MU-NLPC/Calc-aqua_rat huggingface dataset:
A dozen bananas and 5 pounds of apples are currently at the same price. If the price of a dozen bananas rises by 5 percent and the price of apples rises by 5 %. How much more will it cost to buy a dozen bananas and 5 pounds of apples .	Choose the correct answer: A) 5% B) 15% C) 25% D) 14% E) 10%

Solution: Say currently both a dozen bananas and 5 pounds of apples cost $100 (they are at the same price). So, to buy a dozen bananas and 5 pounds of apples we need $200.   After the increase, the price of a dozen bananas will be $105 and the price of 5 pounds of apples will be $105. So after the increase, to buy a dozen bananas and 5 pounds of apples we'll need $210.   Increase = 10%.   Answer: 10%.
Answer: E
If the product of 3 consecutive integers is 210 then the sum of the two smaller integers is:
Pick one.
A) 17
B) 10
C) 11
D) 12
E) 13

Solution: let the numbers be x-1 , x , x+1   (x-1)(x)(x+1)=210   210=2*105   =2*3*35   =2*3*5*7   this can be written as 6*5*7   =>x-1=5   x=6   x+1=7   sum of two smallest numbers=6+5= 
6+5 = 11
11   ANSWER:11
Answer: C
How much more would Rs.20000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly?
Choose the correct choice from the following choices
A) 482  B) 887  C) 299  D) 277  E) 200

Solution: 20000(11/10)4 - 20000(6/5)2 = 482   482nswer: 482
Answer: A
Rajesh can finish 1/4 of his home work in one hour. Seema can finish 3/7 of
her homework in 30 minutes and Ramya can finish 3/4 of her homework in
two and a half hours. If all of them start their home work at 12.00 PM and can
go to play as soon as they all finish their homework. When can they start to
play, if they take a break at 3.30 PM for 30 minutes?	Choices.
A) 4:30pm	B) 5:30pm	C) 6:30pm	D) 5:10pm	E) 3:30pm

Solution: ajesh can finish his work in 4 hr,   ramya can finish her work in 1 hr   seema can finish in 3.3 hr   so 4:30pm they can go to play   4:30pmNSWER:4:30pm
Answer: A
Nine animals of a circus has to be placed inside in nine cages one in each cage. If 4 of the cage are too small for 6 of the animal then find the number of ways of caging the animal.
Select
A) A.6048 B) B.10825 C) C.14400 D) D.784200 E) E.302400

Solution: ways for cages that are too small = 5*4*3*2= 
5*4*3*2 = 120
120   since we have 5 cages and 5 animal left therefore ways for 5 cages=5!   total ways =120*5!=14400   Answer C.14400
Answer: C
Find the average of first 20 natural numbers.
Pick:	A) 20.5	B) 18	C) 19.5	D) 10.5	E) 17

Solution: Exp. Sum of first n natural numbers = n( n+1)/2   So, sum of first 20 natural numbers = 20*21/2= 
20*21/2 = 210
210   Required average=210/20= 
210/20 = 21/2 = around 10.5
10.5   Answer:10.5
Answer: D
Akash leaves Chennai at 6am & reaches Bangalore at 10am. Prakash leaves Hyderabad at 8am and reaches Chennai at 11:30am. At what time do they cross each other?  Choose the correct choice from the following:
A) 6 : 32am B) 7 : 56am C) 8 : 56am D) 3 : 56am E) 5 : 56am

Solution: Time taken by Akash = 4 h   Time taken by Prakash = 3.5 h   For your convenience take the product of times taken by both as a distance.   Then the distance = 14km   Since, Akash covers half of the distance in 2 hours(i.e at 8 am)   Now, the rest half (i.e 7 km) will be coverd by both prakash and akash   Time taken by them = 7/7.5 = 56 min   Thus , they will cross each other at 8 : 56am.   8 : 56am
Answer: C
A certain auto manufacturer sold 5% fewer vehicles in 2007 than in 2006. If the manufacturer sold 2.2 million vehicles in 2006, how many vehicles, to the nearest 10,000, did the manufacturer sell in 2007? Pick one.
A) 17,00,000 B) 18,00,000 C) 19,00,000 D) 21,00,000 E) 21,90,000

