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a ) 322 sec , b ) 14 sec , c ) 11 sec , d ) 13 sec , e ) 34 sec | b | divide(add(156.62, 100), multiply(add(30, 36), const_0_2778)) | two trains of lengths 156.62 and 100 meters are running on parallel lines with respective speeds of 30 km / hr and 36 km / hr . the time of crossing each other , if they run in the opposite direction is | explanation : total distance to be covered = 156.62 + 100 = 256.62 m trains are running in opposite directions , hence relative speed = 30 + 36 = 66 km / hr = 18.33 m / sec time = distance / relative speed = 256.62 / 18.33 sec = 14 sec therefore , the time of crossing each other in the opposite direction is 14 seconds . answer : b | a = 156 + 62
b = 30 + 36
c = b * const_0_2778
d = a / c
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a ) 104345 , b ) 107375 , c ) 108385 , d ) 109395 , e ) 105355 | d | add(add(multiply(add(const_4, const_3), const_10), multiply(add(const_4, const_3), multiply(const_100, const_10))), 102325) | on multiplying a number a by 153 , the result obtained was 102325 . however , it is found that both the 2 ' s are wrong . find the correct result . | the only thing you actually know about the correct number a is that it is divisible by 153 and has 5 as a factor . you should immediately try to find the factors of 153 and look for them in the options . 153 = 9 * 17 divisibility by 9 is easy to check . only ( d ) satisfies . | a = 4 + 3
b = a * 10
c = 4 + 3
d = 100 * 10
e = c * d
f = b + e
g = f + 102325
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a ) 1000 , b ) 1055 , c ) 1428 , d ) 1500 , e ) 1080 | c | add(multiply(11, 98), multiply(7, 50)) | andrew purchased 11 kg of grapes at the rate of 98 per kg and 7 kg of mangoes at the rate of 50 per kg . how much amount did he pay to the shopkeeper ? | "cost of 11 kg grapes = 98 × 11 = 1078 . cost of 7 kg of mangoes = 50 × 7 = 350 . total cost he has to pay = 1078 + 350 = 1428 c" | a = 11 * 98
b = 7 * 50
c = a + b
|
a ) 6,9 , b ) 3,8 , c ) 6,7 , d ) 3,5 , e ) 3,4 | e | multiply(3, 12) | two numbers are in the ratio of 3 : 4 . if 12 be subtracted from each , they are in the ratio of 9 : 8 . find the numbers ? | "( 3 x - 12 ) : ( 4 x - 12 ) = 9 : 8 x = 1 = > 3,4 answer : e" | a = 3 * 12
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a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6000 | c | divide(1380, divide(subtract(60, subtract(const_100, 60)), const_100)) | in an election a candidate who gets 60 % of the votes is elected by a majority of 1380 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1380 20 x / 100 = 1380 x = 1380 * 100 / 20 = 6900 answer is c" | a = 100 - 60
b = 60 - a
c = b / 100
d = 1380 / c
|
a ) 1.004 , b ) 1.006 , c ) 1.008 , d ) 1.012 , e ) 1.016 | d | add(1.003, multiply(divide(3, const_1000), 3)) | rounded to 3 decimal places , 1.003 ^ 4 = | as compared to 1 , 0.003 is a very small quantity . thus , we can write ( 1 + 0.003 ) ^ 4 is nearly equal to ( 1 + 4 * 0.003 ) = 1.012 . as the question asks for approximation to three decimal places , the further terms will anyways not come into picture . d . | a = 3 / 1000
b = a * 3
c = 1 + 3
|
a ) rs . 600 , b ) rs . 500 , c ) rs . 400 , d ) rs . 300 , e ) none of these | a | divide(multiply(const_100, divide(multiply(24, const_100), multiply(2, 10))), multiply(2, 10)) | the banker ' s gain of a certain sum due 2 years hence at 10 % per annum is rs . 24 . what is the present worth ? | "explanation : t = 2 years r = 10 % td = ( bg × 100 ) / tr = ( 24 × 100 ) / ( 2 × 10 ) = 12 × 10 = rs . 120 td = ( pw × tr ) / 100 ⇒ 120 = ( pw × 2 × 10 ) / 100 ⇒ 1200 = pw × 2 pw = 1200 / 2 = rs . 600 answer : option a" | a = 24 * 100
b = 2 * 10
c = a / b
d = 100 * c
e = 2 * 10
f = d / e
|
a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 10,000 | c | divide(50, divide(2, 50)) | in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "total fish = x percentage of second catch = ( 2 / 50 ) * 100 = 4 % so , x * 4 % = 50 x = 1250 answer : c" | a = 2 / 50
b = 50 / a
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a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | multiply(divide(12, const_4.0), divide(12, 14)) | at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 10 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 3 to 14 . how many cooks does the restaurant have ? | "originally there were 3 k cooks and 10 k waiters . 14 k = 10 k + 12 k = 3 there are 9 cooks . the answer is c ." | a = 12 / 4
b = 12 / 14
c = a * b
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a ) 960 , b ) 1060 , c ) 1,200 , d ) 1020 , e ) none of these | d | multiply(divide(add(912, 448), const_2), add(const_1, divide(50, const_100))) | the profit earned by selling an article for 912 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ? | "let the profit or loss be x and 912 – x = 448 + x or , x = 464 ⁄ 2 = 232 \ cost price of the article = 912 – x = 448 + x = 680 \ sp of the article = 680 × 150 ⁄ 100 = 1020 answer d" | a = 912 + 448
b = a / 2
c = 50 / 100
d = 1 + c
e = b * d
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a ) 3 / 15 , b ) 2 / 11 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 | e | divide(subtract(const_1, multiply(4, add(divide(const_1, 10), divide(const_1, 8)))), divide(const_1, 10)) | working individually , julie can peel potatoes in 10 hours and ted can peel potatoes in 8 hours . if julie and ted work together but independently at the task for 4 hours , at which point ted leaves , how many remaining hours will it take julie to complete the task alone ? | in first 4 hrs ted will finish 4 / 8 = 1 / 2 of work and julie will finish 4 / 10 work so total 1 / 2 + 2 / 5 = 9 / 10 work is finished and 1 - 9 / 10 = 1 / 10 work remaining . now julie will take ( 1 / 10 ) * 10 = 1 hrs to finish it . so answer is e . | a = 1 / 10
b = 1 / 8
c = a + b
d = 4 * c
e = 1 - d
f = 1 / 10
g = e / f
|
