Practice Quantitative Aptitude Questions For Upcoming SBI PO Prelims & TMB Exams 2017 (Application Problems):
Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.
1).Some amount out of Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent at 6% per annum was
a) Rs. 2400
b) Rs. 2200
c) Rs. 2000
d) Rs. 1800
e) None of these
2).The speed of a boat in still water is 30 kmph. If it can travel 10 km upstream in 1 hr, what time it would take to travel the same distance downstream?
a) 22 minutes
b) 30 minutes
c) 40 minutes
d) 10 minutes
e) None of these
3).A Pen was sold for Rs. 28.50 with a profit of 10%. If it was sold for Rs. 26.75, what would have been the percentage of profit or loss?
a) 3%
b) 2%
c) 4%
d) 5%
e) None of these
4).A sum of money amounts to Rs.9850 after 5 years and Rs.12010 after 8 years at the same rate of simple interest. The rate of interest per annum is
a) 15%
b) 12%
c) 11.5%
d) 11%
e) None of these
5).Assume that 20 Horses and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 Horse, what will be the cost for keeping 50 Horses and 30 goats for 12 days?
a) Rs.1104
b) Rs.1000
c) Rs.934
d) Rs.1210
e) None of these
6).Walking 6/7th of his usual speed, a man is 15 minutes too late. What is the usual time taken by him to cover that distance?
a) 1 hr 42 min
b) 1 hr 30 min
c) 2 hr
d) 1 hr 12 min
e) None of these
7).The ages of two persons differ by 15 years. 7 years ago, the elder one was 4 times as old as the younger one. What are their present ages of the elder person?
a) 19
b) 27
c) 32
d) 40
e) None of these
8).What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
a) 45%
b) 44%
c) 40%
d) 42%
e) None of these
9).Two pipes A and B can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap B was closed, then it takes total 8 hours to fill up the whole tank. After how many hours B was closed?
a) 4 hours
b) 5 hours
c) 2 hours
d) 6 hours
e) None of the Above
10).Rohan got 75% in English and 55% in Biology and the maximum marks of both papers is 100. What percent does he score in Maths, if he scores 60% marks in all the three subjects?. Maximum Marks of Maths paper is 200.
a) 30%
b) 40%
c) 45%
d) 25%
e) 55%
1).Total simple interest received , I = Rs.1600
Principal , p = 7000
period, n = 5 years
Rate of interest, r = ?
Simple Interest, I=pnr/100
⇒1600=(7000×5×r)/100
⇒r=(1600×100)/(7000×5)=160/35=32/7%
By rule of alligation,
Part1 : Part2 =4/7:1/07=4:10=2:5
Given that total amount is Rs.7000. Therefore, the amount lent at 6% per annum (part1 amount) = 7000 × (2/7) = Rs. 2000
Answer: c)
2).Speed of boat in still water = 30 km/hr
Speed upstream =10/1 = 10 km/hr
Speed of the stream = (30-10) = 20 km/hr
Speed downstream = (30+20) = 50 km/hr
Time taken to travel 10 km downstream = 10/50 hours = (10×60) / 60 = 10 minutes
Answer: d)
3).28.5 =110% of cost price
26.75=(110×26.75) / 28.5 % = 103 % of cost price.
Therefore, required profit percentage = 3%
Answer: a)
4).Simple Interest for 3 years = (Rs.12010 - Rs.9850) = Rs.2160
Simple Interest for 5 years = (2160/3) × 5 = Rs.3600
Principal(P) = (Rs.9850 - Rs.3600) = Rs.6250
R = (100×SI)/PT=(100×3600) / (6125×5)=11.5%
Answer: c)
5).Assume that cost of keeping a Horse for 1 day = h ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 Horses and 40 goats for 10 days = 460
Cost of keeping 20 Horses and 40 goats for 1 day = 460/10 = 46
=> 20h + 40g = 46
=> 10h + 20g = 23 ---(1)
Given that 5g = h
Hence equation (1) can be written as 10h + 4h = 23 => 14h =23
=> h=23/14
cost of keeping 50 Horses and 30 goats for 1 day
= 50h + 30g
= 50h + 6c (substituted 5g = c)
= 56 h = 56×23/14 = 92
Cost of keeping 50 Horses and 30 goats for 12 days = 12×92 = 1104
Answer: a)
6).New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time =7/6 of usual time
Hence, 7/6 of usual time - usual time =15 minutes
⇒1/6 of usual time =15 minutes
Therefore, usual time =15×6 = 90 minutes
=1 hour 30 minutes
Answer: b)
7).Let present age of the elder person = x and
present age of the younger person = x−15
(x - 7) = 4 (x – 15 - 7)
x – 7 = 4x – 88
Answer: b)
8).Let original length =l
original breadth =b
Then original area =lb
Length is increased by 20%
⇒ New length =l×(120/100)=1.2l
Breadth is increased by 20%
⇒ New breadth =b×(120/100)=1.2b
New area =1.2l×1.2b=1.44lb
Increase in area = new area - original area
=1.44lb – lb = 0.44lb
Percentage increase in area
= (increase in area / original area) × 100
= (0.44 lb / lb)×100 = 44%
Answer: b)
9).Pipe A Efficiency = 100/10 = 10%
Pipe B Efficiency = 100/20 = 5%
Net Efficiency = 15%
15x + 10(8-x) = 100
Answer: a)
10).75 + 55 + x = 60% of all three subjects
75 + 55 + x = 60% of 400
x = 240 – 130 = 110
% = (110/200) * 100 = 55%
Answer: e)
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