# CAT Model Paper 6 Questions and Answers with Explanation Part 5

Glide to success with Doorsteptutor material for CTET/Paper-2 : get questions, notes, tests, video lectures and more- for all subjects of CTET/Paper-2.

Q: 23. In a triangle given below, Bisects at P. Also, is the median to the side . If the area of APR is 20 cm2, find the area of the quadrilateral .

(A)

(B)

(C)

(D)

Ans: D

Solution:

The correct choice is (D) .

To solve, draw a line

APR and AQS are similar. Given that (since P is the midpoint of AQ) .

Hence, ratio of areas of and is

Hence, area of . Therefore area of quadrilateral

Since P is the midpoint of , area of area of . Let the area of and Q be ‘x’ .

Hence area of

Since

Area of Eqn (1)

are similar. Hence,

Hence ratio of areas of and is .

(Area of = Area of area of quadrilateral )

Simplifying,

Area of

Q: 24. If p, q, r, s are positive numbers and find find the minimum value of

(A) 3

(B) 2

(C) 4

(D) Cannot be determined

Ans: C

Solution:

SOLUTION

The correct choice is (C) .

The given expression can be simplified as follows:

Since the above equation can be rewritten as:

+

are positive, the minimum value of

and the minimum value of

is 2. Hence the minimum value of the expression is 4.

Q: 25. If a number n can be denoted as in a number system with base m where , find the number n when converted to decimal if m is the greatest single digit prime number

(A)

(B)

(C) 333

(D)

Ans: C

Solution:

The correct choice is (C) . 7 when converted to decimal becomes .

Q: 26. If the roots of are squared, what will be the new equation with the roots?

(A)

(B)

(C)

(D)

Ans: D

Solution:

The correct choice is (D) . (

The roots are

If they are squared, they become

The new equation will be

Which translates to

Q: 27. Manoj repays a loan in three annual instalments which decrease by 1,000 every

year. The initial amount he borrowed was at the rate of Find the second instalment

(A)

(B)

(C)

(D)

Ans: A

Solution:

The correct choice is (A) . Let the instalments be .

End of 1st year, P becomes

End of 2nd year,

End of 3rd year

Therefore,

.

Developed by: