A) 3 or 10 B) 10 C) 5 D) 7 or 10

A) 299 B) 310 C) 321 D) 330

**Q No:- 1, Correct Answer:- A**

Explanation:- It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From 1, h = 0.

From 2, d + e + f = 150 From 3, a = b = c

Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50

From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1

Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7. Since d + e + f = 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7.

Hence, a = 3, 10, 17, . . .

Further, since 3a + g = 50, a must be less than 17. Therefore, only two cases are possible for the value of a, i.e., 3 or 10. We can calculate the values of the other variables for the two cases.

a = 3 or 10

d = 18 or 10

e = 42 or 40

f = 90 or 100

g = 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.

BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.

BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

The number of candidates sitting for separate test for BIE who were at or above 90th percentile in CET (a) is either 3 or 10.

Explanation:- It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From 1, h = 0.

From 2, d + e + f = 150 From 3, a = b = c

Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50

From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1

Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7. Since d + e + f = 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7.

Hence, a = 3, 10, 17, . . .

Further, since 3a + g = 50, a must be less than 17. Therefore, only two cases are possible for the value of a, i.e., 3 or 10. We can calculate the values of the other variables for the two cases.

a = 3 or 10

d = 18 or 10

e = 42 or 40

f = 90 or 100

g = 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.

BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.

BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

From the given condition, g is a multiple of 5. Hence, g = 20. The number of candidates at or above 90th percentile overall and at or above 80th percentile in both P and M = e + g = 60.

**QNo:- 3 ,Correct Answer:- 170**

Explanation:- It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From 1, h = 0.

From 2, d + e + f = 150 From 3, a = b = c

Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50

From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1

Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7. Since d + e + f = 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7.

Hence, a = 3, 10, 17, . . .

Further, since 3a + g = 50, a must be less than 17. Therefore, only two cases are possible for the value of a, i.e., 3 or 10. We can calculate the values of the other variables for the two cases.

a = 3 or 10

d = 18 or 10

e = 42 or 40

f = 90 or 100

g = 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.

BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.

BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

In this case, g = 20. Number of candidates shortlisted for AET = d + e + f + g = 10 + 40 + 100 + 20 = 170

From 1, h = 0.

From 2, d + e + f = 150 From 3, a = b = c

Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50

From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1

Hence, a = 3, 10, 17, . . .

a = 3 or 10

d = 18 or 10

e = 42 or 40

f = 90 or 100

g = 41 or 20

From the given condition, the number of candidates at or above 90th percentile overall and at or above 80th percentile in P in CET

= 104. The number of candidates who have to sit for separate test = 296 + 3 = 299

**Verbal Ability : | Q.01- Q.06 | Q.07- Q.12 | Q.13- Q.18 | Q.19- Q.21 | Q.22- Q.24 | Q.25- Q.29 | Q.30 – Q.34 |**

**Logical Reasoning : | Q.01- Q.04 | Q.05- Q.08 | Q.09- Q.12 | Q.13- Q.16 | Q.17- Q.20 | Q.21- Q.24 | Q.25 – Q.28 | Q.29 – Q.32 |**

**Quantitative Aptitude: | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.25 | Q.26- Q.30 | Q.31 – Q.34 |**

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