Ans .
(c)
From the figure, it is clear that the number of people at the party were 30 + 10 +15 = 55. We can of course solve this mathematically as below: Let n(A) = No. of persons who kissed Sherry = 40 n(B) = No. of persons who shake hands with Sherry = 25 and n(A « B) = No. of persons who shook hands with Sherry and kissed him both = 10 Then using the formula, n(A » B) = n(A) + n(B) – n(A « B) n(A » B) = 40 + 25 – 10 = 55
Ans .
(b)
Ans .
(c)
Ans .
(b)
Ans .
(4)
Let P denote Physics, C denote Chemistry and M denote Maths. % of students who passed in P and C only is given by % of students who passed in P and C – % of students who passed all three = 20% – 8% = 12% % of students who passed in P and M only is given by % of students who passed in P and M – % of students who passed all three = 14% – 8% = 6% % of students who passed in M and C only is: % of students who passed in C and M – % of students who passed all three = 21% – 8% = 13% So, % of students who passed in P only is given by: Total no. passing in P – No. Passing in P & C only – No. Passing P & M only – No. Passing in all threeÆ 52% – 12% – 6% – 8%- = 26% % of students who passed in M only is: Total no. passing in M – No. Passing in M & C only – No. Passing P & M only – No. Passing in all threeÆ 43% – 13% – 6% – 8%- = 16% % of students who passed in Chemistry only is Total no. passing in C – No. Passing in P & C only – No. Passing C & M only – No. Passing in all three Æ 52% – 12% – 13% – 8%– = 19%
Ans .
(4)
The total number of people = 100 + 220 + 130 = 450
Ans .
(1)
The required answer would be 20000 – 5000 – 4000 – 8000 = 3000.
Ans .
(2)
Ans .
(2)
Ans .
(1)
(21 + 21 + 21 + 6 + 4 + 3 + 2)/21 = 78/21.
Ans .
(4)
There is no free capacity in the Avanti Vidisha pipeline.
Ans .
(4)
Avanti-Vaishali flow should be 700 and hence the free capacity is 300.
Ans .
(4)
Since 700 is required at Jhampur, the requirement at Mathura must be 1100, which has to be
supplied from the two pipelines coming into Mathura.
It is clear that since Vaishali to Mathura is only 300, the Vidisha-Mathura pipeline should
carry 800. Hence, Avanti Vidisha should have 1000.
1000.
Ans .
(2)
Refer to the following figure for the solution:
From the figure it would be 700 – (375 + 125) = 200
Ans .
(1)
125 + X – 200 – 675 = 225 Æ X = 975
Ans .
(2)
Refer to the following figure for the solution:
500 + 200 – 350 = 350.
Ans .
(2)
Solutions for Questions 1 and 2: Since there are 14 who are in triathlon and pentathlon, and there are 8 who take part in all three games, there will be 6 who take part in only triathlon and pentathlon. Similarly, Only triathlon and marathon = 12 – 8 = 4 & Only Pentathlon and Marathon = 15 – 8 = 7. The figure above can be completed with values for each sport (only) plugged in: The answers would be: 1. 3 + 6 + 8 + 4 + 5 + 7 + 10 = 43
Ans .
(1)
Option (1) is correct.
Ans .
(2)
The given situation can be read as follows: 115 students are being counted 75 + 65 + 90 = 230 times. This means that there is an extra count of 115. This extra count of 115 can be created in 2 ways. A. By putting people in the ‘passed exactly two subjects’ category. In such a case each person would get counted 2 times (double counted), i.e., an extra count of 1. B. By putting people in the ‘all three’ category, each person put there would be triple counted. 1 person counted 3 times – meaning an extra count of 2 per person. The problem tells us that there are 55 students who passed exactly two subjects. This means an extra count of 55 would be accounted for. This would leave an extra count of 115–55 = 60 more to be accounted for by ‘passed all three’ category. This can be done by using 30 people in the ‘all 3’ category. Hence, the answers are: 3. Option (2)
Ans .
(1)
Option (1)
Ans .
(2)
Solutions for Questions 5 to 8: Based on the information provided we would get the following
figure
The answers could be read off the figure as:
5. [(20 – 10)/10] * 100 = 100%. Option (2) is correct.
Ans .
(1)
15% (from the figure). Option (1) is correct.
Ans .
(3)
10+10+15+15=50%. Option (3) is correct.
Ans .
(2)
Only ice cream is 10% of the total. Hence, 10% of 180 =18. Option (2) is correct.
