CAT 2001 Question Paper

Lets take $$n_1 = n_2 = 10$$ and $$r_1 = r_2 =100$$ .
Using given formula we have $$BA = 20$$ and $$MBA_1 = MBA_2 = 10$$ .
So we have $$BA$$ highest which is not the case in option A,B and C. Hence the correct option is D. 

With no incomplete innings and $$BA$$ of 50 , lets assume $$r_1 = 50$$ and $$n_1 = 1$$.
Now we have $$BA = MBA_2 = 50$$ 
So we have $$r_2 = 45$$ and $$n_2 = 1$$ .
We have $$BA = 95$$ and $$MBA_2 = \frac{95}{2} < 50$$.
Hence $$MBA_2$$ decreases.
So $$BA$$ will increase and $${MBA}_2$$ will decrease. 

At 60 kmph, time taken to reach the destination = 200/60 = 3.33 hours.
Fuel consumed per hour at 60 kmph = 4 liters.
Therefore, total fuel consumed = 3.33*4 = 13.33 liters.

If she drives at 40 kmph, the fuel consumed = 200/40 * 2.5 = 12.5 liters
If she drives at 60 kmph, the fuel consumed = 13.33 liters
If she drives at 80 kmph, the fuel consumed = 200/80 * 7.9 = 19.75 liters
To minimize fuel consumption, Manasa has to decrease the speed.

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

Let the length of each side of the octagon be x.

So, length of the square will be x+2*(x/√2) = 2

=> x(1+√2) = 2 => x = 2/(1+√2)

Take x=3 , z=2 , y=5.

$$y(x-z)^2 = 5(3-2)^2 = 5$$

Option A gives 5 which is odd.

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

-4x = -24, 5y = -30

So none of the options out of a,b or c satisfies .

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min . 

Hence they flash together after every 2 min. So in an hour they flash together 30 times .

Each box contains at least 120 and at most 144 oranges.

So boxes may contain 25 different numbers of oranges among 120, 121, 122, .... 144.

Lets start counting.

1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.

Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.

Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.

Hence the number of boxes containing the same number of oranges is at least 6.

B is the south gate. A is the house and BC = 9 km.

In triangle ABC and triangle AMO

ABC = OMA = 90 degree

BAC = OAM = Common angle 

Triangle ABC is  similar to triangle AMO.

$$\frac{(2r+3)}{\sqrt{(r+3)^2-r^2}} = \frac{9}{r}$$

r = 9/2

Let the length of AB be 3X and the length of AD (and BC) be Y. 

As the length of AB = 3X, AE = EF = FB =X

So, the area of the rectangle ABCD is length * breadth = 3X * Y = 3XY

The area of triangle CEF = 1/2 * base * height = 1/2 * EF * BC = 1/2*X*Y

So, required ratio = 1/2 : 3 = 1:6

A takes 4 days to complete the work.
So, B takes 8 days to complete the same work.
C takes 16 days to complete the work.
D takes 32 days to complete the same work.

In order to measure, let the total work be of 64 units. Hence, the speed of working of each of the four persons is given below.

A - 16 units/hr
B - 8 units/hr
C - 4 units/hr
D - 2 units/hr

From the given options, we need to find two pairs in such a way that their speeds are in the ratio 3:2. Note that A+D=18 while B+C=12 and the ratio is 3:2

Hence, the first pair is A and D and the second pair is B and C

Let the 4 digit no. be xyzw.
 According to given conditions we have x+y=z+w --- Eq 1 , x+w=z --- Eq 2,y+w=2x+2z --- Eq 3
Eq 2 - Eq 3 : x-y = -2x-z --- Eq 4
Eq 1+ Eq 4 :2x = -2x+w
4x=w  --- Eq 5
Substitute w = 4x inEq2
5x = z
Substitute w = 4x inEq3
y+4x=2x+10x
y = 8x
Now the minimum value x can take is 1 so z =5 and the no. is 1854, which satisfies all the conditions.
 Hence option A .

January 1, 1950 to December 31, 1959 is a period of 10 years or 20 half years.
The person X after 1st year gets Rs. 300 in next year he gets Rs. 330 and so on.
So his earning is in AP with 10 300+330+360+...

Similarly earning of Y is in AP with 20 terms 200+215+230+245.... .

So, the total earnings of X equals 12*(300+330+....10 terms) = 52200
The total earnings of Y equals 6*(200+215+230+...20 terms) = 41100

So, the total earnings of the two equals 52200+41100 = 93300

Let the number be X.
From the given information, 53x - 35x = 540 => 18x = 540 => x = 30

So, new product = 35*30 = 1050

Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000/40 = 300

$$xy^2$$ will have it's least value when y=-7 and x=2 and equals 98.
So $$xy^2>98$$

$$x^2y$$ will have it's least value when y=-8 and x=3 and equals -72.
So, $$x^2y > -72$$

$$5xy$$ will have it's least value when y=-8 and x=3 and equals -120
So, $$5xy > -120$$

So, of the three expressions, the least possible value is that of 5xy
 

$$n^3 - 7n^2 + 11n - 5 = (n-1)(n^2 - 6n +5) = (n-1)(n-1)(n-5)$$
This is positive for n > 5
So, m = 6

When the ladder is moved 2 m away from the wall, the top of the ladder rests against the foot of the wall. So, the hypotenuse is 2 m more than the second side of the right triangle. The other non-hypotenuse side is 8 m. So, $$8^2 + x^2 = (x+2)^2$$

=> x = 15 m and length of the ladder = x+2 = 17 m

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)

30 years. The number of leap years is 8 (1972,1976,1980,1984,1988,1992,1996,2000).

