MU-NLPC/Calc-aqua_rat huggingface dataset:
AMOUNT OF BACTERIA PRESENT
Time Amount
1:00 P.M. 8.0 grams
4:00 P.M. x grams
7:00 P.M. 12.4 grams
Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.? Options:	A) 12.0	B) 12.1	C) 12.2	D) 10.0	E) 10.4

Solution: The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.   Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:   Fractional increase from 1 PM to 4 PM = X / 8.0   Fractional increase from 4 PM to 7 PM = 12.4 / X   X \ 8 = 12.4 \ X   X^2 = 12.4 * 8   X = 10   answer : 10.0
Answer: D
A man can swim in still water at 6 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?	Choose the correct choice from the following options.
A) 3.9  B) 4.2  C) 5.3  D) 2  E) 5.2

Solution: M = 5   S = x   2S = 6 + x   US = 6 - x   6 + x = (6 - x)2   6 + x = 12 -2x   3x = 6   x = 2   Answer:2
Answer: D
The residents of Town X participated in a survey to determine the number of hours per week each resident spent watching television. The distribution of the results of the survey had a mean of 24 hours and a standard deviation of 12 hours. The number of hours that Pat, a resident of Town X, watched television last week was between 1 and 2 standard deviations below the mean. Which of the following could be the number of hours that Pat watched television last week? Choose the correct choice from the following answers
A) 30	B) 20	C) 18	D) 12	E) 6

Solution: Given: mean=24 and SD=12   The number of hours that Pat watched television was between 1 and 2 standard deviations below the mean: 1 SD below the mean ismean-1*SD=12and 2 SD below the mean ismean-2*SD=0, so the number of hours that Pat watched television was between 0 and 12 hours.   Answer: 6
Answer: E
Let G represent a set of four distinct prime numbers. If the sum of the numbers in G is even and x is a member of G, then what is the least possible value that x can be?  Choose the correct choice: A) 1 B) 2 C) 3 D) 5 E) 7

Solution: We are told that G is a set of four distinct prime numbers, and the sum of the numbers in G is even. Since the question involves odds, evens and the smallest possible prime, you should immediately think of 2, the smallest prime number and the only even prime number.   If 2 were a member of the set, then we would have one even number and three odd numbers. Since such sum would always be odd, 2 cannot be a member of the set. Thus, all the members of G must be odd to yield an even sum.   Since all the members of the set must be odd prime numbers, the answer is the smallest odd prime number, which is 3.   The correct answer is choice (C).   Note:1 is not a prime number.
Answer: C
How many numbers between 11 and 90 are divisible by 7?  Pick.
A) 10  B) 7  C) 11  D) 19  E) 15

Solution: The required numbers are 14,21,28,...........,84.   This is an A.P with a=14,d=7.   Let it contain n terms   then T =84=a+(n-1)d   =14+(n-1)7   =7+7n   7n=77 =>n=11.   ANSWER 11
Answer: C
The sum of all two digit numbers divisible by 7 is Answers.
A) 910	B) 510	C) 728	D) 628	E) 810

Solution: Required numbers are 14,21,28 . . . . . 98   This is an A.P. in which a=14,d=7 and l=98.   Let the number of terms in it be n.Then t=98   So a+(n-1)d=98.   14+(n-1)*7=98,   14+7n-7=98   7+7n=98   7n=98-7= 
98-7 = 91
91/7= 
91/7 = 13
13.   Required sum=n/2(a+l)   =13/2(14+98)   =728   Answer is 728.
Answer: C
If a2 + b2 + c2 = 390 and ab + bc + ca = 5, then a + b + c is Pick
A) 19
B) 20
C) 25
D) 30
E) 64

Solution: by formula ,   (a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ca),   since, a^2 + b^2 + c^2 = 390 and ab + bc + ca = 5,   (a + b + c)^2 = 390 + 2(5)   = 400   = 20^2   therefore: a + b + c = 20   ANSWER:20
Answer: B
A & B started a partnership business. A's investment was thrice the investment of B and the period of his investment was two times the period of investments of B. If B received Rs 5000 as profit , what is their total profit?
Pick
A) 28000
B) 30000
C) 32000
D) 35000
E) None of these

