Problem
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967
| Rationale
stringlengths 1
2.74k
| options
stringlengths 37
164
| correct
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1.65k
| linear_formula
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925
| category
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|---|---|---|---|---|---|---|---|
if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2080 less than the principal . what was the principal ?
|
"p - 2080 = ( p * 5 * 4 ) / 100 p = 2600 answer : b"
|
a ) 1500 , b ) 2600 , c ) 2507 , d ) 3200 , e ) 11500
|
b
|
divide(2080, subtract(const_1, divide(multiply(4, 5), const_100)))
|
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
|
gain
|
B
|
a dealer offers a cash discount of 10 % and still makes a profit of 20 % when he further allows 45 articles to be sold at the cost price of 40 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ?
|
"given cash discount - 10 % profit - 20 % items sold - 45 price sold at = list price of 40 assume list price = $ 10 total invoice = $ 400 - 10 % cash discount = $ 360 let cost price of 45 items be x so total cost = 45 * x given the shopkeeper had a profit of 20 % 45 * x * 120 / 100 = 360 or x = $ 20 / 3 = $ 6 + ( 2 / 3 ) which means his products were listed at $ 10 which is a 50 % markup over $ 6 + ( 2 / 3 ) answer b"
|
a ) 40 % , b ) 50 % , c ) 55 % , d ) 80 % , e ) 90 %
|
b
|
multiply(subtract(divide(divide(divide(add(const_100, 20), const_100), subtract(const_1, divide(subtract(45, 40), 45))), divide(subtract(const_100, 10), const_100)), const_1), const_100)
|
add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)|
|
gain
|
B
|
in a certain company , the ratio of male to female employees is 7 : 8 . if 3 more men were hired , this ratio would increase to 8 : 9 . how many male employees are there in the company ?
|
another approach is to use two variables . let m = present number of males let f = present number of females the ratio of male to female employees is 7 : 8 so , m / f = 7 / 8 cross multiply to get 7 f = 8 m if 3 more men were hired , this ratio would increase to 8 : 9 so , ( m + 3 ) / f = 8 / 9 cross multiply to get 9 ( m + 3 ) = 8 f expand : 9 m + 27 = 8 f we now have a system of two equations and two variables : 7 f = 8 m 9 m + 27 = 8 f solve to get : m = 189 and f = 216 answer : d
|
a ) 6 , b ) 21 , c ) 27 , d ) 189 , e ) 192
|
d
|
divide(multiply(divide(3, subtract(divide(8, 9), divide(7, 8))), 7), 8)
|
divide(n1,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2)|multiply(n0,#3)|divide(#4,n1)
|
general
|
D
|
for what values of k will the pair of equations 3 x + 4 y = 12 and ( kx + 12 y ) / 2 = 15 does not have a unique solution ?
|
"we have 2 equations 1 . 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . ( kx + 12 y ) / 2 = 15 - - > kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b"
|
a ) 12 , b ) 9 , c ) 3 , d ) 7.5 , e ) 2.5
|
b
|
divide(multiply(2, 4), 3)
|
multiply(n1,n4)|divide(#0,n0)|
|
general
|
B
|
a 6 - liter solution is 30 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ?
|
"let x be the amount of pure alcohol required . 0.3 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 1.8 x = 2.4 liters the answer is c ."
|
a ) 1.8 , b ) 2.1 , c ) 2.4 , d ) 2.7 , e ) 3.0
|
c
|
subtract(6, multiply(divide(multiply(6, 30), const_100), const_2))
|
multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_2)|subtract(n0,#2)|
|
general
|
C
|
a sum of money at simple interest amounts to rs . 815 in 4 years and to rs . 854 in 5 years . the sum is :
|
"sol . s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 4 years = rs . ( 39 * 4 ) = rs . 156 . â ˆ ´ principal = rs . ( 815 - 156 ) = rs . 659 answer a"
|
a ) 659 , b ) 698 , c ) 780 , d ) 910 , e ) none
|
a
|
subtract(815, divide(multiply(subtract(854, 815), 4), 5))
|
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
|
gain
|
A
|
a worker ' s daily wage is increased by 50 % and the new wage is $ 51 per day . what was the worker ' s daily wage before the increase ?
|
"let x be the daily wage before the increase . 1.5 x = $ 51 x = $ 34 the answer is c ."
|
a ) $ 28 , b ) $ 32 , c ) $ 34 , d ) $ 35 , e ) $ 40
|
c
|
divide(51, add(const_1, divide(50, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
physics
|
C
|
in how many ways can a 5 - letter password be chosen , using the letters a , b , c , d , e , f , g , and / or h such that at least one letter is repeated within the password ?
|
total number of four letter passwords = 8 * 8 * 8 * 8 * 8 = 32768 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 8 c 5 * 5 ! = 56 * 120 = 6720 - - - - - - ( 2 ) therefore required value = ( 1 ) - ( 2 ) = 32768 - 6720 = 26048 e
|
a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 26048
|
e
|
subtract(power(add(5, const_3), 5), divide(factorial(add(5, const_3)), factorial(const_3)))
|
add(n0,const_3)|factorial(const_3)|factorial(#0)|power(#0,n0)|divide(#2,#1)|subtract(#3,#4)
|
general
|
E
|
the difference between c . i . and s . i . on an amount of $ 20,000 for 2 years is $ 98 . what is the rate of interest per annum ?
|
"$ 98 is the interest on the first year of interest . let x be the interest rate . the interest after the first year is 20000 * x . the interest on the first year ' s interest is 20000 * x * x 20000 * x ^ 2 = 98 x = 0.07 the answer is c ."
|
a ) 9 % , b ) 8 % , c ) 7 % , d ) 6 % , e ) 5 %
|
c
|
sqrt(98)
|
sqrt(n2)|
|
gain
|
C
|
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 70 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ?
|
"relative speed = 70 + 30 = 100 km / hr . 100 * 5 / 18 = 250 / 9 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 9 / 250 = 36 sec answer : b"
|
a ) 22 , b ) 36 , c ) 48 , d ) 99 , e ) 23
|
b
|
add(70, 30)
|
add(n1,n2)|
|
physics
|
B
|
during the first week of performances of a certain play , 5400 tickets were sold , all at reduced price . during the remaining weeks of performances , 5 times as many tickets were sold at full price as were sold at reduced price . if the total number of tickets e sold was 25200 , how many of them were sold at full price ?
|
given : the question tells us about the number of tickets sold in the first week at reduced price = 5400 . it also tells us that during the remaining weeks , the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price . the total tickets sold was 25200 , and we are asked to find the number of tickets that where sold at full price . approach : the question asks us to find the number of ticket sold at full price . we know that the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price in remaining weeks . hence , assuming tickets sold at reduced price in remaining weeks to be x would give us the no . of tickets sold at full price to be 5 x . tickets sold in first week + tickets sold in remaining weeks = total no . of tickets soldi . e . tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold we know the no . of tickets sold during first week and the total no . of tickets sold . we will use this relation to get to our answer . working out : tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold substituting values in the equation , we get 5400 + 5 x + x = 25200 x = 3300 tickets sold at full price = 5 x = 16500 answer : option c
|
a ) 21000 , b ) 19800 , c ) 16500 , d ) 15800 , e ) 8700
|
c
|
divide(multiply(subtract(25200, 5400), 5), add(const_2, const_4))
|
add(const_2,const_4)|subtract(n2,n0)|multiply(n1,#1)|divide(#2,#0)
|
general
|
C
|
1000 men have provisions for 21 days . if 800 more men join them , for how many days will the provisions last now ?