So, the total number of days = 22*365 + 8*366 = 10958

10958 mod 7 = 3

Since 9/12/2001 is a Sunday, 9/12/1971 should be a Thursday.

The product of 44 and 11 in decimal is 484.
If base is x , then 3*x^3+4*x^2+x+4 = 484 .
Hence, the given base system is of number 5.
Now, we have to convert 3111 (in base 5) to decimal number system.

3111 in base 5 equals $$1*5^0 + 1*5^1 + 1*5^2 + 3*5^3 = 1+5+25+375=406$$

$$12/(R - S) = T$$
$$12/(R + S) = T - 6$$
$$12/(2R - S) = t$$
$$12/(2R + S) = t - 1$$
=> $$12/(R - S) - 12/(R + S) = 6$$ and $$12/(2R - S) - 12/(2R + S) = 1$$
=> $$12R + 12S - 12R + 12S = 6R^2 - 6S^2$$ and $$24R + 12S - 24R + 12S = 4R^2 - S^2$$
=> $$24S = 6R^2 - 6S^2 and 24S = 4R^2 - S^2$$
=> $$6R^2 - 6S^2 = 4R^2 - S^2$$
=> $$2R^2 = 5S^2$$
=> $$24S = 10S^2 - S^2 = 9S^2$$
=> $$S = 24/9 = 8/3$$

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 x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $$\frac{83x+76y+85z}{x+y+z}$$

Substitute the relations in above equation we get, 

$$\frac{83x+76y+85z}{x+y+z}$$  = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

Length of the diagonal of the right triangle is 40.

The height of the isosceles triangle formed, with 40 as its base is 15.
So, area = (1/2 * 32 * 24) + (1/2 * 40 * 15) = 384 + 300 = 684 $$m^2$$

If the number of pages is 44, the sum will be 44*45/2 = 22*45 = 990
So, the number 10 was added twice

Let the number of steps on the escalator be x.

So, by the time Shyama covered 25 steps, the escalator moved 'x-25' steps.

Hence, the ratio of speeds of Shyama and escalator = 25:(x-25)

Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20)

But the ratio is 3:2

Ratio of speeds of Shyama and Vyom = 25(x-20)/20*(x-25) = 3/2

=> 10(x-20) = 12(x-25)

=> 2x = 100 => x = 50

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

Since the product is constant, 

(a+b+c+d)/4 >= $$(abcd)^{1/4}$$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$$(a+1)(b+1)(c+1)(d+1)$$ 
= $$1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$$
We know that $$abcd = 1$$
Therefore, $$a = 1/bcd, b = 1/acd, c = 1/bda$$ and $$d = 1/abc$$
Also, $$cd = 1/ab, bd = 1/ac, bc = 1/ad$$
The expression can be clubbed together as $$1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$$
For any positive real number $$x$$, $$x + 1/x \geq 2$$
Therefore, the least value that $$(a+1/a), (b+1/b)....(ad+1/ad)$$ can take is 2.
$$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$$
=> $$(a+1)(b+1)(c+1)(d+1) \geq 16$$
The least value that the given expression can take is 16. Therefore, option C is the right answer.

Let the time taken working together be t.
Time taken by Arnold = t+1
Time taken by Asit = t+6
Time taken by Afzal = 2t
Work done by each person in one day = $$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}$$
Total portion of workdone in one day $$=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{2-1}{2t}$$
$$2t+7=\frac{(t+1)\cdot(t+6)}{2t}$$
$$3t^2-7t+6=0 \longrightarrow\ t=\frac{2}{3} $$or $$t=-3$$
Therefore total time = $$\frac{2}{3}$$hours = 40mins

Alternatively,
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
From the options, if time $$= 40$$ min, that is, $$t = \frac{2}{3}$$
LHS = $$\frac{3}{5} + \frac{3}{20} + \frac{3}{4} = \frac{(12+3+15)}{20} = \frac{30}{20} = \frac{3}{2}$$
RHS = $$\frac{1}{t}=\frac{3}{2}$$
The equation is satisfied only in case of option C
Hence, C is correct

We know that area = 0.5*h*10 ; we get h = 16=oa.

Using pythagoras we find

$$ob^2+oa^2 = ab^2$$

$$ob^2+16^2 = 20^2$$

ob = 12

Using pythagoras in triangle oam we get

am = $$\sqrt {2^2 + 16^2 } $$ = $$\sqrt {260}$$. 

It is given that in a Fibonacci sequence, from the third term on wards, each term in the sequence is the sum of the previous two terms in that sequence.

Let x and y be the 1st and 2nd term respectively.

3rd term = x+y

4th term = x+2y

5th term = 2x+3y

6th term = 3x+5y

7th term = 5x+8y

We know that difference of the squares of 6th and 7th terms is 517 = 47*11 .

