divelab/Sys2Bench huggingface dataset:
Question: In a certain mathematical activity, we start with seven cards, each with a different prime number written on it. These seven cards are randomly put into three boxes in the following way: one box must get four cards, one must get two, and one gets only one. Then, for each box, we find the product of all the cards in the box, and that’s the “number” of the box. Then, we put those three numbers in order, from lowest to highest, and that is our set. How many different sets E can be created from this process?
Options:
A) 35
B) 105
C) 210
D) 420
E) 630
Rationale: E=7C4 * 3C2 = 105 Answer B.
7C4 ways to choose 4 numbers for Box 1
3C2 ways to choose 2 numbers for Box2
1 way for Box 3 .
every combination will give a different product and can be arranged least to highest only in one way .B
Answer: B
Question: If x^3 > y^2 > z, which of the statements could be true?
I. x > y > z
II. x < z < y
III. y < x < z
Options:
A)I only
B)III only
C)I and II only
D)II and III only
E)I, II and III
Rationale: Given : x^3 > y^2 > z
I. x > y > z is true for x=5, y=2, z=1 i.e. Answers can only be Options A, C or E
II. x < z < y is true for x=3, y=5, z=4 i.e. Answers can only be Options C or E
III. y < x < z is true for x=4, y=3, z=5 i.e. Answers can only be Options E
Answer: option E
Answer: E
Question: A Boeing 757 flies a direct route from Dallas, TX, to Phoenix, AZ and then returns to Dallas, TX. The flight is 900 miles one way. It took 2 hrs and 15 minutes for the flight to Phoenix and 2 hrs for the flight back to Dallas. What was the speed of the wind in mph?
Options:
A)20 mph
B)25 mph
C)30 mph
D)35 mph
E)40 mph
Rationale: Explanation:
Change the times in the problem to minutes first.
Speed of airplane from Dallas to Phoenix: (900/135) X 60 = 400 mph
Speed of airplane from Phoenix to Dallas: (900/120) X 60 = 450 mph
Wind speed: 1/2 (450 - 400) = 50/2 = 25 mph
Answer: Option B
Answer: B
Question: 7% of the total quantity of rice is lost in grinding when a country has to import 6 million tonnes, but when only 7 3/4% is lost, it can import 3 million tonnes. Find the quantity of rice grown in the country.
Options:
A)250
B)300
C)400
D)500
E)600
Rationale: Let x be the total grown quantity of wheat. According to the question
(7%of x)+ 64 31 %of x)+3 7x 31x => 100+ 6 = 400 + 3 1200 x = — 400 million tonnes rice grown
C
Answer: C
Question: Line k is in the rectangular coordinate system. If the v-intercept of k is -2, and the y-intercept is 3, which of the following is an equation of line k?
Options:
A)-3v + 2y = 6
B)3v + 2y = -6
C)3v - 2y = 6
D)2v - 3y = 6
E)-2v - 3y = 6
Rationale: This question can be solved in much simpler way.
To find the v intercept
Put y=0 in the equation of the line.
To find the y-intercept
Put v=0 in the equation of the line.
So
Sub v=0 in the answer choices check whether you are getting y=3.
A. -3v + 2y = 6 --- y = 3
B. 3v + 2y = -6 --- y = -3
C. 3v - 2y = 6 ----- y = -3
D. 2v - 3y = 6 --- y = -2
E. -2v - 3y = 6 --- y =-2
Eliminate answer B,C,D and E.
So the answer is A.
Answer: A
Question: Sum of two numbers prime to each other is 8 and their L.C.M. is 15. What are the numbers?
Options:
A)3 and 5
B)3 and 4
C)4 and 5
D)5 and 10
E)6 and 8
Rationale: As two numbers are prime, only options satisfy ie option A and B and C
but option D will not make the product of numbers i.e 15
ANSWER:A
Answer: A
Question: Meg and Bob are among the 3 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?
Options:
A)24
B)3
C)60
D)90
E)120
Rationale: Total # of ways the race can be finished is 3!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 3!/2=3.
Answer: B.
Answer: B
Question: If 9 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?
Options:
A)9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
B)9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
C)9 × 10
D)36
E)45
Rationale: 1st person will have 9 handshakes, 2nd will have 8, ....8th will have 1 and 9th will have 0
=8+7+6+5+4+3+2+1 = 36
if we generalize this: if there are n people then there will be n(n-1)/2 handshakes.