Solution: Explanation :   Suppose B's investment = x. Then A's investment= 3x   Suppose Bs period of investment = y, then A's period of investment = 2y   A : B = 3x * 2y : xy   = 6 : 1   Total profit * 1/7 3  * 2= 
3  * 2 = 6
6 : 1   Total profit * 1/7= 5000   => Total profit = 5000*7= 
5000*7 = 35_000
35000. Answer : Option 35000
Answer: D
If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 6% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?  Choices
A) 2 B) 5 C) 300/53 D) 110/53 E) 9/4

Solution: Let price be 10/ticket   then 220 tickets for this month's performance at its usual price=2200   which is 10%greater than last month(let X)---->x(1+10/100)=2200----->x=2000   now new revenue this month=200*10.6= 
200*10.6 = 2_120
2120   as per question asked 2000=2120(1-A/100)----->A=300/53%   Ans 300/53
Answer: C
The radius of a cylinder is 2 r units and height is 3 r units. Find the curved surface?	Pick.
A) 862	B) 298	C) 440	D) 278	E) 278

Solution: r = 7   h = 3   2πr(h + r) = 2 * 22/7 * 7(10)   = 440   Answer: 440
Answer: C
A man buys a cycle for Rs.1400 and sells it at loss of 15%. What is the selling price of the cycle? Pick.
A) 1192	B) 1191	C) 1118	D) 1190	E) 1112

Solution: Explanation:   S.P = 85% of Rs.1400 ⇒ Rs.(8510085100 ×1400) = Rs.1190.   Answer: 1190
Answer: D
We can rent a boat for x cents per minute for the first 60 minutes and then y cents for each additional minute. How much will it cost to rent a boat for z minutes, where z>60?  Pick.
A) x + 60y
B) 60x+(z-60)y
C) x + zy
D) 60y+(z-60)x
E) 60x + zy

Solution: The first 60 minutes cost 60x.   The next (z-60) minutes cost (z-60)y.   The total cost is 60x+(z-60)y.   The answer is 60x+(z-60)y.
Answer: B
The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.4 the sixth number is?
Choose the correct choice from the following choices
A) 9.5 B) 10 C) 10.5 D) 11.5 E) 12

Solution: Explanation:   1 to 11 = 11 * 10.9= 
11 * 10.9 = 119.9
119.9   1 to 6= 6 * 10.5= 
6 * 10.5 = 63
63   6 to 11= 6 * 11.4= 
6 * 11.4 = 68.4
68.4   63 + 68.4= 
63 + 68.4 = 131.4
131.4 – 119.9= 11.5   6th number = 11.5   Option 11.5
Answer: D
4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number? Pick one:
A) 311 B) 312 C) 313 D) 314 E) 315

Solution: Let no. be x   4/15*5/7*x - 4/9*2/5*x =8   by further solving   20x/105- 8x/45=8   4x/315=8   x=630   we have to find x/2   =630/2= 
630/2 = 315
315   ANSW315R:315
Answer: E
A certain stock exchange designates each stock with a two-letter or three-letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes without using the letter A?
Answers:  A) 16,250  B) 16,500  C) 16,750  D) 17,000  E) 17,250

Solution: The number of two-letter codes is  25*25= 
 25*25 = 625
625.   The number of three-letter codes is 25*25*25=15,625.   The total number of codes is 625+15,625=16,250.   The answer is 16,250.
Answer: A
The ages of two person differ by 20 years. If 8 years ago, the elder one be 5 times as old as the younger one, their present ages(in years) are respectively
Pick one:
A) 33, 13 B) 25, 5 C) 29, 9 D) 50, 30 E) 20,10

Solution: Let their ages be x and (x + 20) years.   Then, 5(x - 8) = (x + 20 - 8)   => 4x = 52   =>x =13   Their present ages are 33 years and 13 year.   33, 13NSWER:33, 13
Answer: A
If 6 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is? Select the correct option
A) 20:15:15	B) 20:15:18	C) 20:15:12	D) 20:15:14	E) 20:15:11

