divelab/Sys2Bench huggingface dataset:
Question: Set R includes elements {8, 2, 11, x, 3, y} and has a mean of 7 and a median of 5.5. If x < y, then which of the following is the maximum possible value of x?
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: First, reach to the point of getting the equation R : x + y = 18
It should take about 20 seconds.
Then Substitute the answer choices into the equation.
I don't know what the answer choices in this case are.
But I'm sure, you would be able to eliminate at least 2 or 3 answer choices. (about 10 seconds).
Say you are left with 2 answer choices. (If you are short on time, guess One of the two and you'll have a 50% probability of getting it right.)
The Median (of 6 numbers) = 5.5. See if the AVERAGE of any two numbers among (2,3,8,11) results in the median. In this case, it does for 3 and 8. (15 seconds).
Once you know that the numbers that contribute towards Median are 3 and 8, and not x or y, then given x < y, x≤3. (about 10 seconds)=D
Answer: D
Question: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both BG, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations W are there for the children?
Options:
A)240
B)480
C)720
D)1440
E)3600
Rationale: MAGOOSHOFFICIAL SOLUTION:
First, we will consider the restricted elements — children ABG have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?
X X X _ _ _ _
_ X X X _ _ _
_ _ X X X _ _
_ _ _ X X X _
_ _ _ _ X X X
There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the non-restricted elements, the other four. Once ABG are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of ABG, there are 24 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations W of 24*10 = 240.
Answer = A
Answer: A
Question: Two friends are making lemonade to sell at their lemonade stand. First, they make 14 litres of lemonade containing 21% lemon juice, but that doesn't taste quite right to them. They decide to add some store-bought lemonade that is 31% lemon juice until they have a mixture that is 25% lemon juice. How much of the store-bought lemonade should they add?
Options:
A)6.6 litres
B)12 litres
C)9 litres
D)2.6 litres
E)9.3 litres
Rationale: 0.21(14)+0.31x = 0.25(14 + x)
Solve the equation.
0.21(14) + 0.31x = 0.25(14 + x)
2.94 + 0.31x = 3.5 + 0.25x
2.94 + 0.06x = 3.5
0.06x = 0.56
x ≈ 9.3
Answer is E.
Answer: E
Question: Solve for x and check: - 3x = 12
Options:
A)-4
B)-2
C)1
D)3
E)None of these
Rationale: Solution : Dividing each side by -3, we obtain
(-3x/-3)=(12/-3)
Therefore : x=-4
Check :
-3x=12
(-3*-4)=12
12=12
ANSWER : A
Answer: A
Question: Little John had $20.10. He spent $1.05 on sweets and gave to his two friends $1.00 each. How much money was left?
Options:
A)$17.05
B)$17.55
C)$17.15
D)$17.35
E)$17.25
Rationale: John spent and gave to his two friends a total of
1.05 + 1.00 + 1.00 = $3.05
Money left
20.10 - 3.05 = $17.05
Answer :A
Answer: A
Question: Five men and nine women can do a piece of work in 10 days. Six men and twelve women can do the same work in 8 days. In how many days can three men and three women do the work?
Options:
A)33
B)77
C)20
D)88
E)12
Rationale: (5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days.
Answer:C
Answer: C
Question: By selling an article at Rs.800, a shopkeeper makes a profit of 25%. At what price should he sell the article so as to make a loss of 25%?
Options:
A)237
B)480
C)268
D)269
E)171
Rationale: SP = 800
Profit = 25%
CP = (SP)*[100/(100+P)]
= 800 * [100/125]
= 640
Loss = 25% = 25% of 640 = Rs.160
SP = CP - Loss = 640 - 160 = Rs.480
Answer:B
Answer: B
Question: If x/y =2/4, then (x-y)/x?
Options:
A)-1/2
B)-1
C)1/3
D)1
E)5
Rationale: we have this law of fraction:
a/b=c/d= (a-c)/(b-d)
x/y=2/4 ---> x/2=y/4= (x-y)/(2-4)=(x-y)/(-1
2) ---> (x-y)/x= -2/2 = -1
Answer: B
Answer: B
Question: If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 343, what is the common ratio of the G.P?
