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 ) 11 , b ) 10 , c ) 12 , d ) 17 , e ) 19 | b | divide(10, const_1) | the overall age of x and y is 10 year greater than the overall age of y and z . z is how many decades younger that x ? | "b 10 ( x + y ) â € “ ( y + z ) = 10 x â € “ z = 10" | a = 10 / 1
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a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | e | divide(choose(15, 3), choose(add(15, 15), 3)) | a bag contains 15 red jellybeans and 15 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 3 / 10 answer : option e" | a = math.comb(15, 3)
b = 15 + 15
c = math.comb(b, 3)
d = a / c
|
a ) 746.77 , b ) 945.77 , c ) 946.77 , d ) 246.77 , e ) 926.77 | c | subtract(add(add(divide(multiply(divide(700, multiply(divide(12, const_100), 6)), 12), const_100), divide(700, multiply(divide(12, const_100), 6))), divide(multiply(add(divide(multiply(divide(700, multiply(divide(12, const_100), 6)), 12), const_100), divide(700, multiply(divide(12, const_100), 6))), 12), const_100)), divide(700, multiply(divide(12, const_100), 6))) | if the simple interest on a sum of money for 6 years at 12 % per annum is rs . 700 , what is the compound interest on the same sum at the rate and for the same time ? | "sum = ( 700 * 100 ) / ( 6 * 12 ) = rs . 972.22 c . i . on rs . rs . 972.22 for 6 years at 12 % = rs . 1 , 918.99 . = rs . 1 , 918.99 - 972.22 = rs . 946.77 answer : c" | a = 12 / 100
b = a * 6
c = 700 / b
d = c * 12
e = d / 100
f = 12 / 100
g = f * 6
h = 700 / g
i = e + h
j = 12 / 100
k = j * 6
l = 700 / k
m = l * 12
n = m / 100
o = 12 / 100
p = o * 6
q = 700 / p
r = n + q
s = r * 12
t = s / 100
u = i + t
v = 12 / 100
w = v * 6
x = 700 / w
y = u - x
|
a ) 17 years , b ) 20 years , c ) 29 years , d ) 10 years , e ) 12 years | b | divide(multiply(subtract(52, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 52 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 52 5 x = 50 = > x = 10 hence , b ' s age = 2 x = 20 years . answer : b" | a = 52 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 20 , b ) 60 , c ) 50 , d ) 70 , e ) 80 | c | add(divide(10, subtract(divide(125, const_100), const_1)), 10) | a man ' s age is 125 % of what it was 10 years ago , but 83 x 1 / 3 % of what it will be after 10 years . what is his present age ? | c 50 let the age before 10 years = x then , 125 x / 100 = x + 10 ⇒ 125 x = 100 x + 1000 ⇒ x = 1000 / 25 = 40 present age = x + 10 = 40 + 10 = 50 | a = 125 / 100
b = a - 1
c = 10 / b
d = c + 10
|
a ) 17 , b ) 19 , c ) 21 , d ) 24 , e ) 25 | b | subtract(20, const_1) | when average age of 20 members are 0 , how many members greater than 0 ? | "average of 20 numbers = 0 . sum of 20 numbers ( 0 x 20 ) = 0 . it is quite possible that 19 of these numbers may be positive and if their sum is a then 20 th number is ( - a ) answer is 19 ( b )" | a = 20 - 1
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a ) 89136 , b ) 31986 , c ) 68910 , d ) 53113 , e ) 68913 | e | multiply(divide(5301, 13), const_100) | 5301 x 13 = ? | "append 0 before and after : 053010 calculation : 1 x 3 + 0 = 3 ( take 3 as ones digit of the product ) 0 x 3 + 1 = 1 ( take 1 tens digit of the product ) 3 x 3 + 0 = 9 ( take 9 hundreds digit of the product ) 5 x 3 + 3 = 18 ( take 8 as thousands digit of the product , carry over 1 ) 0 x 3 + 5 = 5 ; 5 + 1 = 6 ( take 6 as ten thousands digits of the product ) so , 5301 x 13 = 68913 answer is e ." | a = 5301 / 13
b = a * 100
|
a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 7.2 hrs , e ) 9.2 hrs | a | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 9))) | a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 3 - 1 / 9 = 2 / 9 therefore the cistern will be filled in 9 / 2 hours or 4.5 hours . answer : a" | a = 1 / 3
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 26666 , b ) 26166 , c ) 29666 , d ) 22666 , e ) 26699 | a | divide(multiply(multiply(8, 25), multiply(8, const_1000)), multiply(const_1, const_60)) | a river 8 m deep and 25 m wide is flowing at the rate of 8 kmph the amount of water that runs into the sea per minute is ? | "rate of water flow - 8 kmph - - 8000 / 60 - - 133.33 m / min depth of river - - 8 m width of river - - 25 m vol of water per min - - 133.33 * 8 * 25 - - - 26666 answer a" | a = 8 * 25
b = 8 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 30 , b ) 90 , c ) 120 , d ) 168 , e ) 188 | d | subtract(400, subtract(add(multiply(divide(400, const_100), 56), multiply(divide(400, const_100), 44)), multiply(divide(400, const_100), 42))) | among 400 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 42 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 168 using your analysis : we see that the total number of students who study either maths or sociology = 176 + 224 - 168 = 232 so , in the image we know that the number of students in the zone with the black boundary = 232 let ' s assume the number of students who studyonlybiology to beb ( this is the number that we have to maximize ) and , let ' s assume the number of students who study none of the three subjects , that is the number of students in the white space = w since the total number of students = 400 , we can write : 232 + b + w = 400 or , b + w = 400 - 232 = 168 that is , b = 168 - w so , the maximum value ofbwill happen forw = 0 this is how we get , the maximum value ofb = 168 d" | a = 400 / 100
b = a * 56
c = 400 / 100
d = c * 44
e = b + d
f = 400 / 100
g = f * 42
h = e - g
i = 400 - h
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | e | divide(subtract(add(multiply(10, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))), multiply(20, subtract(20, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))))), multiply(20, 12)), subtract(20, 12)) | each of the 20 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 12 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes must be removed ? | "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 20 boxes , 4 are 10 - pound boxes and 16 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 12 , the ratio of 10 - pound boxes : 20 - pound boxes should be 4 : 1 . the number of 10 pound boxes remains the same so we still have 4 of them . to get a ratio of 4 : 1 , the number of 20 - pound boxes must be 1 . we need to remove 15 of the 20 - pound boxes . the answer is e ." | a = 20 * 20
b = 20 * 18
c = a - b
d = 20 - 10
e = c / d
f = 10 * e
g = 20 * 20
h = 20 * 18
i = g - h
j = 20 - 10
k = i / j
l = 20 - k
m = 20 * l
n = f + m
o = 20 * 12
p = n - o
q = 20 - 12
r = p / q
