26 Sep 2016

Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions with Detailed Solutions

Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions:
Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.


1). In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is:
a)    23/78
b)    19/88
c)    15/92
d)    4/23
e)    7/46

2). Ravi borrowed some money at the rate of 4 p.c.p.a. for the first three years, at the rate of 8 p.c.p.a. for the next two years and at the rate of 9 p.c.p.a. for the period beyond 5 years. If he pays a total simple interest of 19,550 at the end of 7 years, how much money did he borrow?
a)    39,500
b)    42,500
c)    41,900
d)    43,000
e)    None of these

3). Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021?
a)    17
b)    19
c)    21
d)    23
e)    25

4). A dealer sold a radio at a loss of 2.5%. Had he sold it for Rs. 120 more, he would have gained 7.5%. In order to gain 12.5% after a 25% discount, the marked price should be:
a)    1750
b)    1800
c)    1857.75
d)    1925
e)    None of these

5). A father runs after his son, who is 1000 meters ahead. The father runs at a speed of 1 kilometre every 8 minutes, and the son runs at a speed of 1 kilometer every 12 minutes. How much distance has the son covered at the point when the father overtakes him?
a)    2500 meters
b)    2000 meters
c)    1500 meters
d)    1000 meters
e)    1200 meters

6). 24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4
a)    120
b)    140
c)    160
d)    180
e)    None of these

7). Evaluate: 2+√2 + 1/(2√2) + 1/(√2 - 2)
a)    1 + 0.75√2
b)    2 + 1.5√2
c)    2 + √2
d)    2 - √2
e)    None of these

8). Evaluate: 48×52 + 61×59 + 77×83
a)    12486
b)    10996
c)    13406
d)    13206
e)    None of these

9). Evaluate: (0.625 × 0.0729 × 28.9)/(0.0081 × 0.025 × 1.7)
a)    382.5
b)    3725
c)    3625
d)    3825
e)    None of these

10). Evaluate: (4 + 4×18 – 6 - 8) / (123×6 - 146×5)
a)    7.5
b)    7.75
c)    8
d)    8.25
e)    8.5
Answers:
1)c   2)b   3)d   4)b   5)b   6)e   7)a   8)a   9)d   10)b

Solution:
1). 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300
1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300
Difference = (1050 - 675)/2300 = 375/2300 = 15/92
Answer: c)

2). Let the amount borrowed be Rs P
Total amount of simple interest = (P×3×4)/100 + (p×2×8)/100 + (p×2×9)/100 = 19550
0.12P + 0.16P + 0.18P = 19550
0.46p= 19550
P = Rs. 42500
Answer: b)

3). Let the correct ages of Ramesh and Suresh be X in 2015 (since they are twins, they will have the same age)
Given: 2x + 9 = 43
Or x = 17 years in 2015.
Age in 2021 = 17 + 6 = 23
Note: year of birth = 2015 – 17 = 1998. Reversing, Suresh = 1989 or 9 year older.
Answer: d)

4). 120 = 10% of the CP (2.5% loss to 7.5% profit)
Or CP = Rs. 1200
To gain 12.5%, SP = 1200 × 1.125 = Rs. 1350
Rs. 1350 = 25% discounted of Marked Price
Or marked Price = 1350/0.75 = Rs. 1800
Answer: b)

5). Time taken is same for both father and son
Let x be the distance son travelled before being overtaken by his father
Time taken by son = x/5 km/hour
Time taken by father = (1000 + x) / 7.5 km/hr
Since, time taken is same
x/5 = (1000  + x) / 7.5 or x = 2000m
Answer: b)

6). 72÷48×8+14 = (72/48) × 8 + 14 = (3/2) × 8 + 14 = 12 +14 = 26
24 + (72 ÷ 48 × 8 + 14) ×36 ÷ (3 × 4) + 36 ÷ 3 × 4
= 24 + 26 × 36÷12 + 36 ÷ 3 × 4
= 24 + 26 × 3 + 12 × 4
= 24 + 78 + 48 = 150
Answer: e)

7). 1/(√2 - 2) = (√2 + 2)/(2 - 4) = (√2 + 2)/ -2
1/(2√2) = √2/4
The given equation becomes: 2 + √2 +√2/4 – (√2 + 2) / 2
= (4(2 + √2) + √2 – 2(√2 + 2)) / 4
= (4 + 3√2) / 4 = 1 + 0.75√2
Answer: a)

8). 48×52 = (50 - 2) × (50 + 2) = 502 – 22 = 2496
61 × 59 = (60 + 1) × (60 - 1) = 602 – 12 = 3599
77 × 83 = (80 - 3) × (80 + 3) = 802 – 32 = 6391
Sum = 2496 + 3599 + 6391 = 12486
Answer: a)

9). (0.625×0.0729×28.9) / (0.0081×0.025×1.7) = (625/25) × (729/81) × (289/17) = 25×9×17 = 3825.
Answer: d)

10). Numerator = 4 + 72 – 14 = 62
Denominator = 738 – 730 = 8
Required value = 62/8 = 7.75
Answer: b)



 


 

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