CAT Linear Equations Questions

Arranging the equation, we know that there are no solutions when the lines are parallel:
for that the condition had to $$\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$$

Checking through options: 
Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

Therefore, Option B is the correct answer. 

It is given that for some real numbers a and b, the system of equations $$x + y = 4$$ and $$(a+5)x+(b^2-15)y=8b$$ has infinitely many solutions for x and y.

Hence, we can say that 

=> $$\ \frac{\ a+5}{1}=\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}$$

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=> $$\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}\ =>\ b^2-2b-15=0$$

=> $$\left(b-5\right)\left(b+3\right)=0$$

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is $$a+5=b^2-15\ =>\ a\ =b^2-20$$

When $$b=5,a=5^2-20=5$$

When $$b=-3,a=3^2-20=-11$$

The maximum value of $$ab=(-3)\cdot(-11)=33$$

The correct option is A

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $$\dfrac{\left(4a+960\right)}{3}$$

=> $$\dfrac{4a}{3}+320$$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340

Let $$\frac{x}{y}\ be\ t$$

Therefore, $$c=16t\ +\ \frac{49}{t}$$

Applying AM>= GM

$$\frac{\left(16t\ +\ \frac{49}{t}\right)}{2}\ge\ \left(16t\times\frac{49}{t}\right)^{\frac{1}{2}}$$

$$16t\ +\ \frac{49}{t}\ge56$$

When t is positive then c is greater than equal to 56.

When t is negative then c is less than equal to -56.

Therefore $$c\ \in\ \left(-\infty,\ -56\right]\ ∪\ \left[56,\infty\ \right]$$

As -50 is not in the range of c so it is the answer

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 ...... (1)

3x - y + z = 97 ...... (2)

(2)-(1) -> x - y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 - z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.

a(a + b + 1) = 14 ...... (1)

b(a + b + 1) = 28 ...... (2)

$$\frac{a}{b}=\frac{1}{2}$$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$$3a^2+a-14=0$$

$$3a^2-6a+7a-14=0$$

$$3a\left(a-2\right)+7\left(a-2\right)=0$$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = 2(2) + 4 = 8

The answer is option A.

In the question, it is given that the equation $$\mid x + a \mid + \mid x - 1 \mid = 2$$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x - 1 $$\ge\ $$ 0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a $$\ge\ $$ 0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

The answer is option C.

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.

We need to find the maximum value of z.

x + y + z = 100 ...... (1)

100x + 250y + 500z = 15250

2x + 5y + 10z = 305 ...... (2)

2x + 2y + 2z = 200 ....... (1)

(2) - (1), we get

3y + 8z = 105

At z = 12, x = 3

Therefore, maximum value z can take is 12.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Let the number of large shirts be l and the number of small shirts be s.

Let the price of a small shirt be x and that of a large shirt be x + 50.

Now, s + l = 64

l (x+50) = 5000

sx = 1800

Adding them, we get,

lx + sx + 50l = 6800

64x +50l = 6800

Substituting l = (6800 - 64x) / 50, in the original equation, we get

$$\frac{\left(6800-64x\right)}{50}\left(x+50\right)=5000$$

(6800 - 64x)(x + 50) = 250000

$$6800x+340000-64x^2-3200x = 250000$$

$$64x^2-3600x-90000=0$$

Solving, we get, x= $$\frac{225\pm\ 375}{8}=\frac{600}{8}or-\frac{150}{8}$$

SO, x = 75

x + 50 = 125

Answer = 75 + 125 = 200.

Alternate approach: By options.

Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50

Hence, small + small + 50 = option.

SMALL = (Option - 50)/2

LARGE = Small + 50 = (Option + 50)/2

Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later. 

Lets start with Option B.

Large = 150 + 50 / 2 = 100

Small = 150 - 50 / 2 = 50

Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X - This is wrong)

Option C - 

Large = 200 + 50 / 2 = 125

Small = 200 - 50 / 2 = 75

Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64 ( This is the right answer)

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test x >= 0, y < 0,

3x - y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 - 2 = 0.

