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Mathmetics

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

Q21.	The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel ?

1)	450 ft	2)	825 ft
3)	750 ft	4)	900 ft
5)	None of these

Answer : 900 ft
Explanation :

The circumference of the front wheel is 30 ft and that of the rear wheel is 36 feet.

Let, the rear wheel make n revolutions. At this time, the front wheel should have made n+5 revolutions.

As both the wheels would have covered the same distance,

n x 36 = (n+5) x 30

or 36n = 30n + 150

or 6n = 150

or n = 25.

So, distance covered = 25 x 36 = 900 ft

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Q22.	If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter ?

1)	42 cm	2)	48 cm
3)	52 cm	4)	64 cm
5)	None of these

Answer : 52 cm
Explanation :

Let, the diagonals be PR and SQ and O is the centre ,such that PR = 24 cm and SQ = 10 cm

PO = OR = 24/2=12 cm SO = OQ = 10/2=5 cm PQ = QR = RS = SP = 122+52 =√144+25 = √169=13 cm

So, perimeter = 4 � 13 = 52 cm

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Q23.	An error 12% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square ?

1)	20.04 %	2)	24 %
3)	25.02 %	4)	25.44 %
5)	None of these

Answer : 25.44 %
Explanation :

Let, the correct value of the side of the square = 100

Then, the measured value = 100�(100+12)/100=102 (∵ error12% in excess)

Correct Value of the area of the square = 100 � 100 = 10000

Calculated Value of the area of the square = 112 � 112 = 12544

Error = 12544 - 10000 = 2544

So,percentage of Error = (2544/10000)�100 = 25.44%

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Q24.	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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Q25.	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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Q26.	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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Q27.	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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Q28.	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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Q29.	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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Q30.	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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Mathmetics

Q21.

The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel ?

Q22.

If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter ?

Q23.

An error 12% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square ?

Q24.

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 ?

Q25.

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 ?

Q26.

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 ?

Q27.

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 :

Q28.

Each side of a rhombus is 26 cm and one of its diagonals is 48 cm long. The area of rhombus is :

Q29.

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' ?

Q30.

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 ?

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