options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) rs 200 , b ) rs 250 , c ) rs 300 , d ) rs 450 , e ) rs 550 | b | divide(100, 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 100 . 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 = 100 = 250 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 = 100 / n
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a ) 30 , b ) 50 , c ) 60 , d ) 75 , e ) 100 | d | divide(80, const_10) | marginal cost is the cost of increasing the quantity produced ( or purchased ) by one unit . if the fixed cost for n products is $ 10,000 and the marginal cost is $ 80 , and the total cost is $ 16,000 , what is the value of n ? | "total cost for n products = fixed cost for n products + n * marginal cost - - > $ 16,000 = $ 10,000 + n * $ 80 - - > n = 75 . answer : d ." | a = 80 / 10
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a ) a ) 374.80 , b ) b ) 752.804 , c ) c ) 714.642 , d ) d ) 629.906 , e ) of these | a | subtract(multiply(divide(765.765, const_100), 46.82), multiply(divide(const_1, const_3), multiply(divide(765.765, const_100), 46.82))) | 765.765 - ? + 46.82 = 437.785 | "explanation : 374.80 answer : option a" | a = 765 / 765
b = a * 46
c = 1 / 3
d = 765 / 765
e = d * 46
f = c * e
g = b - f
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a ) a ) 165 , b ) b ) 220 , c ) c ) 310 , d ) d ) 185 , e ) e ) 181 | e | multiply(7, divide(subtract(multiply(15000, divide(11.5, const_100)), multiply(10000, divide(9.5, const_100))), multiply(const_10, const_3))) | sam taken a loan rs . 15000 / - from co - operative society with an interest @ 11.5 % per month . at the same time he deposited rs . 10000 / - as fixed deposit with an interest @ 9.5 % per month . after one week sam asked the manager to calculate the interest to be paid . what is the interest amount for 7 days ? | loan amount : rs . 15000 / - @ 11.5 % interest per month = 15000 / - * 11.5 % = rs . 1725 interest for one day = 1725 / 30 = 57.50 interest for 7 days is = 57.50 * 7 = 403 fd amount is = rs . 10000 / - @ 9.5 % interest per month = 10000 * 9.5 % = 950 / - interest for 7 days = 950 / 30 * 7 = 222 interest amount to be paid by sam = 403 - 222 = 181 / - for 7 days answer is e | a = 11 / 5
b = 15000 * a
c = 9 / 5
d = 10000 * c
e = b - d
f = 10 * 3
g = e / f
h = 7 * g
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a ) 37 , b ) 40 , c ) 42 , d ) 45 , e ) 55 | a | divide(150, const_10) | how many integers from 20 to 150 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 20 to 150 , inclusive is ( 147 - 21 ) / 21 + 1 = 7 ; 44 - 7 = 37 . answer : a ." | a = 150 / 10
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | on a two - dimensional coordinate plane , the line d = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | "apparently it ' s d = x ^ 2 - x ^ 3 instead of d = x ^ 2 - d ^ 3 . in this case : the x - intercept is the value ( s ) of x for d = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c ." | a = max(3)
b = a - 1
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a ) 13 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | c | divide(sqrt(100), const_2) | how many even integers from 1 to 100 ( 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 and the corresponding integers are 2 , 4,6 , 8,10 ( more than 4 ) . thus c is the correct answer . | a = math.sqrt(100)
b = a / 2
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a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | c | add(divide(subtract(multiply(multiply(4, 2), 3), multiply(4, 2)), 2), subtract(multiply(multiply(4, 2), 3), multiply(4, 2))) | the ratio of the number of females to males at a party was 1 : 2 but when 4 females and 4 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 4 ) = 2 x - 4 x = 8 there were 3 x = 24 people at the party originally . the answer is c ." | a = 4 * 2
b = a * 3
c = 4 * 2
d = b - c
e = d / 2
f = 4 * 2
g = f * 3
h = 4 * 2
i = g - h
j = e + i
|
a ) 15 , b ) 19 , c ) 12 , d ) 14 , e ) 16 | a | divide(multiply(1500, 5.25), 7875) | if 7875 / 5.25 = 1500 , then 787.5 / 52.5 is equal to ? | "answer given expression 787.5 / 52.5 = 7875 / 525 = 7875 / ( 525 x 100 ) = 1500 / 100 15 correct option : a" | a = 1500 * 5
b = a / 7875
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a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 4 : 5 | a | divide(sqrt(8), sqrt(18)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 18 hours and 8 hours respectively . the ratio of their speeds is | let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 8 : 18 = 4 : 9 . answer : a | a = math.sqrt(8)
b = math.sqrt(18)
c = a / b
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a ) 5 % decrease , b ) 5 % increase , c ) 10 % decrease , d ) 15 % decrease , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 30), add(const_100, 50)), const_100), const_100) | when the price of an article is reduced by 30 % , the sales increases by 50 % . the percentage change in the total amount of receipts is ? | we have , receipts = price x sales therefore , net % change in receipts = ( x + y + xy / 100 ) % = [ - 30 + 50 + ( - 30 x 50 ) / 100 ] % = 5 % increase . answer : b | a = 100 - 30
b = 100 + 50
c = a * b
d = c / 100
e = d - 100
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a ) 2 : 0 , b ) 2 : 3 , c ) 2 : 1 , d ) 2 : 2 , e ) 7 : 9 | e | divide(subtract(6.30, 5.60), subtract(7.20, 6.30)) | find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.60 a kg to produce a mixture worth rs . 6.30 a kg | "by the rule of alligation : cost of 1 kg rice of 1 st kind cost of 1 kg rice of 2 nd kind required ratio = 70 : 90 = 7 : 9 answer : e" | a = 6 - 30
b = 7 - 20
c = a / b
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a ) 17 sec , b ) 12 sec , c ) 16 sec , d ) 14 sec , e ) 18 sec | d | divide(multiply(140, const_2), add(speed(140, 15), speed(140, 10))) | two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 140 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 140 + 140 ) / 20 = 14 sec . answer : d" | a = 140 * 2
b = speed + (
c = a / b
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a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | c | divide(multiply(subtract(multiply(5, 5), multiply(3, 5)), const_100), multiply(5, 5)) | at a special sale , 5 tickets can be purchased for the price of 3 tickets . if 5 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 5 tickets ? | "let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 5 tickets cost 500 5 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 200 , % of 5 tickets = ( 200 / 500 ) * 100 = 40 % answer : c" | a = 5 * 5
