mathewhe/aqua_rat huggingface dataset:
Question: A person starts walking at a speed of 5km/hr through half the distance, rest of the distance he
covers with aspeed 4km/hr. Total time of travel is 36 hours. What is the maximum distance he
can cover?
(A) 20km
(B) 40km
(C) 60km
(D) 80km
(E) 160km
Rationale: t=d/s, so
36=x/2*1/5+x/2*1/4(because half distance with 5 km/ph, and remaining half with 4km/hr)
36=x(18/80)
x=160km
ANSWER:E
Answer: (E) 160km
Question: The average age of three boys is 120 years and their ages are in proportion 2:6:12. What is the age in years of the youngest boy?
(A) 8
(B) 9
(C) 7
(D) 12
(E) 5
Rationale: 2x + 6x + 12x = 120
x =6
2x = 12
Answer:D
Answer: (D) 12
Question: Village X has a population of 76000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year .in how many years will the population of the two villages be equal?
(A) 15
(B) 17
(C) 11
(D) 18
(E) 13
Rationale: Let the population of two villages be equal after p years
Then, 76000-1200p=42000+800p
2000p=34000
p=17
Answer is B.
Answer: (B) 17
Question: In year Y imported machine tools accounted for 25 percent of total machine-tools sales in the United States, and Japanese imports accounted for 65 percent of the sales of imported machine tools. If the total sales of machine tools imported from Japan that year was x billion dollars, then the total sales of all machine tools in the United States was how many billion dollars?
(A) 9x/80
(B) 13x/20
(C) 80/9x
(D) 20x/13
(E) 80x/13
Rationale: Sales of imported tools = 25% of the total sales = 0.25*{total} = {total}/4;
Sales of Japanese tools = 65% of the sales of imported tools = 0.65*{total}/4 = 13/20*{total}/4 = 13*{total}/80.
Since the sales of of Japanese tools was x, then 13*{total}/80 = x --> {total} = 80x/13.
Answer: E.
Answer: (E) 80x/13
Question: Tough and Tricky questions: Combinations.
Six students - 3 Girls and 3 boys - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
(A) 12
(B) 72
(C) 34
(D) 240
(E) 520
Rationale: Six students - 3 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A) 12
B) 36
C) 72
D) 240
E) 720
The boys (B) and girls(G) can either be arranged as BGBGBG or GBGBGB.
Within the arrangement BGBGBG or GBGBGB, the boys can be arranged 3! ways and girls can be arranged 3! ways.
Therefore total number of ways boys and girls can be arranged = 2*3!*3!= 72 ways
Answer:B
Answer: (B) 72
Question: If in a $150 mln portfolio of stock, twice as much is invested in stock A as is invested in all the other stocks, how much is invested in all the other stocks?
(A) 50mln
(B) 60mln
(C) 70mln
(D) 80mln
(E) 90mln
Rationale: Let x denote the amount invested in all the other stocks. Then the amount invested in stock A is 2x. As we have the $150 mln portfolio, x+2x=3x=150 mln from where x=50 mln.
Answer: A
Answer: (A) 50mln
Question: On a sum of money, the S.I. for 2 years is Rs. 660, while the C.I. is Rs. 696.30, the rate of interest being the same in both the cases. The rate of interest is?
(A) 12%
(B) 32%
(C) 72%
(D) 11%
(E) 82%
Rationale: Difference in C.I. and S.I for 2 years = (696.30 - 660) = Rs. 36.30.
S.I for one year = Rs. 330.
S.I. on Rs. 330 for 1 year = Rs. 36.30
Rate = (100 * 36.30)/(330 * 1) = 11%
Answer:D
Answer: (D) 11%
Question: All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
(A) 1289
(B) 1198
(C) 281
(D) 1184
(E) 282
Rationale: Explanation:
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
Answer: Option D
Answer: (D) 1184
Question: How many different pairs of numbers (s, t) such that s = 3t can be obtained if s and t are selected from the set of number {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: Given s=3t
t can take 0,1,2,3, 4
and so s = 0,3,6,9, 12
4 such pairs can be formed.
answer E
Answer: (E) 5
Question: A and B can do a piece of work in 8 days. B and C can do it in 12 days and A and C in 16 days. Working together they will complete the work in how many days?
