Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q11. |
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| 1) | 72 | 2) | 80 |
| 3) | 100 | 4) | 120 |
| 5) | None of these | ||
Answer : 100
Explanation :
Explanation :
Let, the Total Number of Students be x.
20% of Students are below 8 years of age.
The Number of Students above or equal 8 years of age =80% of x............(i)
Number of Students of 8 years of age = 48 .....(ii)
Number of Students above 8 years of age = 2/3 × Number of Students of 8 years of age.
Number of Students of 8 years of age.
Number of Students above age of 8 =2/3 × 48=32---(iii)
From equation (i), (ii) and (iii),
80% of x = 48+2/3 of 48
80% of x = 80
x = 100.
View AnswerQ12. |
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| 1) | Rs.8000 | 2) | Rs.10,000 |
| 3) | Rs.12,000 | 4) | Data Inadequate |
| 5) | None of these | ||
Answer : Rs.8000
View AnswerQ13. | In a triangle ABC, the sides AB and AC have been produced to D and E. Bisectors of angle CBD and angle BCE meet at O. If angle A = 64°, then angle BOC is - |
| 1) | 54° | 2) | 56° |
| 3) | 58° | 4) | 60° |
| 5) | None of these | ||
Answer : 58°
View AnswerQ14. |
|
| 1) | 2 | 2) | 3 |
| 3) | 4 | 4) | 5 |
| 5) | None of these | ||
Answer : 5
Explanation :
Explanation :
Since, DE parallel to BC ![]()
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so, (AD / DB) = (AE / EC)
or {x / (x – 2)} = {(x + 2) / (x – 1)}
or x(x – 1) = (x + 2) (x – 1)
or x2 – x = x2 – 4
or x = 4 .
View AnswerQ15. | If a + b + c = m and 1/a + 1/b + 1/c = 0 , then the average of a2, b2 and c2 is |
| 1) | m2 | 2) | m2/3 |
| 3) | m2/9 | 4) | m2/27 |
| 5) | None of these | ||
Answer : m2/3
Explanation : Given, 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 0
Explanation : Given, 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 0
Or ab + bc + ca = 0
And a + b + c = m
Squaring both sides, we get, (a + b + c)2 = m2
Or a2 + b2 + c2 +2(ab + bc + ca) =m2
Or a2 + b2 + c 2+2(0) = m2
So, the average = ( a2 + b2 + c 2) /3 = m2/3
View AnswerQ16. | If x + y+ z = 9 and x2+y2+z2 = 31 , then what should be the value of x3+y3+z3 |
| 1) | 3 | 2) | 9 |
| 3) | 27 | 4) | 54 |
| 5) | None of these | ||
Answer : 54
Explanation : Given, x + y+ z = 9 and x2+y2+z2 = 31
Explanation : Given, x + y+ z = 9 and x2+y2+z2 = 31
Then, (x + y+ z)2 = 92
Or x2+y2+z2 + 2( xy + yz + zx) = 81
Or 31 + 2( xy + yz + zx) = 81
Or ( xy + yz + zx) = 25
So, x2+y2+z2 - 3xyz = (x + y + z) [x2+y2+z2 -(xy + yz + zx)] = 9[31-25] = 54
View AnswerQ17. | If a + b = 2c, then find the value of a/(a - c) + c/ (b-c) |
| 1) | 0 | 2) | 1 |
| 3) | -1 | 4) | 2 |
| 5) | None of these | ||
Answer : 1
Explanation : Given, a + b = 2c
Explanation : Given, a + b = 2c
Put, a = 1, b = 2, then c = 3
So,1/(1-2) + 2/(3-2) = -1+2 = 1
View AnswerQ18. | If x2+1/x2=1, then find the value of x18+x12+x6+1 |
| 1) | 0 | 2) | 1 |
| 3) | -1 | 4) | -2 |
| 5) | None of these | ||
Answer : 0
Explanation : Given, x2+1/x2 =1, Or x4+1=x2
Explanation : Given, x2+1/x2 =1, Or x4+1=x2
Or x4-x2+1=0
Multiplying both sides by (x2+1), we get
(x2+1)(x4-x2+1) =0x(x2+1)
Or x6+1 =0
So,x18+x12+x6+1 = x12(x6+1)+x6+1 =x12x0+0=0
View AnswerQ19. | If 1/a(a2+1)=3,then what should be the value of (a6+1)/a3 |
| 1) | 1 | 2) | 9 |
| 3) | 18 | 4) | 27 |
| 5) | None of these | ||
Answer : 18
Explanation : Given, 1/a(a2+1) = 3
Explanation : Given, 1/a(a2+1) = 3
Or a+1/a =3
Or a3+1/a3+3a.1/a.(a+1/a) = 33
Or a3+1/a3 +3x3= 27
Or a3+1/a3 =18
Or (a6+1)/a3= 18
View AnswerQ20. | If x+(1/x) = 5, the find the value of 5x/(x2+5x+1) |
| 1) | 1/3 | 2) | 1/4 |
| 3) | 1/2 | 4) | 1/s |
| 5) | None of these | ||
Answer : 1/2
Explanation : Given, x+(1/x) = 5
Explanation : Given, x+(1/x) = 5
or (x2+x)/x = 5
or x2+x = 5x
Now,5x/(x2+5x+1)= 5x/(x2+1+5x) =5x/(5x+5x) =5x/10x=1/2
View Answer