Bank PO Questions

Let C.P. of each Magazine be Rs. 1

C.P. of P articles = Rs. P. 

S.P. of P articles = Rs. 40.

Profit = Rs. (40 - P)

Now, Gain% = ProfitC.P×100 = 40-PP×100 

Here 40-PP×100= 25

-->P = 32.

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

G.C.D is nothing but H.C.F

Given numbers are 1.08 , 0.36 and 0.90

we can take as 108, 36 and 90 

Then required H.C.F is 18.

Therefore, H.C.F of given numbers is 0.18.

Let filling capacity of the pump be 'F' m3/min

Then, the filling capacity of the pump will be (F + 10) m3/min

From the given data,

$$\begin{gathered} 2400F - 2400F+10= 8 \\ ⇒ 2400 F+10-FF(F+10) =8 \\ ⇒ F2 + 10F - 3000 = 0 \\ ⇒ F = 50 m3/min \end{gathered}$$

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts.

Let Q complete that work in 'L' days

=> 1L + 1L-5 = 9100

=> 9L2-245L+500=0

L = 25 days.

If virus is K in 1hr i.e 60 min
then in 57 min it was K/2 as it doubles every 3 minutes
=> In 54 min it was K/4.
At 54 minutes it was K/4.