Solution: Let's assume in 2006 the manufacturer sold X.   In 2007 he sold X-5/100(X)   If sales in 2006,X= 2.2 million find X-5/100(X)   (2.2 x 10^6)-5/100(2.2 x 10^6)   2200000-1,10,000=20,90,000-----> 21,00,000(approx)   Answer-21,00,000
Answer: D
A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability T that both selected dogs are NOT littermates?
Choose the correct option.
A) 1/6
B) 2/9
C) 5/6
D) 7/9
E) 8/9

Solution: We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates.   So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.   Work with probabilities:   Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B).   The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB).   The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate).   The required probability is 1/2 + 1/3 = 5/6.   Find the probability for the complementary event: choose AA or BB.   Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12.   Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12.   Again, we obtain 1 - (1/12 + 1/12) T= 5/6.   Answer: 5/6
Answer: C
The tax on a commodity is diminished by 20% and its consumption increased by 15%. The effect on revenue is?	Pick one: A) 2% B) 8% C) 5% D) 6% E) 1%

Solution: 100 * 100= 
100 * 100 = 10_000
10000   80 * 115= 
80 * 115 = 9_200
9200   -----------   10000-----------800   100-----------?=> 8% decrease   Answer: 8%
Answer: B
What is the lowest positive integer that is divisible by each of the odd integers between 11 and 17, inclusive? Choose the correct choice from the following
A) 3*5*13*17*11
B) 5×17×19×23
C) 7×15×17×19
D) 7×15×19×21
E) 3*4*13*17*11

Solution: We should find the LCM of 11,13,15 = 3*5, 17,-> LCM = 3*5*13*17*11 .   3*5*13*17*11nswer: 3*5*13*17*11.
Answer: A
In digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be?	Choices:
A) 100 cu.m  B) 1000 cu.m  C) 10000 cu.m  D) 2000 cu.m  E) 4000 cu.m

Solution: 20 * 10 * 5= 
20 * 10 * 5 = 1_000
1000   ANSWER:1000 cu.m
Answer: B
By selling 12 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%? Choose the correct option.
A) 8 B) 9 C) 6 D) 5 E) 2

Solution: 80% --- 12   120% --- ?     80/120 * 12= 
  80/120 * 12 = 8
8   8nswer:8
Answer: A
A bag of potatoes weighs 60 lbs divided by half of its weight. How much does the bag of potatoes weight?
Pick
A) 20 lb  B) 11 lb  C) 10 lb  D) 15 lb  E) 5 lb

Solution: Sol.   66÷6 = 11.   Answer : 11 lb
Answer: B
Let F(m,n) =45*m+36*n,where m and n are integers (+ve or -ve) What is the minimum positive value for f(m,n) (This may be achived for various values of m and n)? Choose the correct option:
A) 5 B) 6 C) 7 D) 8 E) 9

Solution: f(m,n)=45*m+36*n   putting m=1,n=-1 in above eqn we get   f(m,n)=45*1+36*(-1)= 
45*1+36*(-1) = 9
9.   ANSW9R:9
Answer: E
The bankerâ€™s discount on Rs. 1600 at 10% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is :
Choose the most appropriate option: A) 2 months B) 4 months C) 6 months D) 7 months E) None

Solution: Sol.   S.I. on Rs. 1600 = R.D. on Rs. 1680.   âˆ´ Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is S.I. on Rs. 1600 at 10%.   âˆ´ Time = [100 * 80 / 1600 * 10] year = 1/2 year = 6 months.   Answer 6 months
Answer: C
A and B are two numbers. Let
@(A,B)= Average of A and B,
/(A,B)= Product of A and B, and
*(A,B)= The result of dividing A by B.
The sum of A and B is given by
Choose the correct option:
A) /(@(A,B),2)	B) @(*(A,B),2)	C) @(/(A,B),2)	D) None of these	E) Cannot be determined

