Quantitative Aptitude - Arithmetic Ability Questions

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

Compound Interest for 1 ¹⁄₂ years when interest is compounded yearly = Rs.(5304 - 5000)

Amount after 1¹⁄₂ years when interest is compounded half-yearly 


Compound Interest for 1 ¹⁄₂ years when interest is compounded half-yearly = Rs.(5306.04 - 5000)

Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000)= 5306.04 - 5304 = Rs. 2.04

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:j=7% compounded semiannually making m=2 and i = j/m= 7%/2 = 3.5%
Let x represent the third payment. Then the second payment must be 2x.
PV1,PV2, andPV3 represent the present values of the first, second, and third payments.

Since the sum of the present values of all payments equals the original loan, then
PV1 + PV2  +PV3
=$4000 -------(1)

PV1
  =FV/(1 + i)^n
=$1000/(1.035)^4=
$871.44

At first, we may be stumped as to how to proceed for
PV2 and PV3. Let’s think about the third payment of x dollars. We can compute the present value of just $1 from the x dollars

pv=1/(1.035)^10=0.7089188

PV2
  =2x * 0.7089188 =
1.6270013x
PV3
  =x * 0.7089188=0.7089188x
Now substitute these values into equation ➀ and solve for x.
$871.442 + 1.6270013x + 0.7089188x
=$4000

2.3359201x
=$3128.558

x=$1339.326
Kramer’s second payment will be 2($1339.326)
=$2678.65, and the third payment will be $1339.33

Given that train K crosses a pole in 30 seconds and train L crosses the same pole in one minute and 15 seconds.

Let the length of train K be Lk and that of train L be Ll

given that Lk = 3/4 Ll

As the train K and L crosses the pole in 30 seconds and 80 seconds respectively,
=> Speed of train K = sk = Lk/30

Speed of train L = sl = Ll/80

Lk = 3/4 Ll
=> sk = 3/4 Ll/(30) = Ll/40

Ratio of their speeds = sk : sl
= Ll/40 : Ll/80

=> 1/40 : 1/80  =  2 : 1

Given total number of students in the class = 21

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

It takes four of those little squares -- each one 1/2 inch on a side -- to cover one square inch. That's because each one is 1/2 x 1/2, or 1/4 of a square inch.
In a rectangle that is 48 inches by 48 inches, there are 2304 square inches.(1 ft = 12 inches)
Since it takes 4 little squares to cover 1 sq inch, then it would need 2304 x 4 squares (1/2 inch on a side each) to cover a rectangle 4 feet by 4 feet.

That is 9,216 of those squares.

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