Breakup G Strategy | Linear Equation + Maximum/Minimum | Moderate
In a school with 1500 students , each students chooses one of the streams out of science , arts and commerce by paying a fee of rs 1100 , 1000 and 800 respectively. The total fees paid by all the students is RS. 15,50000.If the number of science students is not more than the number of arts students , then the maximum possible number of science students in school is : [TITA] _______
Solution & Explanation
Correct Answer: 700
Given S – science students , A – arts students , C – commerce students
S + A + C = 1500 —- I
1100S + 1000A + 800C = 1550000 becomes
11S + 10A + 8C = 15500 ——- II
Where S ≤ A , since we have to maximise S this S = A
Multiplying I by 8 :
8S + 8A + 8C = 12000 —— III
Subtracting III from I
3S + 2A = 3500 , Since S = A
5S = 3500 , S = 700
Answer : 700
Breakup G Strategy | Average + Percentage + Equations | Moderate
The average salary of 5 managers and 25 engineers in a company is Rs 60000. If each of the manager receive 20% hike in the salary while the salary of the engineer remains unchanged , the average salary of all 30 employees increased by 5%.The average salary, of engineer , in rs , is :
1.45000
2.54000
3.50000
4.40000
Solution & Explanation
Correct Answer: 2 (5400)
Total salary of 30 employees (5 managers and 25 engineers):
30 × 60,000 = 1,800,000
After 20% hike for managers, average salary increases by 5%, so new average salary is:
60,000 × 1.05 = 63,000
Total salary after the hike:
30 × 63,000 = 1,890,000
Before hike: 5M + 25E = 1,800,000
After hike: 6M + 25E = 1,890,000
Solve for M (manager’s salary):
(6M + 25E) – (5M + 25E) = 1,890,000 – 1,800,000
M = 90,000
Solve for E (engineer’s salary):
5 × 90,000 + 25E = 1,800,000
450,000 + 25E = 1,800,000
25E = 1,350,000 Thus E = 54,000
Breakup G Strategy: Numbers + Maximum/Minimum | Hard
Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose values lies between 150 and 500.If p is less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is: _____
Solution & Explanation
Correct Answer: 397
Give p + q + r = 900 , 0.3q ≤ p ≤ 0.7q
r lies between 150 to 500 and is a perfect square
So r = 169 , 196 , 225 , 256 , 289 , 324 , 361 , 400 , 441 , 484
For maximum p (0.7q) , r should be minimum, r = 169
0.7q + 1q + 169 = 900 , q = 430 so p = 0.7 × 430 = 301
For minimum p (0.3q) , r should be maximum r = 484
0.3q + 1q + 484 = 900 , q = 320 , so p = 0.3 × 320 = 96
Pmax + P min = 301 + 96 = 397
Answer 397
LCM G Strategy Concept of Time and Work | Hard
Teams A, B, and C consist of five, eight, and ten members, respectively such that the members within each team is equally productive. Working separately teams A , B and C can complete certain job in 40 hours , 50 hours and 4 hours respectively Two members from team A, three members from team B, and one member from team C together starts the work and the member from team C left the work after 23 hours. The number of additional member(s) required from team B , that would be required to replace the member from team C , to finish the job in next one hour is :
1.4
2.2
3.3
4.1
Solution & Explanation
Correct Answer: 2
Total work that has to be completed (LCM of 40,50,4) = 200 units
Efficiencies
Member of team A = 200/40 = 5units per hour , so 1 member (5/5 = 1 unit work per hour)
Member of team B = 200/50 = 4 units per hour , so 1 member (4/8 = 0.5 unit work per hour)
Member of team C = 200/4 = 50 units per hour , so 1 member (50/10 = 5 unit work per hour)
Work Done in 23 hours:
2 + 1.5 + 5 = 8.5 units/hour
Work Done in 23 hours: 8.5 × 23 = 195.5 units
Remaining work = 200 – 195.5 = 4.5 units
Now C left the work , A and B has to complete the remaining work in 1 hour :
2 (1) + (3 + x) × 0.5 = 4.5
On solving x = 2
Thus option 2 is the answer
Breakup G Strategy: TSD + Equation | Moderate
Rahul starts on his journey at 5pm at a constant speed so that he reaches his destination at 11pm same day . However , on his way , he stops for 20 minutes and after that, increases his speed by 3 km per hour to reach on time. If he had to stop for 10 minutes more , he would have had to increase his speed by 5km per hour to reach on time.His initial speed in km per hour was :
1.18
2.15
3.12
4.20
Solution & Explanation
Correct Answer: 2
Let Rahul’s initial speed be V km/h.