a ) 8 / 5 , b ) 6 / 5 , c ) 7 / 5 , d ) 3 / 5 , e ) 4 / 5 | d | divide(3, divide(multiply(8, 12), add(8, 12))) | x is able to do a piece of work in 8 days and y can do the same work in 12 days . if they can work together for 3 days , what is the fraction of work left ? | "explanation : amount of work x can do in 1 day = 1 / 8 amount of work y can do in 1 day = 1 / 12 amount of work x and y can do in 1 day = 1 / 8 + 1 / 12 = 5 / 24 amount of work x and y can together do in 3 days = 3 × ( 5 / 24 ) = 5 / 8 fraction of work left = 1 – 5 / 8 = 3 / 5 answer : option d" | a = 8 * 12
b = 8 + 12
c = a / b
d = 3 / c
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a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 12 % | a | subtract(const_100, divide(multiply(1420, const_100), 1450)) | an article is bought for rs . 1450 and sold for rs . 1420 , find the loss percent ? | "1450 - - - - 30 100 - - - - ? = > 2 % answer : a" | a = 1420 * 100
b = a / 1450
c = 100 - b
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a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 8 | e | subtract(28, add(7, 9)) | in a group of 28 junior high school students , 7 take french , 9 take spanish , and 4 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ? | "e 8 add 7 + 9 + 4 to get 20 . then subtract 21 from the total students ⇒ 28 – 20 = 8 . answer is e" | a = 7 + 9
b = 28 - a
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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 r 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 r ? | "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 * r = 1 = > x = 1 / z 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 / r = 4 / 3 = > 1 / r = ( 8 - 3 ) / 6 = 5 / 6 = > r = 6 / 5 = 1 [ 1 / 5 ] answer b" | a = 4 - 2
b = 2 + 3
c = 1 / b
d = a + c
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a ) 22 , b ) 36 , c ) 60 , d ) 88 , e ) 72 | b | divide(multiply(36, 15), 15) | if 36 men can do a piece of work in 15 hours , in how many hours will 15 men do it ? | "explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 15 : 36 : : 15 : x \ inline \ fn _ jvn \ leftrightarrow ( 15 x x ) = ( 36 x 15 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 15 } { 15 } = 36 hence , 15 men can do it in 36 hours . answer : b ) 36" | a = 36 * 15
b = a / 15
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a ) 36 , b ) 72 , c ) 132 , d ) 144 , e ) 180 | e | multiply(subtract(divide(multiply(const_2, 60), subtract(80, multiply(const_2, 60))), divide(60, subtract(80, 60))), const_60) | buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of 60 miles per hour . one hour later , bus b begins to travel in a straight line in the exact opposite direction at a constant rate of 80 miles per hour . if both buses travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ? | "1 st part : - in 1 hr , bus a covers 30 miles . relative speed of bus abus b is ( 80 - 30 ) = 50 mph . so time required for bus b to cover the exact distance as a is 50 * t = 30 t = 3 / 5 = 36 min 2 nd part 80 * t = 2 d - b has to cover twice the distance 30 * ( t + 1 ) = d - a traveled 1 hr more and has to travel only only d so d / 30 - 2 d / 80 = 1 d = 120 t = 3 hrs = 180 min question asks for + ve difference between part 1 and part 2 in minutes = 180 - 36 = 180 min e" | a = 2 * 60
b = 2 * 60
c = 80 - b
d = a / c
e = 80 - 60
f = 60 / e
g = d - f
h = g * const_60
|
a ) $ 900 , b ) $ 800 , c ) $ 850 , d ) $ 950 , e ) $ 920 | a | divide(13500, 15) | dawson is going with 14 friends on a trip to washington d . c for spring break . airfare and hotel costs a total of $ 13500.00 for the group of 15 friends . how much does each person have to pay for their hotel and airfare ? | answer = a the total cost of the trip ( $ 13500.00 ) divided by 15 equals $ 900.00 . | a = 13500 / 15
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a ) 179 , b ) 367 , c ) 269 , d ) 177 , e ) 199 | e | divide(add(190, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 190 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 â € “ 2552 = 208 runs . let the highest score be x , hence the lowest score = x â € “ 190 x + ( x - 190 ) = 208 2 x = 398 x = 199 runs answer : e" | a = 60 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 190 + d
f = e / 2
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a ) 34.15 , b ) 40.35 , c ) 47.14 , d ) 50.12 , e ) 56.66 | c | multiply(divide(add(divide(30, 30), divide(25, 40)), const_2), const_60) | the point a and point b is 30 miles apart , the point b and point c is 25 miles apart . a car travels from point a to point b in 30 min and point b to point c in 40 min . what is the average speed of the car ? | average speed = total distance covered / total time taken = 55 miles / 70 min = 0.79 miles / min = 47.14 miles / hour answer : c | a = 30 / 30
b = 25 / 40
c = a + b
d = c / 2
e = d * const_60
|
a ) 0.0003 , b ) 0.36 , c ) 0.3 , d ) 36 , e ) 90 | c | divide(0.009, 0.025) | 0.009 / x = 0.025 . find the value of x | "x = 0.009 / 0.36 = 0.25 answer : c" | a = 0 / 9
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a ) 66.67 , b ) 66 , c ) 67 , d ) 65.99 , e ) 67.26 | a | subtract(const_100, 30) | at the moment there are 54210 tagged birds in a certain wildlife refuge . if exactly 30 percent of all birds in the refuge are tagged , what percent of the untagged birds must be tagged so that half of all birds in the refuge are tagged ? | all birds = 54210 currently tagged = 54210 * 30 / 100 = 16263 so untagged birds = 54210 - 16263 = 37947 half of all birds = 27105 the number of birds to be tagged to make half of all birds tagged = 27105 - 16263 = 10842 so now the question remains - 10842 is how much percentage of untagged birds ( 16263 ) = 10842 * 100 / 16263 = 66.67 answer ( a ) | a = 100 - 30
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a ) 288 , b ) 132 , c ) 220 , d ) 592 , e ) 261 | c | multiply(circumface(divide(70, const_2)), 1) | find the cost of fencing around a circular field of diameter 70 m at the rate of rs . 1 a meter ? | "2 * 22 / 7 * 35 = 220 220 * 1 = rs . 220 answer : c" | a = 70 / 2
b = circumface * (