b = 3 * 5
c = a - b
d = c * 100
e = 5 * 5
f = d / e
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a ) 20 , b ) 49 , c ) 22 , d ) 23 , e ) none | b | add(7, divide(multiply(7, subtract(20000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 20000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "sol . let the total number of workers be x . then , 8000 x = ( 20000 × 7 ) + 6000 ( x – 7 ) ‹ = › 2000 x = 98000 ‹ = › x = 49 . answer b" | a = 20000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
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a ) 18 , b ) 20 , c ) 17.5 , d ) 30 , e ) 32 | c | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 6 and 34 which are divisible by 7 . | "solution average = ( 7 + 14 + 21 + 28 ) / 4 ) = 70 / 4 = 17.5 . answer c" | a = 6 + 4
b = 34 - 4
c = a + b
d = c / 2
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a ) 63.16 , b ) 62.6 , c ) 62.1 , d ) 62.7 , e ) 62.2 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 6)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 6 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 94 / 100 ) = 59.375 mp = 63.16 answer : a" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 6
f = d / e
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a ) 1 / 3 , b ) 3 / 2 , c ) 3 / 4 , d ) 3 , e ) 3 / 5 | c | divide(divide(3, 5), divide(4, 5)) | if a mixture is 3 ⁄ 5 alcohol by volume and 4 ⁄ 5 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 3 / 5 } / { 4 / 5 } = 3 / 4 c" | a = 3 / 5
b = 4 / 5
c = a / b
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a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) none of these | c | subtract(divide(4, 5), divide(2, 3)) | what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 6 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 6 = . 833 so biggest is 5 / 6 and smallest is 2 / 3 their difference is 5 / 6 - 2 / 3 = 1 / 6 option c" | a = 4 / 5
b = 2 / 3
c = a - b
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a ) 25.67 , b ) 26.67 , c ) 27.2 , d ) 26.98 , e ) 27.76 | b | multiply(divide(divide(multiply(divide(25, const_100), 800), 10), multiply(divide(25, const_100), 800)), const_100) | a reduction of 25 % in the price of salt enables a lady to obtain 10 kgs more for rs . 800 , find the original price per kg ? | "answer : 800 * ( 25 / 100 ) = 200 - - - 10 ? - - - 1 = > rs . 20 800 - - - 600 ? - - - 20 = > rs . 26.67 . answer : b" | a = 25 / 100
b = a * 800
c = b / 10
d = 25 / 100
e = d * 800
f = c / e
g = f * 100
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a ) 33498 , b ) 17000 , c ) 2665 , d ) 28876 , e ) 313 | b | divide(multiply(const_100, 10200), multiply(divide(multiply(subtract(11730, 10200), const_100), divide(10200, const_2)), const_2)) | the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum . | "the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 answer : b" | a = 100 * 10200
b = 11730 - 10200
c = b * 100
d = 10200 / 2
e = c / d
f = e * 2
g = a / f
|
a ) $ 2587.50 , b ) $ 3587.50 , c ) $ 4187.50 , d ) $ 6187.50 , e ) $ 8587.50 | d | multiply(300, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 300 per sq . metre . | "area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 300 hence , total cost = 5.5 × 3.75 × 300 = $ 6187.50 d" | a = 5 * 5
b = 300 * a
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a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | subtract(subtract(150, 60), 30) | of the 150 people at a party , 70 were women , and 30 women tried the appetizer . if 60 people did not try the appetizer , what is the total number of men who tried the appetizer ? | "total people at party = 150 women = 70 so men 150 - 70 = 80 no . of pple who tried appetizer = 150 - 60 ( given info ) = 90 no of women who tried appetizer = 30 so remaining ppl ( men ) who tried the appetizer = 90 - 30 = 60 correct option c" | a = 150 - 60
b = a - 30
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a ) 10 , b ) 8 , c ) 5 , d ) 2 , e ) 4 | a | divide(add(divide(12, 2), divide(28, 2)), const_2) | a man swims downstream 28 km and upstream 12 km taking 2 hours each time , what is the speed of the man in still water ? | "28 - - - 2 ds = 14 ? - - - - 1 12 - - - - 2 us = 6 ? - - - - 1 m = ? m = ( 14 + 6 ) / 2 = 10 answer : a" | a = 12 / 2
b = 28 / 2
c = a + b
d = c / 2
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a ) 30 % , b ) 51 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(130, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(130, const_100)), const_100))) | in a restaurant , the profit is 130 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 130 , s . p . = rs . 230 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 230 . profit = rs . ( 230 - 112 ) = rs . 118 . required percentage = ( 118 / 230 * 100 ) % = 51 % appox answer : b" | a = 130 / 100
b = 100 * a
c = b + 100
d = 12 / 100
e = 1 + d
f = 100 * e
g = c - f
h = 130 / 100
i = 100 * h
j = i + 100
k = g / j
l = 100 * k
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a ) 1 : 6 , b ) 5 : 6 , c ) 6 : 1 , d ) 5 : 1 , e ) 9 : 1 | d | divide(multiply(5, 3), 6) | a dog takes 5 leaps for every 6 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 3 ) leaps of hare : 6 leaps of hare = 15 : 3 - - > 5 : 1 answer : d" | a = 5 * 3
b = a / 6
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a ) 72.85 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) can not be determined | a | divide(subtract(multiply(35, 72), subtract(66, 36)), 35) | a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 36 instead of 66 then find the correct average marks up to two decimal places . | total marks = 35 x 72 = 2520 corrected total marks = 2520 - 36 + 66 = 2550 correct average = 2550 / 35 = 72.85 answer : a | a = 35 * 72
b = 66 - 36
c = a - b
d = c / 35
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a ) 2 / 7 , b ) 3 / 10 , c ) 5 / 11 , d ) 6 / 13 , e ) 11 / 21 | d | add(multiply(divide(7, add(6, 7)), divide(subtract(7, const_1), subtract(add(6, 7), const_1))), multiply(divide(subtract(6, const_1), subtract(add(6, 7), const_1)), divide(6, add(6, 7)))) | a bag contains 6 green balls and 7 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 13 c 2 = 78 the number of ways to draw two green balls is 6 c 2 = 15 the number of ways to draw two white balls is 7 c 2 = 21 p ( two balls of the same colour ) = 36 / 78 = 6 / 13 the answer is d ." | a = 6 + 7
