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

Q461.
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age ?
1) 2 times 2) 3 times
3) 4 times 4) 5 times
5)None of these
Answer : 2 times
Explanation :
Let, Ronit's present age be x years. 
Then, father's present age = (x + 3x) years = 4x years.
        (4x+8) = (x + 2) 5/2
or 8x + 16 = 5x + 40
or 3x = 24 or  x = 8.
Hence, required ratio = (4x+16) / (x+16) = 48/24 =2.
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Q462.
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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Q463.
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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Q464.
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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Q465.
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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Q466.
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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Q467.
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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Q468.
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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Q469.
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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Q470.
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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