Quantitative Aptitude Quiz for SBI PO | IBPS | RBI – Set 111

In Quantitative Aptitude, Quants & Reasoning, Quiz
May 7, 2016

The following pie-chart shows the percentage distribution of the expenditure incurred in publishing a book. Study the pie-chart and the answer the questions based on it.

Various Expenditures (in percentage) Incurred in Publishing a Book

1) If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, then what will be amount of royalty to be paid for these books?
a) Rs. 19,450
b) Rs. 21,200
c) Rs. 22,950
d) Rs. 26,150

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Answer c) Rs. 22,950
Let the amount of Royalty to be paid for these books be Rs. = r
Then, 20 : 15 = 30600 : r =>
r = Rs. ( 30600 x 15 ) / 20
= Rs. 22,950.

2) What is the central angle of the sector corresponding to the expenditure incurred on Royalty?
a) 15º
b) 24º
c) 54º
d) 48º

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Answer c) 54º
Central angle corresponding to Royalty = (15% of 360)º
=(15 x 360 ) / 100 º
= 54º.

3) The price of the book is marked 20% above the C.P. If the marked price of the book is Rs. 180, then what is the cost of the paper used in a single copy of the book?
a) Rs. 36
b) Rs. 37.50
c) Rs. 42
d) Rs. 44.25

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Answer b) Rs. 37.50
Clearly, marked price of the book = 120% of C.P.
Also, cost of paper = 25% of C.P
Let the cost of paper for a single book be Rs. n.
Then, 120 : 25 = 180 : n
=> n = Rs. (25 x 180) / 120= Rs. 37.50 .

4) If 5500 copies are published and the transportation cost on them amounts to Rs. 82500, then what should be the selling price of the book so that the publisher can earn a profit of 25%?
a) Rs. 187.50
b) Rs. 191.50
c) Rs. 175
d) Rs. 180

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Answer a) Rs. 187.50
For the publisher to earn a profit of 25%, S.P. = 125% of C.P.
Also Transportation Cost = 10% of C.P.
Let the S.P. of 5500 books be Rs. x.
Then, 10 : 125 = 82500 : x
=> x = Rs.(125 x 82500) / 10 = Rs. 1031250.
Therefore S.P. of one book = Rs.(1031250) / 5500= Rs. 187.50 .

5) Royalty on the book is less than the printing cost by:
a) 5%
b) 33 1 / 5 %
c) 20%
d) 25%

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Answer d) 25%
Printing Cost of book = 20% of C.P.
Royalty on book = 15% of C.P.
Difference = (20% of C.P.) – (15% of C.P) = 5% of C.P.
Therefore Percentage difference = (Difference / Printing Cost x 100)%
= (5% of C.P. / Printing Cost x 100)% = 25%.

6) The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
a) 3
b) 4
c) 9
d) Cannot be determined
e) None of these

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Answer b) 4
Let the ten’s digit be x and unit’s digit be y.
Then, (10x + y) – (10y + x) = 36
9(x – y) = 36
x – y = 4.

7) A shopkeeper sells some toys at Rs. 250 each. What percent profit does he make? To find the answer, which of the following information given in Statements I and II is/are necessary?
I. Number of toys sold.
II. Cost price of each toy.
a) Only I is necessary
b) Only II is necessary
c) Both I and II are necessary
d) Either I or II ins necessary
e) None of these

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Answer b) Only II is necessary
S.P. = Rs. 250 each.
To find gain percent, we must know the C.P. of each.

8) A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
a) 230 m
b) 240 m
c) 260 m
d) 320 m
e) None of these

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Answer a) 230 m
Relative speed = (120 + 80) km/hr
=200 x 5/18 m/sec
=500/9 m/sec.
Let the length of the other train be x metres.
Then, x + 270 / 9 = 500/ 9
x + 270 = 500
x = 230.

9) A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
a) 2.91 m
b) 3 m
c) 5.82 m
d) None of these

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Answer b) 3 m
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
Area of the crossroads = (2400 – 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x – x2 = 291
x2 – 100x + 291 = 0
(x – 97)(x – 3) = 0
x = 3.

10) In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
a) 810
b) 1440
c) 2880
d) 50400
e) 5760

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Answer d) 50400
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7! / 2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5! / 3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.