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 ) 5.29 min , b ) 5.28 min , c ) 4.08 min , d ) 9.28 min , e ) 5.988 min | c | multiply(divide(divide(561, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 561 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 561 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 561 m apart , they take ( 100 / 825 * 561 / 1000 ) hrs = ( 561 / 8250 * 60 ) min = 4.08 min . answer : c | a = 561 / 1000
b = 4 + 5
c = a / b
d = c * const_60
|
a ) 13 / 19 , b ) 1 / 3 , c ) 9 / 20 , d ) 7 / 10 , e ) 5 / 7 | a | divide(subtract(divide(60, const_100), multiply(subtract(const_1, divide(76, const_100)), subtract(const_1, divide(2, 3)))), divide(76, const_100)) | at a small company , 76 percent of the employees are women , and 60 percent of the employees are married . if 2 / 3 of the men are single , what fraction of the women are married ? | "lets take total employees are 100 . given that , total women = 76 and total married = 60 . total men = 100 - 76 = 24 and single men = 2 / 3 * 24 = 16 . married men = total men - single men = 24 - 16 = 8 . married women = total married - married men = 60 - 8 = 52 . fraction of women are married = married women / total women = 52 / 76 = 13 / 19 . ans a" | a = 60 / 100
b = 76 / 100
c = 1 - b
d = 2 / 3
e = 1 - d
f = c * e
g = a - f
h = 76 / 100
i = g / h
|
a ) 25500 , b ) 25000 , c ) 40500 , d ) 20000 , e ) 40000 | e | divide(400, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(25, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 25 % profit ans sells the remaining at a loss of 5 % . he incurred an overall profit of rs . 400 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 5 / 4 = 5 x / 20 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 152 x / 200 total sp = 50 x / 200 + 152 x / 200 = 202 x / 200 overall profit = 152 x / 200 - x = 2 x / 200 x / 50 = 400 = > x = 40000 answer : e" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 25 / 100
f = 20 / 100
g = e * f
h = d - g
i = 400 / h
|
a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 13 % | e | multiply(divide(subtract(385, 340), 340), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 340 million , approximately what was the percent increase in sales ? | "new value β old value ) / old value x 100 we are given : february sales this year = 385 million february sales last year = 340 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 385 million and the old value = 340 million . let β s plug them into our percent change formula . ( new value β old value ) / old value x 100 [ ( 385 β 340 ) / 340 ] x 100 65 / 340 x 100 β 13 % . the answer is e ." | a = 385 - 340
b = a / 340
c = b * 100
|
a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6 | a | subtract(add(36, 27), subtract(45, 8)) | each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 36 have long fur , 27 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | "no of dogs = 45 long fur = 36 brown = 27 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 36 + 27 - both both = 26 answer a" | a = 36 + 27
b = 45 - 8
c = a - b
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 10 | d | add(add(const_2, add(const_1, const_4)), const_2) | an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 10000 people who applied for the army and 98 were finally selected , what was the value of n ? | "step ( 1 ) 4000 accepted . step ( 2 ) another 40 % of 40000 = 1600 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 1600 = 800 step ( 4 ) 50 % of 800 = 400 step ( 5 ) 50 % of 400 = 200 . . . here we are done with total 5 steps 70 % of 200 = 140 and last step of accepting 70 % of 140 = 98 ( our target ) total 7 steps required . ans d" | a = 1 + 4
b = 2 + a
c = b + 2
|
a ) 1200 km , b ) 1980 km , c ) 2000 km , d ) 2500 km , e ) 3600 km | b | multiply(30, 55) | a walks at 30 kmph and 30 hours after his start , b cycles after him at 55 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 55 = 30 x = 1980 km answer is b" | a = 30 * 55
|
a ) 25 % , b ) 33 1 / 3 % , c ) 50 % , d ) 66 2 / 3 % , e ) 75 % | a | multiply(divide(subtract(25, 20), 20), const_100) | a certain protective additive increases from 20 days to 25 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? | "general formula for percent increase or decrease , ( percent change ) : percent = change / original β 100 so , the time between maintenance checks increased by 25 β 20 / 20 β 100 = 25 answer : a ." | a = 25 - 20
b = a / 20
c = b * 100
|
a ) 9 % , b ) 15 % , c ) 105 / 8 % , d ) 11 % , e ) 12 % | b | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(13, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | a $ 500 investment and a $ 1,500 investment have a combined yearly return of 13 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1,500 investment have ? | "the equation we can form the question : return on total investment = sum of individual investments ( 500 + 1500 ) ( 13 ) = ( 500 Γ’ Λ β 7 ) + ( 1500 x ) , where x is the return on investment of 1500 . solving the equation , we get x = 15 % ( option b ) answer : b" | a = 3 * 500
b = 500 + a
c = 13 / 100
d = b * c
e = 7 / 100
f = 500 * e
g = d - f
h = 3 * 500
i = g / h
j = i * 100
|
a ) 43 , b ) 144 , c ) 68 , d ) 87 , e ) 92 | b | multiply(add(const_10, const_2), divide(3600, rectangle_area(20, 15))) | the monthly rent of a shop of dimension 20 feet Γ 15 feet is rs . 3600 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 3600 / ( 20 * 15 ) = 12 & annual rent per square feet = 12 * 12 = 144 answer : b" | a = 10 + 2
b = 3600 / rectangle_area
c = a * b
|
a ) 50 mph , b ) 45 mph , c ) 48 mph , d ) 52 mph , e ) 60 mph | e | add(40, multiply(5, const_4)) | liam is pulled over for speeding just as he is arriving at work . he explains to the police officer that he could not afford to be late today , and has arrived at work only four minutes before he is to start . the officer explains that if liam had driven 5 mph slower for his whole commute , he would have arrived at work exactly on time . if liam ' s commute is 40 miles long , how fast was he actually driving ? ( assume that liam drove at a constant speed for the duration of his commute . ) | let t be the number of hours he would need to reach office on time . when he is driving with over speed , he reached office 4 min earlier ! so the equation for this is s ( t - 4 / 60 ) = 40 where s is the speed and 30 is the distance . if he decreases his speed by 5 mph then he would have reached his office on time : ( s - 5 ) t = 40 if you solve above equations , you will arrive at t = 2 / 3 hr and s = 60 mph therefore answer is e | a = 5 * 4
b = 40 + a
|
