MU-NLPC/Calc-aqua_rat huggingface dataset:
It rained as much as on Wednesday as on all the other days of the week combined. If the average rainfall for the whole week was 3.5 cms, How much did it rain on Wednesday?
Choices.
A) 10.9
B) 10.5
C) 12.25
D) 10.1
E) 10.6

Solution: Explanation:   Let the rainfall on wednesday = 6x.   âˆ´ Rainfall on the remaining days = 6x   Given,   (6x + 6x )/7 = 3.5   â‡’12x = 24.5   â‡’6x = 12.25   Answer: 12.25
Answer: C
Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?	Answers
A) 270
B) 288
C) 266
D) 299
E) 126

Solution: Relative speed = (72 - 36) * 5/18 = 2 * 5= 
2 * 5 = 10
10 mps.   Distance covered in 27 sec= 27 * 10= 
27 * 10 = 270
270 m.   The length of the faster train= 270 m.   270nswer: 270
Answer: A
How many positive integers less than 22 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3?  Choose the correct choice from the following.
A) 14  B) 17  C) 12  D) 11  E) 10

Solution: The number = 2a + 3b < 20   When a = 1, b = 1, 2, 3, 4, 5, 6 -> 2a = 2; 3b = 3, 6, 9, 12, 15, 18 -> the number = 5, 8, 11, 14, 17, 20 --> 6 numbers   when a =2, b = 1,2,3,4,5, 6 -> ....--> 6 numbers   when a =3, b = 1,2,3,4, 5 --> ....--> 5 numbers   Total number is already 17. Look at the answer there is no number greater than 17 --> we dont need to try any more   Answer must be 17
Answer: B
a school has 11 maths 8 physics and 5 chemistry teachers each teacher can teach 3 subjects max what is he minimum number of teachers required	Select the correct option.
A) 2	B) 4	C) 6	D) 8	E) 9

Solution: Total subjects = 11+8+5= 
11+8+5 = 24
24   Max subjects by 1 teacher= 3   So, min of teachers required = 24/3= 
24/3 = 8
8   ANSWER:8
Answer: D
Find the odd man out. 16, 36, 121, 152, 169, 225, 289, 324	Pick one:	A) 324	B) 289	C) 225	D) 169	E) 152

Solution: 152xplanation :   152ach of the given numbers except 152 is a perfect square   Answer : Option 152
Answer: E
Harish can paint a wall in just 3 hours. However it takes 6 hours for Ganpat to complete the same job of painting the wall.
How long would it take for both of them to paint the wall , if both work together ?
Choose the correct choice from the following choices:  A) 3 Hours  B) 1 Hours  C) 3.20 Hours  D) 2 Hours  E) 1.35 Hours

Solution: D   Two hours.   Harish who could paint the wall in 3 hours could paint 2/3 of the wall in just two hours. Ganpat who paint the wall in 6 hours could paint 1/3 of the wall in two hours.   2/3 + 1/3= 
2/3 + 1/3 = 1
1.
Answer: D
If 2 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?	Choose the correct option:
A) -7	B) -4	C) -5	D) 1	E) 6

Solution: The phrase “2 is one solution of the equation” means that one value of x is 2. Thus, we first must plug 2 for x into the given equation to determine the value of k. So we have   2^2 + (3)(2) + k = 10   4 + 6 + k = 10   10 + k = 10   k =0   Next we plug 0 into the given equation for k and then solve for x.   x^2 + 3x = 10   x^2 + 3x – 10 = 0   (x+5)(x-2) = 0   x = -5 or x = 2   Thus, -5 is the other solution. Answer -5.
Answer: C
Suzie’s Discount Footwear sells all pairs of shoes for one price and all pairs of boots for another price. On Monday the store sold 22 pairs of shoes and 16 pairs of boots for $650. On Tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $760. How much more do pairs of boots cost than pairs of shoes at Suzie’s Discount Footwear?
Choose the correct choice from the following.
A) $2.50
B) $5.00
C) $5.50
D) $7.50
E) $15.00

Solution: Let X be pair of shoes and Y be pair of boots.   22x+16y = 650 ... eq1   8x +32y = 760....eq 2.   Now multiply eq 1 by 2 and sub eq 2.   44x = 1300   8x = 760.   36x = 540 => x = 15.   Sub x in eq 2.... we get 120 + 32y = 760... then we get 32y = 640 then y = 20   Difference between X and Y is 5.   Answer: option $5.00 is correct answer.
Answer: B
A triangle has a perimeter 16. The two shorter sides have integer lengths equal to x and x + 1. Which of the following could be the length of the other side?  Select:  A) 2  B) 4  C) 6  D) 7  E) 8