(A) 3.33 days
(B) 8.33 days
(C) 3.99 days
(D) 7.38 days
(E) 2.88 days
Rationale: A + B = 1/8
B + C = 1/12
C + A = 1/16
-------------------------------
2(A + B +C) = 1/8 + 1/12 + 1/16 = 13/48
A + B +C = 13/96
96/13 = 7.38 days
Answer:D
Answer: (D) 7.38 days
Question: A gives B as many as rupees B has and C as many as rupees C has.Similarly B gives A as many as rupees A has and C as many as rupees C has, C similarly gives B as many as rupees B has and A as many as rupees A has..Now finally each of them has 32 rupees.how much rupees does A has initially?
(A) 52
(B) 56
(C) 60
(D) 62
(E) 45
Rationale: initially
A
B
C
after A gives
A has ->A-B-C
B has ->2B
C has ->2C
after B gives
A has ->2A-2B-2C
B has ->-A+3B-C
C has ->4C
AFTER C gives
A has ->4A-4B-4C=32---------1
B has ->-2A+6B-2C=32----------2
C has ->-A-B-7C=32-----------3
ON solving 1,2,3 A=52
ANSWER:A
Answer: (A) 52
Question: What least no. must be subtracted from 50248 so that remaining no.is divisible by 20?
(A) 3
(B) 2
(C) 4
(D) 8
(E) 6
Rationale: Explanation:
On dividing 50248 by 20 we get the remainder 8, so 8 should be subtracted
Option D
Answer: (D) 8
Question: If x^2−2ax+a^2=0 get value of x/a
(A) 8
(B) 9
(C) 7
(D) 5
(E) 1
Rationale: a is obviously non-zero. We divide the equation by
a^2
we get (x/a-1)^2=0
x/a=1
Ans: E
Answer: (E) 1
Question: Find the area of a rhombus whose side is 25 cm and one of the diagonals is 30 cm?
(A) 287
(B) 2766
(C) 188
(D) 261
(E) 600
Rationale: Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals bisect at right angles in a rhombus.
BE2 + AE2 = AB2
252 = 152 + AE2 AE = √(625 - 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d1d2
= 1/2 * 40 * 30 = 600 sq.cm.
Answer: E
Answer: (E) 600
Question: On the shelf, there are four books of poetry, four novels, and four reference works. Suppose we picked two books at random. What is the probability that we pick one novel and one reference work?
(A) 2/15
(B) 4/21
(C) 6/25
(D) 8/33
(E) 10/37
Rationale: We could pick one novel and then one reference work, or, we could choose one reference work and then one novel.
P(one novel and one reference work) = 4/12*4/11 + 4/12*4/11 = 8/33
The answer is D.
Answer: (D) 8/33
Question: A, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?
(A) 4 days
(B) 3 days
(C) 6 days
(D) 8 days
(E) 9 days
Rationale: A
4 days
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
Answer: (A) 4 days
Question: Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 10 minutes earlier than usual. How much time did the man spend walking?
(A) 45 minutes
(B) 50 minutes
(C) 55 minutes
(D) 52minutes
(E) 35 minutes
Rationale: As they arrived 10 minutes earlier than usual, they saved 10 minutes on round trip from home to station (home-station-home) --> 5 minutes in each direction (home-station) --> wife meets husband 5 minutes earlier the usual meeting time --> husband arrived an hour earlier the usual meeting time, so he must have spent waking the rest of the time before their meeting, which is hour-5minutes=55 minutes.
Answer: C
Answer: (C) 55 minutes
Question: In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?
(A) 127.7 m
(B) 127.5 m
(C) 197.5 m
(D) 177.5 m
(E) 117.5 m
Rationale: When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000 = 127.5 m.