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | e | multiply(sqrt(divide(264, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 264 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose 12 at first , but then changed to 48 . i ' m not a native speaker , so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) , so is its largest factor . please correct if i ' m wrong here . i know that if n = 12 , n ^ 2 = 144 = 2 * 72 ( satisfy the condition ) . when n = 12 , the largest factor of n is n itself , which is 12 . check : 12 is the largest positive number that must divide 12 - - > true however if n = 48 , n ^ 2 = 48 * 48 = 32 * 72 ( satisfy the condition too ) . when n = 48 , the largest factor of n is n itself , which is 48 . check : 48 is the largest positive number that must divide 48 - - > true so , i also notice that the keyword ismust , notcould . the question is , why is 48 notmust divide 48 , but instead onlycould divide 48 ? i ' m not clear right here . why is 12 must divide 12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ 2 is divisible by 72 . the least value of n for which n ^ 2 is divisible by 72 is 12 , thus nmustbe divisible by 12 ( n is in any case divisible by 12 ) . for all other values of n , for which n ^ 2 is divisible by 72 , n will still be divisible by 12 . this means that n is always divisible by 12 if n ^ 2 is divisible by 72 . now , ask yourself : if n = 48 , is n divisible by 48 ? no . so , n is not always divisible by 48 . e" | a = 264 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 34778 , b ) 26888 , c ) 24475 , d ) 17600 , e ) 12778 | c | divide(multiply(add(const_100, 10), add(divide(multiply(17500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 17500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 10 % if no discount was offered ? | "price at which the tv set is bought = rs . 17,500 discount offered = 20 % marked price = 17500 * 100 / 80 = rs . 21875 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 21875 + 375 = rs . 22250 the price at which the tv should be sold to get a profit of 10 % if no discount was offered = 22250 * 110 / 100 = rs . 24475 answer : c" | a = 100 + 10
b = 17500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
|
a ) a ) 700 , b ) b ) 1065 , c ) c ) 1070 , d ) d ) 1075 , e ) e ) 1080 | a | add(multiply(3, 70), multiply(9, 55)) | harkamal purchased 3 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 3 kg grapes = 70 × 3 = 210 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 210 + 490 = 700 a" | a = 3 * 70
b = 9 * 55
c = a + b
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a ) 25.92 % , b ) 18.75 % , c ) 30 % , d ) 50 % , e ) 100 % | a | multiply(divide(subtract(add(const_100, 60), add(const_100, 35)), add(const_100, 35)), const_100) | the wages earned by robin is 35 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ? | "explanatory answer let the wages earned by erica be $ 100 then , wages earned by robin and charles will be $ 135 and $ 160 respectively . charles earns $ 35 more than robin who earns $ 135 . therefore , charles ' wage is 35 / 135 * 100 = 25.92 % . the correct choice is ( a )" | a = 100 + 60
b = 100 + 35
c = a - b
d = 100 + 35
e = c / d
f = e * 100
|
a ) 60 , b ) 28 , c ) 27 , d ) 26 , e ) 25 | a | add(multiply(const_4, 4.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 4.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 4087.5 + 1 ) = 10 ( 4.9421 + 1 ) = 10 ( 5.9421 ) = 59.421 x = antilog ( 59.421 ) therefore , number of digits in x = 60 . answer : a" | a = 4 * 4
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | d | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 30 percent to 70 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the lowest possible sale price ? | "let the list price be 2 x for min sale price , the first discount given should be 70 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , d is the answer" | a = 80 / 2
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a ) 12 hours , b ) 24 hours , c ) 36 hours , d ) 48 hours , e ) none | d | add(divide(210, add(9, 1.5)), divide(210, subtract(9, 1.5))) | speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph . a man rows to place at a distance of 210 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 7.5 kmph ; speed downstream = 10.5 kmph . ∴ total time taken = [ 210 / 7.5 + 210 / 10.5 ] hours = 48 hours . answer d" | a = 9 + 1
b = 210 / a
c = 9 - 1
d = 210 / c
e = b + d
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a ) 48 , b ) 57 , c ) 66 , d ) 72 , e ) 81 | d | multiply(subtract(multiply(subtract(multiply(3, 4), 3), 3), 3), subtract(4, const_1)) | q ' = 3 q - 3 , what is the value of ( 4 ' ) ' ? | "( 4 ' ) ' = ( 3 * 4 - 3 ) ' = 9 ' = 9 * 9 - 9 = 72 answer d" | a = 3 * 4
b = a - 3
c = b * 3
d = c - 3
e = 4 - 1
f = d * e
|
a ) $ 50 , b ) $ 51.5 , c ) $ 52 , d ) $ 52.5 , e ) $ 51.25 | e | subtract(add(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), 5), const_100)), divide(50, multiply(divide(5, const_100), 2))) | if the simple interest on a sum of money for 2 years at 5 % per annum is $ 50 , what is the compound interest on the same at the same rate and for the same time ? | "sum = 50 x 100 / ( 2 x 5 ) = 500 = 500 x ( 1 + 5 / 100 ) ^ 2 = 500 x 21 / 20 x 21 / 20 = 551.25 . c . i = 551.25 - 500 = 51.25 . answer e ) $ 51.25" | a = 5 / 100
b = a * 2
c = 50 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 50 / g
i = e + h
j = 5 / 100
k = j * 2
l = 50 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 50 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 50 / w
y = u - x
|
a ) 150 meter , b ) 286 meter , c ) 186 meter , d ) 288 meter , e ) 450 meter | e | multiply(divide(multiply(180, const_1000), const_3600), 9) | a train running at the speed of 180 km / hr crosses a pole in 9 seconds . find the length of the train . | "speed = 180 * ( 5 / 18 ) m / sec = 50 m / sec length of train ( distance ) = speed * time ( 50 ) * 9 = 450 meter answer : e" | a = 180 * 1000
b = a / 3600
c = b * 9
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a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 500 | e | subtract(divide(subtract(multiply(3, 1700), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1700), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | the cost of 2 chairs and 3 tables is rs . 1700 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "explanation : 2 c + 3 t = 1700 - - - ( 1 ) 3 c + 2 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 500 = > t - c = 500 answer : e" | a = 3 * 1700
b = 2 * 3
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
h = 3 * 1700
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 ) 4 , b ) 9 , c ) 15 , d ) 20 , e ) 63 | d | subtract(add(power(5, 2), multiply(5, 2)), add(power(3, 2), multiply(3, 2))) | the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 5 seconds than it was moving after 3 seconds ? | "given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 3 ) = 3 ^ 2 + 2 * 3 = 15 z ( 5 ) = 5 ^ 2 + 2 * 5 = 35 therefore z ( 7 ) - z ( 3 ) = 35 - 15 = 20 km / hr option d" | a = 5 ** 2
