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Mathmetics

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

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Q471.

Sachin was twice as old as Ajay 10 years back. How old is Ajay today if Sachin will be 40 yrs. old after 10yrs. ?

1)	10 yrs	2)	12 yrs
3)	16 yrs	4)	20 yrs
5)	None of these

Answer : 20 yrs
Explanation :

Let, age of Ajaya 10 years back was x years.

Then, Sachin�s age 10yrs. was 2x yrs.

So, 2x + 20 = 40

or x = 10 yrs.

So, the present age of Ajay = 10 +10 = 20 yrs.

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Q472.

The sum of the ages of a son and his father is 56yrs. After four yrs, the age of father will be three times that of the son . Find the age of the son ?

1)	6 yrs	2)	10 yrs
3)	12 yrs	4)	20 yrs
5)	None of these

Answer : 12 yrs
Explanation :

Let the age of the son and father be x yrs. and y yrs. respectively.

Then , x + y = 56 yrs. ---- I

After 4yrs. age of both will be 56 + 8 = 64yrs.

But according to hypothesis,

or 3 (x + 4) = y + 4 or 3x + 12 = y + 4 or 3x - y = - 8 -------- II

Solving both equations, we get x = 12.

SHORT-CUT METHOD:

[64/3+1] - 4 = 12 yrs

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Q473.

The age of A's father is four times of his son. If 5 yrs ago, father's age was seven times of his son at that time. What is the present age of A's father ?

1)	28	2)	32
3)	40	4)	45
5)	None of these

Answer : 40
Explanation :

Let age of son be x.Then father's age be 4x.

Five years ago age of boths are x - 5 and 4x - 5 respectly.

or 7 (x - 5 ) = 4x - 5

or 7x - 35 = 4x - 5 or 7x - 4x = 30 or 3x = 30 or x = 10

Therefore, Father's age = 10 x 4 = 40 years

SHORT-CUT METHOD :

Father's age = [5 x(7-1)] /(4-1) x 4 = 40 yrs

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Q474.

A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

1)	90 cm	2)	100 cm
3)	1 m	4)	1.1 cm
5)	None of these

Answer : 100 cm
Explanation :

Let the thickness of the bottom be x cm.

Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000

or 320 x 250 x (110 - x) = (8000 x 1000)

(110 - x) = (8000 x 1000)/(320 x 250) or x = 100

So, x = 100 cm .

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Q475.

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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Q476.

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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Q477.

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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Q478.

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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Q479.

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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Q480.

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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Previous 1 2...45 46 474849 50 51...63 64 Next

Mathmetics

Q471.

Sachin was twice as old as Ajay 10 years back. How old is Ajay today if Sachin will be 40 yrs. old after 10yrs. ?

Q472.

The sum of the ages of a son and his father is 56yrs. After four yrs, the age of father will be three times that of the son . Find the age of the son ?

Q473.

The age of A's father is four times of his son. If 5 yrs ago, father's age was seven times of his son at that time. What is the present age of A's father ?

Q474.

A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

Q475.

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 ?

Q476.

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 :

Q477.

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

Q478.

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' ?

Q479.

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 ?

Q480.

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.

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