**Q1. Anu sold 2 books at Rs. 1.40 each. Her profit on one was 20% and her loss on the other was 20%. Then, she:**

(a) made no gain no loss

(b) Gained 20 paise

(c) Lost 20 paise

**(d) lost 12 paise**

Sol.

S.P of 2 books = Rs. 2.80

Over all percentage change

= 20 + (-20) + (20 × (-20))/100=-4 = 4% loss

∴ C.P. = (2.80 × 100)/(100 – 4) = 2.92

⇒ Loss = Rs. 0.12

**Q2. A house worth Rs. 150000 is sold by X at 5% profit to Y. Y sells the house back to X at a 2% loss. Then, in the entire transaction:**

**(a) X gains Rs. 4350**

(b) X loses Rs. 4350

(c) X gains Rs. 3150

(d) X loses Rs. 3150

Sol.

X’s gain = (5-2-(5 × 2)/100)% of 1,50,000

= Rs. 4350.

**Q3. A businessman marks his goods in such a way that even after allowing 12.5% discount on cash purchase, he gains 20%. If the cost price of the goods is Rs. 140, the marked price is:**

(a) Rs. 162

(b) Rs. 172

**(c) Rs. 192**

(d) Rs. 198

Sol.

Suppose marked price = Rs. K

∴ K – 12.5% of K = 140 + 20% of 140

⇒ 87.5% of K = 140 + 28 = 168

∴ K = 16800/87.50 = 192.

**Q4. 6 men can complete a piece of work in 12 days. 8 women can complete the same price of work in 18 days, whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days. If only men were to complete the remaining work in 1 day how many men would be required?**

**(a) 36**

(b) 24

(c) 18

(d) Cannot be determined

Sol.

Males : Females : Children

6 × 12 : 8 × 18 : 18 × 10

72 : 144 : 180

2 : 4 : 5

So, 2 Males = 4 Females = 5 Children

4 Males + 12 Females + 20 Children

= 4 + 6 + 8 = 18 Males

∵ 6 males finished a piece of work in 12 days.

∴ 18 males finished the work = (12 × 6)/18 = 4 days

Work in 2 days = 2/4=1/2

Rest of the work will be finished in a day by

= 18 × 2 = 36 males

**Q5. 2 men alone or three women alone can complete a piece of work in 4 days. In how many days can 1 woman and 1 man together complete the same piece of work?**

(a) 6 days

**(b) 24/5 days**

(c) 12/1.75 days

(d) Cannot be determined

Sol.

2M = 3W

∴ 1M = 3/2W

∴ 1M + 1W = 3/2W + 1W = 5/2W

Number of days = (3 × 4)/(5/2 )=24/5 days

**Q6. Ram sets out to cycle from Delhi to Mathura and at the same time Suresh starts from Mathura to Delhi. After passing each other, they complete their journeys in 9 and 16 hours, respectively. At what speed does Suresh cycle if Ram cycles at 16 km per hour?**

**(a) 12 km/h**

(b) 16 km/h

(c) 14 km/h

(d) None of these

Sol.

(Ram^’ s speed)/(Suresh^’ s speed)=√(T_2 )/√(T_1 )=√16/√9=4/3.

∴ Suresh’s speed = 3/4 Ram’s speed

=3/4×16 = 12 km/h.

**Q7. A train 300 metres long is running at a speed of 90 km/h. How many seconds will it take cross a 200 metres long train running in the same direction at a speed of 60 km/h?**

(a) 70 seconds

**(b) 60 seconds**

(c) 50 seconds

(d) None of these

Sol.

Here, L1 = 300 m, L2 = 200 m,

s1 = 90 km/h and s2 = 60 km/h

∴ s1-s2 = 90 – 60 = 30 km/h = 30 × 5/18 m/s

∴ Time taken = (L1+ L2)/(s1- s2 )=(300 + 200)/(30 × 5/18)

= (500 × 18)/(30 × 5) = 60 seconds.

**Q8. A person covers half of his journey at 30 km/h and the remaining half at 20 km/h. The average speed for the whole journey is:**

(a) 25 km/h

(b) 28 km/h

(c) 32 km/h

**(d) None of these**

Sol.

Here s1 = 30 and s2 = 20.

∴ Average speed = (2s1 s2)/(s1 + s2 )

= (2 × 30 × 20)/(30 + 20) = 24 km/h.

**Q9. Two trains, one 160 m and other 140 m long, are running in opposite directions on parallel rails. The first train at 77 km/h and the other train at 67 km/h. How long will they take to cross each other.**

(a) 7 seconds

**(b) 7.5 seconds**

(c) 6 seconds

(d) 10 seconds

Sol.

Required time = (160 + 140)/((77 + 67)× 5/18)=(300 × 18)/(144 × 5)

= 7.5 seconds.

**Q10. A car driver leaves Bangalore at 8:30 am and expects to reach a place 300 km from Bangalore at 12:30 pm. At 10:30 he finds that he has covered only 40% of the distance. By how much % he has to increase the speed of the car in order to keep up his schedule?**

(a) 45

(b) 40

(c) 35

**(d) 50**

Sol.

Distance covered by the car in 2 hours

= (300 × 40)/100 = 120 km

Remaining distance = (300 – 120) = 180 km

Remaining time = (4 – 2) = 2 hours

∴ Required speed = (180/2) = 90 km/h

Increase in the speed of car = ((90-60)/60) = 50%