Quantitative aptitude Question for all banking Bank PO,Clerk,IBPS PO,Railway,SSC,PSUs,IAS,OAS ,C-SAT Exams 

Mathmetics

ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

Q21.
A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet ?
1) 96 sq.ft. 2) 100 sq.ft.
3) 126 sq.ft. 4) 144 sq.ft.
5)None of these
Answer : 126 sq.ft.
Explanation :
Given, l = 9 ft.
Then, l + 2b = 37
=> 2b = 37 - l = 37 - 9 = 28
=> b = 28/2 = 14 ft.
So, area = lb = 9 × 14 = 126 sq. ft.
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Q22.
A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required ?
1) 72 feet 2) 84 feet
3) 88 feet 4) 96 feet
5)None of these
Answer : 88 feet
Explanation :
Given, the area of the field = 680 sq. feet
 lb = 680 sq. feet
Length = l = 20 feet
 20 × b = 680
⇒ b=680/20 = 34 feet
So, required length of the fencing = l + 2b  = 20 + (2 × 34) = 88 feet

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Q23.
The length of a plot of land is 4 times it breadth. A playground measuring 1200 sq m occupies one-third of the total area of the plot. What is the length of the plot, in metres ?
1) 90 m 2) 80 m
3) 90 m 4) 120 m
5)None of these
Answer : 120 m
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Q24.
The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :
1) 8 cm 2) 9 cm
3) 12.5 cm 4) 18 cm
5)None of these
Answer : 18 cm
Explanation :
A1=  [(1/2) x 15 x 12]cm2 = 90 cm2 . A2 = 2A1 = 180 cm2.
 = (1/2) x 20 x h = 180  or h = 18 cm.
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Q25.
Each side of a rhombus is 26 cm and one of its diagonals is 48 cm long. The area of rhombus is :
1) 1600 cm2 2) 2400 cm2
3) 3600 cm2 4) 4800cm2
5)None of these
Answer : 2400 cm2
Explanation :
AB = 26 cm and AC = (1/2) x 48 cm   ; OA = 24 cm
OB2 = AB2 - OA2 = (26)2 - (24)2 = (26 + 24) (26 - 24) = 100
or OB = 50 cm   or BD = 2 x OB = (250) cm = 100 cm.
 Therefore , Area =  1/2 x (AC x BD ) =  [(1/2 ) x 48 x100 ] cm2 = 2400 cm2.
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Q26.
A rectangle has width a and length b. If the width is decreased by 20% and the length is increased by 10%, then what is the area of the new rectangle in percentage compared to 'ab' ?
1) 80% 2) 88%
3) 110% 4) Can't be determined
5)None of these
Answer : 88%
Explanation :
According to  AB Methods: 
New area = -20+10 - (20x10)/100 = -12%
So, required percentage 
= (100 - 12 ) = 88%.
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Q27.
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required ?
1) 48 2) 56
3) 76 4) 88
5)None of these
Answer : 88
Explanation :
We have, L = 20 ft and LB = 680 sq.ft 
So, b = 680/20 = 34 ft.
Length of fencing (L+2B) = (20 + 68) ft = 88 ft.

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Q28.
A hall of length 24 cm and breadth 20 m. is to be paved with equal square tiles. What will be the size of the largest tile so that tiles exactly fit and  also find the number of tiles required.
1) 4cm, 20 2) 6cm, 10
3) 2cm, 20 4) 4cm, 30
5)None of these
Answer : 4cm, 30
Explanation :
Size of the largest possible square tile 
= H.C.F ( 24, 20) = 4 m.
So,number of tiles required [(24 x 20)/(4x4) ]=30
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