a ) 106 , b ) 116 , c ) 128 , d ) 136 , e ) 146 | c | subtract(160, divide(multiply(160, 20), const_100)) | an article costing rs . 160 is sold at 20 % discount on a mark - up price . what is the selling price after discount ? | 160 * 80 / 100 = 128 answer : c | a = 160 * 20
b = a / 100
c = 160 - b
|
a ) 160 , b ) 264 , c ) 360 , d ) 354 , e ) 400 | b | divide(110, subtract(divide(75, const_100), divide(1, 3))) | 75 % of x is greater than 1 / 3 rd of x by 110 . what is x ? | 75 x / 100 - x / 3 = 110 5 x / 12 = 110 x = 264 answer : b | a = 75 / 100
b = 1 / 3
c = a - b
d = 110 / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(subtract(21, multiply(8, const_2)), subtract(subtract(12, 8), const_1)) | there are 8 stores in town that had a total of 21 visitors on a particular day . however , only 12 people went shopping that day ; some people visited more than one store . if 8 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visited ? | "8 people visited 2 stores each for 16 visits . to maximize the number of stores that one person visited , let ' s assume that 3 people visited 1 store each . the number of remaining visits is 21 - 16 - 3 = 2 , which is the maximum that one person could have visited . the answer is a ." | a = 8 * 2
b = 21 - a
c = 12 - 8
d = c - 1
e = b - d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(subtract(multiply(2500, power(add(const_1, divide(4, const_100)), 2)), 2500), multiply(multiply(2500, divide(4, const_100)), 2)) | indu gave bindu rs . 2500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "2500 = d ( 100 / 4 ) 2 d = 4 answer : d" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 2500 * c
e = d - 2500
f = 4 / 100
g = 2500 * f
h = g * 2
i = e - h
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | a | divide(subtract(multiply(30, 7), multiply(7, 7)), 7) | 7 people average age is 30 . youngest person age is 7 . find average of the people when youngest was born . | "average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 49 = 161 so average age would be 161 / 7 = 23 answer : a" | a = 30 * 7
b = 7 * 7
c = a - b
d = c / 7
|
a ) 20 , b ) 15 , c ) 40 , d ) 50 , e ) 25 | a | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(35, 40))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | how many pounds of salt at 50 cents / lb must be mixed with 40 lbs of salt that costs 35 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 35 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - salt avg ) / ( salt avg - salt 1 ) = ( 50 - 40 ) / ( 40 - 35 ) = 2 / 1 we know that weight of salt 1 is 40 lbs . weight of salt 2 must be 20 lbs . answer ( a )" | a = 48 * 40
b = 100 + 20
c = b / 100
d = 35 * 40
e = c * d
f = a - e
g = 100 + 20
h = g / 100
i = 50 * h
j = i - 48
k = f / j
|
a ) $ 180,000 , b ) $ 185,000 , c ) $ 170,000 , d ) $ 200,000 , e ) $ 205,000 | c | multiply(add(divide(const_1, const_10), 1), divide(multiply(1000, divide(add(add(multiply(add(multiply(divide(const_3, const_2), multiply(5, 10)), multiply(40, 5)), subtract(const_10, add(const_3, 5))), multiply(multiply(40, 5), 5)), multiply(multiply(5, 10), const_3)), const_10)), 1000)) | a factory that employs 1000 assembly line workers pays each of these workers $ 5 per hour for the first 40 hours worked during a week and 1 Β½ times that rate for hours worked in excess of 40 . what was the total payroll for the assembly - line workers for a week in which 30 percent of them worked 10 hours , 50 percent worked 40 hours , and the rest worked 50 hours ? | 30 % of 1000 = 300 worked for 20 hours payment @ 5 / hr total payment = 300 * 10 * 5 = 15000 50 % of 1000 = 500 worked for 40 hours payment @ 5 / hr total payment = 500 * 40 * 5 = 100000 remaining 200 worked for 50 hours payment for first 40 hours @ 5 / hr payment = 200 * 40 * 5 = 40000 payment for next 10 hr @ 7.5 / hr payment = 200 * 10 * 7.5 = 15000 total payment = 15000 + 100000 + 40000 + 15000 = 170000 hence , answer is c | a = 1 / 10
b = a + 1
c = 3 / 2
d = 5 * 10
e = c * d
f = 40 * 5
g = e + f
h = 3 + 5
i = 10 - h
j = g * i
k = 40 * 5
l = k * 5
m = j + l
n = 5 * 10
o = n * 3
p = m + o
q = p / 10
r = 1000 * q
s = r / 1000
t = b * s
|
a ) 18 , b ) 19 , c ) 10 , d ) 21 , e ) 12 | d | divide(273, subtract(14, const_1)) | along a yard 273 metres long , 14 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 14 trees have 13 gaps between them , required distance ( 273 / 13 ) = 21 option d" | a = 14 - 1
b = 273 / a
|
a ) 11 , b ) 20 , c ) 28 , d ) 24 , e ) 82 | b | multiply(divide(subtract(2926, 2442), 2442), const_100) | a sum of money deposited at c . i . amounts to rs . 2442 in 2 years and to rs . 2926 in 3 years . find the rate percent ? | "explanation : 2420 - - - 484 100 - - - ? = > 20 % answer : option b" | a = 2926 - 2442
b = a / 2442
c = b * 100
|
a ) 60 , b ) 45 , c ) 50 , d ) 48 , e ) none | d | multiply(divide(432, multiply(8, 9)), 8) | a ratio between two numbers is 8 : 9 and their l . c . m . is 432 . the first number is | sol . let the required numbers be 8 x and 9 x . then , their l . c . m . is 72 x . β΄ 72 x = 432 β x = 6 . hence , the first number is 48 . answer d | a = 8 * 9
b = 432 / a
c = b * 8
|
a ) $ 357.67 , b ) $ 375.85 , c ) $ 387.40 , d ) $ 399.30 , e ) $ 433.33 | d | divide(multiply(multiply(subtract(1000, 7), divide(10, const_100)), power(add(1, divide(10, const_100)), 3)), subtract(power(add(1, divide(10, const_100)), 3), const_1)) | jim needs $ 1000 to buy a new flat - screen tv . since he has only $ 7 , he borrows the remaining balance from his sister mary . the loan will be repaid in 3 annual installments at an interest rate of 10 % , compounded annually . the formula for calculating the monthly payment p is p = ( l x c x r ) / ( c β 1 ) where l = amount of the loan , r = annual interest rate , and c = compounding factor = ( 1 + r ) ^ n where n = number of annual payments . how much does jim have to pay mary at the end of each of the next 3 years ( rounded to the nearest penny ) ? | l = 993 r = 0.1 c = 1.1 ^ 3 = 1.331 insert all of these in the formula and we get p = 993 * 1.331 * 0.1 / 0.331 in order to resolve above without a calc , we need to find the approx value of 0.1331 / 0.331 ~ 0.4 ~ ~ 993 * 0.4 ~ 397 answer d | a = 1000 - 7
b = 10 / 100
c = a * b
d = 10 / 100
e = 1 + d
f = e ** 3
g = c * f
h = 10 / 100
i = 1 + h
j = i ** 3
k = j - 1
l = g / k
|
a ) 0 , b ) 4 , c ) 6 , d ) 7 , e ) 10 | d | add(10, 2) | there is a 35 cm line marked at each centimeter and an insect is placed at every centimeter . 9 frogs are trained to jump a constant distance . the first one jumps 2 cm in every leap , the second one jumps 3 cm and so on until the 9 th one jumps 10 cm in every leap and they eat any insect that is available at that spot . if all of them start from start line and finish the entire 35 cm , how many insects were still left after the race was over ? | "only the prime numbers greater than 10 and less than 35 were left . that is 11 , 13 , 17 , 19 , 23 , 29 , and 31 . the total is 7 . the answer is d ." | a = 10 + 2