Solution: 6A = 8B = 1020:15:12   A:B:20:15:12 = 1/6:1/8:1/10   = 20:15:12   Answer: 20:15:12
Answer: C
Maxwell leaves his home and walks toward Brad's house at the same time that Brad leaves his home and runs toward Maxwell's house. If the distance between their homes is 72 kilometers, Maxwell's walking speed is 6 km/h, and Brad's running speed is 12 km/h. What is the distance traveled by Maxwell when they meet in the middle? Choose one.
A) 14 B) 24 C) 16 D) 18 E) 20

Solution: Consider Max starts from Point A and Brad Starts from Point B and move towards each other.   Assume they shall meet at point O after Time 'T'. The question asks us to find OA.   From the question stem we can make out :- Distance OA = 50km - Distance O24   => 6xT = 72 - 12xT (i.e distance =Speed x Time)   => 18T = 72   Hence T = 4   OA = 6 x 4 = 24 Km   Answer : 24
Answer: B
How many prime numbers between 1 and 100 are factors of 67830?
Pick one.
A) 4
B) 5
C) 6
D) 3
E) 2

Solution: factor of 67830 = 2*3*5*7*17*19--- 6 prime numbers   6
Answer: C
What is the sum of 20 consecutive integers from -9 inclusive, in a increasing order?  Choose the correct choice
A) -9 B) 9 C) -10 D) 10 E) 20

Solution: From -9 to -1 --> 9 nos.   zero --> 1 number   from +1 to +9 --> 9 nos.   When we add up nos. from -9 to +9 sum will be zero. Total 19 nos will be added.   20th number will be 10.   Sum of these 20 nos. = 10.   10 is the answer.
Answer: D
The area of a triangle is with base 4m and height 5m? Answers.
A) 15
B) 19
C) 56
D) 10
E) 36

Solution: Explanation:   Area=1/2*b*h    1/2 * 4 * 5= 
 1/2 * 4 * 5 = 10
10 m^2   Answer : 10
Answer: D
Car A runs at the speed of 65km/hr & reaches its destination in 8hr. Car B runs at the speed of 70 km/h & reaches its destination in 4h. What is the respective ratio of distances covered by Car A & Car B? Choose the correct choice from the following:
A) 3 : 7	B) 4 : 9	C) 13 : 7	D) 5 : 7	E) 6 : 11

Solution: Sol. Distance travelled by 13 : 7ar A = 65 × 8 = 520 km   Distance travelled by 13 : 7ar B = 70 × 4 = 280 km   Ratio = 520/280 = 13 : 7   13 : 7
Answer: C
There are 8 periods in each working day of a school. In how many ways can one organize 7 subjects such that each subject is allowed at least one period?
Select.
A) 112456	B) 145678	C) 141120	D) 11456	E) None of these

Solution: 7 subjects can be arranged in 8 periods in 8P7 ways.   Remaining 1 period can be arranged in 7P1 ways.   Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.   Total number of arrangements = (8P7 x 7P1)/2! = 141120   ANSWER:141120
Answer: C
Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.1600? Choose the correct choice from the following:
A) 3000  B) 4200  C) 2025  D) 1205  E) 3251

Solution: 1800* 9/8 * 9/8   = 2025   Answer: 2025
Answer: C
The greatest ratio out of 2:3, 5:8, 3:1, 8:9 and 8:5 is? Select the correct option
A) 2:3
B) 3:1
C) 5:8
D) 8:5
E) 8:9

Solution: 2/3=0.666   5/8= 
5/8 = 5/8 = around 0.625
0.625   3/1= 
3/1 = 3
3   8/9=0.888   8/5= 
8/5 = 8/5 = around 1.6
1.6   ANSWER:3:1
Answer: B
A fruit shake costs R 2,00 more than a chocolate milk shake. If 3 fruit shakes and 5 chocolate milk shakes cost R 78,00, determine the individual prices. Choose one
A) x=6,y=9	B) x=4,y=4	C) x=7,y=5	D) x=5,y=11	E) x=9,y=11

