Quantitative Aptitude - Arithmetic Ability Questions

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

I = (P x T x R) /100

Sum = Rs.(50*100)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

 C.I = Rs.(551.25-500)= Rs.51.25

Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Let Althaf lent Rs. A at 14% per year.
Hence, Money lent at 12% = (1500-A);
Given, total interest = Rs. 186.
{(A x 14x 1)/100} + {[(1500-A) x 12 x 1/100]} = 186;
14A/100 + (18000 -12A)/100 = 186;
14A + 18000 - 12A = 186x100;
2A = 18600-18000;
A = 600/2 = Rs. 300.
Hence, money lent at 12% = 1500-300 = Rs. 1200.

Let t be the usual time. If the usual rate is 1 unit, then the unusual rate is 6/7.
Balance by the distance,
1 x t = (6/7)(t + 30)
7t = 6(t + 30)
we get t as,
t = 180 minutes = 3 hrs.