b = 7 / a
c = 7 - 1
d = 6 + 7
e = d - 1
f = c / e
g = b * f
h = 6 - 1
i = 6 + 7
j = i - 1
k = h / j
l = 6 + 7
m = 6 / l
n = k * m
o = g + n
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 48 | e | divide(multiply(multiply(const_1000, const_100), 0.036), divide(75000, const_1000)) | in the biology lab of ` ` jefferson ' ' high school there are 0.036 * 10 ^ 5 germs , equally divided among 75000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ? | "0.036 * 10 ^ 5 can be written as 3600 75000 * 10 ^ ( - 3 ) can be written as 75 required = 3600 / 75 = 48 answer : e" | a = 1000 * 100
b = a * 0
c = 75000 / 1000
d = b / c
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a ) 12 , b ) 15 , c ) 10 , d ) 20 , e ) 8 | a | add(subtract(12, 1), const_1) | set x consists of the integers from 1 to 12 , inclusive , while set y consists of the integers from 0 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | x = { 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11 , 12 } y = { 0 , 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11,12 , 13,14 , 15,16 , 17,18 , 19,20 } common elements = { 1,2 , 3,4 , 5,6 , 7,8 , 9,10 , 11,12 } = 12 elements answer : option a . | a = 12 - 1
b = a + 1
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a ) 443 m , b ) 354 m , c ) 340 m , d ) 350 m , e ) 250 m | c | subtract(multiply(25, multiply(72, const_0_2778)), 160) | a train 160 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ? | "c 340 c = ( 72 * 5 / 18 ) * 25 - 160 = 340" | a = 72 * const_0_2778
b = 25 * a
c = b - 160
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['a ) 1 : 25', 'b ) 1.225', 'c ) 1 : 52', 'd ) 1 : 522', 'e ) none'] | a | divide(const_4, const_100) | a cube of edge 5 cm is cut into cubes each of edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | sol . required ratio = 6 * 1 * 1 / 6 * 5 * 5 = 1 / 25 = 1 : 25 . answer a | a = 4 / 100
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a ) rs . 15000 , b ) rs . 12000 , c ) rs . 10000 , d ) rs . 20000 , e ) none of these | c | divide(25, subtract(subtract(power(add(divide(divide(const_10, const_2), const_100), 1), const_2), const_1), divide(10, const_100))) | the difference b / w the compound interest payble half yearly and the simple interest on a certain sum lent out at 10 % p . a for 1 year is rs 25 . what is the sum ? | in simple words s . i = ptr / 100 = ( p * 1 * 10 ) / 100 p / 10 here amount a = p + s . i = p + p / 10 = 11 p / 10 . . . . now for compound interest when calculated half yearly amount = p [ 1 + r / 200 ] ^ 2 t = p [ 1 + 10 / 200 ] ^ 2 * 1 = 441 p / 400 here its given in the question dat diff btwn simple interest and compound interest half yearly . . . is 25 = > 441 p / 400 - 11 p / 10 = 25 = > p / 400 = 25 = > p = 400 * 25 = 10000 so sum = 10000 answer : c | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d - 1
f = 10 / 100
g = e - f
h = 25 / g
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a ) 5600 , b ) 6000 , c ) 73 , d ) 7200 , e ) 8600 | c | divide(multiply(multiply(multiply(2, const_100), multiply(3, const_100)), 2), multiply(multiply(25, 11), 6)) | how many bricks , each measuring 25 cm x 11 cm x 6 cm , will be needed to build a wall of 2 m x 3 m x 2 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 200 x 300 x 2 ) / ( 25 x 11 x 6 ) = 72.7 = 73 . answer : option c" | a = 2 * 100
b = 3 * 100
c = a * b
d = c * 2
e = 25 * 11
f = e * 6
g = d / f
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a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | d | multiply(divide(20, 300), multiply(divide(20, 300), 300)) | 60 percent of movie theatres in town x have 2 screens or less . 20 % of those theatres sell an average of more than $ 300 worth of popcorn per showing . 56 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell an average of more than $ 300 worth of popcorn per day ? | "lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 300 = 20 % of 60 = 12 number of movie theaters selling popcorn at $ 300 or less = 56 = > number of movie theaters selling popcorn at more than $ 300 = 100 - 56 = 44 of these 44 theaters , 12 are those with 3 screens or less therefore 40 ( 44 - 12 ) must be those with four screens or more d is the answer" | a = 20 / 300
b = 20 / 300
c = b * 300
d = a * c
|
a ) 20 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | a | subtract(add(floor(divide(subtract(99, 69), 3)), divide(subtract(99, 69), 2)), floor(divide(subtract(99, 69), multiply(2, 3)))) | if w is the set of all the integers between 69 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 69 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 68 = [ 99 / 2 ] - [ 68 / 2 ] = 49 - 34 = 15 multiples of 3 from 69 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 68 = [ 99 / 3 ] - [ 68 / 3 ] = 33 - 23 = 10 multiples of 2 and 3 bothi . e . 6 from 69 to 99 = multiples of 6 from 1 to 99 - multiples of 6 from 1 to 68 = [ 99 / 6 ] - [ 68 / 6 ] = 16 - 11 = 5 these 8 numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3 i . e . total numbers in w = 15 + 10 - 5 = 20 answer option a" | a = 99 - 69
b = a / 3
c = math.floor(b)
d = 99 - 69
e = d / 2
f = c + e
g = 99 - 69
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
|
a ) 160 , b ) 195 , c ) 180 , d ) 190 , e ) 200 | b | divide(multiply(75, multiply(26, const_2)), 20) | 75 boys can complete a work in 26 days . how many men need to complete twice the work in 20 days | "one man can complete the work in 26 * 75 = 1950 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 1950 m = 195 workers answer : b" | a = 26 * 2
b = 75 * a
c = b / 20
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a ) 17 % , b ) 24 % , c ) 12 % , d ) 82 % , e ) 16 % | b | multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 55)), const_100), const_100), const_1), const_100) | a trader bought a car at 20 % discount on its original price . he sold it at a 55 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 80 s = 80 * ( 155 / 100 ) = 124 100 - 124 = 24 % answer : b" | a = 100 - 20
b = 100 + 55
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
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a ) 3 , b ) 8 , c ) 27 , d ) 88 , e ) 99 | b | multiply(power(2, const_2), 2) | how many shots of 1 cm radius can be prepared from a sphere of 2 cm radius ? | "4 / 3 ï € * 2 * 2 * 2 = 4 / 3 ï € * 1 * 1 * 1 * x x = 8 answer : b" | a = 2 ** 2
