divelab/Sys2Bench huggingface dataset:
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. 1000 more than D, what is B's share?
Options:
A)Rs. 500
B)Rs. 1500
C)Rs. 2000
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 = 1000
x = 1000.
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
Answer = C
Answer: C
Question: What is the smallest number Q which when increased by 5 is completely divisible by 8, 11 and 24?
Options:
A)264
B)259
C)269
D)270
E)282
Rationale: This question includes a number of great Number Property shortcuts that you can take advantage of:
1) The question asks for the SMALLEST number that, when increased by 5 is divisible by 8, 11 and 24. Since the answers are numbers, we can TEST THE ANSWERS.
2) Any number that is divisible by 24 is ALSO divisible by 8, so we really just need to consider the 11 and the 24.
3) To be divisible by 24, the end number MUST be EVEN. Since we're adding 5 to get to that end number, the starting number MUST be ODD.
The last shortcut allows us to eliminate Answers A, D and E. Between Answers B and C, TESTING B first will prove that, when increased by 5, the end sum is divisible by both 11 and 24.
Final Answer:
B
Answer: B
Question: In a 600 m race, the ratio of the speeds of two contestants A and B is 5 : 4. A has a start of 100 m. Then, A wins by:
Options:
A)200 m
B)20 m
C)403 m
D)20 m
E)23 m
Rationale: To reach the winning post A will have to cover a distance of (600 - 100)m, i.e., 500 m.
While A covers 5 m, B covers 4 m.
While A covers 500 m, B covers 4 x 500/5	m	= 400 m.
Thus, when A reaches the winning post, B covers 400 m and therefore remains 200 m behind.
A wins by 200 m.
Answer:	A
Answer: A
Question: The ratio of incomes of two person P1 and P2 is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves Rs.2200, then what is the income of P1?
Options:
A)s.800
B)s.2400
C)s.4000
D)s.5500
E)s.4200
Rationale: Let the income of P1 and P2 be Rs. 5x and Rs.4x respectively and let their
expenditures be Rs.3y and 2y respectively.
Then, 5x – 3y = 2200 …(i) and 4x – 2y = 2200 ……..(ii)
On multiplying (i) by 2, (ii) by 3 and subtracting, we get : 2x = 2200 -> x = 1100
P1’s income = Rs 5*1100 = Rs.5500
ANSWER:D
Answer: D
Question: (935421 x 625) = ?
Options:
A)584638125
B)542622125
C)584632125
D)544638125
E)584638125
Rationale: Explanation :
935421×625=935421×5^4=935421×(10/2)^4=(935421×10000)/16=584638125
Answer is A
Answer: A
Question: The number m yields a remainder p when divided by 10 and a remainder q when divided by 7. If p = q + 7, then which one of the following could be the value of m ?
Options:
A)45
B)57
C)72
D)85
E)100
Rationale: I solved this question in the following way:
Q<7 so Q can be from 0 to 6 and P<10 so P can be from 0 to 9, BUT the constraint is P=Q +7 so this will mean that P can be in the range from 7 to 9.
m=10k + P or m= 10k + 7 to 9 and look at the answer as, place different values for k, B will give 57 which is 10*5 + 7, the other answers are out of the range
Answer: B
Question: Which of the following is a parallelogram?
Options:
A)120,60,120,60(angle in deg)
B)120,30,130,20(angle in deg)
C)95,95,90,90(angle in deg)
D)a & c
E)none
Rationale: sum of adjacent angles is 180 degrees.
ANSWER:A
Answer: A
Question: Billy worked for three straight hours on his homework questions. If he solved 132 questions in the third hour, which was twice as many as he solved in the second hour, and three times as many questions as he solved in the first hour, how many questions did he solve total?
Options:
A)242
B)312
C)424
D)525
E)622
Rationale: Total questions solved= questions solved in 3rd hr+ in 2nd hr +in 1st hr
132 +132/2 +132/3= 242
A is the answer
Answer: A
Question: A certain ball team has an equal number of right- and left-handed players. On a certain day, one-third of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of lefthanded players who were not at practice?
Options:
A)1/5
B)2/3
C)5/7
D)7/5
E)3/2
Rationale: Say the total number of players is 18, 9 right-handed and 9 left-handed.
