**Q1. If x=2/3 and y=3/4, then a rational number between (x-y)^(-1) and (x^(-1)-y^(-1) ) is**

(a) 2/3

(b) -71/2

**(c) -71/12**

(d) 1/6

**Sol.**

**Given, x=2/3 and y=3/4**

**∴ x^(-1) = 1/x= 3/2**

**And y^(-1) = 1/y= 4/3**

**Now, (x-y)^(-1) = (2/3-3/4)^(-1)**

**=(8/12-9/12)^(-1)= ((-1)/12)^(-1)**

**= -12 = (-12 × 12)/12 = (-144)/12**

**and x^(-1)-y^(-1) = 3/2-4/3 = (3 × 3)/(2 × 3)-(4 × 2 )/(3 × 2)**

**= 9/6-8/6 = 1/6 = (1 × 2)/(6 × 2) = 2/12**

**Here, (-71)/12 lie between -144/12 and 2/12**

**Q2. If 3(5x-7)-4(8x-13)=2(9x-11)-17, then the value of (7x – 5)/(11x – 9) is**

(a) -14/31

(b) -9/13

**(c) 9/13**

(d) 14/31

**Sol.**

**3(5x-7)-4(8x-13) = 2(9x-11)-17**

**⇒ (15x-21)-(32x-52) = (18x-22)-17**

**⇒ (15x-32x-18x)-21+52+22+17=0**

**⇒ -35x+70=0**

**⇒ -35x=-70**

**⇒ x = 70/35 = 2**

**∴ (7x – 5)/(11x – 9) = (7 (2)- 5)/(11(2)-9)**

**=(14 – 5)/(22 – 9) = 9/13**

**Q3. If (√1058 × √648)/x=18, then the value of x is**

(a) 36

**(b) 46**

(c) 58

(d) 23

**Sol.**

**(√1058 × √648)/x=18**

**⇒ x=(√1058 × √648)/18 = (√1058 × √648)/(18)^2**

**=(√1058 × √648)/√324 = √1058 = √2 = √(1058×2)**

**=√2116 = 46 [∴√a×√b=√(a×b)]**

**∴ x = 46**

**Q4. The value of m such that**

**(2/9)^3×(4/81)^(-6)=(2/9)^(2m-1) is**

(a) –6

(b) 0

**(c) –4**

(d) 3

**Sol.**

**(2/9)^3×(4/81)^(-6) = (2/9)^(2m-1)**

**⇒ (2/9)^3×(81/4)^6 = (2/9)^(2m-1) [As (a/b)^(-1)=b/a]**

**⇒ (2/9)^3×[(81/4)^2 ]^3 = (2/9)^(2m-1)**

**⇒ [2/9×(81/4)^2 ]^3 = (2/9)^(2m-1)**

**⇒ (729/8)^3 = ( 2/9)^(2m-1)**

**⇒ [(9/2)^3 ]^3 = (2/9)^(2m-1)⇒[(2/9)^(-3) ]^3 = (2/9)^(2m-1)**

**⇒ (2/9)^(-9) = (2/9)^(2m-1)**

**⇒ 2m – 1 = – 9 ⇒ 2m=-8**

**∴ m = (-8)/2=-4**

**Q5. Let x = 597×〖10〗^22+73.5×〖10〗^21. When x is expressed in standard from as 6.0435 × 〖10〗^m, then the value of m is**

(a) 22

(b) 23

**(c) 24**

(d) 21

**Sol.**

**x = 594×〖10〗^22+73.5×〖10〗^21**

**= 5970×〖10〗^21+73.5×〖10〗^21**

**=(5970+73.5)×〖10〗^21=6043.5×〖10〗^21**

**On multiply numerator and denominator by 〖10〗^3, we get**

**=6043.5×〖10〗^21×〖10〗^3=6.0435×〖10〗^24**

**On compare this with 6.0435×〖10〗^m, we get m = 24**

**Q6. If 2m-1/2m=2, when m≠0, then the value of m^2+1/(16m^2 ) is**

(a) 2 1/2

(b) 2

**(c) 1 1/2**

(d) 4

**Sol.**

**Given, 2m-1/2m = 2**

**On squaring both side (2m-1/2m)^2 = (2)^2**

**⇒ 4m^2+1/(4m^2 )-2.(2m).(1/2m) = 4**

**⇒ 4m^2+1/(4m^2 )-2 = 4**

**⇒ 4m^2+1/(4m^2 ) = 6**

**On divide both side by 4, we get**

**∴ m^2+1/(16m^2 ) = 6/4 = 1 1/2**

**Q7. The mean of the five observations x, x + 2, x + 4, x + 6, x + 8 is 11. Then, the mean of the first three observations is**

(a) 12

(b) 2

(c) 20

**(d) 9**

**Sol.**

**Given, the mean and the 5 observations**

**x, x + 2, x + 4, x + 6, x + 8 is 11**

**∴(x+(x+2)+(x+4)+(x+6)+(x+8))/5 = 11**

**⇒ 5x + 20 = 55 ⇒ 5x = 55 – 20 = 35**

**x=35/5=7**

**∴Mean of the first three observations**

**=(x+(x+2)+(x+4))/3=(7+(7+2)+(7+4))/3**

**=27/3=9**

**Q8. A bag has 5 red marbles, 4 green marbles and 3 blue marbles. All marbles are identical in all respects other than colour. A marble is taken out from the bag without looking into it. What is the probability that it is non-green marble?**

(a) 1/3

**(b) 2/3**

(c) 7/12

(d) 5/12

**Sol.**

**Given, a bag contains 5 red marbles, 4 green marbles and 3 blue marbles.**

**Total number of marbles = 12**

**Total number of non-green marbles = 8**

**Here, total number of cases = 12**

**And number of favourable casses = 8**

**∴ Probability of getting non-green marbles**

**=(Number of favourable cases )/(Total number of cases )**

**=8/12=2/3**

**Q9. If x, y and z respect the number of faces, number of vertices and number of edges respectively of a polyhedral, then which of the following is true?**

(a) x + 2 = z – y

(b) z – 2 = x – y

(c) y – 2 = z + x

**(d) x – 2 = z – y**

**Sol.**

**Let the given polyhedral is tetrahedron in which number of faces, number of vertices and number or edges are 4, 4 and 6, respectively.**

**∴ x = 4, y = 4 and z = 6**

**In option (d), x – 2 = z – y**

**x – 2 = 4 – 2 = 2**

**and z – y = 6 – 4 = 2**

**∴ x – 2 = z – y (True)**

**Q10. Ravi purchased two articles for Rs. 1500 each. He sold them, gaining 6% on one and losing 4% on the other. His gain/loss per cent in the whole transaction is**

**(a) Gain, 1%**

(b) Loss, 1 1/2%

(c) Gain, 2%

(d) Loss, 1%

**Sol.**

**Given, Ravi purchase two articles for Rs. 1500 each.**

**∴ Total purchase cost = Rs. 3000**

**Ravi get 6% profit on first article, then he got**

**Amount after selling of 1st article**

**= 1500 + 1500 ×6/10 = Rs. 1590**

**After selling, 2nd article, he lose 4%, Then, his amount of 2nd article after selling it**

**=1500-1500×4/100 = Rs. 1440**

**∴ Total amount after selling of both article**

**= 1590 + 1440 = Rs. 3030**

**Which is 1% more than total purchase cost of both article.**