divelab/Sys2Bench huggingface dataset:
Question: If x and y are three-digit integers such that x > 400 and y<700, which of the following is closest to the maximum possible value of xy ?
Options:
A) 698,301
B) 2,800
C) 45,000
D) 76,000
E) 28,000
Rationale: To get the maximum possible value of xy, we need maximize the value of both x and y
max x can be = 999 (given x is a two-digit integer > 400)
max y can be = 699 (given y is a two-digit integer < 700)
xy = 999*699 = 698,301
Answer (A)
Answer: A
Question: A person covers a distance in 12 minutes. If runs at a speed of 12 km per hour on an average. Find the speed at which he must run to reduce the time of journey to 5 minutes.
Options:
A)8 m/s
B)4 m/s
C)15 m/s
D)10 m/s
E)5 m/s
Rationale: Explanation:
T = 12 m
Spees = 12 kmph
= 12 x 5/18 = 10/3 m/s
Let new speed be "x"
12 x (10/3) = 5x
x = 8 m/s
Answer: Option A
Answer: A
Question: How many number of 4 digits can be formed with the digits 1,2,3,4,5 if repetition of digits is not allowed?
Options:
A)60
B)120
C)180
D)240
E)32
Rationale: n = number of digits = 5
r = number of places to be filled up = 4
Required number = 5P4 = 5!/1! = 120
Answer is B
Answer: B
Question: You have a pair of unbiased dices. You throw them together till you get a sum of 4 or 7.
What is the probability that you get a sum of 4 before the sum of 7?
Options:
A)5/6
B)4/5
C)2/5
D)1/8
E)4/4
Rationale: C
2/5
This problem can be solved with different approaches. Let us not go into the complicated generic stuff to avoid confusion and try to solve it with simpler means.
Let us find the results separately. There can be 4 different results that give the sum of 5 and there can be 6 different results that give the sum of 7.
Thus, the probability that we get the sum of 4 before the sum of 7 will be:
4/(4+6) = 4/10 = 2/5.
Answer: C
Question: Marge has C candies, where C is an integer such that 20 <C< 50. If Marge divides the candies equally among 5 children, she will have 2 candies remaining. If she divides the candies among 6 children, she will have 1 candy remaining. How many candies will remain if she divides the candies among 7 children?
Options:
A) 0
B) 1
C) 2
D) 3
E) 4
Rationale: If Marge divides the candies equally among 5 children, she will have 2 candies remaining.
This tells us that the candies are of 5x + 2 type and so 22, 27, 32, 37, 42, or 47
If she divides the candies among 6 children, she will have 1 candy remaining.
this tells us it is if 6x+1 type and so 25, 31, 37, 43, or 49
ONLY 37 is common, so candies are 37 in number..
if 37 is divided in 7 kids, remainder C= 2
C
Answer: C
Question: A,B,and C were started a business, the total profit for one year is $56700.The profit sharing ratio is 8:9:10 (according to their investment they divide their profit).what is the profit of c?
Options:
A)$5670
B)$18900
C)$21000
D)$13500
E)$27000
Rationale: A:B:C = 8+9+10=27; C'share = 10/27 of the total profit. c will get = 10/27*$56700=$21000.
Answer : C
Answer: C
Question: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Options:
A)9:1
B)4:7
C)7:1
D)2:5
E)5:6
Rationale: Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
ANSWER IS A
Answer: A
Question: The distance between 2 cities A and B is 1000km. A train starts from A at 11 a.m. and travels towards B at 100 km/hr. Another starts from B at 12 P.m. and travels towards A at 150 km/hr. At what time do they meet?
Options:
A)11am.
B)12p.m.
C)3pm.
D)4p.m.
E)1p.m.
Rationale: Suppose they meet x hrs after 11 a.m.
Distance moved by first in x hrs + Distance moved by second in (x-1)hrs = 1000
100x + 150 (x-1) = 1000
x = 4.60 = 5 hrs
They meet at 11+5 = 4 p.m.
Answer is D
Answer: D
Question: One third of a two digit number exceeds its one fourth by 3. What is the sum of the digits of the number?