|
"1000 * 21 = 1800 * x x = 11.6 answer : d"
|
a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 11.6 , e ) 12.1
|
d
|
divide(multiply(21, 1000), add(1000, 800))
|
add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|
|
physics
|
D
|
seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 28 % ryegrass , what percent of the weight of the mixture is from mixture x ?
|
"28 % is 3 % - points above 25 % and 12 % - points below 40 % . thus the ratio of mixture y to mixture x is 4 : 1 . the percent of mixture x is 1 / 5 = 20 % . the answer is b ."
|
a ) 10 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 60 %
|
b
|
divide(subtract(28, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100)))
|
add(n2,const_1)|divide(n0,const_100)|divide(#0,const_100)|subtract(n4,#0)|subtract(#1,#2)|divide(#3,#4)|
|
gain
|
B
|
the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 20,000 is made ?
|
"the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 20,000 is made = 96,000 + 20,000 = rs . 1 , 16,000 answer : d"
|
a ) 328897 , b ) 120000 , c ) 877888 , d ) 116000 , e ) 188871
|
d
|
add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100)))
|
add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)|
|
gain
|
D
|
the l . c . m of two numbers is 495 and their h . c . f is 5 . if the sum of the numbers is 10 , then their difference is :
|
"let the numbers be x and ( 100 - x ) . then , x ( 100 - x ) = 5 * 495 x 2 - 100 x + 2475 = 0 ( x - 55 ) ( x - 45 ) = 0 x = 55 or 45 the numbers are 45 and 55 . required difference = 55 - 45 = 10 . answer : a"
|
a ) 10 , b ) 46 , c ) 70 , d ) 90 , e ) 100
|
a
|
divide(multiply(5, 495), 10)
|
multiply(n0,n1)|divide(#0,n2)|
|
general
|
A
|
solution y is 30 percent liquid x and 70 percent water . if 4 kilograms of water evaporate from 8 kilograms of solution y and 4 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ?
|
"in 8 kilograms of solution y there are 0.3 * 8 = 2.4 kilograms of solution x ; after 4 kilograms of water are replaced by 4 kilograms of solution y , to the existing 2.4 kilograms of solution x , 0.3 * 4 = 1.2 kilograms of solution x are added , so in the new solution of 8 kilograms there are 2.4 + 1.2 = 3.6 kilograms of solution x , which is 3.6 / 8 * 100 = 45 % of this new solution . answer : d ."
|
a ) 40 % , b ) 41 1 / 2 % , c ) 44 % , d ) 45 % , e ) 46 %
|
d
|
multiply(const_100, divide(add(multiply(8, divide(30, const_100)), multiply(4, divide(30, const_100))), add(add(multiply(8, divide(30, const_100)), multiply(4, divide(30, const_100))), add(subtract(multiply(8, divide(70, const_100)), 4), multiply(4, divide(70, const_100))))))
|
divide(n0,const_100)|divide(n1,const_100)|multiply(n3,#0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n3,#1)|add(#2,#3)|subtract(#5,n2)|add(#4,#7)|add(#6,#8)|divide(#6,#9)|multiply(#10,const_100)|
|
gain
|
D
|
rs . 100 amounts to rs . 320 in 4 years at simple interest . if the interest is increased by 3 % , it would amount to how much ?
|
"( 100 * 4 * 3 ) / 100 = 12 100 + 12 = 112 answer : c"
|
a ) 246 , b ) 650 , c ) 110 , d ) 240 , e ) 360
|
c
|
multiply(power(add(const_1, divide(3, const_100)), 4), 100)
|
divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|
|
gain
|
C
|
matt and peter can do together a piece of work in 20 days . after they have worked together for 12 days peter stops and peter completes the remaining work in 10 days . in how many days matt complete the work separately .
|
together they complete the job in 20 days means they complete 12 / 20 of the job after 12 days . peter completes the remaining ( 8 / 20 ) of the job in 10 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 10 < = > x = 10 / ( 8 / 20 ) = 25 d
|
a ) 28 , b ) 27 , c ) 26 , d ) 25 , e ) 29
|
d
|
divide(10, subtract(const_1, multiply(12, divide(const_1, 20))))
|
divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2)
|
physics
|
D
|
a batsman makes a score of 95 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning .
|
"let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 95 = 11 x x = 95 - 50 = 45 answer is d"
|
a ) 40 , b ) 50 , c ) 62 , d ) 45 , e ) 56
|
d
|
add(subtract(95, multiply(11, 5)), 5)
|
multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|
|
general
|
D
|
an article is bought for rs . 560 and sold for rs . 340 , find the loss percent ?
|
560 - - - - 220 100 - - - - ? = > 39 % answer : d
|
a ) 36 % , b ) 38 % , c ) 16 % , d ) 39 % , e ) 12 %
|
d
|
subtract(const_100, divide(multiply(340, const_100), 560))
|
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)
|
gain
|
D
|
two boys starts from the same place walking at the rate of 5.5 kmph and 7.5 kmph respectively in the same direction . what time will they take to be 32 km apart ?
|
"explanation : relative speed = 7.5 - 5.5 = 2 kmph ( because they walk in the same direction ) distance = 32 km time = distance / speed = 32 / 2 = 16 hr answer : b"
|
a ) 18 , b ) 16 , c ) 19 , d ) 15 , e ) 34
|
b
|
divide(32, subtract(7.5, 5.5))
|
subtract(n1,n0)|divide(n2,#0)|
|
gain
|
B
|
at an m & m factory , two types of m & ms are produced , red and blue . the m & ms are transported individually on a conveyor belt . anna is watching the conveyor belt , and has determined that 4 out of every 5 red m & ms are followed by a blue one , while one out of every 6 blue m & ms is followed by a red one . what proportion of the m & ms are red ?
|
suppose that the fraction of red m & ms is fr and the fraction of blue m & ms is fb . then the probability that anna sees a red one emerge next is fr . this probabil - ity can also be expressed using conditional probabilities as fr = p ( blue ) p ( red after blue ) + p ( red ) p ( red after red ) , or fr = fb ( 1 = 6 ) + fr ( 1 = 5 ) . likewise , we find fb = fr ( 4 = 5 ) + fb ( 5 = 6 ) . solving the system of equations gives fb = 24 29 and fr = 5 / 29 correct answer e
|
a ) 1 / 29 , b ) 2 / 29 , c ) 3 / 29 , d ) 4 / 29 , e ) 5 / 29
|
e
|
divide(5, add(multiply(4, 5), add(4, 5)))
|
add(n0,n1)|multiply(n0,n1)|add(#0,#1)|divide(n1,#2)
|
other
|
E
|
average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 870 . what was original rent of friend whose rent is increased ?
