started with c . if 40 are made in 8 hours , then 20 are made in 4 hours . so time of z is 4 , and time of b is 2 . rate together : 20 / 4 + 20 / 2 = 5 + 10 = 15 . so in 1 hour , together make 15 widgets . in 3 hours = 45 . way too much . we can eliminate right away c , b , and a - because b and a reduces the time - the total # of widgets made will be even higher . now between d and e - > try only one . . if it does n ' t work , then the other one is the answer . i picked e : 12 h to make 40 widgets , and 6 hours to make 20 . this is the time of z . time of b = 4 hours . 20 / 6 + 20 / 4 = 10 / 3 + 20 / 4 find lcm of 3 and 4 = 12 . multiply first by 4 , and second by 3 : 40 + 60 / 12 = 100 / 12 divide by 4 : 25 / 3 so this is the rate given . e is the correct answer | a = 40 / 20
b = 2 * 3
c = a * b
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a ) 17.69 , b ) 14.85 , c ) 21.3 , d ) 16.5 , e ) 18.51 | e | multiply(const_100, divide(subtract(32.00, multiply(13.50, 2)), multiply(13.50, 2))) | a certain pair of used shoes can be repaired for $ 13.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 32.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? | "1 ) cost of repairing = 13.5 ( for one year ) , therefore for 2 years it would be $ 27 . 2 ) cost of new pair which will last for 2 years is $ 32 . percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoe initial value would be the cost of repairing the old shoe . i . e ( 32 - 27 ) / ( 27 ) * 100 = 18.51 % . ans is e" | a = 13 * 50
b = 32 - 0
c = 13 * 50
d = b / c
e = 100 * d
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a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | e | add(subtract(31, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 31 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 31 kg joins the group , total weight of boys = 20 x + 31 so 20 x + 31 = 21 ( x + 1 ) = > x = 10 . answer : e" | a = 20 + 1
b = 31 - a
c = b + 1
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a ) - 14 , b ) 14 , c ) - 7 , d ) 7 , e ) 0 | d | divide(multiply(2, add(const_4, const_3)), 2) | if x = 55 + 2 q and y = 4 q + 41 , then for what value of q , x is equal to y ? | explanation : x = y < = > 55 + 2 q = 4 q + 41 < = > 2 q = 14 < = > q = 7 . answer : d | a = 4 + 3
b = 2 * a
c = b / 2
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a ) 1.7 , b ) 1.9 , c ) 1.1 , d ) 1.5 , e ) 0.75 | e | divide(4, const_3) | a man can swim in still water at 4 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 4 s = x ds = 4 + x us = 4 + x 4 + x = ( 4 - x ) 2 4 + x = 8 - 2 x 3 x = 4 x = 0.75 answer : e" | a = 4 / 3
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a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 90 | e | divide(multiply(10, subtract(10, const_1)), const_2) | there are 10 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 9 teams . . . so total no of matches = 10 x 9 = 90 . now , each match is played twice = > 90 x 2 but 2 teams play a match = > 90 x 2 / 2 = 90 . answer : e" | a = 10 - 1
b = 10 * a
c = b / 2
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a ) 800 , b ) 600 , c ) 500 , d ) 1000 , e ) 300 | c | divide(divide(multiply(100, const_100), 5), 4) | a sum fetched a total simple interest of rs . 100 at the rate of 5 p . c . p . a . in 4 years . what is the sum ? | "sol . principal = rs . [ 100 * 100 / 5 * 4 ] = rs . [ 10000 / 20 ] = rs . 500 . answer c" | a = 100 * 100
b = a / 5
c = b / 4
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a ) $ 115,000 , b ) $ 160,000 , c ) $ 200,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(18, 10), 10) | an auction house charges a commission of 18 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.18 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 200,000 ( 18 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c ." | a = 18 * 10
