The cost of one helmet, two pairs of gloves, three pairs of pads, four bats and five balls increased by 11%, 22%, 33%, 44% and 55% respectively, over that of the previous year. Last year, if I had spent equal amounts purchasing each of the given varieties of items (i.e., helmets, gloves, pads, bats and balls), by what percentage would the total cost of all the items I purchased increase this year, over that in the previous year?
Let the amount that spent on each variety of items by x.
Hence, total amount I spent last year = 5x.
This year I would have spent x (1.11) + x (1.22) + x (1.33) + x (1.44) + x (1.55)
Hence the total cost increased by
x x 0.11[1 + 2 + 3 + 4 + 5] = x x 0.11 x 15
∴ Percentage increase in cost
A real number x satisfies the following inequation 2 – 1 / y < x ≤ 5 + 1 / y, for each positive integer y. Which of the following represents the best description of x?
2 – 1 / y < x ≤ 5 + 1 / y, for all positive integral values ofy.
Clearly all the numbers between 2 and 5 satisfy the condition. 5 itself satisfies the condition and 2 also satisfies the condition.
But any number greater than 5 or less than 2 doesn’t satisfy the condition for all integral values ofy.
Therefore, the best description is [2, 5] or 2 ≤x ≤ 5.
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If (101110101101110)_{2} = (X)_{16}, then find the number of digits in X each of which is less than 9.
Since 16 = 2^{4}, in order to convert a number from base 2 to base 16, every set of four digits, starting from the left, can be directly replaced with their base 16 equivalent.
∴101110101101110)_{2}= [(101)(1101 )(0110)(1110)]_{2}
= [5DGE],_{16}.
Hence, there are two digits which are less than 9.
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Ajay has a certain number of Rs.1, Rs.2 and Rs.10 notes in his wallet. The number of Rs.1 notes and Rs.2 notes are in the ratio 4 : 3. If the number of Rs.10 notes exceeds the total number of Rs.1 and Rs.2 notes, which of the following is a possible value of the total amount in his wallet?
y > total number of Re.1 and ₹2 notes => y > 7x.
Total value of notes = 1 (4x) + 2(3x) + 10y = 10{x + y).
As x and y are integers this is divisible by 10.
As y > 7x, 10(x + y) > 80x i.e., it is greater than 780.
Going by the choices, both these conditions are satisfied by only Choice (A).
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If the volume of a cylinder decreased by half and its radius decreased by 25%, what is the percentage reduction in its height?
Initial volume = π x r^{2}x h_{1}
After reduction, radius = 3 / 4 r
Final volume = (π x r^{2}x h_{1}) / 2
(π x r^{2}x h_{1}) / 2 = (π x 9 / 16 x r^{2} x h_{2})
⇒h_{1} / h_{2} = 9 / 16 x 2 = 9 / 8
Percentage decrease in height = 1 / 9
=
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The sides of a rightangled triangle are in geometric progression. Find the common ratio, if it is greater than 1.
Let the sides of the triangle be a, ar and ar^{2}.
The largest side of a right angled triangle is the hypotenuse.
Hence it must be ar^{2}
By hypotenuse theorem,
a^{2} + (ar)^{2} = (ar^{2})^{2}
a^{2} + a^{2}r^{2} = a^{2}r^{4}.
=> a^{2}(1 + r^{2}) = a^{2}r^{4}.
=> r^{4} r^{2} 1 =0.
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A party was attended by twenty five persons. There were a lot of handshakes. Which of the following statements can never be true?
(a) A person, say A may have shook hands with 1,3,5 etc people. This is possible.
(b)A person may have shook hands with all the others. This is possible.
(c)If each person has shaken hands with a different number of people, the number of handshakes for the 25 people has to be 0, 1, 2. ....24. But if one person has 0 handshakes, no person can have made 24 handshakes. At the most a person can have made 23 handshakes. This statement can never be true.
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A square sheet of paper has a side of 30 cm. Four squares, each of side x‘ ’ cm, are cut away from each of its corners and the remaining sheet is folded into an open cuboid (i.e., without the top face). What is the maximum possible volume (in cc.) of the cuboid formed?
Height = x cm and length = breadth = 30  2x cm
Now to maximise this volume x(30  7x) (30  2x) must be the maximum.
If x(30  2x)^{2} is maximum, then 2x(30  2x)^{2} must also be maximum.
Now for any expression of the form x(K  x)n to be the maximum, x and (K  x) must be in the ratio 1 : n.
Hence for 2x (30  2x)^{2 }to be maximum 30 – 2x / 2x = 2 / 1
=> x = 5 => 5(30  2 X 5)^{2} = 2000.
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Fourteen children are playing a game on a beach. Initially they begin the game by making one huge pile of sand and then take turns to divide the pile into smaller piles. If each child, in his or her turn, can divide an existing pile into only m or n smaller piles, for what value of (m, n) is it not possible to get 14 piles of sand at any stage in the game?
