Initial salary = 100
After two successive increments, final salary = 187.5
First increment = x%
Second increment =2x%

Going from the options, x=25%

The question is "After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was"

Hence, the answer is '25'

Single digit numbers where none of the digits is repeated = 1 to 9
⇒ n(single digit numbers)= 9
Double digit numbers = AB (where A is any number from 1 to 9 and B is any number except A)
⇒ n(double digit numbers) = 9 × 9 = 81
Triple digit numbers = ABC (where A is any number from 1 to 4, B is anything except A and C is anything except A, B)
⇒ n(triple digit numbers) = 4 × 9 × 8 = 288
Total number of numbers = 288 + 81 + 9 = 378

Total land area =20+30=50 hectares
In the total land area
Wheat part =1120×50=552
Mustard part =920×50=452
In Vimal's land area
Mustard
Wheat part =58×30=754
Mustard part =38×30=454
So, the land area for Rajesh
Wheat part =552−754=354
Mustard part =452−454=454
Ratio, Wheat : Mustard =354:454=7:9

A necessary condition for the system to have no solution for (𝑥,𝑦), is
The condition is a1a2=b1b2≠c1c2
i.e., p3=−4k≠2a
For p3=−4k;pk+12=0
For −4k≠2a;−4a≠2k⇒2a+k≠0
For p3≠2a;pa−6≠0

The question is "For some constant real numbers p,k and a, consider the following system of linear equations in x and y :

A necessary condition for the system to have no solution for (x,y), is "

Hence, the answer is '2a+k≠0'

Put x=1/x,

Solving (1) & (2),
(1) ⇒

Subtracting, we get

So,

Now,

Sum of the roots =−b/a
Answer =−3

The question is "For any non-zero real number x, let f(x)+2f(1x)=3x. Then, the sum of all possible values of x for which f(x)=3, is"

Hence, the answer is '-3'

We are thinking about two scenarios where x≥y and x≥−y. When x≥y,

So, the possibilities are ( 1,1 ), ( 1,0 ) and ( 1,−1 )
When x=0,
The possibilities are (0,1) and (0,−1)
When x=−1
The possibilities are ( −1,1 ), ( −1,0 ) and ( −1,−1 )
So, the number of distinct integer solutions =8

The question is "The number of distinct integer solutions (x,y) of the equation |x+y|+|x−y|=2, is"

Hence, the answer is '8'

Question:
A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is ?

Solution:

Interior angle of a regular octagon = (n−2)/n × 180°
= (8−2)/8 × 180° = 135°

Area of square ACEG = a²

In △BAC, using cosine rule:
a² = b² + c² − 2bc·cosC
a² = 6² + 6² − 2(6)(6)cos135°
a² = 72 − (72 × −√2/2)
a² = 72 + 36√2
a² = 36(2 + √2)

Therefore, Area = 36(2 + √2)

Answer: 36(2 + √2)  

Initial quantity of water in the container =x
So, initial quantity of milk in the container =300−x
Part of mixture taken out =2x300
Part of mixture that is retained =1−2x300
So, part of milk that is retained =(300−x)(1−2x300)
Also, quantity of milk that is retained after the iteration =72% of 300=216
Then,

The total content itself 300 litre. So, x=420 is inadmissible.
Thus, amount of water that was initially poured = 30 litre

Let, CP=x
Then, MP = 1.2 x ;

Also,
Savings = Rs. 15
Savings with respect to percentage =10% of MP=0.12x
Thus, 0.12x=15

Hence,

The question is "Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is"

Hence, the answer is '10'

Choice B is the correct answer.

Question:
If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8 and 8^f = 9, then the value of the product abcdef is ?

