Practice Quantitative Aptitude Questions For Upcoming Exams 2017 (Application Sums)

IPPB Scale I& IBPS SO 2017 - Practice Quantitative Aptitude Questions (Number Series& Quadratic Equation)  Answers Updated
Practice Quantitative Aptitude Questions For Upcoming Exams 2017 (Application Sums):
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.


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1). Five years ago, Somu’s age was 1/3 of Arjun’s age at that time. The ratio of Arjun’s age six years hence to Somu’s age twelve years hence will be 7 : 4. What was Somu’s age three years ago? (in years)

a)  13

b)  29

c)  17

d)  25

e)  27
2). A bag contains 24 eggs, out of which 8 are rotten. The remaining eggs are not rotten. Two eggs are selected at random. What is the probability that one of the eggs is rotten?
a)  11/23
b)  17/23
c)  13/23
d)  32/69
e)  62/69
3). Akbar, Babu and Charles started a business with their investment in the ratio of 1 : 3 : 5. After 4 months, Akbar invested the same amount as before but Babu as well as Charles withdrew half of their investments. The ratio of their profits at the end of the year is
a)  1 : 2 : 3  
b)  3 :4 :15  
c)  3 : 5 : 10  
d)  5:6:10    
e)  4 : 5 : 10
4). Each edge of a cube is decreased by 30%. Find the decrease in its volume.
a)  62.5       
b)  63.5       
c)  67.7       
d)  62.7       
e)  65.7
5). The circumference of the semicircle is 180 cm. If the side of a square is 60% more than the diameter of the circle, then what is the perimeter of the square?
a)  368cm   
b)  464 cm  
c)  486 cm
d)  448 cm  
e)  344 cm
6). The cost price of two Sofas are equal. One Sofa is sold at a profit of 30% and the other one for Rs. 5504 less than the first one. If the overall profit earned after selling both the Sofas is 14%, what is the cost price of each Sofa?
a)  Rs. 17000        
b)  Rs.16800         
c)  Rs.17600 
d)  Rs.17800         
e)  Rs.17200
7). In Jar X, 180 litres milk was with 36 litres water. Some of the mixture was taken out from Jar X and put in jar Y.  If after adding 6 litres of water in the mixture, the ratio  of milk to water in Jar Y was 5 : 2, then what was the amount of mixture that was taken out from Jar X (in litres)?
a)  24
b)  54
c)  30
d)  36
e)  42
8). The ratio of the monthly salary of Omana to that of Prakash is  7 : 9. Omana and Prakash both save 20% and 40% of their respective monthly salary respectively. Omana invests ½ of his savings in PPF and Prakash invests 7/9 of his savings in PPF. If Omana and Prakash together saved Rs. 17500 in PPF, what is Prakash’s monthly salary?
a)  Rs. 72000        
b)  Rs. 36000        
c)  Rs. 45000        
d)  Rs. 35000        
e)  Rs. 54000
9). Raja invested Rs. P in a scheme X offering  simple interest at 10% pa for two years. He invested the whole amount he received from scheme X in another scheme Y offering simple interest at 12% pa for five years. If the difference between the interest earned from schemes X and Y was Rs. 1300, what is the value of P?
a)  Rs. 2500
b)  Rs. 2000
c)  Rs. 3000
d)  Rs. 2800
e)  Rs. 4500
10). The distance between two cities (“A” and “B”) is 569 km. A train from City “A” at 8 am travels towards City “B” at 53 kmph. Another train starts from City “B” at 9 am and travels towards City “A” at 76 kmph. At what time will the trains meet?
a)  12:30 pm
b)  1:00 pm
c)  2 :30 pm
d)  1 30 pm
e)  2:00 pm

Solutions:
1). Five years ago, let Arjun’s age be 3x years. Then, Somu’s age = x
Now, according to the question.
(3x +5 + 6)/( x+5+12) = 7/4
or, (3x+11)/( x+17) =7/4
or, 12x + 44 = 7x + 119
or, 5x = 75
:. x = 15
Somu’s age 3 years ago = 15 + (5 – 3) = 17 years
Answer is: C)