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.'. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Let tom's age = x

=> Dick=3x

=>harry = 6x

Given, 

3x+1 = (x+3x+6x)/3 

=> x= 3

Hence, Harry's age = 18 years

Two linear equations ax+by= c and dx+ ey = f have a unique solution if $$\frac{a}{d}\ne\ \frac{b}{e}$$

Therefore, $$\frac{k}{4}\ne\ \frac{1}{k}$$ => $$k^2\ne\ 4$$

=> k $$\ne\ $$ |2|

Let John buy "m" kg of rice and "p" kg of wheat.

Now let the price of rice be "r" in April. Price in May will be "1.2(r)"

Now let the price of wheat be "w" in April . Price in April will be "1.12(w)".

Now he spent ₹150 more in May , so 0.2(rm)+0.12(wp)=150

Its also given that he had spent ₹450 on rice in April. So (rm)=450 

So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60/0.12 or (wp)=500

Amount spent on wheat in May will be 1.12(500)=₹560

We can write x=z-9 and y=z-1 Now we have x+y< z+5

Substituting we get z-9+z-1<z+5 or z<15

Hence the maximum possible value of z is 14

Maximum value of "x" is 5 and maximum value of "y" is 13

Now 2x+y = 10+13=23

Let the number of pencils bought by Aron be "p" and the cost of each pencil be "a".

Let the number of sharpeners bought Aron be "s" and the cost of each sharpener be "b".

Now amount spent by Aron will be (pa)+(sb)

Aditya bought (2p) pencils and (s-10) sharpeners. Amount spent will be (2pa)+(s-10)b

Amount spent in both the cases is same

pa + sb = 2pa + (s-10)b or pa=10b

Now its given in the question that cost of sharpener is 2 more than pencil i.e. b=a+2

pa= 10a+20 or a=20/(p-10)

Now the number of pencils has to be minimum, for that we have to find smallest "p" such that both "p" and "a" are integers. The smallest such value is p=11 . Total number of pencils bought will be p+2p=11+22=33

Given

A+(B+C)/2=5 => 2A+B+C=10....(i)

(A+C)/2 +B=7 => A+2B+C=14.....(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.

$$5^x - 3^y = 13438$$ and $$5^{x - 1} + 3^{y + 1} = 9686$$

$$5^{x} + 3^{y}*15 = 9686*5$$

$$5^{x} + 3^{y}*15 = 48430$$

16*$$3^y$$=34992

$$3^y$$ = 2187

y = 7

$$5^x$$=13438+2187=15625

x=6

x+y = 13

Let the number of fiction and non-fiction books in 2010 = 100a, 100b respectively

It is given that the total number of books in 2010 = 11500

100a+100b = 11500            -------Eq 1

The number of fiction and non-fiction books in 2015 = 110a, 112b respectively

110a+112b = 12760           -------Eq 2

On solving both the equations we get, b=55, a= 60

The number of fiction books in 2015 = 110*60=6600

$$\left(ax+by\right)^2=65^2$$

$$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$$
$$k = ay - bx$$

$$k^2\ =\ a^2y^2+b^2x^2-2abxy$$

$$(a^2 + b^2)(x^2 + y^2 )= 25* 169$$

$$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$$

$$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$$

k = 0

D is the correct answer.

x - y - z = 25 and $$x\leq40,y\leq12$$, $$z\leq12$$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

$$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$$
=> $$ab = 9(a + b)$$
=> $$ab - 9(a+b) = 0$$
=> $$ ab - 9(a+b) + 81 = 81$$
=> $$(a - 9)(b - 9) = 81, a > b$$
Hence we have the following cases,
$$ a - 9 = 81, b - 9 = 1$$ => $$(a,b) = (90,10)$$
$$ a - 9 = 27, b - 9 = 3$$ => $$(a,b) = (36,12)$$
$$ a - 9 = 9, b - 9 = 9$$ => $$(a,b) = (18,18)$$
Hence there are three possible positive integral values of (a,b)

If two 50 Misos are used, the 107 can be paid in only 1 way.

If one 50 Miso is used, the number of ways of paying 107 is 6 - zero 10 Miso, one 10 Miso and so on till five 10 Misos.

If no 50 Miso is used, the number of ways of paying 107 is 11 - zero 10 Miso, one 10 Miso and so on till ten 10 Misos.