Solution:

From 3^a = 4  ⇒  a = log_3(4) = ln4 / ln3  
From 4^b = 5  ⇒  b = log_4(5) = ln5 / ln4  
From 5^c = 6  ⇒  c = log_5(6) = ln6 / ln5  
From 6^d = 7  ⇒  d = log_6(7) = ln7 / ln6  
From 7^e = 8  ⇒  e = log_7(8) = ln8 / ln7  
From 8^f = 9  ⇒  f = log_8(9) = ln9 / ln8

Multiply:
abcdef = (ln4/ln3)·(ln5/ln4)·(ln6/ln5)·(ln7/ln6)·(ln8/ln7)·(ln9/ln8)

All intermediate terms cancel telescopically, leaving:
abcdef = ln9 / ln3

But ln9 = ln(3^2) = 2 ln3, so
abcdef = (2 ln3) / ln3 = 2

Answer: 2

Distance = XT
From the table, we have:

Also,

Solving (1) & (2),

Adding the equations,

Distance =XT=720km

The question is "A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is "

Hence, the answer is '720'

Solution:

We are given the equation:
10^x + 4/(10^x) = 17/2

Step 1: Let 10^x = y
So, the equation becomes:
y + 4/y = 17/2

Multiply through by y:
2y^2 - 17y + 8 = 0

Step 2: Solve quadratic equation.
2y^2 - 17y + 8 = 0
Factorizing:
(2y - 1)(y - 8) = 0

So, y = 8 or y = 1/2

Step 3: Back-substitute.
10^x = 8  →  x = log10(8) = log10(2^3) = 3 log10(2)
10^x = 1/2  →  x = log10(1/2) = -log10(2)

Step 4: Sum of all possible values.
= 3 log10(2) + (-log10(2))
= 2 log10(2)

Final Answer:
2 log10(2)

Solving this, we get:

Now,

The question is "Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is"

Hence, the answer is '9'

Solution:

We are given:
t1 = 1, t2 = -1, and tn = ((n-3)/(n-1)) * t(n-2) for n ≥ 3.

Step 1: Compute first few terms.
t2 = -1
t4 = (4-3)/(4-1) × t2 = (1/3) × (-1) = -1/3
t6 = (6-3)/(6-1) × t4 = (3/5) × (-1/3) = -1/5
t8 = (8-3)/(8-1) × t6 = (5/7) × (-1/5) = -1/7

So, we see the pattern:
t2 = -1,  t4 = -1/3,  t6 = -1/5,  t8 = -1/7, ...

Step 2: General pattern.
t(2k) = -1/(2k-1)

Step 3: Inverse values.
1/t2 = -1
1/t4 = -3
1/t6 = -5
1/t8 = -7
...
1/t2024 = -2023

Step 4: Sum of series.
1/t2 + 1/t4 + 1/t6 + ... + 1/t2024
= -(1 + 3 + 5 + 7 + ... + 2023)

The series inside is the sum of first 1012 odd numbers.

Sum of first n odd numbers = n²
So, sum = (1012)² = 1024144

Final Answer:
= -1024144

Solution:

Given:
OA = OB = radius (r)
∠AOB = 120°
⇒ ∠OAB = ∠OBA = 30°

Consider OM ⟂ AB.

In right △OMB:
cos30° = MB / OB = (√3)/2 = (5√3)/OB
⇒ OB = 10

Also,
sin30° = OM / OB = 1/2 = OM / 10
⇒ OM = 5

Now, area of smaller region =
Area of sector − Area of △AOB

= (120/360) × π × (10)^2 − (1/2) × (10√3) × 5

= (1/3) × 100π − 25√3

= 25(4π/3 − √3)

Answer:
25(4π/3 − √3)

We are given that Sam completes a piece of work in 20 days. We are also given that Mohit is twice as fast, so
he should take only 10 days. Mohit is thrice as fast as Ayna, so he would take 30 days.
Let's take the total work to be 60 units; this would give the work done per day for Mohit, Sam, and Ayna to be 6,
3 and 2, respectively.
On the first day Sam and Mohit work: doing 9 units
On the second day Sam and Ayan work: doing 5 units
On the third day, Mohit and Ayan work: doing 8 units
Essentially doing 22 units in a 3 days cycle.
After two such cycles, there will be 60-44 = 16 units of work left
On day 7, Sam and Mohit would work 9 units, leaving 7 units
On day 8, Sam and Ayan would work 5 units, leaving 2 units
And on day 9, Ayan and Mohit would complete the remaining work.
So Sam worked for a total of 2+2+2= 6 days and on each day he did 3 units of work, completing 18 units of
work.
The ratio of work done by Sam would be 3/10
Therefore, Option B is the correct answer.