Options:
A)3
B)1/3
C)2
D)7
E)1/7
Rationale: 343=(a1+a2+a3+a4+a5+a6)/(a1+a2+a3)
Factorize the same terms
343=1+(a4+a5+a6)/(a1+a2+a3)
Write every term with respect to r
a1=a1
a2=a1*r^1
a3=a1*r^2
.........
343=1+(a1(r^3+r^4+r^5))/(a1(1+r^1+r^2))
342=(r^3 (1+r^1+r^2))/((1+r^1+r^2))
342=r^3
r=7
D
Answer: D
Question: Find the value of z when the sum of 10z,(15z+20),(20z-4)and (30z+10) is 401?
Options:
A)10
B)15
C)7
D)5
E)12
Rationale: sum = 10z+(15z+20)+(20z-4)+ (30z+10) = 75z+26
75z+26 = 401
x = 375/75 = 5
Answer is D
Answer: D
Question: The average speed of a car is 1 4/5 times the avg speed of a bike. A tractor covers 575km in 23hrs. How much distance will the car cover in 5hrs if the speed of the bike is twice speed of the tractor?
Options:
A)450 km
B)500 km
C)360 km
D)550 km
E)600 km
Rationale: Sol. Average speed of a tractor = 25 km / h
The speed of a bike in an hour = 25 × 2 = 50 km
The speed of a car in an hour = 9/5 * 50 = 90 km
So, the distance covered by car in 5 h is
90 × 5 = 450 km
Ans.(A)
Answer: A
Question: There is a box with chocolates inside it. When we count the chocolates in pairs, one is left. When we count the chocolates in a group of three, two chocolates are left. When we count the chocolates in a group of four, three chocolates are left. When we count the chocolates in a group of five, four are left. When we count the chocolates in a group of six, five are left.
But when the chocolates are counted in a group of seven, no chocolate is left. Can you calculate the minimum number of chocolates that are present in the box?
Options:
A)113
B)116
C)119
D)118
E)112
Rationale: C
119
Let the minimum number of chocolates = C
When divided by 2, remainder = 1
When divided by 3, remainder = 2
When divided by 4, remainder = 3
When divided by 5, remainder = 4
When divided by 6, remainder = 5
When divided by 7, remainder = 0
Therefore, C is divisible by 7.
Whenever C is divided by any number less than 7, the remainder is 1 less than the divisor.
This implies that, C + 1 is the LCM of 2, 3, 4, 5 and 6.
Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7
60 * 2 = 120
120 - 1 = 119 which is divisible by 7
Therefore, C + 1 = 120 or C = 119
Number of minimum number of chocolates in the box = 119
Answer: C
Question: Rahul's Mathematics test had 75 problems, 10 arithmetic, 30 algebra, 35 geometry problems.
Although he answered 70% of arithmetic, 40% of arithmetic and 60% of geometry problems correctly,
still he got less than 64% problems right. How many more questions he would have to answer more to get
passed
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: Explanation:
Number of questions attempted correctly = (70% of 10 + 40% of 30 + 60% of 35)
= 7 + 12 + 21 = 40.
Questions to be answered correctly for 64% = 64% of total quations
= 64 % of 75 = 48.
He would have to answer 48 - 40 = 8
Answer: Option D
Answer: D
Question: Two trains started at the same time, one from A to B and the other from B to A . If they arrived at B and A respectively 16 hours and 49 hours after they passed each other the ratio of the speeds of the two trains was
Options:
A)2:1
B)3:2
C)4:3
D)7:4
E)1:2
Rationale: Explanation:
Note : If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then: (A's speed) : (B's speed) = (b : a)
Therefore, Ratio of the speeds of two trains = = 7 : 4. Answer: D
Answer: D
Question: If 2x + y = 7 and x + 2y = 5, then 8xy/3 =?
Options:
A)A) 1
B)B) 8
C)C) 17/5
D)D) 18/5
E)E) 4
Rationale: 2*(x+2y = 5) equals 2x+4y=10
2x+4y=10
- 2x + y= 7
= 3y=3
Therefore Y = 1
Plug and solve...
2x + 1 = 7
2x=6
x=3
(8*3*1)/3
= 24/3
=8
B
Answer: B
Question: If 2 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is?
