divelab/Sys2Bench huggingface dataset:
Question: 26 +27+28+29=110. How many such 4 consecutive numbers are there less than 2500 when added gives a sum which is divisible by 10?
Options:
A)100
B)200
C)300
D)400
E)500
Rationale: since 1+2+3+4=10 & 6+7+8+9=30
any combination with these no be will give u desirable result...
so total no in 100 will be 20 & that's why in in 2500,it will be 500
ANSWER:E
Answer: E
Question: A palindrome is a number that reads the same forward and backward, such as 196. How many odd, 4-digit numbers are palindromes?
Options:
A)40
B)45
C)50
D)90
E)2500
Rationale: A palindrome is a number that reads the same forward and backward. Examples of four digit palindromes are 1221, 4334, 2222 etc
You basically get to choose the first two digits and you repeat them in opposite order. Say, you choose 45 as your first two digits. The next two digits are 54 and the number is 4554.
Also, you need only odd palindromes. This means that you need an odd digit at the end i.e. 1/3/5/7/9. This means that you need to start the number with an odd digit. Only then will it end with an odd digit.
In how many ways can you pick two digits such that the first one is an odd digit?
The first digit can be selected in 5 ways.(1/3/5/7/9)
The second digit can be selected in 10 ways.(0/1/2/3...8/9)
Total = 5*18 = 90 ways
D
Answer: D
Question: A shopping cart contains only apples, oranges, and pears. If there are four times as many oranges as apples, and four times as many pears as oranges, then the apples are equivalent to what fraction of the pears?
Options:
A)1/2
B)1/4
C)1/8
D)1/16
E)1/32
Rationale: O = 4A
P = 4O = 16A
A = P/16
The answer is D.
Answer: D
Question: Two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 29 kmph and 21 kmph respectively. When they meet, it is found that one train has traveled 60 km more than the other one. The distance between the two stations is?
Options:
A)457 km
B)444 km
C)547 km
D)600 km
E)453 km
Rationale: 1h ----- 5
? ------ 60
12 h
RS = 29 + 21 = 50
T = 12
D = 50 * 12 = 600
Answer:D
Answer: D
Question: A boat goes 100 km downstream in 10 hours, and 75 m upstream in 15 hours. The speed of the stream is?
Options:
A)8 2 ½ kmph
B)2 7 ½ kmph
C)6 2 ½ kmph
D)2 2 ½ kmph
E)2 3 ½ kmph
Rationale: 100 --- 10 DS = 10
? ---- 1
75 ---- 15 US = 5
? ----- 1 S = (10 - 5)/2
= 2 2 ½ kmph
Answer:D
Answer: D
Question: If A got 80 marks and B got 50 marks, then what percent of A's mark is B's mark?
Options:
A)A)62.5%
B)B)80%
C)C)65%
D)D)75%
E)of these
Rationale: A's marks = 80 ; B's marks = 50.
Let x% of A = B => x/100 * 80 = 50
=> x = (50 * 100)/80 = 62.5
B's marks is 62.5% of A's marks.
ANSWER:A
Answer: A
Question: If it takes 8 identical printing presses exactly 4 hours to print 6,000 newspapers, how long would it take 4 of these presses to print 4,000 newspapers?
Options:
A)4 hours, 45 minutes
B)4 hours, 30 minutes
C)4 hours, 50 minutes
D)5 hours, 20 minutes
E)4 hours, 36 minutes
Rationale: 8 presses - 6,000 newspapers - 4 hours ;
4 presses - 3,000 newspapers - 4 hours ; (240 mins)
4 presses - 4,000 newspapers - 240/3000*4000 = 320 mins = 5 hrs 20 mins
Answer : D
Answer: D
Question: Sakshi invests a part of Rs. 12,000 in 12% stock at Rs. 120 and the remainder in 15% stock at Rs. 125. If his total dividend per annum is Rs. 1340, how much does he invest in 12% stock at Rs. 120?
Options:
A)s. 4000
B)s. 4500
C)s. 5000
D)s. 6000
E)s. 6500
Rationale: Let investment in 12% stock be Rs. x.
Then, investment in 15% stock = Rs. (12000 - x).
