Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q461. |
|
| 1) | 2 times | 2) | 3 times |
| 3) | 4 times | 4) | 5 times |
| 5) | None of these | ||
Answer : 2 times
Explanation :
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.
View AnswerQ462. |
|
| 1) | 10 yrs | 2) | 12 yrs |
| 3) | 16 yrs | 4) | 20 yrs |
| 5) | None of these | ||
Answer : 20 yrs
Explanation :
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.
View AnswerQ463. |
|
| 1) | 6 yrs | 2) | 10 yrs |
| 3) | 12 yrs | 4) | 20 yrs |
| 5) | None of these | ||
Answer : 12 yrs
Explanation :
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
View AnswerQ464. |
|
| 1) | 28 | 2) | 32 |
| 3) | 40 | 4) | 45 |
| 5) | None of these | ||
Answer : 40
Explanation :
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
View AnswerQ465. |
|
| 1) | 90 cm | 2) | 100 cm |
| 3) | 1 m | 4) | 1.1 cm |
| 5) | None of these | ||
Answer : 100 cm
Explanation :
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 .
View AnswerQ466. |
|
| 1) | 90 m | 2) | 80 m |
| 3) | 90 m | 4) | 120 m |
| 5) | None of these | ||
Answer : 120 m
View AnswerQ467. |
|
| 1) | 8 cm | 2) | 9 cm |
| 3) | 12.5 cm | 4) | 18 cm |
| 5) | None of these | ||
Answer : 18 cm
Explanation :
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.
View AnswerQ468. |
|
| 1) | 1600 cm2 | 2) | 2400 cm2 |
| 3) | 3600 cm2 | 4) | 4800cm2 |
| 5) | None of these | ||
Answer : 2400 cm2
Explanation :
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.
View AnswerQ469. |
|
| 1) | 80% | 2) | 88% |
| 3) | 110% | 4) | Can't be determined |
| 5) | None of these | ||
Answer : 88%
Explanation :
Explanation :
According to AB Methods:
New area = -20+10 - (20x10)/100 = -12%
So, required percentage
= (100 - 12 ) = 88%.
View AnswerQ470. |
|
| 1) | 48 | 2) | 56 |
| 3) | 76 | 4) | 88 |
| 5) | None of these | ||
Answer : 88
Explanation :
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.
View Answer
