Practice Aptitude Questions For NICL AO Mains 2017 (Data Interpretation)

Practice Aptitude Questions For NICL AO Mains 2017 (Data Interpretation)

Practice Aptitude Questions For NICL AO Mains 2017 (Data Interpretation):

Dear Readers, Important Practice Aptitude Questions for NICL AO Mains 2017 Exams was given here with Solutions. Aspirants those who are preparing for Bank exams& all other Competitive examination can use this material.



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Direction (Q. 1-5): study the following table to answer the given questions.
The number of student appeared and qualified (in thousands) in entrance examination from different schools over the years.

1.If the number of appeared students in 2015and in 2017 from all the schools 376 and 434 thousands respectively, then what is the difference between the number of appeared candidates from school D and that from school E in all the given years (in thousands)?

9
4
8
7
None of these

1). The no. of appeared students from school D,
376 - (70+83+68+80) = 376 – 301 = 75
The no. of appeared students from school E,
434 – (95+78+91+86) = 434 – 350 = 84
∴ required difference = (65+75+83+86) – (72+80+69+84) =309 – 305 = 4.
Answer: B

2.If in 2016 the number of appeared students from school C in 2016 is two thousands less than the number of appeared students from school E in 2015 and the number of qualified students in school E in 2015 is 150% of qualified students from that school in 2016, then the number of qualified students in all the schools in 2015 is what percentage of that of appeared students in all the schools in 2016 (approximately)?

17%
12%
13%
15%
20%
2).The number of appeared students from school C in 2016 = 80 - 2 =78
The number of qualified students in school E in 2015 = (150/100) × 8 = 12
∴ The no.of qualified students in all the schools in 2015 = (12+14+9+13+12) = 60
∴ The no.of appeared students in all the schools in 2016 = (87+76+78+83+69) = 393
∴ Required percentage = (60/393) × 100 = 15.26 = 15%.
Answer: D


3.If the average number of students appeared from school A is 87 thousands and the average number of qualified students from state D is equal to the number of qualified students from school B in 2015, then what is the ratio of appeared students from school A in 2014 to that qualified student from school D in 2016?

8:1
32:5
7:1
7:5
None of these
3).The number of students appeared from school A = 87 × 4 = 348
∴The no.of students appeared from school A in 2014 is,
348 – (70+87+95) = 348 – 252 = 96
The number of qualified students from school B in 2015 is 14
∴ The average number of qualified students from state D is 14
Hence, qualified student from school D = 14×4 =56
∴ The qualified student from school D in 2016 = 56 - (12+13+16) = 56 – 41 = 15
Hence, required ratio = 96 : 15 = 32 : 5.
Answer: B


4.If the average number of qualified students from school B is 14.5 and the average number of qualified student in 2014 is 1/10 more than the average number of qualified students from school B, then what is the number of qualified students from same school in same year (in thousands)?

11
10
15
Cannot be determined
None of these
4). The number of qualified students from school B = 14.5×4 = 58
∴The average no. of qualified student in 2014 = 1/10 + 14.5= 0.1 + 14.5 = 14.6
Hence, total qualified student in 2014 = 14.6 × 5 = 73
∴ The number of qualified students from school B in 2014 is =, 73 - 58 = 15.
Answer: C

5.If in 2013 the number of appeared students from school A, B, C and D are 12%, 12%, 15% and 20% more than that of the year 2014 respectively and the number of appeared students from school E is 25% less than that of 2014. If the number of appeared students in 2014 from school A is 88, then what is average of appeared students in 2013 from all the schools together?

83.08
85
78
76
None of these
5).A : (12/100) × 88 = 10.56= 88 + 10.56 = 98.56
B : (12/100) × 87 = 10.44 = 87 + 10.44 = 97.44
C : (15/100) × 76 = 11.4 = 76 + 11.4 = 87.4
D : (20/100) × 65 = 13 = 65 + 13 = 78
E : (25/100) × 72 = 18 = 72 – 18 = 54
Hence, required average = 415.4 / 5 = 83.08
Answer: A


Direction (Q. 6-10): study the following table to answer the given questions.

6.If an amount becomes Rs. 9600 in given year at given R%, then what is that amount of P?

8000
6000
5500
5000
None of these
6). Amount (A) = P + S.I = P [1+RT/100]
9600 = P [1+ (4×15)/100] = P [8/5]
∴ P = 9600 × 5/8 = Rs. 6000
Answer: B


7.If the rate of interest of Q is given per quarter, then what is the total amount to be paid at end of the given year?

6000
8000
12000
16000
None of these
7). R% = 5% per quarter
∴ R% = 20% per year.
S.I= [PRT/100]
S.I = (10000× 20× 3)/ 100 = 6000
∴ A = P + S.I = 10000 + 6000 = Rs.16000
Answer: D

8.If the difference between the amount and the principal of ‘R’ is Rs.1272, then what is the principal amount?

6000
6500
5000
Cannot be determined
None of these
8). Amount – Principal = C.I
Hence, C.I = 1272;
C.I = P [(1+R/100) ^T – 1]
1272 = P [(1+12/100)^2 - 1]
1272 = P [(112/100)^2 – 1] = P [159/625]
∴ P = 1272 × 625/159 = 5000
Hence, Principal = Rs. 5000.
Answer: C

9. If the difference between Compound interest and Simple interest for S is Rs.48, then what is the Amount of simple Interest?

Rs. 375
Rs. 225
Rs. 600
Data inadequate
None of these
9).The Difference between C.I and S.I (D) for 3 years is,
P = (D × 1003) / (R2 (300+R))
P = (48 × 1003) / (202 (300+20))
P = (48× 100 × 100 × 100) / (20×20× 320)
∴ P = 375
Amount for S.I : S.I = (375 × 20 × 3) / 100 = 225
Hence, A = P + S.I = 375 + 225 = 600
Answer: C

10.If the simple Interest on a sum for T is one- fourth of the sum, then what is the rate of interest per annum?

8%
4%
6%
5%
None of these
10).Let Sum= P,
S.I= [PRT/100]
1/4 × P = [P×R×5] / 100
∴ R = 1/4 × 100/5 = 5
Hence, the rate of interest is 5% per annum.
Answer: D


     
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