On a certain day, two-thirds of the players were absent from practice --> 6 absent and 12 present.
Of the players at practice that day, one-third were left-handed --> 12*1/3=4 were left-handed and 8 right-handed.
The number of right-handed players who were not at practice that day is 9-8=1.
The number of left-handed players who were not at practice that days is 9-4=5.
The ratio = 1/5.
Answer: A.
Answer: A
Question: Find the next number in the series 1, 3 ,7 ,13 ,21 ,31
Options:
A)43
B)33
C)41
D)45
E)51
Rationale: 3-1=2
7-3=4
13-7=6
21-13=8
31-21=10
so the next number is 31+12=43
ANSWER:A
Answer: A
Question: A furniture dealer purchased a desk for $120 and then set the selling price equal to the purchase price plus a markup that was 40% of the selling price. If the dealer sold the desk at the selling price, what was the amount of the dealer's gross profit from the purchase and the sale of the desk?
Options:
A)$40
B)$60
C)$80
D)$90
E)$100
Rationale: Purchase price = 120
Selling price = x
120 + 0.4*x = x
0.6*x = 120
x = 200
Profit = 200 - 150 = 50
ANSWER:A
Answer: A
Question: An investment Q gained an interest payment of $250 each month when the annual interest rate was 8%, how much more should we invest annually if we want to gain 12% more per month with a new annual interest rate of 7.5% ?
Options:
A)$9,360
B)$9,100
C)$8,250
D)$7,300
E)$7,150
Rationale: Answer is D : 7,300
An investment Q gained an interest payment of $250 each month when the annual interest rate was 8%,
Interest per annum with 8% interest rate is = 250* 12 = Principal * Interest Rate(8/100)
This gives Principal as 37,500$
For an revised interest of 12% more and new interest rate of 7.5% , ie. 250* 12 * 1.12 = Principal * Interest Rate (7.5/100)
This gives the new Principal as 44,800.
The question is how much more should we invest, so 44,800 - 37,500 = 7,300.D
Answer: D
Question: An engineer undertakes a project to build a road 10 km long in 60 days and employs 30 men for the purpose. After 20 days, he finds only 2 km of the road has been completed. Find the (approximate) number of extra men he must employ to finish the work in time.
Options:
A)20
B)15
C)30
D)45
E)60
Rationale: 30 workers working already
Let x be the total men required to finish the task in next 40 days
2 km done hence remaining is 8 km
Also, work has to be completed in next 40 days (60 - 20 = 40)
We know that, proportion of men to distance is direct proportion
and, proportion of men to days is inverse proportion
Hence, X = (30 * 8 * 20) / (2 *40)
thus, X = 60
Thus, more men needed to finish the task = 60-30=30
Answer : C
Answer: C
Question: If A and B get profits of Rs. 6000 and Rs. 4000 respectively at the end of year the ratio of their investments are ?
Options:
A)1:4
B)3:2
C)2:3
D)2:5
E)None of these
Rationale: profit=Investment * Time
so,
6000=i1 *1=6000
4000=i2 * 1= 4000
i1/i2=6000/4000=3/2
ANSWER:B
Answer: B
Question: If Y> 0 and Y is divisible by the product of the first nine positive integers, all of the following must be factors of Y except
Options:
A)35
B)40
C)63
D)315
E)704
Rationale: The question states taht Y is divisible by 9!,which is 1*2*3*4*5*6*7*8*9
Y=(9!) * some number
A. 35 = 5*7 so will divide Y
B. 40 =5*8 so will divide Y
C. 63 =9*7 so will divide Y
D. 315 =5*7*9 so will divide Y
E. 704 =8*8*11 This will not divide Y
Ans E
Answer: E
Question: Mr. Karan borrowed a certain amount at 6% per annum simple interest for 9 years. After 9 years, he returned Rs. 8110/-. Find out the amount that he borrowed.
