Welcome to Online Reasoning Section with explanation in AffairsCloud.com. Here we are creating Best question samples for IBPS PO 2017. We have included Some questions that are repeatedly asked in bank exams !!!
Q(1-5). Study the following data and answer the questions
In a class there are six students Raju, Ravi, Ramu, Sita, Gita, Meena. They appeared in college examination, In which ten subjects are divided into five groups as shown in table. Students obtained marks as shown in the table Maximum mark in each subject is 100. Each Group may contain one or two or three subjects.Group mark is the average of all subjects in that group for a student and final score is the average of all group marks he obtained. Final score is represented in the line graph.
- If the marks obtained by Ramu in English is same as marks obtained by Ravi in Telugu then marks obtained by him in Java is
E. NoneAnswer – D. 83
Ramu Marks in English be X then
((x+83+87)/3 + 100 + (87+83+85)/3 +(99+97)/2 +79)/5 = 90
Ramu’s English marks is 94 which is same as Ravi’s telugu marks
Then Ravi’s Java Marks will be
((98+96+94)/3 + 100 + (99+100+95)/3 +(98+98)/2 + java) ) /5 = 95
Java marks = 83
- What is the mark obtained by Raju in Maths if his mark in Maths and Hindi mark are equal?
E. Cannot be DeterminedAnswer – B. 82
((58+X+70)/3 + X + (82+65+60)/3 +(75+71)/2 + 80))/ 5 = 75
- If students are ranked based on the average of Science marks then who among the following will be ranked 3?
D. Both Ramu & Sita
E. NoneAnswer – D. Both Ramu & Sita
- Sita was given a chance to take re-exam on Maths paper now she obtained 95 Marks then what will be her new rank based on final score?
A. 1 st
B. 2 nd
C. 3 rd
D. 4 th
E. NoneAnswer – C. 3 rd
- Who among the following got the highest marks in Java?
E. RamuAnswer – A. Ravi
Q(6-10). Answer the following questions
- Ramesh borrowed a sum of Rs’ X’ from Aruna at the rate of 12% for 2 years. He added 30% more money to the sum borrowed and lent it to Karthik at the rate of 15% for 2 years. If Ramesh gained Rs.750 on the whole transaction then what is sum did he borrowed from Aruna?
E. NoneAnswer – A. Rs.5000
X * 12 *2 /100 = 24X /100
1.3X * 15 * 2/ 100 = 39X/ 100
39X/100 – 24x /100 = 750
Then X = 5000
- Kajal lent sum money at CI rate of 10% per annum for 2 years. He would have got Rs 2753125 more if he lent it for same CI rate but compounded half yearly. Then what is the amount lent by Kajal?
A. Rs.40 Crores
B. Rs.50 Crores
C. Rs.60 Crores
D. Rs.80 Crores
E. NoneAnswer – B. Rs.50 Crores
P ( (1+5/100) 4 – (1+10/100) 2 ) = 2753125
By solving We get P = 50 crores
- A square field is having the area of 58564 Sq m is to be fenced with Four poles & Three wires.Four poles are kept on corners of the Square field. Three wires are fenced at height of 1,2,3 respectively from the ground coiling the Poles. Due to coiling each wire used for fencing is 5% greater than the perimeter of the square. The cost of one-meter wire is Rs. 10 and cost of each Pole are Rs 1127 then the total cost of fencing is?
E. NoneAnswer – D. Rs.35000
S 2 = 58564
4S = 968
Wire used = 3*968*105/100
Cost of wire = 30492
Cost of poles = 4 x 1127 = 4508
Total cost of fencing = 30492+1127 = 35000
- A pipe can fill a tank in 15 minutes and another pipe can fill it in 12 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 min in the beginning and then third pipe is also opened. Time taken to empty the water tank is?
A. 10 mins
B. 25 mins
C. 35 mins
D. 45 mins
E. None of the AboveAnswer – D. 45 mins
x/6 – (x+5)/15 – (x+5)/12 = 0
x = 45 mins
- Two containers A & B are having volumes 140 Litre and 60 litres filled with oil having different prices. Now equal quantities are withdrawn from both containers A & B such oil drawn from A is poured in B and oil drawn from B is poured in A . Then the price per litre becomes equal in both the containers. Then what is amount of oil drawn from B in litres
E. Cannot be determinedAnswer – C. 42
Let price of oil in container A is R and price of oil in container in B is P
Then total price of A is 140*R
Total of Price of B is 60*R
Now suppose L litres are withdrawn and replaced
Then the price of oil in Container A will be (140*R – L*R + L *P)
Price of one litre is (140*R – L*R – L *P)/140
Similarly price of oil in container B is (60*P – L*P + L *R)
Price of one ltre is (60*P – L*P + L *R)/60
Now both prices are equal
Solving equations we get L is 42