b = 5 * 2
c = a + b
d = 3 ** 2
e = 3 * 2
f = d + e
g = c - f
|
a ) 56678 , b ) 16789 , c ) 17576 , d ) 35142 , e ) none of these | c | power(add(add(add(const_10, const_10), const_3), const_3), subtract(5, const_2)) | consider the word rotor . whichever way you read it , from left to right or from right to left , you get the same word . such a word is known as palindrome . find the maximum possible number of 5 - letter palindromes ? | "explanation : the first letter from the right can be chosen in 26 ways because there are 26 alphabets . having chosen this , the second letter can be chosen in 26 ways . = > the first two letters can be chosen in 26 * 26 = 676 wayshaving chosen the first two letters , the third letter can be chosen in 26 ways . = > all the three letters can be chosen in 676 * 26 = 17576 ways . it implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter . answer : c" | a = 10 + 10
b = a + 3
c = b + 3
d = 5 - 2
e = c ** d
|
a ) 7 : 4 , b ) 4 : 7 , c ) 11 : 8 , d ) 8 : 11 , e ) 9 : 12 | a | divide(subtract(36, 8), subtract(24, 8)) | the present ages of reena and usha are 24 and 36 yr . respectively . what was the ratio between the ages of usha and reena respectively 8 yr ago ? | required ratio = ( 36 - 8 ) / ( 24 - 8 ) = 28 / 16 = 7 : 4 answer : a | a = 36 - 8
b = 24 - 8
c = a / b
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a ) 190 , b ) 284.6 , c ) 300 , d ) 292 , e ) 312 | d | multiply(12.2, 24) | a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 24 gallons of gasoline ? | so 12.2 * 24 = 292 . . imo option d is correct answer . . | a = 12 * 2
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a ) 200 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | a | subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 20)) | a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 20 seconds . what is the length of the platform in meters ? | "speed of the train in metres / sec = 72000 / 3600 = 20 distance travelled by train to cross the platform = 30 * 20 = 600 = length of train + length of platform distance travelled by train to cross the man = 20 * 20 = 400 = length of train length of platform = 600 - 400 = 200 answer : a" | a = 72 * 1000
b = a / 3600
c = b * 30
d = 72 * 1000
e = d / 3600
f = e * 20
g = c - f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(multiply(subtract(const_12, const_1), const_2), 20) | what is the 20 - fourth decimal to the right in the fraction 35 / 36 ? | 35 / 36 = 0.97222222222 . . . the answer is b . | a = 12 - 1
b = a * 2
c = b - 20
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a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 10 | e | add(add(2, 3), add(2, 3)) | a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 10 : 00 a . m . ? | sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour and 5 times in next hour . hence , they cross each other 10 times before 10 : 00 a . m . answer e | a = 2 + 3
b = 2 + 3
c = a + b
|
a ) 25 % , b ) 35 % , c ) 45 % , d ) 55 % , e ) 65 % | a | multiply(subtract(const_1, divide(const_3, const_4)), const_100) | a regular hexagon is there . the mid points of the sides were joined and formed another hexagon . then what is the percentage reduction in area . | let abcdef be the regular hexagon with sides ab and ed parallel . p and q are mid points of af and ef resp . ae = root ( 3 ) * s ( s is side of hexagon ) then pq = ( root ( 3 ) * s ) / 2 mid point theorem now u know side of inner hexagon use formula area of hexagon = ( 3 * root ( 3 ) ) 2 * s ^ 2 and find diff in area of two hexagon multiply by 100 and divide the product by area of bigger hexagon ans 25 % answer : a | a = 3 / 4
b = 1 - a
c = b * 100
|
a ) 515 . , b ) 545 . , c ) 606 . , d ) 644 . , e ) 666 . | c | add(226, 423) | in the faculty of reverse - engineering , 226 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 85 % of the total ? | "answer is c : 606 solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 85 % of total is 515 , so 100 % is approx . 606 ." | a = 226 + 423
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a ) 2.5 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | c | divide(20, const_2) | if the length of the longest chord of a certain circle is 20 , 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 20 = 2 * 10 so radius of the circle = 10 correct answer - c" | a = 20 / 2
|
a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 2500 , e ) $ 12000 | d | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(6, const_100)), const_4)), multiply(const_100, const_100)) | carl is facing very difficult financial times and can only pay the interest on a $ 10,000 loan he has taken . the bank charges him a quarterly compound rate of 6 % . what is the approximate interest he pays annually ? | "an easy way too attempt this is by approximation : for the first quarter , he pays 6 % of 10,000 which is $ 600 . so for the four quarters in the year , he will approximately play 600 * 4 = 2400 over the entire year . the compound interest will be more than 2400 . all other options are not right except d . answer : d" | a = 100 * 100
b = 6 / 100
c = 1 + b
d = c ** 4
e = a * d
f = 100 * 100
g = e - f
|
a ) 41 / 50 , b ) 1 / 221 , c ) 1 / 1296 , d ) 1 / 84 , e ) 1 / 42 | c | divide(const_1, power(subtract(divide(26, const_3), const_1), const_2)) | in a certain game of dice , the player ’ s score is determined as a sum of four throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if jim plays this game against 26 other players , what is the probability of the minimum score that will guarantee jim some monetary payoff ? | "toguaranteethat jim will get some monetary payoff he must score the maximum score of 6 + 6 + 6 + 6 = 24 , because if he gets even one less than that so 23 , someone can get 24 and jim will get nothing . p ( 24 ) = 1 / 6 ^ 4 = 1 / 1296 . answer : c ." | a = 26 / 3
b = a - 1
c = b ** 2
d = 1 / c
|
a ) 45 % , b ) 49 % , c ) 50 % , d ) 59 % , e ) 61 % | c | multiply(divide(11628, add(add(4136, 7636), 11628)), const_100) | 3 candidates in an election and received 4136 , 7636 and 11628 votes respectively . what % of the total votes did the winning candidate got in that election ? | "total number of votes polled = ( 4136 + 7636 + 11628 ) = 23400 so , required percentage = 11628 / 23400 * 100 = 50 % c" | a = 4136 + 7636
b = a + 11628
c = 11628 / b
d = c * 100
|
