Q. 1: What is the minimum number of students enrolled in both G and L but not in K?

1. 18

2. 19

3. 17

4. 22

1. 6

2. 7

3. 5

4. 8

The given data can be represented as follows.

f + g + d =10 (given) g + e = b (given)

Since f + g + d = 10, g = 7 = 2g – 1 Therefore, 2g = 8 f = 4

Thus, g = 4, c = 8, a = 7 and f + d = 6

b + e = 39 – (G + c) = 14

therefore g + 2e = 14 Hence, e = 5 and b = 9 Since, L is maximum we get the following cases. Case (i)

G = 17 K = 20 L = 21 d = 2 f = 4

Case (ii)

G = 17 K = 19 L = 22 d = 1 f = 5

Case (iii)

G = 17 K = 18 L = 23 d = 0 f = 6

Question 1:

G and L but not K = f = 4. Ans : 4

Question 2:

The given condition is possible in case (ii). Hence, the number of students enrolled in L = 22.

Ans : 22

question 3:

From g = 4, one person moves to f, one person to d and two persons to e. Then the value of G and K = d + g = 2. Ans : 2

Question 4:

From the above G and L = f = 6. Ans : 6

**Verbal Ability : | Q.01- Q.05 | Q.06- Q.9 | Q.10- Q.14 | Q.15- Q.19 | Q.20- 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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