Options:
A)20:5:15
B)20:5:18
C)20:5:12
D)20:5:4
E)20:5:1
Rationale: 2A = 8B = 10 C
A:B:C = 1/2:1/8:1/10
= 20:5:4
Answer: D
Answer: D
Question: I purchases perfume from a store and gave him a thousand rupee note.
The perfume cost Rs.300.
Since the store person have no change, he gets the change from next shop and return me 800 rupees.
After a while, the next shopkeeper comes and told the 1st shopkeeper that the note is a fraud and takes his money back.
How much loss does the 1st shopkeeper have to bear ?
Options:
A)2004
B)2011
C)5000
D)5400
E)1000
Rationale: E
Rupees 1000
Answer: E
Question: A class of 35 students has an average height of 180 cm. seven students whose average height is 120 cm, left the class and seven others whose average height is 140 cm, joined. Calculate the new average height of the students of the class (in cm) is?
Options:
A)204.6 cm
B)404.6 cm
C)224.6 cm
D)205.6 cm
E)256.6 cm
Rationale: Total height = 180 * 35 = 6300 cm.
Total height of those who left = 120 * 7 = 840
Total height of those who joined = 140 * 7 = 980
New average
= [6300 - 120 + 980 ] / 35 = 7160/35 cm.
= 204.6 cm
Answer:A
Answer: A
Question: David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 12 family members in a row for the picture?
Options:
A)13!
B)11×10!
C)10×11!
D)12!/2!
E)12!
Rationale: The number of ways to arrange 12 people is 12!
We need to subtract the arrangement when these two people are together.
Let's think of these two people as a single unit so there are 11 units.
The number of ways to arrange 11 units is 11!
We need to multiply this by 2 since these two people could switch places.
The total number of valid arrangements is 12! - 2*11! = 11!*(12-2) = 10x11!
The answer is C.
Answer: C
Question: Pipes A and B can fill a tank in 4 and 2 hours. Pipe C can empty it in 8 hours. If all pipes are opened together, then the tank will be filled in?
Options:
A)3/2hr
B)1/2hr
C)8/5hr
D)9/7hr
E)5/3hr
Rationale: Net part filled in 1 hour = 1/4 + 1/2 - 1/8 = 5/8
The tank will be full in 8/5 hr
Answer is C
Answer: C
Question: A and B enter into partnership with capitals in the ratio 3 : 2. At the end of 10 months A withdraws,and the profits now are divided in the ratio of 1 : 3. Find how long B remained in the business?
Options:
A)45 months
B)8 months
C)6 months
D)7 months
E)None of these
Rationale: Initially A’s investment = 3x and B’s investment = 2x
Let B remain in the business for ‘n’ months.
⇒ 3x × 10 : 2x × n = 1 : 3
∴ 3x × 10 × 3 = 2x × n × 1
⇒ n = 45
Answer A
Answer: A
Question: A manufacturer of a certain type of screw rejects any screw whose length is less than 2.47 centimeters or greater than 2.5 + 0.03 centimeters. If k represents the length of a screw, in centimeters, which of the following inequalities specifies all the lengths of screws that are acceptable?
Options:
A)|k + 0.03| > 2.5
B)|k — 0.03| <= 2.5
C)|k — 2.5| > 0.03
D)|k — 2.5| >= 0.06
E)2|k — 2.5| <= 0.06
Rationale: So, let's go through this step by step:
rejects any screw whose length is less than 2.47 centimeters or greater than 2.5 + 0.03 centimeters.
In other words, any screw that is less than: 2.47 or greater than 2.50 + 0.03 =2.53will be rejected.
If k represents the length of a screw
In other words, K is an acceptable screw that must fall within the acceptable range of 2.47 to 2.53, So:
2.47 ≤ K ≤ 2.53
You can rule out answers with < or > as opposed to ≤ or ≥ because the length cannot be LESS than 2.47 or GREATER than 2.53. In other words, 2.47 and 2.53 are acceptable lengths.