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a ) 100.1 m , b ) 223.1 m , c ) 111.1 m , d ) 120.3 m , e ) 133.4 m | c | multiply(divide(multiply(const_1000, 10), multiply(const_60, 15)), 10) | a train takes 10 sec to pass a signal post and covers a distance of 10 km in 15 min . find the length of train ? | explanation : we know , speed = distance / time speed = 10 / 50 x 60 = 40 x 5 / 18 m / sec = 11.1 m / sec length of train = ( speed x time ) = ( 11.11 x 10 ) = 111.1 m answer is c | a = 1000 * 10
b = const_60 * 15
c = a / b
d = c * 10
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a ) s 1200 , b ) s 1500 , c ) s 1600 , d ) s 1900 , e ) s 900 | e | multiply(divide(1500, 5), subtract(6, 3)) | an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 6 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 6 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 1800 . hence , ( r - p ) = ( 1800 - 900 ) = 900 answer : e | a = 1500 / 5
b = 6 - 3
c = a * b
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a ) $ 37.40 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60 | a | add(multiply(34, divide(40, const_100)), 34) | a farmer spent $ 34 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 34 = $ 13.6 on chicken feed , thus he spent the remaining 34 - 13.6 = $ 20.4 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 13.6 - - > x = $ 17 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 17 + 20.4 = $ 37.4 . answer : a ." | a = 40 / 100
b = 34 * a
c = b + 34
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a ) 45 , b ) 49 , c ) 52 , d ) 60 , e ) 74 | d | divide(subtract(312, multiply(2.4, const_100)), subtract(3.6, 2.4)) | a sum of rs . 312 was divided among 100 boys and girls in such a way that each boy gets rs . 3.60 and each girl rs . 2.40 . | let number of boys = x then number of girls = ( 100 - x ) 3.60 x + 2.40 ( 100 - x ) = 312 ¡ ê 1.2 x = 312 - 240 = 72 ¡ ê x = 60 answer : d | a = 2 * 4
b = 312 - a
c = 3 - 6
d = b / c
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a ) 25 , b ) 34 , c ) 50 , d ) 67 , e ) 50 | e | divide(multiply(6, 200), divide(200, const_10)) | according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 6 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 6 oz can = 300 oz / 6 oz = 50 answer e , 50 cans" | a = 6 * 200
b = 200 / 10
c = a / b
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a ) 25 , b ) 60 , c ) 90 , d ) 140 , e ) it can not be determined from the information given . | b | subtract(multiply(60, const_2), multiply(30, const_2)) | if the average ( arithmetic mean ) of a and b is 30 and the average of b and c is 60 , what is the value of c − a ? | - ( a + b = 60 ) b + c = 120 c - a = 60 b . 60 | a = 60 * 2
b = 30 * 2
c = a - b
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a ) 8 , b ) 12 , c ) 40 , d ) 32 , e ) 36 | c | multiply(subtract(const_10, 1), 4) | on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 4 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on november 1 than were in stock on january 1 ? | "jan 1 = c feb 1 = c - 4 march 1 = c - 8 april 1 = c - 12 may 1 = c - 16 june 1 = c - 20 july 1 = c - 24 aug 1 = c - 28 sept 1 = c - 32 oct 1 = c - 36 nov 1 = c - 40 difference between stock on november 1 than were in stock on january 1 will be - c - ( c - 40 ) = 40 hence answer will be ( c )" | a = 10 - 1
b = a * 4
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['a ) a ) 3 : 4', 'b ) b ) 2 : 7', 'c ) c ) 2 : 2', 'd ) d ) 2 : 1', 'e ) e ) 2 : 9'] | a | divide(3, const_4) | the volumes of two cones are in the ratio 1 : 12 and the radii of the cones are in the ratio of 1 : 3 . what is the ratio of their heights ? | the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 12 h 1 / r 22 h 2 = > 1 / 12 = ( 1 ) 2 h 1 / ( 3 ) 2 h 2 = > h 1 / h 2 = 3 / 4 i . e . h 1 : h 2 = 3 : 4 answer : a | a = 3 / 4
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a ) 11.769 % , b ) 10.769 % , c ) 12.769 % , d ) 11.69 % , e ) 11.89 % | b | multiply(divide(subtract(subtract(const_100, 35), subtract(const_100, 42)), subtract(const_100, 35)), const_100) | two numbers are 35 % and 42 % are less than a third number . how much percent is the second number less than the first ? | "explanation : i ii iii 65 58 100 65 - - - - - - - - 7 100 - - - - - - ? = > 10.769 % answer b" | a = 100 - 35
b = 100 - 42
c = a - b
d = 100 - 35
e = c / d
f = e * 100
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a ) 30 days , b ) 48 days , c ) 70 days , d ) 80 days , e ) 90 days | b | divide(16, subtract(const_1, divide(16, 24))) | john and roger can finish the work 24 days if they work together . they worked together for 16 days and then roger left . john finished the remaining work in another 16 days . in how many days john alone can finish the work ? | amount of work done by john and roger in 1 day = 1 / 24 amount of work done by john and roger in 16 days = 16 ã — ( 1 / 24 ) = 2 / 3 remaining work â € “ 1 â € “ 2 / 3 = 1 / 3 john completes 1 / 3 work in 16 days amount of work john can do in 1 day = ( 1 / 3 ) / 16 = 1 / 48 = > john can complete the work in 48 days answer : b | a = 16 / 24
b = 1 - a
c = 16 / b
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a ) 1980 , b ) 1920 , c ) 1990 , d ) 1890 , e ) 1900 | a | multiply(add(43, const_2), subtract(add(43, const_2), const_1)) | there are 43 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 45 from 45 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 45 p 2 ways . 45 p 2 = 45 * 44 = 870 answer : a" | a = 43 + 2
b = 43 + 2
c = b - 1
d = a * c
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a ) 6 , b ) 16 , c ) 18 , d ) 30 , e ) 174 | c | add(subtract(18, 6), subtract(12, 6)) | if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 12 integers , y consists of 18 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ? | "the number of integers that belong to set x only is 12 - 6 = 6 ; the number of integers that belong to set y only is 18 - 6 = 12 ; the number of integers that belong to set x or set y , but not both is 6 + 12 = 18 . answer : c ." | a = 18 - 6
b = 12 - 6
c = a + b