|
"0.2 x = 4 ( 870 - 800 ) 0.2 x = 280 x = 1400 answer b"
|
a ) 800 , b ) 1400 , c ) 1000 , d ) 1100 , e ) 1200
|
b
|
divide(multiply(subtract(870, 800), 4), divide(20, const_100))
|
divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0)|
|
general
|
B
|
heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 40 litres of tomato juice ?
|
"answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 40 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 5 liters . answer : c"
|
a ) 2.8 liters . , b ) 2.5 liters . , c ) 5 liters . , d ) 2.6 liters . , e ) 2.1 liters .
|
c
|
divide(multiply(divide(subtract(const_100, 90), const_100), 40), divide(subtract(const_100, 20), const_100))
|
subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)|
|
gain
|
C
|
the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 47 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ?
|
"20 x + 47 = 21 ( x – 5 ) x = 58 answer : d"
|
a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59
|
d
|
subtract(multiply(add(20, const_1), 5), 47)
|
add(n0,const_1)|multiply(n2,#0)|subtract(#1,n1)|
|
general
|
D
|
what is the smallest integer b for which 27 ^ b > 3 ^ 24 ?
|
"27 ^ b > 3 ^ 24 converting into the same bases : 27 ^ b > 27 ^ 8 therefore for the equation to hold true , b > 8 or b = 9 option c"
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12
|
c
|
add(floor(divide(24, const_3)), const_1)
|
divide(n2,const_3)|floor(#0)|add(#1,const_1)|
|
general
|
C
|
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 ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 4 : 5
|
a
|
divide(sqrt(8), sqrt(18))
|
sqrt(n1)|sqrt(n0)|divide(#0,#1)|
|
physics
|
A
|
change the below lagarithm log 255 = 1 / 2 to exponential form
|
solution : log 255 = 251 / 2 = 5 answer is a
|
a ) 5 , b ) 8 , c ) 7 , d ) 10 , e ) 11
|
a
|
subtract(subtract(subtract(subtract(255, const_100), const_100), multiply(const_10, const_4)), const_10)
|
multiply(const_10,const_4)|subtract(n0,const_100)|subtract(#1,const_100)|subtract(#2,#0)|subtract(#3,const_10)
|
general
|
A
|
in a room with 9 people , 6 people have exactly 1 friend in the room and 5 people have exactly 2 friends in the room ( assuming that friendship is a mutual relationship , i . e . if jane is paul ' s friend , paul is jane ' s friend ) . if two individuals are selected from the room at random , what is the probability that those two individuals are not friends ?
|
"( 6 / 9 ) ( 7 / 8 ) + ( 5 / 9 ) ( 6 / 8 ) if you choose one of the 6 with one other friend , then you have a 7 / 8 chance of not picking their friend 2 nd . if you choose one of the 5 with 2 friends , you have a 6 / 8 chance of not picking one of their friends second . add them up . 42 / 72 + 30 / 72 72 / 72 = 1 e . 1"
|
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 1
|
e
|
divide(subtract(subtract(subtract(9, 1), 1), 1), subtract(9, 1))
|
subtract(n0,n2)|subtract(#0,n2)|subtract(#1,n2)|divide(#2,#0)|
|
general
|
E
|
9 persons went to a hotel for taking their meals . 8 of them spent rs 12 each on their meals and the ninth spent rs . 8 more than the average expenditure of all the 9 . what was the total money spent by them ?
|
exp . let the total expenditure be x , then the average = x / 9 , 8 * 12 + [ x / 9 + 8 ] = x or [ x - x / 9 ] = 104 . 8 x / 9 = 104 , x = 104 * 9 / 8 = 117 . answer : a
|
a ) 117 , b ) 180 , c ) 150 , d ) 200 , e ) 250
|
a
|
multiply(9, divide(add(multiply(8, 12), 8), subtract(9, const_1)))
|
multiply(n1,n2)|subtract(n0,const_1)|add(n1,#0)|divide(#2,#1)|multiply(n0,#3)
|
general
|
A
|
if x is to be chosen at random from the integers between 1 to 7 , inclusive , and y is to be chosen at random from the integers between 8 and 12 , inclusive , what is the probability that x + y will be even ?
|
"x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 3 / 7 * 3 / 5 = 9 / 35 p ( x and y are both odd ) = 4 / 7 * 2 / 5 = 8 / 35 p ( x + y is even ) = 9 / 35 + 8 / 35 = 17 / 35 the answer is d ."
|
a ) 2 / 5 , b ) 3 / 7 , c ) 5 / 7 , d ) 17 / 35 , e ) 24 / 35
|
d
|
divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(7, subtract(7, 1)))
|
multiply(const_2,const_2)|multiply(const_2,const_3)|subtract(n1,n0)|add(#0,#1)|multiply(n1,#2)|divide(#3,#4)|
|
general
|
D
|
if 12 men do a work in 80 days , in how many days will 20 men do it ?
|
"12 * 80 = 20 * x x = 48 days answer : d"
|
a ) 18 days , b ) 38 days , c ) 42 days , d ) 48 days , e ) 44 days
|
d
|
divide(multiply(12, 80), 20)
|
multiply(n0,n1)|divide(#0,n2)|
|
physics
|
D
|
rectangular tile each of size 50 cm by 40 cm must be laid horizontally on a rectangular floor of size 120 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is :
|
"area of tile = 50 * 40 = 2000 area of floor = 120 * 150 = 18000 no of tiles = 18000 / 2000 = 9 so , the no of tile = 9 answer : a"
|
a ) 9 , b ) 15 , c ) 12 , d ) 11 , e ) 10
|
a
|
divide(multiply(120, 150), multiply(50, 40))
|
multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)|
|
geometry
|
A
|
if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 6 years the amount to be paid is ?
|
"a = 7500 ( 26 / 25 ) ^ 6 = 9490 answer : d"
|
a ) 3377 , b ) 2688 , c ) 2688 , d ) 9490 , e ) 1268
|
d
|
multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100)))
|
add(n1,const_100)|divide(#0,const_100)|multiply(#1,#1)|multiply(n0,#2)|
|
gain
|
D
|
there are 18 balls in a jar . you take out 3 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ?
|
there are 18 balls in a jar . you take out 3 blue balls without putting 3 blue balls + 15 / 5 = 6 answer : e
|
a ) 12 , b ) 9 , c ) 8 , d ) 7 , e ) 6
|
e
|
add(divide(subtract(18, 3), 5), 3)
|
subtract(n0,n1)|divide(#0,n3)|add(n1,#1)
|
probability
|
E
|
a rectangle having length 150 cm and width 40 cm . if the length of the rectangle is increased by fifteen percent then how much percent the breadth should be decreased so as to maintain the same area .
