Let Tea A and B be sold at Rs.a and Rs.b respectively 
When A and B are mixed in the ratio 3 : 2 and sold at a profit of 10% at Rs.40 per Kg 
3a+2b / 5 = 40/1.1------(1) 
When A and B are mixed in the ratio 2 : 3 and sold at a profit of 5% at Rs.40 per Kg 
2a+3b/5 = 40/1.05 ------(2) 
Solving equations (1) and (2), we get 
1.1 × (3a+2b/5) = 1.05 × (2a+3b/5) 
11/10 × (3a + 2b) = 21/20 × (2a + 3b) 
66a + 44b = 42a + 63b 
24a = 19b 
a/b = 19/24 
So, the cost prices per Kg of A and B are in the ratio 19 : 24 

Let there be 'n' tests and the Overall Average score be k 
Average (n) = k => Total marks = nk 
So, when we ignore the first 10 questions, 
Average (n-10) = k+1 
Similarly, if we ignore the last 10 questions, 
Average (first (n-10)) = k-1 
It is given that when the first 10 tests are not considered, the overall average increases by 1
(Each question carries 20 marks) 
Total marks – 20×10 = (k+1) (n-10) 
kn – 200 = (k+1) (n-10) ---(1) 
Similarly, If the last 10 tests are not considered, the overall average decreases by 1 (Each
question carries 30 marks) 
kn - 300 = (k-1) (n-10) ---(2) 
Solving (1) and (2), we get 
kn - 200 = (k+1) (n-10) 
(-) kn - 300 = (k-1) (n-10) 
------------------------------------- 
100 = 2(n-10) 
50 = n-10 
n = 60 questions 

Let Narayanan take X hrs to reach B then Partha would take X + 4 hrs 
Its given that Partha reaches the mid-point of A and B two hours before Narayan reaches B 
X+4 / 2 = X – 2 => X=8 hrs 
So, Partha would take 8+4 = 12 hrs to travel 60 Kms at a speed of 60/12 Kmph 
Speed of Partha = 5 Kmph

Given, f(x+2) = f(x) + f(x+1) f(11) = 91, f(15) = 617 
We get 91 + f(12) = f(13) 
Let f(12) be equal to some value ‘a’ 
So, 91 + a = f(13). 
f(12) + f(13) = f(14) 
a + 91 + a = f(14) 
So, f(14) = 2a + 91 f(13) + f(14) = 617 
So, 91 + a + 2a + 91 = 617 
3a + 182 = 617 
a = 145 
Substituting the value of a and f(11), we get 
f(10)+ 91 = 145 
f(10) = 54

. From 9-11: 30, we have 150 minutes. Doctors Ben, Kane and Wayne take 10,15 and 20
minutes respectively for each patient. Therefore Ben, Kane and Wayne can see 150/10=15,
150/15=10 and 150/25=6 respectively. Therefore Ben, Kane and Wayne can see a maximum
of 15,10 and 6 patients respectively every day. Sum =31

Given, on Saturday, the queue is not empty.
=> Each doctor sees the maximum number of patients on a day.
Given Ben, Kane and Wayne charge Rs.100, 200 and 300 respectively.
=> Ben earns 15 (100)= Rs.1500,Kane earns 10(200)
= Rs .2000, Wayne earns 6(300)= Rs .1800
Therefore, Kane earns the maximum amount.
Ans: (Dr. Kane)

Mr. Singh takes maximum duration when he enters Dr. Wayne’s room, who sees each
patient for 25 minutes

He was at the clinic for 85 minutes.
Similarly On Wednesday, he would meet Ben and he would be at the clinic for 70 minutes.
On Friday, he would meet Ben and he would be at the clinic for 70 minutes.
Therefore, Singh stays at the clinic for the maximum duration on Monday
Ans: (Monday)

The above pattern continues.
Hence, there is no time where all the doctors are simultaneously free.
Ans: (0)

Given ∠AOB = 60° 
Area of Sector AOB = 60/360 × π = π/6 ---(1) 
Given OC = OD => ∠OCD = ∠ODC = 60° 
△OCD is an Equilateral Triangle with side = a 
Area(△OCD) = √3/4 × a × a ---(2) 
Its given that Area(OCD) = 1/2 × Area(OAB) 
a 2 (√3/4) = π / ( 6×2) 
a = (π/ 3√3) 1/2

Given, The amount of B in mixture ≤ Amount of A in mixture 
Selling Price = Rs. 264 and Profit = 10 % 
264 = 1.1 × CP 
CP = 264/1.1 = Rs. 240 for 10 litres 
CP per litre = Rs. 24 
We also know that, CP per litre of A = CP per litre of B + 8 Let CP per litre of B = x So, CP per
litre of A = x + 8 