b = a + 10
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a ) 200 km , b ) 212 km , c ) 224 km , d ) 230 km , e ) 256 km | c | add(multiply(21, divide(10, 2)), multiply(24, divide(10, 2))) | if one person completes a journey in 10 hrs . he travels 1 st half of the journey at the rate of 21 km / hr and 2 nd half at therate of 24 km / hr . find the total journey in km . | distance = speed x time let time taken to travel the first half = x hr then time taken to travel the second half = ( 10 - x ) hr distance covered in the the first half = 21 x distance covered in the the second half = 24 ( 10 - x ) but distance covered in the the first half = distance covered in the the second half = > 21 x = 24 ( 10 - x ) = > 21 x = 240 - 24 x = > 45 x = 240 = > 9 x = 48 = > 3 x = 16 ⇒ x = 163 hence distance covered in the the first half = 21 x = 21 × 163 = 7 × 16 = 112 kmtotal distance = 2 × 112 = 224 km c | a = 10 / 2
b = 21 * a
c = 10 / 2
d = 24 * c
e = b + d
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a ) 0 , b ) 1 / 4 , c ) 2 / 5 , d ) 1 / 2 , e ) 3 / 5 | b | divide(2, multiply(4, 2)) | m = { - 6 , - 5 , - 4 , - 3 , - 2 , - 1 } t = { - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above , what is the probability that the product of the two integers will be negative ? | we will have a negative product only if 1 or 2 are selected from set t . p ( negative product ) = 2 / 8 = 1 / 4 the answer is b . | a = 4 * 2
b = 2 / a
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a ) 525 , b ) 528 , c ) 522 , d ) 529 , e ) 521 | b | multiply(circumface(6), 21) | the radius of a cylinder is 6 m , height 21 m . the total surface area of the cylinder is ? | "total surface area of the cylinder is = 2 π r ( r + h ) = 2 × 22 / 7 × 6 ( 6 + 8 ) = 2 × 22 / 7 × 6 ( 14 ) = 44 × 12 = 528 m ( power 2 ) answer is b ." | a = circumface * (
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a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | d | add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5) | an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 2,160 , what was the total number of pairs of contact lenses that she sold ? | "( x + 5 ) * 150 + x * 85 = 2160 = > x = 6 total lens = 6 + ( 6 + 5 ) = 17 answer d" | a = 4 * 100
b = 5 * 10
c = a + b
d = c + 5
e = d + 1000
f = 150 * 5
g = e - f
h = 150 + 85
i = g / h
j = i * 2
k = j + 5
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a ) 1 , b ) 2 , c ) 0 , d ) 6 , e ) 7 | b | floor(divide(divide(subtract(multiply(factorial(5), factorial(5)), multiply(factorial(5), factorial(3))), 5), multiply(const_100, const_10))) | what is the tenth digit of ( 5 ! * 5 ! - 5 ! * 3 ! ) / 5 ? | ( 5 ! * 5 ! + 5 ! * 3 ! ) / 5 = 5 ! ( 5 ! + 3 ! ) / 5 = 120 ( 120 + 6 ) / 5 = 3024 units digit of the above product will be equal to 2 answer b | a = math.factorial(5)
b = math.factorial(5)
c = a * b
d = math.factorial(5)
e = math.factorial(3)
f = d * e
g = c - f
h = g / 5
i = 100 * 10
j = h / i
k = math.floor(j)
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | divide(subtract(570, multiply(75, 2.00)), 30) | 30 pens and 75 pencils were purchased for 570 . if the average price of a pencil was 2.00 , find the average price of a pen . | "since average price of a pencil = 2 ∴ price of 75 pencils = 150 ∴ price of 30 pens = ( 570 – 150 ) = 360 ∴ average price of a pen = 420 ⁄ 60 = 14 answer e" | a = 75 * 2
b = 570 - a
c = b / 30
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a ) 189 , b ) 72 , c ) 144 , d ) 216 , e ) 400 | a | multiply(divide(multiply(42, const_3), 20), 30) | if 20 typists can type 42 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 42 letters , so 30 typists can type = 42 * 30 / 20 42 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 42 * 30 * 60 / 20 * 20 = 189 a is the answer" | a = 42 * 3