|
"explanation : solution : ( 15 / ( 150 + 15 ) * 100 ) % = 9.09 % answer : b"
|
a ) 25 % , b ) 9.09 % , c ) 40 % , d ) 75 % , e ) none of these
|
b
|
multiply(add(const_1, divide(divide(multiply(150, 40), add(150, divide(multiply(multiply(const_3, const_10), 150), const_100))), 40)), const_10)
|
multiply(n0,n1)|multiply(const_10,const_3)|multiply(n0,#1)|divide(#2,const_100)|add(n0,#3)|divide(#0,#4)|divide(#5,n1)|add(#6,const_1)|multiply(#7,const_10)|
|
geometry
|
B
|
murali travelled from city a to city b at a speed of 40 kmph and from city b to city c at 60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is 1 : 2 ? a . 48 kmph
|
let the distances between city a to b and b to c be x km and 2 x km respectively . total time taken to cover from a to c = ( x ) / 40 + ( 2 x ) / 60 = ( 3 x + 4 x ) / 120 = 7 x / 120 average speed = ( x + 2 x ) / ( 7 x / 120 ) = 360 / 7 kmph . answer : b
|
a ) 33 , b ) 360 / 7 , c ) 28 , d ) 27 / 8 , e ) 16
|
b
|
divide(add(40, 60), 2)
|
add(n0,n1)|divide(#0,n3)
|
physics
|
B
|
if jake loses 8 pounds , he will weigh twice as much as his sister . together they now weigh 290 pounds . what is jake ’ s present weight , in pounds ?
|
"lets say j is the weight of jack and s is the wt of his sister . if he loses 8 pounds , he s twice as heavy as his sister . j - 8 = 2 * s also , together they weight 290 pounds j + s = 290 solvong the 2 equation , we get j = 196 pounds ! d"
|
a ) 131 , b ) 135 , c ) 169 , d ) 196 , e ) 212
|
d
|
add(multiply(divide(subtract(290, 8), const_3), const_2), 8)
|
subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)|
|
general
|
D
|
3 candidates in an election and received 1036 , 4636 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ?
|
total number of votes polled = ( 1036 + 4636 + 11628 ) = 17300 so , required percentage = 11628 / 17300 * 100 = 67.2 % e
|
a ) 40 % , b ) 55 % , c ) 57 % , d ) 60 % , e ) 67.2 %
|
e
|
multiply(divide(11628, add(add(1036, 4636), 11628)), const_100)
|
add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)
|
gain
|
E
|
when 242 is divided by a certain divisor the remainder obtained is 8 . when 698 is divided by the same divisor the remainder obtained is 9 . when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 4 . what is the value of the divisor ?
|
"explanatory answer when 242 is divided by a certain divisor the remainder obtained is 8 . let the divisor be d . when 242 is divided by d , let the quotient be ' x ' . the remainder is 8 . therefore , 242 = xd + 8 when 698 is divided by the same divisor the remainder obtained is 9 . let y be the quotient when 698 is divided by d . then , 698 = yd + 9 . when the sum of the two numbers , 242 and 698 , is divided by the divisor , the remainder obtained is 4 . 242 + 698 = 940 = xd + yd + 8 + 9 940 = xd + yd + 17 as xd and yd are divisible by d , the remainder when 940 is divided by d should have been 17 . however , because we know that the remainder is 4 , it would be possible only when 17 / d leaves a remainder of 4 . if the remainder obtained is 4 when 17 is divided by ' d ' , then ' d ' has to be 13 . choice c"
|
a ) 11 , b ) 17 , c ) 13 , d ) 23 , e ) none of these
|
c
|
subtract(add(8, 9), 4)
|
add(n1,n3)|subtract(#0,n6)|
|
general
|
C
|
6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days m do they need to finish the same job ?
|
"let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time m = 3 days to finish the task imo a"
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
|
a
|
divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4))
|
add(n0,n3)|multiply(n0,n1)|multiply(n0,n2)|subtract(#1,#2)|divide(#3,#0)|
|
physics
|
A
|
the cash difference between the selling prices of an article at a profit of 4 % and 6 % is rs . 3 . the ratio of the two selling prices is :
|
"let c . p . of the article be rs . x . then , required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53 answer : b"
|
a ) 51 : 52 , b ) 52 : 53 , c ) 51 : 53 , d ) 52 : 55 , e ) 52 : 56
|
b
|
divide(add(const_100, 4), add(const_100, 6))
|
add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|
|
gain
|
B
|
jaclyn buys $ 20 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter
|
"explanation : i = ( p x r x t ) / 100 = 30000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 475 answer : a"
|
a ) 475 , b ) 234 , c ) 289 , d ) 345 , e ) none of these
|
a
|
divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4)
|
multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_4)|
|
gain
|
A
|
if i earn a profit 50 $ in a day and i have 9 employees . after keeping 10 percent profit for me , and distribute remianing money equally among my employees then how much each employee would get ?
|
since i earned $ 50 sp 10 percent would be 5 $ and 45 $ would remain . dividing 45 by 9 , each employee would get 5 $ each and hince option d is correct .
|
a ) $ 3 , b ) $ 7 , c ) $ 1 , d ) $ 5 , e ) $ 9
|
d
|
divide(multiply(50, subtract(const_100, 10)), multiply(const_100, 9))
|
multiply(n1,const_100)|subtract(const_100,n2)|multiply(n0,#1)|divide(#2,#0)
|
gain
|
D
|
in a office work is distribute between p persons . if 1 / 8 members are absent then work increased for each person is ?
|
"let total % of work is 100 % total person = p 1 / 8 person are absent of total person . so absent person is 1 / 8 p ie p / 8 . left person is , p - p / 8 = 7 p / 8 . p person do the work 100 % 1 person do the work 100 * p % 7 p / 8 person do the work ( 100 * p * 8 ) / 7 p % = 114.28 % work increased for each person is = ( 114.28 - 100 ) % = 14.28 % answer : b"
|
a ) 13.28 % , b ) 14.28 % , c ) 15.28 % , d ) 16.28 % , e ) 17.28 %
|
b
|
subtract(divide(multiply(const_100, 8), subtract(8, 1)), const_100)
|
multiply(n1,const_100)|subtract(n1,n0)|divide(#0,#1)|subtract(#2,const_100)|
|
general
|
B
|
a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , while a and c together can do it in 3 hours . how long will b alone take to do it ?
|
"a ' s 1 hour ' s work = 1 / 4 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 3 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 4 + 1 / 3 ) = 7 / 12 . b ' s 1 hour ' s work = ( 7 / 12 - 1 / 3 ) = 1 / 4 . therefore a alone will take 4 hours to do the work . e"
|
a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 4
|
e
|
divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 3), divide(const_1, 4))))
|
divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)|
|
physics
|
E
|
a , b and c can do a work in 6 , 9 and 12 days respectively doing the work together and get a payment of rs . 1800 . what is b ’ s share ?
|
"wc = 1 / 6 : 1 / 9 : 1 / 12 = > 6 : 4 : 3 4 / 13 * 1800 = 653.8 answer : b"
|
a ) rs . 245.8 , b ) rs . 653.8 , c ) rs . 300 , d ) rs . 400 , e ) rs . 748.5
|
b
|
multiply(1800, divide(inverse(9), add(inverse(12), add(inverse(6), inverse(9)))))
|
inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)|
|
physics
|
B
|
at what rate percent of simple interest will a sum of money double itself in 44 years ?
|
"let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 44 ) = 100 / 44 = 2.27 % answer : e"
|
a ) 6.54 % , b ) 4.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 2.27 %
|
e
|
divide(const_100, 44)
|
divide(const_100,n0)|
|
gain
|
E
|
if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 10 ?