b = a * 2
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a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(660, divide(multiply(multiply(10, 660), divide(add(const_100, 10), const_100)), subtract(multiply(660, divide(add(const_100, 10), const_100)), 660))) | machine a and machine b are each used to manufacture 660 sprockets . it takes machine a 10 hours longer to produce 660 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | b takes x hours a takes x + 10 hours rate of a = 660 / x + 10 rate of b = 660 / x thus , 660 / x = ( 660 / x + 10 ) * 1.10 x = 100 so b = 100 a = 110 sprockets per hour 660 / 110 = 6 answer : a | a = 10 * 660
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 660 * f
h = g - 660
i = d / h
j = 660 / i
|
a ) 450 gms , b ) 100 gms , c ) 300 gms , d ) 240 gms , e ) none of these | d | divide(multiply(50, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 50 % of 40 % of 1200 grams ? | "50 / 100 × 40 / 100 × 1200 = 240 answer : d" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 50 * n
p = o / 100
|
a ) 2480 , b ) 3490 , c ) 6785 , d ) 8225 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1250) | the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1250 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | "you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 15 + 15 ) ^ 2 now we can expand each square ; they are all in the ( x + y ) ^ 2 = x ^ 2 + 2 xy + y ^ 2 pattern . = ( 15 ^ 2 + 2 * 15 + 1 ^ 2 ) + ( 15 ^ 2 + 4 * 15 + 2 ^ 2 ) + ( 15 ^ 2 + 6 * 15 + 3 ^ 2 ) + . . . + ( 15 ^ 2 + 30 * 15 + 15 ^ 2 ) now we have fifteen 15 ^ 2 terms , so adding these gives 15 * 15 ^ 2 = 15 ^ 3 = 3375 . we also have the entire sum 1 ^ 2 + 2 ^ 2 + . . . + 15 ^ 2 , which we know is equal to 1240 . finally adding the middle terms , we have : 2 * 15 + 4 * 15 + 6 * 15 + . . . + 30 * 15 = 15 ( 2 + 4 + 6 + . . . . + 30 ) = 15 * 2 * ( 1 + 2 + 3 + . . . + 15 ) = 15 * 2 * 8 * 15 = 3600 so the sum must be 3375 + 1240 + 3600 = 8215 alternatively , we can use a different factoring pattern . we want to find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 . well if we subtract 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 from this , the answer will be 1240 less than what we want to find . so if we can find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 - ( 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 ) then we can add 1240 to get the answer . now grouping the terms above to get differences of squares , we have = ( 30 ^ 2 - 15 ^ 2 ) + ( 29 ^ 2 - 14 ^ 2 ) + . . . + ( 16 ^ 2 - 1 ^ 2 ) and factoring each of these using x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) , we have = 45 * 15 + 43 * 15 + 41 * 15 + . . . + 17 * 15 = 15 ( 45 + 43 + 41 + . . . + 17 ) in brackets we have an equally spaced sum with fifteen terms , which we can evaluate using the familiar formula . so the above equals 15 * 15 * 62 / 2 = 6975 and adding back the 1250 , we get the answer of 8225 . ( ans d )" | a = 30 + 1
b = 30 * a
c = 2 * 30
d = c + 1
e = b * d
f = 3 + 3
g = e / f
h = g - 1250
|
a ) 776 , b ) 3744 , c ) 299 , d ) 257 , e ) 125 | b | divide(multiply(divide(288, divide(subtract(52, subtract(const_100, 52)), const_100)), 52), const_100) | there were two candidates in an election . winner candidate received 52 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 52 % l = 48 % 52 % - 48 % = 4 % 4 % - - - - - - - - 288 52 % - - - - - - - - ? = > 3744 answer : b" | a = 100 - 52
b = 52 - a
c = b / 100
d = 288 / c
e = d * 52
f = e / 100
|
a ) 190 , b ) 192 , c ) 194 , d ) 196 , e ) 198 | b | divide(const_1, add(divide(const_1, 320), divide(const_1, 480))) | john completes a piece of work in 320 days , rose completes the same work in 480 days . if both of them work together , then the number of days required to complete the work is ? | "if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 320 × 480 / 800 = 192 days answer is b" | a = 1 / 320
b = 1 / 480
c = a + b
d = 1 / c
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | subtract(add(add(18, 19), 2), 30) | in a sports club with 30 members , 18 play badminton and 19 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 18 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 18 - x ) + ( 19 - x ) + x + 2 = 30 39 - 2 x + x = 30 39 - x = 30 x = 9 so 9 members play both badminton and tennis . c" | a = 18 + 19
b = a + 2
c = b - 30
|
a ) 72 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none | a | multiply(multiply(multiply(3, 2), divide(1.2, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1.2 cm when a man gets on it . the mass of man is | "solution volume of water displaced = ( 3 x 2 x 0.012 ) m 3 = 0.072 m 3 . mass of man = volume of water displaced × density of water = ( 0.072 × 1000 ) kg = 72 kg . answer a" | a = 3 * 2
b = 1 / 2
c = a * b
d = c * 1000
|
a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 8 , d ) 3 / 16 , e ) 5 / 16 | e | divide(divide(factorial(5), multiply(factorial(3), factorial(subtract(5, 3)))), power(const_2, 5)) | what is the probability for a family with 5 children to have two boys and 3 girls ( assuming the probability of having a boy or a girl is equal ) ? | one possible case is : girl - girl - girl - boy - boy the probability of this is 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 there are 5 c 3 = 10 such cases so we should multiply by 10 . p ( two boys and three girls ) = 10 / 32 = 5 / 16 the answer is e . | a = math.factorial(5)
b = math.factorial(3)
c = 5 - 3
d = math.factorial(c)
e = b * d
f = a / e
g = 2 ** 5
h = f / g
|
a ) 705 , b ) 825 , c ) 805 , d ) 905 , e ) 810 | c | divide(5915, 5) | the lowest number which should be added to 5915 so that the sum is exactly divisible by 3 , 5 , 8 and 7 is : | "l . c . m . of 3 , 5 , 8 and 7 = 840 . on dividing 5915 by 840 , the remainder is 35 . number to be added = ( 840 - 35 ) = 805 . answer : option ' c '" | a = 5915 / 5
|
a ) 1 1 / 5 min , b ) 4 1 / 5 min , c ) 3 1 / 5 min , d ) 2 1 / 5 min , e ) none of these | a | divide(divide(400, multiply(20, const_0_2778)), const_60) | how many minutes does aditya take to cover a distance of 400 m , if he runs at a speed of 20 km / hr ? | explanation : we know that , time = distance / speedspeed = 20 km / hr = 20 x 5 / 18 m / sec = 509 m / sectime = ( 400 x 9 / 50 ) = 72 sec = 1 1 / 5 min answer : a | a = 20 * const_0_2778