2). Total number of eggs = 24
:. 8 eggs are rotten
:. Remaining eggs = (24 – 8); 16 is not rotten
Now, 2 eggs are randomly selected.
Then n(S) =24C2= 23 x 12
:. n(E)=16C1 ×8C1= 16 × 8
:. P(E)= (16×8)/( 23 ×12) = 32/69
Answer is: D)

3). Ratio of profit
= x × 4 + 2x × 8 : 3x ×4 + (3x/2)× 8 :
5x × 4 + (5x/2) ×8 = 20x: 24x: 40x = 5 : 6 : 10
Answer is: D)

4). We know that increase in volume

=3 × x + (3x2/100) + (x3/1002)
As there is decrease, put the -ve value of x.
:. Percentage decrease in volume
=3 × (-30) +   [3 × (-30)2 / 100] + [(-30)3 /1002 ]
= – 90 + 27 – 2.7
= -65.7
Note: If you don’t recall the formula, compound it like.
Step I. -30 – 30 + [(-30)(-30)/100]= -51
Step II.-51 -30+ [(-51)(-30)/100]
= – 81 + 15.3
= -65.7
Answer is: E

5). Circumference of the semicircle = r (π+2)
:. r[(22/7)+2] =180
or, r(36/7)=180
r = 35cm
Now, side of the square = 2×35 × (160/100)
=2 × 35 × 1.60 = 70 × 1.60 = 112 cm
:. Perimeter of the square = 4 × 112
= 448 cm
Answer is: D

6). Let the cost of each Sofa be Rs.x,
Now, according to the question,
2x×(114/100)= [(x ×130)/100] + [(x ×130)/100] – 5504
or, 2.28x = 2.6x – 5504
or, 0.32x = 5504
:. x= [(5504 × 100)/32] = 172 × 100= Rs. 17200
Answer is: E)

7). The ratio of milk to water in Jar X
= 180 : 36 = 5:1
Now, let 6x litres of mixture be taken out from Jar X and put in Jar Y.
Then, milk in Jar Y = 5x
Water in Jar Y= x
So, 5x/(x+6) = 5/2
or, 10x=(5x + 30)
or, 5x=30,
:. x=6
Hence the mixture that was taken out from Jar X = 6x =6 × 6 = 36 litres
Answer is: D)

8). Let Prakash’s monthly salary be 9x and Omana’s 7x.
Then, 1/2 of 20% of 7x +(7/9) of 40% of 9x = 17500
or, 10% of 7x + 40% of 7x = 17500
or, 0.7x + 2.8x = 17500
or, 3.5x = 17500
:. x =Rs. 5000
Hence Prakash’s monthly salary = 9 × 5000
= Rs. 45000
Shortcut Method:

According to the question, 7 + 28 = 17500
:. 90 = (17500/35)×90 =45000
Answer is: C)

9). Required difference
= [P+(P×10×2)/100] × [(12×5)/100] – [(P×10×2)/100]
or, [(100P+ 20P)/100] × (60/100) – (20P/100) = 1300
or, (120P/100) × (3/5) – (20P/100)=1300
or, 6P/5 × 3/5 – 20P/100 = 1300
or, (72P-20P)/100 =1300
or, 52P = 1300 × 100
:. P=(1300 × 100)/52
= Rs.2500
Shortcut Method:
Suppose Raja invested Rs.100 in scheme X. 
Then,
According to the question:
52 =Rs.1300
:. P = 100 = (1300/52) x 100 =Rs.2500
Answer is: A)

10). The distance covered in (9 am – 8 am) is 1 hour by train “X”=53 × 1 = 53 km Remaining distance 569 – 53 = 516 km
Relative speed 76 + 53 =129 kmph
:. They meet after (516/129) = 4 hours
Hence they meet at (9 am + 4 hours) = 1 pm
Answer is: B)

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