|
"the product of n and 5 = 5 n . . the difference between an integer n and the product of n and 5 = 5 n - n = 4 n . . the outcome of the binary operation of n is less than 10 ? so 4 n < 10 . . . max value that n can take as an integer is 2 answer : b"
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
|
b
|
floor(divide(10, subtract(5, const_1)))
|
subtract(n0,const_1)|divide(n1,#0)|floor(#1)|
|
general
|
B
|
if the average ( arithmetic mean ) of 2 x , 4 x , and 8 x is 140 , what is the value of x ?
|
"am of 2 x , 4 x and 8 x = 2 x + 4 x + 8 x / 3 = 14 x / 3 given that 14 x / 3 = 140 x = 30 answer is b"
|
a ) 20 , b ) 30 , c ) 15 , d ) 33 , e ) 31
|
b
|
subtract(multiply(4, const_2), multiply(2, const_2))
|
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
|
general
|
B
|
when 1 / 10 percent of 7,000 is subtracted from 1 / 10 of 7,000 , the difference is
|
we can break this problem into two parts : 1 ) what is 1 / 10 percent of 7,000 ? 2 ) what is 1 / 10 of 7,000 ? to calculate 1 / 10 percent of 7,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100 = 1 / 1,000 thus , 1 / 10 percent of 7,000 = 1 / 1,000 x 7,000 = 7 . now let ' s concentrate on part 2 . we need to calculate 1 / 10 of 7,000 . to do this we simply multiply 1 / 10 by 7,000 . 1 / 10 x 7,000 = 700 the answer to part 1 is 7 , and the answer to part 2 is 700 . their difference is 700 – 7 = 693 . answer d .
|
a ) 0 , b ) 50 , c ) 450 , d ) 693 , e ) 500
|
d
|
subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000)))
|
add(const_3,const_4)|divide(n0,n1)|divide(#1,const_100)|multiply(#0,const_1000)|multiply(#1,#3)|multiply(#2,#3)|subtract(#4,#5)
|
general
|
D
|
a can do a piece of work in 20 days , and b can do it in 15 days . how long will they take if both work together ?
|
a + b can do the work in 20 * 15 / 20 + 15 = 20 * 15 / 35 = 8.6 days answer is b
|
a ) 7 days , b ) 8.6 days , c ) 13 days , d ) 5.6 days , e ) 10 days
|
b
|
inverse(add(inverse(15), inverse(20)))
|
inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)
|
physics
|
B
|
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 21 adults have their meal , find the total number of children that can be catered with the remaining food .
|
"explanation : as there is meal for 70 adults and 21 have their meal , the meal left can be catered to 49 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 49 adults = 63 children hence , the meal can be catered to 63 children . answer : c"
|
a ) 33 , b ) 54 , c ) 63 , d ) 17 , e ) 01
|
c
|
multiply(subtract(70, 21), divide(90, 70))
|
divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1)|
|
general
|
C
|
a train passes a station platform in 35 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?
|
"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 35 = 15 = > x = 225 m . answer : c"
|
a ) 228 , b ) 240 , c ) 225 , d ) 166 , e ) 1811
|
c
|
multiply(20, multiply(54, const_0_2778))
|
multiply(n2,const_0_2778)|multiply(n1,#0)|
|
physics
|
C
|
if a - b = 3 and a ^ 2 + b ^ 2 = 29 , find the value of ab
|
"2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 29 - 9 = 20 = > ab = 10 . answer : a"
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 19
|
a
|
multiply(multiply(add(3, divide(subtract(sqrt(29), 3), 2)), divide(subtract(sqrt(29), 3), 2)), 2)
|
sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)|
|
general
|
A
|
a train running at 1 / 2 of its own speed reached a place in 8 hours . how much time could be saved if the train would have run at its own speed ?
|
time taken if run its own speed = 1 / 2 * 8 = 4 hrs time saved = 8 - 4 = 4 hrs answer : c
|
a ) 8 hrs , b ) 10 hrs , c ) 4 hrs , d ) 15 hrs , e ) 6 hrs
|
c
|
multiply(divide(1, 2), 8)
|
divide(n0,n1)|multiply(n2,#0)
|
physics
|
C
|
mark and ann together were allocated n boxes of cookies to sell for a club project . mark sold 9 boxes less than n and ann sold 2 boxes less than n . if mark and ann have each sold at least one box of cookies , but together they have sold less than n boxes , what is the value of n ?
|
"if n = 10 mark sold 1 box and ann sold 8 boxes total 9 < 10 answer : b"
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
|
b
|
add(9, const_1)
|
add(n0,const_1)|
|
general
|
B
|
given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 2 ?
|
"p is a positiveeveninteger with apositive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit of p + 2 is 6 + 2 = 8 . answer : b ."
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
|
b
|
add(add(3, 3), 2)
|
add(n0,n0)|add(n3,#0)|
|
general
|
B
|
in an examination , 34 % of total students failed in hindi , 44 % failed in english and 22 % in both . the percentage of these who passed in both the subjects is :
|
explanation : formula n ( a â ˆ ª b ) = n ( a ) + n ( b ) â ˆ ’ n ( a â ˆ © b ) fail in hindi or english = 34 + 44 â € “ 22 = 56 therefore students who passed = 100 â € “ 56 = 44 . answer : d
|
a ) 23 , b ) 37 , c ) 28 , d ) 44 , e ) 81
|
d
|
subtract(const_100, subtract(add(34, 44), 22))
|
add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)
|
general
|
D
|
if 20 % of 30 is greater than 25 % of a number by 2 , the number is :
|
explanation : = > 20 / 100 * 30 - 25 / 100 * x = 2 = > x / 4 = 4 so x = 16 answer : a
|
a ) 16 , b ) 24 , c ) 20 , d ) 32 , e ) none of these
|
a
|
divide(subtract(multiply(divide(20, const_100), 30), 2), divide(25, const_100))
|
divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)
|
gain
|
A
|
an aeroplane covers a certain distance at a speed of 320 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of :
|
"distance = ( 240 x 5 ) = 1280 km . speed = distance / time speed = 1280 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 1280 x 3 / 5 ) km / hr = 768 km / hr answer d ) 768 km / hr"
|
a ) 520 , b ) 620 , c ) 820 , d ) 768 , e ) 720
|
d
|
divide(divide(multiply(320, 4), add(const_1, divide(const_2, const_3))), const_2)
|
divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)|
|
physics
|
D
|
a woman swims downstream 54 km and upstream 6 km taking 6 hours each time , what is the speed of the woman in still water ?
|
"54 - - - 6 ds = 9 ? - - - - 1 6 - - - - 6 us = 1 ? - - - - 1 m = ? m = ( 9 + 1 ) / 2 = 5 answer : d"
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
|
d
|
divide(add(divide(54, 6), divide(6, 6)), const_2)
|
divide(n0,n2)|divide(n1,n2)|add(#0,#1)|divide(#2,const_2)|
|
physics
|
D
|
the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 9 ) , how many units is p from the y - axis ?