|
a ) 5 / 6 , b ) 4 / 5 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 2 | d | subtract(const_1, divide(5, multiply(5, const_3))) | a searchlight on top of the watchtower makes 4 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | the searchlight completes one revolution every 15 seconds . the probability that the man ' s area will be lit up is 5 / 15 = 1 / 3 . the probability that he will stay in the dark is 1 - 1 / 3 = 2 / 3 the answer is d . | a = 5 * 3
b = 5 / a
c = 1 - b
|
a ) 5 , b ) 6 , c ) 7 , d ) 9 , e ) 10 | c | add(subtract(divide(120, 9), 7), const_1) | how many integers are between 7 and 120 / 9 , inclusive ? | "120 / 9 = 13 . xx we are not concerned about the exact value of 120 / 9 as we just need the integers . the different integers between 7 and 120 / 9 would be 7 , 8 , 9 , 10 , 11 , 12,13 total number of integers = 7 option c" | a = 120 / 9
b = a - 7
c = b + 1
|
a ) 7.5 , b ) 10 , c ) 12.5 , d ) 15 , e ) 17.5 | c | divide(multiply(5, 5), const_2) | the two lines y = x and x = - 5 intersect on the coordinate plane . what is the value of the area of the figure formed by the intersecting lines and the x - axis ? | "the point of intersection is ( - 5 , - 5 ) . the triangle has a base of length 5 and a height of 5 . area = ( 1 / 2 ) * base * height = ( 1 / 2 ) * 5 * 5 = 12.5 the answer is c ." | a = 5 * 5
b = a / 2
|
a ) 30 mph , b ) 32 mph , c ) 35 mph , d ) 37 mph , e ) 38 mph | c | divide(70, subtract(divide(100, 40), divide(30, 60))) | in a 100 - mile trip , the first 30 miles were completed at 60 mph . if the average speed for the whole trip was 40 mph , what was the average speed in the last 70 miles ? | the time to complete the first 30 miles was ( 30 miles / 60 mph ) = ( 1 / 2 ) hours . the time to complete the whole trip was ( 100 miles / 40 mph ) = ( 5 / 2 ) hours . the time to complete the last 70 miles was ( 5 / 2 ) - ( 1 / 2 ) = ( 2 ) hours . the average speed in the last 70 miles was ( 70 miles / 2 hours ) = 35 mph . the answer is c . | a = 100 / 40
b = 30 / 60
c = a - b
d = 70 / c
|
a ) $ 62.40 , b ) $ 62.80 , c ) $ 63.20 , d ) $ 63.60 , e ) $ 64.00 | d | add(60, multiply(divide(6, const_100), 60)) | diana took out a charge account at the general store and agreed to pay 6 % simple annual interest . if she charges $ 60 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "1.06 * $ 60 = $ 63.60 the answer is d ." | a = 6 / 100
b = a * 60
c = 60 + b
|
a ) 6000 , b ) 12000 , c ) 18000 , d ) 24009 , e ) 24002 | a | subtract(13000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 13000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 13000 share of b = 6 / 13 ( 13000 ) = rs . 6000 answer : a" | a = const_60 * 100
b = 13000 - a
|
a ) 95 , b ) 97 , c ) 101 , d ) 85 , e ) 87 | c | subtract(multiply(4, const_2), multiply(2, const_2)) | if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 103 , what is the value of x ? | "am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 103 x = 101 answer : c" | a = 4 * 2
b = 2 * 2
c = a - b
|
a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 4000 , 28000 , d ) 12000 , 20006 , e ) 12000 , 20001 | c | multiply(subtract(7, const_2), divide(32000, add(1, subtract(7, const_2)))) | divide rs . 32000 in the ratio 1 : 7 ? | "1 / 8 * 32000 = 4000 7 / 8 * 32000 = 28000 answer : c" | a = 7 - 2
b = 7 - 2
c = 1 + b
d = 32000 / c
e = a * d
|
a ) $ 0.94 , b ) $ 0.96 , c ) $ 0.98 , d ) $ 1.00 , e ) $ 1.20 | c | multiply(add(const_1, divide(10, const_100)), divide(0.80, divide(subtract(const_100, 10), const_100))) | the manager of a produce market purchased a quantity of tomatoes for $ 0.80 per pound . due to improper handling , 10 percent of the tomatoes , by weight , were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal to 10 percent of the cost of the tomatoes . | "assume the manager bought 100 tomatoes . cost price = 80 given : 10 % are damaged - - > available tomatoes to sell = 90 90 * x - 80 = 0.1 * 80 90 x - 80 = 8 90 x = 88 x = 88 / 90 = 44 / 45 = 0.977 x is slightly under 0.977 = 0.98 answer : c" | a = 10 / 100
b = 1 + a
c = 100 - 10
d = c / 100
e = 0 / 80
f = b * e
|
a ) 17.64 % , b ) 33.33 % , c ) 40 % , d ) 75 % , e ) none of these | a | multiply(add(const_1, divide(divide(multiply(140, 40), add(140, divide(multiply(multiply(const_3, const_10), 140), const_100))), 40)), const_10) | a rectangle having length 140 cm and width 40 cm . if the length of the rectangle is increased by thirty percent then how much percent the breadth should be decreased so as to maintain the same area . | explanation : solution : ( 30 / ( 140 + 30 ) * 100 ) % = 17.64 % answer : a | a = 140 * 40
b = 3 * 10
c = b * 140
d = c / 100
e = 140 + d
f = a / e
g = f / 40
h = 1 + g
i = h * 10
|
a ) 312 , b ) 335 , c ) 347 , d ) 361 , e ) 383 | b | subtract(456, reminder(722425, 456)) | which number can we add to 722425 to get a number exactly divisible by 456 ? | "722425 / 456 = 1584 with a remainder of 121 . we need to add 456 - 121 = 335 the answer is b ." | a = 456 - reminder
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a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 2000 meters , e ) none of these | d | multiply(multiply(8, divide(15, const_60)), const_1000) | a man is walking at the rate of 8 km / hr crosses a bridge in 15 minutes . the length of the bridge is | explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . speed = 8 Γ’ Λ β 5 / 18 = 20 / 9 m / sec time = 15 Γ’ Λ β 60 seconds = 900 seconds distance = time Γ’ Λ β speed / distance = 20 / 9 Γ’ Λ β 900 = 2000 meter option d | a = 15 / const_60
b = 8 * a
c = b * 1000
|
a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 13 % | e | multiply(divide(subtract(add(const_1, divide(30, const_100)), add(const_1, divide(15, const_100))), add(const_1, divide(15, const_100))), const_100) | a certain company reported that the revenue on sales increased 15 % from 2000 to 2003 , and increased 30 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 115 and and in 2005 130 , so from 2003 to 2005 it increased by ( 130 - 115 ) / 115 = 15 / 115 = 13 % . answer : e ." | a = 30 / 100
b = 1 + a
c = 15 / 100
d = 1 + c
e = b - d
f = 15 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 30000 , b ) 50000 , c ) 40000 , d ) 20000 , e ) 60000 | e | multiply(divide(40000, 2), 3) | p and q invested in a business . the profit earned was divided in the ratio 2 : 3 . if p invested rs 40000 , the amount invested by q is | let the amount invested by q = q 40000 : q = 2 : 3 β 40000 Γ 3 = 2 q β q = ( 40000 Γ 3 ) / 2 = 60000 answer is e . | a = 40000 / 2
b = a * 3