Solution: Let the price of a chocolate milkshake be x and let the price of a fruitshake be y.   Set up a system of equations   y=x+2.........(1)   3y+5x=78............(2)   Substitute equation (1) into equation (2)   3(x+2)+5x=78   Rearrange and solve for x   3x+6+5x=78   8x=72   x=9   Substitute the value of x back into equation (1) and solve for y   y=x+2   =9+2= 
9+2 = 11
11   One chocolate milkshake costs R 9,00 and one fruitshake costs R 11,00.   Answer is x=9,y=11.
Answer: E
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 17?
Answers:  A) 0  B) 1  C) 2  D) 3  E) 4

Solution: Easier to start with numbers that are of the form 17p+1 ---> 18,35,52,69,86.   Out of these, there is only one (69) is also of the form 4q+1. Thus 1 is the answer. 1 is the correct answer.
Answer: B
In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag- the least chance of drawing a red ball ?
Choose the correct choice from the following options:  A) 1/32  B) 2/32  C) 3/32  D) 4/32  E) 5/32

Solution: Least Chance.   P= 1/2 * 5/16 + 1/2 * 0 = 5/32   ANSW5/32R:5/32
Answer: E
A man sells two articles for Rs.3600 each and he gains 35% on the first and loses 35% on the next. Find his total gain or loss?	Choose the correct choice from the following
A) 9% loss B) 400 C) 12.25%loss D) 324 E) 8% LOSS

Solution: (35*35)/100= 
(35*35)/100 = 49/4 = around 12.25
12.25%loss   ANSWER:12.25%loss
Answer: C
Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?	Answers:
A) Rs. 8600
B) Rs. 8620
C) Rs. 8820
D) Rs. 8000
E) None of these

Solution: Explanation:   Amount   = Rs.[8000x(1+5/100)2]   =Rs.(8000x21/20x21/20)   =Rs.8820.   ANSWER IS Rs. 8820
Answer: C
There are two box,one containing 39 red balls and the other containing 26 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is Choose the most appropriate option.
A) .60	B) .50	C) .70	D) .80	E) .90

Solution: Here we are allowed to transfer the balls between the two boxes, so we keep only one red ball in first box and move all the remaining balls to the second box, so that we get maximum probability of drawing red ball from first box   So first box--- only one red ball   second box--- 38 red and 26 green   Probability of choosing any box is 1/ 2.   probability of choosing the red ball from first box is 1   probability of choosing the red ball from second box is 38/64   So probability of taking one red ball = [(1/2)×(1)]+[(1/2)*(38/64)]≃0.8   ANSWER:.80
Answer: D
If a car went the first third of the distance at 80 kmh, the second third at 30 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip? Choose the correct choice.
A) 36 kmh
B) 40 kmh
C) 42 kmh
D) 45 kmh
E) 50 kmh

Solution: Assume 45 kmh/3 = 240 (this number is convenient because it is divisible by 80, 30 and 48)   So:   240 = 80*T1 = 3 hrs   240 = 30*T2 = 8 hrs   240 = 48*T3 = 5 hrs   T = T1 + T2 + T3 = 16 hrs   45 kmh = RT   (240*3) = R*16   R = 45   ANSWER: 45 kmh
Answer: D
Calculate the speed of a stream if a boat going upstream at
4km/hr and downstream at 8km/hr	Choose the correct choice from the following answers:
A) 1 kmph  B) 2 kmph  C) 3 kmph  D) 6 kmph  E) 8 kmph

Solution: Sol.   Speed of stream = 1/2(8 -4) kmph = 2 kmph.   Answer 2 kmph
Answer: B
A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction R of the cookies that were placed in the other tins were placed in the green tin?
Choose the most appropriate option
A) 15/2
B) 9/4
C) 5/9
D) 7/5
E) 9/7

Solution: Blue tin or red tin: 2/3 (N)   Red tin: (1/3)N   Blue tin: (1/4)N   What the last statment meant, is it wants this fraction: (# of cookies in green tin)/(# of cookies in red and green tin)   # of cookies in green tin = 2N/3 - N/4 = 8N- 3N/12 = 5N/12   # of cookies in red and green tin = N/3 + 5N/12 = 9N/12   Fraction R= 5N/12 * 12/9N = 5/9 (5/9)
Answer: C
Which of the following is closest to the difference between sum Q of all proper fractions (fractions less than 1) in the form 1/x , where x is a positive digit, and the product of all proper fractions in the form y/(y+1), where y is a positive digit?  Select the correct option:	A) 2.82	B) 2.72	C) 1.82	D) 1.72	E) 0.82

