CAT 2017 Question Paper Quant Slot 2

According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15.

Let us consider the matrix as shown below: 

Now we'll try substituting values from 1 to 9 in the exact middle grid shown as 'x'.

If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible.

x cannot be equal to 2.

If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible.

If x=5, we get the grid as shown below:

Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3. 

Hence, 3 is the correct answer. 

By the time A traveled 10 KM, B traveled 9 KM
Hence $$Speed_A : Speed_B = 10:9$$
Similarly $$Speed_B : Speed_C = 10:9$$
Hence $$Speed_A : Speed_B : Speed_C = 100:90:81$$
Hence by the time A traveled 10 KMs , C should have traveled 8.1 KMs
So A beat C by 1.9 KMs = 1900 Mts

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4
Therefore, the proportion of milk in Bottle 1 is $$\frac{7}{9}$$ and the proportion of milk in Bottle 2 is $$\frac{9}{13}$$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $$\frac{7X}{9}+\frac{9}{13}$$
The total volume of water is $$\frac{2X}{9}+\frac{4}{13}$$
They are in the ratio 3:1

Hence, $$\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$$
Therefore, $$91X+81=78X+108$$

Therefore $$X = \frac{27}{13}$$

Let the distance between the home and office be $$2x$$ miles
Time taken for going in the morning = $$\frac{2x}{60}$$ hrs
Time taken for going back in the evening = $$\frac{x}{25} + \frac{x+5}{50}$$. hrs
It is given that he took 30 minutes (0.5 hrs) more in the evening

Hence $$\frac{2x}{60}$$ hrs + 0.5 = $$\frac{x}{25} + \frac{x+5}{50}$$
Solving for x, we get x = 15 miles.
Total distance traveled = 2x + x + x + 5 = 4x + 5 = 65 Miles

Let the total number of shirts be x. Hence number of non defective shirts = x - 15% of x = 0.85x
Number of shirts left for export = No of non defective shirts - number of shirts sold in domestic market
= No of non defective shirts - 20% of No of non defective shirts
= 80% of No of non defective shirts
Hence 8840 = 0.8 * (0.85x) . Solving for x we get, x = 13000

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x - 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

Let the manufacturing price of the table = $$x$$
Hence the price at which the wholesaler bought from the manufacturer = $$1.1 \times x$$
The price at which the retailer bought from the wholesaler = $$1.3 \times 1.1 \times x$$
The price at which the customer bought from the retailer  = $$1.5 \times 1.3 \times 1.1 \times x$$
$$1.5 \times 1.3 \times 1.1 \times x = 4290$$
=> x = 2000

Let the time taken by the outlet pipe to empty = x hours
Then, $$\frac{1}{8} - \frac{1}{x} = \frac{1}{10}$$
=> $$x = 40$$
Hence time taken by the outlet pipe to make the tank half-full = 40/2 = 20 hour

Let the number of dozens of candies he bought of each variety be x
Hence total cost = 12x + 15x = 27x
Total selling price = 16.50*2x = 33x
Profit = 33x - 27x = 6x
Given 6x = 150 => x = 25
Hence he bought 50 dozens of candies in total

Let initial population and production be x,y and final population be z
Final production = 1.4y, final percapita = 1.27 times initial percapita
=> $$\frac{1.4y}{z} $$ = $$ 1.27 \times \frac{y}{x}$$
=> $$\frac{z}{x} = \frac{1.4}{1.27} \approx 1.10$$
Hence the percentage increase in population = 10%

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2
=> a = 3x, b = 4x , c = 2x
=> a + b + c = 9x
=> a + b + c is a multiple of 9.
From the given options only, option C is a multiple of 9

Let the distance traveled by the car be x KMs
Distance traveled by the bike = 168 KMs
Speed of car is double the speed of bike
=> $$\frac{x}{3:40 - 2:00}$$ = $$2 \times \frac{168}{3:40 - 1:00}$$
=> $$\frac{x}{100}$$ = $$2 \times \frac{168}{160}$$
=> x = 210
Hence the distance between A and B is x + 168 = 378 KMs