b = a / 20
c = b * 30
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a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates q apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates q / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a = 24 / 2
b = 2 + 3
c = a * b
d = 24 * 2
e = c + d
f = 3 + 4
g = 24 * f
h = e / g
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a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10 | b | subtract(subtract(add(subtract(90, 11), 18), 44), 18) | in a neighborhood having 90 households , 11 did not have either a car or a bike . if 18 households had a both a car and a bike and 44 had a car , how many had bike only ? | "{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 18 + 11 - - > { bike } = 53 - - > # those who have bike only is { bike } - { both } = 53 - 18 = 35 . answer : b ." | a = 90 - 11
b = a + 18
c = b - 44
d = c - 18
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a ) 30 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(multiply(36, divide(multiply(subtract(42, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 42 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 42 - 36 = 6 km / hr . = 6 * 5 / 18 = 5 / 3 m / sec . 2 x / 36 = 5 / 3 = > x = 30 . answer : a" | a = 42 - 36
b = a * 1000
c = b / 3600
d = 36 * c
e = d / 2
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a ) 50 , b ) 40 , c ) 90 , d ) 60 , e ) 20 | c | add(70, 20) | a customer pays 70 dollars for a coffee maker after a discount of 20 dollars what is the original price of the coffe maker ? | let x be the original price . x - 20 = 70 x - 20 + 20 = 70 + 20 x + 0 = 90 x = 90 answer is c | a = 70 + 20
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a ) 270 , b ) 380 , c ) 200 , d ) 360 , e ) 290 | c | multiply(divide(160, divide(40, const_100)), divide(50, const_100)) | if 40 % of a certain number is 160 , then what is 50 % of that number ? | "explanation : 40 % = 40 * 4 = 160 50 % = 50 * 4 = 200 answer : option c" | a = 40 / 100
b = 160 / a
c = 50 / 100
d = b * c
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a ) 1 kmph , b ) 4 kmph , c ) 98 kmph , d ) 6 kmph , e ) 4 kmph | e | divide(subtract(12, 4), const_2) | a man can row his boat with the stream at 12 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 12 s = ? s = ( 12 - 4 ) / 2 = 4 kmph answer : e" | a = 12 - 4
b = a / 2
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a ) 2996 sq m , b ) 2897 sq m , c ) 4312 sq m , d ) 5768 sq m , e ) 8470 sq m | e | subtract(circle_area(divide(352, multiply(const_2, const_pi))), circle_area(divide(132, multiply(const_2, const_pi)))) | the circumferences of two circles are 132 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 = 132 and 2 ∏ l = 352 s = 132 / 2 ∏ and l = 352 / 2 ∏ difference between the areas = ∏ l ^ 2 - ∏ s ^ 2 = ∏ { 66 ^ 2 / ∏ ^ 2 - 132 ^ 2 / ∏ ^ 2 } = 66 ^ 2 / ∏ - 132 ^ 2 / ∏ = ( 66 - 132 ) ( 66 + 132 ) / ∏ = ( 110 ) ( 242 ) / ( 22 / 7 ) = 8470 sq m answer : e" | a = 2 * math.pi
b = 352 / a
c = circle_area - (
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a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54 | b | divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2) | the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if steve and dinley have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ? | "if steve and dinley submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combinedno discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) ." | a = 80 * 0
b = 2 * a
c = 100 - 25
d = c / 100