Given Amount of B ≤ Amount of A, 
Maximum possible cost of B occurs when B = A 
We need to choose values for A and B in such a way that the quantities remain same.But A should
be 8 more than B 
Therefore, A should have a CP per litre of Rs. 28 and B should have a CP per litre of Rs 20 

B cannot be assigned any more than this as the amount of B would become more than A 
Example: Let us assign B to be Rs. 21 and A to be Rs. 29 which would result in a ratio of 3:5 where
B amounts more than A, which doesn’t satisfy the condition 
Max possible cost of Paint B = Rs. 20 per litre 

The set of students who like Sunita and Ragini are disjoint sets.
Hence, the Venn diagram can be drawn as follows

There are 500 students in all.
From statement (2)
Sunita = 200. Hence, Ragini = 300.
From statement (1) A (Sunita) + A (Ragini) = 250 and B (Sunita) + B (Ragini) = 250.
From (2), A (Sunita) = 160. Hence, A (Ragini) = 90.
From (4), B (Sunita) = 20 % of 250 = 50. Hence, B (Ragini) = 200.
From (6), g (Sunita) = 50 and hence, b (Sunita) = 0 and a (Sunita) = 110. Hence, n (Sunita) =
40.
From (7), n (Ragini) = 60
It is given that 250 support B, hence the other 250 do not support B.
From (5), (a + n) of Ragini = 40 % of 250 = 100. Hence, a (Ragini) = 40.
Thus, the final solution is as follows.

The students who supported proposal B but not A are b (Sunita) and b (Ragini). Among
them those supported Ragini are b (Ragini) 150. Ans: (150)

The set of students who like Sunita and Ragini are disjoint sets.
Hence, the Venn diagram can be drawn as follows

There are 500 students in all.
From statement (2)
Sunita = 200. Hence, Ragini = 300.
From statement (1) A (Sunita) + A (Ragini) = 250 and B (Sunita) + B (Ragini) = 250.
From (2), A (Sunita) = 160. Hence, A (Ragini) = 90.
From (4), B (Sunita) = 20 % of 250 = 50. Hence, B (Ragini) = 200.
From (6), g (Sunita) = 50 and hence, b (Sunita) = 0 and a (Sunita) = 110. Hence, n (Sunita) =
40.
From (7), n (Ragini) = 60
It is given that 250 support B, hence the other 250 do not support B.
From (5), (a + n) of Ragini = 40 % of 250 = 100. Hence, a (Ragini) = 40.
Thus, the final solution is as follows.

The required value is 160/250*100 = 64

The set of students who like Sunita and Ragini are disjoint sets.
Hence, the Venn diagram can be drawn as follows

There are 500 students in all.
From statement (2)
Sunita = 200. Hence, Ragini = 300.
From statement (1) A (Sunita) + A (Ragini) = 250 and B (Sunita) + B (Ragini) = 250.
From (2), A (Sunita) = 160. Hence, A (Ragini) = 90.
From (4), B (Sunita) = 20 % of 250 = 50. Hence, B (Ragini) = 200.
From (6), g (Sunita) = 50 and hence, b (Sunita) = 0 and a (Sunita) = 110. Hence, n (Sunita) =
40.
From (7), n (Ragini) = 60
It is given that 250 support B, hence the other 250 do not support B.
From (5), (a + n) of Ragini = 40 % of 250 = 100. Hence, a (Ragini) = 40.
Thus, the final solution is as follows.

The required answer is 210/250*100 = 84

The set of students who like Sunita and Ragini are disjoint sets.
Hence, the Venn diagram can be drawn as follows

There are 500 students in all.
From statement (2)
Sunita = 200. Hence, Ragini = 300.
From statement (1) A (Sunita) + A (Ragini) = 250 and B (Sunita) + B (Ragini) = 250.
From (2), A (Sunita) = 160. Hence, A (Ragini) = 90.
From (4), B (Sunita) = 20 % of 250 = 50. Hence, B (Ragini) = 200.
From (6), g (Sunita) = 50 and hence, b (Sunita) = 0 and a (Sunita) = 110. Hence, n (Sunita) =
40.
From (7), n (Ragini) = 60
It is given that 250 support B, hence the other 250 do not support B.
From (5), (a + n) of Ragini = 40 % of 250 = 100. Hence, a (Ragini) = 40.
Thus, the final solution is as follows.