b = 400 / a
c = b / const_60
|
a ) 5 kg , b ) 6.625 kg , c ) 6.25 kg , d ) 6.35 kg , e ) 8.8 kg | e | subtract(multiply(divide(9, 25), multiply(divide(4, add(const_1, const_4)), 100)), multiply(divide(const_1, add(const_1, const_4)), 100)) | the ratio of sodium chloride to ammonium in 100 kg of mixed ammonium normally used by 3 chemists is 1 : 4 . the amount of sodium chloride to be added to 100 kg of mixed ammonium to make the ratio 9 : 25 is | sodium chloride in 100 kg = 1 / 5 x 100 = 20 kg ammonium in 100 kg = 4 / 5 x 100 = 80 kg now 80 is 25 parts out of ( 25 + 9 ) = 34 parts . hence 9 parts = 9 / 25 x 80 = 28.8 amount to be added = 28.8 - 20 = 8.8 kg answer : e | a = 9 / 25
b = 1 + 4
c = 4 / b
d = c * 100
e = a * d
f = 1 + 4
g = 1 / f
h = g * 100
i = e - h
|
a ) 3 % decrease , b ) 5 % decrease , c ) 9 % decrease , d ) 4 % decrease , e ) 8 % decrease | c | subtract(const_100, subtract(add(30, const_100), divide(multiply(add(30, const_100), 30), const_100))) | the salary of a worker is first increased by 30 % and afterwards reduced by 30 % . what is net change in his salary ? | ( 30 * 30 ) / 100 = 9 % decrease answer : c : | a = 30 + 100
b = 30 + 100
c = b * 30
d = c / 100
e = a - d
f = 100 - e
|
a ) 24 , b ) 72 , c ) 297 , d ) 267 , e ) 29 | b | multiply(subtract(17, 5), 6) | think of a number , divide it by 6 and add 5 to it . the result is 17 . what is the number thought of ? | "explanation : 17 - 5 = 12 12 x 6 = 72 answer : b" | a = 17 - 5
b = a * 6
|
a ) 560 , b ) 616 , c ) 450 , d ) 900 , e ) 1024 | c | add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 57.40), const_3)) | x and y are both integers . if x / y = 57.40 , then what is the sum of all the possible two digit remainders of x / y ? | "remainder = 0.40 - - > 40 / 100 - - > can be written as ( 40 / 4 ) / ( 100 / 4 ) = 10 / 25 so remainders can be 10 , 20 , 30 , 40 , . . . . . 90 . we need the sum of only 2 digit remainders - - > 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 = 450 answer : c" | a = 3 / 2
b = a * 100
c = 2 + 3
d = c * 57
e = d + 3
f = b + e
|
a ) 3630 , b ) 3690 , c ) 2887 , d ) 9977 , e ) 2212 | b | multiply(divide(6300, add(add(6300, 4200), 10500)), 12300) | a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12300 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12300 = 3690 . answer : b" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 12300
|
a ) $ 208 , b ) $ 216 , c ) $ 224 , d ) $ 232 , e ) $ 240 | d | multiply(divide(406, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | $ 406 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 406 x = 58 4 x = 232 the answer is d . | a = 1 / 2
b = a + 1
c = b + 2
d = 406 / c
e = d * 2
|
a ) a ) 130 , b ) b ) 134 , c ) c ) 148 , d ) d ) 158 , e ) e ) 167 | e | add(multiply(18, 9), 5) | what is the dividend . divisor 18 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 18 * 9 + 5 d = 162 + 5 d = 167 answer e" | a = 18 * 9
b = a + 5
|
a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | c | subtract(add(24, 17), const_1) | at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 17 th highest price and the 24 th lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 16 items sold at a higher price than the radio and 23 items sold at a lower price than the radio . including the radio , there were 16 + 23 + 1 = 40 items sold . the answer is c ." | a = 24 + 17
b = a - 1
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | b | multiply(divide(multiply(divide(20, const_100), 15), add(15, 5)), const_100) | a 15 lt mixture contains water with 20 % of alcohol . if 5 more lts of water added to the mixture what is the percentage of alcohol ? | initially the mixture contains 20 % of 15 = 15 * ( 20 / 100 ) = 3 lts of alcohol after mixing another 5 lts of water . . . the total liquid is 20 lts . . so alcohol percentage = ( 3 / 20 ) * 100 = 15 % answer : b | a = 20 / 100
b = a * 15
c = 15 + 5
d = b / c
e = d * 100
|
a ) 4 , b ) 7 , c ) 6 , d ) 8 , e ) 6 | a | add(subtract(35, add(30, 1)), 1) | the average weight of a group of boys is 30 kg . after a boy of weight 35 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 35 kg joins the group , total weight of boys = 30 x + 35 so 30 x + 35 + 31 ( x + 1 ) = > x = 4 . answer : a" | a = 30 + 1
b = 35 - a
c = b + 1
|
a ) 5 / 2 , b ) 10 / 3 , c ) 7 / 2 , d ) 14 / 3 , e ) 11 / 2 | e | subtract(divide(add(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2)), divide(subtract(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2))) | by how much does the larger root of the equation 2 c ^ 2 + 5 c = 12 exceed the smaller root ? | "for 2 c ^ 2 + 5 c = 12 roots are [ - 5 + sqrt ( 25 + 96 ) ] / 4 or [ - 5 - sqrt ( 25 + 96 ) ] / 4 = 1.5 or - 4 hence larger root 1.5 is 1.5 - ( - 4 ) = 5.5 = 11 / 2 greater than smaller root ( - 4 ) . hence option ( e ) ." | a = 5 ** 2
b = 2 * 12
c = b * 4
d = a + c
e = math.sqrt(d)
f = 5 + e
g = 2 * 2
h = f / g
i = 5 ** 2
j = 2 * 12
k = j * 4
l = i + k
m = math.sqrt(l)
n = 5 - m
o = 2 * 2
p = n / o
q = h - p
|
a ) rs 48000 , b ) rs 47000 , c ) rs 47400 , d ) rs 48400 , e ) none of these | d | multiply(multiply(multiply(multiply(add(2, 2), 2), const_100), const_100), divide(2, add(add(1, 2), 2))) | in business , a and c invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by a and b was 3 : 2 , if rs 157300 was their profit , how much amount did b receive . | "explanation : a : b = 3 : 2 = 6 : 4 = > a : c = 2 : 1 = 6 : 3 = > a : b : c = 6 : 4 : 3 b share = ( 4 / 13 ) * 157300 = 48400 option d" | a = 2 + 2
b = a * 2
c = b * 100
d = c * 100
e = 1 + 2
f = e + 2
g = 2 / f
h = d * g
|
a ) 3 / 25 , b ) 11 / 36 , c ) 5 / 12 , d ) 2 / 3 , e ) 25 / 36 | c | divide(divide(multiply(50, 5), 30), 20) | at a speed of 50 miles per hour , a certain car uses 1 gallon of gasoline every 30 miles . if the car starts with a full 20 gallon tank of gasoline and travels for 5 hours at 50 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | "gas used = ( 5 hours ) * ( 50 miles / hour ) * ( 1 gallon / 30 miles ) = 8 + 1 / 3 gallons portion used = ( 8 + 1 / 3 ) / 20 = 5 / 12 ans c" | a = 50 * 5
b = a / 30
c = b / 20
|
