Tuesday, 24 January 2017

Quantitative Aptitude questions for IBPS, UPSC, CAT and other national exams

Quantitative aptitude questions and solutions for preparation for practice for various exams in 2017 are given below. Try to solve the questions on your own. If you are unable to solve the solution after many attempts, only then look for the solutions. Solutions to the problem are given at the end.

Directions for Questions 1 to 10: Read the following and answer the questions that follow.
A fort contains a granary, that has 1000 tons of grain. The fort is under a siege from an enemy army
that has blocked off all the supply routes.
The army in the fort has three kinds of soldiers:
Sepoys --> 2,00,000.
Mantris --> 1,00,000
Footies --> 1,00,000
100 Sepoys can hold 5% of the enemy for one month.
100 Mantris can hold 10% of the enemy for 15 days.
50 Footies can hold 5% of the enemy for one month.
A sepoy eats 1 kg of food per month, a Mantri eats 0.5 kg of food per month and a footie eats 3 kg of food. (Assume 1 ton = 1000 kg).
The king has to make some decisions based on the longest possible resistance that can be offered to the enemy.
If a king selects a soldier, he will have to feed him for the entire period of the resistance. The king is not obliged to feed a soldier not selected for the resistance.
(Assume that the entire food allocated to a particular soldier for the estimated length of the resistance is redistributed into the king’s palace in case a soldier dies and is not available for the other soldiers.)

1. If the king wants to maximise the time for which his resistance holds up, he should
(a) Select all mantris 
(b) Select all footies
(c) Select all sepoys 
(d) None of these

2. Based on existing resources, the maximum number of months for which the fort’s resistance
can last is
(a) 5 months 
(b) 20 months
(c) 7.5 months 
(d) Cannot be determined

3. If the king makes a decision error, the maximum reduction in the time of resistance could be
(a) 15 months 
(b) 12.5 months
(c) 16.66 months 
(d) Cannot be determined

4. If the king estimates that the attackers can last for only 50 months, what should the king do to ensure victory?
(a) Select all mantris
(b) Select the mantris and the sepoys
(c) Select the footies
(d) The king cannot achieve this

5. If a reduction in the ration allocation by 10% reduces the capacity of any soldier to hold off the enemy by 10%, the number of whole months by which the king can increase the life of the resistance by reducing the ration allocation by 10% is
(a) 4 months 
(b) 2 months
(c) No change 
(d) This will reduce the time

6. The minimum amount of grain that should be available in the granary to ensure that the fort is not lost (assuming the estimate of the king of 50 months being the duration for which the enemy can last is correct) is
(a) 2000 tons 
(b) 2500 tons
(c) 5000 tons 
(d) Cannot be determined

7. If the king made the worst possible selection of his soldiers to offer the resistance, the percentage increase in the minimum amount of grain that should be available in the granary to ensure that the fort is not lost is 
(a) 100%
(b) 500% 
(c) 600%
(d) Cannot be determined 

8.The difference in the minimum grain required for the second worst choice and the worst choice to ensure that the resistance lasts for 50 months is 
(a) 5000 tons
(b) 7500 tons 
(c) 10000 tons
(d) Cannot be determined

9.If the king strategically attacks the feeder line on the first day of the resistance so that the grain is no longer a constraint, the maximum time for which the resistance can last is 
(a) 100 months
(b) 150 months 
(c) 250 months
(d) Cannot be determined 

10.If the feeder line is opened after 6 months and prior to that the king had made decisions based on food availability being a constraint then the number of months (maximum) for which the resistance could last is 
(a) 100 months
(b) 150 months 
(c) 5 months
(d) Cannot be determined

11. Three friends A, B and C start from P to Q that are 100 km apart. A is on a moped while B is riding pillion and C walks on. A takes B to a point R and returns to pick up C on the way and
takes him to point Q. B, on the other hand, walks to Q from point R (R is the mid-point
between P and Q).
If the ratio of speeds of the three people is 5 : 2 : 2, find where will the last person be when the first person reaches Q.
(a) 12.85 km from Q 
(b) 12.75 km from Q
(c) 16.66 km from Q 
(d) 83.33 km from P

12.In his book on Leonardo da Vinci, Sigmund Freud, after a detailed psychoanalysis concluded that Goethe could complete the masterpiece in nine days as he could channelize overly but was more possessed as a result of which he could generate 50% more efficiency than Goethe. The number of days it takes Leonardo da Vinci to do the same piece of work that Goethe completes in nine days is:
(a) 4 (1/2) days
(b) 6 days
(c) 13 (1/2) days
(d) None of these

13.A dealer buys oil at Rs. 100, Rs. 80 and Rs.60 per liter. He mixes them in the ratio 5:6:7 by weight and sells them at a profit of 50%. At what price does he sell oil?
(a) Rs. 80/liter
(b) Rs.116.666/ liter
(c) Rs. 95/liter
(d) None of these

14. The figure contains three squares with areas of 100 m2, 49 m2, and 16 m2 lying side by side. Rahim wishes to reduce the length AB to 19 m by reducing area of only one of the squares without altering its shape, what is the maximum reduction (in %) in the area of that square?