Ans .
(3)
Solutions for Questions 9 to 15: If you try to draw a figure for this question, the figure would be
something like:
We can then solve this as:
x – 10 + 28 – x + x + 30 – x + x + 2 + 32 – x + x – 6 = 94 Æ x + 76 = 94 Æ x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find
the missing value of students who liked all three as follows:
94 = 48 + 54 + 64 – 28 – 32 – 30 + All three Æ All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value – it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:
The answers then are:
8:12 = 2:3 Æ Option (3) is correct.
Ans .
(2)
12 % of 2000 = 240. Option (2) is correct.
Ans .
(1)
30/94 Æ more than 30%. Option (1) is correct.
Ans .
(4)
94%. Option (4) is correct.
Ans .
(3)
Option (3) is correct as the ratio turns out to be 10:20 in that case.
Ans .
(2)
12:12 = 1:1 Æ Option (2) is correct.
Ans .
(1)
14%. Option (1) is correct.
Ans .
(2)
30 = 25 + 20 – x Æ x = 15. Option (2) is correct
Ans .
(3)
Solutions for Questions 17 to 20:
Let people who passed all three be x. Then:
53 + 61 + 60 – 24 – 35 – 27 + x = 95
Æ x = 7.
The venn diagram in this case would become:
Option (3) is correct
Ans .
(3)
33% of 200 = more than 50. Option (3) is correct.
Ans .
(3)
If the number of students is increased by 50%, the number of students in each category would also be increased by 50%. Option (3) is correct.
Ans .
(1)
20:28 = 5:7. Option (1) is correct.
Ans .
(4)
Solutions for Questions 21 to 25: The following figure would emerge on using all the information in the question: The answers would then be: 240/880 = 27.27%. Option (4) is correct.
Ans .
(3)
504/880 = 57.27%. Hence, less than 60. Option (3) is correct
Ans .
(3)
40 + 16 + 56 + 24 = 136. Option (3) is correct.
Ans .
(1)
Option a gives us 16:128 = 1:8. Option (1) is hence correct.
Ans .
(4)
40:160 Æ 1:4. Option (2) is correct.
Ans .
(4)
Solutions for Questions 26 to 30: The following Venn diagrams would emerge: Math Students = 130. English Students =370 130/370 = 35.13%. Option (4) is correct
Ans .
(2)
Number of Female Students = 10 + 8 + 10 + 2 + 10 + 6 + 3 + 1 = 50. Average number of females per course = 50/4 = 12.5. Option (2) is correct.
Ans .
(2)
50:450 = 1:9. Option (2) is correct
Ans .
(3)
40/140 Æ 28.57%. Option (3) is correct.
Ans .
(1)
From the figures, this value would be 150 + 8 + 90 + 10 + 16 + 6 + 37 + 3 = 320. Option(1) is correct.
Ans .
(1)
Solutions for Questions 31 to 34: The following figure would emerge-
Based on this figure we have:
x + x – 13 + 4 + x – 16 = 302 Æ 3x – 25 = 302 Æ x = 327. Hence, x =109.
Consequently the figure becomes:
The answers are:
31. 91 + 93 + 96 = 280. Option (1) is correct.
Ans .
(3)
193/302 @ 64%. Option (3) is correct
Ans .
(2)
6:9:4 is the required ratio. Option (2) is correct.
Ans .
(4)
96 – 4 = 92. Options (4) is correct.
Ans .
(2)
78 = 36 + 48 + 32 – 14 – 20 – 12 + x Æ x = 8. The figure for this question would become: Required ratio is 18:8 Æ 9:4. Option (1) is correct
Ans .
(3)
Option (3) is correct.
Ans .
(2)
There are 30 such people. Option (2) is correct.
Ans .
(2)
Solutions for Questions 38 to 42:
38. Option (2) is correct
39. Option (2) is correct.
40. Option (4) is correct.
41. Option (1) is correct
42. Option (2) is correct
Ans .
(3)
Solutions for Questions 43 to 45:
The answers can be read off the above grid.
43. Option (3) is correct
Ans .
(4)
44. The sum is 208 = 80 + 64 + 36 + 28. Option (4) is correct.
Ans .
(4)
45. There are 2 nodes at the 4th level. Option (4) is correct.
Ans .
(3)
46. The least percentage of people with all 4 gadgets would happen if all the employees who are not having any one of the four objects is mutually exclusive. Thus, 100 – 30 – 25 – 20 – 15 = 10 Option (3) is correct.