(1) (m, n) = (2, 5)
As one pile can be divided into 2, the number of piles can increase by steps of 1. It’ll be possible to get 1, 2, 3,4, 5,….piles.
(2)(m, n) = (3, 6)
In any turn, the number of piles can increase by 2 or 5.
∴It can increase by four 2’s and one 5, i.e., by 13. Hence, from 1 to 14.
Hence it is possible in this case also.
(3)(m, n) = (3, 9)
In any turn, the number of piles can increase only by 2 or 8. Hence, starting from 1 pile, we can only get an odd number (any odd number) of piles, but not 14 piles at any stage. Hence, it is not possible in this case.
(4)(m, n) = (5, 6)
In any turn, the number of piles can increase by 4 or 5.
∴ It can increase by two 4's and one 5, i.e. by 13 (say from 1 to 14)
Hence, it is possible in this case.
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If log_{10} a^{2}b = 5 and log_{10}(a / b^{2}) = 8, then find the value of a.
Given log_{10}a^{2}b= 5 => 2log_{10 }a + log_{10} b = 5
Also, logio a/b^{2} = 8 => log_{10} a  2log_{10} b = 8
Solving for log a, we get log_{10} a = 18/5
=> a = (10)^{18 / 5}
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Two vessels, A and B, contain milk and water respectively. The volume of milk in A and the volume of water in B are in the ratio 4 : 5. Now, x litres of milk from A is poured into B and then y litres of the solution from B is poured into A. If as a result, the concentration of milk in A and that of water in B, both dropped from 100% to 66 x 2/3%, find x : y.
=> Volume of water in B is 5a.
After the transfer of contents from A to B, the concentration of water in B becomes 66 x 2/3%: i.e., 2/3. Hence, the total volume, must have become 3/2 times the initial (i.e., 5a). Hence, the volume of milk transferred to B = 3 / 2 (5a) 5a = 5 / 2 a = 2.5a .
∴Milk remaining in A = 4a  2.5a = 1.5a
Now, the concentration of milk in A and B = 100% and 33 x 1/3%.
If after transferring some quantity of the contents of B to A, the concentration becomes 66 x 2/3, (i.e., midway between 100 and 33 x 1/3%, that means that the volume of contents from B and the volume of milk in A were mixed in equal quantities, i.e., 1.5a of milk and water solution was transferred from B to A.
∴ Required ratio = 2.5a / 1.5a = 5 : 3.
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The i^{th}term, a_{i}, of a sequence is given by the product of the first_{i} prime numbers starting from 7. The sum of the firstj terms of this sequence is denoted by s_{j}. Which of the following is true?
∴All the a's are odd.
s_{i} = a_{i} + a_{2} +….+ a_{i}
If i is even, s_{i} is even and
If i is odd, s_{i} is odd
The converses are also true i.e.
If s_{i} is even, i is even and if s_{i} is odd, i is odd.
∴ (a)is false.
(b)is false. If a_{i} is odd (all as are odd), l_{i} could be odd or even.
(c)is true
(d)is false, as if s_{i} is odd, i is odd and a_{i} is always odd.
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ABCD is a square, of side 7 units, in the coordinate plane, such that the origin lies within the square. If AB is parallel to the yaxis and A is at (−4, 2), what are the coordinates of C?
Therefore, the coordinates of B must be (4. 5), since AB is parallel to yaxis and AB= 7.
Also, since AB is parallel to Y axis, BC must be parallel to X axis.
Now, since B is (4, 6) and BC=7, the coordinates of C must be (3, 5).
DIRECTIONS for questions: Select the correct alternative from the given choices. There are two groups of workers, G_{1} and G_{2}. There are 30 workers in G_{1} and 60 in G_{2}. Each worker in G_{1} works three times as fast as any worker in G_{2}. If all the workers in G_{1} together ploughed a certain field in half a day, then find the time taken by all the workers in G_{1 }and G_{2} together to plough a field whose area is 50% more than that of the first field.
Rate at which each worker in G_{1} works = 3x units/day
Let the area of the first field be w w = 30(3x)(0.5) = 45x
Area of second field = 1,5w = 67.5x
Let the required time be t days
∴ t[30(3x)+60(x)] = 67.5x
⇒ t = 67.5 / 150 = 9 / 20
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Consider two cubes of equal volume. In one of the cubes, the biggest possible sphere is inscribed. The other cube is cut into exactly 64 identical small cubes. In each of them, the biggest possible sphere is placed. The volume of the sphere in the first cube is denoted by V_{1}. The total volume of the spheres in the small cubes is denoted by V_{2}. Find V_{1} : V_{2}.
As the sum of the volumes of all the small cubes is equal to the volume of the first cube, the sum of the volumes of the spheres in these cubes is equal to the volume of the sphere is the first cube.
∴The required ratio is 1 :1.
