### Q.1 Out of 120 applications for a post, 70 are male and 80 have a driver’s license. What is the ratio between the minimum to maximum number of males having driver’s license?

### A) 1 to 2

### B) 2 to 3

### C) 3 to 7

### D) 5 to 7

### Ans. C

### n(A U B) = n(A) + n (B) – n (A n B)

### 120 = 70 + 80 – n (A n B)

### n (A n B)= 150-120

### =30

### Therefore required ratio = 30:70

### =3:7 Ans.

###

### Q.2 In an examination there are three subjects A, B & C.A student has to pass in each subject. 20% students failed in A, 22% students failed in B & 16% failed in C. The total number of students passing the whole examination lies between :

### A) 58% & 84%

### B) 42% & 78%

### C) 42% & 84%

### D) 58% & 78%

### Ans. C

### Total % of students failed individually in each subject : 20+22+16 = 58

### Total % of students failed commonly in subjects: 16

### Therefore total no. of students passing the whole examination lies between 100-58 & 100-16

### i.e. 42% & 84%

###

### Q.3 In an examination 70% of the students passed in the paper I & 60% of the students passed in the paper II. 15% of the students failed in both the papers while 270 students passed in both the papers. What is the total no. of students ?

### A) 560

### B) 580

### C) 540

### D) 600

### Ans. D

### Total % of failed students = 15+15+25 = 55%

### Total % of students passed in both the papers = 45%

### If total no. of students = x

### X*45/100 = 270

### Or 45x=270*100

### X= 270*100 / 45 = 600

###

### Q.4 In how many ways can four books A,B,C & D be arranged one above other in a vertical order such that the books A & B are never in continuous position ?

### A) 14

### B) 18

### C) 12

### D) 9

### Ans. C

### Total no. of arrangements – (no. of arrangements in which A & B are together)

### 4! – 3! * 2! = 24-12 = 12

###

### Q.5 Five balls of different colours are to be placed in three different boxes such that any box contains atleast one ball. What is the maximum no. of different ways in which this can be done ?

### A) 120

### B) 180

### C) 150

### D) 90

### Ans. C

### n= 5

### r=3

### Therefore, no. of ways = rn – rC1 (r-1)n+rC2 (r-2)2 + …

### 35 – 3C1(3-1)5 + 3C2 (3-2)2

### 243-3*32 + 3*1

### =243-96+3 = 150

###

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