Let's look at (E):
2|k — 2.5| <= 0.06 or |k — 2.5| <= 0.03
For the positive case: k - 2.5 ≤ 0.03 ===>k ≤ 2.53
For the negative case: -(k - 2.5) ≤ 0.03 ===> -k +2.5 ≤ 0.03 ===> - k ≤ -2.47 ===>k ≥ 2.47
2.47 ≤ k ≤ 2.53
(E)
Answer: E
Question: We invested a total of $1,000. We invested one part of the money at 3% and the rest of the money at 6%. The total investment with interest at the end of the year was $1,052.50. How much money did we invest at 3%?
Options:
A)$160
B)$190
C)$220
D)$250
E)$280
Rationale: Let x be the money invested at 3%.
1.03x+1.06(1000-x)=1052.50.
0.03x=1060-1052.50.
0.03x=7.50.
3x=750.
x=250.
The answer is D.
Answer: D
Question: If n is an integer greater than 7, which of the following must be divisible by 3?
Options:
A)n(n-5)(n-2)
B)n(n+5)(n+3)
C)n(n+2)(n-1)
D)n(n-4)(n-2)
E)n(n-7)(n+3)
Rationale: Let us start with C -> n(n+2)(n-1) = n(n+2)(n-3+2). Thus we can see that n+2 and (n-3+2) will leave the same remainder. But we need 3 terms with 3 different remainders.
Now let's look at D -> n(n-4)(n-2) = n(n-3 -1)(n-2). This represents 3 terms with 3 different remainders, and one of them will be divisible by 3.
ANSWER:D
Answer: D
Question: A vendor bought bananas at 6 for Rs.10 and sold them at Rs.4 for Rs.6 .Find his gain or loss percent .
Options:
A)10%
B)20%
C)30%
D)40%
E)50%
Rationale: Suppose , number of bananas bought = LCM of 6 and 4=12
CP=Rs.[(10/6)*12]=Rs.20 ; SP= Rs[(6/4)*12]=Rs.18
Loss%=[(2/20)*100]%=10%
ANSWER A
Answer: A
Question: Amar, Bhavan and Chetan divide an amount of Rs. 5600 among themselves in the ratio 3 : 6 : 5. If an amount of Rs. 400 is deducted from each of their shares, what will be the new ratio of their shares of the amount?
Options:
A)2 : 5 : 8
B)2 : 9 : 4
C)2 : 5 : 4
D)2 : 5 : 3
E)2 : 5 : 1
Rationale: Let the shares of Amar, Bhavan and Chetan be Rs. 3x, Rs. 6x and Rs. 5x respectively.
3x + 6x + 5x = 5600 => 14x = 5600 => x = 400.
Required ratio = 3x - 400 : 6x - 400 : 5x - 400
= 3x - x : 6x - x : 5x - x
= 2x : 5x : 4x => 2 : 5 : 4
Answer:C
Answer: C
Question: A man can swim in still water at 4.5 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?
Options:
A)3
B)7.5
C)2.25
D)1.5
E)4:9
Rationale: Explanation:
M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 - x)2
4.5 + x = 9 -2x
3x = 4.5
x = 1.5
Answer: Option D
Answer: D
Question: Private Benjamin is a member of a squad of 10 soldiers, which must volunteer 4 of its members for latrine duty. If the members of the latrine patrol are chosen randomly, what is the probability that private Benjamin will be chosen for latrine duty?
Options:
A)1/10
B)1/5
C)2/5
D)3/5
E)4/5
Rationale: In this case, Excluded probability = probability of Benjamin not being a part of the 4 volunteers. We can choose 4 out of 9 remaining soldiers in 9C4 ways. total ways possible = 10C4.
Thus excluded probability = 9C4/10C4 = 3/5
Thus, the desired probability = 1- 3/5 = 2/5. Thus C is the correct answer
Answer: C
Question: Jane makes toy bears. When she works with an assistant, she makes 62 percent more bears per week and works 10 percent fewer hours each week. Having an assistant increases Jane’s output of toy bears per hour by what percent?
Options:
A)20%
B)80%
C)100%
D)180%
E)200%
Rationale: Let's assume just Jane 40 bears per 40/hrs a week, so that is 1 bear/hr. With an assistant she makes 64.8 bears per 36 hours a week or 1.8 bears/hr ([40 bears * 1.62] / [40 hrs * .90]).