12/120 * x + 15/125 * (12000- x) = 1340
x/10 + 3/25(12000-x) = 1340
5x + 72000 - 6x = 1340 x 50
x = 5000.
Answer = C
Answer: C
Question: A man whose speed is 7 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?
Options:
A)6.7
B)4
C)6
D)9
E)3
Rationale: M = 7.0
S = 1.5
DS = 8.5
US = 5.5
AS = (2 * 8.5 * 5.5) /14 = 6.7
Answer:A
Answer: A
Question: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k+n is a multiple of 38?
Options:
A)3
B)4
C)7
D)32
E)35
Rationale: n = 5p + 1 = 6,11,16,21,26,31
n = 7q + 3 = 3,10,17,24,31
=>n = 38m + 31
To get this, we need to take LCM of co-efficients of p and q and first common number in series.
So we need to add 7 more to make it
38m + 38
Answer - C
Answer: C
Question: A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.
Options:
A)66 m square
B)49 m square
C)77 m square
D)44 m square
E)33 m square
Rationale: Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square
Answer: B
Answer: B
Question: A train passes a man standing on the platform. If the train is 170 meters long and its speed is 72 kmph, how much time it took in doing so?
Options:
A)8 1/8 sec
B)8 1/2 sec
C)8 6/2 sec
D)7 1/2 sec
E)8 1/5 sec
Rationale: D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 1/2 sec
Answer: B
Answer: B
Question: A man started running at a distance of 225 metre from the train. If the speed of the man is 6 km/h, then how much time should the train wait so that the man will be just able to catch it ?
Options:
A)2 1⁄4 min
B)3 min
C)4 1⁄4 min
D)4 1⁄2 min
E)None of these
Rationale: Time = 225/6×518 = 135 sec = 21⁄4 min.
Answer A
Answer: A
Question: If the city centers of Tokyo and Paris are 6035 miles apart, which of the following is closest to the distance between the city centers in inches? (There are 5,280 feet in a mile.)
Options:
A)1.2 × 10^7
B)1.8 × 10^7
C)3.8 × 10^8
D)3.8 × 10^9
E)3.8 × 10^10
Rationale: 6035 miles = 6035 x 5280 feet/mile x 12 inches/foot which is about 3.8 x 10^8 inches.
The answer is C.
Answer: C
Question: What least number must be subtracted from 42739 so that remaining no.is divisible by 15
Options:
A)3
B)4
C)6
D)8
E)9
Rationale: Explanation:
On dividing 42739 by 15 we get the remainder 4, so 4 should be subtracted
Option B
Answer: B
Question: A library branch originally contained 18360 volumes, 30% of which were fiction novels. 1/3 of the volumes were transferred to another location and 1/3 of the volumes transferred were fiction novels. What percent of the remaining collection was fiction novels?
Options:
A)2.5%
B)17.67%
C)28.3%
D)45.2%
E)73.6%
Rationale: As everything is either fraction or percentage, the given figure 18360 is just a false alarm. We can do this by assuming that originally the branch had 100 volumes.
Originally:
Total- 100
Fiction- 30
Transferred:
Total- 33 (One third of original total)
Fiction- 11 (One third of those transferred)
Remaining:
Total- 100−33=67100−33=67
Fiction- 30−11=1930−11=19
To find:
19 is what percent of 67
28.3
Option C
Answer: C
Question: From among the 34 students in a class, one leader and one class representative are to be appointed. In how many ways can this be done?
Options:
A)1122
B)1260
C)1060
D)1160
E)1460
Rationale: There are 34 students and every one has equal chance of being selected as a leader. Hence, the leader can be appointed in 34 ways. When one person is appointed as leader, we are left with 35 students. Out of these 33 teachers, we can select one class representative. So, a class representative can be selected in 33 ways. Hence, the number of ways in which a leader and class representative can be selected = 34 x 33= 1122
ANSWER:A
Answer: A
Question: A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m?
Options:
A)1350
B)2987
C)2369
D)6989
E)2982
Rationale: Area = (l + b +2d) 2d
= (75 + 55 +2.5 * 2) 2*2.5 => 675
675 * 2
= Rs.1350
Answer:A
Answer: A
Question: A store’s selling price of $2500 for a certain computer would yield a profit of 40 percent of the store’s cost for the computer. What selling price would yield a profit of 50 percent of the computer’s cost?
