Ans :16/3 inches
Explanation : Let x be the amount of increase. The volume will increase by the same amount if the radius increased or the height is increased. So, the effect on increasing height is equal to the effect on increasing the radius.
i.e., (22/7)*8*8*(3+x) = (22/7)*(8+x)*(8+x)*3
Solving the quadratic equation we get the x = 0 or 16/3. The possible increase would be by 16/3 inches.
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If s(a) denotes square root of a, find the value of s(12+s(12+s(12+ …… upto infinity.
Ans :4
Explanation : Let x = s(12+s(12+s(12+….. ) We can write x = s(12+x). i.e., x^2 = 12 + x. Solving this quadratic equation, we get x = -3 or x=4. Sum cannot be -ve and hence sum = 4.
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What is the sum of all numbers between 100 and 1000 which are divisible by 14 ?
Ans :35392
Explanation : The number closest to 100 which is greater than 100 and divisible by
14 is 112, which is the first term of the series which has to be summed.
The number closest to 1000 which is less than 1000 and divisible by 14 is 994,
which is the last term of the series.
112 + 126 + …. + 994 = 14(8+9+ … + 71) = 35392.
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A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches is available. What is the length of the circular rod with diameter 8 inches and equal to the volume of the rectangular plate?
Ans :3.5 inches
Explanation : Volume of the circular rod (cylinder) = Volume of the rectangular plate (22/7)*4*4*h = 8*11*2, h = 7/2 = 3.5.
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Out of 80 coins, one is counterfeit. What is the minimum number of weighings needed to find out the counterfeit coin?
Ans :4.
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Gavaskar's average in his first 50 innings was 50. After the 51st innings, his average was 51. How many runs did he score in his 51st innings. (supposing that he lost his wicket in his 51st innings)
Ans :101
Explanation : Total score after 50 innings = 50*50 = 2500 Total score after 51 innings = 51*51 = 2601 So, runs made in the 51st innings = 2601-2500 = 101
If he had not lost his wicket in his 51st innings, he would have scored an
unbeaten 50 in his 51st innings.
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A fast typist can type some matter in 2 hours and a slow typist can type the same in 3 hours. If both type combinely, in how much time will they finish?
Ans :1 hr 12 min
Explanation : The fast typist's work done in 1 hr = 1/2 The slow typist's work done in 1 hr = 1/3 If they work combinely, work done in 1 hr = 1/2+1/3 = 5/6 So, the work will be completed in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min.
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What is the number of zeros at the end of the product of the numbers from 1 to 100?
Ans :127.
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For the following, find the next term in the series
1. 6, 24, 60,120, 210
Ans :336
Explanation : The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, ….. ( '.' means product).
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A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs.- _______ for the horse and Rs.________ for the cart.
Ans :Cost price of horse: Rs. 400 &
Cost price of cart: Rs. 200
Explanation:
Let x be the cost of horse & y be the cost of the cart.
10 % of loss in selling horse = 20 % of gain in selling the cart
Therefore (10 / 100) * x = (20 * 100) * y
x = 2y ———–(1)
5 % of loss in selling the horse is 10 more than the 5 % gain in selling the
cart.
Therefore (5 / 100) * x - 10 = (5 / 100) * y
5x - 1000 = 5y
Substituting (1)
10y - 1000 = 5y
5y = 1000
y = 200
x = 400 from (1).
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There are 3 persons Sudhir, Arvind, and Gauri. Sudhir lent cars to Arvind and Gauri as many as they had already. After some time Arvind gave as many cars to Sudhir and Gauri as many as they have. After sometime Gauri did the same thing. At the end of this transaction each one of them had 24. Find the cars each originally had.
Ans :Sudhir had 39 cars, Arvind had 21 cars and Gauri had 12 cars.
Explanation:
Sudhir Arvind Gauri
Finally 24 24 24
Before Gauri's transaction 12 12 48
Before Arvind's transaction 6 42 24
Before Sudhir' s transaction 39 21 12.
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The number that does not have a reciprocal is ____________.
Ans :1
Explanation:
One is the only number exists without reciprocal because the reciprocal of
one is one itself..
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Which of the following is larger than 3/5? (1) ½ (2) 39/50 (3) 7/25 (4) 3/10 (5) 59/100
Ans :39/50 .
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It takes Mr. Karthik y hours to complete typing a manuscript. After 2 hours, he was called away. What fractional part of the assignment was left incomplete?
Ans :(y - 2) / y.
Explanation:
To type a manuscript karthik took y hours.
Therefore his speed in typing = 1/y.
He was called away after 2 hours of typing.
Therefore the work completed = 1/y * 2.
Therefore the remaining work to be completed = 1 - 2/y.
(i.e.) work to be completed = (y-2)/y.
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The length of the side of a square is represented by x+2. The length of the side of an equilateral triangle is 2x. If the square and the equilateral triangle have equal perimeter, then the value of x is _______.
Ans = 4
Explanation:
Since the side of the square is x + 2, its perimeter = 4 (x + 2) = 4x + 8
Since the side of the equilateral triangle is 2x, its perimeter = 3 * 2x = 6x
Also, the perimeters of both are equal.
(i.e.) 4x + 8 = 6x
(i.e.) 2x = 8 x = 4..
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If every alternative letter starting from B of the English alphabet is written in small letter, rest all are written in capital letters, how the month "September" be written. (1) SeptEMbEr (2) SEpTeMBEr (3) SeptembeR (4) SepteMber (5) None of the above.
Ans :(5).
Explanation:
Since every alternative letter starting from B of the English alphabet is written
in small letter, the letters written in small letter are b, d, f…
In the first two answers the letter E is written in both small & capital letters, so
they are not the correct answers. But in third and fourth answers the letter is written in small letter instead capital letter, so they are not the answers..
