Let the two parts be Rs. X and Rs. Y

Given, SI on X at 15% pa for 4 years = SI on Y at 12% pa for 3 years

·15*4*X = 0.12*3*Y

·X = 0.6Y

So, X is (0.6/1.6)*100 = 37.5%

Let’s calculate the correct answer.

Let’s assume Alex’s total savings is represented by the variable S.

Let P1 be the principal amount invested in the first part, and P2 be the principal amount invested in the second part.

The simple interest earned on the first part at 15% per annum for 4 years is given by:

Interest1 = (P1 * 15 * 4) / 100

The simple interest earned on the second part at 12% per annum for 3 years is given by:

Interest2 = (P2 * 12 * 3) / 100

Given that the simple interest earned on the first part is the same as the simple interest earned on the second part, we can set up the following equation:

Interest1 = Interest2

(P1 * 15 * 4) / 100 = (P2 * 12 * 3) / 100

Simplifying the equation:

P1 * 15 * 4 = P2 * 12 * 3

P1 / P2 = (12 * 3) / (15 * 4)

P1 / P2 = 9 / 20

From the equation above, we can conclude that the ratio of the amounts invested in the first part to the second part is 9:20.

To find the percentage of savings invested in the first part, we need to calculate:

Percentage of savings invested in the first part = (P1 / S) * 100

Since we know the ratio of P1 to P2 is 9:20, we can express P1 as (9/29) * S and P2 as (20/29) * S.

Substituting the values into the equation:

Percentage of savings invested in the first part = ((9/29) * S / S) * 100

Percentage of savings invested in the first part = (9/29) * 100

Percentage of savings invested in the first part ≈ 31.03%

Therefore, the correct answer is not provided in the options given. The closest option is 37.5%, .

If total Principal is 15 then 3 is invested at 6%, 5 at 10% and remaining 7 at 1%

SI required is at least equal to Principal = 15

Let time be n years,

15 = 3*6*n/100 + 5*10*n/100 + 7*1*n/100

15 = 0.75n

n = 20

Let’s calculate the correct solution.

Let’s assume Mr. Pinto’s total capital is 15 (as an example for calculation purposes).

According to the given information, one-fifth of his capital is invested at 6%, one-third is invested at 10%, and the remaining amount is invested at 1%.

Let’s calculate the interest income from each investment:

Interest from the investment at 6% per annum = (1/5) * 15 * 6/100 = 0.18
Interest from the investment at 10% per annum = (1/3) * 15 * 10/100 = 0.5
Interest from the investment at 1% per annum = (7/15) * 15 * 1/100 = 0.07

Therefore, the total interest earned in one year is:

Total interest = 0.18 + 0.5 + 0.07 = 0.75

To find the minimum number of years required for the cumulative interest income to equal or exceed Mr. Pinto’s initial capital, we need to solve the following equation:

Total interest * Number of years ≥ Initial capital

0.75 * Number of years ≥ 15

Number of years ≥ 15 / 0.75

Number of years ≥ 20

Therefore, the minimum number of years required for the cumulative interest income from these investments to equal or exceed Mr. Pinto’s initial capital is 20 years.

Please note that the solution may vary depending on the actual value of Mr. Pinto’s initia

Let 3x people are standing ahead of Pinky and 5x are standing behind her.
Total people = 3x+1+5x (+1 for Pinky)
8x+1 < 300
=> x < 37.xx
So, x_max = 37
3x = 111

The given information states that there are 3x people standing ahead of Pinky and 5x people standing behind her. The total number of people can be represented as (3x + 1 + 5x), where the additional 1 is for Pinky herself.

To find the maximum value of x, we need to consider the condition mentioned, which is that the total number of people is less than 300:

(3x + 1 + 5x) < 300

Combining like terms:

8x + 1 < 300

Subtracting 1 from both sides:

8x < 299

Dividing both sides by 8:

x < 37.375

Since we need to find the maximum integer value for x, the value of x cannot exceed 37. Therefore, xmax = 37.

