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Here is the question for today, please solve it and let us know your approach and answer in the comments section
Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of x%. If both make the same non-zero profit, find x.
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Answer -
SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).
Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000xx2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%
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Here is the question for today, please solve it and let us know your approach and answer in the comments section
If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?
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Answer - B
SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).
Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000xx2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%
Hello Aspirants!
Here is the question for today, please solve it and let us know your approach and answer in the comments section.
Consider a right–angled triangle with inradius 2 cm and circumradius of 7 cm. What is the area of the triangle?
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Hello Aspirants!
Answer - A
r = 2
R = 7 (Half of hypotenuse)
Hypotenuse = 14
r = (a + b - h)/2
2 = (a + b - 14)/2
a + b - 14 = 4
a + b = 18
a2 + b2 = 14 2
a2 + (18 - a)2 = 14 2
a2 + 324 + a2 - 36a = 196
2a2 - 36a + 128 = 0
a2 - 18a + 64 = 0
Now the 2 roots to this equation will effectively be a, 18 – a. Product of the roots = 64.
Area = 1/2 * product of roots (How? Come on, you can figure this out.)
= 32 sq. cms
The question is “What is the area of the triangle?”
Hence, the answer is 32 sq. cms
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Here is the question for today, please solve it and let us know your approach and answer in the comments section.
A circle of radius 5 cm has chord RS at a distance of 3 units from it. Chord PQ intersects with chord RS at T such that TS = 1/3 of RT. Find minimum value of PQ.
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Hello Aspirants!
Answer - B
OM = 3, OS = 5
MS = 4 = RM {Using Pythagoras theorem}
=> RS = 8 cms
TS = 1/3 of RT
TS = 1/4 of RS
TS = 2 cms
RT * TS = PT * TQ
{Intersecting Chords theorem: When there are two intersecting chords, the product of the rectangle formed by the segments of one chord is equal to the product of the rectangle formed by the segments of the other.}
6 × 2 = PT * TQ
PT * TQ = 12
By AM – GM inequality, (PT+TQ)/2 ≥ √(PT * TQ)
(PT+TQ)/2 ≥ √12
PT + TQ ≥ 2√12
=> PQ ≥ 2√12
Or PQ ≥ 4√3
Minimum PQ = 4√3
The question is " Find minimum value of PQ."
Hence, the answer is 4√3.
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CATKing Team
Hello Aspirants!
Here is the question for today, please solve it and let us know your approach and answer in the comments section.
A new game show on TV has 100 boxes numbered 1, 2, …, 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, …, in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.
1- What is the minimum possible number of different types of prizes?
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2- What is the maximum possible number of different types of prizes?
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3-Which of the following is not possible?
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4- You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100. What is the maximum possible number of different types of items?
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Dear Aspirants!
Answer for Question 1
Answer - B
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CATKing Team
Dear Aspirants!
Answer for Question 2
Answer - C
Maximum possible number is when 1 of a, 2 of b, 4 of c, 8 of d, 16 of e, 32 of f,
Now left boxes woud be 100 – (1+2+…32) = 37
Now if one more type is to be added then we need at least 64 which is not available thus maximum possible is 6.
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CATKing Team
Dear Aspirants!
Answer for Question 3
Answer - D
Lets try to prove the given options possible using easy numbers.
A : Never possible
B : 1,30,69 is possible
C : 1,2,4,18.75 is possible
D : 1,9,30,60 is possible.
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CATKing Team
Dear Aspirants!
Answer for Question 4
Answer - A
There have to be then at least 31 + `1 + 43 = 75 gifts of same type,
Thus maximum possible number of boxes = 5 when all types are lest 1,2,4,18,75
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CATKing Team
Hello Aspirants!
Here is the question for today, please solve it and let us know your approach and answer in the comments section.
A supermarket has to place 12 items (coded A to L) in shelves numbered 1 to 16. Five of these items are types of biscuits, three are types of candies and the rest are types of savouries. Only one item can be kept in a shelf. Items are to be placed such that all items of same type are clustered together with no empty shelf between items of the same type and at least one empty shelf between two different types of items. At most two empty shelves can have consecutive numbers.
The following additional facts are known.
1- In how many different ways can the items be arranged on the shelves?
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2- Which of the following items is not a type of biscuit?
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3- Which of the following can represent the numbers of the empty shelves in a possible arrangement?
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4- Which of the following statements is necessarily true?
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Dear Aspirants!
Answer for Question- 1
Answer - D
Total number of Biscuits = 5
Total number of Candies = 3
So Total number of Savouries = 12 – 5 – 3 = 4
From point iii) and iv) it is clear that D, E, F and K are 4 savouries and are kept in shelves numbered 13,14, 15 and 16 as there is no empty shelf between items of the same type.
From point II), V and VII) I, J and L are of the same type L being in the least numbered shelf among 3.
As from point VI,) C is candy so I, J and L must be Biscuits as there are only 3 candies.
From point V) H is not Biscuits so it must be a candy thus A and B must be Biscuits.
Now Item can be placed as given below.
Case 1) when all biscuits are placed after candies.
Case 2) When all candies are placed after biscuits.
Total 8 cases are possible.
All questions can be answered now.
There are two empty shelves between the biscuits and the candies.
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CATKing Team
Dear Aspirants!
Answer for Question- 2
Answer - D
Total number of Biscuits = 5
Total number of Candies = 3
So Total number of Savouries = 12 – 5 – 3 = 4
From point iii) and iv) it is clear that D, E, F and K are 4 savouries and are kept in shelves numbered 13,14, 15 and 16 as there is no empty shelf between items of the same type.
From point II), V and VII) I, J and L are of the same type L being in the least numbered shelf among 3.
As from point VI,) C is candy so I, J and L must be Biscuits as there are only 3 candies.
From point V) H is not Biscuits so it must be a candy thus A and B must be Biscuits.
Now Item can be placed as given below.
Case 1) when all biscuits are placed after candies.
Case 2) When all candies are placed after biscuits.
Total 8 cases are possible.
All questions can be answered now.
All candies are kept before biscuits.
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CATKing Team
Dear Aspirants!
Answer for Question- 3
Answer - C
Total number of Biscuits = 5
Total number of Candies = 3
So Total number of Savouries = 12 – 5 – 3 = 4
From point iii) and iv) it is clear that D, E, F and K are 4 savouries and are kept in shelves numbered 13,14, 15 and 16 as there is no empty shelf between items of the same type.
From point II), V and VII) I, J and L are of the same type L being in the least numbered shelf among 3.
As from point VI,) C is candy so I, J and L must be Biscuits as there are only 3 candies.
From point V) H is not Biscuits so it must be a candy thus A and B must be Biscuits.
Now Item can be placed as given below.
Case 1) when all biscuits are placed after candies.
Case 2) When all candies are placed after biscuits.
Total 8 cases are possible.
All questions can be answered now.
All biscuits are kept before candies.
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CATKing Team
Hello Aspirants!
Here is the question for today, please solve it and let us know your approach and answer in the comments section.
How many factors of 25 * 36 * 52 are perfect squares?
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CATKing Team
Answer - B
Given a prime factorization, how do we find how many factors of the number are perfect squares.
Any factor of this number should be of the form 2a * 3b * 5c.
For the factor to be a perfect square a,b,c have to be even.
a can take values 0, 2, 4.
b can take values 0, 2, 4, 6
and c can take values 0, 2
Total number of perfect squares = 3 * 4 * 2 = 24
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CATKing Team