['a ) 16.665', 'b ) 16.565', 'c ) 15.465', 'd ) 14.365', 'e ) 14.378'] | d | multiply(power(subtract(const_1, divide(35, const_100)), const_2), 34) | a piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse , leaving a smaller triangle . there was a 35 % reduction in the length of the hypotenuse of the triangle . if the area of the original triangle was 34 square inches before the cut , what is the area ( in square inches ) of the smaller triangle ? | explanation : smaller triangle and original triangle will be similar to each other . ratio of area = ( ratio of side ) 2 area of smaller triangle = area of original triangle * ( 65 / 100 ) 2 = 14.365 . answer : d | a = 35 / 100
b = 1 - a
c = b ** 2
d = c * 34
|
a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50 | b | subtract(multiply(subtract(60, 20), const_3), multiply(subtract(60, 30), const_3)) | in x game of billiards , x can give y 20 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 80 ? | "x scores 60 while y score 40 and z scores 30 . the number of points that z scores when y scores 80 = ( 80 * 30 ) / 40 = 60 . in x game of 100 points , y gives ( 80 - 60 ) = 20 points to c . b" | a = 60 - 20
b = a * 3
c = 60 - 30
d = c * 3
e = b - d
|
a ) 28 sec , b ) 16 sec , c ) 35 sec , d ) 18 sec , e ) 17 sec | c | divide(add(240, 110), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 110 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 110 = 350 m . time taken = 350 / 10 = 35 sec . answer : c" | a = 240 + 110
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) f = 8 | e | divide(240, 30) | a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations f are possible ? | "the combinations could be f { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pairs , and we are left only with . { ( 8,30 , ( 10,24 ) , ( 12,20 ) , ( 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) } hence 8 . . e" | a = 240 / 30
|
a ) 23 kmph , b ) 24 kmph , c ) 25 kmph , d ) 26 kmph , e ) 27 kmph | c | divide(subtract(multiply(50, const_3), divide(multiply(multiply(50, const_3), const_2), const_3)), subtract(const_3, const_1)) | by travelling at 50 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "let the time taken to reach the destination be 3 x hours . total distance = 50 * 3 x = 150 x km he covered 2 / 3 * 150 x = 100 x km in 1 / 3 * 3 x = x hours so , the remaining 50 x km , he has to cover in 2 x hours . required speed = 50 x / 2 x = 25 kmph . answer : c" | a = 50 * 3
b = 50 * 3
c = b * 2
d = c / 3
e = a - d
f = 3 - 1
g = e / f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(4, 11) | what is the 29 th digit to the right of the decimal point in the decimal form of 4 / 11 ? | "4 / 11 = 0.363636 . . . the odd numbered positions in the decimal expansion are all 3 . the answer is c ." | a = 4 / 11
|
a ) 100 , b ) 150 , c ) 50 , d ) 200 , e ) 250 | c | subtract(multiply(200, add(const_3, const_2)), add(add(add(300, subtract(250, 100)), 100), subtract(subtract(subtract(multiply(250, const_4), 100), multiply(const_2, const_100)), multiply(const_3, const_100)))) | the average of 100 , 200300 and x is 250 and the average of 300 , 150100 , x and y is 200 . what is the value of y ? | we have : ( 100 + 200 + 300 + x ) / 4 = 250 600 + x = 1000 x = 400 ( 300 + 150 + 100 + x + y ) / 5 = 200 550 + 400 + y = 1000 y = 50 answer is c | a = 3 + 2
b = 200 * a
c = 250 - 100
d = 300 + c
e = d + 100
f = 250 * 4
g = f - 100
h = 2 * 100
i = g - h
j = 3 * 100
k = i - j
l = e + k
m = b - l
|
a ) s 200 , b ) s 1250 , c ) s 300 , d ) s 450 , e ) s 550 | b | divide(500, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4))) | a person spends 1 / 3 rd of the money with him on clothes , 1 / 5 th of the remaining on food and 1 / 4 th of the remaining on travel . now , he is left with rs 500 . how much did he have with him in the beginning ? | "suppose the amount in the beginning was rs ’ x ’ money spent on clothes = rs 1 x / 3 balance = rs 2 x / 3 money spent on food = 1 / 5 of 2 x / 3 = rs 2 x / 15 balance = 2 x / 3 - 2 x / 15 = rs 8 x / 15 money spent on travel = 1 / 4 of 8 x / 15 = rs 2 x / 15 = 8 x / 15 - 2 x / 15 = 6 x / 15 = rs 2 x / 5 therefore 2 x / 5 = 500 = 1250 answer : b" | a = 1 / 3
b = 1 - a
c = 1 / 3
d = 1 - c
e = d / 5
f = b - e
g = 1 / 3
h = 1 - g
i = 1 / 3
j = 1 - i
k = j / 5
l = h - k
m = l / 4
n = f - m
o = 500 / n
|
a ) 35 kmph , b ) 50 kmph , c ) 25 kmph , d ) 30 kmph , e ) 38 kmph | c | multiply(5, 5) | a passenger train takes two hours less for a journey of 300 km if its speed is increased by 5 kmph from its normal speed . the normal speed is | let normal speed = x increased speed = x + 5 initial = > x = 300 / t 1 = > t 1 = 300 / x after = > ( x + 5 ) = 300 / ( t 1 - 2 ) = > ( x + 5 ) = 300 / [ ( 300 / x ) - 2 ] = > x = 25 kmph answer : c | a = 5 * 5
|
a ) 45 , b ) 55 , c ) 65 , d ) 75 , e ) 85 | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 54), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 54 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 54 seconds , the relative difference in distance traveled is 54 * 25 / 9 = 150 meters this distance is twice the length of each train . the length of each train is 150 / 2 = 75 meters the answer is d ." | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 54
e = d / 2
|
a ) 5.29 min , b ) 5.28 min , c ) 5.08 min , d ) 4.8 min , e ) 5.988 min | d | multiply(divide(divide(660, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 660 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 660 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 660 m apart , they take ( 100 / 825 * 660 / 1000 ) hrs = ( 660 / 8250 * 60 ) min = 4.8 min . answer : d" | a = 660 / 1000
b = 4 + 5
c = a / b
d = c * const_60
|
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 1 | e | divide(subtract(subtract(subtract(9, 1), 1), 1), subtract(9, 1)) | in a room with 9 people , 6 people have exactly 1 friend in the room and 5 people have exactly 2 friends in the room ( assuming that friendship is a mutual relationship , i . e . if jane is paul ' s friend , paul is jane ' s friend ) . if two individuals are selected from the room at random , what is the probability that those two individuals are not friends ? | "( 6 / 9 ) ( 7 / 8 ) + ( 5 / 9 ) ( 6 / 8 ) if you choose one of the 6 with one other friend , then you have a 7 / 8 chance of not picking their friend 2 nd . if you choose one of the 5 with 2 friends , you have a 6 / 8 chance of not picking one of their friends second . add them up . 42 / 72 + 30 / 72 72 / 72 = 1 e . 1" | a = 9 - 1
b = a - 1
c = b - 1
d = 9 - 1
e = c / d
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | a | subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 20), 60)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 20 , 60 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 80 = 25 a" | a = 20 + 40
b = a + 60
c = 5 * 3
d = c + 20
e = d + 60
f = b - e
|
a ) 9 : 2 , b ) 3 : 2 , c ) 23 : 5 , d ) 18 : 4 , e ) 17 : 4 | c | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 92,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | a : b = 92000 : 20000 = 92 : 20 = 46 : 10 = 23 : 5 answer : c | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 2
