options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 84 , b ) 96 , c ) 100 , d ) 108 , e ) 120 | d | multiply(add(add(6, subtract(6, 2)), 2), add(subtract(add(6, subtract(6, 2)), divide(6, const_2)), 2)) | roy is now 6 years older than julia and half of that amount older than kelly . if in 2 years , roy will be twice as old as julia , then in 2 years what would be roy ’ s age multiplied by kelly ’ s age ? | "r = j + 6 = k + 3 r + 2 = 2 ( j + 2 ) ( j + 6 ) + 2 = 2 j + 4 j = 4 r = 10 k = 7 in 2 years ( r + 2 ) ( k + 2 ) = 12 * 9 = 108 the answer is d ." | a = 6 - 2
b = 6 + a
c = b + 2
d = 6 - 2
e = 6 + d
f = 6 / 2
g = e - f
h = g + 2
i = c * h
|
a ) 171 , b ) 281 , c ) 391 , d ) 591 , e ) 601 | c | add(multiply(divide(add(30, 40), const_2), add(subtract(40, 30), const_1)), add(divide(subtract(40, 30), const_2), const_1)) | if x is equal to the sum of the integers from 30 to 40 , inclusive , and y is the number of even integers from 30 to 40 , inclusive , what is the value of x + y ? | "sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 30 + ( 11 - 1 ) * 1 } = 11 * 35 = 385 = x number of even number = ( 40 - 30 ) / 2 + 1 = 6 = y x + y = 385 + 6 = 391 c" | a = 30 + 40
b = a / 2
c = 40 - 30
d = c + 1
e = b * d
f = 40 - 30
g = f / 2
h = g + 1
i = e + h
|
a ) $ 9.40 , b ) $ 11.30 , c ) $ 12.14 , d ) $ 17.80 , e ) $ 22.10 | c | 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 $ 850.00 . 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 = 850 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 = 850 / 70 ~ 12.14 c" | 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 ) 4,800 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(100, add(const_2, const_3)), const_3), multiply(divide(100, add(const_2, const_3)), const_4)) | a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 100 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for sides , and knowing c = 100 , you can deduce the following a = 60 b = 80 60 x 80 = 4,800 a is the answer ." | a = 2 + 3
b = 100 / a
c = b * 3
d = 2 + 3
e = 100 / d
f = e * 4
g = c * f
|
a ) 2 / 9 , b ) 4 / 19 , c ) 6 / 29 , d ) 8 / 39 , e ) 10 / 49 | b | multiply(divide(add(const_4, const_4), 19), divide(subtract(add(const_4, const_4), 1), subtract(19, 1))) | 19 balls are numbered 1 to 19 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 9 / 19 p ( 2 nd ball is also even ) = 8 / 18 p ( both balls are even ) = 9 / 19 * 8 / 18 = 4 / 19 the answer is b ." | a = 4 + 4
b = a / 19
c = 4 + 4
d = c - 1
e = 19 - 1
f = d / e
g = b * f
|
a ) 9.55 , b ) 6.25 , c ) 6.75 , d ) 8.05 , e ) 7.25 | d | divide(subtract(360, multiply(12, 4.5)), 38) | in the first 12 overs of a cricket game , the run rate was only 4.5 . what should be the run rate in the remaining 38 overs to reach the target of 360 runs ? | required run rate = 360 - ( 4.5 x 12 ) / 38 = 306 / 38 = 8.05 option d | a = 12 * 4
b = 360 - a
c = b / 38
|
a ) 63.07 kmph , b ) 54.28 kmph , c ) 62.02 kmph , d ) 64.02 kmph , e ) none of these | b | divide(add(add(40, 30), add(60, 60)), add(add(1, divide(30, const_60)), 2)) | a car during its journey travels 1 hour at a speed of 40 kmph , another 30 minutes at a speed of 60 kmph , and 2 hours at a speed of 60 kmph . the average speed of the car is | first car travels 1 hrs at speed of 40 kmph distance = 40 x 1 = 40 m then car travels 30 min at a speed of 60 kmph distance = 30 min at speed of 60 kmph distance = 60 x 1 / 2 = 30 m at last it travels 2 hours at speed of 60 kmph distance = 60 x 2 = 120 m total distance = 40 + 30 + 120 = 190 total time = 1 + 1 / 2 + 2 = 3.50 average speed of the car = 190 / 3.25 = 54.28 answer : b | a = 40 + 30
b = 60 + 60
c = a + b
d = 30 / const_60
e = 1 + d
f = e + 2
g = c / f
|
a ) 1 : 3 , b ) 9 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | b | divide(subtract(27, 26), subtract(26, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 26 seconds , what is the ratio of their speeds ? | "let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 26 s = ( 27 x + 17 y ) / ( x + y ) = 26 = ( 27 x + 17 y ) / = 26 ( x + y ) = x = 9 y = x / y = 9 / 1 answer : b" | a = 27 - 26
b = 26 - 17
c = a / b
|
a ) 272258 , b ) 272358 , c ) 300048 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 56), const_100) | 5358 x 56 = ? | "5358 x 51 = 5358 x ( 50 + 6 ) = 5358 x 50 + 5358 x 6 = 267900 + 32148 = 300048 . c )" | a = 5358 / 56
b = a * 100
|
a ) 12 min , b ) 20 min , c ) 25 min , d ) 30 min , e ) 15 min | e | multiply(const_60, divide(subtract(80, 60), 80)) | excluding stoppages , the speed of a bus is 80 kmph and including stoppages , it is 60 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 20 km less . time taken to cover 20 km = ( 20 / 80 x 60 ) min = 15 min answer : e" | a = 80 - 60
b = a / 80
c = const_60 * b
|
a ) 19 , b ) 35 , c ) 20 , d ) 32 , e ) 23 | e | subtract(multiply(27, const_3), multiply(29, const_2)) | the average age of a , b and c is 27 years . if the average age of a and c is 29 years , what is the age of b in years ? | age of b = age of ( a + b + c ) â € “ age of ( a + c ) = 27 ã — 3 â € “ 29 ã — 2 = 81 â € “ 58 = 23 years e | a = 27 * 3
b = 29 * 2
c = a - b
|
a ) 1192 , b ) 1191 , c ) 1118 , d ) 1190 , e ) 1112 | d | divide(multiply(subtract(const_100, 15), 1400), const_100) | a man buys a cycle for rs . 1400 and sells it at loss of 15 % . what is the selling price of the cycle ? | "explanation : s . p = 85 % of rs . 1400 ⇒ rs . ( 8510085100 × 1400 ) = rs . 1190 . answer : d" | a = 100 - 15
b = a * 1400
c = b / 100
|
a ) 96 min , b ) 72 min , c ) 60 min , d ) 48 min , e ) 40 min | c | multiply(divide(log(const_3), log(add(const_1, divide(25, const_100)))), 12) | a strain of bacteria reproduces @ 25 % every 12 min . in how much time will it triple itself ? ? | the original question is : a strain of bacteria reproduces at the rate of 25 % every 12 min . in how much time will it triple itself ? 1.25 ^ x = 3 - - > x = ~ 5 - - > five 12 minute periods = 60 minutes . answer : c . | a = math.log(3)
b = 25 / 100
c = 1 + b
d = math.log(c)
e = a / d
f = e * 12
|