e = 160 * d
f = e * 0
g = b - f
h = g / 2
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a ) 22 , b ) 21 , c ) 35 , d ) 37 , e ) 29 | c | add(7, divide(multiply(7, subtract(16000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 16000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "explanation : lot the total number of workers be v then , 8 ooov = ( 16000 * 7 ) + 6000 ( v - 7 ) < = > 2000 v = 70000 < = > v = 35 answer : c ) 35" | a = 16000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(reminder(multiply(13, add(5, 2)), 8), 2) | when positive integer w is divided by 13 , the remainder is 2 . when n is divided by 8 , the remainder is 5 . how many such values are less than 180 ? | the equation that can be formed w is 13 x + 2 = 8 y + 5 . . 13 x - 3 = 8 y . . . as we can see x can take only odd values as the rhs will always be even . . also x can take values till 13 as 13 * 14 > 180 . . now we have to substitue x as 1 , 35 , 79 , 1113 . . . once we find 7 fitting in , any other value need not be checked as every 4 th value will give us answer so next value will be 15 . . ans 1 . . b | a = 5 + 2
b = 13 * a
c = reminder - (
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a ) 1 / 11 , b ) 2 / 11 , c ) 2 / 33 , d ) 4 / 55 , e ) 5 / 66 | d | divide(const_4, divide(factorial(11), multiply(factorial(2), factorial(subtract(11, 2))))) | a certain box has 11 cards and each card has one of the integers from 1 to 11 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on the 11 cards ? | the average of the numbers is 6 the total number of ways to choose 2 cards from 11 cards is 11 c 2 = 55 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } , { 1,4 } , { 2,3 } the probability is 4 / 55 the answer is d . | a = math.factorial(11)
b = math.factorial(2)
c = 11 - 2
d = math.factorial(c)
e = b * d
f = a / e
g = 4 / f
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a ) 16 , b ) 18 , c ) 22 , d ) 17 , e ) 20 | d | sqrt(add(281, multiply(4, 2))) | if a 2 + b 2 + c 2 = 281 and ab + bc + ca = 4 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 281 and ab + bc + ca = 4 , ( a + b + c ) ^ 2 = 281 + 2 ( 4 ) = 289 = 17 ^ 2 therefore : a + b + c = 17 answer : d" | a = 4 * 2
b = 281 + a
c = math.sqrt(b)
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a ) 840 , b ) 600 , c ) 800 , d ) 1000 , e ) 900 | a | divide(add(185, 25), divide(25, const_100)) | pradeep has to obtain 25 % of the total marks to pass . he got 185 marks and failed by 25 marks . the maximum marks are | "explanation : let their maximum marks be x . then , 25 % of x = 185 + 25 = > 25 / 100 x = 210 x = ( 210100 / 25 ) x = 840 . answer : a" | a = 185 + 25
b = 25 / 100
c = a / b
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a ) 5 , b ) 6 , c ) 9 , d ) 7 , e ) 8 | a | subtract(subtract(const_100, 98), const_1) | how many digits will be there to the right of the decimal point in the product of 98 and . 08216 ? | "product of 98 and . 08216 is 8.05168 . therefore number of digits to right of decimal point is 5 answer is a ." | a = 100 - 98
b = a - 1
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a ) 29 , b ) 29.25 , c ) 31 , d ) 31.23 , e ) 32 | b | multiply(divide(add(multiply(3, 8), 2), 8), 9) | 9 men went to a theater . 8 of them spent rs . 3 each over their tickets and the ninth spent rs . 2 more than the average expenditure of all the 9 . determine the total money spent by them ? | average of 9 = x 9 x = 8 * 3 + x * 2 x = 3.25 total = 9 * 3.25 = 29.25 b | a = 3 * 8
b = a + 2
c = b / 8
d = c * 9
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a ) 1 , b ) 3 , c ) 4 , d ) 10 , e ) 12 | e | divide(48, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 48 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 48 = 6 k + 4 p = > k = 6 , p = 3 total red ' s in bag a will be 2 k = 12 e" | a = 3 * 2
