Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q81. |
|
1) | 1 : 6 | 2) | 1 : 5 |
3) | 1 : 25 | 4) | 1 : 125 |
5) | None of these |
Answer : 1 : 5
Explanation :
Explanation :
The volume of the larger cube = 53 = 125 cm3.
The volume of each of the smaller cubes = 13 = 1 cm3.
Therefore, one would get 125 smaller cubes.
The surface area of the larger cube = 6a2 = 6(52) = 6 25 = 150
The surface area of each of the smaller cubes = 6 (12) = 6.
Therefore, surface area of all of the 125, 1 cm3 cubes = 125 6 = 750.
Therefore, the required ratio = 150 : 750 = 1 : 5

Q82. |
1) | 8 | 2) | 9 |
3) | 10 | 4) | 12 |
5) | None of these |
Answer : 10
Explanation :
Explanation :
Given, x2 + 9/x = 6x
Or x2 - 6x = 9
Or (x-3)2 = 0
Or x = 3
So, x2 + 9/x2 = 10

Q83. |
|
1) | 221 | 2) | -231 |
3) | -253 | 4) | 0 |
5) | None of these |
Answer : -253
Explanation :
Explanation :
Given, a = 4/3
Or 3a = 4
Or 3a - 4 = 0
Or (3a-4)3 = 27a3 - 108a2 + 144a - 64 = 03
Or 27a3 - 108a2 + 144a - 317 = (27a3 - 108a2 + 144a -64) - 153 = 0 - 253 = - 253

Q84. | A fort has provisions for 250 soldiers for 35 days. After 5 days a reinforcements of 25 soldiers were arrived. Again after 10 days, 25 soldiers left the fort. How long will the remaining provisions lasting ? |
1) | 12 | 2) | 16 |
3) | 18 | 4) | 19 |
5) | None of these |
Answer : 19
Explanation :
Explanation :
Short- Cut Methods:
ND (No. of days) = N1D1 + N2D2 = total amount of food after 5 days.
or ( 250 X 30) = ( 250 + 25) X 10 + 250 X x
or x = 19 days .

Q85. |
|
1) | 33 years | 2) | 40 years |
3) | 43 years | 4) | 48 years |
5) | None of these |
Answer : 43 years
Explanation :
Explanation :
Present age of C = 85 – 7 = 78 years
Present age of B = 78 – 12 = 66 years
Present age of A = 3
11 66 = 18 years
Present age of A’s father = 18 + 25 = 43 years

Q86. |
|
1) | 100.8 sq cm | 2) | 112.4 sq cm |
3) | 104.2 sq cm | 4) | Cannot be determined |
5) | None of these |
Answer : 100.8 sq cm

Q87. | The ratio of the adjacent angles of a parallelogram is 7 : 8. Also, the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12. What is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? |
1) | 136° | 2) | 156° |
3) | 168° | 4) | 208° |
5) | None of these |
Answer : 168°
Explanation :
Explanation :
Sum of the adjacent angles of a parallelogram is 180°.
Smaller angle of the parallelogram = 7/15 × 180 = 84°
Second largest angle of the quadrilateral = 7/30 × 360 = 84°
Required sum = 84 + 84 = 168°

Q88. | The smallest side of a right-angled triangle is 8 cm less than the side of a square of perimeter 56 cm. The second largest side of the right-angled triangle is 4 cm less than the length of the rectangle of area 96 sq cm and breadth 8 cm. What is the largest side of the right-angled triangle ? |
1) | 10 cm | 2) | 12 cm |
3) | 15 cm | 4) | 18 cm |
5) | None of these |
Answer : 10 cm
Explanation :
Explanation :
Side of the square = 56/4 = 14 cm
smallest side of the triangle = 14 – 8 = 6 cm
Length of the rectangle = 96/ 8 = 12 cm
Second largest side of the triangle = 112 – 4 = 8 cm
So, Largest side of the triangle = square root of 62 + 82 = 10 cm

Q89. |
|
1) | 26 | 2) | 27 |
3) | 41 | 4) | 56 |
5) | None of these |
Answer : 41

Q90. |
|
1) | 2420 | 2) | 2880 |
3) | 3260 | 4) | 4840 |
5) | None of these |
Answer : 2880
Explanation :
Explanation :
Treating the group of four vowels (E, I O and U) as one distinct letter then total number of letters will be 8 - 4 + 1 = 5 [i.e. P, R, V, S and (E, I, O, U)].
The five letters can be arranged in 5! = 120 ways The vowels among themselves can be arranged in 4! = 24 ways,
The letters of the word ‘PREVIOUS’ can be arranged in 120 × 24 = 2880 ways such that vowels always come together.