a ) $ 660.67 , b ) $ 646.67 , c ) $ 666.67 , d ) $ 500.56 , e ) $ 600.24 | c | multiply(divide(3, add(add(6, 3), 5)), 1000) | a person want to give his money of $ 1000 to his 3 children a , b , c in the ratio 6 : 4 : 5 what is the a + b share ? | "a ' s share = 1000 * 6 / 15 = $ 400 b ' s share = 1000 * 4 / 15 = $ 266.67 a + b = $ 666.67 answer is c" | a = 6 + 3
b = a + 5
c = 3 / b
d = c * 1000
|
a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12 | a | subtract(10, const_4) | in a group of cows and hens , the number of legs are 10 more than twice the number of heads . the number of cows is : | "let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 10 2 x = 10 x = 5 . answer : a" | a = 10 - 4
|
a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 4 % | a | floor(multiply(subtract(divide(9, 63), divide(7, 60)), const_100)) | a survey was sent to 60 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 60 ) = x / 100 x = 12 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 12 ) % = 2 % answer is a" | a = 9 / 63
b = 7 / 60
c = a - b
d = c * 100
e = math.floor(d)
|
a ) 10 , b ) 15 , c ) 17 , d ) 5 , e ) 9 | e | divide(multiply(subtract(18, 6), 12), add(12, 4)) | 12 persons can complete the work in 18 days . after working for 6 days , 4 more persons added to complete the work fast . in how many days they will complete the work ? | "total work = 12 * 18 = 216 units . after 6 days , work finished = 6 * 12 = 72 units . remaining units = 216 - 72 = 144 units . remaining days = 144 / ( 2 + 4 ) = 9 days the option is e" | a = 18 - 6
b = a * 12
c = 12 + 4
d = b / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(add(add(add(divide(2, 3), divide(divide(2, 3), 3)), divide(divide(divide(2, 3), 3), 3)), divide(divide(divide(divide(2, 3), 3), 3), 3)), divide(divide(divide(divide(divide(2, 3), 3), 3), 3), 3)) | if 2 / 3 of the air in a tank is removed with each stroke of a vacuum pump , how many strokes does it take before less than 1 % of the original amount of air in the tank remains ? | "my approach is to find more or less good number and plug in . lets take 90 as the total volume of an air , and the question asks after how many stokes there will be less than 0,9 air in a tank , if each stroke takes 2 / 3 of an air . so after he 1 strokethe volume will be 30 ( 90 - 60 = 30 ) , and so on 2 stroke - 30 - 20 = 10 3 stroke - 10 - 6,7 = 3,3 4 stroke - 3,3 - 2,2 = 1,1 5 stroke - 1,1 - minus something that gets us less than 0,9 for sure . so the answer is : after 5 strokes there will be less than 1 % of air left in a tank . optiond ." | a = 2 / 3
b = 2 / 3
c = b / 3
d = a + c
e = 2 / 3
f = e / 3
g = f / 3
h = d + g
i = 2 / 3
j = i / 3
k = j / 3
l = k / 3
m = h + l
n = 2 / 3
o = n / 3
p = o / 3
q = p / 3
r = q / 3
s = m + r
|
a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 70 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 35))), 30)) | mahesh can do a piece of work in 35 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 35 days = 20 * 1 / 35 = 4 / 7 remaining work = 1 - 4 / 7 = 3 / 7 3 / 7 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 7 / 3 = 70 days answer is e" | a = 1 / 35
b = 20 * a
c = 1 - b
d = c / 30
e = 1 / d
|
a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 . | e | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.0) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.0 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.0 = 6 * 0 + 2 * 7.2 + 8 - 6.0 = 0 + 14.4 + 2 16.4 answer e" | a = 2 * 7
b = 4 / 10
c = 8 - 4
d = a + c
e = d - 6
|
a ) $ 14 , b ) $ 16 , c ) $ 22 , d ) $ 24 , e ) $ 26 | b | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(42, const_12)) | a parking garage rents parking spaces for $ 10 per week or $ 42 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 42 = 504 $ 520 - 504 = 16 ans b" | a = 3 + 2
b = 10 * a
c = b + 2
d = c * 10
e = 42 * 12
f = d - e
|
a ) 110 , b ) 125 , c ) 75 , d ) 170 , e ) 195 | b | add(lcm(lcm(8, 12), lcm(15, 20)), 5) | what is the least number which when divided by 8 , 12 , 15 and 20 leaves in each case a remainder of 5 | "lcm of 8 , 12 , 15 and 20 = 120 required number = 120 + 5 = 125 answer : option b" | a = math.lcm(8, 12)
b = math.lcm(15, 20)
c = math.lcm(a, b)
d = c + 5
|
a ) 1 / 2 , b ) 2 , c ) 2 / 9 , d ) 3 , e ) 1 / 6 | c | sqrt(divide(4, 81)) | if xy = 4 , x / y = 81 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 4 - - - > x = 4 / y - ( i ) x / y = 81 - - - > replacing ( i ) here - - - > 4 / ( y ^ 2 ) = 81 - - - > y ^ 2 = 4 / 81 - - - > y = 2 / 9 or - 2 / 9 the question states that x and y are positive integers . therefore , y = 2 / 9 is the answer . answer c ." | a = 4 / 81
b = math.sqrt(a)
|
a ) 1050 , b ) 1220 , c ) 1250 , d ) 1060 , e ) 1110 | a | divide(672, subtract(const_1, divide(multiply(6, 6), const_100))) | a fellow borrowed a certain sum of money at 6 % per annum at simple interest and in 6 years the interest amounted to rs . 672 less than the sum lent . what was the sum lent ? | "p - 672 = ( p * 6 * 6 ) / 100 p = 1050 answer : a" | a = 6 * 6
b = a / 100
c = 1 - b
d = 672 / c
|
a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | c | add(subtract(21, 6), subtract(21, 6)) | the 7 th grade french and spanish classes each have 21 students in them , and there are 6 students in the 7 th grade taking both languages . if everyone in the 7 th grade is in at least one of the two language classes , how many students are taking just one of the two classes ? | french class + spanish class = 42 students . six students are in both classes so they were counted twice . the number of students taking just one class is 42 - 2 ( 6 ) = 30 . the answer is c . | a = 21 - 6
b = 21 - 6
c = a + b
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a ) 7 , b ) 8 , c ) 5 , d ) 9 , e ) 6 | c | subtract(divide(factorial(subtract(divide(14, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(14, const_2), const_1)) | a company that ships boxes to a total of 14 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter ) | "back - solving is the best way to solve this problem . you basically need 14 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - c" | a = 14 / 2
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = math.factorial(2)