Ans.D)
Answer: D
Question: Pablo plays 3 rounds of a game, in which his chances of winning each round are 1/4, 1/8, and 1/n, respectively. If n ≠ 0, what is the probability that Pablo wins the first two rounds, but loses the third?
Options:
A)1/16n
B)(n-1)/32n
C)1/8n
D)(n+2)/2n
E)(3n-2)/2n
Rationale: 1/4 * 1/8 * (1-1/n)
= 1/32 * (n-1)/n
= (n-1)/32n
Answer - B
Answer: B
Question: On dividing 199 by a number, the quotient is 11 and the remainder is 1. Find the divisor.
Options:
A)A)12
B)B)14
C)C)16
D)D)18
E)E)22
Rationale: d = (D-R)/Q
= (199 - 1)/11
= 198/11 = 18
D)
Answer: D
Question: Median R of price of the car is 52,000. which could be the median of the other two.
35000, 44000, x, y, 57000.
Options:
A)38000
B)47000
C)48000
D)51,000
E)56,000
Rationale: Median of the car is 52,000 therefore 52,000 is the middle number. Assuming Y >= X, X will have to be the median value i.e. 52,000. Therefore the remaining value would have to be between 52,000 and 57,000 inclusive. The only answer choice that fits is R=56,000.E
Answer: E
Question: Find the L.C.M. of 72, 108 and 2100.
Options:
A)37800
B)37600
C)37200
D)37500
E)None of them
Rationale: 72 = 2^3 x 3^2, 108 = 3^3 x 2^2, 2100 = 2^2 x 5^2 x 3 x 7.
L.C.M. = 2^3 x 3^3 x 5^2 x 7 = 37800.
Answer is A.
Answer: A
Question: A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1100 more than D, what is B's share?
Options:
A)Rs. 500
B)Rs. 1500
C)Rs. 2200
D)Rs. 2500
E)none of the above
Rationale: Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x - 3x = 1100
x = 1100.
B's share = Rs. 2x = Rs. (2 x 1100) = Rs. 2200.
Answer = C
Answer: C
Question: Tarun bought a T.V with 20% discount on the labelled price. Had he bought it with 25% discount? He would have saved Rs. 500. At what price did he buy the T.V?
A. Rs. 5000 B. Rs. 10000
Options:
A)Rs.5000
B)Rs.6000
C)Rs.7500
D)Rs.80000
E)Rs. 10000
Rationale: Explanation:
Labelled price be Rs. 100
S.P in 1st case = Rs. 80
S.P in 2nd case = Rs. 75
Saving is Rs. 5 labelled price = Rs. 100
Saving is Rs. 500 labelled price
= Rs. (100/5 × 500) = Rs. 10000
ANSWER:E
Answer: E
Question: It is known that no more than 7 children will be attending a party. What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies?
Options:
A)220
B)250
C)320
D)420
E)560
Rationale: No more than 7 children attending the party means the number of children could be 1,2,3,4,5,6,7.
We need to find a number that is divisible by each of these. Out of the choices given only 420 is divisible by each of these numbers.
Hence , answer is D.
Answer: D
Question: We had $3500 left after spending 30% of the money that we took for shopping. How much money did we start with?
Options:
A)$4600
B)$4800
C)$5000
D)$5200
E)$5400
Rationale: Let x be the amount of money we started with.
0.7x = 3500
x = 5000
The answer is C.
Answer: C
Question: If y > 0, (2y)/20 + (3y)/10 is what percent of y?
Options:
A)40%
B)50%
C)60%
D)70%
E)80%
Rationale: soln:-
can be reduced to y/10+3y/10 =2y/5=40%
ANSWER:A
Answer: A
Question: If 148/x is an integer and 108/x is an integer, which of the following must be true?
I. 9/x is an integer
II. 12/x is an integer
III. 36/x is an integer
Options:
A)I only
B)III only
C)I and II only
D)II and III only
E)I, II and III
Rationale: I think, here we should use the greatest common factor.
The greatest common factor for 148 and 108 is 36.
So 36/x is an integer.
For the rest of the answers, if x is 36 then 9/x and 12/x leads to fractions.