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | d | divide(add(3, 1), add(const_3.0, 1)) | in the first m games of a team ' s season , the ratio of the team ' s wins to its losses was 3 : 1 . in the subsequent n games , the ratio of the team Β΄ s wins to losses was 1 : 4 . if m : n = 4 : 5 , what was the ratio of the team ' s wins to its losses for all m + n games ? | "m = 4 / 9 of total games n = 5 / 9 of total games wins = 3 / 4 * 4 / 9 + 1 / 5 * 5 / 9 = 3 / 9 + 1 / 9 = 4 / 9 losses = 1 - 4 / 9 = 5 / 9 the ratio of wins to losses is 4 : 5 . the answer is d ." | a = 3 + 1
b = 3 + 0
c = a / b
|
a ) $ 550 , b ) $ 650 , c ) $ 750 , d ) $ 850 , e ) $ 950 | c | subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 65), add(divide(10, const_100), divide(8, const_100)))) | if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 65 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | "0.1 x = 0.08 ( 2000 - x ) + 65 0.18 x = 225 x = 1250 then the amount invested at 8 % is $ 2000 - $ 1250 = $ 750 the answer is c ." | a = 100 * 10
b = a * 2
c = 10 * 8
d = c * 2
e = d + 65
f = 10 / 100
g = 8 / 100
h = f + g
i = e / h
j = b - i
|
a ) 16 % , b ) 17 % , c ) 18 % , d ) 58 % , e ) 23 % | e | multiply(divide(subtract(62900, add(36000, 15000)), add(36000, 15000)), const_100) | ramu bought an old car for rs . 36000 . he spent rs . 15000 on repairs and sold it for rs . 62900 . what is his profit percent ? | total cp = rs . 36000 + rs . 15000 = rs . 51000 and sp = rs . 62900 profit ( % ) = ( 62900 - 51000 ) / 51000 * 100 = 23 % answer : e | a = 36000 + 15000
b = 62900 - a
c = 36000 + 15000
d = b / c
e = d * 100
|
a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years | c | divide(subtract(divide(48, divide(4, 4)), multiply(subtract(4, const_1), 4)), 4) | the sum of the ages of 4 children born at the intervals of 4 years each is 48 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 4 ) , ( x + 4 ) and ( x + 12 ) years . then , x + ( x + 4 ) + ( x + 8 ) + ( x + 12 ) = 48 4 x = 24 = > x = 6 . age of youngest child = x = 6 years . answer : c" | a = 4 / 4
b = 48 / a
c = 4 - 1
d = c * 4
e = b - d
f = e / 4
|
a ) 16 , b ) 8 β 2 , c ) 8 , d ) 7 / β 2 , e ) ( β 2 ) / 3 | d | sqrt(divide(multiply(7, 7), const_2)) | the two lines y = x and x = - 7 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 7 , - 7 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 7 Γ 7 ) = 49 / 2 ; so z = 49 / 2 . the next step requires us to find the length of a side of a cube that has a face area equal to 49 / 2 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 49 / 2 . since the area of a square is equal to s Β² , where s is the length of one of its side , we can write and solve the equation s Β² = 49 / 2 . clearly s = β 49 / 2 = 7 / β 2 , oranswer choice ( d ) ." | a = 7 * 7
b = a / 2
c = math.sqrt(b)
|
a ) 137 , b ) 112 , c ) 287 , d ) 121 , e ) 191 | b | multiply(divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3)) | find the smallest four - digit number which is a multiple of 112 | "explanation : the smallest four digit number is 1000 . if 1000 is divided by 112 , the remainder is 104 . 112 - 104 = 8 , if 8 is added to 1000 , it will become the smallest four digit number and a multiple of 112 . answer : b" | a = 112 / 1000
b = a / 3
c = b / 3
d = c / 3
e = 112 / 1000
f = e / 3
g = f / 3
h = g / 3
i = d * h
|
a ) 230 m , b ) 250 m , c ) 260 m , d ) 270 m , e ) 240 m | b | subtract(multiply(multiply(72, const_0_2778), 26), 270) | a goods train runs at a speed of 72 kmph and crosses a 270 m long platform in 26 seconds . what is the length of the goods train ? | "s = 270 + x / t 72 * 5 / 18 = 270 + x / 26 x = 250 answer : b" | a = 72 * const_0_2778
b = a * 26
c = b - 270
|
a ) 680 , b ) 700 , c ) 240 , d ) 620 , e ) 260 | c | divide(multiply(multiply(multiply(6, const_100), sqrt(const_100)), 6), const_100) | alex takes a loan of $ 4,000 to buy a used truck at the rate of 6 % simple interest . calculate the annual interest to be paid for the loan amount . | "from the details given in the problem principle = p = $ 4,000 and r = 6 % or 0.06 expressed as a decimal . as the annual interest is to be calculated , the time period t = 1 . plugging these values in the simple interest formula , i = p x t x r = 4,000 x 1 x 0.06 = 240.00 annual interest to be paid = $ 240 answer : c" | a = 6 * 100
b = math.sqrt(100)
c = a * b
d = c * 6
e = d / 100
|
a ) 0.35 , b ) 3.5 , c ) 35 , d ) 700 , e ) 3500 | d | multiply(divide(multiply(divide(multiply(multiply(145, 160), 175), multiply(multiply(25, 50), 75)), 40), 200), const_100) | if 145 % of j is equal to 25 % of k , 160 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 40 % of m is equal to what percent of 200 % of j ? | "imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 40 % of 350 , comes out to be 140 . . . . 200 % of 10 is 20 . . . . ( 140 * 100 ) / 20 = 700 . . . . ans : d" | a = 145 * 160
b = a * 175
c = 25 * 50
d = c * 75
e = b / d
f = e * 40
g = f / 200
h = g * 100
|
a ) 8 , b ) 9.52 , c ) 10 , d ) 11.7 , e ) 20 | b | divide(add(multiply(divide(multiply(100, const_2), 5), 5), multiply(divide(multiply(100, const_2), 100), 100)), add(divide(multiply(100, const_2), 100), divide(multiply(100, const_2), 5))) | ron cycles his way up on a hill side and back . he takes the exact same route both ways . on the trip out he cycles an average speed of 5 miles per hour . on the trip back , downhill cycling at an average speed of 100 miles per hour , he is quick to return back . what is his approximate average speed for the round trip in miles per hour ? | average speed = total distance / total time here distance to hill top and back is same as he takes the same route . d = 5 t 1 and d = 100 t 2 therefore , t 1 = d / 5 and t 2 = d / 100 t 1 + t 2 = 21 d / 100 therefore , average speed = 2 d / 21 d / 100 = 2 d * 100 / 21 d answer is 9.52 since , the options are far away we need not solve till the decimal points so correct answer is option b | a = 100 * 2
b = a / 5
c = b * 5
d = 100 * 2
e = d / 100
f = e * 100
g = c + f
h = 100 * 2
i = h / 100
j = 100 * 2
k = j / 5
l = i + k
m = g / l
|
a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability q that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability q = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a = math.comb(4, 1)
b = a * 2
c = math.comb(6, 3)
d = b / c
e = 5 * 5
f = e * 4
g = d * f
|
a ) 128 , b ) 142 , c ) 143 , d ) 141 , e ) 129 | e | divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7) | how many 3 digit positive integers q exist that when divided by 7 leave a remainder of 5 ? | minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e . | a = 100 * 10
b = a - 1
c = 10 + 4
d = c * 7
e = d + 5
f = b - e
g = f / 7
|