Solution: Explanation :   Assume some values for A and B and substitute in the options to get the answer.   Here, Let A = 1 and B = 2.   So, /(@(A,B),2) = /(@(1,2),2)   = /(1.5,2) = 3   = Sum of A and B.   /(@(A,B),2)nswer : /(@(A,B),2)
Answer: A
An bus covers a certain distance at a speed of 240 kmph in 5 hours. To cover the samedistance in 1hr, it must travel at a speed of?  Choose the most appropriate option:
A) 520 km/hr  B) 640 km/hr  C) 656 km/hr  D) 731 km/hr  E) 720 km/hr

Solution: Distance = (240 x 5) = 1200 km.   Speed = Distance/Time   Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]   Required speed = 1200 x	3	km/hr	= 720 km/hr.   5   720 km/hr
Answer: E
If K is a negative integer, which of the following must be a negative integer?	Select the correct option:
A) K + 1
B) K/2+3
C) K^4
D) K – y
E) K^11

Solution: We need to check each option here:   A.K + 1   This will depend on the value of K. Nothing can be said   B. K/2+3   This will depend on the value of K. Nothing can be said   C. K^4   This will always be positive. Even powers of negative numbers are positive and odd powers are negative   D. K – y   This will depend on the value of K. Nothing can be said   E. K^11   This will always be negative. K^11ven powers of negative numbers are positive and odd powers are negative   Correct Option: K^11
Answer: E
Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 7 ½ hours if they are walking in the same direction? Select the correct option: A) 7 km B) 6 km C) 9 3/8 km D) 9 km E) 5 km

Solution: RS = 5 ¾ - 4 ½ = 1 ¼   T = 7 ½ h.   D = 5/4 * 15/2 = 75/8   = 9 3/8 km   Answer:9 3/8 km
Answer: C
21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement.  Select:  A) 2/23  B) 3/25  C) 9/42  D) 1/21  E) 2/25

Solution: The probability that first toy shows the even number,   =1021=1021   Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.   Hence, probability that second toy shows the even number,   =920=920   Required probability,   =(1021)×(920)=(1021)×(920)   =9/42   9/42
Answer: C
If a jogger jogs 16 km/hr instead of 12 km/hr, he would have jogged 10 km more. The actual distance he jogged is	Select:
A) 50 km  B) 40 km  C) 30 km  D) 20 km  E) None of these

Solution: Explanation :   The actual distance jogged be d.   => d/12 = (d+10)/16   => 16d = 12d+120   => 4d = 120   => d = 30   Answer : 30 km
Answer: C
A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 16 cm. What is the length of the wire?  Choose the correct choice from the following answers:
A) 7 cm  B) 2 cm  C) 8 cm  D) 3 cm  E) 9 cm

Solution: Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.   π(16)2 * h = (4/3)π (12)3 => h   = 9 cm   Answer:9 cm
Answer: E
In what time will two trains cross each other completely, which are running on the same parallel lines in opposite directions, each train running with a speed of 60 kmph being 130 m and 120 m in length respectively?
Select the correct option:
A) 6.9 sec B) 7.1 sec C) 7.2 sec D) 7.4 sec E) 7.5 sec

Solution: D = 130m + 120m = 250m * 1/1000 130  + 120= 
130  + 120 = 250
250m * 1/1000 = 0.25 kms   RS 250  * 1/1000= 
250  * 1/1000 = 1/4 = around 0.25
0.25 kms   RS= 60 + 60= 
60 + 60 = 120
120 kmph   T= (0.25/120) * 3600= 
(0.25/120) * 3600 = 7.5
7.5 sec   ANSW7.5 secR:7.5 sec
Answer: E
Printer P can print one lakh books in 8 hours. Printer Q can print the same number of books in 10 hours while Printer R can print the same in 12 hours. All the Printers started printing at 9 A.M. Printer P is stopped at 11 A.M. and the remaining 2printers complete work. Approximately at what time will the printing of 1 lac books be completed?
Choose the most appropriate option.
A) 1 pm B) 2 pm C) 3 pm D) 4 pm E) 5 pm