Options:
A)Rs. 4,900
B)Rs. 5,000
C)Rs. 5,100
D)Rs. 5266
E)None of these
Rationale: Explanation:
Let us assume Mr. Karan borrowed amount is Rs. A. (the principal)
By formula of simple interest, S.I. = PRT / 100
Where P = The principal, R = Rate of interest as a %, T = Time in years
S.I. = (P * 6 * 9)/100 = 54 P/100
Amount = Principal + S.I.
8110 = P + (54P/100)
8110 = (100P + 54P)/100
8110 = 154P/100
P = (8110 * 100)/154 = Rs. 5266.233
ANSWER D
Answer: D
Question: The average of 9 observations was 9, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation?
Options:
A)9
B)8
C)7
D)6
E)4
Rationale: 1 to 9 = 9 * 9 = 81
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5th = 50 + 40 = 90 – 81 = 9
Answer:A
Answer: A
Question: Two horses begin running on an oval course at the same time. One runs each lap in 4 minutes; the other takes 12 minutes to run each lap. How Many minutes after the start will the faster horse have a one lap lead?
Options:
A)36
B)12
C)9
D)4
E)6
Rationale: The rate of the faster horse is 1/4 lap/minute;
The rate of the slower horse is 1/12 lap/minute;
Their relative rate is 1/4-1/12=1/6 lap/minute;
The faster horse to gain one full lap will need time=distance/rate=1/(1/6)=6 minutes.
Answer: E.
Answer: E
Question: Of the 120 passengers on Flight 750, 40% are female. 10% of the passengers sit in first class, and the rest of the passengers sit in coach class. If 1/3 of the passengers in first class are male, how many females are there in coach class?
Options:
A)40
B)48
C)50
D)52
E)56
Rationale: Number of passengers on flight = 120
Number of female passengers= .4*120 = 48
Number of passengers in first class = (10/100)* 120 = 12
Number of passengers in coach class = (90/100)* 120 = 108
Number of male passengers in first class = 1/3 * 12 = 4
Number of female passengers in first class = 12-4 = 8
Number of female passengers in coach class = 48 - 8 = 40
Answer A
Answer: A
Question: How much 70% of 100 is greater than 60% of 80?
Options:
A)18
B)99
C)22
D)26
E)71
Rationale: (70/100) * 100 – (60/100) * 80
70- 48 = 22
Answer: C
Answer: C
Question: In a group of 80 children and 10 youngsters, each child got sweets that are 15% of the total number of children and each youngster got sweets that are 25% of the total number of children. How many sweets were there?
Options:
A)1160
B)1169
C)1162
D)1122
E)1168
Rationale: Number of sweets each child got = 15% of 80 = 15/100 * 80 = 12.
Number of sweets 80 children got = 80 * 12 = 960.
Number of sweets each youngster got = 25% of 80 = 25/100 * 80 = 20.
Number of sweets 10 youngsters got = 10 * 20 = 200.
Total number of sweets = 960 + 200 = 1160.
Answer:A
Answer: A
Question: Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 30 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils?
Options:
A) 0.75
B) 0.8
C) 1.375
D) 1.4
E) 1.85
Rationale: Say the cost price of 80 pencils was $80 ($1 per pencil) and the selling price of 1 pencil was p.
Selling at a loss: 80 - 80p = 30p --> p = 8/11.
(cost price)/(selling price) = 1/(8/11) = 11/8 = 1.375.
Answer: C.
Answer: C
Question: For each positive integer,the quantitySn is defined such thatSn+3=(Sn)^2 -S(n+1).in addition ,S2=1.then which of the following must be true.
Options:
A)a)S4 +S3 +1=0
B)b)(S1)^2 +S4 +2=0
C)c) ifS1=-1thenS4 is also equal to-1.
D)d) ifS4 =-7thenS1=-2.
E)e) ifS1 =-3thenS4=-7
Rationale: Since all the answers include S4, let's start from there.
S4=S2^2-S3
S3=S1^2-S2
Substituting S3 into the first equation we get:
S4=S2^2-(S1^2-S2)
Since S2=1
S4=1-(S1^2-1)
S4=2-S1^2
We need S1^2 in order to calculate S4.
E) provides S1 and from there we can calculate S4:
S4=2-(-3)^2
S4=-7
Correct.