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a ) 3 / 10 , b ) 2 / 5 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | c | divide(4, subtract(multiply(subtract(6, 4), 6), 4)) | when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 6 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 6 ) ( 4 + 4 x ) or , x = 1 / 2 answer c" | a = 6 - 4
b = a * 6
c = b - 4
d = 4 / c
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a ) 360 km , b ) 480 km , c ) 278 km , d ) 297 km , e ) 671 km | a | divide(1, 2) | with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 3 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 3 2 x / 15 - x / 8 = 3 = > x = 360 km . answer : a | a = 1 / 2
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers q are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a = 1 + 1
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a ) 20 % , b ) 25 % , c ) 29 % , d ) 55 % , e ) 100 % | e | subtract(multiply(divide(const_100, 500), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 500 grams per kg , what is his percent ? | "500 - - - 500 100 - - - ? = > 100 % answer : e" | a = 100 / 500
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
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a ) 6846381250 , b ) 6584638130 , c ) 6584638135 , d ) 6584638140 , e ) 6956051792 | e | multiply(7895632, power(add(const_4, const_1), const_4)) | ( 7895632 x 881 ) = ? | 7895632 x 881 = 6956051792 ans e | a = 4 + 1
b = a ** 4
c = 7895632 * b
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a ) 4 , b ) 10 , c ) 7 , d ) 5 , e ) 3 | d | divide(add(40, 5), add(8, 8)) | solve the equation for x : 8 x - 40 + 2 x = 5 + 10 - x | "d 5 10 x + x = 15 + 40 11 x = 55 = > x = 5" | a = 40 + 5
b = 8 + 8
c = a / b
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a ) 9 / 13 , b ) 1 / 6 , c ) 1 / 3 , d ) 4 / 9 , e ) 5 / 9 | a | divide(9, add(9, 4)) | a waitress ' s income consists of her salary and tips . during one week , her tips were 9 / 4 of her salary . what fraction of her income for the week came from tips ? | her tips were 9 / 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 9 / 4 ) ( $ 4 ) = $ 9 so , her total income = $ 4 + $ 9 = $ 13 what fraction of her income for the week came from tips $ 9 $ 13 = 9 / 13 = a | a = 9 + 4
b = 9 / a
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a ) 87 days , b ) 20 days , c ) 16 days , d ) 24 days , e ) 36 days | d | divide(multiply(8, 12), divide(subtract(multiply(8, 12), multiply(add(divide(multiply(8, 12), 8), divide(multiply(8, 12), 12)), 4)), 4)) | a can do a piece of work in 8 days . b can do it in 12 days . with the assistance of c they completed the work in 4 days . find in how many days can c alone do it ? | "c = 1 / 4 - 1 / 8 - 1 / 12 = 1 / 24 = > 24 days answer : d" | a = 8 * 12
b = 8 * 12
c = 8 * 12
d = c / 8
e = 8 * 12
f = e / 12
g = d + f
h = g * 4
i = b - h
j = i / 4
k = a / j
|
a ) 724533811 , b ) 625127481 , c ) 545463251 , d ) 725117481 , e ) 677899932 | b | multiply(subtract(9999, const_4), 62519) | find the value of m 62519 x 9999 = m ? | "62519 x 9999 = 62519 x ( 10000 - 1 ) = 62519 x 10000 - 62519 x 1 = 625190000 - 62519 = 625127481 b" | a = 9999 - 4
b = a * 62519
|
a ) rs . 5000 , b ) rs . 4000 , c ) rs . 6000 , d ) rs . 8000 , e ) rs . 7000 | b | divide(124, subtract(power(add(const_1, divide(3, const_100)), 10), add(const_1, multiply(10, divide(3, const_100))))) | the difference between compound interest and simple interest on a sum for 3 years at 10 % p . a . is rs . 124 find the sum . | "for 3 years , we know c . i - s . i = p ( r / 100 ) ^ 3 + 3 ( r / 100 ) ^ 2 so , putting all values we have , 124 = p ( ( 10 / 100 ) ^ 3 + 3 ( 10 / 100 ) ^ 2 ) = 4000 answer : b" | a = 3 / 100
b = 1 + a
c = b ** 10
d = 3 / 100
e = 10 * d
f = 1 + e
g = c - f
h = 124 / g
|
a ) 12 , b ) 32 , c ) 22 , d ) 15 , e ) 20 | a | divide(8, divide(2, 3)) | an italian sausage is 8 inches long . how many pieces of sausage can be cut from the 8 - inch piece of sausage if each piece is to be 2 / 3 of an inch ? | number of pieces = italian sausage ÷ 2 / 3 of an inch = 8 ÷ 2 / 3 = 8 / 1 ÷ 2 / 3 = 8 / 1 * 3 / 2 = 24 / 2 = 12 . answer is a . | a = 2 / 3
b = 8 / a
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a ) 1978 , b ) 2707 , c ) 7728 , d ) 4200 , e ) 6000 | e | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(2500, add(3500, 2500)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 3500 and b rs . 2500 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "35 : 25 = > 7 : 5 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 7 / 12 = 5040 + 960 = 6000 answer : e" | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 3500 + 2500
e = 2500 / d
f = c * e
g = 9600 - f
|
a ) a ) 145 , b ) b ) 148 , c ) c ) 150 , d ) d ) 153 , e ) e ) 166 | e | add(multiply(20, 8), 6) | what is the dividend . divisor 20 , the quotient is 8 and the remainder is 6 | "d = d * q + r d = 20 * 8 + 6 d = 160 + 6 d = 166 answer : e" | a = 20 * 8
b = a + 6
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a ) 30 , b ) 60 , c ) 25 , d ) 40 , e ) 10 | a | multiply(const_3600, divide(divide(550, const_1000), add(60, 6))) | a train 550 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 550 * 3 / 55 ] sec = 30 sec answer : a | a = 550 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 1.9732 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none | a | multiply(divide(268, const_100), divide(74, const_100)) | given that 268 x 74 = 19732 , find the value of 2.68 x . 74 . | "solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9732 answer a" | a = 268 / 100
b = 74 / 100
c = a * b
|
a ) $ 770 , b ) $ 660 , c ) $ 700 , d ) $ 1100 , e ) $ 840 | d | multiply(divide(70, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 70 more . what was his original selling price ? | "let the original purchase price be x so original selling price at 10 % profit = 1.1 x if product is purchased at 10 % less of original = 0.9 x profit of 30 % on this price = 1.3 ( 0.9 x ) he would have received $ 70 more in second scenario = > 1.3 ( 0.9 x ) - 1.1 x = 70 = > 0.07 x = 70 = > x = $ 1000 original purchase price = $ 1000 hence , original selling price ( at 10 % of profit ) = 1.1 ( 1000 ) = $ 1100 option d" | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 70 / h