Solution: Work done by P in 1 hour = 1/8   Work done by Q in 1 hour = 1/10   Work done by R in 1 hour = 1/12   Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120   Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60   From 9 am to 11 am, all the Printer were operating.   Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60   Pending work = 1- 37/60 = 23/60   Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11   which is approximately equal to 2   Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm   1 pm
Answer: A
Of the three- digit integers greater than 700, how many have two digits T that are equal to each other and the remaining digit different from the other two?	Select:
A) 90
B) 82
C) 80
D) 45
E) 36

Solution: three-digit integers greater than 700: 701 to 999, inclusive.   possible values for hundreds-digit--> 7,8,9   possible values for tens-digit and ones-digit --> 0, 1,2,3,4,5,6,7,8,9   when hundreds-digit and tens-digit are the same: (3x1x10)-3=27 ---> we minus three to exclude 777, 888 and 999   when hundreds-digit and ones-digit are the same: (3x10x1)-3=27 ---> we minus three to exclude 777, 888 and 999   when tens-digit and hundreds-digit are the same:[(3x10x1)-3]-1=26 ---> we minus three to exclude 777, 888 and 999; we minus one to exclude 700   T=27+27+26= 
27+27+26 = 80
80   Answer: 80
Answer: C
Company C sells a line of 25 products with an average retail price of $1,200. If none of these products sells for less than $400, and exactly 10 of the products sell for less than $1,000, what is the greatest possible selling price of the most expensive product? Choose the most appropriate option:	A) 12000	B) 13000	C) 14000	D) 15000	E) 16000

Solution: The average price of 25 products is $1,200 means that the total price of 25 products is 25*1,200=$30,000.   Next, since exactly 10 of the products sell for less than $1,000, then let's make these 10 items to be at $400 each (min possible).   Now, the remaining 14 items cannot be priced less than $1,000, thus the minimum possible price of each of these 14 items is $1,000.   Thus the minimum possible value of 24 products is 10*400+14*1,000=$18,000.   Therefore, the greatest possible selling price of the most expensive product is $30,000-$18,000=$12,000.   12000nswer: 12000.
Answer: A
The edge of three cubes of metal is 3 dm, 4 dm and 5 dm. They are melted and formed into a single cube. Find the edge of the new cube?	Options: A) 7 B) 5 C) 8 D) 6 E) 4

Solution: Explanation:   33 + 43 + 53 = a3 => a = 6   Answer: Option 6
Answer: D
A teacher prepares a test. She gives 8 objective type questions out of which 4 have to be answered. Find the total ways in which they can be answered if the first 5 questions have 3 choices and the last 3 have 4 choices.  Answers.
A) 6255	B) 5280	C) 5192	D) 5100	E) 5144

Solution: Two cases.   First Case: 2 Questions - 3 Choices, 2 Questions - 4 Choices   5C2 * 3C2 ways to select the questions   3C1 * 3C1 * 4C1 * 4C1 ways to answer the questions   = 10 * 3 * 3 * 3 * 4 * 4= 
10 * 3 * 3 * 3 * 4 * 4 = 4_320
4320   Second Case: 1 Question - 3 Choices, 3 Questions - 4 Choices   5C1 * 3C3 ways to select the questions   3C1 * 4C1 * 4C1 * 4C1 ways to answer the questions= 5 * 1 * 3 * 4 * 4 * 4= 
5 * 1 * 3 * 4 * 4 * 4 = 960
960   Total= 4320 + 960= 
4320 + 960 = 5_280
5280   5280
Answer: B
What is the units digit of the product of the first 15 odd numbers?
Choose the correct choice:
A) 0  B) 5  C) 1  D) 2  E) 3

Solution: 1*3*5*7 ................... will end up in 5 in the units place   Answer : 5
Answer: B
If s,u, and v are positive integers and s=2u+2v, which of the following must be true?
i. s=u
ii. u is not equal to v
iii. s > v  Choose the correct choice from the following answers:  A) None  B) I only  C) II only  D) III only  E) II and III