b = a + 4
c = 48 / b
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a ) 200 m , b ) 300 m , c ) 270 m , d ) 160 m , e ) 150 m | a | divide(multiply(add(multiply(1, 3), 2), 80), 2) | a runs 1 2 / 3 times as fast as b . if a gives b a start of 80 m , how far must the winning post be so that a and b might reach it at the same time ? | explanation : ratio of the speeds of a and b = 5 : 1 = 5 : 3 thus , in race of 5 m , a gains 2 m over b . 2 m are gained by a in a race of 5 m . 80 m will be gained by a in race of 5 / 2 x 80 m = 200 m winning post is 200 m away from the starting point . answer is a | a = 1 * 3
b = a + 2
c = b * 80
d = c / 2
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a ) 468 , b ) 444 , c ) 676 , d ) 767 , e ) 663 | a | add(multiply(divide(60, subtract(21, 18)), 18), multiply(divide(60, subtract(21, 18)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 18 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 3 ? - - - - - - 60 12 h rs = 18 + 21 = 39 t = 12 d = 39 * 12 = 468 answer : a" | a = 21 - 18
b = 60 / a
c = b * 18
d = 21 - 18
e = 60 / d
f = e * 21
g = c + f
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a ) one , b ) two , c ) three , d ) seven , e ) ten | a | add(1, 2) | if d = 1 / ( 2 ^ 3 * 5 ^ 6 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | "another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 6 ) = multiply and divide by 2 ^ 3 = 2 ^ 3 / ( 2 ^ 3 * 2 ^ 3 * 5 ^ 6 ) = 2 ^ 3 / 10 ^ 6 = > non zero digits are 8 = > ans a" | a = 1 + 2
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['a ) 12 cm 2', 'b ) 16 cm 2', 'c ) 20 cm 2', 'd ) 24 cm', 'e ) none of these'] | a | divide(subtract(power(divide(14, const_2), const_2), power(5, const_2)), const_2) | if the perimeter and diagonal of a rectangle are 14 and 5 cms respectively , find its area . | in a rectangle , ( perimeter ) 2 / 4 = ( diagonal ) 2 + 2 × area ⇒ ( 14 ) 2 / 4 = 5 ( 2 ) + 2 × area 49 = 25 + 2 × area ∴ area = 49 − 25 / 2 = 24 / 2 = 12 cm 2 answer a | a = 14 / 2
b = a ** 2
c = 5 ** 2
d = b - c
e = d / 2
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a ) 6.1 , b ) 7.2 , c ) 8.3 , d ) 9.4 , e ) 10.5 | b | divide(110, divide(add(50, 5), const_3_6)) | a train 110 m long is running with a speed of 50 km / h . in how many seconds will the train pass a man who is running at 5 km / h in the direction opposite to that in which the train is going ? | "the speed of the train relative to the man = 50 + 5 = 55 km / h . 55000 m / h * 1 h / 3600 s = ( 550 / 36 ) m / s ( 110 m ) / ( 550 / 36 m / s ) = ( 110 * 36 ) / 550 = 36 / 5 = 7.2 seconds the answer is b ." | a = 50 + 5
b = a / const_3_6
c = 110 / b
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a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2 | b | add(subtract(4, 2), divide(const_1, add(2, 3))) | when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in h hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of h ? | "rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * h = 1 = > x = 1 / h when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * 3 / 4 = 1 = > w + x = 4 / 3 = > 1 / 2 + 1 / h = 4 / 3 = > 1 / h = ( 8 - 3 ) / 6 = 5 / 6 = > h = 6 / 5 = 1 [ 1 / 5 ] answer b" | a = 4 - 2
b = 2 + 3
c = 1 / b
d = a + c
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a ) 90 seconds , b ) 150 seconds , c ) 60 seconds , d ) 100 seconds , e ) 120 seconds | b | multiply(divide(100, const_2), const_3) | if two - third of a bucket is filled in 100 seconds then the time taken to fill the bucket completely will be . | 2 / 3 filled in 100 seconds 1 / 3 filled in 50 secs then 2 / 3 + 1 / 3 = 100 + 50 seconds = 150 seconds answer : b | a = 100 / 2