As a side note, I would be very careful with answers in the formif S4=x then S1=y(such as option D ), as we are dealing with S1^2 (not S1), S1 can have two different values and it can not be determined for sure which root is the right answer.
For example:
If S4=-7 then S1=3 :
if we substitute S4 with - 7 we get:
-7=2-S1^2
S1^2=9
now S1 can be +/-3 !!
B
Answer: B
Question: If b is an even integer and b/18 is an odd integer, which of the following is NOT an even integer?
Options:
A)(b^2)/54
B)(b^2)/8
C)(b^2)/9
D)(b^2)/6
E)(b^2)/2
Rationale: since b is an even integer and b/18 is an odd integer. thus maximum power of 2 in a is one. why ?? because if the power of 2 in the expression has been 2 or greater, then the b/18 must have been an even integer. (18=2*3^2. as 18 contain only one 2 in the expression)
now, any expression in which we have 4 at the denominator will be odd. out of the given options only 8 is a multiple of 4. hence answer should be B
Answer: B
Question: If n is a positive integer, which one of the following numbers must have a remainder of 2 when
divided by any of the numbers 4, 5, and 6?
Options:
A)125n + 2
B)240n + 2
C)280n + 2
D)390n + 2
E)520n + 2
Rationale: The numbers 4, 5, and 6 all divide 240 so they all divide 240n.
When we divide (240n + 2) by 4, 5, or 6, the remainder must be 2.
The answer is B.
Answer: B
Question: Set R consists of all even integers between 2 and 100, inclusive. Set X is derived by reducing each term in set R by 50, set Y is derived by multiplying each term in set R by 1.5, and set Z is derived by dividing each term in set R by -4. Which of the following represents the ranking of the three sets in descending order of standard deviation?
Options:
A) X, Y, Z
B) X, Z, Y
C) Y, Z, X
D) Y, X, Z
E) Z, Y, X
Rationale: Set R - {2, 4, ..., 100};
Set X - {-48, -46, ..., 50};
Set Y - {3, 6, ..., 150};
Set Z - {-2/4, -4/4, ..., -100/4} = {-1/2, -1, -3/2, ..., -25}.
If we add or subtract a constant to each term in a set the SD will not change, so sets R and X will have the same SD.
If we increase or decrease each term in a set by the same percent (multiply by a constant) the SD will increase or decrease by the same percent, so set Y will have 1.5 times greater SD than set R and set Z will have 4 times less SD than set R (note SD can not be negative so SD of Z wil be SD of R divided by 4 not by -4).
So, the ranking of SD's in descending order is: Y, R=X, Z.
Answer: D.
Answer: D
Question: Two numbers are in the ratio of 5:7. If 25 be subtracted from each, they are in the ratio of 35:59. Find the numbers?
Options:
A)60,88
B)60,81
C)68,88
D)60,84
E)60,81
Rationale: (5x-25):(7x-25)
= 35:59
x = 12 => 60,84
Answer: D
Answer: D
Question: In a certain city each of 6 Bus lines is represented by a colored flag. Each flag consists of either a single color or a pair of two different colors. If each bus lines has a different flag, what is the minimum number of colors needed for the flags. (Assume that the order of colors in a pair on a flag does not matter.)
Options:
A)2
B)4
C)3
D)1
E)5
Rationale: C.3:
if we have 3 colors we can choose pairs in 3C2 ways = 3 and single color flags are 3. Therefore total number of flags = 6.
ANS C.
Answer: C
Question: Sum of the squares of 3 no.'s is 62 and the sum of their products taken two at a time is 131. Find the sum?
Options:
A)20
B)24
C)26
D)18
E)30
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 62 + 2* 131
a + b + c = √324 = 18
D
Answer: D
Question: The average of three consecutive odd numbers is 12 more than one third of the first of these numbers. What is the last of the three numbers?
Options:
A)15
B)17
C)19
D)21
E)23
Rationale: x,x+2,x+4 are three odd numbers
(x+x+2+x+4)/3=12+(x/3)
x=15
last no.=x+4=19
ANSWER:C
Answer: C
Question: A 9 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point?
Options:
A)1/6
B)1/5
C)2/9
D)1/3
E)2/5
Rationale: A square with an area of 4 has a perimeter of 8.
