Quantitative
Aptitude Practice Questions | IBPS 2017
Dear Aspirants,
Welcome to Mentor for Bank Exams Quantitative
Aptitude Quiz section. The following quiz covers Number Series (5 Questions), Simplifications
(5 Questions), Data Interpretation (5 Questions), Word Problems (5 Questions).
All the best for upcoming IBPS Exams
2017.
Directions (1 – 5): What will come in place of
question mark (?) in the given series:
1. 119
166 221 284
? 434
a) 355
b) 304
c) 329
d) 325
e) 314
2. 95
110 215 530
? 2510
a) 1322
b) 1223
c) 1423
d) 1324
e) 1224
3. 4
5 13 40
104 ? 445
a) 157
b) 227
c) 231
d) 229
e) 193
4. 15
21 39 78
145 ?
a) 243
b) 240
c) 253
d) 245
e) 254
5. 942
942 917 1017
792 1192 ?
a) 497
b) 702
c) 594
d) 567
e) 482
Directions (6 – 10): What value will come in place
of question mark (?) in the following questions? (You are not expected to
calculate the exact value.)
6. 107% of 559.88 – 152% of 164.02 = ?
a) 290
b) 280
c) 300
d) 349
e) 381
7. 28.95 * 7.26 + (34/16) – (23/12) * 2 + (2 5/11)
– (0.34 * 2.11) = ?
a) 310
b) 322
c) 290
d) 125
e) 210
8. √79 + √81 - √15 + √16 + (35.07 * 3.21) = ? +
5.91
a) 124
b) 140
c) 110
d) 130
e) 150
9. 3969 ×
63 – √4225 – 2746251/3 + 35 – 38.042 – 0.981 + 0.63 =
?
a) 215354
b) 292769
c) 250013
d) 249914
e) 285412
10. √783 ÷
? × 11.982 = 252
a) 26
b) 12
c) 8
d) 4
e) 16
Directions
(11 – 15): Study the following table
carefully and answer the questions.
The following table shows the unduplicated (unique)
readership figures (in lakh) of five newspapers A, B, C, D and E, in six states
1, 2, 3, 4, 5 and 6, as per a readership survey done by an independent agency.
11. In state 1, 15% people read sports page of the
newspaper, 11% in state 2 read editorial page, 9% in state 3 read economics
page, while 7% in state 5 read international events page. Which among these
segments accounts for the least number of people?
a) Sports page readers in state 1
b) Economics page in state 3
c) International events page readers
in state 5
d) Editorial page readers in state 2
e) All are same
12. Which newspaper has the maximum percentage
difference in total readership in state 1, 2 and 3 taken together, as compared
to its total readership in states 4, 5 and 6 taken together?
a) A
b) C
c) D
d) E
e) Can’t be determined
13. The population of state 2 is 21 lakh, while
that of state 4 is 27 lakh. By how much is the percentage of newspaper reading
population in state 4 more/less than the percentage of the newspaper reading
population in state 2?
a) 15.14% more
b) 5.3% more
c) 5.3% less
d) 15.14% less
e) None of these
14. In state 3, 11% readers of newspaper B decided
to switch to newspaper E. What per cent of readers of newspaper C, if they
switch over to reading newspaper B, would restore its readership to the
original figure?
a) 11%
b) 4.7%
c) 6.37%
d) 5.88%
e) None of these
15. How much increase/decrease in readership of
newspaper A will make its ratio to the total readership of newspaper C in
states 1, 3 and 5 taken together, equal to 9:10?
a) 51600 increase
b) 51600 decrease
c) 189667 increase
d) 189667 decrease
e) None of these
16. In an alloy, zinc and copper are in the ratio
1:2.In the second alloy, the same elements are in the ratio 2:3. If these two
alloys be mixed to form a new alloy in which two elements are in the ratio 5:8,
in what
ratio of these two alloys in the new alloys?
a) 2:15
b) 3:10
c) 3:7
d) 4:9
e) 2:4
17. In a certain year, the average monthly income
of a person was Rs. 18,000. For the first 8 months of the year, his monthly
income was Rs. 22,000 and for the last 5 months, it was Rs.25,000. What is the
income in the 8th month of the year?
a) 17000
b) 20000
c) 15000
d) 10000
e) 85000
18. The respective ratio between the total number
of students studying in College A and College B is 5:8 . In College B, out of
the total number of students, (5/8)th are boys, out of which 60% study Commerce
and the remaining 800 boys study in other streams. What is the total number of
students in College A?
a) 2400
b) 3000
c) 3500
d) 2500
e) 2000
19. In a market research project 20% opted for
Tide detergent whereas 60% opted for Surf Blue detergent. The rest were unsure.
