Bank PO Questions

Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

 A) 1 B) 2 C) 3 D) 4

Explanation:

Area of the park = (60 x 40) = 2400${m}^{2}$

Area of the lawn = 2109${m}^{2}$

Area of the crossroads = (2400 - 2109) = 291${m}^{2}$

Let the width of the road be x metres. Then,

$60x+40x-{X}^{2}=291$

${x}^{2}-100x+291=0$

(x - 97)(x - 3) = 0
x = 3.

17 15502
Q:

A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum.

 A) 4360 B) 4460 C) 4560 D) 4660

Explanation:

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

45 15149
Q:

The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. if the interest were compounded half yearly , the difference in two interests would be nearly

 A) Rs.24.64 B) Rs.21.85 C) Rs.16 D) Rs.16.80

Explanation:

For 1st year S.I =C.I.

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200

i.e S.I on the principal for 1 year is Rs.200

Principle = $Rs.\frac{100*200}{8*1}$ = Rs.2500

Amount for 2 years, compounded half-yearly

$Rs.\left[2500*{\left(1+\frac{4}{100}\right)}^{4}\right]=Rs.2924.4$

C.I = Rs.424.64

Also, $S.I=Rs.\left(\frac{2500*8*2}{100}\right)=Rs.400$

Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64

24 15091
Q:

Find the ratio of the areas of the incircle and circumcircle of a square.

 A) 1:1 B) 1:2 C) 1:3 D) 1:4

Explanation:

Let the side of the square be x. Then, its diagonal = $\sqrt{2{x}^{2}}=\sqrt{2x}$

Radius of incircle = $\frac{x}{2}$

Radius of circum circle= $\sqrt{2}×\frac{x}{2}=\frac{x}{\sqrt{2}}$

Required ratio = $\frac{{\mathrm{\pi x}}^{2}}{4}:\frac{{\mathrm{\pi x}}^{2}}{2}=\frac{1}{4}:\frac{1}{2}=1:2$

36 15060
Q:

At what rate percent per annum will a sum of money double in 8 years.

 A) 12.5% B) 13.5% C) 11.5% D) 14.5%

Explanation:

Let principal = P, Then, S.I.= P and Time = 8 years

We know that S.I. = PTR/100

Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.

37 14752
Q:

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

 A) 110 B) 120 C) 130 D) 140

Explanation:

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

43 14551
Q:

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

 A) 20% B) 30% C) 40% D) 50%

Explanation:

Let original length = x and original breadth = y.

Original area = xy.

New length = x/2 and New breadth=3y

New area = $\frac{3}{2}xy$

Therefore,  Increase in area = New area-original area = $\frac{3}{2}xy-xy=\frac{1}{2}xy$

Therefore,  Increase % =  %

14 13457
Q:

What is the rate of interest p.c.p.a.?

I. An amount doubles itself in 5 years on simple interest.

II. Difference between the compound interest and the simple interest earned on a certain amount in 2 years is Rs. 400.

III. Simple interest earned per annum is Rs. 2000

 A) I only B) II and III only C) All I, II and III D) I only or II and III only

Answer & Explanation Answer: D) I only or II and III only

Explanation:

$I.\frac{P*R*5}{100}=P⇔R=20$

$II.P{\left(1+\frac{R}{100}\right)}^{2}-P-\frac{P*R*2}{100}=400=>p{R}^{2}=4000000$

$III.\frac{P*R*1}{100}=2000=>PR=200000$

$\frac{P{R}^{2}}{PR}=\frac{4000000}{200000}⇔R=20$

Thus I only or (II and III) give answer.