|
"the x - axis is 9 units from the point p . thus the y - axis is 18 units from the point p . the answer is a ."
|
a ) 18 , b ) 12 , c ) 9 , d ) 4.5 , e ) 3
|
a
|
multiply(9, const_2)
|
multiply(n0,const_2)|
|
general
|
A
|
the average weight of 10 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what might be the weight of the new person ?
|
"explanation : total weight increased = ( 10 x 2.5 ) kg = 25 kg . weight of new person = ( 50 + 25 ) kg = 75 kg . answer : c"
|
a ) 76 kg , b ) 76.5 kg , c ) 75 kg , d ) data inadequate , e ) none of these
|
c
|
add(multiply(10, 2.5), 50)
|
multiply(n0,n1)|add(n2,#0)|
|
general
|
C
|
ms . lopez deposits $ 100 in an account that pays 20 % interest , compounded semiannually . how much money will there be in the account at the end of one year ?
|
"et me explain the concept for those who are not familiar with compounded semi annually / quarterly etc . . it basically means , here , that the interest will be paid twice , after 6 months and at the end of year . . so twice . . but interest is 20 % , which will get distributed semiannually that is 10 % . . . so the q literally means . . after 6 months . . 100 * 10 / 100 + 100 = 110 . . at the end of year 110 * 10 / 100 + 110 = 121 . . or we put it in compound interest formula . . where ci = 10 % , t is 2 and p = 100 . . so amount = 100 ( 1 + 10 / 100 ) ^ 2 = 100 * 1.1 * 1.1 = 121 . . answer : c"
|
a ) $ 118.00 , b ) $ 120.00 , c ) $ 121.00 , d ) $ 122.00 , e ) $ 140.00
|
c
|
multiply(100, power(add(const_1, divide(divide(20, const_100), const_2)), const_2))
|
divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)|
|
gain
|
C
|
truck x is 14 miles ahead of truck y , which is traveling the same direction along the same route as truck x . if truck x is traveling at an average speed of 57 miles per hour and truck y is traveling at an average speed of 63 miles per hour , how long will it take truck y to overtake and drive 4 miles ahead of truck x ?
|
"relative speed = 63 - 57 = 6 miles per hour dist required = 14 + 4 = 18 miles time taken to overtake = 18 / 6 = 3 hours . c is the answer ."
|
a ) 5 hours , b ) 3 hours 20 minutes , c ) 3 hours , d ) 2 hours 45 minutes , e ) 8 hours
|
c
|
divide(add(14, 4), subtract(63, 57))
|
add(n0,n3)|subtract(n2,n1)|divide(#0,#1)|
|
physics
|
C
|
a certain fraction has the same ratio to 1 / 18 , as 2 / 5 does to 2 / 9 . what is this certain fraction ?
|
"x / ( 1 / 18 ) = ( 2 / 5 ) / ( 2 / 9 ) x = 2 * 9 * 1 / 18 * 5 * 2 = 1 / 10 the answer is c ."
|
a ) 1 / 5 , b ) 1 / 7 , c ) 1 / 10 , d ) 1 / 12 , e ) 1 / 15
|
c
|
divide(1, 18)
|
divide(n0,n1)|
|
other
|
C
|
the average of 6 observations is 12 . a new seventh observation is included and the new average is decreased by 1 . the seventh observation is :
|
"explanation : seventh observation = ( 7 × × 11 - 6 × × 12 ) = 5 correct option : c"
|
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 8
|
c
|
subtract(multiply(subtract(12, 1), add(6, 1)), multiply(12, 6))
|
add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)|
|
general
|
C
|
two trains of length 140 m and 280 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 = 140 + 280 = 420 m . the time required = d / s = 420 / 20 = 21 sec . answer : e"
|
a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 21
|
e
|
divide(add(140, 280), multiply(add(42, 30), const_0_2778))
|
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
|
physics
|
E
|
if n is a natural number , then ( 6 n 2 + 6 n ) is always divisible by :
|
"( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . answer c ) 6 and 12 both ."
|
a ) 6 only , b ) 12 only , c ) 6 and 12 both , d ) by 18 only , e ) 15 only
|
c
|
add(multiply(6, const_100), multiply(2, 6))
|
multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|
|
general
|
C
|
car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 42 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
|
"in 72 minutes , car x travels 42 miles . car y gains 7 miles each hour , so it takes 6 hours to catch car x . in 6 hours , car x travels 210 miles . the answer is d ."
|
a ) 105 , b ) 140 , c ) 175 , d ) 210 , e ) 245
|
d
|
multiply(35, divide(multiply(divide(72, const_60), 35), subtract(42, 35)))
|
divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)|
|
physics
|
D
|
the volumes of two cones are in the ratio 1 : 12 and the radii of the cones are in the ratio of 1 : 3 . what is the ratio of their heights ?
|
the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 12 h 1 / r 22 h 2 = > 1 / 12 = ( 1 ) 2 h 1 / ( 3 ) 2 h 2 = > h 1 / h 2 = 3 / 4 i . e . h 1 : h 2 = 3 : 4 answer : a
|
['a ) a ) 3 : 4', 'b ) b ) 2 : 7', 'c ) c ) 2 : 2', 'd ) d ) 2 : 1', 'e ) e ) 2 : 9']
|
a
|
divide(3, const_4)
|
divide(n3,const_4)
|
geometry
|
A
|
the average weight of 4 pigs is 15 kg . weight less than 16 kg is considered as under weight . what maximum no . of pig can be under weight .
|
sum of all four pigs weigh = 60 kg it is possible that all four pigs are underweight . because the average is 15 kg which itself underweight . answer e
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
|
e
|
multiply(subtract(16, 15), 4)
|
subtract(n2,n1)|multiply(n0,#0)
|
general
|
E
|
at a certain organisation , the number of male members went up by 13 % in the year 2001 from year 2000 , and the number of females members went down by 6 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ?
|
men increase by 13 % = = > 1.13 m = males in 2001 women decrease by 6 % = = > 0.94 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.13 m + 0.94 f = 1.012 * ( m + f ) 1.13 m + 0.94 f = 1.012 m + 1.012 f 1.13 m - 1.012 m = 1.012 f - 0.94 f 0.108 m = 0.072 f m / f = ( 0.072 ) / ( 0.118 ) = 72 / 118 = 3 / 5 answer = ( d )
|
a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 5 , e ) 2 : 1
|
d
|
divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(6, const_100)), const_1000)), subtract(multiply(add(const_1, divide(13, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000)))
|
divide(n4,const_100)|divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#2,const_1)|subtract(const_1,#1)|multiply(#3,const_1000)|multiply(#5,const_1000)|multiply(#4,const_1000)|subtract(#6,#7)|subtract(#8,#6)|divide(#9,#10)
|
other
|
D
|
a sum of money at simple interest amounts to rs . 824 in 3 years and to rs . 854 in 4 years . the sum is :
|
"sol . s . i . for 1 year = rs . ( 854 - 824 ) = rs . 30 . s . i . for 3 years = rs . ( 30 * 3 ) = rs . 90 . â ˆ ´ principal = rs . ( 815 - 90 ) = rs . 725 answer e"
|
a ) 500 , b ) 698 , c ) 780 , d ) 910 , e ) 725
|
e
|
subtract(824, divide(multiply(subtract(854, 824), 3), 4))
|
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
|
gain
|
E
|
find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 16 per cubic meter ?