Solution: The SHORTER sides have integral lengths equal to x and x + 1   Let the longest side be 'a'   So, a + x + (x +1) = 16   a + 2x = 15.......eqn (1)   We know that the sum of the lengths of the shorter sides has to be more than the length of the longer one. i.e 2x+1> a   a =6   6
Answer: C
Larry and Daniel are partners in a business, they invested Rs. 50,000 and Rs. 39,000 respectively.If the business yielded a total profit of Rs. 21,000, what is Daniel's share?	Choose the correct choice:
A) 9402.25
B) 9272.25
C) 9982.25
D) 9211.25
E) 9202.25

Solution: Explanation:   Ratio of their shares = 50,000 : 39,000 = 50 : 39.   Daniel's share = Rs. (21000 * 39/89) = Rs. 9202.25.   Answer: 9202.25) 9202.25
Answer: E
If a·b·c·d=360, where a, b, c and d are positive integers, and a<b<c<d, which of the following could be the value of d−a? Choose the correct option.
A) 9 B) 10 C) 11 D) 13 E) 15

Solution: prime factorize : 360 = 2*3*5*12   Hence a = 2 and d = 12 .   Hence answer is  12 - 2= 
 12 - 2 = 10
10   Answer 10
Answer: B
The maximum number of students among them 1008 pens and 928 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils?	Choose the correct choice:
A) 16 B) 18 C) 36 D) 19 E) 32

Solution: number of pens = 1008   number of pencils = 928   Required number of students = H.C.F. of 1008 and 928 = 16   16nswer is 16
Answer: A
Every student in a room is either a junior or a senior. There is at least one junior and at least one senior in the room. If 1/2 of the juniors is equal to 2/3 of the seniors, what fraction of the students in the room are juniors? Choose the correct choice from the following
A) 4/7  B) 1/3  C) 5/3  D) 12/7  E) 17/20

Solution: Let total number of juniors= J   total number of seniors =S   (1/2) J = (2/3) S   => S = 3/4 J   Total number of students = J+S = (7/4) J   Fraction of the students in the room are juniors = J/(J+S) = J/[(7/4) J]   =4/7   4/7nswer 4/7
Answer: A
For any non-zero a and b that satisfy |ab| = ab and |a| = -a, |b-5| + |ab+b| = Choose the correct choice from the following options.
A) ab-4	B) 2b-ab-4	C) ab+4	D) ab-2b+4	E) 4+2ab

Solution: Given: |ab| = ab and |a| = -a   Question: |b-5| + |ab+b| = ?   **** Looking at |ab| = ab tells us that a and b are either both positive or negative   **** Looking at |a| = -a tells us that a must be negative   **** Combine two observations: a and b are both negative values   Let a=-1 and b=-1   |b-5| + |ab+b| = |-1-5| + |1+(-1)| = 6   Test a) ab-4 = (-1)(-1)-4 = -3   Test b) 2b-ab-4 = (2)(-1) - (1) - 4 = -7   Test c) ab+4 = 1 + 4= 
1 + 4 = 5
5   Test d) ab-2b+4= 1-2(-1)+4=7   test e) 4+2ab= 4+2(-1)(-1)= 6   Answer: 4+2ab
Answer: E
Rani bought more apples than oranges. She sells apples at ₹23 apiece and makes 15% profit. She sells oranges at ₹10 apiece and makes 25% profit. If she gets ₹653 after selling all the apples and oranges, find her profit percentage.  Choose the correct choice from the following answers:
A) 16.8%
B) 17.4%
C) 17.9%
D) 18.5%
E) 19.1%

Solution: Given: Selling price of an Apple = 23 --> Cost price = 23/1.15= 
23/1.15 = 20
20   Selling price of an orange= 10 --> Cost price = 10/1.25= 
10/1.25 = 8
8   A > O   23*(A) + 10*(O)= 653   653 - 23*(A) has to be divisible by 10 --> Units digit has to be 0   Values of A can be 1, 11, 21, 31, .... --> 1 cannot be the value   17.4%etween 11 and 21, If A = 11, O = 30 --> Not possible   If A = 21, O = 17 --> Possible   Cost price = 20*21 + 8*17 = 420 + 136= 
420 + 136 = 556
556   Profit= 653 - 556= 
653 - 556 = 97
97   Profit%= (97/556)*100 = 17.4%   Answer: 17.4%
Answer: B
The C.P of 10 pens is equal to the S.P of 12 pens. Find his gain % or loss%? Choose the correct option
A) 16 2/7%	B) 16 6/3%	C) 16 2/3%	D) 16 1/3%	E) 12 2/3%