f = d * e
g = c / f
h = 14 / 2
i = h - 1
j = g - i
|
a ) 160 , b ) 180 , c ) 220 , d ) 225 , e ) none | a | multiply(520, divide(multiply(const_4, const_2), add(add(add(const_4, const_3), multiply(const_4, const_2)), 11))) | a , b , c hired a car for rs . 520 and used it for 7,8 and 11 hours respectively . hire charges paid by b were : | sol . a : b : c = 7 : 8 : 11 . hire charges paid by b = rs . [ 520 * 8 / 26 ] = rs . 160 answer a | a = 4 * 2
b = 4 + 3
c = 4 * 2
d = b + c
e = d + 11
f = a / e
g = 520 * f
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | c | multiply(add(add(power(3, const_3), multiply(add(const_1, const_4), 3)), multiply(3, const_2)), inverse(multiply(3, const_2))) | maximun number of identical pieces ( of same size ) of a cake by making only 3 cuts ? | since cake is a 3 d object , it will have length , breadth , height . . . first cut across length now we will get 2 pieces then cut across breadth we will get 4 parts in total now . . . now cut across height , finally we will get 8 parts . . . answer : c | a = 3 ** 3
b = 1 + 4
c = b * 3
d = a + c
e = 3 * 2
f = d + e
g = 3 * 2
h = 1/(g)
i = f * h
|
a ) 30 prime number , b ) 29 prime number , c ) 27 prime number , d ) 31 prime number , e ) none of these | a | multiply(const_4, 67) | how many prime numbers exist in 67 x 353 x 1110 ? | "solution : = ( 2 x 3 ) 7 x ( 5 x 7 ) 3 x 1110 = 27 x 37 x 53 x 73 x 1110 thus , there are ( 7 + 7 + 3 + 3 + 10 ) = 30 prime number answer : a" | a = 4 * 67
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a ) 50 miles , b ) 60 miles , c ) 72 miles , d ) 22 miles , e ) 62 miles | b | multiply(12, 5) | a car travels at a speed of 12 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 12 + 12 + 12 + 12 + 12 = 5 * 12 = 60 miles correct answer b" | a = 12 * 5
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a ) 30 , b ) 45 , c ) 50 , d ) 60 , e ) 65 | b | divide(multiply(250, 36), subtract(250, 50)) | a hostel had provisions for 250 men for 36 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 36 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 36 250 × 36 = 200 x 5 × 36 = 4 x x = 5 × 9 = 45 answer b" | a = 250 * 36
b = 250 - 50
c = a / b
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a ) 22 , b ) 34 , c ) 27 , d ) 32 , e ) 25 | b | subtract(negate(10), multiply(subtract(4, 6), divide(subtract(4, 6), subtract(3, 4)))) | 3 , 4 , 6 , 10 , 18 , ( . . . ) | "explanation : 3 3 × 2 - 2 = 4 4 × 2 - 2 = 6 6 × 2 - 2 = 10 10 × 2 - 2 = 18 18 × 2 - 2 = 34 answer : option b" | a = negate - (
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a ) 87 , b ) 16 , c ) 10 , d ) 76 , e ) 59 | e | subtract(multiply(add(10, const_1), add(4, 15)), multiply(10, 15)) | the average of runs of a cricket player of 10 innings was 15 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | average after 11 innings = 19 required number of runs = ( 19 * 11 ) - ( 15 * 10 ) = 209 - 150 = 59 . answer : e | a = 10 + 1
b = 4 + 15
c = a * b
d = 10 * 15
e = c - d
|
a ) 26.7 days , b ) 77 days , c ) 20 days , d ) 88 days , e ) 44 days | a | divide(multiply(25, 16), 15) | 16 men can complete a piece of work in 25 days . in how many days can 15 men complete that piece of work ? | "16 * 25 = 15 * x = > x = 26.7 days answer : a" | a = 25 * 16
b = a / 15
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a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | a | divide(subtract(sqrt(add(multiply(multiply(280, 5), const_4), power(5, const_2))), 5), const_2) | a bus trip of 280 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ? | "the time is the distance / speed . the time difference is 1 hour . 280 / v - 280 / ( v + 5 ) = 1 280 ( v + 5 ) - 280 v = ( v ) ( v + 5 ) 1400 = ( v ) ( v + 5 ) 35 * 40 = ( v ) ( v + 5 ) v = 35 mph the answer is a ." | a = 280 * 5
b = a * 4
c = 5 ** 2
d = b + c
e = math.sqrt(d)
f = e - 5
g = f / 2
|
a ) 12 , b ) 15 , c ) c is elder than a , d ) data inadequate , e ) none | b | multiply(15, const_1) | the total age of a and b is 15 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 15 â € ¹ = â € º a - c = 15 . answer b" | a = 15 * 1
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['a ) 127 cm ^ 2', 'b ) 125 cm ^ 2', 'c ) 120 cm ^ 2', 'd ) 102 cm ^ 2', 'e ) none of them'] | c | divide(subtract(power(divide(46, const_2), const_2), power(17, const_2)), const_2) | if the diagonal of a rectangle is 17 cm long and its perimeter is 46 cm , find the area of the rectangle . | let length = x and breadth = y . then , 2 ( x + y ) = 46 or x + y = 23 and x ^ 2 + y ^ 2 = ( 17 ) ^ 2 = 289 . now , ( x + y ) ^ 2 = ( 23 ) ^ 2 < = > ( x ^ 2 + y ^ 2 ) + 2 xy = 529 < = > 289 + 2 xy = 529 = xy = 120 area = xy = 120 cm ^ 2 answer is c . | a = 46 / 2
b = a ** 2
c = 17 ** 2
d = b - c
e = d / 2
|
a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | e | add(multiply(divide(multiply(divide(11, multiply(subtract(0.75, 0.3), const_2)), const_2), const_10), const_2), multiply(divide(11, multiply(subtract(0.75, 0.3), const_2)), const_2)) | the toll for crossing a certain bridge is $ 0.75 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 11.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day , what is the least value of x for which this driver can save money by using the sticker ? | option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than 0.6 x 14 + 13 ( 21.4 ) . . . it ' s pretty close so let ' s see if one more day will make it worth it . . . if we raise the number of days to 15 , the sticker option looks like a better deal . . . 1.5 x 15 ( 22.5 ) vs 0.6 x 15 + 13 ( 22 ) . answer : e | a = 0 - 75
b = a * 2
c = 11 / b
d = c * 2
e = d / 10
f = e * 2
g = 0 - 75
h = g * 2
i = 11 / h
j = i * 2
k = f + j
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a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 40 | e | subtract(60, divide(subtract(160, 60), 5)) | mother , her daughter and her grand child weighs 160 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | "mother + daughter + child = 160 kg daughter + child = 60 kg mother = 160 - 60 = 100 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 100 = 20 kg so now daughter = 120 - ( mother + child ) = 160 - ( 100 + 20 ) = 40 kg answer : e" | a = 160 - 60