Options:
A)	$2678
B)	$2464
C)	$2650
D)	$2732
E)	$2800
Rationale: 1.4x = 2500
x=2500/1.4
so, 1.5X = 2500*1.5/1.4= 2478
Answer:- A
Answer: A
Question: A wooden box of dimensions 8 m x 3 m x 6 m is to carry rectangularboxes of dimensions 8 cm x 3 cm x 2 cm. The maximum number ofboxes that can be carried in the wooden box, is
Options:
A)9800000
B)1000000
C)7500000
D)3000000
E)None of these
Rationale: Explanation: Number = (800*300*600)/8*3*2 = 3000000 Answer: D
Answer: D
Question: What is the rate percent when the simple interest on Rs.800 amount to Rs.160 in 4 Years?
Options:
A)5%
B)7%
C)3%
D)9%
E)1%
Rationale: 160
= (180*4*R)/100
R = 5%
Answer:A
Answer: A
Question: A pipe can fill a cistern in 20 minutes whereas the cistern when full can be emptied by a leak in 28 minutes. When both pipes are opened, find when the cistern will be full?
Options:
A)17
B)16
C)70
D)13
E)12
Rationale: 1/20 - 1/28 = 1/70
70 minutes
Answer: C
Answer: C
Question: A large box contains 18 small boxes and each small box contains 28 chocolate bars. How many chocolate bars are in the large box?
Options:
A)A)220
B)B)490
C)C)380
D)D)450
E)E)504
Rationale: the only thing you have to do is a simple multiply
we have:
18*28
=504
correct answer is:
E) 504
Answer: E
Question: Three maths classes: A, B and C take an algebra test. The average score of class A is 83. The average score of class B is 76. The average score of class C is 85. The average score of class A and B is 79 and average score of class B and C is 81. What is the average score of classes A, B, C ?
Options:
A)91.5
B)21.5
C)61.5
D)81.5
E)11.5
Rationale: Explanation:
Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5
Answer: D
Answer: D
Question: If X = 2 + 1/(1+1/(1-1/2))then the value of 3x+7/3?
Options:
A)9 1/3
B)10 1/3
C)11
D)11 2/3
E)12
Rationale: X = 2 + 1/(1+1/(1-1/2))
= 2 + 1/(1+2)
= 2 + 1/3
X = 7/3 ==> 3X + 7/3 => 3(7/3) + 7/3
==> 21/3 + 7/3
==> 28/3
==> 9 1/3.
A)
Answer: A
Question: The relationship between quantities m and n is expressed by the equation 25m = 5(n - 42). If the difference between the two chosen values of n is 30, what is the difference in the corresponding values of m?
Options:
A)30/11
B)42/11
C)6
D)150/11
E)14
Rationale: Another Simple way
25m=5(n-42)
Assume n = 42 to make one side equal to 0
then 25m=0
and m = 0
Repeat it by assuming n as 42+30 i.e 72
25m= 5(72-42)
25m=150
m=150/25
diff = 6 - 0 = 6
Answer : C
Answer: C
Question: By selling 99 pencil, a trader gains the cost of 33 Pencil. Find his gain percentage?
Options:
A)26 1/3%
B)51 1/3%
C)33 1/3%
D)53 1/3%
E)34 1/3%
Rationale: C
33 1/3%
Let the CP of each pencil be Rs. 1.
CP of 99 pens = Rs. 99
Profit = Cost of 33 pencil = Rs. 33
Profit% = 33/99 * 100 = 33 1/3%
Answer: C
Question: An anticraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and the fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane.
Options:
A)0.3024
B)0.6976
C)0.44
D)0.868
E)0.839
Rationale: Explanation :
Let E ≡ the gun hits the plane i.e at least one of the 4 shots hits the plane.
P(E) = 1 - P( none hits the plane)
= 1 - P( ~S1 ~S2 ~S3 ~S4).
( S1 S2 S3 S4 are the probabilities of shots)
= 1 - P( ~S1) x P(~S2) x P(~S3) x P(~S4).
= 1 - (1-0.4)(1-0.3)(1-0.2)(1-0.1)
=1 - [0.6 x 0.7 x 0.8 x 0.9 ]
= 1 - 0.3024.