Solution: Sum Q of all proper fractions (fractions less than 1) in the form 1/x, where x is a positive digit:   1/1 + 1/2 + 1/3 +.....+ 1/9   This is a harmonic progression. Harmonic progression is inverse of arithmetic progression.   Approximate sum of a harmonic progression with even number of terms = number of terms * (average of middle 2 terms)   Approximate sum of a harmonic progression with odd number of terms = number of terms * (middle term)   The actual sum will be slightly more than the approximation.   Here we have 9 terms (odd).   Sum = 9 * 1/5   = 9/5   Product of all proper fractions in the form y/(y+1), where y is a positive digit:   1/2 * 2/3 * 3/4 *.....*9/10   We will be left with 1/10.   Required = 9/5 - 1/10   = 1.8 - 0.1= 
1.8 - 0.1 = 1.7
1.7   Closest is 1.72   Answer (1.72).
Answer: D
A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat?  Choose the correct option.
A) 400 m	B) 500 m	C) 250 m	D) 100 m	E) 150m

Solution: Explanation:   A can give B (5 min - 4 1/2 min) = 30 sec start.   The distance covered by B in 5 min = 1000 m.   Distance covered in 30 sec = (1000 * 30)/300= 
(1000 * 30)/300 = 100
100 m.   A can give B 100m start.   ANSWER 100 m
Answer: D
If x > 3000, then the value of (3x)/(2 x^1.21 - 1111^1.3) is closest to?  Select.
A) 1/6 B) 1/3 C) 10/21 D) 4/10 E) 3/2

Solution: assume x = 3002   (3x)/(2 x^1.21 - 1111^1.3) = 9006 / (2 * (3002)^1.21 - 1111^1.3)   =9006 / 23154   = = 4 /10   4/10
Answer: D
Q. How many 9-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6, 6, 7? Select:  A) 180  B) 140  C) 210  D) 270  E) 300

Solution: the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2   i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _   case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6   when the units digit is 2 i.e 1 _ _ _ _ _ 2   number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60   when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30   case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)   number of ways this can be done would be 5! / (2!) for 2 =60   and 5! / 2! for 6 ... = 60   adding all of these ... gives the answer 140 ...   the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat   140
Answer: B
Find the 100th place digit of a number forming from digits 1 to 9 without repeatation and number passes the divisibility test of 11 means difference of sum of alternative digits is 0 or divisible by 11.	Select
A) 1  B) 2  C) 3  D) 4  E) 5

Solution: Largest no is 987652413   So digit will be 4   ANSWER:4
Answer: D
n is a whole number which when divided by 4 gives 3 as remainder. What will be the remainder when 2*n is divided by 4 ?
Choose the correct answer: A) 8 B) 2 C) 6 D) 7 E) 1

Solution: Explanation:   Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.   Thus when 2*n is divided by 4, the reminder is 2.   Answer: 2) 2
Answer: B
A candidate got 30% of the votes polled and he lost to his rival by 4000 votes. How many votes were cast? Choose the correct choice from the following answers
A) 7500 B) 2028 C) 2775 D) 10000 E) 6851

Solution: 30%-----------L   70%-----------W   ------------------   40%----------4000   100%---------? => 10000   Answer:10000
Answer: D
The average of first 10 odd numbers is?  Choose the correct choice from the following options
A) 12
B) 10
C) 87
D) 23
E) 34

Solution: Sum of 10 odd no.   = 100   Average = 100/10= 
100/10 = 10
10   Answer:10
Answer: B
A cricketer scored 138 runs which included 12 boundaries and 2 sixes. What percent of his total score did he make by running between the wickets.  Choose the correct choice from the following answers:
A) 46.2%
B) 54.54 %
C) 63%
D) 56.52%
E) None of these