a ) 0 , b ) 12 , c ) 24 , d ) 26 , e ) 18 | a | subtract(add(20, 20), add(1, 1)) | a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the addition between the maximum and minimum possible values of ' a - b ' ? | "possible values of a and b can be 5,4 ; 4,5 and 1,20 ; 20,1 maximum possible value for a - b is 20 - 1 = 19 minimum possible value for a - b is 1 - 20 = - 19 19 - 19 = 0 ans is a" | a = 20 + 20
b = 1 + 1
c = a - b
|
a ) 3 / 140 , b ) 1 / 28 , c ) 3 / 56 , d ) 3 / 35 , e ) 7 / 40 | e | add(add(multiply(multiply(divide(const_4, 5), divide(3, 8)), divide(2, 7)), multiply(multiply(divide(1, 5), divide(5, 8)), divide(2, 7))), multiply(multiply(divide(1, 5), divide(3, 8)), divide(5, 7))) | alice , benjamin , and carol each try independently to win a carnival game . if their individual probabilities for success are 1 / 5 , 3 / 8 , and 2 / 7 , respectively , what is the probability that exactly two of the 3 players will win but one will lose ? | p = p ( a wins , b wins , c loses ) + p ( a wins , b loses , c wins ) + p ( a loses , b wins , c wins ) = 1 / 5 * 3 / 8 * 5 / 7 + 1 / 5 * 5 / 8 * 2 / 7 + 4 / 5 * 3 / 8 * 2 / 7 = 7 / 40 . answer : e . | a = 4 / 5
b = 3 / 8
c = a * b
d = 2 / 7
e = c * d
f = 1 / 5
g = 5 / 8
h = f * g
i = 2 / 7
j = h * i
k = e + j
l = 1 / 5
m = 3 / 8
n = l * m
o = 5 / 7
p = n * o
q = k + p
|
a ) 50 minutes , b ) 20 minutes , c ) 40 minutes , d ) 60 minutes , e ) 80 minutes | c | multiply(10, divide(subtract(const_100, 20), 20)) | a person takes 10 minutes more to cover a certain distance by decreasing his speed by 20 % . what is the time taken to cover the distance at his original speed ? | let the distance and original speed be d km and k kmph respectively . d / 0.8 k - d / k = 10 / 60 = > 5 d / 4 k - d / k = 1 / 6 = > ( 5 d - 4 d ) / 4 k = 1 / 6 = > d = 4 / 6 k = 2 / 3 k time taken to cover the distance at original speed = d / k = 2 / 3 hours = 40 minutes . answer : c | a = 100 - 20
b = a / 20
c = 10 * b
|
a ) 334 , b ) 500 , c ) 200 , d ) 288 , e ) 271 | c | divide(184, divide(92, const_100)) | victor gets 92 % marks in examinations . if these are 184 marks , find the maximum marks . | "let the maximum marks be m then 92 % of m = 184 β 92 / 100 Γ m = 184 β m = ( 184 Γ 100 ) / 92 β m = 18400 / 92 β m = 200 therefore , maximum marks in the examinations are 200 . answer : c" | a = 92 / 100
b = 184 / a
|
a ) 8 , b ) 9 , c ) 2 , d ) 6 , e ) 4 | c | multiply(divide(const_1, multiply(add(const_100, 60), divide(const_1, subtract(const_100, 60)))), 8) | by selling 8 pencils for a rupee a man loses 60 % . how many for a rupee should he sell in order to gain 60 % ? | "40 % - - - 8 160 % - - - ? 40 / 160 * 12 = 2 answer : c" | a = 100 + 60
b = 100 - 60
c = 1 / b
d = a * c
e = 1 / d
f = e * 8
|
a ) 5 , 500,000 , b ) 3 , 200,000 , c ) 55,000 , d ) 28,000 , e ) 280 | b | subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 3.5), 3.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 3)), 3)), volume_cube(100)) | a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 3.5 grams per cubic centimeter and mineral y weighs 3 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 3.5 gms in 400,000 cm ^ 3 , x = 3.5 * 400,000 = 1 , 400,000 grams in 1 cm ^ 3 , y is 3 gms in 600,000 cm ^ 3 , y = 3 * 600,000 = 1 , 800,000 gms total gms in 1 m ^ 3 = 1 , 400,000 + 1 , 800,000 = 3 , 200,000 answer : b" | a = volume_cube / (
b = a * 10
c = b * 3
d = c + 3
e = volume_cube / (
f = e * 10
g = 2 * 3
h = f * g
i = d - h
|
a ) 5 , b ) 6 , c ) 15 , d ) 16 , e ) 18 | d | add(add(4, 3), add(add(4, 4), 1)) | for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 1000 , what is the maximum possible sum of the length of x and the length of y ? | basically the length of an integer is the sum of the powers of its prime factors . for example the length of 24 is 4 because 24 = 2 ^ 3 * 3 ^ 1 - - > 3 + 1 = 4 . given : x + 3 y < 1,000 . now , to maximize the length of x or y ( to maximize the sum of the powers of their primes ) we should minimize their prime bases . minimum prime base is 2 : so if x = 2 ^ 9 = 512 then its length is 9 - - > 512 + 3 y < 1,000 - - > y < 162.7 - - > maximum length of y can be 7 as 2 ^ 7 = 128 - - > 9 + 7 = 16 . answer : d . | a = 4 + 3
b = 4 + 4
c = b + 1
d = a + c
|
a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total t of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) t 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) ." | a = 3000 + 3
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a ) 3.6 sec , b ) 18 sec , c ) 36 sec , d ) 72 sec , e ) none | c | multiply(multiply(divide(divide(add(240, 120), const_1000), subtract(45, 9)), const_60), const_60) | a jogger running at 9 kmph along side a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? | category formulae examples directions : problems based on trains . 11 . a train 280 m long , running with a speed of 63 km / hr will pass a tree in a . 15 sec . b . 16 sec . c . 18 sec . d . 20 sec . solution speed = ( 63 x 5 / 18 ) m / sec = 35 / 2 msec time taken = ( 280 x 2 / 35 ) m / sec = 16 sec . 12 . a train running at the speed of 60 km / hr crosses a pole in 9 seconds . what is ihe length of the train ? a . 120 metres b . 180 metres c . 324 metres d . none of these solution 13 . a jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? a . 3.6 sec b . 18 sec c . 36 sec d . 72 sec solution speed of train relative to jogger = ( 45 - 9 ) km / hr = 36 km / hr = ( 36 x 5 / 18 ) m / sec = 10 m / sec distance covered = ( 240 + 120 ) m = 360 m time taken = ( 360 / 10 ) sec = 36 sec . answer c | a = 240 + 120
b = a / 1000
c = 45 - 9
d = b / c
e = d * const_60
f = e * const_60
|
a ) 280 , b ) 400 , c ) 540 , d ) 500 , e ) 840 | d | divide(divide(divide(100, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(2, 3)) | of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 100 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 100 geese survived the first year : 1 / 5 * ( total ) = 100 - - > ( total ) = 500 . answer : d ." | a = 3 / 5
b = 1 - a
c = 100 / b
d = 3 / 4
e = c / d
f = 2 / 3
g = e / f
|
a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | b | floor(multiply(const_100, divide(subtract(2524.36, 2400), 2400))) | a shopkeeper sold an article for rs 2524.36 . approximately what was his profit percent if the cost price of the article was rs 2400 | "explanation : gain % = ( 125.36 * 100 / 2400 ) = 5.2 % = 5 % approx option b" | a = 2524 - 36
b = a / 2400
c = 100 * b
d = math.floor(c)
|
a ) 18 , b ) 28 , c ) 40 , d ) 68 , e ) 70 | c | divide(multiply(divide(multiply(subtract(subtract(300, divide(multiply(70, 300), const_100)), 30), const_100), subtract(const_100, 40)), 40), const_100) | at a particular graduation party with 300 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 30 males did not bring gifts to the party , how many females did bring gifts ? | "the correct method total = 300 . . 70 % of 300 = 210 got gifts . . 90 did not get gift , out of which 30 are males , so remaining 90 - 30 = 60 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 60 , 100 % = 60 * 100 / 60 = 100 . ans 40 % of 100 = 40 c" | a = 70 * 300