Answer: B
Answer: (B) 127.5 m
Question: The mass of 1 cubic meter of a substance is 200 kilograms under certain conditions. What is the volume, in cubic centimeters, of 1 gram of this substance under these conditions? (1 kilogram = 1,000 grams and 1 cubic meter = 1,000,000 cubic centimeters)
(A) 0.8
(B) 1.25
(C) 8.0
(D) 12.5
(E) 5.0
Rationale: Density is mass divided by volume. So density of the given substance will be mass/volume=200kg/1m^3=800 kg/m^3 or 1g/5cm^3=0.2 g/cm^3.
Next, ask yourself if 200,000 g is equivalent to 1,000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters? --> 1 g - 1,000,000/800,000 = 10/2 = 5 cubic centimeters.
Answer is E
Answer: (E) 5.0
Question: It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car?
(A) 3:4
(B) 3:0
(C) 3:8
(D) 3:2
(E) 3:1
Rationale: Let speed of the train = x kmph and speed of the car = y kmph
Time needed for traveling 600 km if 120 km by train and the rest by car = 8 hr
⇒120x+(600−120)y=8⇒120x+480y=8.............(Equation 1)⇒15x+60y=1.............(Equation 1)
Time needed for traveling 600 km if 200 km by train and the rest by car = 8 hr 20 min
⇒200x+(600−200)y=82060=813=253⇒200x+400y=253⇒8x+16y=13⇒24x+48y=1.............(Equation 2)Solving Equation1 and Equation2Here Equation1 = Equation2 = 1⇒15x+60y=24x+48y⇒12y=9x⇒4y=3x⇒xy=34⇒x:y=3:4
Answer : A
Answer: (A) 3:4
Question: In a group of cows and hens, the number of legs are 14 more than twice the number of heads. The number of cows is
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
Rationale: Explanation:
Let the number of cows be x and the number of hens be y.
Then, 4x + 2y = 2 (x + y) + 14 4x + 2y = 2x + 2y + 14 2x = 14 x = 7.
Answer: B
Answer: (B) 7
Question: Two trains, each 150 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?
(A) 90
(B) 98
(C) 60
(D) 88
(E) 12
Rationale: Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(150 + 150)/8 = 3x => x = 25/2.
So, speed of the faster train = 50/2 = 25 * 18/5 = 90 km/hr.
Answer:A
Answer: (A) 90
Question: The sum of three consecutive odd numbers is always divisible by:
I. 2 II. 3 III. 5 IV. 6
(A) Only I
(B) Only II
(C) Only I and II
(D) Only II and IV
(E) None of these
Rationale: Let the three consecutive odd numbers be (2x+1), (2x+3) and (2x+5). Their sum = (6x=9)= 3 (2x+3), which is always divisible by 3.
Answer: B
Answer: (B) Only II
Question: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130
The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 on of meters. If the standard devastation of the 10 running times is 27.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?
(A) one
(B) two
(C) three
(D) four
(E) five
Rationale: The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake.
it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90;
Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-27,4=72,6. There are only two numbers below 72,6. The answer is B
Answer: (B) two
Question: The average of first 10 odd numbers is?
(A) 11
(B) 10
(C) 19
(D) 16
(E) 13
Rationale: Sum of 10 odd no. = 100
Average = 100/10 = 10
Answer: B
Answer: (B) 10
Question: What percent is 2 minutes 24 seconds of an hour?
(A) 6%
(B) 2%
(C) 4%
(D) 5%
(E) 7%
Rationale: %=144/60×60 = 4%
Answer: C
Answer: (C) 4%
Question: What approximate value should come in place of the question mark (?) in the following equation?
1.542 × 2408.69 + 2134.632 = ?
(A) 4600
(B) 4800
(C) 5800
(D) 6400
(E) 3600
Rationale: Explanation 3700 + 2100 = 5800
Answer C
Answer: (C) 5800
Question: Solve for x and check: 15x = 165
(A) 98
(B) 165
(C) 180
(D) 253
(E) None of these
Rationale: Solution : Dividing each side by 15, we obtain
(15x/15)=(165/15)
Therefore : x=11
Check :
15x=165
(15*11)=165
165=165
ANSWER : B
Answer: (B) 165
Question: If in a certain sequence of consecutive multiples of 50, the median is 825, and the greatest term is 1000, how many terms that are smaller than 825 are there in the sequence?