j = 10 / 100
k = 1 + j
l = i * k
|
a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | divide(subtract(subtract(subtract(50, multiply(50, divide(3, 5))), multiply(subtract(50, multiply(50, divide(3, 5))), divide(3, 5))), 3), subtract(subtract(subtract(50, multiply(50, divide(3, 5))), multiply(subtract(50, multiply(50, divide(3, 5))), divide(3, 5))), 3)) | george baked a total of 50 pizzas for 5 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ? | 3 / 5 of the 50 pizzas cooked on saturday = 30 pizzas 3 / 5 of the remaining pizzas on sunday = 12 pizzas we ' re left with ( 50 - 30 - 12 ) = 8 pizzas for the remaining 3 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w 5 2 1 = 8 w = 1 e | a = 3 / 5
b = 50 * a
c = 50 - b
d = 3 / 5
e = 50 * d
f = 50 - e
g = 3 / 5
h = f * g
i = c - h
j = i - 3
k = 3 / 5
l = 50 * k
m = 50 - l
n = 3 / 5
o = 50 * n
p = 50 - o
q = 3 / 5
r = p * q
s = m - r
t = s - 3
u = j / t
|
a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number ? | "explanatory answer let the original number be x . let a be the value obtained when x is increased by 12.5 % . therefore , a = x + 12.5 % of x let b be the value obtained when x is decreased by 25 % . therefore , b = x - 25 % of x the question states that a - b = 30 i . e . , x + 12.5 % of x - ( x - 25 % of x ) = 30 x + 12.5 % of x - x + 25 % of x = 30 37.5 % of x = 30 x = 30 / 37.5 * 100 = 80 the correct choice is ( b )" | a = 12 / 5
b = 1 + a
c = 25 / 100
d = 1 - c
e = b - d
f = 30 / e
|
a ) 1521 , b ) 1492 , c ) 1667 , d ) 2500 , e ) 1112 | d | multiply(divide(multiply(10, const_1000), const_60), 15) | a man walking at a rate of 10 km / hr crosses a bridge in 15 minutes . the length of the bridge is ? | "speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 15 minutes = 50 / 18 * 15 * 60 = 2500 m answer is d" | a = 10 * 1000
b = a / const_60
c = b * 15
|
a ) $ 60,000 , b ) $ 65,000 , c ) $ 70,000 , d ) $ 75,000 , e ) $ 80,000 | c | subtract(divide(divide(subtract(multiply(12, 47500), add(multiply(5, 36000), multiply(4, 45000))), 3), const_1000), 5) | in a company of 12 employees , 5 employees earn $ 36000 , 4 employees earn $ 45000 , and the 3 highest - paid employees earn the same amount . if the average annual salary for the 12 employees is $ 47500 , what is the annual salary for each of the highest - paid employees ? | 5 * 36,000 + 4 * 45,000 + 3 x = 12 * 47,500 3 x = 570,000 - 180,000 - 180,000 3 x = 210,000 x = 70,000 the answer is c . | a = 12 * 47500
b = 5 * 36000
c = 4 * 45000
d = b + c
e = a - d
f = e / 3
g = f / 1000
h = g - 5
|
a ) a ) 40 % , b ) b ) 10 % , c ) c ) . 4 % , d ) d ) 50 % , e ) e ) 12 % | a | multiply(divide(2000, 5000), const_100) | dames school has 2000 boys and 5000 girls . what is the percentage increase from boys to total attendance and girls to total attendance ? | ratio of boys to total attendance ( 2 / 7 ) ratio of girls to total attendance ( 5 / 7 ) percentage increase is ( difference / initial quantity ) * 100 ( 2 / 7 ) / ( 5 / 7 ) * 100 = 40 % correct answer is a | a = 2000 / 5000
b = a * 100
|
a ) 3 kmph , b ) 1.5 kmph , c ) 3.5 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph | c | divide(add(divide(48, 4), divide(76, 4)), const_2) | a man swims downstream 76 km and upstream 48 km taking 4 hours each time ; what is the speed of the current ? | "explanation : 76 - - - 4 ds = 19 ? - - - - 1 48 - - - - 4 us = 12 ? - - - - 1 s = ? s = ( 19 - 12 ) / 2 = 3.5 answer : option c" | a = 48 / 4
b = 76 / 4
c = a + b
d = c / 2
|
a ) 20 , b ) 21 , c ) 35 , d ) 23 , e ) 24 | 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 | "solution let the toatl number of workers be x . then 8000 x = ( 16000 x 7 ) + 6000 ( x - 7 ) x = 35 . answer c" | a = 16000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
a ) 7 / 12 , b ) 1 / 6 , c ) 1 / 8 , d ) 5 / 17 , e ) 12 / 7 | a | add(divide(1, 3), divide(1, 4)) | n = 1 / 3 + 1 / 4 what is the value of n ? | to sum 1 / 3 and 1 / 4 multiply both sides by 12 12 * n = 12 * 1 / 3 + 12 * 1 / 4 12 * n = 4 + 3 n = 7 / 12 ans : a | a = 1 / 3
b = 1 / 4
c = a + b
|
a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(1101, 549), multiply(55, const_2)), const_1) | how many even multiples of 55 are there between 549 and 1101 ? | "550 = 10 * 55 1100 = 20 * 55 the even multiples are 55 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ." | a = 1101 - 549
b = 55 * 2
c = a / b
d = c + 1
|
a ) 9 and 8 , b ) 5 and 10 , c ) 7 and 9 , d ) 7 and 8 , e ) 10 and 8 | d | add(56, 15) | sum of two numbers prime to each other is 15 and their l . c . m . is 56 . what are the numbers ? | "as two numbers are prime , only options satisfy ie option a and d and c but option d will make the product of numbers i . e 56 answer : d" | a = 56 + 15
|
a ) 264 . , b ) 428 , c ) 0 , d ) 462 , e ) 642 | c | multiply(divide(add(84, multiply(divide(add(28, 84), 6), 6)), 3), divide(add(28, 84), 6)) | if - 3 x + 2 y = 28 and 3 x + 6 y = 84 , what is the product of x and y ? | "given - 3 x + 2 y = 28 - - - eq 1 3 x + 6 y = 84 - - eq 2 sum both eqns we get 8 y = 112 = > y = 14 sum 2 y in eq 1 = > - 3 x - 28 = 28 . = > x = 0 now xy = 0 * 14 = 0 . option c is correct answer ." | a = 28 + 84
b = a / 6
c = b * 6
d = 84 + c
e = d / 3
f = 28 + 84
g = f / 6
h = e * g
|
a ) 1 / 3 , b ) 1 / 30 , c ) 1 / 10 , d ) 1 / 45 , e ) 1 / 60 | e | divide(const_1, multiply(divide(5, 5), 60)) | before leaving home for the town of madison , pete checks a map which shows that madison is 5 inches from his current location , gardensquare . pete arrives in madison 5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ? | "pete covered 5 * 60 = 300 miles which correspond to 5 inches on the map - - > scale in inches per mile is 5 / 300 = 1 / 60 . answer : e ." | a = 5 / 5
b = a * 60
c = 1 / b
|
