**Q1. When x^m is multiplied by x^n, product is 1. The relation between m and n is**

(a) mn = 1

(b) m + n =

(c) m = n

(d) m = -n

**Q2. The reciprocal of x + (1/x) is**

(a) x/(x^2+1)

(b) x/(x+1)

(c) x-(1/x)

(d) (1/x)+x

**Q3. The simplest form of the expression (p^2-p)/(2p^3+p^2 ) + (p^2-1)/(p^2+3p) + p^2/(p+1)**

(a) 2p^2

(b) 1/(2p^2 )

(c) p + 3

(d) 1/(p+3)

**Q4. If x = -2k and y = 1 – 3k, then for what value of k, will be x = y?**

(a) 0

(b) 1

(c) -1

(d) 2

**Q5. Ifa/(a + b)=17/23, what is (a + b)/(a – b) equal to?**

(a) 11/23

(b) 17/32

(c) 23/11

(d) 23/17

**Q6. A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is**

(a) 9 hours

(b) 7 hours

(c) 412 hours

(d) 8 hours

**Q7. A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?**

(a) 230 m

(b) 245 m

(c) 260 m

(d) 275 m

**Q8. Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:**

(a) 10.8 sec

(b) 9.5 sec

(c) 7.4 sec

(d) 8.9 sec

**Q9. Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:**

(a) 18 km/hr

(b) 26 km/hr

(c) 36 km/hr

(d) 42 km/hr

**Q10. Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?**

(a) 8 sec

(b) 12 sec

(c) 15 sec

(d) 10 sec

**Solutions**

S1. Ans.(d)

Sol. x^m × x^n=1

x^(m+n) = x^0 (∵ x^0 = 1)

m + n = 0

m = -n

S2. Ans.(a)

Sol. Reciprocal of (x+(1/x)) = 1/(x+(1/x) )

= x/(x^2+1)

S3. Ans.(b)

Sol. (p^2-p)/(2p^3+p^2 ) + (p^2-1)/(p^2+3p) + p^2/(p+1)

In such type of question assume values of p.

∴ Let p = 1

∴ (1-1)/(2+1) + (1-1)/(1+3) + 1/(1+1)

= 0 + 0 + (1/2) = 1/2

Now check options (b)

1/(2p^2 ) = 1/2

Hence option (b) is Answer.

S4. Ans.(b)

Sol. y = 1 – 3k and x = –2k (given)

∴ for x = y

–2k = 1 – 3k

k = 1

S5. Ans. (c)

Sol. Given that a/(a+b)=17/23

i.e. if a= 17, then a + b = 23, or b = 6

a – b = 17 – 6 = 11

∴ (a+b)/(a-b)=23/11

S6. Ans. (d)

Sol. Suppose the man covers first distance in x hrs.and second distance in y hrs. Then,

4x+5y=35 and 5x+4y = 37

Solving the equations,

we get x = 5 and y = 3

Total time taken = (5+3)hrs = 8 hrs

S7. Ans. (a)

Sol. Relative speed = (120 + 80) km/hr

= 200 x 5/18

= 500/9 m/sec

Then, (x+270)/9 = 500/9

–> x + 270 = 500

–> x = 230.

S8. Ans. (a)

Sol. Relative speed = (60 + 40) km/hr = 100 x 5/18 = 250/9 m/ sec.

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = 300 x 9/250 = 54/5 = 10.8 sec.

S9. Ans. (c)

Sol. Let the speed of each train be x m/sec.

Then, relative speed of the two trains = 2x m/sec.

So, 2x = (120 + 120)/12

–> 2x = 20

–> x = 10.

–> Speed of each train = 10 m/sec = 10 x 18/5 km/hr = 36 km/hr.

S10. Ans. (b)

Sol. Speed of the first train = 120/10 m/sec = 12 m/sec.

Speed of the second train = 120/15 m/sec = 8 m/sec.

Relative speed = (12 + 8) = 20 m/sec.

Required time = (120 + 120)/20 sec = 12 sec.