So, the total number of ways is 18

Let the value of cheque be x Rs and y ps and the amount she received is y Rs and x ps.

After 50 ps is deducted she has the amount which is 3 times the amount on cheque,

So 100y+x-50=3(100x+y)    (After converting the amount in paise)

y= (299x+50)/97 = 3x+ (8x+50)/97

Since x is an integer, 3x will definitely be an integer. Now, (8x+50)/97 has to be an integer for y to be an integer.

So, 8x+50 has to be a multiple of 97. 

8x+50= 97

$$\Rightarrow$$ x = 47/8, this value of x is not an integer. 

8x+50= 97*2 = 194

8x = 144

$$\Rightarrow$$ x = 144/8 = 18. Here, we got an integer.

So, the amount has to be 18 rupees and some paise. Hence, option D is correct.

Price of Darjeeling tea on 100th day= 100+(0.1*100)=110
Price of Ooty tea on nth day= 89+0.15n
Let us assume that the price of both varieties of tea would become equal on nth day where n<=100
So
89+0.15n=100+0.1n
n=220 which does not satisfy the condition of n<=100
So the price of two varieties would become equal after 100th day.
89+0.15n=110
n=140
140th day of 2007 is May 20 (Jan=31,Feb=28,March=31,April=30,May=20)

Let the number be xy
10y + x = 10x + y + 18
=> 9y - 9x = 18
=> y - x = 2
So, y can take values from 9 to 4 (since 3 is already counted in 13)
Number of possible values = 6

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
So the charge for the excess of (20-$$\frac{2x}{3}$$) = k(20-$$\frac{2x}{3}$$) = 1200  ....(1)
Also, the charge for the excess of 60-x = k(60-x) = 5400 .....(2)
Dividing (1) by (2), we get
=>$$\frac{\left(60-2x\right)}{3\times\ \left(60-x\right)}=\frac{1200}{5400}$$ 
Solving this, x = 15 kg
So, Praja's luggage = 35 kg

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
=> (60-2x)/3*(60-x) = 1200/5400
Solving this, x = 15 kg

Let x be no. of male and y be no. of female operators.
We have 40x+50y=1000 .
So x = 25-(5*y/4) also 7<=y<=12.
So y can be 8 or 12.
If y=8 then x=15 and y=12 then x=10 .
Then we have to find total cost incurred in both the cases.
We find that cost is minimum in 2nd case when no. of males are 10.

Let C and R be no. of columns and rows respectively.

The number of red coloured tiles would be given by (R-2)(C-2). This is because two outer rows made of white tiles and the two outer columns made up of outer columns are removed.

Similarly the number of white tiles would be given by R*2 + (C-2)*2. Two tiles are removed from columns because the corner tiles would have already been included while considering the rows.

So according to given condition we have (C-2)*2 + 2*R = (C-2)(R-2).
Now start putting value of c from options into this equation. Only for one option B we get an integer value of R .

Machine A takes 15 min to produce 1000 nuts with clean time. machine b takes 30 min to make 1500 nuts with clean time . So B is slower. So with B 900 nuts will be made in 180 mins but at last round cleaning time of 10 min no need to count hence 170 mins

xy = x + y 

=> xy - x - y = 0

=> xy - x - y + 1 = 1

=> (x - 1) (y - 1) = 1

=> Both x - 1 and y - 1 have to be equal to 1 or -1.

So, values taken by (x,y) are (2,2) and (0,0).

=> 2 solutions

Group B contains 23 questions => Marks of group B = 46
Let the number of questions in A be x and in C be 77-x.
Marks of group A = x
So, x/(x+46+3*77-3x) >= 60%
=> 5x >= 3(277-2x)
=> 11x >= 831
=> x >= 75.54
=> x = 76 (min)
So, the possible number of questions in group C = 1.

Let the number of questions in group B be x.
So, number of questions in group A = 92-x
Marks of group B = 2x
2x/(92-x+2x+24) >= 20%
=> 10x >= 116+x
=> 9x >= 116
=> x >= 12.88
From the options, x can be 13 or 14

Substitute value of p,q,r in the options only option A satisfies .