Solution: 10 16 2/3%P = 12 SP   12 --- 2 16 2/3%P loss   100 --- ? => 16 2/3%   Answer: 16 2/3%
Answer: C
Jen has a jar of jam, and she ate 1/3 of the jam for lunch. If Jen ate 1/7 of the remaining jam for dinner, what fraction of the jam was left after dinner?	Choose the correct choice from the following options:
A) 2/7
B) 3/7
C) 4/7
D) 9/14
E) 11/14

Solution: Let x be the amount of jam at the beginning.   After lunch, the remaining jam was (2/3)x.   After dinner, the remaining jam was (6/7)(2/3)x = (4/7)x.   The answer is 4/7.
Answer: C
At what price must an article costing Rs.47.50 be marked in order that after deducting 5% from the list price. It may be sold at a profit of 25% on the cost price?
Choose the correct choice from the following answers:
A) 62.5
B) 62.6
C) 62.1
D) 62.7
E) 62.2

Solution: CP = 47.50   SP = 47.50*(125/100)= 
47.50*(125/100) = 59.375
59.375   MP*(95/100)= 59.375   MP = 62.5   62.5nswer: 62.5
Answer: A
In a two digit number, the digit in the units place is two times the digit in ten's place and sum of the digits is equal to 12. What is the number? Choose the most appropriate option:
A) 25 B) 28 C) 32 D) 36 E) 48

Solution: Solution   Let the ten digits be x. Then,units digit=2x   Therefore x + 2x = 12   ‹=›3x = 12   ‹=›x= 4.   So, ten's digit = 4, unit's digit = 8.   Hence, the required number is 48.   Answer 48
Answer: E
Two whole numbers whose sum is 40 cannot be in the ratio	Options.
A) 4:10	B) 2:3	C) 1:3	D) 2:4	E) 3:5

Solution: 2:4) 2:4
Answer: D
The integers from 1 to 1851 are entered sequentially in 12 columns (numbered 1 to 12); i.e. Row 1 contains the numbers 1 to 12, Row 2 contains the next 12 numbers and so on. If the number 1369 occurs in row p and column q, then the value of p + q is	Pick:
A) 114  B) 115  C) 116  D) 117  E) 118

Solution: 1369=114*12+1   1368=114*12 mean 1368 occurs in 114 row and 12 column.   so 1369 occur in 115 row and 1st column.as elements are entered sequentially.   so p=115 q=1.p+q=116   ANSWER:116
Answer: C
For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *6 + 3 and *6 + 6, inclusive?  Choose the correct choice from the following answers:
A) One
B) None
C) Two
D) Three
E) Four

Solution: Generally *p or p! will be divisible by ALL numbers from 1 to p. Therefore, *6 would be divisible by all numbers from 1 to 6.   => *6+3 would give me a number which is a multiple of 3 and therefore divisible (since *6 is divisible by 3)   In fact adding anyprimenumber between 1 to 6 to *6 will definitely be divisible.   So the answer is none (A)!   Supposing if the question had asked for prime numbers between *6 + 3 and *6 + 11 then the answer would be 1. For *6 +3 and *6+ 13, it is 2 and so on...   None
Answer: B
A man swims downstream 51 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water?
Select: A) 2 B) 8.5 C) 9.5 D) 6.5 E) 11.5

Solution: 51 --- 3 DS = 17   ? ---- 1   18 ---- 3 US = 6   ? ---- 1 M = ?   M = (17 + 6)/2= 
(17 + 6)/2 = 23/2 = around 11.5
11.5   Answer: 11.5
Answer: E
Convert 400 miles into meters?  Choices:
A) 784596
B) 845796
C) 804670
D) 784596
E) 643736

Solution: 1 mile = 1609.34 meters   400mile = 400*1609.34= 
400*1609.34 = 643_736
643736 meters   Answer is 643736
Answer: E
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream?	Choose the correct choice from the following.
A) 3:7  B) 3:6  C) 3:1  D) 3:2  E) 3:5