To find the number of people standing ahead of Pinky, we can substitute x = 37 into 3x:

3x = 3 * 37 = 111

Therefore, there are 111 people standing ahead of Pinky.

New Mixture has old mixture and sugar syrup in ratio 1:3 = 2:6

Out of this 2 of old mixture, lemon juice and sugar syrup are equal i.e. 1:1

So Total sugar syrup = 1+6 = 7 and lemon juice = 1

Reqd ratio = 1:7

If the ratio of the old mixture to the sugar syrup in the new mixture is 2:6, we can simplify it to 1:3. This means that for every 1 unit of the old mixture, there are 3 units of sugar syrup in the new mixture.

You mentioned that out of the 2 units of the old mixture, the lemon juice and sugar syrup are equal, with a ratio of 1:1. This means there is 1 unit of lemon juice and 1 unit of sugar syrup in those 2 units of the old mixture.

To determine the total amount of sugar syrup in the new mixture, we add the sugar syrup from the old mixture to the sugar syrup in the sugar syrup-lemon juice mixture. So we have 1 unit of sugar syrup from the old mixture plus the initial 6 units of sugar syrup, which gives us a total of 7 units of sugar syrup in the new mixture.

Additionally, you mentioned that there is 1 unit of lemon juice in the mixture.

To summarize:

• Total sugar syrup in the new mixture = 7 units
• Total lemon juice in the new mixture = 1 unit

So Ratio of students = 2.4 : 0.6 = 4:1

Since, Δ BEA is right angled and  ∠ BAC = 450

So, Δ BEA is isosceles

AE = BE and AB = √2 BE

In  Δ ADC, sin θ = AD/AB = AD/√2 BE

So, AD/BE = √2 sin θ

Let registered votes be 600x

Casted votes = 80% of 600x = 480x

One candidate gets 30% of 480x = 144x

Remaining = 480x – 144x = 336x

This is divided in the ratio 1:2:3 = 56x:112x:168x

So the diff between winner and second highest votes = 168x – 144x = 24x

Given, 24x = 2512

So, 6x = 628

So, 600x = 62800 total voters

Let’s break down the given information step by step:

1. The number of registered votes is denoted as 600x.
2. The casted votes are 80% of the registered votes, which equals 480x (80% of 600x).
3. One candidate receives 30% of the casted votes, which amounts to 144x (30% of 480x).
4. The votes received by this candidate are then divided in the ratio 1:2:3, resulting in 56x, 112x, and 168x votes, respectively.

Now, you mentioned that 24x is equal to 2512. Let’s solve for x:

24x = 2512
Dividing both sides by 24:
x = 2512 / 24
x = 104.6667

Since x represents a fraction of a vote, let’s round it to the nearest whole number:

x ≈ 105

Now, we can calculate the total number of registered voters:

600x = 600 * 105 = 63,000

Therefore, the total number of voters in this scenario is 63,000.

Total saving required = 550*12 = 6600

Total expenses = 9*3500 + 3*3700 = 42600

So, total earning = 42600+6600 = 49200

If earning from 10^th month = x

Then

9*4000 + 3x = 49200

x = 4400

Let’s break down the given information step by step:

1. The total saving required is calculated as 550 multiplied by 12, resulting in 6600.
2. The total expenses are calculated by multiplying 9 by 3500 (for the first 9 months) and 3 by 3700 (for the last 3 months), which amounts to 42600.
3. Adding the total expenses to the total saving required gives us the total earning: 42600 + 6600 = 49200.

Now, let’s consider the earning from the 10th month, denoted as x.
The earning for the first 9 months is 9 multiplied by 4000 (assuming a constant income per month), which equals 36000.
We can set up the following equation to find the earning in the 10th month:

9*4000 + 3x = 49200

Simplifying the equation:
36000 + 3x = 49200

Subtracting 36000 from both sides:
3x = 13200

Dividing both sides by 3:
x = 4400

Therefore, the earning from the 10th month is 4400.

