Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q41. |
|
1) | Rs.6400 | 2) | Rs.6500 |
3) | Rs.7200 | 4) | Rs.7500 |
5) | None of these |
Answer : Rs.6400
Explanation :
Explanation :
Explanation:
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
Then, (x x 14 x 2)/100 + ((13900 - x) x 11 x 2) /100= 3508
28x - 22x = 350800 - (13900 x 22)
or 6x = 45000
or x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Q42. |
|
1) | I only | 2) | I and II only |
3) | II and III only | 4) | I and III only |
5) | None of these |
Answer : I only
Explanation :
Explanation :
Explanation:
P = Rs. 5000 & T = 2 years.
I. S.I. on Rs. 5000 in 5 years is Rs. 2000.
(5000 x R x 5)/100 = 2000 or R = 8.
Thus I only gives the answer.
Correct answer is (1).

Q43. |
|
1) | I and II only | 2) | II and III only |
3) | All I, II and III | 4) | Any two of the three |
5) | None of these |
Answer : All I, II and III
Explanation :
Explanation :
Explanation:
I. Material cost = Rs. 2.50 per m2
II. Labour cost = Rs. 3500.
III. Total cost = Rs. 14,500.
Let the area be A sq. metres.
Material cost = Rs. (14500 - 3500) = Rs. 11,000.
5A/2 = 11000 or A = (11000 x 2 )/5 = 4400 m2.
Thus, all I, II and III are needed to get the answer.
Correct answer is (3).

Q44. | A person invested some amount at the rate of 12% simple interest and a certain amount at the rate of 10% simple interest. He received yearly interest of Rs. 130. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest. How much did he invest at 12% simple interest ? |
1) | Rs.400 | 2) | Rs.500 |
3) | Rs.700 | 4) | Rs.800 |
5) | None of these |
Answer : Rs.700
Explanation :
Explanation :
Explanation :
Let the amount invested at 12% be x and the amount invested at 10% be y respectively
( x x 12 x 1)/100 +( y x 10 x 1)/100 = 130
1.2x + y = 1300
Similarly, 1.2y + x = 1340
or (1.2x) x 1.2 + 1.2y = 1300 x 1.2
x +1.2y = 1340
or 0.44x = 1560 - 1340 = 220
or x = (220/0.44) - Rs. 500
or y = 1300 - 1.2 x 500 = Rs. 700

Q45. | A person invested some amount at the rate of 12% simple interest and a certain amount at the rate of 10% simple interest. He received yearly interest of Rs. 130. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest. How much did he invest at 12% simple interest ? |
1) | Rs.400 | 2) | Rs.500 |
3) | Rs.700 | 4) | Rs.800 |
5) | None of these |
Answer : Rs.700
Explanation :
Explanation :
Explanation :
Let the amount invested at 12% be x and the amount invested at 10% be y respectively
( x x 12 x 1)/100 +( y x 10 x 1)/100 = 130
1.2x + y = 1300
Similarly, 1.2y + x = 1340
or (1.2x) x 1.2 + 1.2y = 1300 x 1.2
x +1.2y = 1340
or 0.44x = 1560 - 1340 = 220
or x = (220/0.44) - Rs. 500
or y = 1300 - 1.2 x 500 = Rs. 700

Q46. |
|
1) | 36 pie m3 | 2) | 48 pie m3 |
3) | 50 pie m3 | 4) | 54 pie m3 |
5) | None of these |
Answer : 36 pie m3
Explanation :
Explanation :
Explanations:
Volume of the oil = 22/7x 62 x 14 = 504 x 22/7 m3.
Base radius of all the containers = 7 m
Volume of the spherical can = (4/3) x (22/7)x 63 m3.
Volume of the conical can = (1/3) x (22/7) x 62 x 6 m3 = 72 x (22/7)
Balance oil = Volume of the cylindrical can
= 504 (22/7) - 288 (22/7)- 72(22/7) = 144(22/7) m3.
Oil dropped = 1/4 th of the cylindrical can
= 1/4x 144 = 36 (22/7) m3.

Q47. |
|
1) | 2 : 3 | 2) | 1 : 3 |
3) | 1 : 2 | 4) | 1 : 4 |
5) | None of these |
Answer : 1 : 2

Q48. | If 2x2 -7xy +3y2 = 0 ,then the value of x:y = ? |
1) | 3:2 | 2) | 2:3 |
3) | 3:1 and 1:2 | 4) | 5:6 |
5) | None of these |
Answer : 3:1 and 1:2
Explanation : Given, 2x2 -7xy +3y2 = 0
Explanation : Given, 2x2 -7xy +3y2 = 0
or 2(x/y)2-7(x/y)+ 3 = 0
So, x/y=3/1 and x/y=1/2

Q49. | If a,b and c are real numbers such that a-b+c <1 , a+b+c >1 and 9a+3b+c < -4, then what is the sign of 'a'? |
1) | + only | 2) | - only |
3) | + or - | 4) | Can't be said |
5) | None of these |
Answer : - only

Q50. | If a,b and c are the two roots of a quadratic equation such that a+b =24 and a-b = 8, the quadratic equation having a and b as its roots is ---. |
1) | x2+2x+8=0 | 2) | x2-4x+8=0 |
3) | x2-24x+128=0 | 4) | 2x2+8x+9=0 |
5) | None of these |
Answer : x2-24x+128=0
Explanation : Given, a+b =24 and a-b = 8
Explanation : Given, a+b =24 and a-b = 8
So,solving from two equations,we get a =16 and b=8
Therefore,ab=16 x 8 = 128
Quadratic equations
x2 - ( sum of roots)x + products of the roots = 0
or x2-24x+128 = 0