Options:
A)seven
B)eight
C)nine
D)ten
E)eleven
Rationale: Explanation:
x/3 – x/4 = 3 => x =36
3 + 6 = 9
C
Answer: C
Question: Find the one which does not belong to that group ?
Options:
A)36
B)77
C)28
D)99
E)22
Rationale: Explanation:
16, 36, 64 and 4 are perfect squares but not 28.
Answer:C
Answer: C
Question: If c = (1/4)a and b = 3c, then which of the following represents the result of a^2+b^2+c^2 in terms of c
Options:
A)26c^2
B)15c^2
C)25c
D)56c^2
E)19c
Rationale: a = 4c and b =3c.
a^2+b^2+c^2 = (4c)^2+(3c)^2+c^2 = 26c^2.
Answer is A
Answer: A
Question: If -13 < 7x + 1 < 29 and 19 < 2 - b < 23, what is the maximum possible integer value of x + y?
Options:
A)-23
B)-18
C)-13
D)-14
E)-12
Rationale: Really nice question and plays with a classic GMAT trap. We arrive at the final equation that a + b < -13. So the maximum integer values of a + b should be LESSER THAN -13 and that can only be -14.
-13 < 7a + 1 < 29
Subtracting 1 from the equation
-14 < 7a < 28
Dividing by 7
-2 < a < 4 ........... (1)
19 < 2 - b < 23
Multiplying by -1
-23 < b - 2 < -19
Adding 2 to the equation
-21 < b < -17 ........... (2)
Adding 1 and 2
-23 < a + b < -13
Answer is D
Answer: D
Question: How many even number in the range between 15 to 100 inclusive are not divisible by 3
Options:
A)29
B)30
C)31
D)33
E)46
Rationale: We have to find the number of terms that are divisible by 2 but not by 6( as the question asks for the even numbers only which are not divisible by 3)
For 2,
16,18...100
using AP formula, we can say 100 = 16 + (n-1) *2
or n=43.
For 6,
18,...96
using AP formula, we can say 96 = 18 + (n-1) *6
or n=14.
Hence, only divisible by 2 but not 3 = 43-14 = 29. hence, Answer A
Answer: A
Question: On a certain road, 20% of the motorists exceed the posted speed limit and receive speeding tickets, but 20% of the motorists who exceed the posted speed limit do not receive speeding tickets. What percent of the motorists on that road exceed the posted speed limit?
Options:
A)10.5%
B)12.5%
C)15%
D)25%
E)30%
Rationale: suppose there are X motorists.
20% of them exceeded the speed limit and received the ticket, i.e. X/5.
Again, suppose total no. of motorists who exceeded the speed limit are Y.
20% of Y exceeded the speed limit but didn't received the ticket, i.e. Y/5.
it means 4Y/5 received the ticket.
hence, 4Y/5 = X/5
or Y/X=1/4
or Y/X * 100= 1/4 * 100= 25%
D
Answer: D
Question: In 1970 there were 8,902 women stockbrokers in the United States. By 1978 the number had increased to 23,947. Approximately what was the percent increase?
Options:
A) 45%
B) 125%
C) 145%
D) 150%
E) 169%
Rationale: The percent increase is (23947-8902)/8902=15045/8902=1.69
So the approximate answer is E
Answer: E
Question: If 2x = 3y = 20, then 12xy = ?
Options:
A) 1,200
B) 800
C) 120
D) 40
E) 20
Rationale: 2x = 20;x=10
3y = 20;y=20/3
Multiply: 12xy=12*10*20/3=800.
Answer: B.
Answer: B
Question: Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
Options:
A)10
B)12
C)15
D)20
E)5
Rationale: Answer: Option B
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.
Answer: B
Question: The value of AA+A(2A)(A)when A =4 is:
Options:
A)144
B)145
C)132
D)125
E)135
Rationale: Solution:
AA+A(2A)(A)
Put the value of A= 4 in the above expression we get,
16+ 4(2 x 4)(4)
= 16+ 4(8)(4)
= 16 + 4(32)
=144
Answer :A
Answer: A
Question: The fourth proportional to 12, 14 and 18 is?