Solution: The ratio of the times taken is 2:1.   The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1)= 
(2+1)/(2-1) = 3
3/1= 
3/1 = 3
3:1   Speed of the stream= 42/3= 
42/3 = 14
14 kmph.Answer:3:1
Answer: C
Rani started a business investing Rs. 10,000. Roja joined him after six months with an amount of Rs. 35,000 and Sachin joined them with Rs. 70,000 after another six months. The amount of profit earned should be distributed in what ratio among Rani, Roja and Sachin respectively, 3 years after started the business ?
Pick.
A) 7:6:10  B) 12:15:16  C) 42:45:56  D) None of these  E) 12 : 35 : 56

Solution: 12 : 35 : 56xplanation :   Akash : Roja : Sachin = (10,000 x 36) : (35,000 x 30) : (70,000 x 24)   = 360 : 1050 : 1680 => 12 : 35 : 56   Answer : 12 : 35 : 56
Answer: E
How many different 6-letter sequences are there that consist of 1A, 20B's and 15C's ?
Choose the correct option:
A) 6 B) 60 C) 120 D) 360 E) 720

Solution: How many different 6-letter sequences are there that consist of 1 A, 20B's, and 15 C's?   (A) 6   (B) 60   (C) 120   (360) 360   (E) 720 (ANS 360)
Answer: D
In an examination 60% failed in Math and 40% failed in French. If 10% failed in both. What % of students passed in both.
Select the correct option.
A) 25%
B) 15%
C) 20%
D) 30%
E) 10%

Solution: Passed in Math = 100 - 60= 
100 - 60 = 40
40%   Passed in French= 100 - 40= 
100 - 40 = 60
60%   Students who passed Math or French= 100 -10= 
100 -10 = 90
90%   Students passed in Math and French= (Passed in Math) + (Passed in French) - (Students who passed Math or French) = 60+40-90= 
60+40-90 = 10
10%   ANSW10%R:10%
Answer: E
Which of the following is(are) true for a@b=b@a?
I. a@b=ab+ba
II. a@b=(a+b)(a-b-1)
III. a@b=(a/b)+(b/a) Choose the correct choice from the following.
A) only Ⅰ	B) only Ⅱ	C) only Ⅲ	D) only ⅠⅢ	E) only ⅡⅢ

Solution: -> If I. a@b=ab+ba, a@b=ab+ba=ba+ab=b@a (O).   If II. a@b=(a+b)(a-b), a@b=(a+b)(a-b-1)≠(b+a)/(b-a)=b@a (O).   If III.a@b=(a/b)+(b/a), a@b=(a/b)+(b/a)=(b/a)+(a/b)=b@a (X).   Thus, ⅠⅢ are the answers. Therefore, only ⅡⅢ is the answer.
Answer: E
The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas. Choose the correct choice.
A) 3:7  B) 2:7  C) 3:5  D) 5:2  E) 4:25

Solution: Let the diagonals of the squares be 2x and 5x respectively.   Ratio of their areas = (1/2)*(2x) 2:(1/2)*(5x) 2   = 4x2: 25x2   = 4 : 25.   Ans: 4:25
Answer: E
A volume of 10976 l water is in a container of sphere. How many hemisphere of volume 4l each will be required to transfer all the water into the small hemispheres? Choose the correct choice from the following choices.
A) 2812 B) 8231 C) 2734 D) 2744 E) 4254

Solution: a volume of 4l can be kept in 1 hemisphere   therefore, a volume of 10976l can be kept in (10976/4) hemispheres   ans.2744   ANSWER:2744
Answer: D
576, 529, 484, 441, 400, ?	Options
A) 324	B) 361	C) 289	D) 256	E) 225

Solution: Explanation :   The pattern is 24^2, 23^2, 22^2, 21^2, 20^2, ...   So next number is 19^2 = 361   Answer : Option 361
Answer: B
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?  Choices.
A) 9:1	B) 9:8	C) 9:0	D) 9:5	E) 9:2

Solution: Explanation:   Milk = 3/5 * 20= 
3/5 * 20 = 12
12 liters, water= 8 liters   If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.   Remaining milk = 12 - 6= 
12 - 6 = 6
6 liters   Remaining water= 8 - 4= 
8 - 4 = 4
4 liters   10 liters of pure milk are added, therefore total milk= (6 + 10)= 
(6 + 10) = 16
16 liters.   The ratio of milk and water in the new mixture= 16:4 = 4:1   If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10= 
4/5 * 10 = 8
8 liters.   Amount of water removed= 2 liters.   9:1nswer: Option 9:1
Answer: A
In how many ways can 5 Indians and 5 Englishmen be seated along a circle so that they are alternate?
Choose one
A) 6!8!  B) 4!5!  C) 3!9!  D) 8!5!  E) 7!9!

