### CAT 2019 – Slot 1 – Quantitative Ability – A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper

Q. 1: A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is
1. 8464π
2. 928π
3. 1026(1 + π)
4. 1044(4 + π)

As cylinders have the same volume and each of these has radius 3 cm. So volume of each cylinder will be equal to HCF of (405, 783 and 351) which is 27.

So volume of each cylinder = 27

No of cylinders = [ 405/27 ] + [783/27] + [351/27] = 15 + 29 + 13 =57

Using V = πr^2 h

27 = 22/7 *9 *h

So h = 3/ π cm

total surface area of all these cylinders = 57*(Surface area of one cylinder ) = 57*2* πr(h+r)=57*2π*3 (3/π+3)=1026(1+π)

##### Q. 2: Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is1. 102. 8√23. 6√24. 5

Let AB = a and AC = b

A^2 + b^2 = 400 by pythagoras

Let AP = x and BP = y, so CP= 20-y

So, by Pythagoras

X^2 + y^2 = a^2 and b^2 = x^2 + (20-y)^2

Maximum possible value of AP occurs when a = b, so a= b=10 root 2

On solving, we get AP = 10 units

Q. 3: The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is
1. 8√3
2. 10√2
3. 12
4. 5√5

As we know that the square pyramid of edge length a has height

h=1/2 sqrt(2)* a,

So, h = ½ * root 2* 20 = 10 root 2

Q. 4: Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals

Given f (mn) = f (m) f (n)and (1), f (2) and f (3) are positive integers.

As we know F(2*1) = f(2)= f(2)*f(1), so f(1) = 1

F(4) = f(2) ^ 2

F(6) = f(3) * f(2)

F(24) = f(4) * f(6) = f(2)^3 * f(3) = 54, only f(2) = 3 and f(3) =2 satisfies the equation , so we get

F(18) = f(9) * f(2) = f(3)^2 * f(2) = 4*3 = 12

Q. 5:Sequence and series – If (2n+1)+(2n+3)+(2n+5)+…+(2n+47)=5280, then what is value of 1+2+3+…+n?

On solving, we get

48n + 576 = 5280

N = 98

Hence, n(n+1)/2 = 98*99/2 = 4851

Inspiring Education… Assuring Success!!

Ghatkopar | Borivali | Andheri | Online