Question:
If 10^68 is divided by 13, what is the remainder?

Solution:

We calculate remainders of powers of 10 modulo 13:

10^1 mod 13 = 10  
10^2 mod 13 = 9  
10^3 mod 13 = 12  
10^4 mod 13 = 3  
10^5 mod 13 = 4  
10^6 mod 13 = 1  

So, after every power of 6, the remainders repeat in a cycle: {10, 9, 12, 3, 4, 1}.

Now, 68 ÷ 6 = 11 remainder 2.  
Thus, 10^68 mod 13 corresponds to the 2nd element in the cycle, which is 9.

Answer: 9

Let the numbers be a (smallest), b (middle), c (largest).
Given: (a + b + c)/3 = 28  ⇒  a + b + c = 84.         ...(1)

After changes:
 smallest → a + 7
 middle → b
 largest → c − 10

New mean:
[(a + 7) + b + (c − 10)] / 3 = (a + b + c − 3)/3.
But a + b + c = 84, so new mean = (84 − 3)/3 = 81/3 = 27.

Given: new mean = b + 2 ⇒ 27 = b + 2 ⇒ b = 25.       ...(2)

Given difference after changes:
(c − 10) − (a + 7) = 64
⇒ c − a − 17 = 64
⇒ c − a = 81.                                         ...(3)

From (1) and (2): a + c = 84 − b = 84 − 25 = 59.      ...(4)

Solve (3) and (4):
Add: (c − a) + (a + c) = 81 + 59 ⇒ 2c = 140 ⇒ c = 70.

Answer:
The largest number originally is 70.

Question:
If (a + b√3)^2 = 52 + 30√3, where a and b are natural numbers, find a + b.

Solution:

Expand the left side:
(a + b√3)^2 = a^2 + 2ab√3 + 3b^2

Equate real and irrational parts with the right side 52 + 30√3:

Real part:    a^2 + 3b^2 = 52
Irrational:   2ab = 30  ⇒  ab = 15

Find natural-number factor pairs of 15:
(ab = 15) ⇒ (a,b) ∈ {(1,15), (3,5), (5,3), (15,1)}

Check each pair against a^2 + 3b^2 = 52:

(1,15): 1^2 + 3·15^2 = 1 + 675 = 676 ≠ 52  
(3,5):  3^2 + 3·5^2  = 9 + 75  = 84  ≠ 52  
(5,3):  5^2 + 3·3^2  = 25 + 27 = 52  ✓  
(15,1): 15^2 + 3·1^2 = 225 + 3 = 228 ≠ 52

Only (a,b) = (5,3) satisfies both equations.

Therefore:
a + b = 5 + 3 = 8

Answer: 8

Here, with respect to boys
Swimming + Running = 50% + 70% = 120%
The extra 20% is meant for the boys who opted both swimming and running.
Similarly, with respect to girls
Swimming + Running = 80% + 60% = 140%
The extra 40% is meant for the boys who opted both swimming and running.
In pictorial representation:

So, percentage of people who opted for both swimming and running = 20% boys + 40% girls

The question is "In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are"

Hence, the answer is '72 and 80, respectively'

Side note: We cannot consider the temperatures outside and inside at 23:00 and 2:00 since the time limit does
not include those.)
The maximum difference between the elements of the outside and inside is maximum at 0:00 when the outside
temperature is 36, and the inside temperature is 26.
Giving the maximum difference as 10
Therefore,10 is the correct answer.