And $$a^2-b^2=(a+b)(a-b)$$.

Applying above formula we get (8x+13y)(2x+3y) = 47*11.

So only possible way is (8x+13y)=47 and

2x+3y=11 .

Solving we get x=1 and y=3 .

Using the concept that every term is the sum of the previous two terms, as used in the beginning of the solution, we get 10th term as 21x+34y, which gives 10th term as 123.

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg. 

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

Distance between A-B , A-C, C-B is 180, 120 and 60 km respectively.

Let x be the distance from A where the 2 trains meet.

According to given condition we have

$$\frac{x}{70}=\frac{60}{50} + \frac{1}{4} + \frac{120-x}{50}$$.

Solving the equation we get x around 112 km.

Since the number starts from 1 if there are n numbers then initial average = $$\dfrac{n+1}{2}$$.

Average of N natural number can be either an integer {ab} or {ab.50} type. For example average of first 10 number = 5.5 whereas the average of first 11 natural numbers is 6. 
Even if we erased the largest number change in average will be always less than 0.5. 
Here we are given the average is 602/17 or 35$$\frac{7}{17}$$ Hence we can say that average must have been 35.5 or 35 before. 
Case 1: If the average was 35.5 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35.5 = $$\dfrac{N+1}{2}$$
N = 70. 
Sum of these 70 numbers = 70*71/2 = 35*71 = 2485.
Sum of the 69 numbers which we are left with after removing a number = (602/17)*69 = 2443.41. Which is not possible as the sum of natural numbers will always be an integer. Hence, we can say that case is not possible. 

Case 1: If the average was 35 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35 = $$\dfrac{N+1}{2}$$
N = 69. 
Sum of these 69 numbers = 69*70/2 = 35*69 = 2415.
Sum of the 68 numbers which we are left with after removing a number = (602/17)*68 = 2408. 
Hence, we can say that the erased number = 2415 - 2408 = 7. 

Let angle EAC = x, so angle AEC = x and angle ACE = 180-2x
Let angle FBC = y, so angle BFC = y and angle BCF = 180-2y
So, angle ACB = 180-(180-2x+180-2y) = 2(x+y) - 180
x+y = 180 - 40 = 140
So, angle ACB = 280 - 180 = 100

Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $$x^2/100 * P = 441$$
Let the new price be N => $$P - x^2/100 * P = N$$
Price after the second cycle = $$N - x^2/100 * N$$ = 1944.81
=> $$(P - x^2/100 * P)(1 - x^2/100) = 1944.81$$
=> $$(P - 441)(1 - 441/P) = 1944.81$$
=> $$P - 441 - 441 + 441^2/P = 1944.81$$
=> $$P^2 - (882 + 1944.81)P + 441^2 = 0$$
=> $$P^2 - 2826.81P + 441^2 = 0$$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25

Let x be the required distance.

Let a,b,c be speed of the A,B and C respectively.

From the given conditions we have, 

$$\frac{a}{b}=\frac{x}{x-12}$$ and $$\frac{a}{c}=\frac{x}{x-18}$$ and $$\frac{b}{c}=\frac{x}{x-8}$$ . From first 2 equations we can deduce $$\frac{b}{c}=\frac{x-12}{x-18}$$. 

$$\frac{b}{c}=\frac{x-12}{x-18} = \frac{x}{x-8}$$

x = 48 satisfy the equation.

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ =12.5

Approach 2:

$$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$

Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$\sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM).

Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = y +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ = 12.5

We will solve this question by taking the options.

Suppose x = 11

x+4 = 15, x-3 = 8

Hyptenuse = $$\sqrt{225+64} = 17$$

17-11 = 6

$$6^2+(x-3)^2 = y^2$$

x-3 = 8

y = 10 which is similar to what is given in the question.

Hence x = 11

If the width of the walkway is x, then its area = (60+2x)(20+2x) - 1200 = 516
Solving this, we get x = 3m

We know that a=$$b^2-b$$.

So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1)

The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)

In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).

Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24.

Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

The possible arrangements are 1,multiples of 2 ,remaining. So we have 1+2+4+8+16+32+64+31 = 158. Hence minimum no. of bags required is 8. 

We have a +c=e so possible summation 6+4=10 or 4+2 = 6.
Also b=2d so possible values  4=2*2 or 10=5*2.
So considering both we have b=10 , d=5, a=4 ,c=2, e=6.
Hence the correct option is B .

We know that quadratic equation can be written as $$x^2$$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $$x^2$$-(7)*x+(12)=0 where the constant term is wrong.

Keshab got the roots as (3, 2) so the equation is $$x^2$$-(5)*x+(6)=0 where the coefficient of x is wrong .

So the correct equation is $$x^2$$-(7)*x+(6)=0. The roots of above equations are (6,1).

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

The distinct paths are:

A -> D -> C -> F
A -> D -> E -> F
A -> D -> C -> E -> F
A -> B -> D -> C -> F
A -> B -> D -> E -> F
A -> B -> D -> C -> E -> F
A -> B -> C -> F
A -> B -> E -> F
A -> B -> F
A -> B -> C -> E -> F

So, the total number of distinct paths is 10

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $$^4C_3 $$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192

Manually counting non-zero value number of fabrics in yellow fabric category. We get 14. Hence option D. 