[(1.8 - 1)/1] * 100% = 80%
ANSWER:B
Answer: B
Question: If the cost price of 22 pens is equal to the selling price of 20 pens, the gain percent is:
Options:
A)9%
B)26%
C)50%
D)80%
E)10%
Rationale: Solution :
Let cost price of the price be Rs.1
20 pens C.P. = 20
22 pens S.P. = 22
Gain = 2/20*100 = 10 %
Answer :E
Answer: E
Question: How much is 70% of 40 is greater than 4/5 of 25?
Options:
A)22
B)67
C)88
D)12
E)8
Rationale: (70/100) * 40 – (4/5) * 25
= 8
Answer: E
Answer: E
Question: A hall 36m long and 15m broad is to be paved with stones, each measuring 6dm by 5 dm. The number of stones required is:
Options:
A)180
B)1800
C)18
D)18000
E)1.8
Rationale: Area of the hall= 3600*1500
Area of each stone= (60*50)
Therefore, number of stones= (3600*1500/ 60*50)= 1800
ANSWER:B
Answer: B
Question: What is the average of first 21 multiples of 7 ?
Options:
A)22
B)77
C)88
D)21
E)65
Rationale: Required average = 7(1+2+....+21) / 21
(7/21)X ( (21x22) / 2 ) (because sum of first 21 natural numbers)
= 77
Answer:B
Answer: B
Question: A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his
average after 17th match
Options:
A)36
B)37
C)38
D)39
E)40
Rationale: Explanation:
Let the average after 17th match is x
then the average before 17th match is x-3
so 16(x-3) + 87 = 17x
=> x = 87 - 48 = 39
Answer: Option D
Answer: D
Question: A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?
Options:
A)150
B)145
C)100
D)160
E)140
Rationale: x(10%) y(15%)
38%
23% 28%
the protein parts in the final mixture for X:Y::23:28
hence
23 X
----------- = --------------
(23+28) 300
23/51 = x/300
x = 23 * 300
------------
51
hence x is approximately equal to 139.8
hence the answer is E
Answer: E
Question: If X can run 48m and Y 42m, then in a race of 1km, X beats Y by:
Options:
A)140m
B)125m
C)100m
D)110m
E)None of these
Rationale: Explanation:
When X runs 48m, Y runs 42m.
Hence, when X runs 1000m, Y runs (1000 * 42) / 48 = 875m
X beats Y by 125m.
ANSWER:B
Answer: B
Question: A 240 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
Options:
A)230m
B)240m
C)250m
D)260m
E)270m
Rationale: speed = (120 + 80)km/h (because direction is opposite hence relative velocity is added)= 500/9 m/s
time= 9sec
let the lenght of second train is x
total distance covered= 240 + x
therefore, D= speed*time
thus 240 + x = 500/9 * 9
x= 500-240=260 m
ANSWER:D
Answer: D
Question: In a store, the total price for 25 shirts is $400 and the total price for 75 sweaters is $1500. By how much does the average (arithmetic mean) price of a sweater exceed that of a shirt in this store?
Options:
A)$1
B)$2
C)$3
D)$4
E)$5
Rationale: The average price of a shirt is: $400 / 25 = $16.
The average price of a sweater is: $1500 / 75 = $20.
The difference in price is: $20 - $16 = $4.
The answer is D.
Answer: D
Question: There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 19 days more?
Options:
A)98
B)40
C)28
D)27
E)29
Rationale: 760 ---- 22
760 ---- 20
x ----- 19
x*19 = 760*20
x = 800
760
-------
40
Answer: B
Answer: B
Question: In a group of 95 students, 36 are taking History, and 32 are taking Statistics. If 59 students are taking History or Statistics or both, then how many students are taking History but not Statistics?
Options:
A)9
B)19
C)23
D)45
E)27
Rationale: Number of students taking history =H=36
Number of students taking statistics =S= 32
Total number of students =T= 90
Number of students taking History or Statistics or both =B= 59
Number of students taking neither history nor statistics =N= 95-59=36
LetXbe the number of students taking both history and statistics.
Then
T = H + S + N - X
or 95 = 36 + 32 +36 - X
or X = 9
Now,
Number of students taking only history will be
H-X
or 36-9 = 27
Answer:- E
Answer: E
Question: The speed of a subway train is represented by the equation z=s^2+2s for all situations where 0≤s≤7, where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving. In kilometers per hour, how much faster is the subway train moving after 7 seconds than it was moving after 2 seconds?