= 0.6976.
Hence, the required probabilities is 0.6976.
Answer : B
Answer: B
Question: Evaluate: 20 - 12÷12×2 =
Options:
A)A)4
B)B)8
C)C)12
D)D)16
E)E)20
Rationale: According to order of operations, 12÷12×2 (division and multiplication) is done first from left to right
12÷12×2 = 1 × 2 = 2
Hence
20 - 12÷12×2 = 20 - 12 = 8
correct answer is B) 8
Answer: B
Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
Options:
A)1/2
B)3/8
C)3/4
D)5/16
E)9/16
Rationale: In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E	= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) =	n(E)/n(S)	=	27/36	=	3/4	.
Answer : C.
Answer: C
Question: In how many seconds will a train 150 meters long pass an oak tree, if the speed of the train is 36 km/hr?
Options:
A)11
B)13
C)15
D)17
E)19
Rationale: speed = 36 * 5/18 = 10 m/s
time = 150/10 = 15 seconds
The answer is C.
Answer: C
Question: If the cost price of 50 articles is equal to the selling price of 40 articles, then the gain or loss percent is?
Options:
A)22
B)65
C)78
D)33
E)25
Rationale: Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%
Answer:E
Answer: E
Question: How many plants will be there in a circular bed whose outer edge measure 32 cms, allowing 4 cm2 for each plant ?
Options:
A)18
B)750
C)24
D)20
E)None of these
Rationale: Circumference of circular bed = 32 cm
Area of circular bed = (32)2â„4Ï€
Space for each plant = 4 cm2
âˆ´ Required number of plants
= (32)2â„4Ï€ Ã· 4 = 20.36 = 20(approx)
Answer D
Answer: D
Question: On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year?
Options:
A)150 cm
B)160 cm
C)170 cm
D)140 cm
E)150 cm
Rationale: Explanation:
Given that height on 15th birthday = 159 cm and growth = 6%
Let the previous year height = x
Then height on 15th birthday = x × (100 + 6)/100 = x × 106/100
⇒159 = x × 106/100
⇒x = 159 × 100/106 = 1.5 × 100 = 150 cm
Answer: Option A
Answer: A
Question: If in a cricket one day match, there were no wide balls, no wides, no extras and no overthrows.
What is the maximum number of runs that a batsman can score in an ideal case scenario?
Options:
A)2663
B)1653
C)3693
D)4253
E)2653
Rationale: B
1653
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653
Answer: B
Question: A sum of money lent out at S.I. amounts to Rs. 780 after 2 years and to Rs. 1020 after a further period of 5 years. The sum is?
Options:
A)s. 684
B)s. 500
C)s. 540
D)s. 740
E)s. 840
Rationale: S.I for 5 years = (1020 - 780) = Rs. 240.
S.I. for 2 years = 240/5 * 2 = Rs. 96.
Principal = (780 - 96) = Rs. 684
ANSWER:A
Answer: A
Question: Three numbers are in the ratio 4 : 5 : 6 and their product is 76680. the largest number is
Options:
A)30
B)24
C)26
D)28
E)22
Rationale: Explanation:
Let the number be 4x, 5x and 6x. Then,
(4x)(5x)(6x) = 2880
=> 120x3 = 7680
=> x3 = 64 => x = 4
So, largest number is 6x = 24
Answer: Option B
Answer: B
Question: A present value of a machine is $1100. Its value depletion rate is 10% per annum then find the machine value after 2 years?
Options:
A)$900
B)$891
C)$915
D)$715
E)$795
Rationale: P= $1100
R = 10%
T = 2 years
Machine value after 2 years = P[(1-R/100)^T]
=1100*9/10 * 9/10
=$891
Answer is B
Answer: B
Question: Three beakers namely A, B and C each contain 100 ml of spirit in water solution. The ratio of spirit to water in the beakers A, B and C is 1 : 3, 1 : 4 and 2 : 3 respectively. 40 ml of solution is transferred from beaker A to beaker C and than 28 ml of solution is transferred from beaker C to beaker B. Find the final ratio of spirit in the beakers A, B and C.