b = a * 3
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a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 14 | a | power(add(multiply(9, 2), 2), 2) | if a is a positive integer , and if the units digit of a ^ 2 is 9 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units digit of ( a + 2 ) ^ 2 ? | "for unit digit of a ^ 2 to be 9 . . . unit digit of a has to be 3 or 7 . . . now for unit digit of ( a + 1 ) ^ 2 to be 4 . . unit digit of a has to be 1 or 7 . . . . from the above two conditions , unit value of a has to be 7 , which will satisfy both the conditions . . . now id unit digit of a is 7 , unit digit of ( a + 2 ) ^ 2 hast to be 1 . answer : a" | a = 9 * 2
b = a + 2
c = b ** 2
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a ) 22.45 , b ) 23.45 , c ) 32.45 , d ) 23.54 , e ) 23.55 | b | multiply(divide(subtract(const_100, 81), 81), const_100) | if the cost price is 81 % of selling price then what is the profit percentage . | selling price = rs 100 : then cost price = rs 81 : profit = rs 19 . profit = { ( 19 / 81 ) * 100 } % = 23.45 % answer is b . | a = 100 - 81
b = a / 81
c = b * 100
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a ) 15 % , b ) 16 % , c ) 0.125 % , d ) 0.2083 % , e ) none | b | divide(multiply(multiply(divide(480, 5000), const_100), const_100), 60) | farm tax is levied on the 60 % of the cultivated land . the tax department collected total $ 5000 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 5000 ) x 100 = 9.6 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 9.6 % x ( 100 / 60 ) = 16 % the correct answer is ( b ) ." | a = 480 / 5000
b = a * 100
c = b * 100
d = c / 60
|
a ) $ 450 , b ) $ 500 , c ) $ 600 , d ) $ 650 , e ) $ 750 | b | divide(divide(30000, 30), const_2) | in company j , the total monthly payroll for the 15 factory workers is $ 30000 and the total monthly payroll for the 30 office workers is $ 75000 . by how much does the average ( arithmetic mean ) monthly salary of an office worker exceed that of a factory worker in this company ? | the average monthly salary of a factory worker is : $ 30000 / 15 = $ 2000 . the average monthly salary of an office worker is : $ 75000 / 30 = $ 2500 . the difference in average salary is : $ 2500 - $ 2000 = $ 500 . the answer is b . | a = 30000 / 30
b = a / 2
|
a ) 20 , b ) 15 , c ) 10 , d ) 12 , e ) 25 | c | divide(subtract(multiply(90, 5), multiply(5, 50)), subtract(90, 70)) | the average mark of the students of a class in a particular exam is 70 . if 5 students whose average mark in that exam is 50 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 70 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 70 x - ( 5 * 50 ) = 90 ( x - 5 ) 200 = 20 x = > x = 10 answer : c" | a = 90 * 5
b = 5 * 50
c = a - b
d = 90 - 70
e = c / d
|
a ) 5 , b ) 5 ^ 2 , c ) 30 , d ) 30 ^ 2 , e ) 30 ^ 4 | e | add(const_3, const_4) | what is the smallest positive integer nn such that √ 6,480 ∗ n is a perfect cube ? | "sol : let ' s factorize 6480 and we get 6480 = 3 ^ 4 * 2 ^ 4 * 5 now we need to see for what minimum value of n √ n * 6480 = a ^ 3 where a is an integer so from 6480 we already have 2 ^ 4 * 3 ^ 4 * 5 * n √ n = ( 2 ^ 2 ) ^ 3 * ( 3 ^ 2 ) ^ 3 * ( 5 ) ^ 3 why cause a is an integer which will need to be have the same factors which are in lhs solving for n √ n = ( 2 ^ 6 * 3 ^ 6 * 5 ^ 3 ) / 2 ^ 4 * 3 ^ 4 * 5 and we get n √ n = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 or n = 2 ^ 4 * 3 ^ 4 * 5 ^ 4 or 30 ^ 4 answer is e" | a = 3 + 4
|
a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 3 / 25 , e ) none of these | d | divide(circle_area(divide(12, const_2)), const_2) | what will be the fraction of 12 % | "explanation : 12 * 1 / 100 = 3 / 25 . option d" | a = 12 / 2
