29 Oct 2016

IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions

 IBPS Clerk 2016 - Practice Quantitative Aptitude Questions
IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions Set-26:
Dear Readers, Important Practice Aptitude Questions for IBPS Clerk and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.


1). The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A at 75 km/hr. At what time do they meet?
a)    10 am. 
b)    10 : 30 am.
c)    11 am.  
d)    11 : 30 am.
e)    None of these

2). Busses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes?
a)    3 km/hr 
b)    4 km/hr
c)    5 km/hr  
d)    7 km/hr
e)    None of these

3). In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is?
a)    5 kmph   
b)    6 kmph
c)    6.25 kmph
d)    7.5 kmph
e)    None of these

4). It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is?
a)    2 : 3
b)    3: 2
c)    3 : 4
d)    4 : 3
e)    None of these

5). If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is?
a)    13  
b)    15
c)    19  
d)    21
e)    None of these

Directions (Q. 6-10): what approximate value should come in place of question mark (?) in the following question?
6). (32.3)2 ÷ 4 + √361 = ?2 + 50
a)    15
b)    13
c)    11
d)    17
e)    None of these

7). 256×256 + 173×173 = ?
a)    96432
b)    94465
c)    95465
d)    90510
e)    None of these

8). √ (191×7+231-839) = ?
a)    27
b)    33
c)    23
d)    37
e)    None of these

9). [3/2 + 2(1/5) – 7/10] of ? =  1098
a)    463
b)    326
c)    276
d)    366
e)    None of these

10). (1675÷5) + (5328÷12) * (8430÷15) = ?2 – 235
a)    152
b)    138
c)    158
d)    142
e)    None of these

Answers:                    
1).c)  2).c)  3).a)  4).c)  5).c)  6).a)  7).c)  8).a)  9).d) 10). c)

SOLUTION:
1).  Distance travelled by first train in one hour
= 60 x 1 = 60 km
Therefore, distance between two train at 9 a.m.
= 330 – 60 = 270 km
Now, Relative speed of two trains = 60 + 75 = 135 km/hr
Time of meeting of two trains =270/135=2 hrs.
Therefore, both the trains will meet at 9 + 2 = 11 A.M.
Answer: c)

2). Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20+x) kmph
20× 10/60=8/60(20+x)
200 = 160 + 8x
8x = 40
x=40/8=5 kmph
Answer: c)

3). Let Abhay's speed be x km/hr.
Then, 30/x-30/2x= 3
 6x = 30
 x = 5 km/hr.
Answer: a)

4). Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x+480/y= 8        1/x+4/y=1/15 ....(i)  
And, 200/x+400/y=25/3      1/x+2/y=1/24 ....(ii)    
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
Answer: c)

5). If the distance be x km, then x/40-x/50=6/60
 x/4-x/5=1
 x=20 km.
 Required time = (20/40) hour – 11 minutes
= (1/2×60-11) minutes = 19 minutes
Answer: c)

6). 32*32 = 1024
1024/4 + 19 = 256+19= 275
Answer: a)

7). a2+b2 = [(a+b)2 +(a-b)2 ] / 2
? = 184041 + 6889
? = 95465
Answer: c)

8). 1337+231-839 = 729
27*27 = 729
Answer: a)

9). [15+22-7/10] * x = 1098
X = 1098/3 = 366
Answer: d)

10). 335+444*56 = 24864+335 = 25199 – 235 = 24964
158*158 = 24964
Answer: c)




 

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