1, 3, 4, 6, 7, 8, 9, 11, 12, 15, 21, 24, 25, 27

Manually counting lays for which number of lays produced is non-zero . We get 15 lays. Hence option A. 

Manually counting  lays which are are used to produce XL yellow or XL white fabrics. We get 15 number of such lays. 

Counting the number of variety of fabrics which have positive non-zero surplus, we get 4 such varieties. Hence option B. 

1 Million = 10 lakhs. Using the conversion and manually counting the  international airports of type ‘A’ accounting for more than 40 million passengers, we get 5 such airports. Hence option B . 

There are 6 airports from USA which are in top 10 busiest aiports. Hence 600/10 = 60%. Hence option A.

Total passengers handled in top 5 airports are 336648008 . So percentage of passengers handled by heathrow airport is (62263710*100)/(336648008) which is approximately equal to 20%. 

All the top 20 busiest airports handle more than 30 million passengers. So, we have to just count the airports which are not located in the USA, out of these 20 airports. The airports ranked 4, 6, 7, 8, 11, 18 are located outside the USA. Thus, there are 6 such airports.
Hence, option B is the correct answer.

The number of man-hours spent in coding = 425(offshore) + 100(onsite) = 525
The number of man-hours spent in offshore design and offshore coding together = 100+425 = 525
So, option a) is the correct answer

Total number of man-hours needed for the entire work to be done = 100+75+425+100+300+150 = 1150
Man-hours spent onsite = 75+100+150 = 325
So, required percentage = 325/1150 * 100% = 13/46 * 100% = 28.26% = 30% approx

Total number of man-hours spent onsite = 75+100+150 = 325
The sum of the estimated and actual effort for offshore design = 200 (approx)
The estimated man-hours of offshore coding = 425
The actual man-hours of offshore testing = 300 (approx)
Half of the man-hours of estimated offshore coding = 210(approx)
So, option c) is the closest

Total working hours = 1150
So, we have to find the task where the number of working hours is approximately 575
Coding = 425 + 100 = 525
Design = 100 + 75 = 175
Offshore testing = 300
Offshore testing + design = 300 + 100 = 400
So, option a) is the correct answer

Total number of hours needed for testing = 300+150 = 450
New number of hours spend offshore for testing = 300/2 = 150
So, required percentage = 150/450 * 100% = 33.33%
From the options, the closest answer is option b)
 

Total offshore work = 100+420+300 = 820. If 50% of the offshore work were to be carried out onsite i.e 410 carried out onsite. 

So new work carried out onsite is in design is [80 + (100*410)/(825)] which is around 130. Also new work carried out onsite is in coding is [100 + (425*410)(825)] which is around 315. 

And new work carried out onsite is in testing is [150+(300*410)(825)] which is around 300. Hence the amount of coding done is greater than that of testing. 

We know that quantity between Vaishali and jyotishmati is 300,

So quantity in avanti-vaishal route should be 700.

Now at jyotishmati the required quantity is 400+700 = 1100.

But through vaishali only 300 comes , so  800 should come through Vidisha-jyotishmati route.

Now demand at vidisha is 200. So total quantity required in avanti - vidisha route is 800+200=1000. Hence option D. 

We know that quantity between Vaishali and jyotishmati is 300,

So quantity in avanti-vaishal route should be 400+300=700.

So free capacity is 1000-700=300. Hence option D . 

We know that quantity between Vaishali and jyotishmati is 300, So quantity in avanti-vaishal route should be 700. Now at jyotishmati the required quantity is 400+700 = 1100.

But through vaishali only 300 comes , so  800 should come through Vidisha-jyotishmati route.

Now demand at vidisha is 200. So total quantity required in avanti - vidisha route is 800+200=1000.

Hence, free capacity available = 1000-1000=0. Hence option D. 

If suppose effort allocation is inter-changed between operations B and C, then C and D, and then D and E then operation E will have value of B . So arranging effort in E in ascending order we have company 4 ,company 5 , company 3, ... Hence company 3 is ranked third. Hence option B. 

Now from B through F total work done is 81.7%. Dividing it in 5 we get 16.34. So difference in operation E is 28.6-16.34 = 12.3 %. Hence there is a reduction of 12.3 .

Operations B, C and D have more effort weightage in companies 1,3,5,6.

Also we notice that among them operation E has highest already, and for E company 5 has already the highest percentage. So redistributing the share of E would increase further more. Hence option D.

Total tonnes of transportation by both rail and road is about  31*12/100 = 3.72 million tonnes and total cost incured 18*30/100 = 5.4 .

Hence required value is 5.4/3.72 which is about 1.5.

Hence option B . 

Cheapest mode of transport will be the one which will have highest transport volume and comparatively lowest cost. We can figure out from the graph that Road have wide gap with very less costs. 

For Road, Cost = 6/22

For Rail, Cost = 12/9

For Pipeline, = 65/49

For Ship, Cost = 10/9

Lowest cost is for road.

If the costs per tonne of transport by ship, air and road are represented by P, Q and R respectively.

P = 10/9

Q = 7/11

R = 6/22

We can see that P>Q and Q>R  and P>R. Hence option C. 