|
"22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 16 = 1584 answer : e"
|
a ) 2998 , b ) 2799 , c ) 2890 , d ) 1485 , e ) 1584
|
e
|
multiply(volume_cylinder(divide(3, const_2), 14), 16)
|
divide(n1,const_2)|volume_cylinder(#0,n0)|multiply(n2,#1)|
|
physics
|
E
|
working simultaneously and independently at an identical constant rate , 10 machines of a certain type can produce a total of x units of product p in 4 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 3 x units of product p in 6 days ?
|
"the rate of 10 machines is rate = job / time = x / 4 units per day - - > the rate of 1 machine 1 / 10 * ( x / 4 ) = x / 40 units per day ; now , again as { time } * { combined rate } = { job done } then 6 * ( m * x / 40 ) = 3 x - - > m = 20 . answer : d ."
|
a ) 14 , b ) 15 , c ) 18 , d ) 20 , e ) 22
|
d
|
multiply(multiply(10, 3), divide(4, 6))
|
divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)|
|
general
|
D
|
if p and q are prime numbers , how many divisors does the product p ^ 2 * q ^ 5 have ?
|
"when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is e . 3 * 6 = 18"
|
a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 18
|
e
|
multiply(add(2, const_1), add(5, const_1))
|
add(n0,const_1)|add(n1,const_1)|multiply(#0,#1)|
|
general
|
E
|
in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 11 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture .
|
"given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 11 ) = 36 / 20 = 9 / 5 . answer : d ."
|
a ) 2 / 1 , b ) 4 / 1 , c ) 2 / 3 , d ) 9 / 5 , e ) 3 / 2
|
d
|
divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 11))
|
add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)|
|
general
|
D
|
a boy is travelling from his home to school at 5 km / hr and reached 7 min late . next day he traveled at 10 km / hr and reached 8 min early . distance between home and school ?
|
let the distance be x t 1 = x / 5 hr t 2 = x / 10 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 5 - x / 10 = 1 / 4 x / 10 = 1 / 4 x = 2.5 km answer is a
|
a ) 2.5 km , b ) 3.5 km , c ) 4.5 km , d ) 5.5 km , e ) 6.5 km
|
a
|
divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 10))
|
divide(n1,const_60)|divide(n3,const_60)|divide(const_1,n2)|add(#0,#1)|divide(#3,#2)
|
physics
|
A
|
shopkeeper rise price by 32 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ?
|
"let d initial price be 100 32 % rise now price = 132 / 100 * 100 = 132 10 % discount then price = 132 * 90 / 100 = 118.8 15 % discount then price = 118.8 * 85 / 100 = 100.98 so gain = 100.98 - 100 = 0.98 gain % = gain * 100 / cp = = > 0.98 * 100 / 100 = 0.98 % answer : a"
|
a ) 0.98 % , b ) 4.98 % , c ) 3.95 % , d ) 6.98 % , e ) 7.98 %
|
a
|
subtract(multiply(multiply(add(const_100, 32), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100)
|
add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)|
|
gain
|
A
|
one - third less than 15 % of 180 is equal to :
|
"lots of ways to tackle this . 15 % of 180 = 27 1 / 3 of 27 = 9 so , 1 / 3 less than 27 is equal to 27 - 9 = 18 answer : a"
|
a ) 18 , b ) 30 , c ) 35 , d ) 40 , e ) 45
|
a
|
subtract(multiply(divide(15, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(15, const_100), 180)))
|
divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)|
|
general
|
A
|
mike took a bus from home to market , that travels at 30 kmph . while walking back at 6 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 20 kmph . find the average speed .
|
let the distance be 2 x ( one way ) time taken by bus = 2 x / 30 , by walking = x / 6 , by cycling = x / 20 hours : . average speed = total distance / total time = 6 x / x / 15 + x / 6 + x / 20 = 6 * 60 / 4 + 10 + 3 = 21.2 answer : a
|
a ) 21.2 kmph , b ) 12.1 kmph , c ) 23.4 kmph , d ) 20.4 kmph , e ) none of these
|
a
|
divide(add(divide(add(6, 20), const_2), 30), const_2)
|
add(n1,n2)|divide(#0,const_2)|add(n0,#1)|divide(#2,const_2)
|
physics
|
A
|
65 bags of nuts are to be divided among 13 students . each bag contains 15 nuts . how many nuts will each student receive ?
|
65 ÷ 13 = 5 bags per student 5 x 15 = 75 nuts per student correct answer a
|
a ) 75 , b ) 65 , c ) 43 , d ) 52 , e ) 78
|
a
|
multiply(divide(65, 13), 15)
|
divide(n0,n1)|multiply(n2,#0)
|
general
|
A
|
all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 150 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ?
|
"a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 150 liters was transferred from container c to container b , then both the containers would have equal quantities of milk . 5 k - 150 = 3 k + 150 = > 2 k = 300 = > k = 150 the initial quantity of milk in a = 8 k = 8 * 150 = 1200 liters ."
|
a ) 1200 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372
|
a
|
divide(150, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100))))
|
divide(n0,const_100)|inverse(const_2)|subtract(const_1,#0)|subtract(#1,#2)|divide(n1,#3)|
|
general
|
A
|
an engineer undertakes a project to build a road 15 km long in 300 days and employs 40 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time .
|
"40 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 40 * 12.5 * 100 ) / ( 2.5 * 200 ) thus , x = 100 thus , more men needed to finish the task = 100 - 40 = 60 hence answer is e"
|
a ) a . 43 , b ) b . 45 , c ) c . 55 , d ) d . 68 , e ) e . 60
|
e
|
subtract(divide(multiply(multiply(40, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 40)
|
subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)|
|
physics
|
E
|
82 a 7 + 6 b ____ 190 if a and b represent positive single digits in the correctly worked computation above , what is the value of a + 2 b ?
|
adding the digits in unit ' s place , 2 + 7 + b = 10 ( since a and b are positive single digits ) = > b = 1 now adding the digits in ten ' s place , 8 + a + 6 + 1 = 19 ( 1 has been carried over from unit ' s place addition ) = > a = 4 a + 2 b = 4 + 2 * 1 = 6 answer b
|
a ) 2 , b ) 6 , c ) 9 , d ) 11 , e ) 12
|
b
|
add(multiply(subtract(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), multiply(floor(divide(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), const_10)), const_10)), 2), floor(divide(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), const_10)))
|
multiply(n2,const_10)|subtract(n3,#0)|subtract(#1,n0)|subtract(#2,n1)|divide(#3,const_10)|floor(#4)|multiply(#5,const_10)|subtract(#3,#6)|multiply(n4,#7)|add(#5,#8)
|
physics
|
B
|
a boat running up stram takes 6 hours to cover a certain distance , while it takes 9 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ?
