Twisted Ones Quantitative Aptitude Questions for NABARD Grade B 2017

Dear Readers,


Practice is the key to perfection in Quant, so brush up your skills and test yourself with these 15 questions of Quantitative Aptitude for NABARD Grade B 2017.

Q1. A man bought 30 defective machines for Rs. 1000. He repaired and sold them @Rs. 300 per machine. He got profit of Rs. 150 per machine. How much did he spend on repairs?
(a) 5500
(b) 4500
(c) 3500
(d) 2500
(e) None of these

S1. Ans.(c)
Sol.
Total Machine = 30
Total amount of machine = 30 × 30 = 9000
Total profit = 150 × 30 = 4500

Amount spent on repairs = (9000 – 4500 – 1000) = 3500 Rs.

Q2. If Aman borrowed same amount as Babu from Arunoday at same rate of interest for 3 years at simple interest while Babu borrowed at compound interest compounded annually at same rate and time period and the difference between their interest is Rs. 992.25 then find the rate pcpa if Aman borrowed Rs. 14,000.
(a) 15%              
(b) 10%              
(c)  5%                
(d) 12%              
(e) None of these



Q3. In an examination the percentage of students qualified to the number of students appeared from school ‘A’ is 70%. In school ‘B’ the number of students appeared is 20% more than the students appeared from school ‘A’ and the number of students qualified from school ‘B’ is 50% more than the students qualified from school ‘A’. What is the percentage of students qualified to the number of students appeared from school ‘B’?
(a) 87.5%
(b) 75%
(c) 85.5%
(d) 77.5%
(e) None of these


Q4. Pipe A, B and C are kept open and fill a tank in t minutes. Pipe A is kept open throughout, Pipe B is kept open for the first 10 minutes and then closed. Two minutes after Pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills as equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes. How long will C alone take to fill the tank?
(a) 18
(b) 36
(c) 27
(d) 24
(e) 20


Q5.  The average age of a couple was 24 years. After their 1st and 2nd children(twins) were born, the average age of the family became 13.5 years. The average age of the family just after 3rd child was born was 13.2 years. The average age of the family after 4th child was born was 16 years. The current  average age of the family is 19 years. What is the current age of the twin children?
(a) 14 years 
(b) 15 years 
(c) 11 years
(d) 12 years
(e) None of these

S5. Ans.(d)
Sol.
Sum of the ages of the couple = 24 × 2 = 48
After Ist and 2nd children, sum = 13.5 × 4 = 54
Difference in sum = 54 – 48 = 6 yrs.
or, after 6/2 = 3 yrs,
twins were born to the couple
After 3rd child, sum = 13.2 × 5 = 66 yrs
Difference = 66 – 54 = 12
∴ after 12/4 = 3 yrs, third child was born.
Now, after 4th child, sum = 16 × 6 = 96 yrs.
Difference = 96 – 66 = 30
∴ after 30/5=6 yrs, 4th child was born
Current sum = 19 × 6 = 114 yrs
Difference = 18 yrs
∴ After 18/6=3 yrs, we are calculating current age.
Hence, the gap between the children are as follows:
3 yrs, 6 yrs, 3 yrs
∴ Age of eldest ones = 3 + 6 + 3 = 12 yrs.

Q6. There are 500 rooms in a multi-floored hotel. However, due to a change in rule, the hotel was to decrease the number of floors by 5. However, the management is able to put 5 more rooms in each floor. Overall, the number  of rooms in the hotel decreases by 10%. Find the number of floors and the number of rooms/floor  the hotel originally had?
(a) 10 floors    50 rooms
(b) 20 floors   20 rooms
(c) 20 floors     25 rooms
(d) 50 floors    25 rooms
(e) Cannot be determined

S6. Ans.(c)
Sol.
Let the no. of floors be x& rooms per floor be y
So, xy=500
After the change,
(x-5)(y+5)=450
⇒xy-5y+5x-25=450
⇒500-5(y-x)-25=450   [∵xy=500]
or, 5(y-x)=25
from option (c), condition satisfies.