The total distance D is D = V × 6.
When he stops for 20mins (6-⅓ = 17/3) , t – l time
6V = (v × t) + (v + 3) × (17/3 – t)
On solving we have v/3 = 17 – 3t , v = 51 – 9t —- I
When he stops for 30 mins (6-½) = 11/2
6v = (v × t) + (v + 5) (11/2 – t)
On solving V/2 = 55/2 – 5t , v = 55 – 10t —— II
On solving I and II , we have t = 4 , then v = 15km/hr
Thus option 2 is the answer
Breakup G Strategy: Time Speed + Percentage | Moderate
Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. The total time taken is 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, required to walk from A to B and run from B to C is: [TITA]
Solution & Explanation
Correct Answer: 444
Assume walking speed = s , Running speed = 1.4s
Distance from A to B = d1 , Distance from B to C = d2
Now based on the information given in the question:
d2/v = 3.5 hrs , d1/1.4v = 3.5 hrs ,
d1/v = 3.5 × 1.4
Now total time required to walk from A to B and run from B to C.
Total time = d1/v + d2/1.4v = (3.5 × 1.4) + (3.5/1.4) =
4.9 hrs + 2.5 hrs = 7.4 hrs = 7hrs 24 mins
7hrs 24 mins = (7 × 60) + 24 = 444 mins
Answer 444
Breakup G Strategy : Mixture + Equation | Moderate
Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After mixing well, an equal volume is taken out from B and poured into A. If the ratio of alcohol and water in A is 15:4, then the volume, in litres, initially taken out from A, is [TITA]
Solution & Explanation
Correct Answer: 16
Given vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively.
Assum x litres of alcohol from A is transferred to B
After the first transfer (from A to B):
Vessel A: 60 – x liters of alcohol, 60 liters of water.
Vessel B: x liters of alcohol, 60 liters of water.
After the second transfer (from B to A):
Alcohol transferred back to A: x * (x / (60 + x)).
Water transferred to A: x * (60 / (60 + x)).
Alcohol and water in A after the second transfer:
Alcohol in A: 60 – x + (x² / (60 + x)).
Water in A: 60 + (60x / (60 + x)).
Set up ratio equation (15:4):
(60 – x + (x² / (60 + x))) / (60 + (60x / (60 + x))) = 15 / 4
Solve for x: x = 16 liters
LCM G Strategy | Time and Work | Easy
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled by pipe A in 15 hours. If all pipes are used simultaneously to fill up this empty tank, the time, in minutes, taken to fill up the tank entirely tank completely is nearest to :
1.71
2.78
3.76
4.73
Solution & Explanation
Pipe A fills the tank in 15 hours.
Pipe A’s rate = 1/15 of the tank per hour.
Let the flow rates be:
Pipe A’s rate = 4x , Pipe B’s rate = 9x ,Pipe C’s rate = 36x
From 4x = 1/15, we find x = 1/60. Now putting x = 1/60 , we have pipe A , B , C rates as 4/60 , 9/60 , 36/60 respectively
Now, add the rates:
Total rate = 4/60 + 9/60 + 36/60 = 49/60
The time to fill the tank is the reciprocal of the total rate:
Time = 1 / (49/60) = 60 / 49 hours.
Convert this to minutes:
Time in minutes = (60 / 49) × 60 ≈ 73.47 minutes ≈ 73