Let the time take by kamal to complete the task be x days.
Hence we have $$\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$$
=> x = 40 days.
Ratio of the work done by them = $$\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 * 1000 = 100

The proportion of water in the first mixture is $$\frac{1}{3}$$
The proportion of Liquid A in the first mixture is $$\frac{2}{3}$$

The proportion of water in the second mixture is $$\frac{1}{4}$$
The proportion of Liquid B in the second mixture is $$\frac{3}{4}$$

The proportion of water in the third mixture is $$\frac{1}{5}$$
The proportion of Liquid C in the third mixture is $$\frac{4}{5}$$

As they are mixed in the ratio 4:3:2, the final amount of water is $$4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$$
The final amount of Liquid A in the mixture is $$4\times\frac{2}{3} = \frac{8}{3}$$
The final amount of Liquid B in the mixture is $$3\times\frac{3}{4} = \frac{9}{4}$$
The final amount of Liquid C in the mixture is $$2\times\frac{4}{5} = \frac{8}{5}$$

Hence, the ratio of Water : A : B : C in the final mixture is $$\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$$

From the given choices, only option C  is correct.

The length of the diagonals of a regular hexagon with side s are $$\sqrt{3}s$$.
Here length of AC = $$\sqrt{3}s$$ = $$\sqrt{3}$$ cms
Hence area of the square = $$\sqrt{3}^2$$ = 3 sq cm

See the diagram below

Length of side AD = $$\sqrt{12^2 + 5^2 } = 13$$
Area of the trapezium = 12 * (10 + 20)/2 = 180
Perimeter of the trapezium = 10 + 20 + 13 + 13 = 56
Area of the sides of the pillar = 56 * height  = 56 * 20 = 1120.
Total are of the pillar = 1120 + area of base + area of the top = 1120 + 180 + 180 = 1480

The midpoint of one diagonal lies on the other diagonal.

Midpoint is ((2+6)/2, (5+3)/2) = (4,4)
Hence 4 = 3 * 4 + c => c = -8

$$\angle COD = 120$$ => $$\angle CAD = 120/2 = 60$$ (The angle subtended by the chord DC at the major arc is half the angle subtended at the centre of the circle.)
$$\angle BAC = 30$$
$$\angle BAD = \angle BAC + \angle CAD$$ = 30 + 60 = 90.
$$\angle BCD = 180 - \angle BAD$$ = 180 - 90 = 90

Let the length and breadth of the park be l,b, l > b
Case 1: 2l + b = 400
Area = lb. Area is maximum when 2l * b is maximum, which is maximum when 2l = b (using AM $$ \geq $$ GM inequality) => l = 100, b = 200. Which can't happen since l > b

Case 2: l + 2b = 400
Area = lb. Area is maximum when l *2 b is
maximum, which is maximum when l = 2b (using AM $$ \geq $$ GM
inequality) => l = 200, b = 100.

Hence length of the longer side is 200 ft

Let the length of non-hypotenuse sides of the right angled triangle be $$a$$. Then the hypotenuse h = $$\sqrt{2}a$$
P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.
In a right angled triangle, inradius = $$\frac{a + b - h}{2}$$
=> $$\frac{a + a - \sqrt{2}a}{2} = 4(\sqrt{2}-1)$$
=> $$ \sqrt{2}a( \sqrt{2} - 1) = 8(\sqrt{2} -1)$$
=> $$ a = 4\sqrt{2}$$
Area of the triangle = $$\frac{1}{2}a^2$$ = 16 sq cm

$$(x -1)x(x+1) = 15600$$
=> $$x^3 - x= 15600 $$
The nearest cube to 15600 is 15625 = $$25^3$$
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

$$1 < \log_{3}5 < 2$$
=> $$ 1 < \log_{5}(2 + x) < 2 $$
=> $$ 5 < 2 + x < 25$$
=> $$ 3 < x < 23$$

$$f[f(g(1)) + g(f(1))]$$ 
= $$f[f(2^1) + g(1^2)]$$
= $$f[f(2) + g(1)]$$
= $$f[2^2 + 2^1]$$
= $$f(6)$$
= $$6^2 = 36$$