Ans is C.
Answer: C
Question: Find the no.of ways of arranging the boy and 8 guests at a circular table so that the boy always sits in a particular seat?
Options:
A)7!
B)8!
C)9!
D)10!
E)11!
Rationale: Ans.(B)
Sol. Total number of persons = 9 Host can sit in a particular seat in one way. Now, remaining positions are defined relative to the host. Hence, the remaining can sit in 8 places in 8P8 = 8! Ways ... The number of required arrangements = 8! x 1= 8! = 8! ways
Answer: B
Question: A person covered one-fourth of the total distance at 26 kmph and remaining distance at 24 kmph. What is the average speed for the total distance?
Options:
A)91 (1/3) kmph
B)21 (1/3) kilometre per hour
C)81 (1/3) kmph
D)21 (1/7) kmph
E)23 (1/3) kmph
Rationale: Let the total distance be x km
total time taken
= (x/4)/16 + (3x/4)/ 24
= x/64 + x/32 = 3x/64
Average speed = x/(3x/64)
= 64/3kmph = 21 (1/3)kmph.
Answer:B
Answer: B
Question: Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
Options:
A)10 min. 20 sec.
B)11 min. 45 sec.
C)12 min. 30 sec.
D)14 min. 40 sec.
E)15 min. 40 sec.
Rationale: Take LCM of 15 and 20 and get the total capacity of tank as 60 .
Then A will fill 4li in 1 min and B will fill 3 li in 1 min so total will be 7 . A left after 4 min so they worked together for 4 min so 7*4= 28 . now subtract it from 60.(60-38=32). This will be filled by b so time taken would be..10 min 20 sec.
ANSWER:A
Answer: A
Question: The probability that a man will be alive for 10more yrs is 1/4 & the probability that his wife will alive for 10 more yrs is 1/3. The probability that none of them will be alive for 10 more yrs, is
Options:
A)1/2
B)2
C)2/3
D)4/5
E)5/7
Rationale: Sol. Required probability = Pg.) x P(B) = (1— D x(1 —I) =: x 1=1/2
Ans.(A)
Answer: A
Question: The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?
Options:
A)15 kmph
B)17 kmph
C)18 kmph
D)19 kmph
E)12 kmph
Rationale: 90 * 10/60
= 15 kmph
Answer: A
Answer: A
Question: The simple interest accrued on an amount Rs.10,000 at the end of two years is same as the compound interest on Rs.8,000 at the end of two years. The rate of interest is same in both the cases. What is the rate of interest?
Options:
A)57%
B)60%
C)58%
D)50%
E)20%
Rationale: Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I - S.I
=> 8,000 { [1 + r/100]2 - 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r2 / (100)2 - 1} = r/5
=> 16r/100 + 8r2/(100)2 = 20r/100
=> 4r/10 = 8r2/(100)2
=> 8[r/100]2 - 4r/100 = 0
=> r/100 {8r/100 -4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50%
Answer:D
Answer: D
Question: The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 16 years more than twice the age of the son. Find the present age of the father.
Options:
A)33
B)38
C)29
D)51
E)48
Rationale: if the present age be x years.
father's will be (3x+3)years..
so, (3x+3+3)=2(x+3)+16
or, x=16
so the fathers present age =(3x+3)=(3*16+3)years =51years..
ANSWER : OPTION D
Answer: D
Question: For every order, a certain mail-order company charges a shipping fee of $3 plus an additional $2 if the value of the order is over $50 but not over $100 or an additional $3 if the value of the order is over $100. How much greater are the total shipping fees for 2 orders of $75 each than the total shipping fee for 2 order of $150 ?
Options:
A) $1
B) $2
C) $3
D) $4
E) $5
Rationale: Total cost involved for each 75 $ shipment: 3+2 = 5 $. Thus, for two such orders = 10$
Total cost involved for a 150$ shipment: 3+3 = 6$. Thus, for two such orders = 12$
The difference = 2$.
B.
Answer: B
Question: The average height of 60 pupils in a class is 150 cm. Five of them whose height is 146 cm, leave the class and five others whose average height is 156 cm, join. The new average height of the pupils of the class (in cm) is ?