b = a / 5
c = 60 - b
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a ) 39 , b ) 35 , c ) 20 , d ) 40.5 , e ) 41.5 | c | add(subtract(100, multiply(17, 5)), 5) | a batsman makes a score of 100 runs in the 17 th inning and thus increases his averages by 5 . what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 100 then average run after 17 th inning = ( 16 x + 100 ) / 17 ( 16 x + 100 ) / 17 = x + 5 = > 16 x + 100 = 17 x + 85 = > x = 15 x = 15 ; average after 17 th inning = 15 + 5 = 20 answer : c" | a = 17 * 5
b = 100 - a
c = b + 5
|
a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(5, const_2), choose(5, const_3)), choose(5, const_4)), choose(5, 5)), power(const_2, 5)) | a fair coin is tossed 5 times . what is the probability of getting more heads than tails in 5 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ." | a = math.comb(5, 2)
b = math.comb(5, 3)
c = a + b
d = math.comb(5, 4)
e = c + d
f = math.comb(5, 5)
g = e + f
h = 2 ** 5
i = g / h
|
a ) 14,15 , b ) 15,16 , c ) 17,18 , d ) 18,19 , e ) none | c | divide(add(35, const_1), const_2) | the difference between the squares of two consecutive numbers is 35 . the numbers are | "explanation : let the numbers be a and ( a + 1 ) ( a + 1 ) 2 − a 2 = 35 ⇒ a 2 + 2 a + 1 − a 2 = 35 ⇒ 2 a = 34 ⇒ 2 a = 34 or a = 17 the numbers are 17 & 18 . correct option : c" | a = 35 + 1
b = a / 2
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 % | c | subtract(100, 60) | john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 60 . what is the percent decrease ? | the is always the difference between our starting and ending points . in this case , it ’ s 100 – 60 = 40 . the “ original ” is our starting point ; in this case , it ’ s 100 . ( 40 / 100 ) * 100 = ( 0.4 ) * 100 = 40 % . c | a = 100 - 60
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a ) s . 300 , b ) s . 360 , c ) s . 396 , d ) s . 368 , e ) s . 323 | c | divide(561, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | if rs . 561 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 561 * 2 / 17 = rs . 66 b ' s share = 561 * 3 / 17 = rs . 99 c ' s share = 561 * 12 / 17 = rs . 396 . answer : c" | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 561 / f
|
a ) 52 , b ) 65 , c ) 78 , d ) 91 , e ) 104 | d | divide(multiply(divide(multiply(56, 3.8), 1.6), 3.9), 5.7) | a certain car can travel 56 kilometers on a liter of fuel . if the fuel tank ’ s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers ) | fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 / 1.6 mile speed ( km / hour ) = d / t = 56 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 56 x 3.9 x 3.8 / 5.7 x 1.6 = 78 miles / hour ans : d ps : i felt the factors were easy to cancel out , so did n ' t require much rounding off = 56 x 3.9 x 3.8 / 5.7 x 1.6 = 91 d | a = 56 * 3
b = a / 1
c = b * 3
d = c / 5
|
a ) 2.5 , b ) 5 , c ) 10 , d ) 7 , e ) 9 | d | divide(14, const_2) | if the length of the longest chord of a certain circle is 14 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 14 = 2 * 7 so radius of the circle = 7 correct answer - d" | a = 14 / 2
|
a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 5 , e ) 2 : 1 | d | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(6, const_100)), const_1000)), subtract(multiply(add(const_1, divide(13, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | at a certain organisation , the number of male members went up by 13 % in the year 2001 from year 2000 , and the number of females members went down by 6 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ? | men increase by 13 % = = > 1.13 m = males in 2001 women decrease by 6 % = = > 0.94 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.13 m + 0.94 f = 1.012 * ( m + f ) 1.13 m + 0.94 f = 1.012 m + 1.012 f 1.13 m - 1.012 m = 1.012 f - 0.94 f 0.108 m = 0.072 f m / f = ( 0.072 ) / ( 0.118 ) = 72 / 118 = 3 / 5 answer = ( d ) | a = 1 / 2
b = 1 + a
c = b * 1000
d = 6 / 100
e = 1 - d
f = e * 1000
g = c - f
h = 13 / 100
i = 1 + h
j = i * 1000
k = 1 / 2
l = 1 + k
m = l * 1000
n = j - m
o = g / n
|
a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 1,000 | e | divide(40, divide(2, 50)) | in a certain pond , 40 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 ? | "this is a rather straight forward ratio problem . 1 . 40 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 40 / x thus , x = 1000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . e" | a = 2 / 50
b = 40 / a
|
a ) 20 , b ) 26 , c ) 23 , d ) 25 , e ) 21 | e | subtract(35, add(add(8, const_2), 8)) | set a of 8 positive integers may have the same element and have 35 . and set b of 8 positive integers must have different elements and have 35 . when m and n are the greatest possible differences between 35 and other elements ’ sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 35 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 28 and 35 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 7 . so , 28 - 7 = 21 . the correct answer is e . | a = 8 + 2
b = a + 8
c = 35 - b
|
a ) 23.89 , b ) 72.9 , c ) 33.62 , d ) 78.3 , e ) 79.3 | c | subtract(subtract(50, multiply(const_3, const_3)), multiply(divide(subtract(50, multiply(const_3, const_3)), 50), multiply(const_3, const_3))) | a vessel of capacity 50 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 50 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 50 - 9 ) / 50 ] ^ 2 ( 50 ) = 33.62 litres . answer : option c" | a = 3 * 3
b = 50 - a
c = 3 * 3
d = 50 - c
e = d / 50
f = 3 * 3
g = e * f
h = b - g
|
a ) $ 600 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | c | divide(multiply(10, subtract(60, 40)), divide(20, const_100)) | on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 20 percent . what was the total rent the motel actually charged for that night ? | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 20 % . 2 x = 200 = > x = 1000 answer c" | a = 60 - 40
b = 10 * a
c = 20 / 100
d = b / c
|
