Quantitative Aptitude - Arithmetic Ability Questions

Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms

Let the Charminar Express takes 't' hrs to catch Chennai express

=> 80 x t = 60 x (2 + t)

=> 80 t = 120 + 60t

=> t = 6 hrs

Therefore, the distance away from Hyderabad the two trains meet = 80 x 6 = 480 kms.

Let cost of each cell phone = Rs.100 & Sale = 100 phones.

Money receipt = Rs.(100 x 100) = Rs.10,000

New cost per cellphone = Rs.120 and New sale = 70 phones

New Money Receipt = Rs.(70 x 120) = Rs.8400

Then the effect of sales = decrease in money receipt = 

160010000 x 100 = 16%.

Cost of 4 mangoes = Rs. 6 

Cost of 1 mango = Rs. 6/4 = Rs. 1.5

Now, Selling Price of 4 mangoes = Rs. 4

Selling Price of 1 mango = Rs. 4/4 = Rs. 1.0

• We observe that, S.P < C.P

Hence, the shopkeeper incurs a loss.

Required loss% = loss/C.P x 100

Loss = 1.5 - 1 = 0.5

C.P = 1.5

Loss% = (0.5/1.5) x 100

 = (5/15) x 100

 = 1/3 x 100

 = 100/3

 = 33.333%.

$$\begin{gathered} 78 x 448 + 67 x 3374 \\ => 392 + 2892 \\ => 3284 \end{gathered}$$

As these three strategies P, Q and R are independent so these will cut cost one after the other.
If initial cost is Rs 100, then
20% cost is cut after initializing strategy P, then cost will remain 80% = 80
further 30% cost is cut after strategy Q, then cost will remain 70% of 80 = 56
further 10% cost is cut after strategy R, then cost will remain 90% of 56 = 50.4

Thus final cost remains 50.4 % of the original cost. Hence net saving is 100 - 50.4 = 49.6 %.

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

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14

= C414 ways = 14×13×12×114×3×2×1= 1001

Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

  ⇒C15+ C44+ C15 = 5+1+5 =11

 P(E) = n(E)n(S)=111001=191  

 Required probability = 1-191=9091

K = 100 L = 130
130------ 30
100------- ? => 23 1/13 %