b = a / 100
c = 300 - b
d = c - 30
e = d * 100
f = 100 - 40
g = e / f
h = g * 40
i = h / 100
|
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | d | add(divide(subtract(36, 1), 5), const_1) | how many multiples of 5 are there between 1 and 36 , exclusive ? | "7 multiples of 5 between 1 and 36 exclusive . from 5 * 1 upto 5 * 7 , ( 1,2 , 3,4 , . . . , 7 ) . hence , 7 multiples ! correct option is d" | a = 36 - 1
b = a / 5
c = b + 1
|
a ) 10 minutes , b ) 12 minutes , c ) 18 minutes , d ) 19 minutes , e ) 16 minutes | b | subtract(multiply(divide(12, 4), 5), const_3) | a cave is going to explode . a family is trapped inside that cave . the only way to get outside is a narrow tunnel . but the tunnel is weak and thus only two people can go at one time . also , everybody is afraid of going alone . the father takes one minute to go through till the end , the mother takes two , the son takes 4 and the daughter takes 5 . the cave and tunnel will explode in 12 minutes . can they make it to the other end in that time frame ? ps : keep in mind that when two people are moving , the total time taken will be as per the slower one . | b 12 minutes explanation : yes , surely they can all make it out of the cave . first , father and mother will go . time taken = 2 minutes . father comes back . time taken = 2 + 1 = 3 minutes . both children go . time taken = 3 + 5 = 8 minutes . mother comes back . time taken = 8 + 2 = 10 minutes . mother and father go . time taken = 10 + 2 = 12 minutes . thus in 12 minutes , all of them will be out of the cave through the tunnel . | a = 12 / 4
b = a * 5
c = b - 3
|
a ) 72 / 100 , b ) 3 / 25 , c ) 4 / 5 , d ) 1 / 6 , e ) 1 / 9 | b | divide(divide(multiply(multiply(10, 9), add(const_4, const_4)), add(const_4, const_2)), power(const_10, const_3)) | a bag contains 10 balls numbered from 0 to 9 . the balls are such that the person picking a ball out of the bag is equally likely to pick anyone of them . a person picked a ball and replaced it in the bag after noting its number . he repeated this process 2 more times . what is the probability that the ball picked first is numbered higher than the ball picked second and the ball picked second is numbered higher than the ball picked third ? | explanation : let , the number of the ball picked first = a , second = b and third = c and these three numbers a , b and c are distinct . three distinct ball can be picked in ( 10 Γ 9 Γ 8 ) ways . the order of a , b and c can be as follows : - ( i ) a > b > c . ( ii ) a > c > b . ( iii ) b > c > a . ( iv ) b > a > c . ( v ) c > a > b . ( vi ) c > b > a . they will occur equal number of times . so , the number of ways in which ( a > b > c ) are : - = > 1 / 6 x 10 x 9 x 8 . = > 120 . hence , the required probability is : - = > 120 / ( 10 x 10 x 10 ) . = > 3 / 25 . answer : b | a = 10 * 9
b = 4 + 4
c = a * b
d = 4 + 2
e = c / d
f = 10 ** 3
g = e / f
|
a ) 263 , b ) 500 , c ) 367 , d ) 368 , e ) 666.7 | e | divide(70, divide(multiply(3, add(3, divide(1, 2))), const_100)) | what sum of money will produce rs . 70 as simple interest in 3 years at 3 1 / 2 percent ? | "70 = ( p * 3 * 7 / 2 ) / 100 p = 666.7 answer : e" | a = 1 / 2
b = 3 + a
c = 3 * b
d = c / 100
e = 70 / d
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | divide(subtract(const_1, multiply(add(add(divide(const_1, 5), divide(const_1, 20)), divide(const_1, 30)), 2)), divide(const_1, 30)) | a person x working alone can complete a work in 5 days . a person y completes the same amount of work in 20 days , and a person z when working alone can complete the same amount of work in 30 days . all 3 people work together for 2 days but then x and y leave . how many more days does z need to complete the work after x and y leave ? | the fraction of work completed in two days is 2 / 5 + 2 / 20 + 2 / 30 = 17 / 30 the fraction of work remaining is 1 - 17 / 30 = 13 / 30 the number of days required for z is 13 days . the answer is a . | a = 1 / 5
b = 1 / 20
c = a + b
d = 1 / 30
e = c + d
f = e * 2
g = 1 - f
h = 1 / 30
i = g / h
|
a ) 46 , b ) 50 , c ) 40 , d ) 36 , e ) 41 | e | divide(divide(subtract(100, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 100 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = 100 / 10 m / sec = 10 m / sec . = 10 x 18 / 5 km / hr = 36 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 36 = 41 km / hr . answer : e" | a = 5 * const_0_2778
b = a * 5
c = 100 - b
d = c / 5
e = d / const_0_2778
|
a ) 7 , b ) 9 , c ) 5 , d ) 10 , e ) 3 | b | divide(200, multiply(80, const_0_2778)) | in what time will a train 200 meters long completely pass an electric pole , if it travels at a speed of 80 kmph . | sol . speed = [ 80 x 5 / 18 ] m / sec = 22.2 m / sec . time taken = ( 200 / 22.2 ) sec = 9 sec . answer b | a = 80 * const_0_2778
b = 200 / a
|
a ) 68 , b ) 69 , c ) 70 , d ) 67 , e ) 76 | a | divide(multiply(add(const_1, const_4), add(24, 17)), const_3) | a company organzied a help desk . there have been done 24 hrs work on software , 17 hrs on help user nd 40 % on other services . find the total no . of hrs have been worked on that day ? | let total no of hrs = x they have done 24 + 17 = 41 40 % of x is 40 / 100 = 2 / 5 x = 41 + 2 x / 5 x = 68.3 answer : a | a = 1 + 4
b = 24 + 17
c = a * b
d = c / 3
|
a ) rs . 432 , b ) rs . 425 , c ) rs . 412 , d ) rs . 442 , e ) none of these | b | multiply(divide(360, subtract(2360, 360)), 2360) | the true discount on a bill of rs . 2360 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2360 td = rs . 360 pw = f - td = 2360 - 360 = rs . 2000 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2000 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2360 for unexpired time = 360 / 2000 Γ 2360 = 0.18 Γ 2160 = rs . 425 answer : option b" | a = 2360 - 360
b = 360 / a
c = b * 2360
|
a ) 120 sec , b ) 176 sec , c ) 360 sec , d ) 187 sec , e ) 189 sec | c | divide(600, subtract(multiply(48, const_0_2778), multiply(18, const_0_2778))) | a and b go around a circular track of length 600 m on a cycle at speeds of 18 kmph and 48 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 18 * 5 / 18 ) , 600 / ( 48 * 5 / 18 ) } = 360 sec . answer : c" | a = 48 * const_0_2778
b = 18 * const_0_2778
c = a - b
d = 600 / c
|