a ) 25 , b ) 30 , c ) 45 , d ) 60 , e ) 65 | b | add(multiply(divide(subtract(90, add(add(10, 10), 10)), add(add(3, 2), 1)), 2), divide(subtract(90, add(add(10, 10), 10)), add(add(3, 2), 1))) | total of the ages of a , b ahd c at present is 90 years . 10 years ago , the ratio of their ages was 1 : 2 : 3 . what is the age of b at present | explanation : let their ages 10 years ago is x , 2 x and 3 x years . 10 + 2 x + 10 + 3 x + 10 = 90 hence x = 10 b � s present age = ( 2 x + 10 ) = 30 years answer : option b | a = 10 + 10
b = a + 10
c = 90 - b
d = 3 + 2
e = d + 1
f = c / e
g = f * 2
h = 10 + 10
i = h + 10
j = 90 - i
k = 3 + 2
l = k + 1
m = j / l
n = g + m
|
a ) - 0.53 , b ) 1.0 , c ) 1.07 , d ) 1.71 , e ) 2.71 | a | divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 8.0) | ( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) / 8.0 = ? | "dove straight into calculation ( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) / 7.0 = - 0.53 answer a" | a = 1 * 9
b = negate - (
c = 2 * 6
d = b / c
|
a ) 542 , b ) 540 , c ) 675 , d ) 829 , e ) 279 | c | multiply(multiply(inverse(subtract(add(add(divide(const_1, 12), divide(const_1, 20)), divide(const_1, 45)), divide(const_1, 15))), const_3), 20) | two pipes a and b can separately fill a tank in 12 and 20 minutes respectively . a third pipe c can drain off 45 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ? | 1 / 12 + 1 / 20 - 1 / x = 1 / 15 x = 15 15 * 45 = 675 answer : c | a = 1 / 12
b = 1 / 20
c = a + b
d = 1 / 45
e = c + d
f = 1 / 15
g = e - f
h = 1/(g)
i = h * 3
j = i * 20
|
a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability e that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . total of 15 favorable outcomes e = 15 / 36 . answer : c ." | a = 12 + 3
b = 1 / 6
c = b ** 2
d = a * c
|
a ) 1900 , b ) 19,303 , c ) 19 , 356.732 , d ) 19,502 , e ) 19,909 | a | add(100, const_1) | if 100 < x < 199 and 20 < y < 100 , then the product xy can not be equal to : | "correct answer : ( a ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 100 * 20 . the largest value must be less than 199 * 100 . this means that 2000 < xy < 19,900 . ( a ) is outside of this range , so it is not a possible product of xy ." | a = 100 + 1
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a ) - 3 , b ) - 2 , c ) - 1 , d ) 0 , e ) 1 | b | subtract(2, multiply(2, 2)) | if d | | d ' , d : 2 x - y = 1 and d ' : ( a - 1 ) x + 2 y = x - 2 , then find the value of a . | d : 2 x − y = 1 , d ’ : ( a − 1 ) x + 2 y = x − 2 d : 2 x − y = 1 , d ′ : ( a − 1 ) x + 2 y = x − 2 d ∥ d ′ ⇒ md = m ′ dd ∥ d ′ ⇒ md = md ′ 2 x − y = 1 → y = 2 x − 1 → md = 22 x − y = 1 → y = 2 x − 1 → md = 2 ( a − 1 ) x + 2 y = x − 2 → 2 y = x ( 1 − a + 1 ) − 2 → 2 y = x ( 2 − a ) − 2 → ( a − 1 ) x + 2 y = x − 2 → 2 y = x ( 1 − a + 1 ) − 2 → 2 y = x ( 2 − a ) − 2 → y = x × ( 2 − a ) 2 − 1 → m ′ d = ( 2 − a ) 2 y = x × ( 2 − a ) 2 − 1 → md ′ = ( 2 − a ) 2 md = m ′ d → ( 2 − a ) 2 = 2 → 2 − a = 4 → a = − 2 answer : b - 2 | a = 2 * 2
b = 2 - a
|
a ) $ 3,750 , b ) $ 5,600 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 22,500 | e | multiply(2500, power(const_3, divide(28, divide(112, 8)))) | money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 2500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | solution : money compounded annually at x % triples in 112 / x years . we need to find the final amount of 2500 at the end of the 28 years . . compounded annually at 8 % by putting the value in formula it will give us = 2500 ( 1.08 ) ^ 28 there must be a relation between these two conditions ? x = 8 % so money will triple in 112 / 8 = 14 years so money will be 3 ^ 2 = 9 times in 28 years so 2500 * 9 = 22500 answer : e | a = 112 / 8
b = 28 / a
c = 3 ** b
d = 2500 * c
|
a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) 1 / 2 | b | multiply(divide(subtract(3, const_1), multiply(subtract(3, const_1), 3)), divide(multiply(subtract(3, const_1), const_2), multiply(subtract(3, const_1), 3))) | let a be the event that a randomly selected two digit number is divisible by 3 and let b be the event that a randomly selected two digit number is divisible by 2 . what is p ( a and b ) ? | "p ( a and b ) = 1 / 3 * 1 / 2 = 1 / 6 the answer is b ." | a = 3 - 1
b = 3 - 1
c = b * 3
d = a / c
e = 3 - 1
f = e * 2
g = 3 - 1
h = g * 3
i = f / h
j = d * i
|
a ) 30 , b ) 29 , c ) 34 , d ) 21 , e ) 20 | c | add(25, subtract(14, 5)) | replace x with the appropriate number in 4 , 5 , 14 , 15 , 24 , 25 , x | c list of consecutive numbers that has an ' f ' in the spelling | a = 14 - 5
b = 25 + a
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a ) 32.8 , b ) 32.4 , c ) 32.1 , d ) 18 , e ) 32.9 | d | add(divide(circumface(3.5), const_2), multiply(3.5, const_2)) | the radius of a semi circle is 3.5 cm then its perimeter is ? | "diameter = 7 cm 1 / 2 * 22 / 7 * 7 + 7 = 18 answer : d" | a = circumface / (
b = a + 2
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a ) $ 10 , b ) $ 12 , c ) $ 13.20 , d ) $ 15 , e ) $ 16.80 | d | divide(subtract(multiply(71, 3), multiply(69, 2)), subtract(multiply(3, 3), multiply(2, 2))) | if bill can buy 3 pairs of jeans and 2 shirts for $ 69 or 2 pairs of jeans and 3 shirts for $ 71 , how much does one shirt cost ? | "3 j + 2 s = 69 2 j + 3 s = 71 - - - - - - - - - - - - - - - - 5 j + 5 s = 140 - - - - ( divide by 5 ) - - - > j + s = 28 3 j + 2 s = j + 2 ( j + s ) = j + 56 = 69 - - - > j = 13 3 * 13 + 2 s = 69 39 + 2 s = 69 2 s = 30 s = 15 answer : d" | a = 71 * 3
b = 69 * 2
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
|
a ) 94.2 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 165), 7) | 3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 165 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 165 / 7 = 94.2 kgs answer : a" | a = 1 + 3
b = a * 165
c = b / 7
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | divide(subtract(divide(120, const_2), sqrt(subtract(multiply(divide(120, const_2), divide(120, const_2)), multiply(const_4, 800)))), const_2) | the area of a rectangular field is equal to 800 square meters . its perimeter is equal to 120 meters . find the width of this rectangle . | "l * w = 800 : area , l is the length and w is the width . 2 l + 2 w = 120 : perimeter l = 60 - w : solve for l ( 60 - w ) * w = 800 : substitute in the area equation w = 20 and l = 40 correct answer d" | a = 120 / 2