5(x+2y-3z)-2(2x+6y-11z)-(x-2y+7z) = 5x+10y-15z-4x-12y+22z-x+2y-7z  = 0

Let the number of questions answered correctly be x and the number of questions answered wrongly be y.

So, number of questions left unattempted = (50-x-y)

So, x - y/3 - (50-x-y)/6 = 32

=> 6x - 2y - 50 + x + y = 192 => 7x - y = 242 => y = 7x - 242

If x = 35, y = 3

If x = 36, y = 10

So, min. value of y is 3.

The number of wrongly answered questions cannot be less than 3.

Total to be paid = 1248 Baht

Each has to pay 1248/3 = 416 Baht

R paid 92 Baht

M paid 184+27 = 211 Baht

So, R owes S 416 - 92 = 324 Baht

Total to be paid = 1248 Baht

Each has to pay 1248/3 = 416 Baht

R paid 92 Baht

M paid 184+27 = 211 Baht

So, R owes S 416 - 92 = 324 Baht

M owes S 416-211 Baht = 205 Baht = 5 US Dollars

Suppose the thief stole 'x' diamonds. \

After giving the share to the first watchman, the thief has (x/2)-2 diamonds.

After giving to the second watchman, the thief has (x/4)-3 diamonds.

After giving to the third watchman, the thief has (x/8)-(7/2) diamonds.

This is equal to 1. So, (x/8) - 7/2 = 1

Solving this equation, we get x = 36

Let the amount paid by Mayank be x. So, amount paid by the other three = 2x

=> Total bill = x+2x = 3x = 60 => x = 20. So, Mayank paid 20

Similarly, amount paid by Mirza + 3*Amount paid by Mirza = 60

=> Amount paid by Mirza = 15

Amount paid by Little + 4*Amount paid by Little = 60

=> Amount paid by Little = 12

So, amount paid by Jaspal = 60 - (20+15+12) = 60 - 47 = $13

Suppose Akil drove the car for less than 5 hrs. In this case, by distance basis, Rs 360 should be charged. This is not the case.

So he dove for more than 5 hrs. Cost comes more using time basis; which is Rs 300, i.e. he used the car for 6 hours.

Let the longest piece be x
Shortest piece = x - 23
Middle-sized piece = x/3
So, x + x - 23 + x/3 = 40 => 7x/3 = 63 => x = 27
Shortest piece = 27 - 23 = 4

Suppose A, B and C have 5 pieces of bread, 3 pieces of bread and 8 coins respectively. Since in total there are 8 pieces of bread, each one should get around 2.66 bread. So A must give 2.33 part of his bread to C and B must give 0.33. Distributing the amount in the same ratio of bread contribution, A must get 7 coins and B must get 1 coin.

Let the total number of mints in the bowl be n
Sita took n/3 - 4. Remaining = 2n/3 + 4
Fatim took 1/4(2n/3 + 4) - 3. Remaining = 3/4(2n/3 + 4) + 3
Eswari took 1/2(3/4(2n/3+4)+3) - 2
Remaining = 1/2(3/4(2n/3+4)+3) + 2 = 17
=> 3/4(2n/3+4)+3 = 30 => (2n/3+4) = 36 => n = 48
So, the answer is option d)

Let the number of males in Mota Hazri = x
No. of males in Chota Hazri = x - 4522
Let the number of females in Mota Hazri = y
No. of females in Chota Hazri = y - 2910
(y - 2910) = 2(x - 4522) => y = 2x - 9044 + 2910 = 2x - 6134
Also y = x + 4020
So, x + 4020 = 2x - 6134 => x = 10154
So, number of males in Chota Hazri = 10154 - 4522 = 5632

Let the price of 1 burger be x and the price of 1 shake be y and the prize of 1 french fries be z
3x + 7y + z = 120
4x + 10y + z = 164.5
=> x + 3y = 44.5
=> x = 44.5 - 3y
=> 3(44.5 - 3y) + 7y + z = 120 => z = 120 - 133.5 + 2y
So, x+y+z = 44.5 - 3y + y -13.5 + 2y = 31
So, the cost of a meal consisting of 1 burger, 1 shake and 1 french fries = Rs 31

Let the number of coins of the three denominations be x, y and z respectively.
x+y+z = 300  
x+2y+5z = 960
2x+y+5z = 920
=> 3(x+y) + 10z = 1880
=> 3(300 - z) + 10z = 1880
=> 900 + 7z = 1880 => z = 980/7 = 140
So, the number of 5 rupee coins is 140

Let the current capacity of conical flask be C. So, cylinder = C+500.