Options:
A)287 cm
B)150.8 cm
C)276 cm
D)287 cm
E)297 cm
Rationale: Total height = 150 * 60 = 9000 cm.
New average = [9000 - 5 * 146 + 5 * 156 ] / 60
= 150.8 cm.
Answer:B
Answer: B
Question: The sale price shirts listed for Rs.400 after successive discount is 10% and 5% is?
Options:
A)280
B)290
C)Rs.342
D)250
E)253
Rationale: 400*(90/100)* (95/100) = 342
C
Answer: C
Question: If the price of gasoline increases by 25% and a driver intends to spend only 10% more on gasoline, by how much percent should the driver reduce the quantity of gasoline that he buys?
Options:
A)5%
B)8%
C)10%
D)12%
E)15%
Rationale: Let x be the amount of gasoline the driver buys originally.
Let y be the new amount of gasoline the driver should buy.
Let P be the original price per liter.
(1.25*P)y = 1.1(P*x)
y = (1.1/1.25)*x = 0.88x which is a reduction of 12%.
The answer is D.
Answer: D
Question: A bus travel 5/7 of its speed covers 42 km in 1 hr 40 min48 sec. What is theactual speed of the bus ?
Options:
A)20 km/hr
B)35 km/hr
C)40 km/hr
D)53 km/hr
E)55 km/hr
Rationale: Time = 1 hr 40 min 48 sec = 1hr +4060hr+483600hr=1+23+175=12675hrdistance = 42 kmspeed=distancetime=42(12675)=42×75126⇒57 of the actual speed = 42×75126⇒actual speed = 42×75126×75=42×1518=7×153=7×5=35 km/hr
B
Answer: B
Question: A tank is 25% full of oil. The oil consists of 25% WTI Crude Oil and 75% Brent Crude Oil. 2/3 rd of the remainder of the tank is filled with 400 barrels of WTI Crude Oil. How much WTI Crude Oil is in the tank?
Options:
A)425 barrels
B)437.5 barrels
C)450 barrels
D)500 barrels
E)650 barrels
Rationale: The tank was 25% full. Then, 2/3 rd of the remainder of the tank, so 2/3 rd of 75% of the tank, was filled with 400 barrels of WTI Crude Oil. 2/3 rd of 75% is 50%, thus we are told that 50% of the tank amounts for 400 barrels, which makes the capacity of the tank 800 barrels.
Therefore, initially, the tank had 0.25∗0.25∗800=500 barrels of WTI Crude Oil. After 400 barrels of WTI Crude Oil were added the amount of WTI Crude Oil became 50+400=450 barrels.
Answer: C.
Answer: C
Question: What is the maximum value of x3y3 + 3 x*y when x+y = 8?
Options:
A)4144
B)4122
C)455
D)4189
E)4156
Rationale: The question probably be x3.y3+3x∗yx3.y3+3x∗y
Sustitute x = 4 and y = 4
Option A
Answer: A
Question: A sum of money at simple interest amounts to Rs. 845 in 3 years and to Rs. 854 in 4 years. The sum is:
Options:
A)s. 650
B)s. 690
C)s. 698
D)s. 700
E)s. 728
Rationale: S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (845 - 117) = Rs. 728.
Answer: Option E
Answer: E
Question: In the set of positive integers from 1 to 70, what is the sum of all the odd multiples of 5?
Options:
A)180
B)245
C)240
D)200
E)190
Rationale: reduce 1 - 70
5 - 15 - 25 - 35 - 45 +55+ 65 are valid multiples 0f 5 . Add them --> 245
B
Answer: B
Question: If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be?
Options:
A)4 days
B)8 days
C)6 days
D)5 days
E)3 days
Rationale: Let 1 men's 1 day work = x and 1 boy's 1 day work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2
Solving these two equations, we get:
x = 1/100 and y = 1/200
(15 men + 20 boys)'s 1 day work = (15/100 + 20/200) = 1/4
15 men and 20 boys can do the work in 4 days.
Answer: A
Answer: A
Question: David has d books, which is 3 times as many as Jeff and 2/3 as many as Paula. How many books do the three of them have altogether, in terms of d?