b = 120 / 2
c = 120 / 2
d = b * c
e = 4 * 800
f = d - e
g = math.sqrt(f)
h = a - g
i = h / 2
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | divide(20, subtract(3, const_1)) | lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 20 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 20 3 c = c + 20 c = 10 the answer is c ." | a = 3 - 1
b = 20 / a
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a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 % | a | multiply(const_100, divide(subtract(subtract(const_100, 48), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | in a company , 48 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | "the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 52 % of all employees are women , so non - union women are 52 % - 18 % = 34 % 40 % of all employees are non - union . the percent of non - union employees who are women is 34 % / 40 % = 85 % the answer is a ." | a = 100 - 48
b = 70 / 100
c = 60 * b
d = 60 - c
e = a - d
f = 100 - 60
g = e / f
h = 100 * g
|
a ) 1.66 % , b ) 1.96 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(500, 0.8), multiply(700, 1.8)), const_1000) | by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 500 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 500 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 500 ) / 100 + ( 1.8 * 700 ) / 100 = 16.6 gms % of liquid x in resultant mixture = ( 16.6 / 1000 ) * 100 = 1.66 % a" | a = 500 * 0
b = 700 * 1
c = a + b
d = c / 1000
|
a ) 45 , b ) 50 , c ) 88 , d ) 52 , e ) 12 | a | multiply(9, 5) | what number has a 5 : 1 ratio to the number 9 ? | "5 : 1 = x : 9 x = 45 answer : a" | a = 9 * 5
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a ) 1 , b ) 2 , c ) 46 , d ) 50 , e ) 97 | b | divide(subtract(multiply(100, divide(99, const_100)), multiply(100, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | there are 100 employees in a room . 99 % are president . how many managers must leave the room to bring down the percentage of president to 98 % ? | "we have 99 presidents and 1 director . that 1 director to compose 2 % of the total number of people , there must be 50 people in the room , hence 50 presidents must leave . answer : b ." | a = 99 / 100
b = 100 * a
c = 98 / 100
d = 100 * c
e = b - d
f = 98 / 100
g = 1 - f
h = e / g
|
a ) 900 , b ) 930 , c ) 990 , d ) 993 , e ) none of these | d | subtract(multiply(3000, power(add(const_1, divide(10, const_100)), 3)), 3000) | what is the compound interest paid on a sum of rs . 3000 for the period of 3 years at 10 % per annum . | "solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 / 100 = 11 interest % for 3 rd year = 10 + 10 % ( 10 + 11 ) = 10 + 2.1 = 12.1 total % of interest = 10 + 11 + 12.1 = 33.1 total interest = 33.1 % 3000 = 3000 * ( 33.1 / 100 ) = 993 answer d" | a = 10 / 100
b = 1 + a
c = b ** 3
d = 3000 * c
e = d - 3000
|
a ) 35 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 15 | a | subtract(50, multiply(divide(50, const_100), 10)) | how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 10 - percent solution ? | "3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 10 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 10 x = 15 since you need a 15 l solution , you must evaporate 35 of the original 50 l solution to get a 10 % solution . answer is a ." | a = 50 / 100
b = a * 10
c = 50 - b
|
a ) 120 rs , b ) 150 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | a | divide(subtract(divide(multiply(multiply(6000, 6), 2), const_100), divide(multiply(multiply(6000, 4), 2), const_100)), 2) | a person borrows rs . 6000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 6000 * 6 * 2 ) / 100 ] - [ ( 6000 * 4 * 2 ) / 100 ] 720 - 480 = 240 gain in 1 year = ( 240 / 2 ) = 120 rs answer : a" | a = 6000 * 6
b = a * 2
c = b / 100
d = 6000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 4.1 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 7.2 | e | divide(divide(1440, const_1000), divide(multiply(12, const_60), const_3600)) | a person crosses a 1440 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 1440 / ( 12 x 60 ) m / sec = 2 m / sec . converting m / sec to km / hr = 2 x ( 18 / 5 ) km / hr = 7.2 km / hr . answer : e" | a = 1440 / 1000
b = 12 * const_60
c = b / 3600
d = a / c
|
a ) 62 , b ) 68 , c ) 76 , d ) 84 , e ) 96 | d | add(multiply(negate(14), power(subtract(add(3, 2), 3), 2)), 140) | an object thrown directly upward is at a height of h feet after t seconds , where h = - 14 ( t - 3 ) ^ 2 + 140 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ? | "we see that h will be a maximum h = 140 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 14 ( 5 - 3 ) ^ 2 + 140 = - 14 ( 4 ) + 140 = 84 the answer is d ." | a = negate * (
b = 3 + 2
c = b - 3
d = c ** 2
e = a + d
|
a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100) | rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage x . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % x = ( 97 / 556 ) * 100 = 17.4 % answer : b" | a = 2 * 10
b = 2 * 10
c = b + 1
d = a * c
e = 25 / 100
f = e + 1
g = 10 / f
h = 15 + 2
i = g * h
j = d + i
k = 653 - j
l = 2 * 10
m = 2 * 10
n = m + 1
o = l * n
p = 25 / 100
q = p + 1
r = 10 / q
s = 15 + 2
t = r * s
u = o + t
v = k / u
w = v * 100
|
a ) 18 hours , b ) 22 hours , c ) 25 hours , d ) 26 hours , e ) 20 hours | e | subtract(multiply(divide(90, add(add(2, 3), 4)), 4), multiply(divide(90, add(add(2, 3), 4)), 2)) | the amount of time that three people worked on a special project was in the ratio of 2 to 3 to 4 . if the project took 90 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 2 * 90 / 9 = 20 hours b = 3 * 90 / 9 = 30 hours c = 4 * 90 / 9 = 40 hours c is the hardest worker and a worked for the least number of hours . so the difference is 40 - 20 = 20 hours . answer : e" | a = 2 + 3
b = a + 4
c = 90 / b
d = c * 4
e = 2 + 3
f = e + 4
g = 90 / f
h = g * 2
i = d - h
|
a ) 14 , b ) 16 , c ) 21 , d ) 24 , e ) 27 | d | divide(336, divide(subtract(462, 336), 9)) | a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 9 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 9 ) x solve for r , r = 33 33 - 9 = 24 mpg d" | a = 462 - 336