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Which of the following statements drawn from the given statements are correct? Given: All watches sold in that shop are of high standard. Some of the HMT watches are sold in that shop.
a) All watches of high standard were manufactured by HMT.
b) Some of the HMT watches are of high standard.
c) None of the HMT watches is of high standard.
d) Some of the HMT watches of high standard are sold in that shop.
Ans :(b) & (d).
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All men are vertebrates. Some mammals are vertebrates. Which of the following conclusions drawn from the above statement is correct.
1. All men are mammals
2. All mammals are men
3. Some vertebrates are mammals.
4. None
Ans :Some vertebrates are mammals..
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If point P is on line segment AB, then which of the following is always true? (1) AP = PB (2) AP > PB (3) PB > AP (4) AB > AP (5) AB > AP + PB
Ans :(4)
Explanation:
P
A B
Since p is a point on the line segment AB, AB > AP.
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If a light flashes every 6 seconds, how many times will it flash in ¾ of an hour?
Ans :451 times.
Explanation:
There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will flashes 451 times in ¾ of an
hour..
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Five boys were climbing a hill. J was following H. R was just ahead of G. K was between G & H. They were climbing up in a column. Who was the second?
Ans :G.
Explanation:
The order in which they are climbing is R - G - K - H - J.
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Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards. Last year, each of them discovered that every tree in their own orchard bore exactly the same number of apples. Further, if the third farmer gives one apple to the first, and the fifth gives three to each of the second and the fourth, they would all have exactly the same number of apples. What were the yields per tree in the orchards of the third and fourth farmers?
Ans :11 & 9 apples per tree.
Explanation:
Let a, b, c, d & e be the total number of apples bored per year in A, B, C, D &
E 's orchard. Given that a + 1 = b + 3 = c - 1 = d + 3 = e - 6
But the question is to find the number of apples bored per tree in C and D 's orchard.
If is enough to consider c - 1 = d + 3.
Since the number of trees in C's orchard is 11 and that of D's orchard is 13.
Let x and y be the number of apples bored per tree in C & d 's orchard respectively.
Therefore 11 x - 1 = 13 y + 3
By trial and error method, we get the value for x and y as 11 and 9.
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A tennis marker is trying to put together a team of four players for a tennis tournament out of seven available. males - a, b and c; females - m, n, o and p. All players are of equal ability and there must be at least two males in the team. For a team of four, all players must be able to play with each other under the following restrictions: b should not play with m, c should not play with p, and a should not play with o. Which of the following statements must be false? 1. b and p cannot be selected together 2. c and o cannot be selected together 3. c and n cannot be selected together.
Ans :3.
Explanation:
Since inclusion of any male player will reject a female from the team. Since
there should be four member in the team and only three males are available, the girl,
n should included in the team always irrespective of others selection..
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A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs.- _______ for the horse and Rs.________ for the cart.
Ans :Cost price of horse = Rs. 400 & the cost price of cart = 200.
Explanation:-
Let x be the cost price of the horse and y be the cost price of the cart.
In the first sale there is no loss or profit. (i.e.) The loss obtained is equal to the gain.
Therefore (10/100) * x = (20/100) * y
X = 2 * y —————–(1)
In the second sale, he lost Rs. 10. (i.e.) The loss is greater than the profit by Rs. 10.
Therefore (5 / 100) * x = (5 / 100) * y + 10 ——-(2)
Substituting (1) in (2) we get
(10 / 100) * y = (5 / 100) * y + 10
(5 / 100) * y = 10
y = 200
From (1) 2 * 200 = x = 400.
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what is a percent of b divided by b percent of a?
Ans :1
Explanation:
a percent of b : (a/100) * b
b percent of a : (b/100) * a
a percent of b divided by b percent of a : ((a / 100 )*b) / (b/100) * a )) = 1.
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A contractor agreeing to finish a work in 150 days, employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by ________ each working now for 10 hours daily, the work can be completed in time.
Ans :150 men.
Explanation:
One day's work = 2 / (7 * 90)
One hour's work = 2 / (7 * 90 *
One man's work = 2 / (7 * 90 * 8 * 75)
The remaining work (5/7) has to be completed within 60 days, because the
total number of days allotted for the project is 150 days.
So we get the equation
(2 * 10 * x * 60) / (7 * 90 * 8 * 75) = 5/7 where x is the number of men
working after the 90th day.
We get x = 225
Since we have 75 men already, it is enough to add only 150 men..
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A man was engaged on a job for 30 days on the condition that he would get a wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for each day of his absence. If he gets Rs. 216 at the end, he was absent for work for … days.
Ans :7 days
Explanation:
The equation portraying the given problem is:
10 * x - 2 * (30 - x) = 216 where x is the number of working days.
Solving this we get x = 23
Number of days he was absent was 7 (30-23) days..
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A software engineer has the capability of thinking 100 lines of code in five minutes and can type 100 lines of code in 10 minutes. He takes a break for five minutes after every ten minutes. How many lines of codes will he complete typing after an hour?
Ans :250 lines of codes.
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A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of 950gm. for a kg. His gain is____%.
Ans :5.3 %
Explanation:
He sells 950 grams of pulses and gains 50 grams.
If he sells 100 grams of pulses then he will gain (50 / 950) *100 = 5.26.
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A student divided a number by 2/3 when he required to multiply by 3/2. Calculate the percentage of error in his result.
Ans :0 %
Explanation:
Since 3x / 2 = x / (2 / 3).
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It was calculated that 75 men could complete a piece of work in 20 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work?
Ans :30 days.
Explanation:
Before:
One day work = 1 / 20
One man's one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 = 30