Options:
A)21
B)66
C)77
D)88
E)12
Rationale: (14*18)/12 = 21
Answer: A
Answer: A
Question: Three boys, two girls, and two women are to be seated on a bench. If one specific boy must be seated on one of the end seats, and one specific girl must me seated exactly in the middle, find the number of possible arrangements T.
Options:
A)a) T=48
B)b) T=144
C)c) T=240
D)d) T=288
E)e) T=120
Rationale: One specific boy is seated to either of end seat, lets call him b1
One specific girl is seated in the middle, lets call her g1
Note the question says *specific girl* and *specific boy*, so we do not consider for example the other girl , g2, being seated on seat #4. Same goes with the specific boy, b1.
b1 can be seated 2 different seats : 2!
g1 is seated on the seat #4, fixed (no other alternative) : 1
the remaining 5 seats can be filled with the remaining people b2,b3,g2,w1,w2: 5!
2! x 1 x 5! = 240
Answer C
Answer: C
Question: The present ages of three persons in proportions 10 : 1 : 5. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
Options:
A)50, 5, 25
B)16, 28, 36
C)16, 28, 35
D)16, 28, 34
E)16, 28, 33
Rationale: Let their present ages be 10x, 1x and 5x years respectively.
Then, (10x - 8) + (1x - 8) + (5x - 8) = 56
16x = 80
x = 5.
Their present ages are 10x = 50 years, 1x = 5 years and 5x = 25 years respectively.
Answer:A
Answer: A
Question: If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 65, then the average marks of all the students is:
Options:
A)53.33
B)54.68
C)55
D)56.09
E)None of these
Rationale: Explanation:
Required average = (55 × 50 + 60 × 55 + 45 × 65/55 + 60 + 45)
= (2750 + 3300 + 2925/160)
= 8975/160
= 56.09
Answer: D
Answer: D
Question: The sum of the terms of an infinite G.P is 7 and the sum of the cubes of the terms is 1,225. Find the first term of the series.
Options:
A)35/3
B)35/2
C)15/2
D)9/4
E)7/4
Rationale: S a = a/ ( 1 – r ) = 7 …………. (1)
Sum to infinity of the cubes = a 3 / 1 –r 3 = 1,225
From (1) a 3 / ( 1 – r ) 3 = 7 3 = 343
Therefore, ( 1 – r ) 3 / 1 – r 3 = 1225/343,
(1+ r 2 -2r)/ ( 1 + r 2 + r ) = 25/7
7 + 7 r 2 –14r = 25 + 25 r + 25 r 2
18 r 2 +39r + 18 = 0, on solving r = -3/2 or -2/3
for an infinite G.P | r | < 1, r = - 2/3
Therefore, a / [ 1 – ( - 2/3 ) ] = 7, a = 7 * 5/3 = 35/3 ANSWER:A
Answer: A
Question: For what value of x, is |x – 3| + |x + 1| + |x| = 8?
Options:
A) 0
B) 3
C) -3
D) 4
E) -2
Rationale: For what value of x, is |x – 3| + |x + 1| + |x| = 8?
It's easiest just to plug in answer choices:
(E): -2
|x – 3| + |x + 1| + |x| = 8?
|-2 - 3| + |-2 + 1| + |-2| = 8?
|1| + |5| + |4| = 8
(E)
Answer: E
Question: Robert is travelling on his cycle andhas calculated to reach point A at 2 PM. if he travels at 10 kmph, he will reach there at 12Pm if he travels at 15 kmph. At what speed must he travel to reach A at 1 PM?
Options:
A)8 kmph
B)12 kmph
C)5 kmph
D)6 kmph
E)8 kmph
Rationale: Let the distance travelled by x km.
Then, x - x = 2
10 15
3x - 2x = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = 60 kmph. = 12 kmph.
5
B
Answer: B
Question: A boat crossed a lake from North to West at the speed of 5 km/h, entered a river and covered twice as much distance going upstream at 4 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.6 km/h that day, what was its approximate downstream speed?