b = a / 9
c = 336 / b
|
a ) 1 : 7 , b ) 2 : 9 , c ) 1 : 9 , d ) 4 : 49 , e ) 3 : 4 | d | divide(circle_area(2), circle_area(7)) | the ratio of the radius of two circles is 2 : 7 , and then the ratio of their areas is ? | "r 1 : r 2 = 2 : 7 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 4 : 49 answer : d" | a = circle_area / (
|
a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many q miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance traveled by tom when karen is 4 miles ahead of him : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) q = 24 miles . d is the correct answer ." | a = 45 / 60
b = a * 4
c = b + 4
d = 60 / 60
e = 45 / 60
f = d - e
g = c / f
h = g - 4
|
a ) 1560 , b ) 1561 , c ) 1559 , d ) 1557 , e ) none of the above | b | multiply(add(add(multiply(multiply(4, 5), add(7, 3)), multiply(4, 5)), 3), 7) | there are 7 thieves . they stole diamonds from a diamond merchant and ran away . while running , night sets in and they decide to rest in the jungle . when everybody was sleeping , two of them woke up and decided to divide the diamonds equally among themselves . but when they divided the diamonds equally , one diamond is left . so they woke up the 3 rd thief and tried to divide the diamonds equally again but still one diamond was left . then they woke up the 4 th thief to divide the diamonds equally again , and again one diamond was left . this happened with the 5 th and 6 th thief – one diamond was still left . finally , they woke up the 7 th thief and this time the diamonds were divided equally . how many diamonds did they steal in total ? | we need a number that is a multiple of 7 that will give a remainder of 1 when divided by 2 , 3 , 4 , 5 , and 6 . the least common multiple of these numbers is 60 . so , we need a multiple of 7 that is 1 greater than a multiple of 60 . answer b | a = 4 * 5
b = 7 + 3
c = a * b
d = 4 * 5
e = c + d
f = e + 3
g = f * 7
|
a ) 1 / 140 , b ) 1 / 180 , c ) 11 / 24 , d ) 10 / 15 , e ) 57 / 120 | c | add(add(divide(1, 4), divide(1, 8)), divide(1, 12)) | in a race where 15 cars are running , the chance that car x will win is 1 / 4 , that y will win is 1 / 8 and that z will win is 1 / 12 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 4 + 1 / 8 + 1 / 12 = 11 / 24 answer : c" | a = 1 / 4
b = 1 / 8
c = a + b
d = 1 / 12
e = c + d
|
a ) 2532.93 , b ) 2552.26 , c ) 2524.23 , d ) 4098.25 , e ) 2512.23 | d | subtract(multiply(26,000, power(add(const_1, divide(9, const_100)), 20)), 26,000) | find the compound interest on $ 26,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 26000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 26000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 30098.25 ci . = $ ( 30098.25 - 26000 ) = $ 4098.25 answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 26 * 0
e = d - 26
|
a ) 1 / 3 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | a | divide(subtract(divide(40, const_100), divide(50, const_100)), subtract(divide(20, const_100), divide(50, const_100))) | some of 50 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 40 % . what fraction of the original paint was replaced ? | "40 % is 20 % - points above 20 % and 10 % - points below 50 % . thus the ratio of 20 % - solution to 50 % - solution is 1 : 2 . 1 / 3 of the original paint was replaced . the answer is a ." | a = 40 / 100
b = 50 / 100
c = a - b
d = 20 / 100
e = 50 / 100
f = d - e
g = c / f
|
a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46 | e | divide(subtract(multiply(const_3, 50), 50), const_3) | shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 50 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 46 hence answer is e" | a = 3 * 50
b = a - 50
c = b / 3
|
a ) 17 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(78, 78)), multiply(const_4, const_2))), const_10) | how many three digit numbers r are divisible by 78 or 91 ? | "the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as well - - 13 * 7 * 6 = 91 * 6 = 546 so total number of multiples r - - 11 + 9 - 1 = 19 . hence answer is 19 . b" | a = 1000 - 10
b = 78 * 78
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 1800 , b ) 1000 , c ) 1200 , d ) 1400 , e ) 1600 | c | divide(multiply(divide(80, const_100), multiply(multiply(60, 150), 10)), 60) | the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 60 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | "volume of pool = 60 * 150 * 10 cu . ft , 80 % full = 60 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 60 * 150 * 10 * 0.8 / 60 min = 1200 min c" | a = 80 / 100
b = 60 * 150
c = b * 10
d = a * c
e = d / 60
|
a ) 35 % , b ) 20 % , c ) 17 % , d ) 18 % , e ) none of these | a | subtract(subtract(add(const_100, 50), multiply(add(const_100, 50), divide(10, const_100))), const_100) | a shopkeeper labeled the price of his articles so as to earn a profit of 50 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ? | "explanation : let the cp of the article = rs . 100 . then labeled price = rs . 150 . sp = rs . 150 - 10 % of 150 = rs . 150 - 15 = rs . 135 . gain = rs . 135 â € “ rs . 100 = rs . 35 therefore , gain / profit percent = 35 % . answer : option a" | a = 100 + 50
b = 100 + 50
c = 10 / 100
d = b * c
e = a - d
f = e - 100
|
a ) 191 , b ) 193 , c ) 195 , d ) 212 , e ) 213 | b | divide(770, 4) | to be considered for “ movie of the year , ” a film must appear in at least 1 / 4 of the top - 10 - movies lists submitted by the cinematic academy ’ s 770 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for “ movie of the year ” ? | "total movies submitted are 770 . as per question we need to take 1 / 4 of 770 to be considered for top 10 movies = 192.25 approximate the value we 193 imo option b is the correct answer . . ." | a = 770 / 4
|
a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 2 / 3 | d | divide(add(divide(const_12, 3), divide(const_12, 2)), add(const_12, const_12)) | 1 / 3 of girls , 1 / 2 of boys go to canteen . what factor and total number of classmates go to canteen . | let total girls be 3 . . . and total boys be 2 . . . so total girls going to canteen = 1 boys going to canteen = 1 total no . of classmates = 5 ( only suppose . . . . . . to find out the factor of students going to canteen . . . ) factor of classmates going to canteen = 2 / 5 . answer : d | a = 12 / 3