If the difference between those who opted for Surf Blue and those who were
uncertain is 720. How many respondents were covered in the survey?
a) 2400
b) 2000
c) 1800
d) 1500
e) 2100
20. Two equal sum of money are lent at the same
time at 8% and 7% per annum simple interest. The former is recovered in 6
months earlier than the latter and the amount in each case is Rs. 2560. Find
the sum and the time for which the sum of money are lent out.
A)
Rs.3500, 2.5 yrs. and 3 yrs.
B) Rs.3000, 3 yrs. and 5 yrs.
C) Rs.2000, 2 yrs. and 4 yrs.
D) Rs.2000, 3.5yrs. and 4 yrs.
E) Rs.2500, 2yrs. and 6yrs.
Solutions:
1. A)
Series
Pattern
|
Given Series
|
112 =
121 – 2
|
119
|
132 =
169 – 3
|
166
|
152 =
225 – 4
|
221
|
172 =
289 – 5
|
284
|
192 =
361 – 6
|
355
|
212 =
441 – 7
|
434
|
Hence, there should be
355 in place of question mark (?).
2. B)
Series Pattern
|
Given Series
|
95
|
95
|
95 + 15 (=1×3×5)
= 110
|
110
|
110 + 105 (=3×5×7) =
215
|
215
|
215 + 315 (=5×7×9) =
530
|
530
|
530 + 693 (=7×9×11) = 1223
|
1223
|
1223 +
1287 (=9×11×13) = 2510
|
2510
|
Hence, there should be
1223 in place of question mark (?).
3. D) The series is +13, +23,
+33, +43 ...
Hence, there should be
229 in place of question mark (?)
4. D)
Series I
|
15
|
21
|
39
|
78
|
145
|
?
|
|||||
Series II
|
6
|
18
|
39
|
67
|
?
|
||||||
Series III
|
12
|
21
|
28
|
?
|
|||||||
Series IV
|
9
|
7
|
5
|
||||||||
2
|
2
|
Clearly, the pattern in
series III is +9, +7, +5.
So, the missing term in
series III = 28 + 5 = 33;
∴ missing term in series II = 67 + 33 = 100;
∴ missing term in series I = 100 + 145 = 245.
Finally the series will
become as follows:
Series I
|
15
|
21
|
39
|
78
|
145
|
245
|
|||||
Series II
|
6
|
18
|
39
|
67
|
100
|
||||||
Series III
|
12
|
21
|
28
|
33
|
|||||||
Series IV
|
9
|
7
|
5
|
||||||||
2
|
2
|
Hence, there should be
245 in place of question mark (?)
5. D)
Series Pattern
|
Given Series
|
942
|
942
|
942 + (0)2 =
942
|
942
|
942 – (5)2 =
917
|
917
|
917 + (10)2 =
1017
|
1017
|
1017 –
(15)2 = 792
|
792
|
792 + (20)2 =
1192
|
1192
|
1192 –
(25)2 = 567
|
567
|
Hence, there should be
567 in place of ?
6. D) 107% of 559.88 – 152% of 164.02 = ?
? ≈ (100 + 7)% of 560 –
(100 + 50 + 2)% of 164
? ≈ (560 + 39.2) – (164 +
82 + 3.28)
? = 599.2 – 249.82
≈ 599 – 250 = 349
7. E) 28.95
* 7.26 + (34/16) – (23/12) * 2 + (2 5/11) – (0.34 * 2.11) = ?
≈ 30 * 7 + (32/16) – (24/12)
* 2 + (27/11) – (0.34 * 2.11) = ?
≈ 210 + 2 – 4 + 2.5 – 0.34 × 2.11
= 209.8 ≈ 210
8. A) √79
+ √81 – √15 + √16 + (35.07 × 3.21) = ? + 5.91
≈ √81 + √81 – √16 + √16 + (35.07 × 3.21) = ? +
5.91
or, ≈ 9 + 9 – 4 + 4 + 112 = ? + 6
? ≈ 9 + 9 + 112 – 6 ≈ 124.
9. D) 3969
× 63 – √4225 – 2746251/3 + 35 – 38.042 – 0.981 + 0.63
= ?