Options:
A) 5/6*d
B) 7/3*d
C) 17/2*d
D) 7/2*d
E) 9/2*d
Rationale: Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.
number of books David has = d
number of books Jeff has = j
number of books Paula has = p
We are given that David has 3 times as many books as Jeff. We can now express this in an equation.
d = 3j
d/3 = j
We are also given that David has ½ as many books as Paula. We can also express this in an equation.
d = (2/3)p
3d/2 = p
Notice that we immediately solved forj in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:
d + d/3 + 3d/2
=17d/2
The answer is C.
Answer: C
Question: A can do a piece of work in 10 days. He works at it for 4 days and then B finishes it in 9 days. In how many days can A and B together finish the work?
Options:
A)6
B)8
C)12
D)15
E)20
Rationale: 4/10+9/X = 1/x = 15
1/10+1/15 = 1/6 = 6 days
ANSWER A
Answer: A
Question: A bat is bought for Rs.400 and sold at a gain of 20% find its selling price
Options:
A)Rs.480/-
B)Rs.500/-
C)Rs.520/-
D)Rs.540/-
E)Rs.560/-
Rationale: 100 % ------> 400 (100 * 4 = 400)
120 % ------> 480 (120 * 4 = 480)
Selling price = Rs.480/-
A
Answer: A
Question: If the number is divided by 3, it reduced by 34. The number is
Options:
A)51
B)50
C)45
D)40
E)36
Rationale: Explanation:
Let the number be x. Then,
x - (x/3) = 34 => 2x/3 = 34
=> x = 51
Answer: Option A
Answer: A
Question: As a treat for her two crying children, a mother runs to the freezer in which she has three cherry ice pops, three orange ice pops, and four lemon-lime ice pops. If she chooses two at random to bring outside to the children, but realizes as she runs out the door that she cannot bring them different flavors without one invariably being jealous of the other and getting even more upset, what is the probability that she has to return to the freezer to make sure that they each receive the same flavor?
Options:
A)1/9
B)1/6
C)5/18
D)11/120
E)5/6
Rationale: Probability of not getting the same flavor ->
Favorable -> Cherry - orange [ 3C1 * 3C1 [or simply 3*3 ] or Cherry - lemon [ 3*4] or orange - lemon [3*4]
Prob = (3*3 + 3*4 + 3*4 ) / 9C2 = 33/360 = 11/120
Answer -> D
Answer: D
Question: If 2^z is a factor of 30!, which of the following could be the value of z?
I. 23
II. 26
III. 24
Options:
A)I ONLY
B)II ONLY
C)BOTH I AND II
D)III ONLY
E)NONE OF THE ABOVE
Rationale: 2^z is a factor of 30!, what values can z take.
Powers of 2 in 30! = [30/2] + [30/4] +[30/8] + [30/16] , where [x] is the integral part of x
Powers of 2 in 30! = 15 + 7 + 3 + 1 = 26
Hence the maximum value of 2^z in 30! = 2^26
Only Statement II satisfies
Correct Option: B
Answer: B
Question: If 4 (P's Capital) = 6 (Q's Capital) = 10 (R's Capital), then out of the total profit of Rs 4650, how much R will receive?
Options:
A)300
B)600
C)900
D)800
E)500
Rationale: Let P's capital be p,
Q's capital be q,
and R's capital be r
Then
4p=6q=10r2p=3q=5r⋯(A)
From (A),
q=2p/3 ⋯(1)
r=2p/5 ⋯(2)
p:q:r=p:2p/3:2p/5
=15:10:6
R's share =4650×6/31=150×6=900
Answer is C.
Answer: C
Question: How many four letter words are formed using the letters of the word ENTERTAINMENT?
Options:
A)1716
B)1616
C)1816
D)1516
E)1761
Rationale: The number of letters in the given word is thirteen.
The number of four letter words that can be formed using these thirteen letters is 13P4
= 13 * 12 * 11 = 1716.
Answer: A
Answer: A
Question: What will be the difference between simple and compound interest at 10% per annum on a sum of Rs. 1000 after 4 years?
Options:
A)Rs.64.18
B)Rs.64.67
C)Rs.64.08
D)Rs.64.10
E)Rs.64.22
Rationale: S.I. = (1000 * 10 * 4)/100 = Rs. 400
C.I. = [1000 * (1 + 10/100)4 - 1000]
= Rs. 464.10 Difference
= (464.10 - 400)
= Rs. 64.10
Answer: D
Answer: D
Question: In a certain animal shelter, the ratio of the number of catsto the number of dogs is 15 to 7. If 16 additional dogs were to be taken in by the shelter, the ratio of the number of cats to the number of dogs would be 15 to 11. How many cats are in the shelter?
