RBI GRADE ‘B’ OFFICER’S EXAM Held on: 11.10.2009 (Part-II)

Total Questions: 50

1. In these questions, a question is given followed by information in three statements. You have to consider the information in all the three statements and decide the information in which of the Statement(s) is not necessarily required to answer the question and therefore can be dispensed with. Indicate your answer accordingly.

How many students from Institute 'A' got placement ?

I. Number of students studying in Institutes A and B are in the ratio of 3:4 respectively,
II. Number of students who got placement from Institute B is 120% of the number of students who got placement from Institute A.
III. 80% of the students studying in Institute B got placement.

Correct Answer: (5) Question cannot be answered even with the information in all three statements
Solution:From statement I. Let the number of students in the institutes A and B be 3x and 4x respectively. However we get no conclusive answer by using data given in all the statements.

2. What is the monthly income of Mr. X?

I. Mr. X spends 85% of his in- come on various items and remaining amount is saved.
II. Monthly savings of Mr. X are Rs.4,500
III. Out of the total money spent by Mr. X in a month, one-fifth is spent on food and remaining amount of Rs.20, 400 on other items.

Correct Answer: (3) Only either II or III
Solution:From statements I and II,
Let Mr. X's monthly income
= Rs. x. Then,

From statements I and III,

x ×  = 20400

3. What is Suchitra's present age?

I. Suchitra's present age is double the age of her son.
II. Ratio between present ages of Suchitra and her mother is 2:3 respectively.
III. Four years hence the ratio between 'Suchitra's age and her son's age will be 13: 24 respectively.

Correct Answer: (1) Only II
Solution:From statements I and III.
Let Suchitra's son's present age
be x years.
.. Suchitra's present age
= 2x years
After 4 years,

We can get the required answer by this relation.

4. What is Neeta's share in the profit earned at the end of 2 years in a joint business run by Neeta, Seeta and Geeta ?

I. Neeta invested Rs. 85,000 to start the business.
II. Seeta and Geeta joined Neeta's business after six months investing amounts in the ratio of 3: 5 respectively.
III. Totat amount invested by Seeta and Geeta is Rs.2.3 lakhs.

Correct Answer: (4) Information in all the three statements is required for answering the question.
Solution:From all three statements, Seeta's investment

= Rs. ( x 25) lakh

= Rs. 93750

Geeta's investment = Rs. 156250

Ratio of Neeta's, Seeta's and Geeta's profit

= 85000 x 24 : 93750 x 18 :
156250 x 18
We can get the share of Neet if total profit is given.

5. What is the labelled price of the article ?

I. Cost price of the article is Rs.500.
II. Selling price after offering 5% discount on the labelled price is Rs. 608.
III. Profit earned would have been 28% if no discount was offered.

Correct Answer: (4) Only I and III
Solution:Let the marked price.of th
article be Rs. x.
From statement II,

6. Study the following information carefully to answer these questions:

For an examination consisting of three subjects-Maths, Physics and Chemistry, 280 students appeared. When the results were declared, 185 students had passed in Maths, 210 had passed in Physics and 222 had passed in Chemistry.

All those except 5 students who passed in Maths, passed in Physics.
All those except 10 students who passed in Maths, passed in Chemistry.
47 students failed in all the three subjects.
200 students who passed in Physics also passed in Chemistry.

How many students passed in Chemistry only ?

Correct Answer: (1) 18
Solution:Number of students = 280
Cleared Maths = 185
Cleared Physics= 210
Cleared Chemistry = 222
Students who passed in Maths
and Physics = 180
Students who passed in Maths
and Chemistry = 175
Students who passed in Physics and Chemistry = 200
Number of students who
passed in atleast one subject
= 280 - 47 = 233

x + w = 180 ....(i)
w + z = 200 ....(ii)
w + y = 175 ....(iii)
M + x + w + y = 185 ....(iv)
P + x + w + z = 210 ....(v)
C + y + w + z = 222 ....(vi)
M + P + C + y + x + z + w = 233
....(vii)
By (iv) + (v) + (vi) - (vii),
x + y + z + 2w = 384 ....(viii)
by (i) + (ii) + (iii),
x + y + z + 3w = 555 ....(ix)
By (ix) - (viii),
w = 171
From equation (i),
x=9

From equation (ii),
z = 200 - 171 = 29
From equation (iii),
y = 175 - 171 = 4
M = 185 - 4 - 9 - 171 = 1
P = 210 - 9 - 171 - 29 = 1
C = 222 - 4 - 29 - 171 = 18

