CAT Model Paper 3 Questions and Answers with Explanation Part 2

Glide to success with Doorsteptutor material for CTET/Paper-1 : get questions, notes, tests, video lectures and more- for all subjects of CTET/Paper-1.

Download PDF of This Page (Size: 165K)

Q: 8. Consider a triangle with sides and PR having lengths , and units respectively. Altitudes PL and QM are dropped on sides and PR respectively. It is given that PL and intersects at X. Calculate

(A) 3:5

(B) 4:5

(C) 7:9

(D) Cannot be determined

Ans: A

Solution:

Consider the figure given below

Q 8 Image of Triangle PQR with sides PQ, QR and PR having Le …

Triangle PQR with Sides PQ, QR and PR Having Lengths

We have, PL and QM are perpendiculars on QR and PR respectively. Hence triangles PLR and QMR are right angled triangles sharing the common acute angle R. Hence they must be similar. Further, we note that right angled triangles MQR and LQX share the common acute angle XQL. Hence triangle XQL must be similar to triangle MQR and thereby also similar to triangle RLP.

Therefore

(1)

Now, PQ, QR, and PR are of lengths 13, 14 and 15 units respectively .Hence , using Heron’s formula, we get that the area of the triangle is 84 square units. But area of triangle

Hence units. By Pythagoras theorem we get square units. Hence .

Thus using (A) and the values of LR and PR, we get, the required ratio as 3:5 as given in option (A).

Q: 9. Given that p, q, x and y are integers such that How may ordered quadruplets (p, q, x, y) satisfy the given equation?

(A) Only one

(B) Eight

(C) Sixteen

(D) Infinitely many

Ans: B

Solution

The given equation is

or

or

From the above equation we can conclusively say that

pair 1

Or

— pair 2

Solving pair 1, we get

Since p, q, x, y are integers, the only possible values are tabulated below.

Q_9_Table of the Only Possible Values Are Tabulated
Q_9_Table of the Only Possible Values are Tabulated

P

Q

X

Y

1

0

1

0

-1

0

-1

0

0

1

0

-1

0

-1

0

1

Similarly, by solving the 2nd pair, we get

Q_9_1_Table of the Only Possible Values Are Tabulated
Q_9_1_Table of the Only Possible Values are Tabulated

P

Q

X

Y

0

1

0

1

0

-1

0

-1

1

0

-1

0

-1

0

1

0

Hence the number of possible Solutions in integers is 8 as given by option (B).

Q: 10. Vyom, a student of Jago-Re Higher Secondary School calculated the value of as . Then which of the following is true?

(A) He was correct

(B) He made an error of

(C) He made an error of

(D) He made an error of

Ans: B

Solution:

By applying method of induction,

For ,

Thus , we can write .

Hence, option (B).

Q: 11. Let us consider a positive integer N such that the factors of N, 1 and N inclusive, are given by a1, a2 ….. an. Also let a1 + a2 + …. an = p. Then which among is the value of ?

(A)

(B)

(C)

(D) None of these

Ans: C

Solution

The given expression is =

This is because the denominator comprises of the of the factors of N which is N and the numerator comprises of the sum of the complements of in N, which basically boils down to finding the sum of the factors of N .

Hence the given expression has a value as given by option (C).

Q: 12. A quadratic function has two values ‘α’ and ‘β’ between 4 and 7 such that at ‘α’ and ‘β’,. Find the probable value of among the following :

(A)

(B)

(C)

(D)

Ans: B

Solution:

As at ‘α’ and that means they are the root of the quadratic function f(x) so the quadratic function can be rewritten as

Moreover, all are positive quantities. (As α and β lie between 4 and 7). Applying Arithmetic Mean Geometric Mean , we have

...... Eqn (1)

and

...... Eqn (2)

Taking Eqn (1),

or, ),

or,

Similarly from Eqn (2),

Combining both

or,

or,

or,

or,

Clearly options (a), (c), and (d) are greater than 81/16 so the only possible value of

Option (B)

Q: 13. Find the minimum value of the polynomial

(A)

(B)

(C)

(D)

Ans: A

Solution

The given polynomial may be rewritten as

Or,

Thus the given polynomial consists of the sum of three non-negative definite terms and one. Hence the minimum value 1 is attained when each of the other three non negative definite terms become zero at .Hence the answer is option (A).

Developed by: