CAT 2023 Slot 2 Question Paper

Since BC is the diameter of the circle, which implies angle BAC is 90 degrees. Let AB = a cm, which implies AC = b cm. Hence, $$BC\ =\ \sqrt{\ a^2+b^2}$$, which is diameter of the circle (2r).

Hence, $$2r\ =\sqrt{\ a^2+b^2}$$

=> $$4r^2=a^2+b^2$$

The area of the triangle is $$\frac{1}{2}\times\ a\times\ b$$, which can be written as 

=> $$\ \frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ \left(a^2+b^2\right)$$

=> $$\frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ 4r^2$$

=> $$\frac{\ a\cdot b}{a^2+b^2}\times\ 2r^2$$

The correct option is B

It is given that AB = 9 cm, BC = 6 cm.

It is also known that the areas of the figures ABP, APQ, and AQCD are in geometric progression.

Hence, the area of the ABP, APQ, and AQCD are k, 2k, and 4k respectively.

The ratio of BP, PQ, QC will be the ratio of the respective triangles. Hence, we can draw a line from point A to point C.

Let the area of triangle AQC be x, which implies the area of triangle ADC = ADQC - AQC = 4k -x, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.

Therefore, 4k-x = 3k+x 

=> x = k/2

Hence the ratio of BP: PQ: CQ = k:2k: k/2 =2:4:1

From the inequality and nature of x, and y, we get the given diagram:

We need to find the area of the quadrilateral ABDE = area of rectangle ABCD + area of triangle CDE

=> Area of ABCD = (7-3)*5 = 20 units, and the area of triangle CDE = (1/2)*10*5 = 25 units.

Hence, the area of the quadrilateral ABDE = (20+25) = 45 units.

Let the first term of both series be $$a_1$$, and $$b_1$$, respectively, and the common difference be $$d_1$$, and $$d_2$$, respectively.

It is given that $$a_5=b_9$$, which implies $$a_1+4d_1\ =b_1+8d_2$$

=> $$a_1-b_1\ =8d_2-4d_1$$  ..... Eq(1)

Similarly, it is known that a_{19}=b_{19}, which implies $$a_1+18d_1=b_1+18d_2$$

=> $$a_1-b_1=18d_2-18d_1$$  ...... Eq(2)

Equating (1) and (2), we get:

=> $$18d_2-18d_1=8d_2-4d_1$$

=> $$10d_2=14d_1$$

=> $$5d_2=7d_1$$

Since, $$d_1,\ d_2$$ are the prime numbers, which implies $$d_1=5,\ d_2=7$$.

It is also known that $$b_{2}=0$$, which implies $$b_1+d_2=0\ =>\ b_1=-d_2\ =\ -7$$

Putting the value of $$b_1,d_1,\ \text{and, }d_2$$ in Eq(1), we get:

$$a_1=8d_2-4d_1+b_1=56-20-7\ =\ 29$$

Hence, $$a_{11}=a_1+10d_1\ =\ 29+10\cdot5\ =\ 29+50\ =\ 79$$

The correct option is B

Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)

Now, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2-2pq=9$$ => $$\left(p-q\right)^2=9$$ => $$p-q=\pm\ 3$$

Also, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2=2pq+9=2\left(18\right)+9=45$$

Now, $$p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$$ = $$\left(p-q\right)\left(45+18\right)$$ = $$\left(p-q\right)\left(63\right)$$

=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.

=> The difference = 189 - (-189) = 378.

It is given that $$a_{n}=13+6(n-1)$$, which can be written as $$a_n=13+6n-6\ =\ 7+6n$$

Similarly, $$b_{n}=15+7(n-1)$$, which can be written as $$b_n=15+7n-7\ =\ 8+7n$$

The common differences are 6, and 7, respectively, The common difference of terms that exists in both series is l.c.m (6, 7) = 42

The first common term of the first two series is 43 (by inspection)

Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.

$$t_m=a+\left(m-1\right)d\ <\ 1000$$

=> $$43+\left(m-1\right)42\ <\ 1000$$

=> $$\left(m-1\right)42\ <\ 957$$

=> m-1 < 22.8 => m < 23.8 => m = 23

Hence, the 23rd term is $$43+22\times\ 42\ =\ 967$$