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What best can be concluded about the number of times the AC must have either been turned on or the AC temperature setting been altered between 11:01 pm and 1:59 am?"

Solution: We know that the AC got turned ON after 11:01 PM at 12:30 AM and 1:30 AM.

Hence, the answer is 'Exactly 3'

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What was the temperature outside, in degree Celsius, at 9 pm?"

Solution: From the table we can infer that the outside temperature drops 1 degree Celsius for every 30 minutes. Temperature at 23:00 is 38 degrees. Hence at 9 PM it will be 38+1+1+1+1= 42 degrees Celsius.

Hence, the answer is '42'

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What was the temperature outside, in degree Celsius, at 1 am?"

Solution: From the table we can infer that the temperature outside, in degree Celsius, at 1 am is 34.

Hence, the answer is '34'

From this alone, we can answer the first question:
The AC was turned off for only two instances between 11:01 pm and 1:59 am:
Once at 0:00 and once at 1:00
Therefore, Option B is the correct answer.

We are asked to find the number of countries where the GDP per capita is lower in 2027 than it was in 2024.
For the GDP per capita to be lower in the consequent years, the population growth rate has to exceed the GDP
growth rate, since GDP per capita is nothing but GDP divided by the population. That means if the population
growth rate is lesser than that of the GDP growth rate, the GDP per capita will only increase.
We can clearly see that none of the following countries fall in the scenario where GDP growth rate is lesser than
that of the population growth rate. Countries 1, 2 and 7 have actually decreasing population rates thereby
definitely increasing the GDP per capita.
The rest of the countries have GDP growth rates larger than the population growth rates.
Hence, we can conclude that, none of the countries will have a smaller GDP per capita in 2027 when compared
to 2024.

Country GDP GDP per capita GDP growth rate Population growth rate Country 1 0.15x 0.41y 0.20% -0.12% Country 2 0.14x 0.25y 0.90% -0.41% Country 3 0.13x 0.02y 6.50% 0.70% Country 4 0.12x 0.38y 0.50% 0.49% Country 5 0.10x 0.36y 0.70% 0.31% Country 6 0.08x 0.08y 3.20% 0.61% Country 7 0.08x 0.30y 0.70% -0.11% Country 8 0.07x 0.41y 1.20% 0.71% Country 9 x Country 10 y

The question is "Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?"

Answer: Population of country 1 in 2027 = * (1 - 0.0012)³ = 0.364 (highest)

Population of country 4 in 2017 = * (1 + 0.0049)³ = 0.320

Population of country 5 in 2017 = * (1 + 0.0031)³ = 0.280

Population of country 7 in 2017 = (1 - 0.0011)³ = 0.261

Hence, the answer is 'Country 1'

The question is "Which one among the countries 1 through 8, has the smallest population in 2024?"

Answer: GDP per capita = GDP / Population. Therefore, Population of a country = GDP / GDP per capita.

• Population in country 3 = 6.5 x/y
• Population in country 5 = 0.277 x/y
• Population in country 8 = 0.17 x/y (lowest)
• Population in country 7 = 0.266 x/y

Hence, the answer is 'Country 8'

Country GDP GDP per capita GDP growth rate Population growth rate Country 1 0.15x 0.41y 0.20% -0.12% Country 2 0.14x 0.25y 0.90% -0.41% Country 3 0.13x 0.02y 6.50% 0.70% Country 4 0.12x 0.38y 0.50% 0.49% Country 5 0.10x 0.36y 0.70% 0.31% Country 6 0.08x 0.08y 3.20% 0.61% Country 7 0.08x 0.30y 0.70% -0.11% Country 8 0.07x 0.41y 1.20% 0.71% Country 9 x Country 10 y

The question is "The ratio of Country 4's GDP to Country 5's GDP in 2026 will be closest to"

Answer: The GDP growth rates, and population growth rates of the countries will remain constant for the next three years.