The required answer is 50/250*100 = 50

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

 The Low monsoon states are Gujarat, Karnataka, Rajasthan, MP, Assam, WB, Jharkhand,
Delhi and Manipur. The deviation from LPA for these states are 25,20,15, 10,-10,-30,-35,-40
and -60. The median value is -10

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The states which have a negative deviation from LPA and have an actual rainfall of 600 mm
or less are Assam, WB, Jharkhand, Delhi and Manipur. The average rainfall in these states is
2300/5 = 460 mm

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The actual rainfall in Gujarat in 2019 is 600 mm. The rainfall in Gujarat in 2009 was 500 mm.
As the value of 500 is replaced by 600 in calculating the LPA, the LPA would increase by 10
as it is the average of 10 years.
Ans: (490 mm)

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The heavy monsoon states are Maharashtra, Sikkim, Mizoram, Goa, Arunachal, Kerala and
Meghalaya. Among these, Arunachal, Kerala and Meghalaya have a negative deviation from
respective LPAs in 2019. The required percentage = 3/7*100 = 42.86

From (1) and (5), the persons in Team 1 speak English. Chinese. Arabic and French. (Robert
speaks both Arabic and French).
From (1) and (5), the persons in Team 3 speak. English, Chinese and Dutch. (Quentin speaks
Dutch and English). Since each person speaks two languages and each team speaks exactly
four languages, we need to find one person for Team 3. who speaks one language among
English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in
exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from (5) and the above, Paula, Quentin and Sally, (Basque. Chinese. Dutch and
English) are in Team 3. Since there are three persons in Team 3. Teams 1 and 2 should also
have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic
and French) is one of the team members. Now, two more persons, who speak languages
among the above four are to be selected. It is possible only with Paula and Terence.
From (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two
languages. Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to
find one more person, who speaks one of these three languages and French. It is possible
with only Terence.

From (1) and (5), the persons in Team 1 speak English. Chinese. Arabic and French. (Robert
speaks both Arabic and French).
From (1) and (5), the persons in Team 3 speak. English, Chinese and Dutch. (Quentin speaks
Dutch and English). Since each person speaks two languages and each team speaks exactly
four languages, we need to find one person for Team 3. who speaks one language among
English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in
exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from (5) and the above, Paula, Quentin and Sally, (Basque. Chinese. Dutch and
English) are in Team 3. Since there are three persons in Team 3. Teams 1 and 2 should also
have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic
and French) is one of the team members. Now, two more persons, who speak languages
among the above four are to be selected. It is possible only with Paula and Terence.
From (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two
languages. Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to
find one more person, who speaks one of these three languages and French. It is possible
with only Terence.

From (1) and (5), the persons in Team 1 speak English. Chinese. Arabic and French. (Robert
speaks both Arabic and French).
From (1) and (5), the persons in Team 3 speak. English, Chinese and Dutch. (Quentin speaks
Dutch and English). Since each person speaks two languages and each team speaks exactly
four languages, we need to find one person for Team 3. who speaks one language among
English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in
exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from (5) and the above, Paula, Quentin and Sally, (Basque. Chinese. Dutch and
English) are in Team 3. Since there are three persons in Team 3. Teams 1 and 2 should also
have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic
and French) is one of the team members. Now, two more persons, who speak languages
among the above four are to be selected. It is possible only with Paula and Terence.
From (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two
languages. Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to
find one more person, who speaks one of these three languages and French. It is possible
with only Terence.

From (1) and (5), the persons in Team 1 speak English. Chinese. Arabic and French. (Robert
speaks both Arabic and French).
From (1) and (5), the persons in Team 3 speak. English, Chinese and Dutch. (Quentin speaks
Dutch and English). Since each person speaks two languages and each team speaks exactly
four languages, we need to find one person for Team 3. who speaks one language among
English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in
exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from (5) and the above, Paula, Quentin and Sally, (Basque. Chinese. Dutch and
English) are in Team 3. Since there are three persons in Team 3. Teams 1 and 2 should also
have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic
and French) is one of the team members. Now, two more persons, who speak languages
among the above four are to be selected. It is possible only with Paula and Terence.
From (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two
languages. Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to
find one more person, who speaks one of these three languages and French. It is possible
with only Terence.

If n is even, f(n) = n (n + 1)

So, f (2) = 2(2+1) = 2 (3) = 6

If n is odd, f(n) = n + 3

f (1) = 1 + 3 = 4

It is given that, 8 x f(m+1) - f(m) =2

So, m can either be even or odd

Case-1: If m were even and m+1 odd

So, 8 x f(m+1) - f(m) =2
8(m + 4) - m (m + 1) = 2
8m + 32 - m2
- m = 2
m2
- 7m - 30 = 0
(m-10) (m+3) = 0
m = 10 or -3
m = 10, since m is positive

Case-2: If m were even and m+1 odd

8 x f(m+1) - f(m) =2
8 (m +1) (m + 2) - (m + 3) = 2
Now, let us substitute m = 1 which is the minimum possible value
8 (1 + 1) (1 + 2) - (1 + 3) 3

Case 2 does not work

Hence the answer is 10.

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%

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

=> Number of students passed = x 150 = 102 students

=> Since the number of boys passed is 30,

=> Number of girls passed = 102 - 30 = 72

Total number of girls = x 150 = 90

Total number of girls = Girls who passed + Girls who did not pass

Girls who did not pass = 90 - 72 = 18

Percentage of girls who did not pass = 18/90 x 100 = 20 %