= 250047 – 65 – 65 + 35 –
38 – 1 + 0.6
= 249913.6 ≈ 249914
10. E) √783
÷ ? × 11.982 = 252
=> √783 ÷ ? × 12 × 12= 252
=> 28/? × 12 × 12 ≈ 252
=> ? ≈ (28 * 12 *
12)/252 = 16
11. C) Number
of people who read sports page = 15% of population of state 1
= 15% of (1.5 + 1.65 +
2.73 + 0.98 + 1.36) lakh = (15/100) × 8.22 lakh = 1.233 lakh
Number of people who read
editorial page = 11% of population of state 2
= 11% of (1.15 + 1.82 +
1.36 + 3.15 + 0.92) lakh = (11/100) × 8.4 lakh = 0.924 lakh
Number of people who read
economics page = 9% of population of state 3
= 9% of (2.3 + 1.23 +
2.85 + 0.78 + 1.62) lakh = (9/100) × 8.78 = 0.79 lakh
Number of people who read
international events page = 7% of population of state 5
= 7% of (1.57 + 0.87 +
0.96 + 2.78 + 3.25) = (7/100) × 9.43 = 0.66 lakh
As we can see from the
calculated numbers, the number of international events page readers in state 5
is the least among all four given numbers.
7. D)
∴ Newspaper E has the maximum percentage difference in total
readership in state 1, 2 and 3 taken together, as compared to its total
readership in states 4, 5 and 6 taken together
8. C) Population of state 2 = 21 lakh.
Number of readers in
state 2 (in lakh) = 1.15 + 1.82 + 1.36 + 3.15 + 0.92 = 8.4
Population of state 4 =
27 lakh.
Number of readers in
state 4 (in lakh) = 0.75 + 2.62 + 3.15 + 1.63 + 1.23 = 9.38
Percentage of newspaper
reading population in state 2 = 8.4/21 * 100 = 40%
Percentage of newspaper
reading population in state 4 = 9.38/27 * 100 = 34.74%
difference between
percentage of newspaper reading population in state 2 to that of state 4 = (40
– 34.74)% = 5.26% ≈ 5.3%
9. A) In
state 3, 11% readers of newspaper B decided to switch to newspaper E, i.e.
11% of readers of
newspaper B in state 3 = 11/100 * 123000 = 13530
Let the percentage of
readers of newspaper C in state 3 who switch to newspaper B be x.
After x% of readers of
newspaper C in state 3 switch to newspaper B, the readership of newspaper B
gets restored to its original figure.
∴ x% of 285000 = 13530
⇒ (x/100) × 285000 =
13530
⇒ x = 13530 × 100/285000 =
4.7%
10. A) Suppose
there is increase in readership of A and that is x (in lakh).
∴ Total readership of newspaper A (in lakh) in state 1, 3 and
5 = 1.5 + 2.3 + 1.57 + x = 5.37 + x
Total readership of
newspaper C (in lakh) in state 1, 3 and 5 = 2.73 + 2.85 + 0.96 = 6.54
Thus, ratio of total
readership of newspaper A to total readership of newspaper C in states 1, 3 and
5 taken together = (5.37 + x)/6.54
According to the
question, this ratio should be 9 : 10
∴ (5.37 + x)/6.54 = 9/10
⇒ 10 × (5.37 + x) = 9 × 6.54
⇒ 53.7 + 10x = 58.86
⇒ 10x = (58.86 – 53.7)
⇒ x = (5.16/10)
⇒ x = 0.516 (in lakh)
∵ value of x is positive, increase in readership = x = 0.516
lakh = 51600
16. B)
Qty. of zinc in both the
alloys ,
A—————B
(1/3)————(2/5)
———-(5/13)
(1/65)————-(2/39)
Required ratio = 3 : 10
17. E) Person’s
income in the 8th month
= (8*22000 + 5*25000
-12*18000) = 85000
18. E) Total
number of students in College A and College B be 5x and 8x resp.
In college B , Boys =
(5/8)*8x = 5x
Boys who study commerce =
(5x*60)/100 = 3x
Boys in other streams =
5x – 3x = 2x
Therefore , 2x = 2 *800 =
400
Total number of students in college A = 5x = 5*400 = 2000
19. C) Unsure
persons = (100 -60-20)% = 20%
If the number of persons
included in the survey be x , then
x* (60 – 20)% = 720
=>( x*40)/100 = 720
=> x = 1800
20. D) Let
each amount be Rs. x and time be t yr.
x + (t –(1/2))*x*(8/100)
= x + [( x+ t + 7)/100]
=> 2xt/25 – x/25
= 7xt/100
=> t = 4 yrs.
First money was lent for
4- (1/2) =3.5 yrs.
Therefore, Amount = x +
(3.5*x*8)/100 = 2560
=> x = Rs.2000