a ) 16.67 % , b ) 23 % , c ) 18.3 % , d ) 17.64 % , e ) 21.23 % | d | multiply(divide(subtract(add(15, 2), add(multiply(divide(subtract(const_100, 20), const_100), 15), 2)), add(15, 2)), const_100) | 15 litres of mixture contains 20 % alcohol and the rest water . if 2 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 15 litres of mix . = 20 % of 15 litres = ( 20 * 15 / 100 ) = 3 litres water in it = 15 - 3 = 12 litres new quantity of mix . = 15 + 2 = 17 litres quantity of alcohol in it = 3 litres percentage of alcohol in new mix . = 3 * 100 / 17 = 17.64 % answer is d" | a = 15 + 2
b = 100 - 20
c = b / 100
d = c * 15
e = d + 2
f = a - e
g = 15 + 2
h = f / g
i = h * 100
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | e | multiply(subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 15)), 2)), 15) | two pipes p and q can fill a cistern in 10 and 15 minutes respectively . both are opened together , but at the end of 2 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | "let x be the total time it takes for the cistern to fill . 2 / 10 + x / 15 = 1 x / 15 = 4 / 5 x = 12 after the first pipe is turned off , it takes 10 more minutes to fill the cistern . the answer is e ." | a = 1 / 10
b = 1 / 15
c = a + b
d = c * 2
e = 1 - d
f = e * 15
|
a ) 750 , b ) 900 , c ) 1050 , d ) 1200 , e ) 1350 | d | divide(multiply(multiply(16, subtract(16, const_1)), 10), const_2) | there are 16 teams in the hockey league , and each team faces all the other teams 10 times each . how many games are played in the season ? | the number of ways to choose two teams is 16 c 2 = 16 * 15 / 2 = 120 the total number of games in the season is 10 * 120 = 1200 . the answer is d . | a = 16 - 1
b = 16 * a
c = b * 10
d = c / 2
|
a ) 30 % , b ) 70 % , c ) 80 % , d ) 87.5 % , e ) none | d | add(subtract(70, 20), subtract(subtract(50, multiply(const_0_25, const_2)), const_12)) | in an examination , the percentage of students qualified to the number of students appeared from school a is 70 % . in school b , the number of students appeared is 20 % more than the students appeared from school a and the number of students qualified from school b is 50 % more than the students qualified from school a . what is the percentage of students qualified to the number of students appeared from school b ? | solution let number of students appeared from school a = 100 . then , number of students qualified from school a = 70 . number of students appeared from school b = 120 . number of students qualified from school b = [ 150 / 100 x 70 ] = 105 . required percentage = [ 105 / 120 x 100 ] % = 87.5 % answer d | a = 70 - 20
b = const_0_25 * 2
c = 50 - b
d = c - 12
e = a + d
|
a ) 52 kmph . , b ) 62 kmph . , c ) 72 kmph . , d ) 92 kmph . , e ) none | d | subtract(multiply(divide(280, 9), const_3_6), 20) | a man sitting in a train which is travelling at 20 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ? | "solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 20 ) kmph = 92 kmph . answer d" | a = 280 / 9
b = a * const_3_6
c = b - 20
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(subtract(52, subtract(multiply(const_4, divide(6, 2)), 20)), 6) | the area of one square is x ^ 2 + 6 x + 9 and the area of another square is 4 x ^ 2 β 20 x + 25 . if the sum of the perimeters of both squares is 52 , what is the value of x ? | "the areas are ( x + 3 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 3 and 2 x - 5 . if we add the two perimeters : 4 ( x + 3 ) + 4 ( 2 x - 5 ) = 52 12 x = 60 x = 5 the answer is c ." | a = 6 / 2
b = 4 * a
c = b - 20
d = 52 - c
e = d / 6
|
a ) 39.2 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) none of these | a | multiply(divide(add(divide(130, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 130 m long ? | "speed = 45 km / hr = 45 Γ£ β ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 130 = 490 meter time taken to cross the platform = 490 / ( 25 / 2 ) = 490 Γ£ β 2 / 25 = 39.2 seconds answer : a" | a = 130 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
|
a ) 1 , b ) 5 , c ) 8 , d ) 4 , e ) 3 | a | divide(multiply(const_4, const_3), multiply(const_4, const_3)) | if g is to be chosen at random from the set { 5,8 , 7,1 } and k is to be chosen at random from the set { 14 , 8,3 } , what is the probability that gk will be even ? | method - 1 gk will be even when 1 ) g is even and k is odd , probability of g even is ( 1 / 4 ) and probability of k odd is ( 1 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 1 / 3 ) = ( 1 / 12 ) 2 ) g is odd and k is even , probability of g odd is ( 3 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 2 ) = ( 3 / 4 ) * ( 2 / 3 ) = ( 6 / 12 ) 3 ) g is even and k is even , probability of g even is ( 1 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 2 / 3 ) = ( 2 / 12 ) total favorable probability = ( 1 / 12 ) + ( 6 / 12 ) + ( 2 / 12 ) = ( 12 / 12 ) = 1 answer : option a | a = 4 * 3
b = 4 * 3
c = a / b
|
a ) 7.82 km , b ) 6 km , c ) 5.76 km , d ) 5.66 km , e ) 10 km | a | multiply(divide(multiply(add(8, 1.2), subtract(8, 1.2)), add(add(8, 1.2), subtract(8, 1.2))), const_2) | a man can row 8 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 8 s = 1.2 ds = 9.2 us = 6.8 x / 9.2 + x / 6.8 = 1 x = 3.91 d = 3.91 * 2 = 7.82 answer : a" | a = 8 + 1
b = 8 - 1
c = a * b
d = 8 + 1
e = 8 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 0 , b ) 4905 , c ) 49050 , d ) 490500 , e ) none of them | d | divide(add(86510529, const_1), const_2) | the difference between the place values of two fives in the numerical 86510529 is | "required difference = ( 500000 - 500 ) = 490500 . answer is d" | a = 86510529 + 1
b = a / 2
|
a ) - 29 , b ) - 19 , c ) 17 , d ) 29 , e ) 39 | c | subtract(56, subtract(45, add(subtract(28, 37), 15))) | if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 56 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 56 = > 45 - [ 28 - { 37 - 15 + * } ] = 56 45 - [ 28 - 37 + 15 - * ] = 56 = > 45 [ 43 - 37 - * ] = 56 45 - [ 6 - * ] = 56 = > 45 - 6 + * = 56 39 + * = 56 = > * = 56 - 39 = 17 answer : c" | a = 28 - 37
b = a + 15
c = 45 - b
d = 56 - c
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | 150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration w of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration w would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a = 150 * 30
b = a / 100
c = 400 * 12
d = c / 100
e = b + d
f = 150 + 400
g = e / f
|
a ) 16.5 , b ) 14.33 , c ) 12.5 , d ) 15.8 , e ) 16.66 | e | divide(multiply(80, const_100), 480) | how much percentage 80 is of 480 ? | 80 * 100 / 480 = 16.66 % answer e | a = 80 * 100
b = a / 480
|
a ) $ 3125 , b ) $ 2945 , c ) $ 3012 , d ) $ 2545 , e ) $ 3150 | a | multiply(2000, power(subtract(const_1, divide(20, const_100)), 2)) | a present value of a machine is $ 2000 . its value depletiation rate is 20 % per annum then find the machine value before 2 years ? | "p = $ 2000 r = 20 % t = 2 years machine value before 2 years = p / [ ( 1 - r / 100 ) ^ t ] = 2000 * 5 / 4 * 5 / 4 = $ 3125 answer is a" | a = 20 / 100