Quantitative Aptitude questions for IBPS, UPSC and other national exams

(a) (Approx) 33%
(b) 50%
(c) (Approx) 66%
(d) 75%

15. The village of Hara Kiri has two-thirds of its male population married to 90% of its women population. Each of such man is married to exactly two of such women. What is the minimum possible 4 digit population of the village (Children are not counted as part of the populationunless they are married)?
(a) 1005
(b) 1026
(c) 1080
(d) Cannot be determined/None of these

Answers and Solutions to problems above: 
Sol 1. If all sepoys are chosen, the food requirement will be 200 tons/month. The resistance will last for 1000/200 = 5 months.
If footies are chosen, the food will last for 1000/100×3 = 3.33 months.
If mantris are chosen, the food will last for 1000/100×0.5 = 20 months
So, answer is (a)

Sol 2. B) 20 months

Sol 3. c) 20-3.33 = 16.66 months

Sol 4. d) The king cannot achieve this based on existing resources

Sol 5. No change will occur

Sol 6. for 50 months, mantris will require 1000×2.5/100 ×0.5 = 50 months
Mantris eat 50 tons of food per month. If 2500 tons of food is available only then mantris can last for 50 months. Sepoys and footies eat more food than Mantris so minimum amount required is 2500 tons

Sol 7. The worst decision will occur for footies
For footie , 1000×15/100×3 = 50 months
So, 15000 tons required for footies
% increase from the minimum amount= (15000 - 2500)/2500 = 5 ×100 = 500%

Sol 8. In worst case the minimum grain required is 15000 tons and the 2nd worst case occurs for soldiers where the minimum grain required is 10000 tons for 50 months.
So, the difference is 15000-10000 = 5000 tons

Sol 9. For questions 19 & 20, when food is not a constraint, the number of lives become constraint. 
So, to offer 100% resistance
for sepoys, 100  sepoys offer 5%
so, 20×100 sepoys offer 100% resistance
Similarly 1000 mantris offer 100% resistance for 15 days 
Or 2000 mantris offer 100% resistance for 1 month
& 50 footies offer 5% resistance for 1 month 
1000 footies offer 100% resistance for 1 month
So, length of resistance = 200000/2000 + 100000/2000 + 100000/1000 = 250 months

Sol 10. If decision based on food are made before 6 months, then all the sepoys and footies will die in 5 months as the maximum period for survival of sepoys is 5 months. After 6th month, when feeder line is opened food no longer is a constraint so resistance becomes constraint.
Number of mantris that will die in 6 months are  2000×6 = 12000 mantris will be lost.

Sol 11. the question may look lengthy or confusing but the solution is easy if you tske a step by step approach. Let's solve the problem:
let the speeds of A,B and C be 5, 2 and 2 km/hr
I. A and B goes from P to R. PR = 50km
T(A+B) = 50/5 =10
In the same time C covers a distance = 2×10 =20

II. now A returns from R to pick C and C moves ahead of 20 to meet eachother.
SO, time taken by C ahead of 20 km to meet A = time taken by A from R to meeting point,S 
time = x/2 = (30 - x)/5 => x = 60/7
time taken by C  = 60/7×2 = 30/7
SO total distance covered by C to point S = 20+60/7 =200/7

distance covered by B in the same time= 30×2/7 =60/7
distance covered by B = 50+60/7 =410/7
III. after the point of meeting of C and A, both move towards Q at speed 5 km/hr.
SO distance that needs to be covered by C and A is 100 - 200/7 = 500/7
and this distance is covered in 500/7×5 =100/7 hrs

Distance that needs to be covered by B is 100 - 410/7 = 290/7
And this distance is covered by B in 290/7×2 = 145/7 hrs

SO, C and A reach point Q earlier than B. SO, distance that needs to be covered by B to reach point Q is (145-100)×2/7 = 90/7 = 12.85km. SO B is at a distance of 12.85 km from Q when A and C reach Q.

Sol 12. Since efficiency is 50% more. 
Efficiency of leonardo da vanci =1.5 times the efficiency of goathe 
So, the number of days to do the same work =9/1.5 days = 6 days

Sol 13. Price of the mixed oil =
5/18 X 100 + 6/18 X 80 + 7/18 X 60
= 1400/18 = 77.777
Now he sells at a profit of 50%
Therefore , selling price = 77.77 x 1.5 = 116.66

Sol 14. The given squares have lengths of 10, 7 and 4. The sum of lengths =10+7+4=21
Now we have to reduce 2 m from side of one of the squares so that the total length becomes equal to 19. But the length must be reduced from that square for which the reduction in area is maximum. This Will happen when the length is reduced from smallest square . so (4^2 -2^2) = 4^2 =12/16 =3/4 which corresponds to 75% reduction in area .

Sol 15. Let us write the conditions given:
Married Male(mm) = 2/3 x total male 
Total Female x 0.9 =married female (MF)

Total population = total male + total Female 
= 3/2 x married Male + MF/0.9
= 3/2 x married Male + 2/0.9 x MF 
= 67/18 x married Male
So, mM = 18/67 x total population 
Now try the values in options . The smallest 4 digit value that results in natural number is 1005

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