Number of students passed
in Chemistry only = 18

7. How many students passed in all the three subjects ?

Correct Answer: (4) 171
Solution:Number of students = 280
Cleared Maths = 185
Cleared Physics= 210
Cleared Chemistry = 222
Students who passed in Maths
and Physics = 180
Students who passed in Maths
and Chemistry = 175
Students who passed in Physics and Chemistry = 200
Number of students who
passed in atleast one subject
= 280 - 47 = 233

x + w = 180 ....(i)
w + z = 200 ....(ii)
w + y = 175 ....(iii)
M + x + w + y = 185 ....(iv)
P + x + w + z = 210 ....(v)
C + y + w + z = 222 ....(vi)
M + P + C + y + x + z + w = 233
....(vii)
By (iv) + (v) + (vi) - (vii),
x + y + z + 2w = 384 ....(viii)
by (i) + (ii) + (iii),
x + y + z + 3w = 555 ....(ix)
By (ix) - (viii),
w = 171
From equation (i),
x=9

From equation (ii),
z = 200 - 171 = 29
From equation (iii),
y = 175 - 171 = 4
M = 185 - 4 - 9 - 171 = 1
P = 210 - 9 - 171 - 29 = 1
C = 222 - 4 - 29 - 171 = 18

Number of students who
have passed in all three subjects = 171

8. How many students failed in Physics and Maths ?

Correct Answer: (1) 65
Solution:Number of students = 280
Cleared Maths = 185
Cleared Physics= 210
Cleared Chemistry = 222
Students who passed in Maths
and Physics = 180
Students who passed in Maths
and Chemistry = 175
Students who passed in Physics and Chemistry = 200
Number of students who
passed in atleast one subject
= 280 - 47 = 233

x + w = 180 ....(i)
w + z = 200 ....(ii)
w + y = 175 ....(iii)
M + x + w + y = 185 ....(iv)
P + x + w + z = 210 ....(v)
C + y + w + z = 222 ....(vi)
M + P + C + y + x + z + w = 233
....(vii)
By (iv) + (v) + (vi) - (vii),
x + y + z + 2w = 384 ....(viii)
by (i) + (ii) + (iii),
x + y + z + 3w = 555 ....(ix)
By (ix) - (viii),
w = 171
From equation (i),
x=9

From equation (ii),
z = 200 - 171 = 29
From equation (iii),
y = 175 - 171 = 4
M = 185 - 4 - 9 - 171 = 1
P = 210 - 9 - 171 - 29 = 1
C = 222 - 4 - 29 - 171 = 18

Required answer
= 47 + 18 = 65

9. How many students passed in Maths but failed in both Physics and Chemistry ?

Correct Answer: (4) 1
Solution:Number of students = 280
Cleared Maths = 185
Cleared Physics= 210
Cleared Chemistry = 222
Students who passed in Maths
and Physics = 180
Students who passed in Maths
and Chemistry = 175
Students who passed in Physics and Chemistry = 200
Number of students who
passed in atleast one subject
= 280 - 47 = 233

x + w = 180 ....(i)
w + z = 200 ....(ii)
w + y = 175 ....(iii)
M + x + w + y = 185 ....(iv)
P + x + w + z = 210 ....(v)
C + y + w + z = 222 ....(vi)
M + P + C + y + x + z + w = 233
....(vii)
By (iv) + (v) + (vi) - (vii),
x + y + z + 2w = 384 ....(viii)
by (i) + (ii) + (iii),
x + y + z + 3w = 555 ....(ix)
By (ix) - (viii),
w = 171
From equation (i),
x=9

From equation (ii),
z = 200 - 171 = 29
From equation (iii),
y = 175 - 171 = 4
M = 185 - 4 - 9 - 171 = 1
P = 210 - 9 - 171 - 29 = 1
C = 222 - 4 - 29 - 171 = 18

Required answer = 1

10. A student has to pass in at least two subjects to get promoted. How many students get promoted ?

Correct Answer: (2) 23
Solution:Number of students = 280
Cleared Maths = 185
Cleared Physics= 210
Cleared Chemistry = 222
Students who passed in Maths
and Physics = 180
Students who passed in Maths
and Chemistry = 175
Students who passed in Physics and Chemistry = 200
Number of students who
passed in atleast one subject
= 280 - 47 = 233

x + w = 180 ....(i)
w + z = 200 ....(ii)
w + y = 175 ....(iii)
M + x + w + y = 185 ....(iv)
P + x + w + z = 210 ....(v)
C + y + w + z = 222 ....(vi)
M + P + C + y + x + z + w = 233
....(vii)
By (iv) + (v) + (vi) - (vii),
x + y + z + 2w = 384 ....(viii)
by (i) + (ii) + (iii),
x + y + z + 3w = 555 ....(ix)
By (ix) - (viii),
w = 171
From equation (i),
x=9

From equation (ii),
z = 200 - 171 = 29
From equation (iii),
y = 175 - 171 = 4
M = 185 - 4 - 9 - 171 = 1
P = 210 - 9 - 171 - 29 = 1
C = 222 - 4 - 29 - 171 = 18

Required answer

= x + y + z + w
= 9 + 29 + 4 + 171 = 213