f = 3 * 3
g = f + 1
h = 2 * g
i = e / h
|
a ) 15 % , b ) 20 % , c ) 40 % , d ) 13 % , e ) 150 % | d | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 25), subtract(const_100, 15)), const_100)), multiply(divide(subtract(const_100, 25), subtract(const_100, 15)), const_100)), const_100) | the charge for a single room at hotel p is 25 percent less than the charge for a single room at hotel r and 15 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "let rate in r = 100 x then p = 75 x g = 100 y p = 85 y thus 75 x = 85 y or x = 1.13 y ans r = 113 y so increase = 13 % answer : d" | a = 100 - 25
b = 100 - 15
c = a / b
d = c * 100
e = 100 - d
f = 100 - 25
g = 100 - 15
h = f / g
i = h * 100
j = e / i
k = j * 100
|
a ) 20 % , b ) 36 % , c ) 44 % , d ) 50 % , e ) 64 % | c | multiply(subtract(power(add(const_1, divide(20, const_100)), const_2), const_1), const_100) | if a large pizza has a radius that is 20 % larger that that of a medium pizza , what is the percent increase in area between a medium and a large pizza ? | "let the radius of medium pizza be r . then the radius of large pizza is 1.2 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.2 * r ) ^ 2 = 1.44 * pi * r ^ 2 , an increase of 44 % . the answer is c ." | a = 20 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 36 , b ) 232 , c ) 27 , d ) 48 , e ) none of these | c | divide(volume_cube(15), volume_cube(divide(5, const_100))) | how many cubes of 5 cm edge can be cut out of a cube of 15 cm edge | "explanation : number of cubes = ( 15 x 15 x 15 ) / ( 5 x 5 x 5 ) = 27 answer : c" | a = volume_cube / (
|
a ) 5 % , b ) 20 % , c ) 320 % , d ) 200 % , e ) 500 % | c | multiply(divide(160, 50), const_100) | 160 is what percent of 50 ? | "50 * x = 160 - - > x = 3.2 - - > 3.2 expressed as percent is 320 % . answer : c ." | a = 160 / 50
b = a * 100
|
a ) rs . 5750 , b ) rs . 6000 , c ) rs . 5500 , d ) rs . 5000 , e ) rs . 7000 | d | subtract(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), add(multiply(divide(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), power(35, const_2)), power(multiply(5, 2), const_2)), multiply(divide(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), power(35, const_2)), power(multiply(5, 5), const_2)))) | a precious stone weighing 35 grams worth rs . 12,250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5 . if the price varies as the square of the weight then find the loss incurred . | the price varies as the square of the weight , p = k x w 2 12250 = k x 352 k = 10 thus , p = 10 w 2 when the two pieces are in the ratio 2 : 5 ( weight wise ) then we know that their weights must be 10 g and 25 g respectively . their values should be : 10 g piece : 10 x 102 = rs . 1000 ; 25 g piece : 10 x 252 = rs . 6250 . total value of stone now , = 1000 + 6250 = rs . 7250 loss incurred = 12250 - 7250 = rs . 5000 answer : d | a = 100 * 100
b = 100 * 100
c = b / 2
d = c / 2
e = a + d
f = e - 100
g = f - 100
h = 100 / 2
i = g - h
j = 100 * 100
k = 100 * 100
l = k / 2
m = l / 2
n = j + m
o = n - 100
p = o - 100
q = 100 / 2
r = p - q
s = 35 ** 2
t = r / s
u = 5 * 2
v = u ** 2
w = t * v
x = 100 * 100
y = 100 * 100
z = y / 2
A = z / 2
B = x + A
C = B - 100
D = C - 100
E = 100 / 2
F = D - E
G = 35 ** 2
H = F / G
I = 5 * 5
J = I ** 2
K = H * J
L = w + K
M = i - L
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.25 , e ) 0.5 | d | power(divide(1, 4), 1) | what is the decimal equivalent of ( 1 / 4 ) ^ 1 ? | "( 1 / 4 ) = 0.25 answer : d" | a = 1 / 4
b = a ** 1
|
a ) 3 , b ) 5 , c ) 4 , d ) 7 , e ) 2 | d | subtract(5474827, multiply(floor(divide(5474827, 12)), 12)) | find the least number must be subtracted from 5474827 so that remaining no . is divisible by 12 ? | "on dividing 5474827 by 12 we get the remainder 7 , so 7 should be subtracted d" | a = 5474827 / 12
b = math.floor(a)
c = b * 12
d = 5474827 - c
|
a ) 13 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | e | divide(sqrt(200), const_2) | how many even integers from 1 to 200 ( both inclusive ) have even number of factors ? | "integers having even number of factors will be perfect squares . even numbers will have even perfect squares . thus , the possible values for the perfect squares are : 4 , 16,36 , 64,100 , 144,196 and the corresponding integers are 2 , 4,6 , 8,10 , 12,14 ( more than 6 ) . thus e is the correct answer ." | a = math.sqrt(200)
b = a / 2
|
a ) 192 , b ) 200 , c ) 208 , d ) 216 , e ) 224 | d | multiply(inverse(add(divide(const_1, 6), divide(const_1, 9))), const_60) | working alone , pump a can empty a pool in 6 hours . working alone , pump b can empty the same pool in 9 hours . working together , how many minutes will it take pump a and pump b to empty the pool ? | "pump a can empty ( 1 / 6 ) of the pool per hour . pump b can empty ( 1 / 9 ) of the pool per hour . together the pumps can empty 1 / 6 + 1 / 9 = 5 / 18 of the pool per hour . 1 pool / ( 5 / 18 ) pool per hour = 18 / 5 hours = 216 minutes . the answer is d ." | a = 1 / 6
b = 1 / 9
c = a + b
d = 1/(c)
e = d * const_60
|
a ) $ 19,200 , b ) $ 19,500 , c ) $ 19,800 , d ) $ 20,000 , e ) $ 20,400 | c | divide(add(add(multiply(add(add(add(const_10, const_10), 1), 1), 3), multiply(subtract(add(const_10, const_10), 1), 2)), add(add(const_10, const_10), 1)), 6) | a bakery has 6 employees . it pays annual salaries of $ 18,000 to each of 2 employees , $ 20,000 to 1 employee , and $ 21,000 to each of the remaining 3 employees . which number is closest to the average ( arithmetic mean ) annual salary of these employees ? | "average = 2 ( 18000 ) + ( 20000 ) + 3 ( 21000 ) / 6 = $ 19,833 the answer is c ." | a = 10 + 10
b = a + 1
c = b + 1
d = c * 3
e = 10 + 10
f = e - 1
g = f * 2
h = d + g
i = 10 + 10
j = i + 1
k = h + j
l = k / 6
|
a ) 24 / 286 , b ) 25 / 709 , c ) 25 / 286 , d ) 24 / 268 , e ) 24 / 226 | c | divide(add(add(3, 6), add(const_10, add(3, 3))), multiply(add(const_10, 3), 6)) | a box contains 4 black , 3 red and 6 green marbles . 3 marbles are drawn from the box at random . what is the probability that both the marbles are of the same color ? | "explanation : total marbles in a box = 4 black + 3 red + 6 green marbles = 13 marbles 3 marbles are drawn from 13 marbles at random . therefore , n ( s ) = 13 c 3 = 286 ways let a be the event that 2 marbles drawn at random are of the same color . number of cases favorable to the event a is n ( a ) = 4 c 3 + 3 c 3 + 6 c 3 = 4 + 1 + 20 = 25 therefore , by definition of probability of event a , p ( a ) = n ( a ) / n ( s ) = 25 / 286 answer : c" | a = 3 + 6
b = 3 + 3
c = 10 + b
d = a + c
e = 10 + 3
f = e * 6
g = d / f
|