['a ) 10 ( √ 3 - 1 )', 'b ) 5', 'c ) 10 ( √ 2 - 1 )', 'd ) 5 ( √ 3 - 1 )', 'e ) 5 ( √ 2 - 1 )'] | d | divide(subtract(multiply(10, sqrt(const_3)), 10), const_2) | a sphere is inscribed in a cube with an edge of 10 . what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere ? | as sphere is inscribed in cube then = > the edge of the cube equals to the diameter of a sphere - - > diameter = 10 diagonal of a cube equals to diagonal = square _ root ( 10 ^ 2 + 10 ^ 2 + 10 ^ 2 ) = 10 √ 3 . half of diagonal - diameter is gap between the vertex of a cube and the surface of the sphere = ( 10 √ 3 - 10 ) / 2 = 5 ( √ 3 - 1 ) answer : d | a = math.sqrt(3)
b = 10 * a
c = b - 10
d = c / 2
|
a ) 80 , b ) 60 , c ) 50 , d ) 30 , e ) 10 | d | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 20) | by selling 20 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 20 120 % - - - ? 80 / 120 * 20 = 30 answer : d" | a = 100 + 20
b = 100 - 20
c = 1 / b
d = a * c
e = 1 / d
f = e * 20
|
a ) 1.012526 , b ) 0.012625 , c ) 0.12526 , d ) 0.12625 , e ) 0.12725 | b | divide(25.25, 2000) | 25.25 / 2000 is equal to : | "25.25 / 2000 = 2525 / 200000 = 0.012625 answer : b" | a = 25 / 25
|
a ) 10 % , b ) 11 % , c ) 12 % , d ) 10.5 % , e ) none | b | divide(multiply(const_100, subtract(subtract(696.30, divide(660, 2)), divide(660, 2))), divide(660, 2)) | on a sum of money , the simple interest for 2 years is rs . 660 , while the compound interest is rs . 696.30 , the rate of interest being the same in both the cases . the rate of interest is | "solution difference in c . i and s . i for 2 years = rs ( 696.30 - 660 ) = rs . 36.30 . s . i for one years = rs 330 . s . i on rs . 330 for 1 year = rs . 36.30 rate = ( 100 x 36.30 / 330 x 1 ) % = 11 % . answer b" | a = 660 / 2
b = 696 - 30
c = 660 / 2
d = b - c
e = 100 * d
f = 660 / 2
g = e / f
|
a ) $ 9.00 , b ) $ 9.50 , c ) $ 10.50 , d ) $ 13.50 , e ) $ 20.00 | c | divide(subtract(multiply(15000, 20), add(22500, multiply(8, 15000))), 15000) | a certain tire company can produce tires at a cost of $ 22500 per batch plus $ 8 per tire . the company can sell tires to the wholesaler at a cost of $ 20 per tire . if a batch of 15000 tires is produced and sold , what is the company ’ s profit per tire ? | cp ( 15000 tires ) = $ 22500 + ( $ 8 × 15000 ) = $ 142500 sp ( 15000 tires ) = $ 20 × 15000 = $ 300000 profit = sp - cp = $ 300000 - $ 142500 = $ 157500 profit / tire = $ 157500 / 15000 = $ 10.50 answer c | a = 15000 * 20
b = 8 * 15000
c = 22500 + b
d = a - c
e = d / 15000
|
a ) 7 / 5 , b ) 6 / 11 , c ) 2 / 5 , d ) 1 / 3 , e ) 4 / 15 | c | subtract(add(subtract(const_1, divide(const_1, const_3)), divide(const_2, add(const_1, const_4))), subtract(const_1, divide(const_1, const_3))) | at a certain company , two - thirds of the employees had cell phones and two - fifths had pagers . if one - third of the employees had neither cell phones nor pagers then what fraction of the employees had both cell phones and pagers ? | let there be 15 employees cell phone = 10 pagers = 6 neither pager nor cell phone = 5 so , 10 = 10 + 6 - x or , x = 6 so , fraction of the employees had both cell phones and pagers = 6 / 15 = > 2 / 5 answer : c | a = 1 / 3
b = 1 - a
c = 1 + 4
d = 2 / c
e = b + d
f = 1 / 3
g = 1 - f
h = e - g
|
a ) $ 182 , b ) $ 191 , c ) $ 200 , d ) $ 209 , e ) $ 219 | a | subtract(add(200, divide(multiply(200, 30), const_100)), divide(multiply(add(200, divide(multiply(200, 30), const_100)), 30), const_100)) | the original price of a suit is $ 200 . the price increased 30 % , and after this increase , the store published a 30 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 30 % off the increased price , how much did these consumers pay for the suit ? | "given the foregoing discussion , it may be obvious now the trap - mistake answer is ( c ) . even if you can ’ t remember the correct thing to do , at the very least , learn to spot the trap ! the multiplier for a 30 % increases is 1 + 0.30 = 1.3 , and the multiplier for a 30 % decrease is 1 – 0.30 = 0.70 , so the combined change is 1.3 * 0.7 = 0.91 , 91 % percent of the original , or a 9 % decreases . now , multiply $ 200 * 0.91 = $ 182 . answer = ( a ) ." | a = 200 * 30
b = a / 100
c = 200 + b
d = 200 * 30
e = d / 100
f = 200 + e
g = f * 30
h = g / 100
i = c - h
|
a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) none of these | c | divide(81, add(22, 5)) | a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 81 km downstream | "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 81 km time taken = distance / speed = 81 / 27 = 3 hours . answer : option c" | a = 22 + 5
b = 81 / a
|
a ) 2 days , b ) 6 days , c ) 12 days , d ) 20 days , e ) none of these | d | divide(multiply(5, 4), subtract(5, 4)) | a man can do a piece of work in 5 days , but with the help of his son he can do it in 4 days . in what time can the son do it alone ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 4 − 1 / 5 ) = ( 5 − 4 ) / 20 = 1 / 20 so son will do whole work in 20 days answer : d" | a = 5 * 4
b = 5 - 4
c = a / b
|
a ) 27 , b ) 64 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(149, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(149, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 149 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 149 x = 4 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 4 = 64 answer b" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 149 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 149 / n
p = i * o
q = h - p
|
a ) 60 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 150 % | c | multiply(subtract(divide(divide(divide(add(const_100, 36), const_100), subtract(const_1, divide(subtract(25, 20), 25))), divide(subtract(const_100, 15), const_100)), const_1), const_100) | a dealer offers a cash discount of 15 % and still makes a profit of 36 % when he further allows 25 articles to be sold at the cost price of 20 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | given cash discount - 15 % profit - 36 % items sold - 25 price sold at = list price of 20 assume list price = $ 10 total invoice = $ 200 - 15 % cash discount = $ 170 let cost price of 25 items be x so total cost = 25 * x given the shopkeeper had a profit of 36 % 25 * x * 136 / 100 = 170 or x = $ 5 which means his products were listed at $ 10 which is a 100 % markup over $ 5 answer c | a = 100 + 36