E = 3Y
Z = W/2
Y > Z
So, Y > W/2 => 2Y > W => 2E/3 > W => E > 3W/2 => E > W
So, Elle is older than W

E = 3Y
Z = W/2
Y > Z

Using statement A alone, Z and W can be determined but not E

Using statement B, Y = W. So, using this statement alone, we can get a relation between all the four ages but cannot determine the absolute values.

Using both the statements, we can determine the value of E.

10 = W/2

W = 20 = Y

E = 3*20 = 60

So, c) is the correct answer.

Shyam and Rahim have to be selected together. This rules out options a) and c). Ram will be in the group only if Peter is also there. This rules out option d). Option b) is a feasible group

David and Peter cannot be selected together. So, option d) is not possible.
David and Fiza have to be selected together. So, option b) is ruled out.
Shyam and Rahim have to be selected together. So, option a) is also ruled out.
Option c) is the correct answer

If Ram is selected, Peter also has to be selected. If Kavita is selected, David also has to be selected. But, Peter and David cannot be selected together. So, option a) is false.

If both the women are selected, David has to be selected. If David is selected, Peter cannot be selected. So, Ram also cannot be selected. Shyam and Rahim have to be selected together but there is room for only one more person. So, option b) cannot be true.

If David is selected in the group, Fiza has to be there. So, David cannot be in a group of all men. But, Ram and Shyam cannot be selected together. So, a group of all men is not possible. So option c is false.

Using either statement alone, we cannot answer the question.
The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. From the information given in both statements, the value of m is 15, and n is 2. So, option c).

This is a previous year's CAT question. We can also consider the negative values, which gives us other possibilities. So, a single answer can't be obtained. 

Even by using both the statements, we do not know whether country X has a higher GDP than country Y because we do not know the absolute values of the GDPs 5 years back. So, the question cannot be answered.

If we take the first statement alone,
X and Y can be among 2, 4, 6, and 8.
It is given that X/Y is an odd integer. So, X must be greater than Y.
If we take different pairs of (X,Y) among (8, 2), (8, 4), (8, 6), (6, 2), (6, 4), (4, 2), only (6, 2) satisfies the above condition of X/Y being an odd integer.
So, we can uniquely determine the values of X and Y which are 6 and 2 respectively.

If we take the second statement alone,
12 = 2 * 2 * 3
We want 12 to be the product of two even numbers.
The only possibility is when the numbers are 2 and 6.
However, we don't know any relation between X and Y.
So, both of them can be either 2 or 6.
Therefore, statement II alone is not sufficient to determine the value of X.

Hence, option A is the correct answer.

The question asked is about the number of round trips made in 12 hours. This can be answered if we know the time of travel and the time of waiting. So, the question can be answered by using statement 2 alone but not by using statement 1 alone.

We cannot answer the question using either of the statements alone. If we consider both the statements i.e. transfer rate is 6 kilobytes per second and the size of the software is 4.5*1024 kilobytes = 4608 kilobytes then download time = 4608 / 6 seconds. Hence, the question can be answered by using both the statements together, but cannot be answered by using either statement alone.

In these type of questions, we don't need to find out the exact answer but just check if the answer can be calculated.

When a square is inscribed inside a circle, the diagonal of the square is a diameter of the circle. Hence, if either the circle diameter or side of the square is given, then we can determine the areas of both the circle and square as well as the difference between their areas.

Hence, the question can be answered by using either statement alone.

Consider the first statement. We just know the relation between number of apples of bought by the two guys. Hence, the question cannot be answered using this statement.

Based on the total number of available apples, we cannot determine how many of them were bought by Ram and Gopal. 

Using both statements together, we don't know how many apples the two of them bought together. Hence, we cannot answer the question using even both the statements together.

According to given conditions , following time table is possible:

hence option C.

According to given conditions ,we get

According to given conditions we have -

We know that  Shopkeeper 1 said: “The dog had black hair and a long tail.”. Lets consider the first part as false and other as true. So we have - dog didn't have black hair and the dog had a long tail. Shopkeeper 2 said: “The dog had a short tail and wore a collar.” Here first part has to be false and other consequently will be true. So the dog wore a collar. Shopkeeper 3 said: “The dog had white hair and no collar.” Here 2nd part has to be false and hence 1st part has to be true. Thus, our first assumption satisfies all required conditions. Hence, in this case, we have - The dog had white hair, long tail and a collar. Hence option b.

6+x+22+x+6y/5 = 50
=> 5x + 3y = 55
Since y/5 is even, y should be a multiple of 10
The only possible value is 10
So, y = 10 and x = 5
No. of oak leaves = 17

It is given in the statements, that C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F, or H, or D. By formulating the table we get

Q is with A and G is with Q => G and Q are travelling together on motorcycle M3
F travels on M1 and E travels on M2 motorcycles.
D is travelling with P on M2 => D and E are traveling together on M2
B cannot go with R => F and B go together on M1
Therefore, C and H go together on M4

So, the table can formed as below :

Hence, C would be travelling with H.

The bare minimum requirement for the single grandfather and grandmother are as follows:

GF implies Grandfather, GM Grandmother, F Father, M Mother, GS Grandson and GD Granddaughter. The family trees above account for 2 fathers, 2 mothers, 1 GF and 1 GM. Hence, we need 1 more GF, 1 GM, 2 fathers and 2 mothers and two married couples.