After pumping out 200 liters, C+300 = 2(C-200) => C = 700

So, full capacity of cylinder = 700+500 = 1200

When numbers of chocolates, biscuits and apples are integers.
Now let's say number of chocolates taken 1 , then biscuits will be 2 and apples can be 4,5,6,7
Hence minimum money that should be spent = 1+1+8 = 10 (Hence option C is cancelled)
Now when number of chocolates are 4
Biscuits will be 8
And apples can be 13,14,15....
Now total money spent can be 4+4+26 = 34 and more of it.
Hence answer will be A

Let the cost of pen, pencil and eraer be x,y,z respectively

5x+7y+4z = A

6x+8z+14y = 3A/2

4x + 16/3 z + 28/3 y = A

Comparing two equations

5x+7y+4z = 4x+16/3 z + 28/3 y

x = 7/3 y + 4/3 z

3x = 7y+4z

Now required percentage = $$\frac{5x}{5x+7y+4z}\times100=\frac{5x}{5x+3x}=62.5$$%

Total matches played = 17+3 = 20

Total matches = 20*$$\frac{3}{2}$$ = 30

Number of wins required = 75% of 30 = 22.5 = 23 wins

23-17 = 6 more wins are required out of 10 matches to maintain 75% win recor which means there would be 4 losses.

For points to be coincident, value of determinant should not be equal zero, so that they have a unique value of system.

Here value of determinant is not equal to zero, simultaneously not any two lines are parallel or perpendicular.

So system has a unique value

Hence points are coincident.

Let's say Iqbal has x cards initially and Mushtaq has y number of cards initially.

So first Mushtaq gave t cards to Iqbal, hence (x+t) = 4(y-t)

Now second time, Iqbal gave t cards to Mushtaq, hence x-t = 3(y+t)

Solving above two equations we will get x=31t and y=9t

And we know x+y<52 hence 40t<52 

because t should be a whole number it will be 1 here and x=31 and y=9

let's say integers are x and y
so x+y = 10 => y = 10 - x
and $$\frac{1}{x} + \frac{1}{y} = \frac{5}{12}$$
$$\frac{1}{x} + \frac{1}{10-x} = \frac{5}{12}$$

=> (10 - x + x)*12 = 5*x(10-x)

=> $$120 = 50x - 5x^2$$

=> $$24 = 10x - x^2$$

=> x = 4, 6

=> y = 6 or 4

The bigger of the two numbers is 6.

Let the number of sticks assigned to each boy be N.

Let the number of boxes be M.

So, number of sticks per box = N/M

Now, if he reduces the number of sticks in each box, the equation becomes N/(M+3) = N/M - 25

So, 25 = N/M - N/(M+3)

From the options, if N = 150, then, we get 25 = 150 [ 1/M - 1/(M+3) ]

=> 1/6 = 1/M - 1/(M+3) => M = 3

So, the number of sticks assigned to each boy = 150

1. x=1 ; y=2

2. x=2 ; y=3

3. x=6 ; y=4

4. x=24 ; y=5

Hence when y=5 , x will be 24

According to question Rahul put exact number of 10 rs. notes, hence total price will be a multiple of 10.
And Rahul wants 30 cards, where he took 5 each of two kind and 10 each of other two kind.
So summation of (price of one type card multiplied by number of that type of card) should be a multiple of 10.
By looking at the prices of cards and to make the sum a multiple of 10, we can say that two 5 cards were of rs. 3.5 and 4.5
and two 10 cards were of prices 2 rs.and 5 rs each respectively.
Hence total sum will be 5 $$\times$$ (3.5+4.5) + 10 $$\times$$ (2+5) = 110
So rahul gave 11 notes of 10 rs.