Solution: Explanation :   Number of runs made by running = 138 - (12 x 4 + 2 x 6)   = 138 - (60)= 
138 - (60) = 78
78   Now, we need to calculate 72 is what percent of 138.=> 78/138 * 100 = 56.52 %   Answer : 56.52%
Answer: D
If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it ?	Choose the correct choice from the following answers
A) 10 hours  B) 40 hours  C) 20 hours  D) 60 hours  E) None of them

Solution: Let the required number of hours be x. Then,   Less men, More hours	(Indirect Proportion)   15 : 36 : : 25 : x = (15 x x) = (36 x 25) = (36 x 25)/15 = 60   Hence, 15 men can do it in 60 hours.   Answer is 60 hours.
Answer: D
Calculate the value of q from the below equation: q^3âˆ’25q+5=173
Choose the correct answer
A) 7
B) 4
C) 6
D) 9
E) 8

Solution: Use elimination method to find the correct option. you find that of all the options 7 is the correct value for q   7nswer: 7
Answer: A
If 45% of z is 90% of y and y is 75% of x, what percent of x is z?
Select the correct option:
A) 200
B) 160
C) 100
D) 65
E) 150

Solution: (45/100)z = (90/100)y   and y = (75/100)x i.e. y = (3/4)x   i.e. (45/100)z = (90/100)*(3/4)x   i.e. z = (90*3)x/(45*4)   i.e. z = (1.5)x = (150/100)x   i.e. z is 150% of x   Answer: Option 150
Answer: E
The area of a rectangular field is equal to 200 square meters. Its perimeter is equal to 60 meters. Find the width of this rectangle.	Choose one:  A) 5  B) 10  C) 15  D) 20  E) 25

Solution: L * W = 200 : area , L is the length and W is the width.   2 L + 2 W = 60 : perimeter   L = 30 - w : solve for L   (30 - W) * W = 200 : substitute in the area equation   W = 10 and L = 20 :   correct answer 10
Answer: B
417 x 417 + 383 x 383=?  Choices:	A) 320578	B) 80578	C) 80698	D) 81268	E) None of them

Solution: =(417)^2 + (383)^2   =(400+17)^2 + (400-17)^2   =2[(400)^2+(17)^2]   =2[160000+289]   =2 x 160289 = 320578   320578nswer is 320578
Answer: A
The average weight of a group of boys is 30 kg. After a boy of weight 38 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?  Choose one:	A) A)4	B) B)8	C) C)6	D) D)2	E) E)7

Solution: Let the number off boys in the group originally be x.   Total weight of the boys = 30x   After the boy weighing 38 kg joins the group, total weight of boys = 30x + 38   So 30x + 38 = 31(x + 1) = > x = 7.   Answer:7
Answer: E
Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negative integers. What is the value of (a + b) + 5, when a # b = 100?	Pick:
A) 5	B) 8	C) 10	D) 13	E) 17

Solution: OFFICIAL SOLUTION:   (B) We know that a # b = 100 and a # b = 4a² + 4b² + 8ab. So   4a² + 4b² + 8ab = 100   We can see that 4a² + 4b² + 8ab is a well-known formula for (2a + 2b)². Therefore   (2a + 2b)² = 100.   (2a + 2b) is non-negative number, since both a and b are non-negative numbers. So we can conclude that 2(a + b) = 10. (a + b) + 5 = 10/2 + 5= 
10/2 + 5 = 10
10.   The correct answer is 10.
Answer: C
If the median of a list of numbers is m, the first quartile of the list is the median of the numbers in the list that are less than m. What is the first quartile of the list of numbers 42, 24, 30, 28, 26, 19, 33 and 35 ? Choose the most appropriate option:  A) 33  B) 28  C) 27  D) 25  E) 23

Solution: It is given that a quartile is the middle number of all numbers less than median..   so lets arrange the number in ascending order- 42, 24, 30, 28, 26, 19, 33 and 35   19, 24, 26, 28, 30 , 33, 35, 42...   numbers less than median are 19, 24, 26, 28..   the median of these numbers = center of 24 and 26 = 25   25
Answer: D