Q7. A, B and C start from the same place and travels in same direction at speeds of 20, 30 and 40 kmph respectively. B starts 3  hour after A. If B and C overtake A at the same instant. How many hours after A did C start?
(a) 4
(b) 3.25
(c) 4.5
(d) 5.5
(e) None of these

S7. Ans.(c)
Sol.
Speed of A, B and C are 20 kmph, 30 kmph &40 kmph resp.
B start when A already travelled for 3 hours
∴ Distance covered by A = 3 × 20 = 60 km
∴  Time taken by B to overtake A = 60/((40-30) )=6 hr.
II means, when B overtake A, A has travelled for 9 hr and B for 6 hr.
Acc. to question,
B & C will overtake A at same instant
Let C take t hours to cover the same distance as covered by A in 9 hr
∴ t × 40 = 9 × 20
t = 4.5 hr
So, C started after 9 – 4.5 = 4.5 hours after A started

Q8. Sameer went from Delhi to Kanpur via Agra  by car. The distance from Delhi to Agra is 1/2 times  the distance  from Agra to Kanpur. The average speed from Delhi to Agra was one –third time as much as that from Agra to Kanpur. The average speed for the entire journey was 60 kmph. What was the average speed from Agra to Kanpur?
(a)1.15 kmph
(b)90 kmph
(c)120 kmph
(d)100 kmph
(e)None of these

S8. Ans.(d)
Sol.
Let total distance from Delhi to Kanpur=3d/2  km
Distance from Delhi to Agra=  d/2  km
and distance from Agra to Kanpur = d km
Let Avg. speed from Agra to Kanpur = x
Avg.speed from Delhi to Agra =  1/3 x
∴Avg.speed=(3d/2)/(3d/2x+d/x)=60
⇒x=100 kmph

Q9. Out of the 75 students enrolled in the Economics class, 12% did not qualify to take  the final examination on account of lack of attendance. Of those who qualified to take the test, 9.09% absented themselves from the exam. Two thirds of those who took the exam passed the examination.  75% of those who passed secured a first class in the paper. How many students who passed the exam scored less than first class?
(a) 15
(b) 10
(c) 12
(d) 16
(e) None of these

S9. Ans.(b)
Sol.
Out of 75 students, 12% did not qualify for final
∴ Remaining = 75 – 12 % of 75 = 66
Also 9.09% out of 66 were absent
∴Present=66-66×(1/11)=60
Passed=2/3×60=40
No. of students who got 1st class = 40 × 75% = 30
So, No. of students who passed the exam & did not get 1st class = 40 – 30 = 10

Q10. P, Q and R are three towns on a river which flows uniformly. Q  is equidistant from P and R. A man rows from P to Q and returns in 10 h. He can row from P to R in 4 h. What is the ratio of speed of the man in still water to the speed of the current?
(a)  5 : 3
(b) 3 : 5
(c) 2 : 5
(d) 1 : 2
(e) None of these


Q11. A, B and C can walk at the rates of 3,4 and 5 km an hour respectively. They start from Poona at 1,2,3 o’clock respectively. When B catches A, B sends him back with a message to C. When will C get the message?
(a) 5:15
(b) 5:20
(c) 5 : 30
(d) 6:00
(e) None of these


Q12. At Gorakhnath Temple there are some magical bells which toll 18 times a day simultaneously. But every bell tolls at a different interval of time, but not in fraction of minutes. What is the maximum number of bells in the temple?
(a) 18
(b) 10
(c) 24
(d) 29
(e) None of these


Q13. A student of 5th standard started writing down the counting numbers as 1,2,3,4,…… and then he added all those numbers and got the result 500. But when Varsha checked the result she found that he had missed a number. What is the missing number?
(a) 30
(b) 32
(c) 28
(d) 25
(e) None of these

S13.Ans(c)
Sol.
Here, we will first find the sum of n natural numbers  which is just greater than 500.
Sum of first 31 numbers = 496
And Sum of first 32 natural numbers  = 496 +32 = 528
Since, the student has got exact 500 as sum of natural numbers then
He must have skipped (528 – 500) = 28 in between the process.

Q14. Anjali and Shivani fired 45 shots each. Total 66 bullets hit the target and the remaining bullets missed it. How many times does the Anjali hit the target if it is known that the number of hits per one miss shown by the Anjali is twice that of Shivani ?
(a) 30 
(b) 36
(c) 40
(d) 35
(e) None of these


Q15. One day very early morning, Anjali went to temple to offer some flowers as a part of puja. She purchased some flowers but the seller offered her that if she would give him all his Rs. 2, she could get all the remaining  6 flowers and thus could gain Rs. 60 paise per dozen. If each time the transaction is possible only in rupees then how many flowers did Anjali purchase initially?
(a) 6
(b) 3
(c) 4
(d) 12
(e) None of these

S15.  Ans (c)
Sol.
Initially  Anjali has only Rs. 2 with her. 
Since, we know that
Anjali can pay the sum in Rupees only ; it means she cannot pay in paise.
By hit and trial method, Consider option C
∴She must have purchased initially the flowers of exactly Rs.1.=4
Later on,she could purchased the number of flowers for total Rs. 2 = 4 + 6 = 10
Thus, the initial cost of one dozen flowers = Rs. 3. 
And the changed cost of one dozen flowers = Rs. 2.40
Hence,she couldgain 60 paisa per dozen.