Options:
A)4
B)55
C)15
D)48
E)63
Rationale: Given: z=s^2+2s for 0≤s≤7
z(2) = 2^2 + 2*2 = 8
z(7) = 7^2 + 2*7 = 63
Therefore z(7) - z(3) = 63 - 8 = 55 km/hr
Option B
Answer: B
Question: A crew can row a certain course up the stream in 84 minutes; they can row the same course down stream in 9 minutes less than they can row it in still water. How long would they take to row down with the stream
Options:
A)45 or 23 minutes
B)63 or 12 minutes
C)60 minutes
D)19 minutes
E)25 minutes
Rationale: speed of a crew in still water = 1/2 ( speed upstream + speed downstream)
approximating this for time,
t = 1/2(84+ (t-9))
giving t = 66 close to B
ANSWER:B
Answer: B
Question: A reduction of 10% in the price of salt enables a lady to obtain 20kgs more for Rs.400, find the original price per kg?
Options:
A)2.0
B)2.8
C)2.2
D)2.5
E)3.0
Rationale: Answer:	400*(10/100) = 40 --- 20
? --- 1 => Rs.2
400 --- 360
? --- 2 => Rs.2.22.Answer: C
Answer: C
Question: In how many different ways can 4 boys and 2 girls sit on a bench such that the girls always sit together?
Options:
A)240
B)280
C)320
D)360
E)400
Rationale: We can think of the girls as one unit.
The number of ways to arrange 5 units in order is 5!
The girls can have different arrangements, so we multiply this by 2!
The number of ways to sit is 5!*2!=240
The answer is A.
Answer: A
Question: A and B walk around a circular track. They start at 1 p.m from the same point in the opposite directions. A and B walk at a speed of 4 rounds per hour and 6 rounds per hour respectively. How many times shall they cross each other before 4 p.m
Options:
A)15
B)20
C)25
D)30
E)35
Rationale: Explanation:
Relative speed = (4+6) =10 rounds per hour
So, they cross each other 30 times in three hours hence, they cross 30 times before 4 p.m
Answer: Option D
Answer: D
Question: Between 1800 and 1850 a population of a country increased from 12 million to 15 million. Between 1850 and 1900 its population increased from 15 million to 22 million. By what percent did the rate of population growth increase from the period 1800-1850 to the period 1850-1900?
Options:
A)75%-100%
B)101%-125%
C)126%-150%
D)151%-175%
E)176%-200%
Rationale: Solution:
This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.
We first set up the ratios of population growth to initial population for the two periods. The first ratio will be for the period 1800-1850, and the second ratio will be for the period 1850-1900. Because the populations are in millions, we do not have to express all the trailing zeros in our ratios.
1800-1850
growth/initial population = 3/12 = 1/4
1850-1900
growth/initial population = 7/15
Because each ratio is not an easy number to use, we can simplify each one by multiplying each by the LCM of the two denominators, which is 60. Keep in mind that we are able to do this only because our answer choices are expressed in percents.
1800-1850
growth/initial population = (1/4) x 60 = 15
1850-1900
growth/initial population = (7/15) x 60 = 28
We can plug 15 and 28 into our percent change formula:
(new – old)/old x 100
[(28 – 15)/15] x 100
13/15 x 100
At this point we can stop and consider the answer choices. 13/15 x 100 is greater than 75% but less than 100%.
Answer A.
Answer: A
Question: Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6
Options:
A)3/18
B)5/18
C)18/7
D)18/5
E)7/18
Rationale: Clearly n(S)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then
E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),
(6,6)}
n(E)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18
Option: E
Answer: E
Question: Sam invested Rs. 8000 @ 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be?
Options:
A)3200.0
B)11520.0
C)8820.0
D)7354.0
E)16537.11
Rationale: P = Rs. 8000 ; R = 10% p.a. = 5% per half-year; T = 1 year
= 2 half-year
Amount = [8000 * (1 + 5/100)2]
= (8000 * 21/20 * 21/20)
= Rs.8820.00
Answer:C
Answer: C