Solution: 5 Indians can be seated along a circle in 4! ways   Now there are 5 places for 5 Englishmen   5 Englishmen can be seated in 5! ways   Required number = 4!5!   Answer is 4!5!
Answer: B
The sum of the first 50 positive even integers is 2,000. What is the sum of the odd integers from 101 to 200, inclusive?  Answers:
A) 5,050	B) 7,500	C) 8,050	D) 15,000	E) 19,600

Solution: 101+103+.......199   If we remove 100 from each of these it will be sum of 1st 100 odd numbers.   so 101+103+.......199 = 50 * 100 + (1+3+5+7+......)   sum of 1st 100 natural numbers = (100 * 101) / 2= 
(100 * 101) / 2 = 5_050
5050   Sum of 1st 50 positive even integers= 2000   sum of 1st 100 odd numbers = 5050 - 2000= 
5050 - 2000 = 3_050
3050   so 101+103+.......199= 50 * 100 + (1+3+5+7+......) = 5000 + 3050= 
5000 + 3050 = 8_050
8050   8,050 is the answer.
Answer: C
The tax on a commodity is diminished by 20% but its consumption is increased by 10%. Find the decrease percent in the revenue derived from it?
Choose the correct choice from the following choices:
A) 20%	B) 18%	C) 15%	D) 12%	E) 10%

Solution: Answer: Option 12%    100 * 100= 
 100 * 100 = 10_000
10000   80 * 110= 
80 * 110 = 8_800
8800   10000------- 1200   100 ------- ?= 12%
Answer: D
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 40% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of	Choose the most appropriate option
A) 18%  B) 20%  C) 21%  D) 33%  E) Cannot be determined

Solution: Answer: 33%.
Answer: D
What is the total number of integers between 100 and 500(exclusive) that are divisible by 5? Pick one.
A) 51
B) 63
C) 79
D) 66
E) 42

Solution: 105, 110, 115, ..., 490,495   This is an equally spaced list; you can use the formula:   n = (largest - smallest) / ('space') + 1 = (495 - 105) / (5) + 1 = 78 + 1= 
78 + 1 = 79
79   Answer is 79
Answer: C
Rs.770 were divided among A,B,C in such a way that A had Rs.40 more than B and C had Rs 30 more than A . How much was Bâ€™s share?	Choose the correct answer
A) Rs.260	B) Rs.270	C) Rs.280	D) Rs.290	E) Rs.300

Solution: Let B gets Rs x. Then We can say A gets Rs (x + 40 ) and C gets Rs ( x + 70) .   x + 40 + x + x + 70 = 770   3x = 660   x = 220 .   Bâ€™s share = Rs ( 220 + 70 ) = Rs.290   Rs.290
Answer: D
A student gets 60% in one subject, 70% in the other. To get an overall of 70% how much should get in third subject. Select.
A) 75%	B) 80%	C) 45%	D) 55%	E) 65%

Solution: Let the 3rd subject % = x   60+70+x = 3*70   130+x =210   x = 210-130= 
210-130 = 80
80   Answer :80%
Answer: B
In how many ways can the letters L, I, M, I, T, S be arranged so that the two I's are not next to each other?	Choose one
A) 36
B) 48
C) 240
D) 296
E) 428

Solution: 1 L   2 I   1 M   1 T   1 S   Number of ways these letters can be arranged = 6!/2! (2! to account 2 same Is) = 360   240onsider 2 Is as 1 entity and thus the number of arrangements for (II)LMTS = 5! = 120   Total allowed cases = 360-120= 
360-120 = 240
240   240 is the correct answer.
Answer: C
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? Choose the correct choice.
A) 3.46%
B) 3.96%
C) 3.49%
D) 4.46%
E) 3.66%

Solution: Explanation:   Let the original rate be R%. Then, new rate = (2R)%.   Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.1/3 year(s).   (2175 + 725) R = 33.50 x 100 x 3   (2175 + 725) R = 10050   (2900)R = 10050   Original rate = 3.46%   3.46%nswer: 3.46%) 3.46%
Answer: A
Solve the given two equations and find the inequality of X and Y. (a) x^2 = 729 and (b) Y= (729)^(1/2)
Choose one:
A) If x>y  B) If x>=y  C) If x<y  D) If x<=y  E) x=y or relationship cannot be established