a ) $ 260 , b ) $ 360 , c ) $ 160 , d ) $ 460 , e ) $ 560 | b | add(add(add(multiply(multiply(const_3, 10), const_2), multiply(const_2, 10)), add(add(multiply(multiply(const_3, 10), const_2), multiply(const_2, 10)), 10)), subtract(multiply(add(add(multiply(multiply(const_3, 10), const_2), multiply(const_2, 10)), 10), const_3), add(multiply(multiply(const_3, 10), const_2), multiply(const_2, 10)))) | a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 10 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 10 a = 10 + b ( 1 ) = = = > c = 30 + 3 b - b = 2 b + 30 = = > 2 b - c = - 30 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 10 = 2.5 b - 10 = = > 2.5 b - c = 10 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 40 b = $ 80 a = $ 90 c = 270 - 80 = $ 190 total amount = 90 + 80 + 190 = $ 360 answer : b" | a = 3 * 10
b = a * 2
c = 2 * 10
d = b + c
e = 3 * 10
f = e * 2
g = 2 * 10
h = f + g
i = h + 10
j = d + i
k = 3 * 10
l = k * 2
m = 2 * 10
n = l + m
o = n + 10
p = o * 3
q = 3 * 10
r = q * 2
s = 2 * 10
t = r + s
u = p - t
v = j + u
|
a ) 52 , b ) 56 , c ) 60 , d ) 64 , e ) 68 | c | add(add(multiply(divide(multiply(1, 20), subtract(multiply(3, 3), multiply(1, 5))), 5), 20), multiply(divide(multiply(1, 20), subtract(multiply(3, 3), multiply(1, 5))), 3)) | in a can , there is a mixture of milk and water in the ratio 5 : 3 . if the can is filled with an additional 20 liters of milk , the can would be full and the ratio of milk and water would become 3 : 1 . find the capacity of the can ? | "let c be the capacity of the can . ( 5 / 8 ) * ( c - 20 ) + 20 = ( 3 / 4 ) * c 5 c - 100 + 160 = 6 c c = 60 the answer is c ." | a = 1 * 20
b = 3 * 3
c = 1 * 5
d = b - c
e = a / d
f = e * 5
g = f + 20
h = 1 * 20
i = 3 * 3
j = 1 * 5
k = i - j
l = h / k
m = l * 3
n = g + m
|
a ) 44 % increase , b ) 44 % decrease , c ) 66 % increase , d ) 75 % increase , e ) none of these | a | add(add(20, 20), divide(multiply(20, 20), const_100)) | when the price of a radio was reduced by 20 % , its sale increased by 80 % . what was the net effect on the sale ? | let the original price be x and sale be of y units . then , the revenue collected initially = x × y now , new price = 0.8 x , new sale = 1.8 y then , new revenue collected = 1.44 xy % increase in revenue = 0.44 xy / xy × 100 = 44 % increase answer a | a = 20 + 20
b = 20 * 20
c = b / 100
d = a + c
|
a ) 18 sec , b ) 15 sec , c ) 21 sec , d ) 20 sec , e ) 19 sec | b | divide(add(100, 200), multiply(add(42, 30), const_0_2778)) | two trains of length 100 m and 200 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 200 = 300 m . the time required = d / s = 300 / 20 = 15 sec . answer : b" | a = 100 + 200
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 3 , b ) 7 , c ) 11 , d ) 21 , e ) 24 | b | divide(divide(105, const_3), add(const_1, const_4)) | the greatest number that exactly divided 105 , 1001 and 2436 is : | solution h . c . f . of 2436 and 1001 is 7 . also , h . c . f . of 105 and 7 is 7 . ∴ h . c . f . of 105 , 1001 and 2436 is 7 . answer b | a = 105 / 3
b = 1 + 4
c = a / b
|
a ) 10,300 , b ) 10,030 , c ) 1,353 , d ) 1,352 , e ) 1,539 | e | subtract(478,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 478,600 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 324,700 and 478,600 is 478,600 - 324,700 = 153,900 - one number per each hundred gives 153,900 / 100 = 1,539 numbers . answer : e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 478 - 600
|
a ) 14700 , b ) 14500 , c ) 14900 , d ) 14300 , e ) 15540 | e | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 37,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 ⇒ 3 x + 14000 = 50000 ⇒ 3 x = 50000 – 14000 = 36000 ⇒ x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit × 21 / 50 = 37000 × 21 / 50 = 15540 answer is e" | a = 5000 * 10
b = 4000 + 5000
c = b + 5000
d = a - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 3 * 12
i = h * 1000
j = 5000 * 10
k = i / j
l = g * k
|
a ) 1,980 , b ) 2,020 , c ) 2,100 , d ) 2,200 , e ) 2,500 | c | multiply(const_2, const_1) | john would make the 3 - letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and both are consonant . how many cases of the codes are there ? | no of vowels : 5 ( a , e , i , o , u ) no of consonants : 21 ( 26 - 5 ) now , the middle letter can be filled up in 5 ways . the first letter can be filled in 21 ways ( as only consonants are allowed ) as one of the consonants is used in the first letter , the third letter can be filled up in 20 ways , so that the letter must be different from each other . so total number of ways = 21 x 5 x 20 = 2100 answer : c | a = 2 * 1
|
a ) 277 , b ) 209 , c ) 121 , d ) 200 , e ) 128 | d | subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000)) | ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 10 percent . overall he make a . ? | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 800 = rs . 200 as this is positive , an overall profit of rs . 200 was made . answer : d" | a = 4 / 100
b = 1 - a
c = 15000 * b
d = 10 / 100
e = 1 + d
f = 8000 * e
g = c + f
h = 15000 + 8000
i = g - h
|
a ) 9.1 % , b ) 6.23 % , c ) 7 % , d ) 8.12 % , e ) 10 % | a | multiply(divide(subtract(6000, add(4700, 800)), add(4700, 800)), const_100) | alfred buys an old scooter for $ 4700 and spends $ 800 on its repairs . if he sells the scooter for $ 6000 , his gain percent is ? | "c . p . = 4700 + 800 = $ 5500 s . p . = $ 6000 gain = 6000 - 5500 = $ 500 gain % = 500 / 5500 * 100 = 9.1 % answer is a" | a = 4700 + 800
b = 6000 - a
c = 4700 + 800
d = b / c
e = d * 100
|
a ) 33300 , b ) 34500 , c ) 38500 , d ) 39800 , e ) 45200 | a | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 3,5), const_1), 1), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | what is the sum of all 3 digit nos formed by { 1 , 3,5 , 7,9 } with nos not repeated | "here 3 digit no . can be formed in 5 * 4 * 3 = 60 . thus we hv to find the sum of 60 nos . now 60 / 5 = 12 ; we hv 12 no ' s . starting with 1 , 12 no ' s . with 3,5 , 7,9 . . . . so 12 * 100 * ( 1 + 3 + 5 + 7 + 9 ) + 12 * 10 * ( 1 + 3 + 5 + 7 + 9 ) + 12 * 1 * ( 1 + 3 + 5 + 7 + 9 ) = 30000 + 3000 + 300 = 33300 answer : a" | a = 1000 - 3
b = 3 * 3
c = b * 10
d = 3 * 3
e = c + d
f = e + 3
g = a - f
h = g / 3
i = h + 1
j = i / 1
k = 1000 - 3
l = 3 * 3
m = l * 10
n = 3 * 3
o = m + n
p = o + 3
q = k + p
r = j * q
|
a ) 5.5 , b ) 10.45 , c ) 7 , d ) 8 , e ) 9.5 | b | divide(180, multiply(add(55, 7), const_0_2778)) | a train 180 m long is running with a speed of 55 km / hr . in what time will it pass a man who is running at 7 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 55 + 7 = 62 km / hr . = 62 * 5 / 18 = 155 / 9 m / sec . time taken to pass the men = 180 * 9 / 155 = 10.45 sec . answer : option b" | a = 55 + 7