a ) 3.625 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | b | add(divide(subtract(divide(45, 10), 2.5), const_2), 2.5) | annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.5 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 45 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exactly 45 minutes , for how many kilometers total did she hike east ? | "set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.5 from which t = 25 mins . we know total journey time now is 45 + 25 = 70 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 70 = 7 km now the total journey would be halved as distance would be same in each direction . 7 / 2 = 3.5 b ." | a = 45 / 10
b = a - 2
c = b / 2
d = c + 2
|
a ) 1800 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2789 | a | multiply(multiply(divide(18, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 18 of money ? | "p = 18 ( 100 / 10 ) 2 = > p = 1800 answer : a" | a = 10 * 2
b = 18 / a
c = b * 100
d = 10 * 2
e = c * d
|
a ) 404 , b ) 302 , c ) 108 , d ) 115 , e ) 210 | a | subtract(negate(439), multiply(subtract(509, 474), divide(subtract(509, 474), subtract(544, 509)))) | 544 , 509 , 474 , 439 , . . . ? | "each number is 35 less than the previous number . 544 544 - 35 = 509 509 - 35 = 474 474 - 35 = 439 439 - 35 = 404 answer : a" | a = negate - (
|
a ) 15 , b ) 26 , c ) 28 , d ) 91 , e ) none of these | e | floor(sqrt(182)) | if the sum of a number and its square is 182 , what is the number ? | "let the number be x . then , x + x 2 = 182 ( x + 14 ) ( x - 13 ) = 0 x = 13 answer : e" | a = math.sqrt(182)
b = math.floor(a)
|
a ) 33772 , b ) 26782 , c ) 26788 , d ) 13500 , e ) 27761 | d | subtract(divide(multiply(multiply(multiply(multiply(184, 3), 3), 3), 6), 2), const_1000) | the difference between compound and simple interest on a certain sum of money for 3 years at 6 2 / 3 % p . a is rs . 184 . find the sum ? | p = ( 184 * 106 ) / [ 6 2 / 3 * 6 2 / 3 * ( 300 * 6 2 / 3 ) ] p = 13500 answer : d | a = 184 * 3
b = a * 3
c = b * 3
d = c * 6
e = d / 2
f = e - 1000
|
['a ) 5.6 %', 'b ) 8.4 %', 'c ) 11.2 %', 'd ) 14.8 %', 'e ) 17.5 %'] | a | multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100) | a block of wood has dimensions 10 cm x 10 cm x 80 cm . the block is painted red and then cut evenly at the 40 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? | the area of each half is 100 + 4 ( 400 ) + 100 = 1800 the area that is not painted is 100 . the fraction that is not painted is 100 / 1800 = 1 / 18 = 5.6 % the answer is a . | a = 4 * 100
b = a * 4
c = b + 100
d = c + 100
e = 100 / d
f = e * 100
|
a ) 25 / 3 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32 | b | divide(divide(3200, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2) | the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 3200 meters per second , what was the speed , in meters per second , when the stopwatch was started ? | "let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 3200 x = ( 2 ^ 7 * 25 ) / 2 ^ 9 = 25 / 4 the answer is b ." | a = const_60 / 2
b = a + const_60
c = b / 10
d = c - 1
e = 2 ** d
f = 3200 / e
g = f / 2
|
a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 33 % . , e ) 51 % . | d | multiply(divide(add(multiply(divide(30, const_100), 2), multiply(divide(45, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 30 % of alcohol and another vessel of capacity 6 litre had 45 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "30 % of 2 litres = 0.6 litres 45 % of 6 litres = 2.7 litres therefore , total quantity of alcohol is 3.3 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 33 % d" | a = 30 / 100
b = a * 2
c = 45 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
|
a ) 66 % , b ) 75 % , c ) 89 % , d ) 116 % , e ) 150 % | c | multiply(divide(add(100, 60), add(100, divide(add(100, 60), const_2))), 100) | company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 60 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw would be approximately what percent of these combined assets ? | "let the price of company a ' s assets be 100 price of assets of kw is 60 % more than company a ' s assets which is 160 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 80 a + b = 180 kw = 160 kw / ( a + b ) * 100 = 160 / 180 * 100 = 88.88 % or 89 % c" | a = 100 + 60
b = 100 + 60
c = b / 2
d = 100 + c
e = a / d
f = e * 100
|
a ) 12 by 2 , b ) 11 by 3 , c ) 10.5 by 3.5 , d ) 10 by 4 , e ) 9 by 3 | d | add(const_10, 2) | a certain rectangular window is ( 2 / 5 ) times as long as it is wide . if its perimeter is 28 feet , what are its dimensions in terms of length by width ? | "2 x + 2 y = 28 x + y = 14 x + ( 2 / 5 ) x = 14 1.4 x = 14 x = 10 answer d" | a = 10 + 2
|
a ) 5 km , b ) 3 km , c ) 7 km , d ) 9 km , e ) 2 km | a | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 5), divide(const_1, 6)), divide(const_1, 7))), const_3), const_1000) | a person travels equal distances with speeds of 5 km / hr , 6 km / hr and 7 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 5 + x / 6 + x / 7 = 27 / 53 27 x / 53 = 47 / 60 = > x = 1.54 total distance = 3 * 1.54 = 4.61 km . answer : a" | a = 47 / const_60
b = 1 / 5
c = 1 / 6
d = b + c
e = 1 / 7
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 24.5 , b ) 12.5 , c ) 17.25 , d ) 6.5 , e ) 13.62 | a | multiply(divide(subtract(490, 465.50), 490), const_100) | if the a radio is sold for rs 490 and sold for rs 465.50 . find loss value . | "cp = rs 490 , sp = 465.50 . loss = rs ( 490 - 465.50 ) = rs 24.50 . ans : a" | a = 490 - 465
b = a / 490
c = b * 100
|
a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | d | add(add(10, 14), 14) | set s consists of integers { 4 , 7 , 10 , 14 , 15 } . if integer n is included in the set , the average ( arithmetic mean ) of set s will increase by 50 % . what is the value of integer n ? | "the average of the numbers in set s is 10 . if we increase the mean by 50 % , the new mean is 15 . thus , on average , 5 numbers increase by 5 . therefore n = 15 + 25 = 40 the answer is d ." | a = 10 + 14
b = a + 14
|
a ) 1 / 51 , b ) 4 / 51 , c ) 7 / 51 , d ) 3 / 44 , e ) 7 / 44 | c | divide(subtract(51, 44), 51) | a ’ s speed is 51 / 44 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 51 meters , b covers 44 meters in that time . so if the race is 51 meters long , when a reaches the finish line , b would be 7 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 7 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 51 meters ) would be the same as the time required by b ( to cover 44 meters ) to reach the finish line . so b should get a head start of 7 / 51 th of the race . answer ( c )" | a = 51 - 44
b = a / 51
|