The bare minimum tree that would provide this structure is as follows:

Using these three structures, we have fit all of our requirements. Thus, if GF1, GF2 and GM1 are siblings with the following family trees present, we would have the required number of people in attendance. Hence, the minimum is 12 people.

According to given condition we find average costs for products bought.
If D is bought then - D +2*B . So we spend 40+180 =220 Rs for 3000 points. Hence, cost per 1000 points is 73.33
If A is bought then - A+C . So we spend 110 + 70 = 180 for 2000 points. Hence, cost per 1000 points is 90
If E is bought then - E+2*D+4*B+B . So we spend 45 + 80 + 360 + 90 = 575 for 8000 points. Hence, cost per 1000 points is 71.875.

If B is bought then I spend Rs 90 per 1000 points.

If C is bought then I spend Rs 70 per 1000 points.

To maximise points we need to select the item that costs the least per 1000 points. Hence, C costs the least per 1000 points. Hence, we should buy as many C items as possible.
Maximum C that can be bought is [1000/70] = 14 items and Rs 20 would be left over. However, the unspent money would attract a penalty of 30000 points. Hence, instead of buying 14 C items, we should buy 13 C items and 1 B item so that total money spent is 1000.

According to given conditions Bashir started with 500Rs Chirag should have 700 + something , which when subtracted from 500 get money which Deepak has. Also Ashok should have < 1500 and his money should be 3 times that of deepak . Thus there;s only 1 possibility that Ashok has 1200rs and deepak has 400rs. So the costliest item that deepak can buy is shawl worth 400 rs.

According to given conditions, the groups are as follows:

Case 1:

W1 : Sonali, Shalini, Shubhra ,Shahira and Rupa
W4: Somya , Ruchika, Jyotika, sweta
: Renuka, Rupali, Komal
: Ritu and Tara
: Radha

Case 2:

W1 : Sonali, Shalini, Shubhra ,Shahira and Rupa
W4: Somya , Ruchika
: Renuka, Rupali, Komal
: Ritu, Tara, Jyotika, sweta
: Radha

Hence, Radha must be alone in a group . Therefore, her trainer is Elina.

Looking at each option and the table. Only Option D satisfies the given condition  that each selected person shares a liking with at least one of the other three selected. The selected persons must also hate at least one of the likings of any of the other three persons selected.

It is given that $$X = M \cap D$$ is such that $$X = D$$, which means D is a subset of M . Which means all dogs are mammals. Hence , option A. 

We know that all dogs are vertebrates. Hence $$D \cap V)$$ = D . Ahead it is given that $$Y = F \cap D $$ is not a null set  which implies that some fish are dogs. Hence , option C. 

From $$P \cap D$$ is a set containing just pluto ,as pluto is subset of dog D, hence $$P \cap D$$ = P . Also $$P \cup M$$ would be set of pluto and all mammals together. 

We know that $$P \cap A = \phi$$ means that the pluto is not an alsatian and $$P \cup A = D$$ means that the pluto along with alsatian makes the set D . hence option D - Both A and C.

Take a look at the sentence F where it is mentioned  "Their accomplishment exceeded our expectations." This means exceed refers toB. So B goes with F. 

Now take a look at the sentence G.It is written that "He exceeded his authority....". Here "exceed " refers to crossing the limit which is mentioned in D. Hence D will go with G.

Only option A goes with B-F, D-G

The sentence in E goes like this:"We see smoke and infer fire." It means we are infering fire based on some reason. Hence A will go with E.

"Surmise" ,means "to suppose something is true without having proof." In F, the usage of infer is similar where it is mentioned that the user may infer something without any proof. Hence B goes with F.

In  G, infer means to point out something which relates option C.

Hence the correct option is D

In E mellow means "freed from rashness" which appropriately matches with B.

In H, mellow soils means "soils of soft and loamy consistency" which matches with C.

In G, mellow means "to be adequately aged so as to exhibit softness". This matches with A.

Hence the correct option is C.

Point E says that the release of a sentry after the morning shift is followed by a ceremony => D-E
In point F, removing shows helped to lighten his distress => A-F
Clearly, playing cards acts as a diversion to that person => C-G
In point H, disaster victims were offered an aid => B-H

In option E, the opposition was removed after the coup, so C-E is a pair.
In option F, he removed his name from the charge of heresy, hence A-F is a pair.
In option G, causing evacuation of interior parts is bad for the brain, hence D-G.
In option H, cleaning water by distillation is recommended. B-H fits best.

Read these lines "Yet, we would humbly submit that if globalising our markets are thought good for the ‘national’ pocket, globalising our social inequities might not be so bad for the mass of our people. After all, racism was as uniquely institutionalised in South Africa as caste discrimination has been within our society; why then can’t we permit the world community to express itself on the latter with a fraction of the zeal with which, through the years, we pronounced on the former?"
The issue at the hand is to openly discuss the social inequities.Hence option A is correct.
Option C is incorrect as delimitation means to fix the limits which is nowhere mentioned in the passage.