b = a * const_0_2778
c = 180 / b
|
a ) 4 / 13 , b ) 1 / 52 , c ) 1 / 4 , d ) 1 / 5 , e ) none of above | a | divide(multiply(const_4, const_3), 52) | from a pack of 52 cards , 1 card is drawn at random . find the probability of a face card drawn | "explanation : total number of cases = 52 total face cards = 16 [ favourable cases ] so probability = 16 / 52 = 4 / 13 answer : a" | a = 4 * 3
b = a / 52
|
a ) 2372 , b ) 2572 , c ) 4303 , d ) 2343 , e ) 3972 | c | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), const_3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), const_3))) | there is 60 % increase in an amount in 6 years at si . what will be the ci of rs . 13,000 after 3 years at the same rate ? | let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 13000 . t = 3 years and r = 10 % p . a . c . i . = rs . 13000 x 1 + 10 3 - 1 100 = rs . 13000 x 331 1000 = 4303 . c | a = 100 * 100
b = 60 / 6
c = 100 * b
d = c * 3
e = a + d
f = 60 / 6
g = f / 100
h = 1 + g
i = 60 / 6
j = i / 100
k = 1 + j
l = h * k
m = 60 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 60 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
|
a ) 648 , b ) 300 , c ) 252 , d ) 225 , e ) 26 | c | add(add(divide(subtract(1000, 150), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | how many times digit 6 is used while writing numbers from 150 to 1000 ? | "there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' so far we have calculated 190 . the total now comes to 280 . the nearest to which is 252 . hence c ." | a = 1000 - 150
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 6 * 2
h = f + g
|
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | a | divide(396, add(multiply(8, const_3), multiply(6, const_2))) | sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 396 per week . how much does she earn in dollars per hour ? | "let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 396 per week 36 x = 396 x = 11 correct option : a" | a = 8 * 3
b = 6 * 2
c = a + b
d = 396 / c
|
a ) 24 , b ) 77 , c ) 88 , d ) 29 , e ) 12 | e | divide(multiply(18, 36), 54) | 36 men can complete a piece of work in 18 days . in how many days will 54 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 54 : 36 : : 18 : x x = 12 answer : e ) 12 days" | a = 18 * 36
b = a / 54
|
a ) 16 % , b ) 17 % , c ) 18 % , d ) 41 % , e ) 13 % | d | multiply(divide(subtract(65000, add(34000, 12000)), add(34000, 12000)), const_100) | ramu bought an old car for rs . 34000 . he spent rs . 12000 on repairs and sold it for rs . 65000 . what is his profit percent ? | "total cp = rs . 34000 + rs . 12000 = rs . 46000 and sp = rs . 65000 profit ( % ) = ( 65000 - 46000 ) / 46000 * 100 = 41 % answer : d" | a = 34000 + 12000
b = 65000 - a
c = 34000 + 12000
d = b / c
e = d * 100
|
a ) 8.5 kmph , b ) 16.0 kmph , c ) 22.5 kmph , d ) 18.6 kmph , e ) none of these | d | add(const_0_25, divide(35, const_2)) | john took a bus from home to market , that travels at 50 kmph . while walking back at 5 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 35 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 50 , by walking = x / 5 , by cycling = x / 35 hours : . average speed = total distance / total time = 5 x / x / 25 + x / 5 + x / 35 = 5 * 60 / 2.4 + 12 + 1.7 = 18.6 answer : d | a = 35 / 2
b = const_0_25 + a
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | multiply(divide(subtract(const_1, multiply(add(inverse(3), inverse(9)), 2)), inverse(3)), const_60) | working at a constant rate , p can finish a job in 3 hours . q , also working at a constant rate , can finish the same job in 9 hours . if they work together for 2 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | "each hour they complete 1 / 3 + 1 / 9 = 4 / 9 of the job . in 2 hours , they complete 2 ( 4 / 9 ) = 8 / 9 of the job . the time for p to finish is ( 1 / 9 ) / ( 1 / 3 ) = ( 1 / 3 ) hour = 20 minutes the answer is a ." | a = 1/(3)
b = 1/(9)
c = a + b
d = c * 2
e = 1 - d
f = 1/(3)
g = e / f
h = g * const_60
|
a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | b | divide(add(divide(subtract(4, 6), 2), 2), divide(add(2, 6), 2)) | line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 6 ) . what is the slope of line m ? | "ans : b solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 1 ) as we can see that the y axis of intercept point is ( 0 , - 2 ) means line m is parallel to x axis slope m = - 1 ans : b" | a = 4 - 6
b = a / 2
c = b + 2
d = 2 + 6
e = d / 2
f = c / e
|
a ) 5 , b ) 10 , c ) 12 , d ) 40 , e ) 50 | c | divide(subtract(multiply(15.5, 12), multiply(12, 15)), subtract(16, 15.5)) | the average age of a group of persons going for picnic is 16 years . 12 new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years . the number of persons initially going for picnic is | solution let the initial number of persons be x . then 16 x + 12 x 15 = 15.5 ( x + 20 ) = 0.5 x = 6 x = 12 . answer c | a = 15 * 5
b = 12 * 15
c = a - b
d = 16 - 15
e = c / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(divide(multiply(const_3, multiply(6, 3)), add(6, 3)), 3) | from an island , it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for 3 hours at 6 kilometers per hour , while ferry q takes a route that is three times longer . if ferry p is slower than ferry q by 3 kilometers per hour , how many hours longer is the journey of ferry q compared with the journey of ferry p ? | "the distance traveled by ferry p is 18 km . then the distance traveled by ferry q is 54 km . ferry q travels at a speed of 9 kph . the time of the journey for ferry q is 54 / 9 = 6 hours , which is 3 hours more than ferry p . the answer is c ." | a = 6 * 3
b = 3 * a
c = 6 + 3
d = b / c
e = d - 3
|
a ) 21375 , b ) 25750 , c ) 26550 , d ) 26750 , e ) 25725 | a | multiply(900, multiply(5, 4.75)) | the length of a room is 5 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre . | "area = 5 × 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 5 × 4.75 × 900 = 5 × 4275 = rs . 21375 answer is a ." | a = 5 * 4
b = 900 * a
|
a ) rs . 2535 , b ) rs . 2646 , c ) rs . 2546 , d ) rs . 2656 , e ) rs . 2565 | b | add(divide(246, subtract(power(add(const_1, divide(5, const_100)), const_2), const_1)), 246) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 5 % p . a . was rs . 246 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned . | "let the sum be rs . p p { [ 1 + 5 / 100 ] 2 - 1 } = 246 p ( 5 / 100 ) ( 2 + 5 / 100 ) = 246 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 246 / ( 0.05 ) ( 2.05 ) = 2400 . amount = rs . 2646 answer : b" | a = 5 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 246 / d