For the area to be >4, the longer piece must be >8.
The wire must be cut within 1 meters from either end.
The probability of this is 2/9.
The answer is C.
Answer: C
Question: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 80%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
Options:
A)100% decrease
B)50% decrease
C)40% decrease
D)45% increase
E)50% increase
Rationale: The rate of reaction is invesrsly proportional to the concentration of chemical B present. It used to have B=1 . Now that B is increased by 80%. So the new equation would be 2B=(1/2). In order for the rate of reaction to still be 1, we need to change the concentration of A to yield a 2. It used to be A^2=1, now the new equation should be (sqrt(2)*A)^2=2. The change in the concentration of A can be calculated as (sqrt(2) -1)/1 or approximately 40% increase. Answer D.
Answer: D
Question: A man sells a horse for Rs.800 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse?
Options:
A)860
B)900
C)875
D)950
E)1000
Rationale: CP = SP + 1CP = SP - g
800 + x = 980 - 2x
3x = 180 => x = 60
CP = 800 + 60 = 860
ANSWER A
Answer: A
Question: The ages of two persons differ by 12 years. If 5 years ago, the elder one be 5 times as old as the younger one, their present ages (in years) are respectively
Options:
A)20,20
B)20,8
C)25,15
D)30,10
E)None of these
Rationale: Explanation:
Let their ages be x and (x + 12) years.
5 (x - 5) = (x + 12 - 5) or 4x = 32 or x = 8.
Their present ages are 20 years and 8 years
Option B
Answer: B
Question: The area of a triangle is with base 10m and height 10m?
Options:
A)11
B)10
C)50
D)122
E)12
Rationale: 1/2 * 10 *10= 50 m2
Answer: C
Answer: C
Question: 8 men can do a piece of work in 12 days. 4 women can do it in 48 days and 10 children can do it in 24 days. In how many days can 18 men, 4 women and 10 children together complete the piece of work?
Options:
A)5 days
B)15 days
C)28 days
D)4 days
E)7 days
Rationale: Explanation:
1 man’s 1 day’s work = 1/8 × 12 = 1/96
18 men’s 1 day’s work = 1 × 18/96 = 3/16
1 woman’s 1 day’s work = 1/192
4 women’s 1 day’s work = 1/192 × 4 = 1/48
1 child’s 1 day’s work = 1/240
10 children’s 1 day’s work = 1/24
Therefore, (18 men + 4 women + 10 children)’s 1 day’s work = 3/16 + 1/48 + 1/24
= 1/4
The required No. of days = 4 days
Answer: Option D
Answer: D
Question: The cost of a one-family home was $ 120,000 in 1980. In 1988, the price had increased to $ 198,000. What was the percent increase in the cost of the home?
Options:
A)65%
B)50%
C)55%
D)40%
E)33.3%
Rationale: Increase = 198000-120000 = 78000
% increase = 78000*100/120000 = 65%
Answer: option A
Answer: A
Question: A person takes 5hrs 45 min in walking to a place and riding back. He would have gained 5 hrs by riding both ways. The time he would take to walk both ways is?
Options:
A)5 hours 45 min
B)5 hours 55 min
C)6 hours 45 min
D)10 hours 45 min
E)11 hours
Rationale: ----------------------------------------------------------------
Solution 1
----------------------------------------------------------------
Given that time taken for riding both ways will be 5 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 5 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min +5 hours = 10 hours 45 min
D
Answer: D
Question: Xavier starts from P towards Q at a speed of 50 kmph and after every 12 mins increases his speed by 10 kmph. If the distance between P and Q is 52km, then how much time does he take to cover the distance?
Options:
A)40
B)60
C)48
D)70
E)80
Rationale: first 12min = 50 *12/60 =10 km
2nd 12min = 60*12/60 = 12km
3rd 12 min = 70*12/60 = 14 km
4th 12 min = 80 *12/60 = 16 KM
TOTAL time 12.4 =48 min
C
Answer: C
Question: If 1/(a – 2) = 1/(a + 2) + 1/(a – 1), which of the following is a possible value of a?