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Mrs. Ahluwalia spent less than Rs.10,000 to buy some cotton sarees and some silk sarees. If Mrs. Ahluwalia had spent the amount of money she actually spent on cotton sarees to buy silk sarees and vice versa, she could have purchased two more sarees. Find the maximum number of sarees that Mrs. Ahluwalia could have originally purchased, if each cotton saree and silk saree costs Rs.300 and Rs.400 respectively.
∴300x + 400y< />
Now, if 300x rupees are spent to buy silk sarees, the number of silk sarees that could be purchased is 300x / 400 or 3 / 4x. Similarly if 400y rupees are spent to buy cotton sarees, the number of cotton sarees that could be purchased is 400y / 300 or 4 / 3 y
Given : x + y + 2 = 4 / 3y + 3 / 4x
=>4y3x = 24. Tabulating the possible values of x and y, we have
The higher values of x and y are ignored as 300x + 400y < 10000.="" hence="" mrs.="" ahluwalia="" could="" have="" bought="" a="" maximum="" of="" 27="" />
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How many distinct rectangles of area 900 sq.cm are possible with both their length and breadth being integers, when expressed in cm?
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If S_{1} is a set comprising ten integers and S_{2} is a subset of S_{1} comprising nine elements, which of the following statements are true?
I.The range of S_{2} can be more than that of S_{1}
II.The median of S_{2} can be less than that of S_{1}
III.The mean of S_{2} can be more than that of S_{1}
Statement I
S_{2} is a subset of Si and it has one element less than S_{1}.
If this element was the greatest or the least element of S_{1}, range of S_{2}< range of S_{1}
(Range = Max element Min element)
If this element was neither the greatest element nor the least element of S_{1}, range of S_{2} = range of S_{1}.
In any case, range of S_{2} cannot be > range of S_{1}
Statement I is false.
Statement II
Let S_{1} = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} and
S_{2} = {1,2, 3,4, 5,6, 7, 8,9}
Median of S_{1} = 5^{th} element + 6^{th} element / 2 = 5 + 6 / 2 = 5.5
Median of S_{2} = 5th element = 5
Statement II is true.
Statement III
Let S_{1} = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} and
S_{2} = {2, 3, 4, 5, 6,7, 8, 9, 10}
MeanofS_{1}= 55 / 10 = 5.5 and mean of S_{2} = 54 / 9 = 6
Statement III is true.
Only statements II and III are true.
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Sam and Ram together started a business with initial investments in the ratio of 3 : 4 but Ram left the business aftexr months. If at the end of the year, Sam and Ram shared the profit in the ratio 3 : 2, what is the value of x?
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If a Ψb = ab − (a + b) + 1
a π b = LCM (a, b) / a
a Λ b = ab / a + b
what is the value of (6 Ψ 13)π (12 Λ 3)?
(6 Ψ 13) = 78 – 19 + 1 = 60
(12 Λ 3) = 36 / 15 = 12 / 5
Fresh dates contain 80% water by weight while dry dates contain 20% water by weight. What is the weight (in kg) of dry dates available from 300 kg of fresh dates?
300 kg of fresh dates contain 240 kg water
=> 60 kg is nonwater.
=> 60 kg is 80% of dried grape.
Weight of dried grapes
= 60 / 80% = 60 x 5 / 4 = 75 Kg
In an obtuse triangle, the difference between the obtuse angle and one of the acute angles of a triangle is 20°, while the difference between the two acute angles is 56°. Find the measure (in degrees) of the smallest angle of the triangle.
Now, given that
a  b = 20 => a = 20 + h and
b  c = 56 => c = b  56
Also, a + b + c = 180 => (20 + b) + b + (b  56) = 180
=> 3b = 216 or b = 72 => a = 20 + 72 = 92
and c = 72  56 = 16
If x is an integer and x^{3} − 2(x + 2)^{2} = 993, find the value of x.
=> x^{3} + 23  2(x + 2)^{2} = 1001
=> (x + 2) [x^{2}  2x + 4  2x  4] = 1001
=> (x + 2) (x) (x  4) = 13 x 11 x 7 = 1001
∴ x = 11
In a school, there are three houses, A, B and C. If 170 students leave house B and join house C, the ratio of the number of students in house A and house C would be reversed. Instead, if 40 students leave house A and join house C, then the number of students in house A and house C will be equal. Find the number of students in house A and house C together.
Let the number of students in house A be x.
=> the initial number of students in house C be x  80.
=> x / x – 80 = (x – 80) + 170 / x
=> x^{2} = (x – 80)^{2} + 170(x  80)
=> x^{2} = x^{2}  160x + 6400 + 170x  13600
=> 10x = 7200
∴ x = 720
∴ The number of students in house A and house C is
x + x  80 = 2(720)  80 = 1360.
The product of three numbers is 1620. If the HCF of any two out of the three numbers is 3, find the LCM of the three numbers.
= HCF (3b, 3c) = HCF (3a, 3c) = 3
=> The LCM of the three numbers = 3abc
The product of the numbers = 27abc = 9(3abc) = 1620
∴ 3abc= 1620 / 9 =180
∴ The LCM of the three numbers is 180.0
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