b = a / 100
c = 25 - 20
d = c / 25
e = 1 - d
f = b / e
g = 100 - 15
h = g / 100
i = f / h
j = i - 1
k = j * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | sqrt(divide(32, 2)) | if x + y = 2 and x 2 y 3 + y 2 x 3 = 32 , what is the value of xy ? | "xy = 4 as x + y = 2 x 2 y 3 + y 2 x 3 = 32 x 2 y 2 ( y + x ) = 32 substituting x + y x 2 y 2 = 16 xy = 4 answer : d" | a = 32 / 2
b = math.sqrt(a)
|
a ) 1.5 , b ) 3.5 , c ) 4.5 , d ) 2.5 , e ) 5.5 | c | divide(add(divide(45, 3), divide(72, 3)), const_2) | a man swims downstream 72 km and upstream 45 km taking 3 hours each time ; what is the speed of the current ? | "72 - - - 3 ds = 24 ? - - - - 1 45 - - - - 3 us = 15 ? - - - - 1 s = ? s = ( 24 - 15 ) / 2 = 4.5 answer : c" | a = 45 / 3
b = 72 / 3
c = a + b
d = c / 2
|
a ) a ) 40 , b ) b ) 45 , c ) c ) 50 , d ) d ) 55 , e ) e ) 111 | e | subtract(choose(4, 4), choose(6, 4)) | there are 4 books on a shelf , of which 3 are paperbacks and 6 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ? | "approach 1 at - least 1 paper back = total - no paper back 9 c 4 - 6 c 4 = 111 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 3 c 1 * 6 c 3 + 3 c 2 * 6 c 2 + 3 c 3 * 6 c 1 = 111 answer is e" | a = math.comb(4, 4)
b = math.comb(6, 4)
c = a - b
|
a ) 10 , b ) 12 , c ) 37 , d ) 29 , e ) 22 | b | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 20), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 10), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 20 % and then increased by 10 % , then the net change in the price will be : | explanation : let the original price be rs . 100 . decreased by 20 % = 80 then increased 10 % on rs 80 = 80 + 8 = 88 net change in price = 100 - 88 = 12 answer : b | a = 100 * 20
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 10
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) $ 9.40 , b ) $ 12.57 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10 | b | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent computer b . computer b would , however , require 20 hours more than computer a to do the job . if either computer a , or computer b were rented the total cost to rent the computer would be $ 880.00 . what would be the approximate hourly charge to rent computer b ? | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 880 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 880 / 70 ~ 12.57 b" | a = 40 / 100
b = 4 + 3
c = 40 / 100
d = 20 / 4
e = d / 100
f = c + e
g = b + f
h = a + g
|
a ) 7 km , b ) 6 km , c ) 9 3 / 8 km , d ) 9 km , e ) 5 km | c | add(multiply(add(7, divide(const_1, const_2)), subtract(add(5, divide(const_3, 4)), add(4, divide(const_1, const_2)))), const_2) | two men a and b start from place x walking at 4 ½ kmph and 5 ¾ kmph respectively . how many km apart they are at the end of 7 ½ hours if they are walking in the same direction ? | "rs = 5 ¾ - 4 ½ = 1 ¼ t = 7 ½ h . d = 5 / 4 * 15 / 2 = 75 / 8 = 9 3 / 8 km answer : c" | a = 1 / 2
b = 7 + a
c = 3 / 4
d = 5 + c
e = 1 / 2
f = 4 + e
g = d - f
h = b * g
i = h + 2
|
a ) $ 16.32 , b ) $ 18.00 , c ) $ 21.60 , d ) $ 34.56 , e ) $ 28.80 | d | multiply(divide(subtract(const_100, 20), const_100), multiply(0.60, 72)) | the regular price per can of a certain brand of soda is $ 0.60 . if the regular price per can is discounted 20 percent when the soda is purchased in 24 - can cases , what is the price of 72 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.8 ) ( $ 0.60 ) , or $ 0.48 therefore , the price of 72 cans of soda at the discounted price would be ( 72 ) ( $ 0.48 ) = 34.56 answer : d ." | a = 100 - 20
b = a / 100
c = 0 * 60
d = b * c
|
a ) 22.2 % , b ) 36 % , c ) 80 % , d ) 122.2 % , e ) none | d | multiply(subtract(divide(subtract(const_100, 20), 36), const_1), const_100) | the cost price of an article is 36 % of the marked price . calculate the gain percent after allowing a discount of 20 % . | "sol . let marked price = rs . 100 . then , c . p . = rs . 36 . s . p = rs . 80 . â ˆ ´ gain % = [ 44 / 36 * 100 ] % = 122.2 % . answer d" | a = 100 - 20
b = a / 36
c = b - 1
d = c * 100
|
a ) 50 , b ) 60 , c ) none of these , d ) 70 , e ) 100 | c | subtract(divide(subtract(multiply(3, 1300), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1300), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | the cost of 2 chairs and 3 tables is rs . 1300 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "c 100 2 c + 3 t = 1300 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 100 = > t - c = 100" | a = 3 * 1300
b = 2 * 3
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
h = 3 * 1300
i = 2 * 3
j = h - i
k = 3 * 3
l = 2 * 2
m = k - l
n = j / m
o = 2 * n
p = 3 - o
q = p / 3
r = g - q
|
a ) 2 : 6 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 5 | d | inverse(subtract(const_1, divide(const_3.0, 5))) | he ratio between the sale price and the cost price of an article is 5 : 1 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . x and s . p . = rs . 5 x . then , gain = rs . 4 x required ratio = 2 x : x = 2 : 1 answer : d" | a = 3 / 0
b = 1 - a
c = 1/(b)
|
a ) 6 , b ) 8 , c ) 9 , d ) 15 , e ) 7 | d | add(5, const_1) | the average of first five multiples of 5 is ? | "average = 5 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 75 / 5 = 15 . answer : d" | a = 5 + 1
|
a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 1584 , e ) 1625 | d | multiply(divide(divide(19008, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 19008 , how much amount per month would he have spent on clothes and transport ? | "∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 19008 ⁄ 12 = 1584 answer d" | a = 100 - 60
b = a * 50
c = b / 100
d = c / 100
e = 19008 / d
f = 3 * 4
g = e / f
h = 100 - 60
i = h * 50
j = i / 100
k = j / 100
l = g * k
|
a ) 22 , b ) 88 , c ) 90 , d ) 21 , e ) 54 | e | multiply(divide(15, const_1000), const_3600) | express 15 mps in kmph ? | "15 * 18 / 5 = 54 kmph answer : e" | a = 15 / 1000
b = a * 3600
|
