Quants for CAT2007 - 8

Q1. The length of common chord of two circles of radii 15cm and 20cm, whose centers are 25cm apart, is
a. 24 cm                           b. 25 cm                   c. 15 cm                       d. 20 cm

Ans.

There is a right-triangle with sides 15, 20 and 25 – now, altitude of this triangle from the vertex containing right angle is the half of chord = 15*20/25 = 12

So, length of chord = 2*12 = 24

Q2.if x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx =3, what is the largest value that x can have ?
a. 5/3                    b. √19              c. 13/3               d. None of these

Ans.

When x is maximum y=z

So, x+2y=5 => 2y= (5-x)

xy+yz+zx = 2xy+y² = x(5-x)+{(5-x)/2}² = 5x – x² + 25/4 – 5x/2 + x²/4= 25/4+5x/2 -3x²/4

So, 25/4+5x/2 -3x²/4 = 3

ð    25+10x - 3x² = 12

ð    3x² – 10x – 13 = 0

x = (10+√(100+12*13))/6 (note – other value of x will be negative)

= [5+√(25+39)]/3 = [5+√64]/3 = 13/3

Directions for Q3 and Q4.
2 dices are thrown simultaneously for each throw. Points equal to sum of numbers of the two dices are awarded. A bonus of 15, 10 & 5 point is awarded for sum less than 4, more than equal to 4 but less than 8, and more than or equal to 8 but less than 12, respectively. A penalty of 2 points is awarded for two consecutive throws having same sum or two consecutive throws, which fall under same category.

Q3.  Maximum number of points possible in two throws is
a. 33                 b. 35                c. 37                      d. 40

Ans.

For category sum < 4 à max point = 3+15 = 18

For category 4<= sum < 8 à max point = 7+10 = 17

For category 8<= sum < 12 à max point = 11+ 5 = 16

For category sum = 12 à max point = 12

Now, to gain max point one throw should be in (category sum < 4) and other in (4<= sum < 8) and max points = 18+17 = 35

Also, minimum points

Case 1 = 2+15 = 17

Case 2 = 4+10 = 14

Case 3 = 8+5 = 13

Case 3 = 12

So, minimum of 2 throws = 12+13 = 25

Q4. Score at the end of three throws is 41. which of the following cannot be the sum of numbers in anyone of three throws?
a. 3                  b. 4                 c. 5                    d. 6

Ans.

Let’s try options.

a.   other two throw sum = 41-(3+15) = 23 – NOT possible as min points in two throws is 25

no need to check other options.

Q5.On  a straight road XY, 100m long , five heavy stones are placed 2 m apart beginning at the end X. A worker, starting at X, has to transport all the stones to Y, by carrying only one stone at a time.
The minimum distance he has to travel is
a. 472m                b. 422m                   c. 744m                  d. 860m

Ans.

He will go like take stone#1 to 2. Then transport stone#2 to stone#3 and again come back and take stone#1 (currently at loc#2) to stone#3 and so on…

So, distance = 2+(3*2)+(5*2)+(7*2)+(9*92) = 2+6+10+14+828 = 860m

Q6. In how many ways can you put 120 apples of different shape and size in three bags containing 30, 40 and 50 apples?
a. 120!         b. 30x40x50       c. 30!x40!x50!        d. None of these

Ans.

For 1^st bag you can choose 30 apples from 120 in 120C3 ways and then from remaining 90 apples you can choose 40 apples for 2^nd bag in 90C40 ways.

So, total ways = 120C30 * 90C40 ways = (120!*90!)/(30!*90!*40!*50!) = 120!/(30!*40!*50!)

Q7. The distance of the point (1, 4) from the line 3x – 4y + 3 = 0 is
a. 1/5 units         b. 2/5 units         c. 8/5 units           d. 2 units

Ans.

The distance is = |(3*1 – 4 *4 + 3)/√(9+16)| = 10/5 = 2

Q8. From a cone of height 10cm, a smaller cone of height 4cm has been cut out such that the smaller cone has the same vertex and vertical axis as the original cone. What is the ratio of the smaller cone to that of the remaining part of the bigger cone?
a. 4: 21              b. 8: 117                c. 8 : 125             d. Cannot be determined

Ans.

Ratio = 4³ : (10³-4³) = 64:936 = 8:117