The author says inverted representations have "often been deployed in human histories as balm for the forsaken - religion being the most persistent of such inversions." The word "balm" might seem positive, but the author's framing is critical - these inversions are deployed to pacify the oppressed, not to genuinely help them. 

The author is pointing out that such representations serve to keep people oppressed by offering false comfort rather than real change. Option A is wrong because the author isn't endorsing this as "good." Option D misreads the meaning — it's not about inverting the status quo but about maintaining it through ideological deception.

Option C is out. Also there is nothing related to racial pride. Out of others only A and  E are the broad areas that fall under UN preview of discussion. 

The passage says the sociologist is "to be complimented now for admitting, however tangentially, that caste discrimination is a reality." This maps directly to option B. 

Option A is wrong  - the author notes the sociologist "earlier fiercely opposed the Mandal Commission Report," placing him in opposition to the author's orientation. 

Option D goes too far: the sociologist acknowledges caste discrimination tangentially but is arguing the two are incompatible, not affirming both equally. 

Option C distorts the argument - the sociologist speaks of conceptual incompatibility between race and caste, not between people.

Option A is out . Option D is also out as caste is completely social construct. Out of B and C , B can be inferred from the last part of the passage. 

Option A: This is true; for example, the monosyllabic word pray has only one onset /pr/ while the trisyllabic word val:en:tine has two onsets - /v/ and /t/, and three rimes corresponding to the selling patterns ‘al’, ‘en’ and ’ine’. Therefore, Option A is a valid inference. 

Options B and C: We do not know if these are necessarily true.  

It is clearly stated in the passage phoneme emerges only at the age of  6. Hence option D. 

It can be inferred from the passage that phonological skills in young children can be measured at a number of different levels. Hence phonological deficit can be classified in any one or more of the options . Hence D. 

We can clearly infer from the passage that 4-5 years old found the onset-rime version to be easier, it was only for the 6 years old who were able to perform both the versions with equal success. Hence option B. 

It is stated in the passage that ' rimes correspond to rhymes in single-syllabus words ' . Hence option B. The onset parts would be different. For example, for cat and hat the onsets would be /c/ and /h/. 

The term , survive many who shall receive longer obituaries , means that her name and fame would continue to live longer than most people. 

From the passage it can be inferred that her claim to fame was her creation of the Blues. Option B and D cannot be inferred from the passage. The passage does not state that she had  a mellow voice as well. Instead it states that she had a thin and gritty voice. Hence, option A is the correct answer.

The passage explains that her physical death has been called for as relief rather than sorrow. Hence option C is clearly the answer.

You can also refer to the following lines:"What sort of middle age would she have faced without the voice to earn money for her drinks and fixes, without the looks — and in her day she was hauntingly beautiful — to attract the men she needed, without business sense, without anything but the disinterested worship of ageing men who had heard and seen her in her glory?" This implies option c.

Following sentences from the passage shows that option D is not what author said ' Suffering was her profession; but she did not accept it.'. 

The question asks which of the following cannot be inferred with respect to the passage. 

From the last paragraph , we can understand that she had a honest producer to work with, that she had the looks, and also that at her prime she had religious fan following.

But, the fact that Bessie and Louis Armstrong accompanied her cannot be inferred from the passage as it is only mentioned that she was inspired by the singing styles of Bessie and Louis Armstrong.

In the first para and 2nd sentence itself the author mentions that Kurosawa uses 3 frames and shift them. Hence option C. 

We can see 'We see all this at work in the enormously evocative prologue...' in 2nd para starting sentences where Arseniev’s search for Dersu’s grave is described . Since it is part of prologue it;s in start of film. Hence option A. 

In the last para it is described that 'The hallucinatory dreams and visions of Dodeskaden are succeeded by nostalgic, melancholy ruminations..' . Option D is perfect. 

IN the 3rd para we can find the sentence '....and to delineate the code of ethics by which Dersu lives and which permits him to survive in these conditions...' Which is part of 1st section after prologue. Hence option C. 

We can easily find option B and C reffered in last para. hence it has to be all of these. 

The film clearly doesn't expresses its arguments in cirtuitous way. Also it doesn't highlight insularity of anyone. Option C is correct as Dersu - the main protagonist is not in the beginning part. Also refer to the following liines of the paragraph:"The first section of the film has two purposes: to describe the magnificence and in human vastness of nature and to delineate the code of ethics by which Dersu lives and which permits him to survive in these conditions". These lines confirm that the protagonist does not appear in the first part.

From the following text, which is  picked up from passage '..as greater the urge for a change in society the stronger is the appeal for a dynamic leadership..' Hence option C. 

We can infer from the passage that equality before law or formal equality is a hindrance to the establishment of real or substantive equality. . Refer to the lines in the last paragraph:"From the argument that formal equality or equality before the law is but a limited good, it is often one short step to the argument that it is a hindrance or an obstacle to the establishment of real or substantive equality."This substantiates the answer.

Following sentences from the passages are in line with the statements A,B and D. In statement A, if modern age is preoccupied with scientific rationality then it is also no less preoccupied with change. In B, a world preoccupied with scientific rationality have advantages system based on impersonal rule of laws should be recommendation with everybody. Statement D,states that democracy guarantees formal equality beyond this it can only what people’s appetite for substantive equality.
Option C is a broad statement. In the passage, it has been mentioned that tradition is preferred in music, dance and drama. But in society, modernisation and development have been chosen, instead of tradition. Thus, option C cannot be inferred. So option A is the correct answer.  