(A) 6
(B) 4
(C) 8
(D) 12
(E) 13
Rationale: Since the median is 825 we know there must be a even number of integers because 50 is not a multiple of 825.
So the list around 825 must go. 700 750 800 (825) 850 900 950 1000
Since we know there are 4 numbers greater than 825 then there must be 4 numbers less then 825.
answer:B
Answer: (B) 4
Question: A, B, C are three consecutive positive integers (A>B>C). What is the value of the expression 2A +B +3C?
(A) 6A+7
(B) 5A+1.
(C) 5A-1.
(D) 6A-5.
(E) 6A-7.
Rationale: A= C+2
=> C = A-2
A= B+1
=> B = A-1
Since the answer options contain variable A , we have to write B and C in terms of A .
2A +B +3C = 2A + (A-1) +3(A-2)
=2A + A-1 + 3A -6
=6A -7
Answer E
Answer: (E) 6A-7.
Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 4320 interest only for the period. The principal amount borrowed by him was:
(A) s. 12,000
(B) s. 10,000
(C) s. 15,000
(D) s. 18,000
(E) s. 20,000
Rationale: Principal = Rs. (100 x 4320)/(12X3) = Rs. 12000. Answer: Option A
Answer: (A) s. 12,000
Question: Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
(A) 8
(B) 7
(C) 6
(D) 3
(E) 4
Rationale: 9261 = 8000(21/20)N
(21/20)3 = (21/20)N => N = 3
Answer:D
Answer: (D) 3
Question: The total of company C's assets in 1994 was 500% greater than the total in 1993, which in turn was 400% greater than the total in 1992. If the total of company C's assets in in 1992 was N dollars, which one of the following represents company C's assets in 1994:
(A) 7N
(B) 8N
(C) 19N
(D) 22N
(E) 30N
Rationale: Let's use actual numbers. Starting in 1992, let's say company C had $100 in assets.
In 1993, the total assets were 400% GREATER, which means 1992 PLUS 400% of 1992:
$100 + 4x$100 = $100 + $400 = $500
In 1994, the total assets were 500% GREATER than they were in 1993, which means 1993 PLUS 500% of 1993:
$500 + 5x$500 = $500 + $2500 = $3000
This is 30 times the 1992 number, so the correct answer is 30N.
Answer: (E) 30N
Question: If 6^k + 6^k = (6^9)^(6^9) - 6^k, then k = ?
(A)  11/3
(B)  11/2
(C)  242
(D)  6^11 - 1
(E)  3^11 - 1
Rationale: 6^k + 6^k = (6^9)^6^9 - 6^k
6*(6^k) = 6 ^ (36 * 6^9 ) = 6^(6^2 *6^9) = 6^(6^11)
6^k+1 = 6^(6^11)
So k+1 = 6^11
So k = 6^11 -1
Answer is D
Answer: (D)  6^11 - 1
Question: The cash difference between the selling prices of an article at a profit of 2% and 4% is Rs 3. The ratio of two selling prices is
(A) 51:52
(B) 52:53
(C) 53:54
(D) 54:55
(E) 51:52
Rationale: Explanation:
Let the Cost price of article is Rs. x
Required ratio = (102% of x) / (104% of x)
=102/104
=51/52 = 51:52.
Answer: E
Answer: (E) 51:52
Question: In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
(A) 100
(B) 150
(C) 50
(D) 250
(E) 75
Rationale: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs. ((25x/100) + (10 x 2x/100)+(5 x 3x)/100= Rs. 60x/100
Therefore, 60x/100 = 30
x = 30 x 100/60 = 50.
Hence, the number of 5 p coins = (3 x 50) = 150.
Answer is B.
Answer: (B) 150
Question: A 300 m long train crosses a platform in 30 sec while it crosses a signal pole in 18 sec. What is the length of the platform?
(A) 286 m
(B) 350 m
(C) 200 m
(D) 278 m
(E) 267 m
Rationale: Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/30 = 50/3
x = 200 m.