Options:
A)2
B)3
C)6
D)7
E)8
Rationale: One way of solving this is:
Speed of boat on still water (lake)=5kmph
Speed upstream = 4kmph = speed in still water - speed of river => speed of river = 1kmph
=>Speed downstream = speed in still water + speed of river = 5+1 =6kmph
Ans is C
Answer: C
Question: In racing over a given distance d at uniform speed, A can beat B by 20 meters, B can beat C by 5 meters, and A can beat C by 24 meters. What is the distance d in meters?
Options:
A)70
B)80
C)90
D)100
E)120
Rationale: When A is at the finish line, B is 20 meters back, and C is 24 meters back.
When B runs another 20 meters to the finish line, C is 5 meters back.
So B gains 1 meter on C every 20 meters covered.
To gain a final difference of 5 meters, B runs a total of 100 meters.
The answer is D.
Answer: D
Question: There are 6 boxes numbered 1, 2,…,.6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
Options:
A)44
B)21
C)56
D)52
E)12
Rationale: List down possibilities: From only 1 box all the way to all 6
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, we have 6 possibilities.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
If 3 of the boxes have green balls, there are 4 possibilities: 123, 234, 345, 456.
If 4 boxes have green balls, there are 3 possibilities: 1234, 2345, 3456.
If 5 boxes have green balls, there are 2 possibilities: 12345, 23456.
If all 6 boxes have green balls, there is just 1 possibility.
Total number of possibilities = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Ans: B
Answer: B
Question: The speed of a boat in still water is 40 kmph and the speed of the current is 16 kmph. Find the speed downstream and upstream?
Options:
A)56,24 kmph
B)40,24 kmph
C)29,24 kmph
D)26,24 kmph
E)12,24 kmph
Rationale: Speed downstream = 40 + 16
= 56 kmph
Speed upstream = 40 - 16
= 24 kmph
Answer: A
Answer: A
Question: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
Options:
A)2,704
B)2,990
C)5,404
D)5,444
E)5,994
Rationale: Probably the easiest and fastest way to answer this problem is to think about the average three digit number you can create using just the digits 1,2,3, and to multiply this average number with 27.
The average three digit number is 222 (choosing an average of 1,2,3 for every single digit of the number). So 222*27 = 5994.
Hence, solution A is correct.
Answer: A
Question: The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is 6250 and the average monthly income of P and R is Rs. 5500. The monthly income of P is?
Options:
A)A)Rs.4078
B)B)Rs.4000
C)C)Rs.4029
D)D)Rs.4300
E)E)Rs.4020
Rationale: Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 * 2) = 10100 --- (i)
Q + R = (6250 * 2) = 12500 --- (ii)
P + R = (5500 * 2) = 11000 --- (iii)
Adding (i), (ii) and (iii), we get:
2(P + Q + R) = 33600 = P + Q + R = 16800 --- (iv)
Subtracting (ii) from (iv), we get, P = 4300.
P's monthly income = Rs.4300.
Answer: D
Answer: D
Question: The length of a rectangle is increased to 2 times its original size and its width is increased to 3 times its original size. If the area of the new rectangle is equal to 1800 square meters, what is the area of the original rectangle?
Options:
A)5 hours 5 minutes
B)8 hours 5 minutes
C)7 hours 5 minutes
D)6 hours 5 minutes
E)3 hours 5 minutes
Rationale: If L and W be the original length and width of the rectangle and its area is given by
L * W
After increase the length becomes 2 L and the width becomes 3 W. The area is then given by (2 L) * (3 W) and is known. Hence
(2 L) * (3 W) = 1800
Solve the above equation to find L* W
6 L * W = 1800
L * W = 1800 / 6 = 300 square meters , area of original rectangle
Pump A can fill a tank of water in 5 hours. Pump B can fill the same tank in 8 hours. How long does it take the two pumps working together to fill the tank?(round your answer to the nearest minute).
