Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q591. |
|
| 1) | 76 kg | 2) | 76.5 kg |
| 3) | 85 kg | 4) | Data Inadequate |
| 5) | None of these | ||
Answer : 85 kg
Explanation :
Explanation :
Total weight increased = (8 x 2.5) kg = 20 kg.
So,weight of new person = (65 + 20) kg = 85 kg.
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Q592. |
|
| 1) | 35 years | 2) | 40 years |
| 3) | 50 years | 4) | Data Inadequate |
| 5) | None of these | ||
Answer : 40 years
Explanation :
Explanation :
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
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Q593. |
|
| 1) | 34 | 2) | 31 |
| 3) | 36 | 4) | 32 |
| 5) | None of these | ||
Answer : 36
Explanation :
Explanation :
Let the batsmans average runs up to 12th innings was x.
Total run = 12 x.
Now total run up to 13th innings = (12 x + 96)
According to equation = (12 x + 96)/13 = x + 5 or x = 31.
So, after 13th innings average will be 31 + 5 = 36.
Shortcut Method :
96 - (135) + 5 = 36
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Q594. |
|
| 1) | 500 | 2) | 600 |
| 3) | 550 | 4) | 480 |
| 5) | None of these | ||
Answer : 550
Explanation :
Explanation :
By alligantion method :
12000 3750
6000
6000 - 3750 = 2250 12000 - 6000 = 6000
So, final ratio = 2250 : 6000 = 3 : 8
So,total number of employees = (150/3) x 11 = 550
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Q595. |
|
| 1) | 3:5 | 2) | 1:1 |
| 3) | 4:5 | 4) | D.I |
| 5) | None of these | ||
Answer : 1:1
Explanation :
Explanation :
With the help of alligation method :
50 40
45
45 - 40 = 5 50 - 45 = 5
So, final ratio = 5 : 5 = 1 : 1
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Q596. |
|
| 1) | 279 | 2) | 283 |
| 3) | 308 | 4) | 318 |
| 5) | None of these | ||
Answer : 308
Explanation : We know that H.C.F x L.C.M = Product of two numbers
Explanation : We know that H.C.F x L.C.M = Product of two numbers
So,other number = (11 x 7700)/275 = 308.
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Q597. |
|
| 1) | 1677 | 2) | 1683 |
| 3) | 2523 | 4) | 3363 |
| 5) | None of these | ||
Answer : 1683
Explanation :
Explanation :
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
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Q598. |
|
| 1) | 196 | 2) | 630 |
| 3) | 1260 | 4) | 2520 |
| 5) | None of these | ||
Answer : 630
Explanation :
Explanation :
L.C.M. of 12, 18, 21 30 = 1260.
Required number = (1260 χ 2) = 630.
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Q599. |
|
| 1) | 1 | 2) | 2 |
| 3) | 3 | 4) | 4 |
| 5) | None of these | ||
Answer : 2
Explanation :
Explanation :
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
So, the required numbers are (13 x 1, 13 x 12) and
(13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
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Q600. |
|
| 1) | 20 | 2) | 23 |
| 3) | 169 | 4) | 400 |
| 5) | None of these | ||
Answer : 23
Explanation :
Explanation :
Let the numbers be x and y.
Then, xy = 120 and x2 + y2 = 289.
(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529
x + y = 23.
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