Solution: Notice two things: 1. we are asked to find out which of the following MUST be true, not COULD be true and 2. s, u, and v are positive integers.   Given: s=2u+2v --> s/2=u+v. Now, since s, u, and v are positive integers then s is more than either u or v, so I is never true and III is always true. Nones for II: it's may be necessarily true,   Nonenswer: None
Answer: A
If n divided by 9 has a remainder of 8, what is the remainder when 2 times n is divided by 9?	Select the correct option:
A) 1
B) 7
C) 3
D) 5
E) 6

Solution: As per question => N=9P+8 for some integer p   hence 2N => 18Q + 16   7ut again, 16 can be divided by 9 to get remainder 7 for some integer Q   hence 7
Answer: B
On a certain scale of intensity, each increment of 20 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125? Choices:
A) 40	B) 100	C) 400	D) 1000	E) 10 000

Solution: Increase of 40 in magnitude corresponds to 10^4 increase in intensity:   If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 165 it'll be 10*10*10*10*x=400*x.   Answer: 100.
Answer: B
The present value of a bill due at the end of 2 years is Rs.1250. If the bill were due at the end of 2 years and 11 months, its present worth would be Rs.1200. Find the rate of interest and the sum.  Choose the correct choice.
A) rs.1175  B) rs.1375  C) rs.1475  D) rs.1575  E) None of these

Solution: Explanation:   PW=100xAmount/100+(Rx T)   Answer: rs.1375
Answer: B
A five-character password consists of one letter of the alphabet and four different digits between 0 and 9, inclusive. The letter must appear as the second or third character of the password. How many different passwords are possible?
Choose the correct choice:
A) 5,040	B) 18,720	C) 262,000	D) 137,440	E) 262,080

Solution: 4 character password: DLDD   First digit can be chosen in 10 ways.   The letter can be chosen in 26 ways.   Next digit can be chosen in 9 ways.   Next digit can be chosen in 8 ways.   Next digit can be chosen in 8 ways.   This gives us 10*26*9*8*7 ways   Now, the letter can also be DDLDD so there will be another 10*9*26*8*7 ways   Total = 10*26*9*8*7*2 = 262,080 ways   262,080
Answer: E
Average of 13 results is 65. If the average of first six results is 61 and average of last six results is 59 Then find the seventh result?
Pick one.
A) 120  B) 123  C) 125  D) 128  E) 130

Solution: 125   Option '125'
Answer: C
Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $195 million, which represented 10 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars?	Answers.
A) 17,010  B) 2,100  C) 1,890  D) 1,950  E) 210

Solution: 10% ---- 195 millions   for 100% => ( 195 * 100% )/10% = 1950.   Option 1,950.( 195 * 100  )/10= 
( 195 * 100  )/10 = 1_950
1950.   Option 1,950.
Answer: D
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3? Choose one
A) 10,300  B) 10,030  C) 1,353  D) 1,352  E) 1,339

Solution: There is one number in hundred with 1 in th tens digit and 3 in the units digit: 13, 113, 213, 313, ...   The difference between 324,700 and 458,600 is 458,600-324,700=133,900 - one number per each hundred gives 133,900/100=1,339 numbers.   Answer: 1,339
Answer: E
A man can row downstream at 20 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?	Choices: A) 13, 5 B) 15, 2 C) 15, 5 D) 14, 5 E) 12, 5

Solution: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.   Given x + y = 20--- (1)   and x - y = 10 --- (2)   From (1) & (2) 2x = 30 => x = 15, y = 5.   Answer:15, 5
Answer: C
A 24 month project had a total budget of $60,000. After eleven months, the project had spent $26,300. At this point, how much was the project under budget? Choices.
A) $800
B) $1000
C) $1200
D) $1400
E) $1600

Solution: Each month, the project should spend $60,000 / 24 = $2500.   In 11 months, the project should spend 11*$2500=$27,500.   The project is under budget by $27,500 - $26,300 = $1200.   The answer is $1200.
Answer: C
At a loading dock, each worker on the night crew loaded 1/2 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
Choose the correct choice from the following options
A) 1/2 B) 2/5 C) 3/5 D) 4/5 E) 5/7