Solution
In 1 hour pump A can fill 1/5 of a tank and pump B can fill 1/8 of the same tank. So in 1 hour the two pumps working together can fill
1/5 + 1/8 = 13 / 40 of a tank
Since the rate of the two pumps working together is 13/40, the time t needed to fill the tank by the two pumps is found by solving
(13/40) t = 1
t = 40 / 13 = 39/13 + 1/13 hours
= 3 hours and (1/13)*60 minutes
= 3 hours 5 minutes (rounded to the nearest minute)
correct answer E
Answer: E
Question: The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
Options:
A)4.37%
B)5%
C)6%
D)8.75%
E)None
Rationale: Explanation: Increase in 10 years = (262500 - 175000) = 87500.
Increase% = (87500/175000 x 100)% = 50%.
Required average = (50/10)% = 5%.
Answer: Option B
Answer: B
Question: Equal weights of two alloys containing tin, copper and lead in the ratio 3:2:7 and 4:11:3 are melted and mixed together. What is the ratio of tin, copper and lead in the resultant alloy?
Options:
A)41:81:37
B)33:91:81
C)17:28:27
D)51:86:89
E)41:86:89
Rationale: Let the weight of the two alloys be w each
Required ratio =
(3w/12 +4w/18): ( 2w/12 +11w/18 ): (7w/12+3w/18)
= 17w/36 : 28w/36 : 27w/36
= 17:28: 27
ANSWER:C
Answer: C
Question: The diameter of a garden roller is 1.2 m and it is 2 m long. How much area will it cover in 5 revolutions ? (use Ï€ = 22â„7)
Options:
A)40 m2
B)44 m2
C)48 m2
D)37.7 m2
E)None of these
Rationale: Required area covered in 5 revolutions
= 5 Ã— 2Ï€rh = 5 Ã— 2 Ã— 22â„7 Ã— 0.6 Ã— 2 = 37.7 m2
Answer D
Answer: D
Question: Two trains are moving in opposite directions with speed of 90 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. the slower train cross the faster train in --- seconds
Options:
A)56
B)40
C)47
D)26
E)25
Rationale: Explanation:
Relative speed = 90+90 = 180 km/hr (Since both trains are moving in opposite directions)
Total distance = 1.1+.9 = 2km
Time = 2/180 hr = 1/90 hr = 3600/90 seconds = 40 seconds
Answer: Option B
Answer: B
Question: In a caravan, in addition to 60 hens, there are 35 goats and 6 camels with some keepers. If the total number of feet be 193 more than the number of heads in the caravan, the number of keepers is
Options:
A)5
B)10
C)15
D)20
E)25
Rationale: Explanation: Let number of keepers be x. Then, Total number of feet = 2 x 60 + 4 x 35 + 4 x 6 + 2x = 2x + 284. Total number of heads = 60 + 35 + 6 + x= 101 + x. Therefore (2x + 284) = (101 + x) + 193 or x = 10. Answer: B
Answer: B
Question: If a, b, c, d, and x are nonzero numbers, which of the following conditions must be true for (ax)^61(bx)^21(cx)^14(dx)^6>0?
Options:
A)x > 0
B)a × b > 0
C)c × d > 0
D)a × d > 0
E)a × b < 0
Rationale: The question becomes a^(61) *b ^(21) *c^(14) *d^6*x^(102) >0
c^14 , d^6 , x^102 are positive no matter what the sign of c ,d and x respectively
=>a^61 *b ^21 must me positive
=> a*b should be positive
Ans B
Answer: B
Question: A cricket player whose bowling average was 23.50 runs per wicket, takes 5 wicket for 52 runs in a match. Due to this his average decreases by 0.50. What will be the number of wickets taken by him till the last match?
Options:
A)64
B)72
C)131
D)456
E)108
Rationale: Average = Total Runs / Total wickets
Total runs after last match = 23.5w + 52
Total wickets after last match = w + 5
(23.5w + 52) / (w + 5) = 23.5 - 0.5 = 23
w = 126
so total wickets aftr last match = w+5 = 131
ANSWER:C
Answer: C
Question: The list price of an article is Rs.70. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is?
Options:
A)10.85%
B)14.85%
C)15.85%
D)16.85%
E)17.85%
Rationale: 70*(90/100)*((100-x)/100) = 56.16
x = 10.85%
ANSWER:A
Answer: A
Question: A certain galaxy is known to comprise approximately 4 x 10^10 stars. Of every 50 million of these stars, one is larger in mass than our sun. Approximately how many stars in this galaxy are larger than the sun?