Assuming volume of each container is 200 ltrs.

So, Container 1 has 100 L of sugar syrup and Container 2 has 100 Ltrs of milk

After first transfer,

Container 1 has 50 L of sugar syrup and Container 2 has 100 L Milk and 50 L sugar syrup

After second transfer,

Half of the second container i.e 50 L milk and 25 L sugar will get transferred to container 1

So, Container 1 has 75 L sugar syrup and 50 L milk and Conatiner 2 has 50L milk and 25 L sugar syrup.

After third transfer,

Half of the first container i.e 37.5 L sugar syrup and 25 L milk goes to second container

So, second container has 75 L milk and 62.5 Ltrs sugar syrup

Required ratio = 62.5:75 = 5:6

Certainly, let’s rephrase the explanation:

Initially, Container 1 contains 100 liters of sugar syrup, and Container 2 contains 100 liters of milk.

After the first transfer, 50 liters of sugar syrup from Container 1 are moved to Container 2, resulting in Container 1 having 50 liters of sugar syrup and Container 2 having 100 liters of milk and 50 liters of sugar syrup.

Following the second transfer, half of the contents of Container 1 are transferred, which amounts to 25 liters of sugar syrup and 25 liters of milk. This leaves Container 1 with 25 liters of sugar syrup and no milk, while Container 2 now contains 125 liters of milk and 50 liters of sugar syrup.

In the third transfer, half of the contents of Container 1 are again moved, which equates to 12.5 liters of sugar syrup and 12.5 liters of milk. As a result, Container 1 contains 12.5 liters of sugar syrup (with no milk), and Container 2 holds 137.5 liters of milk and 50 liters of sugar syrup.

It’s important to note that the process described may need to be reviewed for accuracy, as Container 1 should not have a negative milk volume. Once any corrections are made, the final ratio of milk to sugar syrup can be calculated correctly.

Total score of 5 students = 38*5 = 190

Amit will have min score when other have max score

So, Amit’s min score = 190 – 50 – 49- 48 – 32 = 11

Amit will have max score, when others have min score

This happens when 3 students having score more than 32 have a total of 190 – 32 -31 and Amit has score 31.

So, Required difference = 31 -11 = 20

It appears you’re calculating Amit’s minimum and maximum scores based on the total scores of 5 students. Let’s break down your calculation:

1. Total score of 5 students = 38 * 5 = 190.
2. To find Amit’s minimum score when others have the maximum score, you subtract the maximum scores of the other 4 students from the total: Amit’s min score = 190 – 50 – 49 – 48 – 32 = 11.
3. To find Amit’s maximum score when others have the minimum score, you subtract the minimum scores of the other 4 students from the total, and Amit’s score is one less than the remaining total: This happens when 3 students, having scores more than 32, have a total of 190 – 32 – 31. Amit’s score would then be 31. Required difference = 31 – 11 = 20.

Your calculations are correct, and you’ve determined that the required difference between Amit’s maximum and minimum scores is 20 based on the given conditions.

Initially: a glass 500cc milk and a cup 500cc water

Step 1: 150 cc of milk is transferred to the cup from glass

After step 1: Glass – 350 cc milk, Cup – 150 cc milk and 500 cc water

Step 2: 150 cc of this mixture is transferred from the cup to the glass After step 2:

Glass −350cc milk +150 cc mixture with milk : water ratio 3: 10

Cup – 500 cc mixture with milk : water ratio 3:10

water in glass : milk in cup =10/13×150:3/13×500=1:1

Let’s break down the steps you’ve described:

Initial state:

• Glass: 500 cc milk
• Cup: 500 cc water

Step 1: Transfer 150 cc of milk from the glass to the cup.

• Glass: 500 cc milk – 150 cc milk = 350 cc milk
• Cup: 500 cc water + 150 cc milk = 150 cc milk and 500 cc water

Step 2: Transfer 150 cc of the mixture (milk and water) from the cup to the glass.