Answer:C
Answer: (C) 200 m
Question: If 4 men and 1 boy working together, can do two times as much work per hour as a man and a boy together. Find the ratio of the work done by a man and that of a boy for a given time?
(A) 3:2
(B) 2:1
(C) 3:3
(D) 1:2
(E) 6:1
Rationale: 4M + 1B =2(1M + 1B)
4M + 1B = 2M + 2B
2M = 1B
The required ratio of work done by a man and a boy = 1:2
D
Answer: (D) 1:2
Question: What annual installment will discharge a debt of Rs.1092 due in 3 years at 12% simple interest?
(A) 442
(B) 441
(C) 440
(D) 320
(E) 532
Rationale: Let the installment be Rs. X
If he keeps the total amount for 3 years, he have to pay the interest
of Rs. (1092*12*3)/100.
But he paid Rs. X after 1 year.
So he need not to pay interest for Rs. X for remaining 2 years. i.e he
need not to pay Rs. (X*12*2)/100.
And he paid another Rs. X after 2 years.
So he need not to pay interest for for remaining 1 years. i.e. Rs.
(X*12*1)/100.
So The total interest he has to pay is Rs. ((1092*12*3)/100) - [
((X*12*2)/100) + ((X*12*1)/100) ].
So, The total amount he has to pay is Rs. 1092 + Rs.((1092*12*3)/100) -
[ ((X*12*2)/100) + ((X*12*1)/100) ].
The total amount he paid = 3X. ( because he paid Rs.X for three times)
1092 + ((1092*12*3)/100) - [ ((X*12*2)/100) + ((X*12*1)/100) ] = 3X
BY solving, we will get X = 442.
ANSWER:A
Answer: (A) 442
Question: Divide Rs. 2379 into 3 parts so that their amounts after 2, 3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is?
(A) 3378
(B) 627
(C) 276
(D) 828
(E) 112
Rationale: Let the parts be x, y and [2379 - (x + y)]
x + (x * 2 * 5/100) = y + (y * 3 * 5/100) = z + (z * 4 * 5/100)
11x/10 = 23y/20 = 6z/5 = k
x = 10k/11, y = 20k/23, z = 5k/6
But x + y + z = 2379
10k/11 + 20k/23 + 5k/6 = 2379
k = (2379 * 11 * 23 * 6)/3965 = (3 * 11 * 23 * 6)/5
x = [10/11 * (3 * 11 * 23 * 6)/5] = 828
Hence, the first part is Rs. 828.
Answer: D
Answer: (D) 828
Question: How many numbers are there between 200 and 800 which are divisible by both 5 and 7?
(A) 35
(B) 16
(C) 17
(D) can’t be determined
(E) None of these
Rationale: Answer C
Solution
In the given range, the last number which is divisible by both 5 and 7. i.e., 35 is 210 and the highest number is 770. So the total number of numbers between 200 and 800 which are divisible by both 5 and 7 is
(770-210/35)+ 1 = 17
Hence option C
Answer: (C) 17
Question: What is the greatest value of n such that 16^n is a factor of 18! ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: Another way to solve this question :
We know that 18 = 2 *3^2.
Hence, we need to find total number of 3s in 18! ( As out of 3 and 2 , we will have count of 3 least)
No. of 3s = 18/3 =6/3 =2.
Add the numbers in Bold, you will get total number of 3s = 8.
Hence, total number of 2^1 = 2/2 =1
Hence, answer is 1. A
Answer: (A) 1
Question: The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is?
(A) 51.4
(B) 55
(C) 56.4
(D) 52.2
(E) 60.8
Rationale: Let the required mean score be x
20*80 + 25*31 + 55*x = 52*100
1600+775+55x = 5200
x = 51.4
Answer is A
Answer: (A) 51.4
Question: The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#33?
(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
Rationale: First, let's translate the expression 21#33, using the definition given:
21#33 = 21×24×27×30×33
We need the prime factorization of this product.
Let's factor out 3 from each multiple.
