Among all these 2018 pairs, one of the integer is less than or equal to the other. (Equal in the case of (2017,2017))
We assume that a is the least one of the two…
Hence there are 2018 pairs that satify this condition.
Alternate Method:
x + y = 2×2017
We have 2 conditions to deal with…
x ≤ y and x + y = 2×2017
Let’s start with x = y:
Here, x = y = 2017
From here on, we decrement x and increment y to maintain the conditions x ≤ y and x + y = 2×2017
We can keep doing this until x = 0, because if x is negative, a which is 2x will not remain an integer.
x
y
2017
2017
2016
2018
2015
2019
⁝
⁝
⁝
⁝
⁝
⁝
2
4033
1
4034
0
4035
Hence x can range from 2017 to 0; and thereby x can take 2018 values.
In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is [TITA]
Given that there are 60 % girls and 40 % boys
It is also given that there are 30 more girls than boys.
So, (60 % - 40 %) of total class strength = 30 students
=> 20 % of total class strength = 30 students
=> Total class strength = 30 x 5 = 150 students
It is also given that 68% of students pass the the examination, which includes 30 boys
So, Number of students passed = 68% of Total students
=> Since the number of boys passed is 30,
=> Number of girls passed = 102 - 30 = 72
Total number of girls = Girls who passed + Girls who did not pass
Girls who did not pass = 90 - 72 = 18
With rectangular axes of coordinates, the number of paths from (1,1) to (8,10) via (4,6), where each step from any point (x,y) is either to (x,y+1) or to (x+1,y) is [TITA]
Let us first consider travelling from (1,1) to (4, 6)
This means, Travelling from 1 to 4 units in the x axis → 3 horizontal movements (h h h)
And travelling from 1 to 6 units in the y axis → 5 vertical movements (v v v v v)
No matter how we proceed, reaching from (1,1) to (4,6) requires 5 vertical movements and 3 horizontal movements.
So, Number of paths to travel from (1,1) to (4,6) = Number of ways of arranging (h h h v v v v v)
Similarly, travelling from (4, 6) to (8, 10) requires 4 horizontal movements and 4 vertical movements
A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is
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I am getting 95 Percentile in CL PRIME CAT. Is it good to score 99.5Percentile in Actual CAT22??
From observing the data given, we find that it is a closed 3 set Venn diagram.
Let the three sports be F, T and C for Football, Tennis and Cricket respectively
n(FUTUC) = 256 , n(F) = 144, n(T) = 123, n(C) = 132, n(F T) = 58, n(C ∩T) = 25, n(F C) = 63
We know that (AUBUC) = n(A) + n(B) +n(C) - n(A B) - n(B ∩C) - n(C A) + n(A ∩B C)
So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n (F T ∩C)
n (F T C) = 256 - 144 + 123 +132 - 146
n (F T C) = 256 - 253 = 3
Now, it is easy to calculate the number of students who only play tennis using a Venn diagram.
n (Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80
n (Students who play only Tennis) = 43 students
In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is
The idea that we use here is remarkably simple.
We know, the product of rectangles formed by two intersecting chords are always equal
So, AE x EB = CE x ED
AE x EB = 7 x 15
Also, we know that AB = 20.5 cms = AE + EB
So, the Sum of AE and EB must be 20.5 and their product must be equal to 7 x 15
7 x 15 = 105
The numbers must be close to each other, for their sum to be 20.5
From trial and error, we find their values to be 10 and 10.5 cms respectively
So, AE x EB = 7 x 15
10 x 10.5 = 7 x 15
Difference = 10.5 - 10 cms = 0.5 cms
Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is
Meena scores 40 % in an exam
After review, she scores 50 % more => Increase of 50 % from 40 % = 40% + 20% = 60%
She fails by 35 marks, by scoring 60%
60% score = Pass mark - 35 ----- (1)
If her post review score is increased by 20%, she would have 7 more than the pass mark.
20% of 60% = 12 %
So, 60% + 12% = 72% of marks = Pass mark + 7 ------ (2)
So, 12% marks = 35 + 7 (5: 1 ratio)
So, similarly 12% can be re written as 10 % and 2 % (maintaining the 5:1 ratio)
Hence the pass percentage = 60 % + 10 % = 70%
(or)
Pass percentage = 72 % - 2 % = 70%
Construct an Equilateral triangle and cut of equal lengths from all three sides. So we obtain a hexagon and three equilateral triangles of side ‘a’ in return.
We know that a hexagon comprises of 6 Equilateral Triangles. Since the side of the hexagon is also ‘a’, we obtain 6 equilateral triangles in return.
Thus, we have a total of 6 + 3 = 9 equilateral triangles of side ‘a’
Ratio of area of Hexagon with Area of Triangle = 6: 9 = 2: 3
Let T be the triangle formed by the straight line 3x + 5y - 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is [TITA]
We know that the equation of the straight line is 3x + 5y = 45
The intercepts are (15,0) and (0,9) respectively
Since it’s a right-angled triangle, we know that Circumradius
We know that is approximately equal to 6
So, from trial and error to find the closest number, we find that the value of Circumradius is very close to 9
So, the integer closest to L = 9
For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) - f(m) = 2, then m equals [TITA]
T) = 58, n(C ∩T) = 25, n(F 
is approximately equal to 6