b = circle_area / (
|
a ) 4 / 25 , b ) 8 / 23 , c ) 2 / 5 , d ) 8 / 14 , e ) 2 / 3 | d | divide(divide(multiply(40, 40), const_100), add(divide(multiply(40, 40), const_100), divide(multiply(20, subtract(const_100, 40)), const_100))) | at joel ’ s bookstore , the current inventory is 40 % historical fiction . of the historical fiction books , 40 % are new releases , while 20 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? | "let there be 100 books in all historic fiction books = 40 % of total = 40 other books = 60 new historic fiction = 40 % of 40 = 16 other new books = 20 % of 60 = 12 total new books = 28 fraction = 16 / 28 = 8 / 14 ans : d" | a = 40 * 40
b = a / 100
c = 40 * 40
d = c / 100
e = 100 - 40
f = 20 * e
g = f / 100
h = d + g
i = b / h
|
a ) 26 , b ) 28 , c ) 80 , d ) 26 , e ) 21 | c | divide(40, subtract(add(multiply(40, add(const_1, divide(10, const_100))), multiply(50, add(const_1, divide(20, const_100)))), multiply(90, add(const_1, divide(15, const_100))))) | a person purchases 90 clocks and sells 40 clocks at a gain of 10 % and 50 clocks at a gain of 20 % . if he sold all of them at a uniform profit of 15 % , then he would have got rs . 40 less . the cost price of each clock is ? | "let c . p . of clock be rs . x . then , c . p . of 90 clocks = rs . 90 x . [ ( 110 % of 40 x ) + ( 120 % of 50 x ) ] - ( 115 % of 90 x ) = 40 44 x + 60 x - 103.5 x = 40 0.5 x = 40 = > x = 80 answer : c" | a = 10 / 100
b = 1 + a
c = 40 * b
d = 20 / 100
e = 1 + d
f = 50 * e
g = c + f
h = 15 / 100
i = 1 + h
j = 90 * i
k = g - j
l = 40 / k
|
a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 7.69 % , e ) none of these | d | multiply(add(const_1, divide(divide(multiply(120, 50), add(120, divide(multiply(multiply(const_3, const_10), 120), const_100))), 50)), const_10) | a rectangle having length 120 cm and width 50 cm . if the length of the rectangle is increased by ten percent then how much percent the breadth should be decreased so as to maintain the same area . | "explanation : solution : ( 10 / ( 120 + 10 ) * 100 ) % = 7.69 % answer : d" | a = 120 * 50
b = 3 * 10
c = b * 120
d = c / 100
e = 120 + d
f = a / e
g = f / 50
h = 1 + g
i = h * 10
|
a ) 65 degrees , b ) 73 degrees , c ) 31 degrees , d ) 34 degrees , e ) 74 degrees | c | subtract(39, subtract(multiply(48, const_4), multiply(46, const_4))) | the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 39 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 39 tu + w + th = 192 - 39 = 153 f = 184 – 153 = 31 answer : c" | a = 48 * 4
b = 46 * 4
c = a - b
d = 39 - c
|
a ) 1 , b ) 6 , c ) 13 , d ) 14 , e ) 15 | b | divide(2, subtract(divide(multiply(const_2, 2), 3), 1)) | it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 3 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 2 miles an hour let the other half be x miles an hour now , avg speed = 3 mile an hour 2 * 2 * x / 2 + x = 3 4 x = 3 x + 6 = > x = 6 b" | a = 2 * 2
b = a / 3
c = b - 1
d = 2 / c
|
a ) 21 years , b ) 14 years , c ) 16 years , d ) 18 years , e ) 26 years | b | divide(subtract(16, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 16 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 16 . = > 2 ( x + 2 ) = ( x + 16 + 2 ) = > 2 x + 4 = x + 18 = > x = 14 years option b" | a = 2 * 2
b = a - 2
c = 16 - b
d = 2 - 1
e = c / d
|
a ) 10 , b ) 20 , c ) 50 , d ) 30 , e ) 40 | a | subtract(subtract(15, 2.5), 2.5) | a man ' s speed with the current is 15 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 15 - 2.5 = 12.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km / hr answer is a ." | a = 15 - 2
b = a - 2
|