21#33 = 3^6(7×8×9×10×11)
Now let's replace each consecutive integer with its prime factorization:
21#33 = 3^5(7×2^3×3^2×(2×5)×11)
Let's group the prime bases:
21#33 = 2^4×3^7×5×7×11
The sum of the exponents is 4 + 7 + 1 + 1 + 1 = 14
The answer is C.
Answer: (C) 14
Question: In a class, the average age of 40 boys is 15 years and the average of 30 girls is 13 years. what is the average age of the whole class?
(A) 11 Years
(B) 12.6 Years
(C) 13 Years
(D) 13.5 Years
(E) 14.1 Years
Rationale: Total age of 70 students
(40X 15+30 X 13) = 990
Average = 990/70 = 14.1 Years
E)
Answer: (E) 14.1 Years
Question: In how many ways can 3 postcards can be posted in 5 postboxes?
(A) 100
(B) 150
(C) 175
(D) 200
(E) 125
Rationale: First card can go into any of the five boxes,second can go into any of the five boxes.
Third card also can go into any of the five boxes
Number of ways=5*5*5=125 ways
Option E
Answer: (E) 125
Question: What is the equation of a circle of radius 8 units centered at (3, 2)?
(A) x^2 + y^2 + 6x – 4y = 23
(B) x^2 + y^2 - 6x + 4y = 23
(C) x^2 + y^2 + 6x + 4y = 23
(D) x^2 + y^2 - 6x – 4y = 51
(E) x^2 + y^2 - 6x – 4y = - 23
Rationale: the equation of a circle with center at (a,b) and radius R is (X-a)^2+(Y-b)^2=R^2
=> answer = (X-3)^2+(Y-2)^2=8^2
solving,,we get the equation in answer choice D
Answer: (D) x^2 + y^2 - 6x – 4y = 51
Question: John and Jane went out for a dinner and they ordered the same dish. Both used a 10% discount coupon. John paid a 15% tip over the original price of the dish, while Jane paid the tip over the discounted price for the coupon. If John paid $0.63 more than Jane, what was the original price of the dish?
(A) 24
(B) 34.8
(C) 37.8
(D) 42
(E) 84
Rationale: The difference between the amounts John paid and Jane paid is the deference between 15% of p and 15% of 0.9p:
0.15p - 0.15*0.9p = 0.63 --> 15p - 13.5p = 63 --> p = 42.
Answer: D.
Answer: (D) 42
Question: Two pipes A and B can fill a cistern in 20 and 10 minutes respectively, and a third pipe C can empty it in 40 minutes. How long will it take to fill the cistern if all the three are opened at the same time?
(A) 19 1/7 min
(B) 8 min
(C) 17 1/7 min
(D) 7 1/7 min
(E) 7 2/7 min
Rationale: 1/20 + 1/10 - 1/40 = 5/40
40/5 = 8
ANSWER:B
Answer: (B) 8 min
Question: If x and y are positive integers and 5 + x + y + xy = 19, what is the value of x + y?
(A) 3
(B) 5
(C) 6
(D) 8
(E) 9
Rationale: Try each answer choices.
For A: 5 + 3 + xy = 19; xy = 11 (impossible, 11 prime number. 1 + 11 doesn't equal 3)
For B: 5 + 5 + xy = 19; xy = 9 (no combination of xy = 9 and x + y = 5)
For C: 5 + 6 + xy = 19; xy = 8 (x + y = 6; x = 2, y = 4 or x = 4, y = 2)
For D: 5 + 8 + xy = 19; xy = 6 (no combination of xy = 6 and x + y = 8)
For E: 5 + 9 + xy = 19; xy = 5 (impossible, 5 prime number. 1 + 5 doesn't equal 9)
Therefore, answer C.
Answer: (C) 6
Question: George's drawer has 10 loose black socks, 14 loose blue socks, and 8 loose white socks. If George takes socks out of the drawer at random, how many would he need to take out to be sure that he had at least two matching pairs of socks in two different colors?
(A) 13
(B) 14
(C) 15
(D) 16
(E) 17
Rationale: The worst case scenario is choosing 14 blue socks. Then he would need 3 more socks to be sure of a matching pair of black socks or white socks.
The answer is E.
Answer: (E) 17