b = 1 - a
c = b ** 2
d = 2000 * c
|
a ) $ 1250 , b ) $ 1060 , c ) $ 1350 , d ) $ 900 , e ) $ 1000 | b | multiply(subtract(10000, 2000), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 25000 sheets of xerox paper ? | "for first 2000 pages - cost 5 cents / paper for 2001 - 10000 - cost 4.5 cents / paper for 10000 + - 4 cents / paper so cost = ( 5 cents * 2000 ) + ( 4.5 cents * 8000 ) + ( 4 cents * 15000 ) = $ 1060 alternatively , ( 0.05 x 2000 ) + ( 0.045 x 8000 ) + ( 0.04 x 15000 ) = 1,060 answer : b" | a = 10000 - 2000
b = 1 / 10
c = 1 - b
d = 1 / 10
e = c * d
f = a * e
|
a ) 400 , b ) 800 , c ) 600 , d ) 1200 , e ) 1000 | b | divide(multiply(divide(multiply(const_100, add(multiply(3, 1000), 1000)), 9000), 1800), const_100) | 3 persons invested rs . 9000 in a joint business . the second person invested rs . 1000 more than the first and the third rs . 1000 more than second . after two years , they gained rs . 1800 . how much third person will get ? | first persons investment = x second persons investment = x + 1000 third persons investments = x + 2000 x + x + 1000 + x + 2000 = 9000 3 x = 6000 x = 2000 ratio = 2000 : 3000 : 4000 2 : 3 : 4 4 / 9 * 1800 = 800 answer : b | a = 3 * 1000
b = a + 1000
c = 100 * b
d = c / 9000
e = d * 1800
f = e / 100
|
a ) 120 km , b ) 360 km , c ) 480 km , d ) 500 km , e ) none | c | multiply(80, divide(multiply(60, const_2), subtract(80, 60))) | bombay express left delhi for bombay at 14.30 hrs , travelling at a speed of 60 kmph and rajdhani express left delhi for bombay on the same day at 16.30 hrs , travelling at a speed of 80 kmph . how far away from delhi will the two trains meet ? | sol . suppose they meet x hours after 14.30 hrs . then , 60 x = 80 ( x - 2 ) or x = 8 . β΄ required distance = ( 60 x 8 ) km = 480 km . answer c | a = 60 * 2
b = 80 - 60
c = a / b
d = 80 * c
|
a ) 30 , b ) 60 , c ) 70 , d ) 80 , e ) 20 | a | divide(divide(multiply(600, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 600 , what is the reduced price for kg ? | "a 600 * ( 25 / 100 ) = 150 - - - - 5 ? - - - - 1 = > rs . 30" | a = 600 * 25
b = a / 100
c = b / 5
|
a ) 97 , b ) 151 , c ) 209 , d ) 211 , e ) 219 | d | add(add(add(7, multiply(7, const_10)), add(multiply(7, const_10), 3)), add(multiply(6, const_10), 1)) | the numbers { 1 , 3 , 6 , 7 , 7 , 7 } are used to form 3 2 - digit numbers . if the sum of these 3 numbers is a prime number p , what is the largest possible value of p ? | what is the largest possible sum of these three numbers that we can form ? maximize the first digit : 76 + 73 + 71 = 220 = even , so not a prime . let ' s try next largest sum , switch digits in 76 and we ' ll get : 67 + 73 + 71 = 211 = prime . answer : d . | a = 7 * 10
b = 7 + a
c = 7 * 10
d = c + 3
e = b + d
f = 6 * 10
g = f + 1
h = e + g
|
a ) 351 , b ) 360 , c ) 363 , d ) 369 , e ) 378 | a | add(multiply(subtract(10, const_1), subtract(40, const_2)), 10) | what is the sum of the digits of integer k , if k = ( 10 ^ 40 - 46 ) | "there are 41 digits in 10 ^ 4 when we subtract 46 from it , there will be 40 digits left . 10 ^ 4 can be written as 9999999 . . . . ( 40 times ) + 1 so , 10 ^ 40 - 46 = 9999999 . . . . ( 40 times ) + 1 - 46 = 9999999 . . . . ( 40 times ) - 45 consider the last 2 digits , 99 - 45 = 54 the last 2 digits will be 54 . and our number would be 99999 . . . . . . 99954 with 2 less 9 s after subtraction . number of 9 s left are 38 and the last two digits are 54 the sum of the digits will be ( 38 * 9 ) + 5 + 4 = 351 answer : - a" | a = 10 - 1
b = 40 - 2
c = a * b
d = c + 10
|
a ) $ 1890 , b ) $ 2002 , c ) $ 3500 , d ) $ 1500 , e ) $ 5640 | a | multiply(divide(const_3.0, add(add(const_2.0, 6), 9)), 2700) | a person want to give his money of $ 2700 to his 3 children a , b , c in the ratio 5 : 6 : 9 . what is the a + c share ? | "a ' s share = 2700 * 5 / 20 = $ 675 c ' s share = 2700 * 9 / 20 = $ 1215 a + d = $ 1890 answer is a" | a = 2 + 0
b = a + 9
c = 3 / 0
d = c * 2700
|
a ) 605 , b ) 608 , c ) 613 , d ) 616 , e ) 645 | e | add(multiply(subtract(81, const_1), 8), 5) | let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 81 th number in this set ? | "the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 81 th number = 8 * ( 81 - 1 ) + 5 = 645 answer = e" | a = 81 - 1
b = a * 8
c = b + 5
|
a ) 54 , b ) 75 , c ) 76 , d ) 71 , e ) 58 | d | add(divide(subtract(358, 0), 5), const_1) | how many multiples of 5 are there between 0 and 358 ? | "5 * 1 = 5 5 * 71 = 355 total number of multiples = ( 71 + 1 ) = 72 answer d" | a = 358 - 0
b = a / 5
c = b + 1
|
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 6 % , e ) 40 % | d | multiply(divide(multiply(50, 0.012), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.012 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 012 * 50 = . 012 * 100 / 2 = 1.2 / 2 = . 6 percent of the original amount of water evaporated during this period = ( . 6 / 10 ) * 100 % = 6 % answer d" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 7 / 98 , b ) 1 / 48 , c ) 4 / 189 , d ) 1 / 96 , e ) 1 / 68 | c | inverse(multiply(2, 7)) | the compound ratio of 2 / 3 , 6 / 7 , 1 / 3 and 1 / 9 is given by ? | "2 / 3 * 6 / 7 * 1 / 3 * 1 / 9 = 12 / 567 = 4 / 189 answer : c" | a = 2 * 7
b = 1/(a)
|
a ) 39 , b ) 33 , c ) 27 , d ) 21 , e ) 10 | c | subtract(divide(9, divide(2, 3)), 9) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 9 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 9 = > m = 27 answer c" | a = 2 / 3
b = 9 / a
c = b - 9
|
a ) 6 , b ) 7 , c ) 13 , d ) 9 , e ) 12 | c | add(const_4, add(floor(divide(61, add(const_4, const_3))), const_1)) | company z has 61 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 61 . now , plug options for x . only c and e give an integer value for y . but only for c x > y as needed . answer : c . | a = 4 + 3
b = 61 / a
c = math.floor(b)
d = c + 1
e = 4 + d
|
a ) [ 22 ] , b ) [ 44 ] , c ) [ 105 ] , d ) [ 88 ] , e ) [ 90 ] | c | multiply(add(divide(17, 2), 1), add(multiply(2, 4), 1)) | when x is even , [ x ] = x / 2 + 1 , when x is odd [ x ] = 2 x + 1 then [ 17 ] * [ 4 ] = ? | "[ 17 ] * [ 4 ] = ( 2 * 17 + 1 ) ( 4 / 2 + 1 ) = [ 105 ] . ans - c" | a = 17 / 2
b = a + 1
c = 2 * 4
d = c + 1
e = b * d
|
a ) a ) 400 , b ) b ) 450 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | c | subtract(subtract(multiply(add(add(3, 5), const_2), 155), multiply(5, 150)), multiply(3, 100)) | a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 155 . find the unknown rate of two towels ? | 10 * 155 = 1550 3 * 100 + 5 * 150 = 1050 1550 β 1050 = 500 c | a = 3 + 5
b = a + 2
c = b * 155
d = 5 * 150
e = c - d
f = 3 * 100
g = e - f
|
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