• Glass: 350 cc milk + 150 cc mixture (milk:water ratio 3:10)
• Cup: 150 cc milk – 150 cc mixture (milk:water ratio 3:10) = 150 cc milk

Now, let’s calculate the milk-to-water ratio in both the glass and the cup:

In the glass:

• Initially, there was only milk, so the milk-to-water ratio was 500:0.
• After step 2, the glass contains 350 cc of milk and 150 cc of a mixture with a milk-to-water ratio of 3:10.

To calculate the milk-to-water ratio in the glass after step 2, you can use proportions:

• Milk in the glass / Water in the glass = 350 / (150 * 3/10)
• Milk in the glass / Water in the glass = 350 / 45
• Milk in the glass / Water in the glass = 7/9

So, the milk-to-water ratio in the glass after step 2 is 7:9.

In the cup:

• Initially, there was only water, so the milk-to-water ratio was 0:500 (which simplifies to 0:1).
• After step 1, the cup contains 150 cc of milk and 500 cc of water.

To calculate the milk-to-water ratio in the cup after step 1, you can use proportions:

• Milk in the cup / Water in the cup = 150 / 500
• Milk in the cup / Water in the cup = 3/10

So, the milk-to-water ratio in the cup after step 1 is 3:10.

Your calculations are correct, and the final milk-to-water ratio in the glass is 7:9, while in the cup, it is 3:10.

Let the six numbers be a, b, c, d, e, f in ascending order a+b=28

e + f = 56 If we want to maximise the average then we have to minimise a, b, maximise c, d ,e and minimise f

(do remember the numbers are in ascending order)

a+b= 28 = 13+ 15

if=5627 +29

As a, be and ƒ are distinct natural numbers

Therefore c and d will be 25 and 26 respecitively

So average a+b+c+d+e+f = 6 28+25+26+56 6 = 22.5

General term = 38+ (n − 1)17 = 17n + 21 = 17(n + 1) + 4 = 17k +4
Each term is in the form of 17k + 4
Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106
Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990 Average of an A.P.
first term + last term/2 = 106 +990 /2 =548

Both the cars take 1.5 hrs to meet when they travel towards each other.
It is given, speed of slower car is 60 km/hr
Therefore, distance covered by slower car when they meet =60×1.5=90 km

In this scenario, when two cars are traveling towards each other and they meet after 1.5 hours, you can calculate the distance covered by the slower car using the formula:

Distance = Speed × Time

Given:

• Speed of the slower car = 60 km/hr
• Time taken to meet = 1.5 hours

Distance covered by the slower car = 60 km/hr × 1.5 hrs = 90 km

So, the slower car covers a distance of 90 kilometers when they meet.

Since the total number of students, when divided by either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9, 10, 12, 25)k + 4 = 900k + 4.

It is given that the value of 900k + 4 is less than 5000.

Also, it is given that 900k + 4 is divided by 11.

It is only possible when k = 2 and total students = 1804.

So, the number of 12 students group = 1800 12 = 150

5.5% of 8000 + 5.6% of 5000 + x% of 7000=5% of 20000,

x=4,

4% of 20000=800

Let the unit of work done by 1 man in 1 hour and 1 day be 1 MDH unit (Man Day Hour).
Thus, in 7 hours per day for 10 days, the work done by N people = N x 7 x 10 MDH units. = 70N
Since this is equal to 35% of the total work,
35% of the total work = 70 N
Total work = (70N × 100) 35 =200N
The work left = 200 N – 70 N = 130 NMDH units.
Now, 10 people left the job. So, the number of people left = (N-10) Since (N-10) people completed the rest of work in 14 days by working 10 hours a day,
(N-10) × 14 × 10 = 130N
10N = 1400
N = 140

Let the number of students in section A and B be n and n + 10

32n+60(n+10)/n+n+10

= 92n +600 /2n+10

= 46n+300 /n+5

= 46+(70/n+5)

i=(an integer)

n can take values 2, 5, 9, 30, 65

Difference= 65-2= 63