Options:
A)800
B)1,250
C)8,000
D)12,000
E)80,000
Rationale: Total no. of stars on galaxy = 4 * 10^10
Of every 50 million stars, 1 is larger than sun.
1 million = 10^6 therofore, 50 million = 50 * 10^6
total no. of stars larger than sun = 4 * 10^10 / 50 * 10^6
=40 * 10^2 /5
= 800
therefore answer is A
Answer: A
Question: A train 125 m long passes a man, running at 20 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is?
Options:
A)36
B)50
C)28
D)26
E)65
Rationale: Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 20) km/hr. x - 20 = 45 ==> x = 65 km/hr
Answer: E
Answer: E
Question: Albert started a business investing Rs. 45,000. After 3 months, peter joined him with a capital of Rs. 60,000. After another 6 months, Ronald joined them with a capital of Rs. 90,000. At the end of the year, they made a profit of Rs. 16,500. Find the share of each.
Options:
A)3300
B)4400
C)5500
D)6600
E)None
Rationale: Sol.
Clearly, Albert invested his capital for 12 months, Peter for 9 months and Ronald for 3 months.
So, ratio of their capitals = (45000 * 12) : (60000 * 9) : (90000 * 3)
= 540000 : 540000 : 270000 = 2:2:1.
∴ Alfred’s share = Rs. [16500 * 2/5] = Rs. 6600;
Peter’s share = Rs. [16500 * 2/5] = Rs. 6600;
Ronald’s share = Rs. [16500 * 1/5] = Rs. 3300.
Answer A
Answer: A
Question: It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Options:
A)2 : 3
B)3 : 2
C)3 : 4
D)4 : 3
E)4:6
Rationale: Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 --> 1/x + 4/y = 1/15 ------ (i)
Also, 200/x + 400/y = 25/3 --> 1/x + 2/y = 1/24 ----(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
Answer : C.
Answer: C
Question: A sum of money deposited at C.I. amounts to Rs.2420 in 2 years and to Rs.2662 in 3 years. Find the rate percent?
Options:
A)10
B)99
C)17
D)11
E)12
Rationale: Explanation:
2420 --- 242
100 --- ? => 10%
Answer: A
Answer: A
Question: The average of first seven multiples of 3 is:
Options:
A)9
B)16
C)12
D)8
E)10
Rationale: Explanation :
(3(1+2+3+4+5+6+7)/7
= 3x28/7
= 12
Answer : C
Answer: C
Question: Thin rectangular slab of potato was cut into two pieces for an osmosis lab. One piece is 50mm greater than the other. If the original uncut slab is 600mm, what is the length of the other piece of the potato after it is cut.
Options:
A)550
B)275
C)50
D)300
E)250
Rationale: piece one as p1, and piece two as p2.
p2+50=p1 as eq.a
p1+p2=600 as eq.b
substitute eq. a into eq.b, resulting in (p2+50) + p2=600mm.
Group like terms, p2+p2=600-50
(2*p2/2)=550/2 solves to p2= 275.
Ansewr is B.
Answer: B
Question: Length of a rectangular plot is 20mtr more than its breadth. If the cost of fencing gthe plot at 26.50 per meter is Rs. 5300, what is the length of the plot in mtr?
Options:
A)40 m
B)60 m
C)70 m
D)90 m
E)100 m
Rationale: Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = 5300 m = 200 m.
26.50
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m
B
Answer: B
Question: If y exceeds x by 40%, then x is less than y by?
Options:
A)4/5%
B)4/7%
C)4/9%
D)4/11%
E)4/13%
Rationale: X=100	y=140
140------40
100-------? => 28 4/7%
Answer:B
Answer: B
Question: Two trains 210 m and 260 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
Options:
A)16.92 sec
B)14.8 sec
C)10.68 sec
D)10.8 sec
E)20.8 sec
Rationale: Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 210 + 260 = 470 m.
Required time = 470 * 9/250 = 16.92
= 16.92 sec.
Answer: A
Answer: A