f = e + 246
|
a ) s . 800 , b ) s . 2400 , c ) s . 4000 , d ) s . 3200 , e ) s . 4500 | e | multiply(subtract(multiply(divide(1800, 2), 3), 1800), 5) | the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 1800 , then what is the income of p 1 ? | "let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x – 3 y = 1800 … ( i ) and 4 x – 2 y = 1800 … … . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 1800 - > x = 900 p 1 ’ s income = rs 5 * 900 = rs . 4500 answer : e" | a = 1800 / 2
b = a * 3
c = b - 1800
d = c * 5
|
a ) 278 , b ) 366 , c ) 349 , d ) 335 , e ) 295 | c | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(4, const_100)), multiply(add(100, 4), 100)), 100) | a computer manufacturer produces a certain electronic component at a cost of $ 100 per component . shipping costs for delivering the components are $ 4 per unit . further , the manufacturer has costs of $ 24,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 100 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 24500 is a fixed cost each component is $ 104 ( $ 100 to produce , $ 4 to ship ) manufacturer will be producing and selling 100 components so therefore the equation to find price would be 100 * p = 24500 + ( 100 * 100 ) + ( 100 * 4 ) p = ( 24500 + 10000 + 400 ) / 100 p = 349 answer : c" | a = 4 * 4
b = a * 1000
c = 4 * 100
d = b + c
e = 100 + 4
f = e * 100
g = d + f
h = g / 100
|
a ) a ) 15 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7 | a | divide(multiply(45, 432), 21) | a number when divided by 45 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 21 | "explanation : p ÷ 45 = 432 = > p = 432 * 45 = 19440 p / 31 = 19440 / 21 = 925 , remainder = 15 answer : option a" | a = 45 * 432
b = a / 21
|
a ) 39 ° c , b ) 44 ° c , c ) 37 ° c , d ) 42 ° c , e ) 38 c | e | subtract(multiply(45, const_3), subtract(multiply(50, const_3), 53)) | the average temperature for tuesday , wednesday and thursday was 45 ° c . the average temperature for wednesday , thursday and friday was 50 ° c . if the temperature on friday be 53 ° c , what was the temperature on tuesday ? | "explanation : t + w + t = 45 × 3 = 135 ° c w + t + f = 50 × 3 = 150 ° c also , temperature on friday = 53 ° c temperature on tuesday = 135 + 53 - 150 = 38 ° c answer : option e" | a = 45 * 3
b = 50 * 3
c = b - 53
d = a - c
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 6 | e | multiply(subtract(5, 2), 2) | if ( c - a ) / ( c - b ) = 1 , then ( 5 b - 2 a ) / ( c - a ) = | let ' s say c = 3 , b = 2 , a = 2 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 6 answer e ( hopefully ) ) ) | a = 5 - 2
b = a * 2
|
a ) 22 , b ) 24 , c ) 26 , d ) 30 , e ) 44 | d | divide(add(87, const_1), const_2) | the sum of three consecutive numbers is 87 . the greatest among these three numbers is : | "sol . let the numbers be x , x + 1 and x + 2 . then , x + ( x + 1 ) + ( x + 2 ) = 87 ⇔ 3 x = 84 ⇔ x = 28 . greatest number = ( x + 2 ) = 30 . answer d" | a = 87 + 1
b = a / 2
|
a ) 18 , b ) 20 , c ) 21 , d ) 22 , e ) 25 | a | subtract(add(13, 15), add(subtract(subtract(22, 3), 15), subtract(subtract(22, 3), 13))) | in a group of 22 boys , 13 play basketball and 15 play football . 3 of the boys play neither . how many play both ? | 18 boys play something . if b play both , then 13 + 15 - b = 18 answer is a | a = 13 + 15
b = 22 - 3
c = b - 15
d = 22 - 3
e = d - 13
f = c + e
g = a - f
|
['a ) 11', 'b ) 9', 'c ) 7', 'd ) 5', 'e ) 3'] | a | divide(add(sqrt(add(power(const_3, const_2), multiply(multiply(44, const_2), const_4))), const_3), const_2) | a polygon has 44 diagonals , then the number of its sides are | let the number of sides be n . the number of diagonals is given by nc 2 - n therefore , nc 2 - n = 44 , n > 0 n ( n - 1 ) / 2 - n = 44 n 2 - 3 n - 88 = 0 n 2 - 11 n + 8 n - 88 = 0 n ( n - 11 ) + 8 ( n - 11 ) = 0 n = - 8 or n = 11 . answer : a | a = 3 ** 2
b = 44 * 2
c = b * 4
d = a + c
e = math.sqrt(d)
f = e + 3
g = f / 2
|
a ) 51 : 53 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | a | divide(add(const_100, 2), add(const_100, 6)) | the cash difference between the selling prices of an article at a profit of 2 % and 6 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 102 % of x ) / ( 106 % of x ) = 102 / 106 = 51 / 53 = 51 : 53 . answer : a" | a = 100 + 2
b = 100 + 6
c = a / b
|
a ) 1.0 , b ) 8.2 , c ) 7.7 , d ) 2.6 , e ) 8.3 | e | multiply(multiply(multiply(const_0_2778, subtract(60, 60)), 50), inverse(multiply(const_0_2778, add(60, 60)))) | two trains of equal length , running with the speeds of 60 and 60 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 50 * 100 / 18 = 2500 / 9 rs = 60 + 60 = 120 * 5 / 18 t = 2500 / 9 * 18 / 600 = 8.3 sec . answer : e" | a = 60 - 60
b = const_0_2778 * a
c = b * 50
d = 60 + 60
e = const_0_2778 * d
f = 1/(e)
g = c * f
|
a ) 1 , b ) 20 , c ) 2 , d ) 200 , e ) 2.5 | a | divide(2000, multiply(200, 10)) | in a certain village , 200 litres of water are required per household per month . at this rate , if there are 10 households in the village , how long ( in months ) will 2000 litres of water last ? | "i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 10 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 10 = 2000 ( m * n ) water available is 2000 ( p ) it will last 1 months ( p / m * n ) answer ( a )" | a = 200 * 10
b = 2000 / a
|
a ) 62 % , b ) 68 % , c ) 52 % , d ) 22 % , e ) 42 % | b | subtract(const_100, add(multiply(80, divide(20, const_100)), multiply(divide(80, const_100), 20))) | in a factory , there are 80 % technicians and 20 % non - technicians . if the 80 % of the technicians and 20 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ? | "total = 100 t = 80 nt = 20 80 * ( 20 / 100 ) = 24 80 * ( 20 / 100 ) = 24 16 + 16 = 32 = > 100 - 32 = 68 % answer : b" | a = 20 / 100
b = 80 * a
c = 80 / 100
d = c * 20
e = b + d
f = 100 - e
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | add(add(const_4, const_3), const_2) | what is the units digit of ( 147 ^ 21 ) ^ 48 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 21 has the form 4 n + 1 so the units digit is 7 inside the bracket . the number 48 has the form 4 n , so the units digit is 1 . the answer is a . | a = 4 + 3
b = a + 2
|
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