Country 4’s GDP in 2026 = 0.12x * (1.005) * (1.005) = 0.1212x

Country 5’s GDP in 2026 = 0.10x * (1.007) * (1.007) = 0.1014x

The ratio of Country 4's GDP to Country 5's GDP in 2026 = 1.195.

Hence, the answer is '1.195'

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of
55x, 33x used one app.
Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x
- 15000)
Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000
In clue 2, we were given that 33000 kids and elders out of 55000 kids and elders use one app, showing that
22000 use multiple apps. To minimize the total usage of multiple apps, we can assume that all 55000 other
subscribers use one app.
Giving the percentage of multiple app users as: 20
Therefore, Option B is the correct answer.

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of
55x, 33x used one app.
Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x
- 15000)
Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000
In 2023, there were 20,000 elders; in 2024, there were 33,000 elders.
The increase in percentage would be 65
Therefore, Option C is the correct answer.

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of
55x, 33x used one app.
 VIDEO SOLUTION
 VIDEO SOLUTION
Downloaded from Cracku.in For MBA/CAT Courses:  6303239042 68/84
Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x
- 15000)
Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000
There are 15x, that is, 15,000 kids in 2023, of which we are given that 10,000 use one app.
So, 5,000 use multiple apps.
The percentage of kids using multiple apps in 2023 would hence be 33.33 percent
Therefore, Option A is the correct answer.

Reading the bar graphs, we can find the distribution of the subscribers as:
Taking clue one into consideration, let's take the total number of subscribers in 2023 as 100x and in 2204 as
110x
This would give the split of subscribers as follows:
34.C
Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of
55x, 33x used one app.
Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x
- 15000)
Equating these two we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000
The first question asks about the number of others in 2024, which is 55x, or simply 55,000
Therefore, Option C is the correct answer.

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater
than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we
right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats
are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24
grams of others in M3.

The proteins in the foodgrains, when arranged in ascending order, are: 8, 8, 10, 12, 12,14, 16
Giving the median as 12.

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater
than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.

Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we
right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats
are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24
grams of others in M3

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater
than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we
right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0

The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats
are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24
grams of others in M3.
We can see that there were 12 grams of protein in M2.

The set's starting point is that the sum of each row must be 100
Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that
the missing elements of the other three columns are multiples of 4.
Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo
cereal.
We have the reference point of P1 at 66
Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95.
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a
multiple of 4.
Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated.
Leaving only 80 grams of carb in C1 and 8 grams of protein.

Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are
eliminated.
Taking 85 carbs would give 2 grams of protein, which can be eliminated.
80 and 70 would also not work for the same reason.
Leavin has only 75 grams of carbs, leaving 12 grams of protein.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity
being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.

Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and
RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
ATMs that can be uniquely determines are the ATMs with cash 9L, 11L and 12L. Hence the answer is 3.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

Question:
The sum of the infinite series
(1/5)*( (1/5) − (1/7) )
+ (1/5)^2*( (1/5)^2 − (1/7)^2 )
+ (1/5)^3*( (1/5)^3 − (1/7)^3 )
+ ...... 
is equal to ?

Solution:

Let a = 1/5 and b = 1/7. The series is
S = Σ_{n=1 to ∞} a^n( a^n − b^n )
  = Σ_{n=1 to ∞} ( a^{2n} − (ab)^n )

Separate the sums (both geometric, |a|<1, |ab|<1):
S = Σ_{n=1 to ∞} a^{2n}  −  Σ_{n=1 to ∞} (ab)^n
  = (a^2)/(1 − a^2)  −  (ab)/(1 − ab)

Substitute a = 1/5, b = 1/7:
a^2 = 1/25,   ab = 1/35

S = (1/25)/(1 − 1/25)  −  (1/35)/(1 − 1/35)
  = (1/25)/(24/25)  −  (1/35)/(34/35)
  = 1/24  −  1/34
  = (34 − 24) / (24·34)
  = 10 / 816
  = 5 / 408

Answer:
S = 5/408