a ) 46457 , b ) 465745 , c ) 3000 , d ) 3456 , e ) 35346 | c | divide(800, subtract(multiply(divide(const_2, const_3), divide(700, add(700, 300))), multiply(divide(const_2, const_3), divide(300, add(700, 300))))) | john and mike enter into a partnership by investing $ 700 and $ 300 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if john received $ 800 more than mike did , what was the profit made by their business in that year ? | say the profit was $ x . john share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.7 mike ' s share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.3 thus ( x - x / 3 ) * 0.7 - ( x - x / 3 ) * 0.3 = 800 - - > x = 3000 . answer is c | a = 2 / 3
b = 700 + 300
c = 700 / b
d = a * c
e = 2 / 3
f = 700 + 300
g = 300 / f
h = e * g
i = d - h
j = 800 / i
|
a ) 89.7 minutes , b ) 90 minutes , c ) 90.3 seconds , d ) 91 seconds , e ) none of these | b | divide(multiply(multiply(const_60, const_10), divide(18, const_2)), const_60) | two men are walking towards each other alongside a railway track . a freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction . the train passes this man is 18 seconds . assume the velocities are constant throughout . how long after the train has passed the second man will the two men meet ? | explanation : let ‘ l ’ be the length of train , ‘ x ’ be the speed of the first man , ‘ y ’ be the speed of the second man and ‘ z ’ be the speed of the train . = > 20 = 1 / ( z − x ) and = > 18 = 1 / ( z + x ) = > z = 10 x + 9 y . distance between the two men = 600 ( z + y ) . time = ( 600 ( z + y ) - 600 ( x + y ) ) / ( x + y ) . = > 600 ( 9 x + 9 y ) / ( x + y ) . = > 90 minutes . answer : b | a = const_60 * 10
b = 18 / 2
c = a * b
d = c / const_60
|
a ) 500 , b ) 334 , c ) 531 , d ) 664 , e ) 5598 | c | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 16)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 16 % , then the cost price of the article is : | "110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 116 = rs . 531 answer : option c" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 16
g = e / f
|
a ) 2 , b ) 5 , c ) 9 , d ) 7 , e ) 6 | d | divide(add(sqrt(add(multiply(4, 21), power(4, const_2))), 4), const_2) | a positive number when decreased by 4 is equal to 21 times the reciprocal of the number . the number is : | let the number be x x - 4 = 21 / x x ^ 2 - 4 x - 21 = 0 ( x - 7 ) ( x + 3 ) = 0 x = 7 correct answer d | a = 4 * 21
b = 4 ** 2
c = a + b
d = math.sqrt(c)
e = d + 4
f = e / 2
|
a ) s . 380 , b ) s . 600 , c ) s . 420 , d ) s . 440 , e ) s . 480 | d | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3520) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3520 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 3520 × ( 1 / 8 ) = 440 answer : option d" | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 3520
|
a ) 23 , b ) 22 , c ) 45 / 2 , d ) 36 , e ) 48 | c | multiply(15, divide(const_3, const_2)) | a is half good a work man as b and together they finish a job in 15 days . in how many days working alone b finish the job ? | "c 45 / 2 wc = 1 : 2 2 x + x = 1 / 15 = > x = 1 / 45 2 x = 2 / 45 = > 45 / 2 days" | a = 3 / 2
b = 15 * a
|
a ) 8 , b ) 12 , c ) 6 , d ) 4 , e ) 5 | b | divide(subtract(subtract(add(multiply(subtract(30, 10), 10), multiply(10, 20)), 240), multiply(gcd(30, 10), 20)), 10) | bookman purchased 30 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 240 , how many paperback copies were sold ? | "the bookman had 10 hardback ad 30 - 10 = 20 paperback copies ; 14 copies were sold , hence 30 - 14 = 16 copies were left . let # of paperback copies left be p then 10 p + 20 ( 16 - p ) = 240 - - > 10 p = 80 - - > p = 8 # of paperback copies sold is 20 - 8 = 12 answer : b" | a = 30 - 10
b = a * 10
c = 10 * 20
d = b + c
e = d - 240
f = math.gcd(30, 10)
g = f * 20
h = e - g
i = h / 10
|
a ) 450 m , b ) 350 m , c ) 150 m , d ) 250 m , e ) 240 m | d | multiply(divide(multiply(80, const_1000), const_3600), 10) | a train running at the speed of 80 km / hr crosses a pole in 10 seconds . what is the length of the train ? | speed = ( 90 * 5 / 18 ) m / sec = 25 m / sec length of the train = ( speed x time ) = ( 25 * 10 ) m = 250 m . answer : d | a = 80 * 1000
b = a / 3600
c = b * 10
|
a ) 240 , b ) 120 , c ) 280 , d ) 160 , e ) 200 | a | subtract(subtract(multiply(divide(1360, const_10), const_2), const_12), const_12) | a sum of rs . 1360 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ' s share is : | "let c ' s share = rs . x then , b ' s share = rs . x / 4 , a ' s share = rs . ( 2 / 3 x ( x / 4 ) ) = rs . x / 6 therefore , x / 6 + x / 4 + x = 1360 = 17 x / 12 = 1360 x = 1360 x 12 / 17 = rs . 960 hence , b ' s share = rs . ( 960 / 4 ) = rs . 240 . answer is a ." | a = 1360 / 10
b = a * 2
c = b - 12
d = c - 12
|
a ) rs . 200 , b ) rs . 250 , c ) rs . 240 , d ) rs . 220 , e ) rs . 230 | a | divide(242, power(add(divide(10, const_100), const_1), 2)) | the present worth of rs . 242 due in 2 years at 10 % per annum compound interest is : | "explanation : present worth of rs . x due t years hence is given by present worth ( pw ) = x / ( 1 + r / 100 ) t present worth ( pw ) = 242 / ( 1 + 10 / 100 ) 2 = 242 / ( 11 / 10 ) 2 = rs . 200 answer : option a" | a = 10 / 100
b = a + 1
c = b ** 2
d = 242 / c
|
a ) 28 miles , b ) 30 miles , c ) 26 miles , d ) 18 miles , e ) 22 miles | c | sqrt(add(power(add(6, 18), const_2), power(10, const_2))) | beginning in town a , biker bob rides his bike 20 miles west , 6 miles north , 10 miles east , and then 18 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | using pythagoras we have one side i , e total distance traveled in north direction = 18 + 6 = 24 m other being the base ie distance traveled west - distance traveled eat = 20 - 10 = 10 m now this third side or the distance between town a and town b = 24 ^ 2 + 10 ^ 2 = sq root 676 = 26 m answer : c | a = 6 + 18
b = a ** 2
c = 10 ** 2
d = b + c
e = math.sqrt(d)
|
a ) 22377 , b ) 27782 , c ) 27891 , d ) 25000 , e ) 23000 | e | divide(divide(divide(multiply(multiply(multiply(23, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5) | a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 23 m * 2 m * 0.75 m ? | "23 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 23 = 1 / 100 * x = > x = 23000 answer : e" | a = 23 * 100
b = 2 * 100
c = a * b
d = 0 * 75
e = c * d
f = e / 20
g = f / 10
h = g / 7
|
a ) 76 , b ) 60 , c ) 65 , d ) 48 , e ) 52 | c | add(40, divide(subtract(1340, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1340 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 1340 - 640 = 700 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 700 / 28 = 25 total hrs = 40 + 25 = 65 answer c 65" | a = 16 * 40
b = 1340 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