Solution: Solution:-   X=+27, -27 and Y= +27   Comparing we get X<=Y   If x<=y
Answer: D
A light has a rating of 80 watts, it is replaced with a new light that has 25% higher wattage. How many watts does the new light have?
Choose one
A) 105
B) 95
C) 80
D) 60
E) 100

Solution: Final number = Initial number + 25%(original number) = 80 + 25%(80) = 80 + 20= 
80 + 20 = 100
100.   Answer 100
Answer: E
A box contains 25 mangoes out of which 5 are spoilt. If five mangoes are chosen at random, find the probability that A number is selected at random from first fifty natural numbers. What is the chance that it is a multiple of either 7 or 9? Choose the correct choice.
A) 17/25	B) 1/25	C) 6/25	D) 7/25	E) 9/25

Solution: The probability that the number is a multiple of 7 is 7/50. (Since 7*7= 
7*7 = 49
49).   Similarly the probability that the number is a multiple of 9 is 5/50. {Since   9*5= 
  9*5 = 45
45).   Neither 7 nor 9 has common multiple from 1 to 50. Hence these events are mutually exclusive events. Therefore chance that the selected number is a multiple of 7 or 9 is (7+5)/50= 6/25.   ANSWER:6/25
Answer: C
The average speed of a car decreased by 3 miles per hour every successive 8-minutes interval. If the car traveled 3.6 miles in the seventh 8-minute interval, what was the average speed of the car, in miles per hour, in the first 8 minute interval?  Answers.
A) 33
B) 36
C) 39
D) 42
E) 45

Solution: (3.6 miles / 8 minutes)*60 minutes/hour = 27 mph   Let x be the original speed.   x - 6(3) = 27   x = 45 mph   The answer is 45.
Answer: E
7 members have to be selected from 12 men and 3 women ,Such that no two women can come together .In how many ways we can select them ?
Options
A) 12c6*3c1 + 12c7	B) 12c6*3c1 + 2c7	C) 1c6*3c1 + 12c7	D) 11c6*3c1 + 12c7	E) 12c6*3c1 + 11c7

Solution: since no two women can come together,therfore women can be selected in 3c1 ways ...and men will be chosen in 12c6 so to complete the 7 group members...   =12c6*3c1   and only selection of men also possible = 12c7   so final ans : 12c6*3c1 + 12c7   12c6*3c1 + 12c7NSWER:12c6*3c1 + 12c7
Answer: A
In a class, the average age of 30 boys is 13 years and the average of 20 girls is 12 years. what is the average age of the whole class? Options
A) 12 Years  B) 12.6 Years  C) 13.6 Years  D) 14 Years  E) 14.5 Years

Solution: Total age of 50 students   (30X 13+20 X 12) = 630   Average = 630/50= 
630/50 = 63/5 = around 12.6
12.6 Years \   12.6 Years
Answer: B
In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way?
Answers.
A) 6 sec  B) 9 sec  C) 7 sec  D) 11 sec  E) 13 sec

Solution: Explanation:   T = 60/36 * 18/5= 
60/36 * 18/5 = 6
6 sec   6 sec)
Answer: A
Lucy bought a 1-year,10000 certificate of deposit that paid interest at an annual rate of 8 percent compunded semiannually. What was the total amount of interest paid on this certificate at maturity ?
.	Choose the correct choice from the following:  A) 10464  B) 864  C) 816  D) 800  E) 480

Solution: Solution:   It says, 8 percent per annum per compounded semi annually.   Therefore r = 8/2= 
8/2 = 4
4% .   So, P= 10000   r = 4   n = 2   Net Amount A = 10816, I = 800.   ANSWER:800
Answer: D
The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector so formed has a central angle of 40°, what fraction of the wheel’s area is represented by the smallest sector?	Select
A) 5/36 B) 1/36 C) 1/18 D) 1/12 E) 1/9

Solution: Its an AP question .... it is given clearly in the question .   Let the smallest angle be a   and the circle has 8 sectors and hence 8 angle with a common difference d   hence all the angles can be written in 5/36P form with Cd as d ,   a, a+d, a+2d, a+3d ,a+4d, a+5d, a+6d ,a+7d,   given that a+7d = 40 --------1   also   a + a+d + a+2d + a+3d +   a+4d + a+5d + a+6d + a+7d = 360 ( as sum of all the angle is 360)   which is 8a + 28d = 360 --------2   solving 1 and 2   we get a=50   We are almost done ,   now the question ask what fraction of the wheel’s area is represented by the smallest sector ?   (50/360)( pie r*r)/ (pie r*r) = 50/360= 5/36   5/36 ans ....
Answer: A