It can be inferred from the passage that tocqueville believed that unlike aristocratic societies there was no proper place in democracy for heroes and hence if they arose they would sooner or later turn to despots. Hence option A is the correct answer. Also refer to the following lines:"Tocqueville viewed this with misgiving because he believed, rightly or wrongly, that unlike in aristocratic societies there was no proper place in a democracy for heroes and, hence, when they arose they would sooner or later turn into despots."

It can be inferred from the passage that impersonal rules can ensure stability but it can’t create any shinning version of a future in formal equality. Hence option D. 

A can be inferred from 1st para and  D can be inferred as continued preoccupation with plans schemes will help to bridge the gap between ideals of equality and reality. Hence only A and D can be inferred. 

Option B is a distortion of what is given in the passage. The author says that a need for impersonal rules is not felt in a small society. He is not mentioning it as solution. 

Option C is only partly true. Impersonal rules are necessary in large and amorphous societies. Thus, a large society with a clear structure would not require impersonal rules.

Refer to the given lines "Sir Martin Rees, Britain’s astronomer royal, named the long interval between these two enlightenments the cosmic “Dark Age”. The name describes not only the poorly lit conditions, but also the ignorance of astronomers about the period."

Refer these line"The main problem that plagued previous efforts to study the Dark Age was not the lack of suitable telescopes but rather the lack of suitable things at which to point them. Because these events took place over 18 billion years ago, if astronomers are to have any hope of unraveling them they study objects that are at least 13 billion light years away."
This indicates B.

Option D is incorrect as by referring to these lines "However at the distances required for the study of Dark Age, even quasars are extremely rare and faint." , it is clear that the quasars are not faint to begin with but the distance which is required to study them is too large, so they appear faint.

Refer to the given lines "Recently some members of Dr. Becker’s steam announced their discovery of the four most distant quasars known, all the new quasars are terribly faint, a challenge that both teams overcame by peering at them through one of the twin Keck telescope in Hawaii. These are the world’s largest and can therefore collect the most light."

Refer to the given lines "The fog prolonged the period of darkness until the heat from the first stars and quasars had the chance to ionize the hydrogen (breaking it into its constituent parts, protons and electrons). Ionized hydrogen is transparent to ultraviolet radiation, so at that moment the fog lifted and the universe became the well-lit place it is today."

Of the given options starting with E or A, E seems to be a better introduction since it highlights the main idea of the passage. Between E- A or E - C, we observe that E- A  forms a mandatory pair since it moves from the general “India” to specific “regional variations”. We also notice that D-B also forms a pair with D discussing the "arrival" of employees at the office and then B adding about activities post-arrival (during office hours). F continues on the topic of mingling with relatives mentioned in B and hence, forms a block B - F. Thus, we are left with the arrangement EADBFC.

Option C is the correct answer. 

F introduces the theme of the passage. This is followed by a question in statement D. E follows with an answer to the question. This answer is further described in B, then A and finally C (chronologically in that order).

E starts the para by introducing the subject. A comes next and it introduces Michael Hofman. Statements C and D describe Michael Hofman and what he feels about the subject. D is the best concluding statement. Hence, the correct order is EACBD.

A is the best opening sentence. It introduces the subject. B follows by giving an example of where passivity might not be universal. D-E is a link as D talks about the problem of the peasants and E talks about the environment in which the problem occurs. The logical sequence of sentences is ABCDE.

A is the best opening sentence. It talks about the nature of violence being clearly defined. This is followed by C, which talks about an exception to the theory mentioned in A. C is followed by D and B is the best concluding statement. The correct order of sentences is, therefore, ACDB.

We can identify the correct choice using the second blank: 'practices' do not wither or trade or die 'away'; they 'fade away.' Options A, B and C don't fit in the second blank and hence, can be eliminated. 

"Combines" is a better fit for the 2^nd blank as it has to deal with two phrases, hence eliminating options A and B.

The word in the 1^st blank and the word "attributes" must be in the same tense. Hence option D is eliminated and the answer is option C.

"Diagnose" is a wrong word for the 2^nd blank. Hence option C eliminated.
"Make-up" and "scars" are bad fit-in for the 1^st blank.
As her face was expressionless, there was no way to find if she appreciated what had happened. This sounds more meaningful. Hence option B is the answer.

Clearly, affair and activity don't fit in the first blank, hence can be eliminated. In the second blank, "significant" is the best fit complementing "experience" in the first blank. Hence option C.

Even though both "cliques" and "societies" fit into the second blank, clearly "ancestors" fits into first blank the best.

Specious means 'deceitful' whereas credible means 'convincing'.
So, option c) is the correct answer.

Obviate means 'to prevent' whereas bolster means 'to strengthen'. So, the answer is option d).

Disuse means being discarded, whereas prevalent is being in use. So, option a) is the most inappropriate word in this context.

Parsimonious means frugal, whereas altruistic is the exact opposite of it. So, option d) is the correct answer.

Facetious is being mischievous. Jovian means relating to god Jove (or Jupiter)
So, jovian is the correct answer.