b = 12 / 2
c = a + b
d = 12 + 12
e = c / d
|
a ) 149 , b ) 190 , c ) 128 , d ) 178 , e ) 190 | c | divide(subtract(200, multiply(9, 8)), subtract(10, 9)) | suraj has a certain average of runs for 9 innings . in the 10 th innings he scores 200 runs thereby increasing his average by 8 runs . what is his average after the 10 th innings ? | "to improve his average by 8 runs per innings he has to contribute 9 x 8 = 72 runs for the previous 8 innings . thus , the average after the 9 th innings = 200 - 72 = 128 . answer : c" | a = 9 * 8
b = 200 - a
c = 10 - 9
d = b / c
|
a ) 18 , b ) 36 , c ) 42 , d ) 68 , e ) 70 | b | divide(multiply(divide(multiply(subtract(subtract(220, divide(multiply(70, 220), const_100)), 12), const_100), subtract(const_100, 40)), 40), const_100) | at a particular graduation party with 220 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 12 males did not bring gifts to the party , how many females did bring gifts ? | "the correct method total = 220 . . 70 % of 220 = 154 got gifts . . 66 did not get gift , out of which 12 are males , so remaining 60 - 12 = 54 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 54 , 100 % = 54 * 100 / 60 = 90 . . ans 40 % of 90 = 36 b" | a = 70 * 220
b = a / 100
c = 220 - b
d = c - 12
e = d * 100
f = 100 - 40
g = e / f
h = g * 40
i = h / 100
|
a ) 4 , b ) 3 , c ) 1 , d ) 6 , e ) 8 | b | subtract(add(multiply(divide(1, 4), 2), multiply(divide(1, 5), 2)), divide(5, 2)) | if ( 1 / 5 ) ^ m * ( 1 / 4 ) ^ 3 = 1 / ( 2 * ( 10 ) ^ 3 ) , then m = | "1 / 5 ^ m * 1 / 4 ^ 3 = 2 * 1 / 2 ^ 3 * 5 ^ 3 * 8 2 ^ 3 * 5 ^ 3 * 8 / 4 ^ 3 = 5 ^ m or 5 ^ 3 = 5 ^ m m = 3 b" | a = 1 / 4
b = a * 2
c = 1 / 5
d = c * 2
e = b + d
f = 5 / 2
g = e - f
|
a ) 16 % , b ) 25 % , c ) 32 % , d ) 34 % , e ) 52 % | d | multiply(divide(subtract(64, 42), 64), const_100) | in town x , 64 percent of the population are employed , and 42 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 42 are employed males , hence 22 % are employed females . ( employed females ) / ( total employed people ) = 22 / 64 = 34 % answer : d ." | a = 64 - 42
b = a / 64
c = b * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | c | multiply(3, 1) | if n divided by 7 has a remainder of 1 , what is the remainder when 3 times n is divided by 7 ? | "as per question = > n = 7 p + 1 for some integer p hence 3 n = > 21 q + 3 = > remainder = > 6 for some integer q alternatively = > n = 2 > 3 n = > 3 = > 3 divided by 7 will leave a remainder 3 hence c" | a = 3 * 1
|
a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | c | divide(34, 53) | how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 34 / 53 ? | "there is no certain way to solve 2 unknown with 1 equation . the best way is to look at the question and retrospect the most efficient way . in this question , a and b are only positive integers . so that is a big relief . now , we can start with putting a = 1,2 , . . and so on till the time we are confident about one of the options . so , we start with a = 1 , we get b as - ve . out a = 2 , we get b as 6 . yes ( now ( a , b ) = ( 2,6 ) . we can directly see that ( a , b ) = ( 6,2 ) will also satisfy . so we have 2 possible solutions ) a = 3 , we get b as 3 . yes ( now we have 3 possible solutions ) a = 4 , we get b as fraction . out a = 5 , we get b again as some fraction . out a = 6 already taken . we have a , b options left . c , d , e are out . a is 6 . to have 6 as the answer , we will need one more pair like 2,6 and one more solution where a = b . when a = b , we have only 1 solution = 3 . so , one more solution , where a = b is not possible . so , answer will be c ." | a = 34 / 53
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(divide(18, 3), const_1) | in a house , there are 4 birds , 3 dogs , and 18 cats living with the human family . if the total number of feet in the house are 74 more than the number of heads in the house , how many family members are there ? | explanation : let number of family members be x . then , total number of feet = 2 x 4 + 4 x 3 + 4 x 18 + 2 x = 2 x + 92 . total number of heads = 4 + 3 + 18 + x = 25 + x . therefore ( 2 x + 92 ) = ( 25 + x ) + 74 or x = 7 . answer : d | a = 18 / 3
b = a + 1
|
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 3 , d ) 1 / 8 , e ) none of above | c | divide(circle_area(divide(33.33, const_2)), const_2) | what will be the fraction of 33.33 % | "explanation : it will 33.33 * 1 / 100 = 1 / 3 answer : option c" | a = 33 / 33
b = circle_area / (
|
a ) 168 ° , b ) 228 ° , c ) 156 ° , d ) 224 ° , e ) none of these | c | multiply(divide(multiply(add(multiply(multiply(6, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 9) | the ratio of the adjacent angles of a parallelogram is 6 : 9 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 6 x + 9 x = 180 ° or , 15 x = 180 ° or , x = 12 ° hence the angles of the parallelogram are 72 ° and 108 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y + 6 y + 7 y + 12 y = 360 ° or , 30 y = 360 ° or , y = 12 ° hence the angles of the quadrilateral are 60 ° , 72 , 84 ° and 144 ° will be 72 ° + 84 ° = 156 ° answer : c" | a = 6 * 2
b = a * 10
c = b + 100
d = c * 2
e = 5 + 6
f = e + 7
g = f + 12
h = d / g
i = h * 9
|
a ) 66 , 78 , b ) 70 , 82 , c ) 94 , 106 , d ) 84 , 96 , e ) none | d | multiply(12, add(const_3, const_4)) | the h . c . f . of two numbers is 12 and their difference is 12 . the numbers are | "solution out of the given numbers , the two with h . c . f . 12 and difference 12 are 84 and 96 answer d" | a = 3 + 4
b = 12 * a
|
['a ) 8', 'b ) 2', 'c ) 6', 'd ) 4', 'e ) 3'] | d | divide(16, const_4) | if the length and breadth of a room are increased by y feet each , the perimeter increases by 16 feet . find y | 2 ( l + b ) = x ; 2 ( l + y + b + y ) = x + 16 ; 2 ( l + b ) + 4 y = x + 16 ; x + 4 y = x + 16 ; y = 4 answer : d | a = 16 / 4
|
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