### CAT 2019 – Slot 1 – Quantitative Ability – Raju and Lalitha originally had marbles in the ratio 4:9

Q. 1: Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?
1. 1/4
2. 7/33
3. 1/5
4. 6/19

Let original number of marbles with Raju and Lalitha are 4x and 9x respectively. Let Lalitha gave n marbles to Raju and the ratio becomes 5:6

So (4x+n)/(9x-n)=5/6

24x+6n=45x-5n

21x=11n

n=21x/11

Fraction of marbles given by Lalitha to Raju = n/9x=(21x/11)/9x=7/33

##### Q. 2: Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?1. 24, 102. 25, 93. 24, 124. 25, 10

As ABCD is a rectangle angles A,B,C and D will be 90°. Thus AC will be diameter of circle of length 13*2 = 26

So length , width and 26 will form a Pythagorean triplet . From given options ,

only option 1. 24, 10 satisfy this condition as

10^2+24^2=26^2

Q. 3: In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is?
1. 24√3
2. 12√3
3. 32√3
4. 18√3

As given, Area of parallelogram ABCD = 72

So AB×AP=72

9AP=72

AP = 8 cm

in right-angled triangle APD, AP^2 + PD^2 = AD^2

So PD^2 = AD^2-AP^2 = 16^2-8^2 = 64×3

PD=8√3

Thus Area of triangle APD = 1/2 AP×PD = 1/2×8×8√3 = 32√3

Q. 4: In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is?
1. (π/6)^(1/2)
2. (π/(4√3))^(1/2)
3. (π/(3√3))^(1/2)
4. (π/4)^(1/2)

As OC =OD so angle OCD = angle ODC =(180 -60)/2 = 60

So triangle OCD is an equilateral triangle,

Area of OCD = √3/4 OC^2 = 1/6×π×1^2

OC^2 = π/(3√3)

So OC= (π/(3√3))^(1/2)

Q. 5: How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

As all given digits are different natural numbers and we have to use them at most once, ways of arrangements for each set of digits will be exactly once.

Thus Total such numbers = 9(baseC2) + 9(baseC3) +9(baseC4) +9(baseC5) +9(baseC6) + 9(baseC7) +9(baseC8) +9(baseC9) = 2^9 – 2 = 510

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