Quant Test 5 Practice

Arun's present age in years is 40% of Barun's. And after few years, Arun’s age will be half of Barun’s. We have to find by what percentage will Barun’s age increase during this period.
So Arun’s present age is 2x and Barun’s present age will be equal to 5x.
2(2x + y) = 5x + y
4x + 2y = 5x + y
4x + 2y = 5x + y
y = x
Hence 2x and 5x becomes 3x and 6x respectively.
The percentage by which Barun's age increase during this period is that x increases of 5x so it increases by 20%.

He works alone on Day 1.
On Day 2, he is joined by another person who also can complete the job in exactly 120 days.
On Day 3, they are joined by another person of equal efficiency.
Each day a new person with the same efficiency joins the work
Thus by first day 1 person
and by second day 2
Third day 3 and it goes on.......until it makes a total of 120
We have to find how many days are required to complete the job
So 15×162 = 120 [by n(n+1)2]
So 1 + 2 + 3 + 4 ...... till 15 days are required to complete the Job.
It will get completed in 15 days.

Given that a man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed.
If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure.
d/12 is one time taken and d/15 is another time taken.
The difference between these two times is 20 minutes
We have to find the distance (in km) from his home to the railway station.
20 mins is 13^rd of an hour.
d/12 = d/15 + 13
d/12 – d/15 = 13
(5d−4d)/60 = 13
d/60 = 13
d = 20 km
The distance (in km) from his home to the railway station is 20 km.
Important point to remember is that we don’t need to take the actual time here, we can take that difference of 10 minutes before and 10 minutes after which is equal to 20 minutes i.e. 13^rd of an hour.

The elevator has a weight limit of 630 kg.
It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg.
We have to find the maximum possible number of people in the group.
We can take one person to be 57 kg and since they have asked for the maximum possible number of people it can accommodate in the elevator we can take more people with lightest weight.

So, at least one guy with 57 kg.
630 – 57 = 573
57353

= 10.8 or at most there can 10 people.
Hence, 10 + 1 = 11 people
The maximum possible number of people in the group is 11.

Given that if a seller gives a discount of 15% on retail price, she still makes a profit of 2%.
Let us assume marked price to be x. So, selling price to be 0.85x on giving 15% discount on retail price she still makes a profit of 2%.

We have to find which of the following options ensures that she makes a profit of 20% so the discount should be much lower.

0.851.02 × 1.20 = 1
Hence by selling at retail price she makes a profit of 20%.

The profits made by C1, C2, and C3 are in the ratio 9 : 10 : 8 while the profits made by C2 , C4 and C5 are in the ratio 18 : 19 : 20.
Now we can simplify this and take all of them into same variable so we need to take LCM for 10 and 18 which happens to be 90.

With 90 as LCM,

C1 = 81
C2 = 90
C3 = 72
C4 = 95
C5 = 100
We have to find the total profit (in Rs) made by all five companies
C5 is 19 crores more than C1 which works out as such. i.e C1 = 81 , C5 = 100.
So to get the total profit we can add them altogether = 81 + 90 + 72 + 95 + 100 = 438
Hence Rs.438 crore is the total profit made by all five companies.

Ravi invests 50% of his monthly savings in fixed deposits so from the remaining 50%, 30% percent of the rest of his savings = 15% is invested in stocks.
And the rest  50 – 15 = 35% goes into Ravi's savings bank account.
If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500
i.e. 50 + 35 = 85 % of total savings = 59500
We have to find the Ravi's total monthly savings (in Rs)
Total savings = 5950085 × 100 = Rs.70000
Ravi's total monthly savings is 70000 Rupees.

We have to find the number of solutions (x, y, z) to the given equation.
x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12.
x – y – z = 25, so the maximum value y and z can take is 12 and x can take is 40.
From 40 we can subtract some values to get 25 i.e.15 can be subtracted,
If x = 40 then y + z = 15
y ≤ 12 so y can take all the corresponding values 12,11,10,......3 it is a set of 10 values.
z ≤ 12, so z can take corresponding other values to get with 15 when added with x
Similarly,
If x = 39, then y + z = 14 here y can take all the value from 12 ,11,...till 2
So there are 11 values.
If x = 38, then y + z = 13 here y can take all values from 12,11,10,....till 1
So there are 12 values.
If x = 37 then y + z = 12 here y can take 11,10,9,.....,1 so there are 11 values
If x = 36 then y + z = 11 here y can take 10,9,......,1 so there are 10 values
If x = 35 then y + z = 10 here y can take 9,8,.......,1 so there are 9 values
If x = 34 then y + z = 9 here y can take 8,.......,1 so there are 8 values
If x = 33 then y + z = 8 here y can take 7,........,1 so there are 7values
If x = 32 then y + z = 7 here y can take 6,........,1 so there are 6 values
If x = 31 then y + z = 6 here y can take 5.........,1 so there are 5 values
If x = 30 then y + z = 5 here y can take 4,......,1 so there are 4 values
If x = 29 then y + z = 4 here y can take 3,...,1 so there are 3values
If x = 28 then y + z = 3 here y can take 2,1 so there are 2 values
If x = 27 then y + z = 2 here y can take 1 so there is 1 value
x cannot be less than 27 because it is given that y and z are positive integers so it has to be at least one cannot be less than one
So, 12×132 + 21 = 78 + 21 = 99.
The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is 99.

Let the average marks scored by boys during the mid semester exam be n.
Then, the girls’ average mark will be n + 5.
On calculation, the average of the entire class will be n + 3.
This is in the ratio of 2 : 3 since there are 20 boys and 30 girls in the class.

During the final exam, the average score of the girls dropped by 3.
So, n + 5 becomes n + 2 while the average score of the entire class increased by 2 or it becomes n + 5.
Using alligation we can say that the difference between the average mark of entire class and average mark of girls is 3 which is 25^th of the total.
Hence 35^th of the total is 4.5.
Thus the average of boys = (n + x) – (n + 5) = 4.5,
On solving we get x = 9.5.

Given that a solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. We have to find the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube.
Let us take the volume of the 5 cubes to be 1x³ , 1x³ , 8x³ , 27x³ , 27x³
Volume of the original cube = x³(1 + 1 + 8 + 27 + 27)
Volume of the original cube = 64x³
Sides of the original cube = 64x³ = 4x
Similarly, sides of the 5 smaller cubes = x , x , 2x , 3x , 3x .
Surface Area of a cube = 6a²
Surface Area of the original cube = (4x) × (4x) = 16x²
Surface area of the smaller cubes = x² , x² , 4x² , 9x² , 9x²
Sum of the surface areas of the smaller cubes = x²(1 + 1 + 4 + 9 + 9) = 24x²
Change in surface area = 24x² – 16x² = 8x²
%change = 8x216x2 × 100 = 50%
Hence the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to 50%.