Dear Students,

Quantitative Aptitude Questions for Syndicate Bank PO

Quantitative Aptitude is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. These two weeks are very important as Syndicate Bank PO and IBPS Clerk Mains are lined up. So, these 15 questions can help you practice three very important topics of Quant Section.

S1. Ans.(b)
Sol.
?=1664/64=26

S3. Ans.(e)
Sol.
? = 0.0018

S4. Ans.(d)
Sol.
70 × 155 + 385.85 = ? + 5245.45
⇒ ? = 5990.4

S5. Ans.(e)
Sol.
25 – ? = 0.948
⇒ ? = 24.052
Q6. In a class 60% of the students pass in Hindi and 45% pass in Sanskrit. If 25% of them pass in both subjects, what percentage of the students fail in both the subjects?
(a) 80%
(b) 75%
(c) 20%
(d) 25%
(e) 30%

Q7. The ratio of the number of boys to that of girls in a school is 4 : 1. If 75% of boys and 70% of the girls are scholarship holders, then the percentage of students who do not get scholarship is :
(a) 50%
(b) 28%
(c) 75%
(d) 26%
(e) 36%

Q8. The strength of a school increases and decreases in every alternate year by 10%. It started in 2000 and  increases in 2001, then the strength of the school in 2003 as compared to that in 2000 was:
(a) increased by 8.9%
(b) decreased by 8.9%
(c) increased by 9.8%
(d) decreased by 9.8%
(e) increase by 7.8%

Q9. A spider climbed 62 1/2% of the height of the pole in one hour and in the next hour it covered 12 1/2% of the remaining height. If pole’s height is 192 m, then the distance climed in second hour is
(a) 3 m
(b) 5 m
(c) 7 m
(d) 9 m
(e) 12 m

Q10. The red blood cells in a blood sample grows by 10% per hour in first two hours, decreases by 10% in next one hour, remains constant in next one hour and again increases by 5% per hour in next two hours. If the original count of the red blood cells in the sample is 40000, find the approximate red blood cell count at the end of 6 hours.
(a) 40000
(b) 45025
(c) 48025
(d) 50025
(e) 52025

Direction for questions 11 to 14 : Read the passage below and solve the questions based on it.
India is supposed to send its boxing team for Olympics in each of the following 10 weight group divisions.
A (48 kg – 52 kg) B (52 kg – 56 kg)
C (56 kg – 60 kg) D (60 kg – 64 kg)
E (64 kg – 68 kg) F (68 kg – 72 kg)
G (72 kg – 76 kg) H (76 kg – 80 kg)
I (80 kg – 84 kg) J (84 kg – 88 kg)
After selecting one player from each group, their average weight comes out to be 68 kg. If one of the players named X leaves the team, their average weight comes down to 66.5 kg.

Q11. Player X is from group
(a) A
(b) E
(c) I
(d) H
(e) J

Q12. If X leaves the team, and two new players join the group, then their average weight increases to 68 kg. These players can be from group …………
(a) A, C
(b) D, J
(c) Both from G
(d) Either (b) or (c)
(e) Both from I

Q13. What is the average weight (in kg) of all the players taken together?
(a) 56.7
(b) 58.8
(c) 61.4
(d) Cannot be determined
(e) 67.5

Q14. In the average of all the group together, which group contributes most in overall average?
(a) F
(b) G
(c) H
(d) Cannot be determined
(e) J

Q15. The average of 5 positive number is A. The average of the average of all the possible distinct triplets of these 5 numbers is B. Which of the following is true?
(a) A > B
(b) A < B
(c) A = B
(d) Cannot be determined
(e) None of these

S11. Ans.(c)
Sol. Total weight of all players initially = 68 × 10 = 680 kg
Total weight of players when 1 players left the team = 66.5 × 9 = 598.5 kg
Difference in weight = weight of X = (680 – 598.5) kg = 81.5

S12. Ans.(d)
Sol. Total weight of 11 players (68 × 11) kg = 748 kg
Increase in weight = (748 – 598.5) kg = 149.5
From the given information we can easily say that both come from either D and J or from group G

S13. Ans.(d)
Sol. Since the exact weight is not known, we cannot find out the average weight of all the players taken together.

S14. Ans.(d)
Sol. Exact weight of players are not known; hence, option (d) is the answer.

S15. Ans.(c)
Sol. Let the five numbers be 1, 2, 3, 4 and 5
Average of these five numbers = (1+2+3+4+5)/5=3
Distinct triplets are (1, 2, 3), (2, 3, 4), (3, 4, 5), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 4, 5), (2, 3, 5).
[Total number of distinct triplets = 5C3 = 10]
In the above triplets, each of the numbers occur 6 times. For example, 1 will occur six times in total. Instead of calculating the average of all the triplets, what we can do is – calculate the average of all the ten triplets at one go.
There is a total of 30 numbers [3 numbers in each triplets × 10 triplets]
Average of all the ten triplets = (6(1+2+3+4+5))/30
= 90/30=3
Hence, A = B.