Options:
A)3 : 6 : 8
B)6 : 15 : 20
C)15 : 28 : 42
D)25 : 28 : 42
E)None of these
Rationale: Initial quantity of spirit and water in the beakers.
Beaker A: spirit = 25 ml and Water=75 ml
Beaker B: spirit = 20 ml and Water = 80 ml
Beaker C: spirit = 40 ml and Water= 60 ml
After 40 ml is transferred from beaker A to beaker C, the quantity of spirit and water in the beakers is as follows:
Beaker A : spirit = 25 - 10=15 ml and
Water=75 - 30 = 45 ml
Beaker B : spirit = 20 ml and Water = 80 ml
Beaker C: spirit=40 + 10 = 50ml and
Water=60 + 30 = 90 ml
Now, spirit : Water in Beaker C = 5:9
After 28 ml is transferred form beaker C to beaker B:
Beaker A: spirit= 15 and Water = 40mI
Beaker B: spirit = 20 + l0 = 30ml and
Water=80 + 18 = 98 ml
Beaker C: spirit= 50 - 10 = 40ml and
Water=90 - 18 = 72 ml
Required Ratio = 15 : 30 : 40 = 3 : 6 : 8
ANSWER:A
Answer: A
Question: The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.2 the sixth number is?
Options:
A)10.3
B)11.3
C)11.8
D)11.5
E)11.1
Rationale: 1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.2 = 67.2
63 + 67.2 = 130.2 – 119.9 = 10.3
6th number = 10.3
Answer: A
Answer: A
Question: Jim drove 215 miles of a 1200 miles journey. How many more miles does he need to drive to finish his journey?
Options:
A)113 miles
B)432 miles
C)456miles
D)887 miles
E)985 miles
Rationale: The number of miles to drive to finish his journey is given by
1200 - 215 = 985 miles
correct answer E
Answer: E
Question: M = {-6, -5, -4, -3, -2}
T = {-2, -1, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
Options:
A)0
B)1/3
C)2/5
D)1/2
E)3/5
Rationale: Answer E.
Total # of outcomes: 5*5= 25
# of outcomes where Product is -ve : (-6,1) ,(-6,2), (-6,3)... Hence, total: 15
Probability: 15/25 = 3/5
Answer: E
Question: When positive integer x is divided by positive integer y, the result is 59.32. What is the sum T of all possible 2-digit remainders for x/y?
Options:
A)560
B)616
C)672
D)728
E)784
Rationale: ans B 616...
remainders = .32=32/100=8/25=16/50 and so on..
so two digit remainders are 16+24+32+....+96..
T=8(2+3+4....+12)=616.B
Answer: B
Question: A tank has both an inlet pipe and an outlet pipe. Working alone, the inlet pipe can fill up the tank in 4 hours. Working alone, the outlet pipe can empty out the tank in 10 hours. If it is desired that the tank should be filled, from empty, exactly 6 hours after the inlet pipe is turned on at 9:30am, then at what time should the outlet pipe be opened ?
Options:
A)8:30am
B)9:30am
C)10:00am
D)10:30am
E)11:30am
Rationale: In 6 hours the inlet pipe will pump in water equal to 6*1/4 = 6/4 of the pool. So, 2/4 more than necessary, which should be compensated by the outlet pipe. To pump out 2/4 of the water the outlet pipe needs 20/4 = 5 hours.
Therefore, the outlet pipe should be opened at 9:30am + 6 hours - 5 hours = 10:30am.
Answer: D
Answer: D
Question: In a mixture of 60 liters, the ratio of milk and water is 2:1. What amount of water must be added to make the ratio of milk and water as 1:2?
Options:
A)60
B)88
C)77
D)66
E)44
Rationale: 2:1 --- 60
1:2
2:4
---------------
3
3 ----------- 60
3 ----------- ? => 60 liters
Answer:A
Answer: A
Question: If st > 0, which of the following must be negative?
Options:
A)-s/t
B)st
C)-s*-t
D)-s/-t
E)s+t
Rationale: Given st> 0 So both will be the positive integers. If both are positive, then B,C,D,E also positive.
Answer :A
Answer: A
Question: If the cost price is 92% of sp then what is the profit %
Options:
A)8.70%
B)8.60%
C)8.10%
D)8.30%
E)8.27%
Rationale: Sol. sp=Rs100 : then cp=Rs 92:profit =Rs 8.