Solution: Method: X = No. of boxes loaded by day crew.   Boxes by night crew = 1/2 * 4/5 X = 2/5 X   % loaded by day crew = X/(X + 2/5X) = 5/7   Answer 5/7
Answer: E
Of the science books in a certain supply room, 60 are on botany, 65 are on zoology, 90 are on physics. 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same science?  Choose the correct answer
A) 81	B) 159	C) 166	D) 285	E) 334

Solution: I solve it using the tough luck technique:   According to question: What is the least number of books you should pick so as to get at least 80 books of the same science subject.   80 books of the same science subjects is possible only for two subjects: Physics=90>80 OR Chemistry=110>80   Now, we need to be certain that out of the books we picked, there are either at least 80 physics books or 80 chemistry books   What if we pick the first 80 books and none of them is either Physics or Chemistry. Possible. Thus, we first count all our negatives.   We picked:   60 Botany Books   65 Zoology Books   50 Geology Books   Now, any book we pick will be either Chemistry or Physics. But unfortunately, we can't be lucky enough to pick 80 books and all of them will be Physics, right!!   Thus, in order to make sure that we have 80 books of either of these 2 subjects, we must pick   79*2+1 books   Because, we could have picked the books in following order;   1st book picked: Physics   2nd book picked: Chemistry   3rd book picked: Physics   Thus, Total= 60+65+50+79*2+1=175+1+158= 
175+1+158 = 334
334   Ans:334
Answer: E
Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 2:5. The first number is:
Answers:
A) A)33	B) B)98	C) C)34	D) D)35	E) E)62

Solution: (3x-9):(5x-9) = 2:5   x = 11 => 3x = 33   33nswer:33
Answer: A
A person spends 40% of his salary on food, 25% on house rent, 15% on entertainment and 15% on conveyance. If his savings at the end of the month is Rs. 1200, then his salary per month in rupees is: Pick one:  A) 4000  B) 24000  C) 8000  D) 10000  E) None of these

Solution: Total expenditure = 40 + 25 + 15 + 15= 
40 + 25 + 15 + 15 = 95
95%   Saving= (100 - 95)= 
(100 - 95) = 5
5%   5/100 × Salary= 1200, Salary = 24000 Rs.   Answer:24000
Answer: B
Five machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 20 hours to fill a certain production order, how many fewer hours does it take all five machines, operating simultaneously, to fill the same production order? Pick one.
A) 4
B) 5
C) 6
D) 16
E) 24

Solution: Total work = 4*20 machine hrs   Time taken = 4* 20 /5= 
4* 20 /5 = 16
> 16 hours...   Thus all five machines, operating simultaneously will take ( 20 - 16 ) 4 hours..   4nswer will be (4) 4
Answer: A
Four years ago, the average age of a family of 4 members was 21 years.A baby having been born,the average age of the family is the same today.The present age of the baby? Choose the correct option.
A) 5 years B) 7 years C) 2 years D) 3 years E) 6 years

Solution: Total age of the 4 members, 4 years ago = (21*4)= 
(21*4) = 84
84 years.   Total age of the 4 members now=( 84+ 4*4)= 
( 84+ 4*4) = 100
100 years.   Total age of the 5 members now= (21*5)= 
(21*5) = 105
105 years.   5 yearsge of the baby is= 105- 100= 
105- 100 = 5
5 years   5 yearsnswer :5 years
Answer: A
How many 1/10s are there in 37 1/2?
Choose the correct option:	A) 375	B) 475	C) 500	D) 670	E) 700

Solution: Required number = (75/2)/(1/10)   = (75/2 x 10/1)   = 375.   375NSWER:375
Answer: A
a light flashes every 5seconds, how many times will it flash in ? of an hour?
Choices.
A) 720
B) 600
C) 650
D) 700
E) 750

Solution: 1flash=5sec   for 1 min=12flashes   so for 1 hour=12*60= 
12*60 = 720
720 flashes.   720NSWER:720
Answer: A