Options:
A)15
B)25
C)30
D)45
E)60
Rationale: This ratio question can be solved in a couple of different ways. Here's an Algebraic approach...
We're told that the ratio of the number of cats to the number of dogs is 15:7. We're then told that 16 more dogs are added to this group and the ratio becomes 15:11. We're asked for the number of cats.
Algebraically, since the number of cats is a multiple of 15 and the number of dogs is a multiple of 7, we can write this initial relationship as...
15X/7X
When we add the 16 cats and factor in the 'ending ratio', we have an equation....
15X/(7X+16) = 15/11
Here we have 1 variable and 1 equation, so we can solve for X....
(15X)(11) = (7X+16)(15)
(X)(11) = (7X+16)(1)
11X = 7X + 16
4X = 16
X = 4
With this X, we can figure out the initial number of dogs and cats...
Initial dogs = 15X = 15(4) = 60
Final Answer: D
Answer: D
Question: In the new budget the price of ghee rose by 50%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase?
Options:
A)7.5%
B)9.1%
C)10.9%
D)12.6%
E)33.33%
Rationale: Reduce in consumption = R/(100+R) *100% = 50/150 *100 =33.33%
Answer is E
Answer: E
Question: A train 400 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Options:
A)298 m
B)279 m
C)500 m
D)289 m
E)900 m
Rationale: Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (400 + x)/60 = 65/3
x = 900 m.
Answer: E
Answer: E
Question: The compound interest earned on a sum for the second and the third years are $1400 and $1498 respectively. What is the rate of interest?
Options:
A)3%
B)5%
C)7%
D)9%
E)11%
Rationale: 1498 - 1400 = 98 is the rate of interest on $1400 for one year.
The rate of interest = (100 * 98) / (1400) = 7%
The answer is C.
Answer: C
Question: Ashok secured average of 78 marks in 6 subjects. If the average of marks in 5 subjects is 74, how many marks did he secure in the 6th subject?
Options:
A)66
B)74
C)78
D)98
E)None of these
Rationale: Explanation:
Number of subjects = 6
Average of marks in 6 subjects = 78
Therefore total marks in 6 subjects = 78 * 6 = 468
Now, No. of subjects = 5
Total marks in 5 subjects = 74 * 5 = 370
Therefore marks in 6th subject = 468 – 370 = 98
ANSWER D
Answer: D
Question: Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. How many times will they ring together in 60 minutes ?
Options:
A)5
B)11
C)21
D)31
E)32
Rationale: LCM of 4, 8, 10, 12, 15 and 20 = 120 seconds = 2 minutes
That means All the six bells will ring together in every 2 minutes
Number of times they will ring together in 60 minutes =1+60/2=31
Answer :D
Answer: D
Question: y = 248 - 390x
Which of the following values of x gives the greatest value of y in the equation above?
Options:
A) 200
B) 100
C) 0.5
D) -2
E) -1
Rationale: Clearly, the negative value of x will result in maximum value of y.
Answer is D
Answer: D
Question: 21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement.
Options:
A)1/42
B)3/41
C)9/42
D)5/44
E)5/47
Rationale: The probability that first toy shows the even number,
=1021=1021
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number,
=920=920
Required probability,
=(1021)×(920)=(1021)×(920)
=9/42
C
Answer: C
Question: A man whose speed is 4.5 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
B)4
C)8
D)2
E)3
Rationale: M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9
= 4
Answer: B
Answer: B
Question: A, Band C are three contestants in a kmrace. If A can give B a start of 40 m and A can give C a start of 64m how many metre's start can B give C ?.
Options:
A)540 m
B)178 m
C)936 m
D)234 m
E)713 m
Rationale: While A covers 1000 m, B covers (1000 - 40) m = 960 m and
C covers (1000 - 64) m or 936 m.
When B covers 960 m, C covers 936 m
Option C
Answer: C
Question: A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is
Options:
A)100
B)120
C)130
D)150
E)140
Rationale: Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.
Answer: Option D
Answer: D