|
"explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 9 ( x - y ) 6 x + 6 y = 9 x - 9 y 15 y = 3 x 5 y = x x / y = 5 / 1 5 : 1 answer : option b"
|
a ) 2 : 3 , b ) 5 : 1 , c ) 4 : 5 , d ) 7 : 1 , e ) 8 : 1
|
b
|
subtract(9, 6)
|
subtract(n1,n0)|
|
physics
|
B
|
at a certain university , 68 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ?
|
total women = 68 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 22 % but question wants to know the percent of men that are tenured 22 % / 40 % = 55 % c
|
a ) 25 , b ) 37.5 , c ) 55 , d ) 62.5 , e ) 75
|
c
|
add(subtract(const_100, 68), subtract(90, 68))
|
subtract(const_100,n0)|subtract(n2,n0)|add(#0,#1)
|
gain
|
C
|
there are 6 equally spaced points a , b , c , d , e and f marked on a circle with radius r . how many convex pentagons of distinctly different areas can be drawn using these points as vertices ?
|
solution : since , all the points are equally spaced ; hence the area of all the convex pentagons will be same . answer : option d
|
['a ) 6 p 5', 'b ) 1', 'c ) 5', 'd ) none of these', 'e ) can not be determined']
|
d
|
divide(factorial(6), multiply(factorial(subtract(6, const_2)), factorial(const_2)))
|
factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)
|
geometry
|
D
|
a fruit drink is made of grapefruit , lemon , and orange juice , where 25 percent of the drink is grapefruit juice and 35 percent is lemon juice . if the drink is made with 20 ounces of orange juice , how many ounces is the drink total ?
|
let the total number of ounces in the drink be x % of orange = 25 % % of watermelon = 35 % % of grape = 100 % - 60 % = 40 % but this number is given as 20 ounces so 40 % of x = 20 and x = ( 20 ) ( 100 / 40 ) = ( 20 ) ( 5 / 2 ) x = 50 therefore there a total of 50 ounces in the drink . correct answer - c
|
a ) 60 , b ) 80 , c ) 50 , d ) 100 , e ) 40
|
c
|
divide(20, divide(subtract(const_100, add(25, 35)), const_100))
|
add(n0,n1)|subtract(const_100,#0)|divide(#1,const_100)|divide(n2,#2)
|
general
|
C
|
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 11 per item . if the manufacturer ’ s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ?
|
"let the items manufactured or sold bex 130000 + 11 x = 15 x - 150000 4 x = 280000 x = 70000 ans : e"
|
a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 70,000
|
e
|
divide(multiply(divide(divide(multiply(add(add(const_3, const_10), 15), power(const_100, const_2)), subtract(15, 11)), multiply(const_4, const_2)), const_3), const_1000)
|
add(const_10,const_3)|multiply(const_2,const_4)|power(const_100,const_2)|subtract(n2,n1)|add(n2,#0)|multiply(#4,#2)|divide(#5,#3)|divide(#6,#1)|multiply(#7,const_3)|divide(#8,const_1000)|
|
general
|
E
|
if a = 105 and a ^ 3 = 21 × 25 × 45 × d , what is the value of d ?
|
"a = 105 = 3 * 5 * 7 a ^ 3 = 21 × 25 × 45 × d = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x d = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x d = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x d d = 7 ^ 2 = 49 answer d"
|
a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54
|
d
|
divide(power(105, 3), multiply(multiply(21, 25), 45))
|
multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)|
|
general
|
D
|
if ( 2 ^ 14 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ?
|
"given 2 ^ 14 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 14 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans e on comparing the power of 2 = > m = 14"
|
a ) 7 , b ) 8 , c ) 15 , d ) 16 , e ) 14
|
e
|
power(divide(10, 5), const_4)
|
divide(n4,n3)|power(#0,const_4)|
|
general
|
E
|
a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 50 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 2 hours ?
|
"the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.5 ( 6 ) = 3 gallons at the start , the amount of water is 0.5 ( 6 ) = 3 gallons after 2 hours , the amount of water is 3 - 0.5 ( 2 ) = 2 gallons the concentration of water is 2 / ( 3 + 2 ) = 2 / 5 = 40 % the answer is c ."
|
a ) 32 % , b ) 36 % , c ) 40 % , d ) 44 % , e ) 48 %
|
c
|
multiply(divide(subtract(divide(multiply(2, subtract(const_100, 50)), const_100), multiply(0.5, 2)), subtract(2, multiply(0.5, 2))), const_100)
|
multiply(n2,n3)|subtract(const_100,n0)|multiply(n3,#1)|subtract(n3,#0)|divide(#2,const_100)|subtract(#4,#0)|divide(#5,#3)|multiply(#6,const_100)|
|
gain
|
C
|
a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 1300 . how many workers were part of the latter production run ?
|
"the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1300 . . thus for the current period = 1300 - 500 = 800 ( 5 n + 45 ) k = 800 k = 800 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 800 / 5 n + 45 500 ( 5 n + 45 ) = 800 ( 5 n + 20 ) 25 n + 225 = 40 n + 160 - 15 n = - 65 n = 4 thus n + 5 = 9 hence c"
|
a ) a ) 6 , b ) b ) 10 , c ) c ) 9 , d ) d ) 15 , e ) e ) 23.5
|
c
|
add(20, 5)
|
add(n0,n1)|
|
general
|
C
|
if a person buy radio worth rs 2468 and pay 7 % sales . how much price of radio should reduce to pay only rs 2468 .
|
( x + . 07 x ) = 2468 1.07 x = 2468 x = 2306 reduce = 2468 - 2306 = 162 answer : a
|
a ) 162 , b ) 158 , c ) 159 , d ) 130 , e ) 160
|
a
|
subtract(2468, divide(2468, add(const_1, divide(7, const_100))))
|
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)
|
gain
|
A
|
find the circumference and area of radius 11 cm .
|
"area of circle = π r ² = 22 / 7 × 11 × 11 cm ² = 154 cm ² answer : c"
|
a ) 124 cm ² , b ) 144 cm ² , c ) 380 cm ² , d ) 284 cm ² , e ) 394 cm ²
|
c
|
circle_area(11)
|
circle_area(n0)|
|
geometry
|
C
|
the sides of a square region , measured to the nearest centimeter , are 9 centimeters long . the least possible value of the actual area of the square region is
|
"though there might be some technicalities concerning the termnearest ( as 7.5 is equidistant from both 7 and 8 ) the answer still should be : 7.5 ^ 2 = 56.25 . answer : c"
|
a ) 60.25 sq cm , b ) 66.25 sq cm , c ) 56.25 sq cm , d ) 72.25 sq cm , e ) 58.25 sq cm
|
c
|
power(subtract(subtract(9, const_0_25), const_0_25), const_2)
|
subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2)|
|
geometry
|
C
|
a candidate got 30 % of the votes polled and he lost to his rival by 1760 votes . how many votes were cast ?
|
"let x be the total number of votes . 0.3 x + 1760 = 0.7 x 0.4 x = 1760 x = 1760 / 0.4 = 4400 the answer is e ."
|
a ) 4000 , b ) 4100 , c ) 4200 , d ) 4300 , e ) 4400
|
e
|
divide(1760, subtract(subtract(const_1, divide(30, const_100)), divide(30, const_100)))
|
divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|
|
gain
|
E
|
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