a ) 9.2 , b ) 10.3 , c ) 11.5 , d ) 12.4 , e ) 13 | d | subtract(subtract(21, 4.3), 4.3) | a man ' s speed with the current is 21 km / hr and the speed of the current is 4.3 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 21 - 4.3 ) km / hr = 16.7 km / hr . man ' s rate against the current = ( 16.7 - 4.3 ) km / hr = 12.4 km / hr . answer : d" | a = 21 - 4
b = a - 4
|
a ) 300 , b ) 315 , c ) 330 , d ) 345 , e ) 360 | b | add(add(multiply(subtract(45, subtract(80, divide(add(85, subtract(80, 45)), const_2))), 3), multiply(subtract(80, divide(add(85, subtract(80, 45)), const_2)), 3)), multiply(divide(add(85, subtract(80, 45)), const_2), 3)) | martin buys a pencil and a notebook for 80 cents . at the same store , gloria buys a notebook and an eraser for 85 cents , and zachary buys a pencil and an eraser for 45 cents . how many cents would it cost to buy 3 pencils , 3 notebooks , and 3 erasers ? ( assume that there is no volume discount . ) | pencil + notebook = 80 notebook + eraser = 85 pencil + eraser = 45 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 210 cents the cost to buy 3 of each would be ( 3 / 2 ) ( 210 ) = 315 the answer is b . | a = 80 - 45
b = 85 + a
c = b / 2
d = 80 - c
e = 45 - d
f = e * 3
g = 80 - 45
h = 85 + g
i = h / 2
j = 80 - i
k = j * 3
l = f + k
m = 80 - 45
n = 85 + m
o = n / 2
p = o * 3
q = l + p
|
a ) s . 1050 , b ) s . 1160 , c ) s . 1190 , d ) s . 1202 , e ) s . 1204 | a | divide(multiply(subtract(const_100, 25), 1400), const_100) | a man buys a cycle for rs . 1400 and sells it at a loss of 25 % . what is the selling price of the cycle ? | "since , c . p = 1400 loss % = ( c . p - s . p ) / c . p * 100 25 = ( 1400 - s . p ) / 1400 * 100 so , after solving answer = 1050 . answer : a" | a = 100 - 25
b = a * 1400
c = b / 100
|
a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | c | multiply(add(multiply(5, 5), 1), 5) | 1 + 3 = 2 2 + 3 = 10 3 + 3 = 30 4 + 3 = 68 5 + 3 = ? ? | 1 ^ 3 + 3 - 2 = 2 , 2 ^ 3 + 3 - 1 = 10 , 3 ^ 3 + 3 - 0 = 30 , 4 ^ 3 + 3 + 1 = 68 , 5 ^ 3 + 3 + 2 = 130 answer : c | a = 5 * 5
b = a + 1
c = b * 5
|
a ) 160 , b ) 240 , c ) 360 , d ) 495 , e ) 560 | d | add(90, multiply(divide(multiply(90, 3), const_2), 3)) | a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 90 square feet of the wall , and no tiles overlap , what is the area of the entire wall ? | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 90 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 90 = 405 . total area = 90 + 405 = 495 answer : d ." | a = 90 * 3
b = a / 2
c = b * 3
d = 90 + c
|
a ) 160 , b ) 165 , c ) 180 , d ) 190 , e ) 540 | e | subtract(multiply(multiply(2, const_3), const_100), multiply(multiply(2, const_3), const_10)) | what is the difference between the place value of 2 in the numeral 7669 ? | answer : option ' e ' 600 - 60 = 540 | a = 2 * 3
b = a * 100
c = 2 * 3
d = c * 10
e = b - d
|
a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 12 | b | divide(770, divide(multiply(multiply(10, 770), divide(add(const_100, 10), const_100)), subtract(multiply(770, divide(add(const_100, 10), const_100)), 770))) | machine a and machine b are each used to manufacture 770 sprockets . it takes machine a 10 hours longer to produce 770 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | "machine b : takes x hours to produce 770 sprockets machine a : takes ( x + 10 ) hours to produce 770 sprockets machine b : in 1 hour , b makes 770 / x sprockets machine a : in 1 hour , a makes 770 / ( x + 10 ) sprockets equating : 1.1 ( 770 / ( x + 10 ) ) = 770 / x 847 / ( x + 10 ) = 770 / x 847 x = 770 x + 7700 77 x = 7700 x = 100 a makes 770 / ( 110 ) = 7 sprockets per hour answer : b" | a = 10 * 770
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 770 * f
h = g - 770
i = d / h
j = 770 / i
|
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