Profit={(8/92)*100}%=8.70%
Answer is A.
Answer: A
Question: What is the are of an equilateral triangle of side 10 cm?
Options:
A)66âˆš3 cm2
B)25âˆš3 cm2
C)64âˆš3 cm2
D)64âˆš5 cm2
E)14âˆš3 cm2
Rationale: Area of an equilateral triangle = âˆš3/4 S2
If S = 10, Area of triangle = âˆš3/4 * 10 * 10 = 25âˆš3 cm2;
Answer:B
Answer: B
Question: The product of HCF and LCM of 18 and 15 is
Options:
A)120
B)150
C)175
D)270
E)190
Rationale: HCF of 18 and 15
18 = 2 x 3 x 3
15 = 3 x 15 = 3
LCM of 18 and 15 = 2 x 3 x 3 x 5 = 90
∴ Product of HCF and LCM of both numbers = 3 x 90 = 270
ANSWER :D
Answer: D
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are drawn what is the probability that one is yellow and two are red?
Options:
A)12/455
B)12/458
C)12/487
D)12/456
E)12/4512
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When three marbles are drawn, the probability that one is yellow and two are red
= (2 * 4 * 3 * 3 * 2)/(1 * 2 * 15 * 14 * 13)
= 12/455
Answer:A
Answer: A
Question: When a student Joe, weighing 45 kg, joins a group of students whose average weight is 30 kg, the average weight goes up by 1 kg. Subsequently, if two students, excluding Joe, leave the group the average weight comes back to 30 kg. What is the difference between the average weight of the two students who left and the weight of Joe?
Options:
A)7.5 kg
B)11 kg
C)30 kg
D)36.5 kg
E)71 kg
Rationale: After two persons leave the group the average remains the same. That means the weight of the two persons = 45+30 = 75
So, the average the two persons = 37.5
That gives the answer 45-37.5 = 7.5
ANswer A
Answer: A
Question: A number when divided by a certain divisor left remainder 251, when twice the number was divided by the same divisor, the remainder was 112. Find the divisor?
Options:
A)370
B)365
C)390
D)456
E)460
Rationale: Easy solution:
N= Dq1+251
2N = 2Dq1 + 502 - (1)
2N = Dq2 + 112 - (2)
As (1) = (2) = 2N
D*(q2-2q1) = 390
D * Some integer = 390
Checking all options only (C) syncs with it.
Answer C
Answer: C
Question: A man sold 20 articles for $60 and gained 20%. How many articles should he sell for $60 to incur a loss 20%?
Options:
A)30
B)36
C)40
D)50
E)48
Rationale: Production cost per article: $60*(100%-20%) / 20 = $2.40
Required production costs for a loss of 20%: $60*(100% + 20%) = $72
Number of articles to be sold for $72 to incur a 20% loss: $72 / $2.40 = 30
Thus, solution A is correct.
Answer: A
Question: How many shots of 1cm radius can be prepared from a sphere of 4cm radius?
Options:
A)33
B)88
C)27
D)64
E)99
Rationale: 4/3 Ï€ * 4 * 4 * 4 = 4/3 Ï€ * 1 * 1 * 1 * x
x = 64
Answer: D
Answer: D
Question: A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:
Options:
A)0%
B)1%
C)2%
D)3%
E)4%
Rationale: C.P. = Rs. 3000.
S.P. = Rs.	(3600 x 100)/(100 + (10 x 2))= Rs. 3000.
Gain = 0%.
answer :A
Answer: A
Question: Four usual dice are thrown on the ground. The total of numbers on the top faces of these four dice is 13 as the top faces showed 4, 3, 1 and 5 respectively. What is the total of the faces touching the ground?
Options:
A)22
B)37
C)15
D)99
E)26
Rationale: Explanation:
In a usual dice, the sum of the numbers on any two opposite faces is always 7. Thus, 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4.
Consequently, when 4, 3, 1 and 5 are the numbers on the top faces, then 3, 4, 6 and 2 respectively are the numbers on the face touching the ground. The total of these numbers = 3 + 4 + 6 + 2 = 15.
Answer: C) 15
Answer: C