CAT 2020 Question Paper Slot 3

Let the total marks be 100x

Marks obtained by Bishnu = 52x

Marks obtained by Asha = 64x

Marks obtained by Ramesh = 52x+23

Marks obtained by Ramesh = 64x-34

=> 52x+23 = 64x-34

=> x = $$\frac{19}{4}$$

Marks obtained by Geeta =84x = 84*19/4 = 399

Let  $$14^a=36^b=84^c$$ = k

=> a = $$\log_{14}k$$ , b = $$\log_{36}k$$, c=$$\log_{84}k$$

$$6b(\frac{1}{c}-\frac{1}{a})$$ = $$6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$$ = 3

Given, 

Rate of interest = 10%

Since it is compounded half-yearly, R=5%

n=3

We know, A = $$P\left(1+\frac{R}{100}\right)^{^n}$$

18522 = $$P\left(1+0.05\right)^3$$

=> P = 16000

Let the cost price of 1kg of sugar = Rs 100

The total cost price of 35 kg = Rs3500

Marked up price per kg = Rs 120

GIven, the final profit is 15% => Final SP of 35 kg = 3500 *1.15 = Rs 4025

First 5 kg's are sold at 20% marked up price => $$SP_1=5\cdot100\cdot1.2$$ = Rs 600

Next 15 kgs are sold after giving 10% discount => $$SP_2=15\cdot100\cdot1.2\cdot0.9\ =\ 1620$$

3kgs of sugar got wasted

=> 23 kg of sugar was sold at Rs (600 +1620) = Rs  2220

Remaining 12kg should be sold at Rs 4025 - 2220 = Rs1805

=> SP of 1kg = 1805/12 $$\simeq\ 150$$

Hence, the seller should further mark up by $$\frac{\left(150-120\right)}{120}\cdot100\ =\ 25\%$$

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on $$x+9y+c=0$$

midpoint of (2,1) and (-3,-4) = ($$\frac{2-3}{2},\frac{1-4}{2}$$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

The distance travelled by A between 9:00 Am and 10:30 Am is 3/2*40 =60 km.

Now they are separated by 30 km

Let the time taken to meet =t 

Distance travelled by A in time t + Distance travelled by B in time t = 30

40t + 20t =30 => t=1/2 hour

Hence they meet at 11:00 AM

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

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|

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41

$$ab\ =\ 4^{2017}=2^{4034}$$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.'. many pairs(a, b) of positive integers are there such that $$a\leq b$$ and $$ab=4^{2017}$$ = 2018.