['a ) 34 cm ²', 'b ) 38 cm ²', 'c ) 54 cm ²', 'd ) 56 cm ²', 'e ) 64 cm ²'] | c | divide(multiply(sqrt(subtract(power(15, const_2), power(12, const_2))), 12), const_2) | find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm . | ab ² = ac ² - bc ² = 15 ² - 12 ² = 225 - 144 = 81 therefore , ab = 9 therefore , area of the triangle = ¹ / ₂ × base × height = ¹ / ₂ × 12 × 9 = 54 cm ² answer : c | a = 15 ** 2
b = 12 ** 2
c = a - b
d = math.sqrt(c)
e = d * 12
f = e / 2
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 21 | b | divide(multiply(24, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 15), multiply(divide(add(24, 40), 4), 3)), multiply(divide(8, 2), 3))) | 3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 24 days . if 40 more cows and 15 more buffaloes are brought in , how many days will the fodder last ? | "2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 24 days . 40 cows + 15 buffaloes = 20 oxen + 10 oxen = 30 oxen the new total is equal to 60 oxen instead of 30 oxen . 30 * 24 / 60 oxen = 12 days the answer is b ." | a = 24 / 4
b = a * 3
c = 15 + b
d = 8 / 2
e = d * 3
f = c + e
g = 24 * f
h = 15 + 15
i = 24 + 40
j = i / 4
k = j * 3
l = h + k
m = 8 / 2
n = m * 3
o = l + n
p = g / o
|
a ) 34 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(add(negate(10), sqrt(subtract(power(negate(10), 2), multiply(const_4, negate(divide(multiply(300, 10), 2)))))), 2) | the time it took car p to travel 300 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | "let speed of car r be = x then speed of car p = x + 10 a / q , ( 300 / x ) - ( 300 / ( x + 10 ) ) = 2 solving for x = 34 miles \ hr . a" | a = negate + (
b = negate ** (
c = b - 2
d = 300 * 10
e = d / 2
f = 4 * negate
g = math.sqrt(c)
h = a / g
|
a ) 140 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none of these | a | multiply(multiply(multiply(7, 2), divide(1, const_100)), const_1000) | a boat having a length 7 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "explanation : volume of water displaced = ( 7 x 2 x 0.01 ) m 3 = 0.14 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.14 x 1000 ) kg = 140 kg . answer : a" | a = 7 * 2
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 52 , b ) 65 , c ) 78 , d ) 91 , e ) 104 | e | divide(multiply(divide(multiply(64, 3.8), 1.6), 3.9), 5.7) | a certain car can travel 64 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 = 64 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 64 x 3.9 x 3.8 / 5.7 x 1.6 = 78 miles / hour ans : e ps : i felt the factors were easy to cancel out , so did n ' t require much rounding off = 64 x 3.9 x 3.8 / 5.7 x 1.6 = 104 e" | a = 64 * 3
b = a / 1
c = b * 3
d = c / 5
|
a ) s 1800 , b ) s 1500 , c ) s 1600 , d ) s 1900 , e ) s 1700 | a | multiply(divide(1500, 5), subtract(9, 3)) | an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 9 . 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 = 9 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 2700 . hence , ( r - p ) = ( 2700 - 900 ) = 1800 answer : a" | a = 1500 / 5
b = 9 - 3
c = a * b
|
a ) rs . 5870 , b ) rs . 5991 , c ) rs . 6020 , d ) rs . 6850 , e ) none of these | a | subtract(multiply(add(5, const_1), 6000), add(add(add(add(5420, 5660), 6200), 6350), 6500)) | a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6000 ? | "explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6000 * 6 ) – 30,130 ] = rs . ( 36000 – 30,130 ) = rs . 5870 answer a" | a = 5 + 1
b = a * 6000
c = 5420 + 5660
d = c + 6200
e = d + 6350
f = e + 6500
g = b - f
|
a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 4.8 % | e | multiply(divide(add(multiply(60, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 60 percent on clothing , 10 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2.4 tax on others = 2.4 percentage = 4.8 / 100 = 4.8 % answer : e" | a = 4 / 100
b = 60 * a
c = 8 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 3 , b ) 6 , c ) 8 , d ) 12 , e ) 24 | d | multiply(add(const_3, const_1), add(const_2, const_1)) | x ^ 2 + sx + 72 = 0 has two distinct integer roots ; how many values are possible for s ? | for a quadratic equation ax ^ 2 + sx + c = 0 , we know that - s / a is sum of roots and c / a is product of roots . the quadratic equation here is x ^ 2 + sx + 72 = 0 , where product of roots is 72 . if we find all the factors of 72 , we have the answer . by prime factorization , we get 72 = 2 ^ 3 * 3 ^ 2 . we know that total factors are ( 3 + 1 ) * ( 2 + 1 ) = 12 ( reason : with 2 ^ n , we have n + 1 possibilities . n ^ 0 to n ^ n . so n + 1 ) = d | a = 3 + 1
b = 2 + 1
c = a * b
|
a ) 25.25 sq cm , b ) 20.25 sq cm , c ) 30.25 sq cm , d ) 36.25 sq cm , e ) 40.25 sq cm | b | power(subtract(subtract(5, const_0_25), const_0_25), const_2) | the sides of a square region , measured to the nearest centimeter , are 5 centimeters long . the least possible value of the actual area of the square region is | "though there might be some technicalities concerning the termnearest ( as 4.5 is equidistant from both 4 and 5 ) the answer still should be : 4.5 ^ 2 = 20.25 . answer : b" | a = 5 - const_0_25
b = a - const_0_25
c = b ** 2
|
a ) 8600 litres , b ) 6400 litres , c ) 12800 litres , d ) 11200 litres , e ) 13200 litres | b | multiply(divide(multiply(multiply(4, const_60), add(10, 6)), 6), 10) | an outlet pipe empties a tank which is full in 10 hours . if the inlet pipe is kept open , which lets water in at the rate of 4 litres / min then outlet pipe would take 6 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 4 litres / min , or 4 * 60 = 240 liters / hour ; net outflow rate when both pipes operate would be x - 240 liters / hour . capacity of the tank = x * 10 hours = ( x - 240 ) * ( 10 + 6 ) hours 10 x = ( x - 240 ) * 16 - - > x = 640 - - > capacity = 10 x = 6400 liters . answer : b ." | a = 4 * const_60
b = 10 + 6
c = a * b
d = c / 6
e = d * 10
|
a ) 10150 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | a | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(350, subtract(multiply(4, 5), multiply(3, 6)))) | a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 350 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 350 total earning = \ inline \ frac { 58 xy } { 100 } = 10150 answer : a ) rs . 10150" | a = 3 * 6
b = 4 * 5
c = a + b
d = 5 * 4
e = c + d
f = 4 * 5
g = 3 * 6
h = f - g
i = 350 / h
j = e * i
|
a ) 3 sec . , b ) 7 sec . , c ) 9 sec . , d ) 11 sec . , e ) none | a | divide(50, multiply(add(68, 8), const_0_2778)) | a train 50 m long is running with a speed of 68 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going ? | "solution speed of the train relative to man = ( 68 - 8 ) = 60 kmph = 60 x 5 / 18 = 50 / 3 m / sec . time taken by it to cover 50 m at ( 50 / 3 ) m / sec = ( 112 - 50 ) kmph = ( 50 x 3 / 50 ) sec = 3 sec . answer a" | a = 68 + 8
b = a * const_0_2778
c = 50 / b
|
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