Day-10 | Bank Aptitude Questions and Answers with Explanation

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-1[/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

13/7 *3/8* 4/4 of what =4240.03

Options :

a)3599
b)3479
c)4599
d)5566

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Answer:   b) 3479

((4240.03*7*8*4)/(13*3*7)=3479

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-2 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

(0.07*0.07+0.06*0.06*0.06)/0.07*0.07-0.070.04-0.04 0.04=?

Options :

a)0.11
b)0.09
c)0.14
d)0.13

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Answer:   d)0.13

(0.07*0.07+0.06*0.06*0.06)/0.07*0.07-0.070.04-0.04 0.04=?
This expression is similar(a^3+b^2)/(a^2-a*b-b^2)=
(a+b)
Where a=0.07 b=0.06=0.13

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-3 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

0.857*0.8570.857*0.143+0.143*0.143/0.857*0.857*0.857*0.143*0.143*0.143=?

Options :

a)0.1
b)0.01
c)1
d)0.0001

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Answer:   c)1

This expression is similar to (a^2-ab+b^2)/(a^2+b^2)=1/a+b
Here a=0.857 and b=0.143 =1/1=1.

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-4 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

Coromandel express train -812 metre long crosses a telephone pole in 28 seconds. There is a platform of length which Coromandel Express train covers in 32 seconds. Mr. Kuppusamy a regular train passenger crosses the same platform in 9 minutes and 15 seconds. Identify the speed of Mr. Kuppusamy in metre/second?

Options :

a)1.87 m/sec
b)1.37 m/sec
d)1.99 m/sec
c)1.67 m/sec

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Answer:   d) 1.67 m/sec

A train crosses a telephone pole taking the time it takes to cross its own length.
The speed of coromandel Express train =812/28= 29 m/sec
The length of the platform is the distance the train takes to cover in 32 sec.
Platform length =32*29=928m.
This distance is covered by Kuppusamy in 9 min and 15 sec=555 sec.
So speed of kuppuswamy = 928/555=1.67 m/sec.

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-5 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

In an elementary school in Pozhichalur near Chennai there are 60 students out of whom 15 per cent are girls. Each girl’s monthly fee is Rs. 250 and each boy’s monthly fee is 34 per cent more than a girl’s fee. Calculate the total monthly fee of girls and boys together?

Options :

a)Rs.21335
b)Rs.20335
c)Rs.19335
d)Rs.19355

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Answer:  c) Rs. 19335

Total number of students=60.
Number of girls=15%=60*15/100=9
Number of boys=60-9=51
Fees paid by the girls=250×9=Rs.2250
Fees paid by boys is 34% more than that paid by girls that is (250*1.34/100)
Multiplied by the base figure.)
Fees paid by boys=51*250*1.34=Rs.17085
Total fees paid=17085+2250=Rs.19335.

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-6 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

Adhavaitha Playground is circular in shape and its circumference is twice the perimeter of Govinda Park-a circumference I stwice the perimeter of govinda Park-a rectangular park. Ara of Adhvaitha Playground is 5544 sq.m. The length of Govinda Park rectangle is 40 meters. Calculate the area of Govinda park.

Options :

a)1040 sq.m
b)1140 sq.m
c)1240 sq.m
d)1880 sq.m.

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Answer:  a) 1040 sq.m

Area of Advaitha playground (circular in shape)+ 5544 sq.m.
Radius of adhvaitha playground=22/7*r^2=5544
(where R is the radius)
R^2=5544*7/22=1764
R=radius=√1764=42m
Circumference of Adhvaitha Playground=2*22/7*42=264
It is given that the circumference is twice the area of perimeter of rectangular park
So perimeter of govinda Park=264/2=132m
Perimeter=2(length+breadth)
It is given that the length is 40 metres
So breadth of govinda Park={132-(2*40)/2=52/2=26m
Area of govinda park=length*breadth=40*26=1040 sq.m

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-7 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

√(12^2*16/24+193+7*5)=(?)^2

Options :

a)3√2
b)4√2
c)5√2
d)18

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Answer:  d)18

Apply BODMAS principle
√(12^2*16/24+193+7*5)
=√(12^2*2/3+193+7*5)
=√(12*4*2+193+35)
=√(96+193+35)
=√(324)=18^2
So answer is d) 18.

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-8 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

31.36/0.64*252=(?)^2*36

Options :

a)81
b)64
c)-7
d)-11

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Answer: c) -7

Applying BODMAS principle
√31.36/√0.64*252
=5.6/0.8*252=(?)^@*36
(?)^2=(5.6*252)/0.8*36
=Therefore?=√49=+or-7
Hence answer is c)-7.

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-9 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

Mr. Srinvasaraghavan bought a land in Perunagalathur near Tambaram and constructed a factory for making toys. He made toys Triangular in shape. The sides of the triangle are 50 cm, 78 cm and 112 cm. identify the smallest altitude of such triangle shaped toys?

Options :

a)20cm
b)45cm
c)75cm
d)30cm

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Answer: d) 30 cm

Semi-perimeter of triangle=(50+78+112)/2=120cm
Area of triangle=√s(s-a)(s-b)(s-c)=√[120(120-50)120-78)(120-112)] =√(120*70*42*8)
=1680 sq.cm.
There fore the altitude will be smallest when base is largest—112 cm
½ * 112*H=1680
So H=(1680*2)112=30cm

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[divider] [tabs type=”horizontal”][tabs_head][tab_title]Question-10 [/tab_title][tab_title]Explanation[/tab_title][/tabs_head] [tab]

Mr. Venkatachalapathy of Tirupati bought a land in Srikalhasti rectangular in shape of length 18 m and width 15m. A pit 7.5 m long, 6m broad and 0.8 m deep is dug in a corner of the field and the earth taken out is evenly spread over the remaining area of the field. The level of the field raised is how much?

Options :

a)10cm
b)16cm
c)12cm
d)24cm

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Answer: b) 16cm

Rectangular plot measures 18m length and width 15 cm.
The dimensions of pit dug in a corner of the field= 7.5m*6m*0.8m
Volume of earth taken out of the pit=(7.5m*08m)
Cubic metre=36 cu. Metre
Area of the remaining field (18*15-7.5*6)sq.metre=270-45=225 sq.metre
Therefore level of the field raised=36/225 metre=16cm.

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