Let the roots of the equation $$x^2+(a+3)x-(a+5)=0 $$ be equal to $$p,q$$

Hence, $$p+q = -(a+3)$$ and $$p \times q = -(a+5)$$

Therefore, $$p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$$

As $$(a+4)^2$$ is always non negative, the least value of the sum of squares is 3

It is given that $$9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$$

Let us try to reduce them to powers of $$3$$ and $$2$$
The given equation can be reduced to $$3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$$

Hence, $$3^{2x-3} \times 10 = 2^{2x-2} \times 5$$
Therefore, $$3^{2x-3} = 2^{2x-3}$$

This is possible only if $$2x-3=0$$ or $$x=3/2$$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $$

$$3log(2^{a}\times3^{b}\times5^{c} )$$ = $$ log ( 2^{9}\times3^{6}\times5^{12}) $$
Hence, 3a = 9 or a = 3

Sum of the sequence of even numbers is $$2a_{3} + (2a_{3} - 2) + (2a_{3} - 4)$$ $$+ (2a_{3} - 6) + (2a_{3} - 8) = 450$$
=> $$10a_{3} - 20 = 450$$
=> $$a_{3} = 47$$
Hence $$a_{5} = 47 + 4 = 51$$

$$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$$
=> $$ab = 9(a + b)$$
=> $$ab - 9(a+b) = 0$$
=> $$ ab - 9(a+b) + 81 = 81$$
=> $$(a - 9)(b - 9) = 81, a > b$$
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)

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $$^{2+3-1}C_{3-1}$$ = $$^4C_2$$  = 6

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6
Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.
Hence total number of ways = 3*6 = 18

Case II: 0, 2, 3, 4
case II a: 0 is in the units place => 3! = 6 ways
case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8
Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6
case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12
Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14  + 18 = 50 ways

f(1 * 1) = f(1)f(1)
=> f(1) = f(1)f(1)
=> f(1) = 0 or f(1) = 1
Hence maximum value of f(1) is 1

$$|f(x)+ g(x)| = |f(x)| + |g(x)|$$ if and only if

case 1: $$f(x) \geq 0$$ and $$g(x) \geq 0$$
<=> $$ 2x-5 \geq 0 $$ and $$7-2x \geq 0$$
<=> $$ x \geq \frac{5}{2}$$ and $$ \frac{7}{2} \geq x$$ 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

case 2: $$f(x) \leq 0$$ and $$g(x) \leq 0$$

<=> $$ 2x-5 \leq 0 $$ and $$7-2x \leq 0$$
<=> $$ x \leq \frac{5}{2}$$ and $$ \frac{7}{2} \leq x$$
 So x<=5/2 and x>=7/2 which is not possible.

Hence, answer is 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

Let the common ratio of the G.P. be r.
Hence we have $$a_n= 3(a_{n+1}+ a_{n+2} + ...)$$

The sum up to infinity of GP is given by $$\frac{a}{1-r}$$ where a here is $$a_{n+1}$$

=> $$a_n= 3(\frac{a_{n+1}}{1-r})$$
=> $$a_n= 3(\frac{a_{n}\times r}{1-r})$$
=> $$ r = \frac{1}{4}$$
Now, $$a_1+a_2+a_2...+=32$$
=> $$\frac{a_1}{1-r} = 32$$
=> $$\frac{a_1}{3/4} = 32$$
=> $$a_1 = 24$$

$$a_5 = a_1 \times r^4$$
$$a_5 = 24 \times (1/4)^4 = \frac{3}{32}$$

$$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$$

$$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$

$$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$

$$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$
....

$$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

Hence $$a_{1}+a_{2}+a_{3}+...+a_{100}$$ = $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$ + $$\frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$ + $$\frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$ + ... + $$\frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

= $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{302})$$

= $$\frac{25}{151}$$