# Quant with Kamal Lohia - Part 3

• Q1) There is a 4-digit number ‘abcd’ that satisfies the following property. ‘abcd’ = ab ^2 + cd ^2. Find abcd.

Let the four digit number abcd = XY such that XY = X^2 + Y^2
=> 100X + Y = X^2 + Y^2
=> (X – 50)^2 + (Y – 1/2)^2 = 10001/4
=> (X – 50)^2 + {(2Y – 1)^2}/4 = 10001/4
=> 4(X – 50)^2 + (2Y – 1)^2 = 10001

Now sum of last two digits of two perfect squares is 01, it is only possible for the following two combinations; (00 + 01) or (76 + 25).
It can be easily checked that first combination doesn't help and in the second one also there is only one favourable case i.e. 76^2 + 65^2 = 5776 + 4225 = 10001.
So X = 12 and Y = 33 and the four digit number = abcd = XY = 1233 = 12^2 + 33^2.

Q2) If 12 men and 16 boys can do a piece of work in 5 days and 13 men and 24 boys can do the same piece of work in 4 days. How long will 7 men and 10 boys take to complete the same work?

Let a man's daily work is x units and that of a boy is y units.
So total amount of work, W = 5(12x + 16y) = 4(13x + 24y)
=> 60x + 80y = 52x + 96y
=> 8x = 16y
=> x = 2y i.e. W = 5(24y + 16y) = 200y.
Now 7men and 10 boys' work of a day is = 7x + 10y = 14y + 10y = 24y.
So total days required to finish the work is = 200y/24y = 25/3 = 8 and 1/3 days.

Q3) What is the length of the line segment which divides a non-isosceles trapezium, with parallel sides of lengths 1 and 7, in two equal area trapeziums?

Let parallel sides of the trapezium are a and b in length.
Line joining the mid points of non-parallel sides has length AM(a, b) = (a+b)/2
Line passing through intersection of diagonals and parallel to the parallel sides has length HM(a, b) = 2ab/(a+b)
Line segment, drawn parallel to parallel sides such that it divides the trapezium in two equal area trapeziums, has length RMS(a, b) = sqrt[(a^2 + b^2)/2]
Here, Length of the line segment which divides a trapezium with parallel sides of lengths 1 and 7 in equal area trapeziums is root-mean-square of 1 and 7 i.e. = sqrt[(1^2 + 7^2)/2] = 5.

Q4) The two diagonals of a trapezium are drawn to divide the trapezium in four triangles. If area of smallest triangle and largest triangle are 10 and 90 square units, then find the sum of areas of remaining two triangles in the trapezium.

The two triangles formed along non-parallel sides of trapezium have equal area.
Also the three different area triangles have their area in geometric progression.
So area of the triangles along non-parallel sides is = sqrt(10*90) = 30 square units.
And the required sum of area of two triangles is = 30 + 30 = 60 square units.

Q5) ABCD is a quadrilateral. The diagonals of ABCD intersect at the point P. The area of the triangles APD and BPC are 27 and 12, respectively. If the areas of the triangles APB and CPD are equal then the area of triangle APB is
A. 12
B. 15
C. 16
D. 21
E. 18 [ XAT 2008 ]

Ans : E

Q6) ABCD is a square with sides of length 10 units. OCD is an isosceles triangle with base CD. OC cuts AB at point Q and OD cuts AB at point P. The area of trapezoid PQCD is 80 square units. The altitude from O of the triangle OPQ is:
A. 12
B. 13
C. 14
D. 15 [ XAT 2009 ]

PQ=6 and then apply similarity in triangle OPQ and OCD
Answer is D ( 15 )

Q7) Two circles with radii 15cm and 20cm intersect each other at two points A and B such that AB is 25cm. Find the difference of non-common area of two circles.

It is simply the difference of areas of the two circles i.e. pi * 20^2 - pi * 15^2 = 175 * pi = 550 sq cm.

Q8) A point P lies inside unit square ABCD such that angle(APB) is acute while angle(BPC) is obtuse. Find the area of the locus of point P.

If we draw two semicircles on AB and BC as diameters, then P must lie inside the semicircle on BC but outside the semicircle on AB.
So the required area comes out to be 1/4.

Q9) Arc AB subtends an angle of 120 degrees on the center of the circle, O.
Q is a point in the major sector of arc AB. Which of the following may be true?
I. angle(AQB) > 120
II. angle(AQB) = 120
III. angle(AQB) < 120
a. III only
b. both II and III
c. both I and II
d. I, II and III

Answer : D i.e. I, II and II all may be true.

Q10) S is a set of first 100 natural numbers. 13 numbers are selected randomly. At least 'n' of 13 numbers give same remainder when divided by 5. What is the least value of 'n'?

There are 5 possible different remainders i.e. 0, 1, 2, 3, 4 and to get the least number with same remainders, we should distribute the 13 numbers as widely as possible. In WORST scenario, we'll have at least ONE REMAINDER which is being shared by three numbers. Hence the answer will be 3.

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