a ) 1 , b ) - 2 , c ) 12 , d ) - 3 , e ) 4 | c | subtract(subtract(subtract(100, 30), add(100, 30)), 30) | if | 5 x - 30 | = 100 , then find the sum of the values of x ? | "| 5 x - 30 | = 100 5 x - 30 = 100 or 5 x - 30 = - 100 5 x = 130 or 5 x = - 70 x = 26 or x = - 14 sum = 26 - 14 = 12 answer is c" | a = 100 - 30
b = 100 + 30
c = a - b
d = c - 30
|
a ) 30 , b ) 31 , c ) 45 , d ) 40 , e ) 89 | d | multiply(2, 5) | the class mean score on a test was 60 , and the standard deviation was 5 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ? | "1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 65 to 55 , where 65 is within sd above the mean and 55 within 1 sd below the mean 2 sd = 5 twice = 10 from the the mean , which is 70 to 40 , where 70 is within 2 sd above the mean and 40 is within 2 sd below the mean . answer = d" | a = 2 * 5
|
a ) 10 % , b ) 60 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(80, 50), 50), const_100) | a man buy a book in rs 50 & sale it rs 80 . what is the rate of profit ? ? ? | "cp = 50 sp = 80 profit = 80 - 50 = 30 % = 30 / 50 * 100 = 60 % answer : b" | a = 80 - 50
b = a / 50
c = b * 100
|
a ) s . 8500 , b ) s . 8900 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | b | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 71,200 at 16 2 / 3 % per annum for 9 months . | "p = rs . 71200 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 71,200 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8900 answer is b ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) 4 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(51, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if resultant is trebled , it becomes 51 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 51 = > 2 x + 9 = 17 = > x = 4 answer : option a" | a = 3 * 9
b = 51 - a
c = 3 * 2
d = b / c
|
a ) 113.33 . , b ) 103.33 , c ) 93.33 , d ) 83.33 , e ) 73.33 | a | divide(add(multiply(divide(40, add(40, 40)), 80), multiply(divide(40, add(40, 40)), 60)), divide(add(40, 40), const_60)) | a car was driving at 80 km / h for 40 minutes , and then at 60 km / h for another 40 minutes . what was its average speed ? | "driving at 80 km / h for 40 minutes , distance covered = 80 * 2 / 3 = 53.33 km driving at 90 km / h for 40 minutes , distance covered = 90 * 2 / 3 = 60 km average speed = total distance / total time = 113.33 / 1 = 113.33 km / h answer : 113.33 km / h" | a = 40 + 40
b = 40 / a
c = b * 80
d = 40 + 40
e = 40 / d
f = e * 60
g = c + f
h = 40 + 40
i = h / const_60
j = g / i
|
a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50 | d | subtract(multiply(subtract(45, 20), const_3), multiply(subtract(45, 30), const_3)) | in x game of billiards , x can give y 20 points in 45 and he can give z 30 points in 45 . how many points can y give z in x game of 100 ? | "x scores 45 while y score 40 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 25 = 60 . in x game of 100 points , y gives ( 100 - 60 ) = 40 points to c . d" | a = 45 - 20
b = a * 3
c = 45 - 30
d = c * 3
e = b - d
|
a ) 20 , b ) 25 , c ) 28 , d ) 30 , e ) 35 | e | divide(subtract(680, multiply(multiply(2, 5), 5)), add(multiply(2, 5), multiply(const_2, const_4))) | bill spends two days driving from point a to point b . on the first day , he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day . if during the two days he drove a total of 680 miles over the course of 18 hours , what was his average speed on the second day , in miles per hour ? | first recognize that we have two pieces of information regarding the time bill spent driving each day . on day 1 , bill drove 2 hours longer than he drove on day 2 . so , let x = # of driving hours on day 2 then x + 2 = # of driving hours on day 1 bill drove a total of 18 hours so , x + ( x + 2 ) = 18 simplify : 2 x + 2 = 18 solve , x = 8 so , bill drove 10 hours on day 1 and he drove 8 hours on day 2 now let ' s solve the question by starting with a word equation . let x = speed driven on day 2 so , x + 5 = speed driven on day 1 ( distance traveled on day 1 ) + ( distance traveled on day 2 ) = 680 distance = ( rate ) ( time ) we get : ( x + 5 ) ( 10 ) + ( x ) ( 8 ) = 680 expand : 10 x + 50 + 8 x = 680 simplify : 18 x + 50 = 680 18 x = 630 x = 35 ( mph ) answer : e | a = 2 * 5
b = a * 5
c = 680 - b
d = 2 * 5
e = 2 * 4
f = d + e
g = c / f
|
a ) 2107 , b ) 2197 , c ) 2187 , d ) 2177 , e ) 2167 | b | multiply(power(add(const_1, divide(30, const_100)), 3), 1000) | the expenditure of an woman increase consistently by 30 % per year . if her present expenditure is rs . 1000 then what will her expenditure be after 3 years ? | explanation : expenditure = 1000 x 1.3 x 1.3 x 1.3 = 2197 answer : option b | a = 30 / 100
b = 1 + a
c = b ** 3
d = c * 1000
|
a ) 39 , b ) 66 , c ) 288 , d ) 19 , e ) 17 | a | multiply(divide(6.3, 25), const_100) | what percent of 6.3 kg is 25 gms ? | "explanation : required percentage = ( 25 / 63000 * 100 ) % = 39 / 100 % = 0.39 % answer : a ) . 39 %" | a = 6 / 3
b = a * 100
|
a ) 515 , b ) 480 , c ) 750 , d ) 650 , e ) 560 | a | subtract(subtract(multiply(45, 9), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 9 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 9 * 45 - 15 * 22 + 20 * 22 = 450 - 330 + 440 = 515 answer is a" | a = 45 * 9
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
|
a ) 20.0 % , b ) 13.5 % , c ) 11.5 % , d ) 14.5 % , e ) 21.5 % | a | divide(const_100, 5) | at what rate percent per annum will a sum of money double in 5 years . | let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 5 ) ] % = 20.0 % per annum . answer : a | a = 100 / 5
|
a ) 30 % , b ) 44 % , c ) 50 % , d ) 55 % , e ) 60 % | b | multiply(subtract(multiply(divide(add(const_100, 20), const_100), divide(add(const_100, 20), const_100)), const_1), const_100) | the percentage increase in the area of rectangle , if each of its side is increased by 20 % is ? | "let original length = x meters original breadth = y meters original area = xy m ^ 2 new length = 120 x / 100 new breadth = 120 y / 100 = 6 y / 5 new area = 6 x / 5 * 6 y / 5 = 36 xy / 25 m ^ 2 increase percent = 11 xy / 25 * 1 / xy * 100 = 44 % answer is b" | a = 100 + 20
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
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