Options:
A)1
B)2
C)-3
D)3
E)0
Rationale: Multiply by (a-2)*(a+2)*(a-1) throughout to get:
(a+2)*(a-1) = (a-2)*(a-1) + (a+2)(a-2)
Of the given options, only a=0 satisfies this.
Answer : E
Answer: E
Question: What is the sum of all the even numbers between 31 and 99, inclusive?
Options:
A)3,000
B)2,562
C)2,500
D)2,180
E)2,766
Rationale: To figure this problem out, you just have to add all the even numbers, starting with 32.
32+34+36+38 and so on, all the way up to 98. You'll get 2,180, so
Final Answer:
D
Answer: D
Question: The daily high temperatures were recorded at an Antarctic weather station. If a temperature of -38.2 degrees Celsius is 2 units of standard deviation below the mean temperature, and a temperature of -22.6 degrees Celsius is 4 units of standard deviation above the mean, which of the following temperatures is 2 unit of standard deviation above the mean?
Options:
A)-35.6 degrees Celsius
B)-33.0 degrees Celsius
C)-30.4 degrees Celsius
D)-27.8 degrees Celsius
E)-25.2 degrees Celsius
Rationale: Let us suppose mean = M and SD = d.
So, we are given M - 2d = -38.2
and M + 4d = -22.6
solving both the equations, we will get M = -33 an d = 2.6
So, we need to find out M + 2d = -33 + 2*2.6 = -27.8. Hence, D
Answer: D
Question: In a box of 11 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?
Options:
A)28/55
B)19/33
C)7/11
D)4/7
E)3/5
Rationale: P(neither pen is defective) = 8/11*7/10 = 28/55
The answer is A.
Answer: A
Question: A bag is filled with pennies and nickels in the ratio of 5to3. There are 30 nickles in the bag, how many coins are there in that bag?
Options:
A)25
B)34
C)45
D)50
E)80
Rationale: A ratio of pennies to nickels of 5 to 3 means that we can write the number of pennies and nickels in the form
number of pennies = 5x and number of nickels = 3x
But we know the number of nickels, 30. Hence
3x = 30
Solve for x
x = 10
The total number of coins is given by
5x + 3x = 8x = 8 × 10 = 80
E
Answer: E
Question: Express a speed of 18 kmph in meters per second?
Options:
A)10 mps
B)06mps
C)09 mps
D)05 mps
E)11 mps
Rationale: D
5 mps
18 * 5/18 = 5 mps
Answer: D
Question: Bus 1 and Bus 2 run between cities A and B. Bus 1 leaves city A at the same time Bus 2 leaves city B,
each at a constant rate of speed. Their first meeting is 70 miles from city A. After reaching their
respective destinations and immediately turning around, their second meeting is 30 miles from city B.
What is the distance in miles between city A and city B?
Options:
A)90
B)120
C)125
D)150
E)180
Rationale: let d=distance between cities A and B
bus 1 distance to first meeting=70 miles
bus 1 distance from first to second meeting=(d-70)+30=d-40 miles
bus 2 distance to first meeting=d-70 miles
bus 2 distance from first to second meeting=70+(d-30)=d+40 miles
because both buses take the same time for the pre-meeting leg, and
the same time for the inter-meeting leg, and their speeds remain
constant, their ratios between pre-meeting distance and inter-meeting
distance should be the same
therefore, 70/(d-40)=(d-70)/(d+40)
d^2=180d
d=180 miles
E
Answer: E
Question: Two friends Simpson & Stewie decide to get together
So they start driving cars towards each other. They plan to meet halfway. Each is riding at 6 Km/hr. They live 36 kms apart. Stewie has a pet "eagle" and it starts flying the instant they start traveling. The eagle flies back and forth at 18 Km/hr between them until the two meet.
How many kms does the eagle travel ?
Options:
A)54
B)55
C)86
D)45
E)62
Rationale: A
54
It takes 3 hours for the Simpson & Stewie to meet.
so the eagle flies for 3 hours at 18 Km/hr = 54 kms
Answer: A
Question: The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Options:
A)4 years
B)8 years
C)6 years
D)5 years
E)2 years
Rationale: Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20 => x = 4.
Age of youngest child = x = 4 years.
Answer: A
Answer: A