**Q1. A and B started a partnership business investing some amount in the ratio of 3:5. C joined them after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?**

(a) 3:5:2

(b) 3:5:5

(c) 6:10:5

(d) Data inadequate

Ans.(c)

Sol.

Let the initial investments of A and B be 3x and 5x.

A : B : C = (3x × 12) : (5x × 12) : (5x × 6) = 36 : 60 : 30 = 6 : 10 : 5

**Q2. A began a business with Rs. 85,000. He was joined afterwards by B with Rs. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3:1?**

(a) 4 months

(b) 5 months

(c) 6 months

(d) 8 months

Ans.(d)

Sol.

Suppose B joined for x months.

Then, (85000 × 12)/(42500 × x)=3/1 or x = (85000 × 12)/(42500 × 3) = 8.

So, B joined for 8 months.

**Q3. An amount of Rs. 2430 is divided among A, B and C such that if their shares be reduced by Rs. 5, Rs. 10 and Rs. 15 respectively, the remainders shall be in the ratio of 3 : 4 : 5. Then, B’s share was:**

(a) Rs. 605

(b) Rs. 790

(c) Rs. 800

(d) Rs. 810

Ans.(d)

Sol.

Remainder = Rs. [2430 – (5 + 10 + 15)] = Rs. 2400.

∴ B’s share = Rs. [(2400×4/12)+10] = Rs. 810.

**Q4. At what price should a shopkeeper mark a radio that costs him Rs. 1200 in order that he may offer a discount of 20% on the marked price and still make a profit of 25%?**

(a) Rs. 1675

(b) Rs. 1875

(c) Rs. 1900

(d) Rs. 2025

Ans.(b)

Sol.

C.P. = Rs. 1200. S.P. = 125% of Rs. 1200 = Rs. (125/100×1200) = Rs. 1500.

Let marked price be Rs. x. Then, 80% of x = 1500 ⇒ x = ((1500 × 100)/80) = 1875.

∴ Marked price = Rs. 1875.

**Q5. Two shopkeepers announce the same price of Rs. 700 for a sewing machine. The first offers successive discounts of 30% and 6% while the second offers successive discounts of 20% and 16%. The shopkeeper that offers better discount, charges ………… less than the other shopkeeper.**

(a) Rs. 9.80

(b) Rs. 16.80

(c) Rs. 22.40

(d) Rs. 36.40

Ans.(a)

Sol.

S.P. in 1st case = 94% of 70% of Rs. 700 = Rs. (94/100×70/100×700) = Rs. 460.60

S.P. in 2nd case = 84% of 80% of Rs. 700 = Rs. (84/100×80/100×700) = Rs. 470.40

∴ Difference = Rs. (470.40 – 460.60) = Rs. 9.80

**Q6. Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.**

(a) 12 sec

(b) 24 sec

(c) 48 sec

(d) 60 sec

Ans.(c)

Sol.

Relative speed (45 + 30) km/hr = (75×5/18) m/sec = (125/6) m/sec.

Distance covered = (500 + 500) m = 1000 m.

Required time = (1000×6/125) sec = 48 sec.

**Q7. A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?**

(a) 69.5 km/hr

(b) 70 km/hr

(c) 79 km/hr

(d) 79.2 km/hr

Ans.(d)

Sol.

Let the length of the train be x metres and its speed by y m/sec.

They, x/y = 8 ⇒ x = 8y

Now, (x + 264)/20 = y ⇔ 8y + 264 = 20y ⇔ y = 22.

∴ Speed = 22 m/sec = (22×18/5) km/hr = 79.2 km/hr.

**Q8. A train travelling at a speed of 75 mph enters a tunnel 7/2 miles long. The train is 1/4 mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?**

(a) 2.5 min

(b) 3 min

(c) 3.2 min

(d) 3.5 min

Ans.(b)

Sol.

Total distance covered = (7/2+1/4) miles = 15/4 miles.

∴ Time taken = (15/(4 × 75)) hrs = 1/20 hrs = (1/20×60) min. = 3 min.

**Q9. A man rows to a place 48 km distant and back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:**

(a) 1 km/hr

(b) 1.5 km/hr

(c) 1.8 km/hr

(d) 3.5 km/hr

Ans.(a)

Sol.

Suppose he moves 4 km downstream in x hours. Then,

Speed downstream = (4/x) km/hr, Speed upstream = (3/x) km/hr.

∴ 48/((4/x))+48/((3/x)) = 14 or x = 1/2.

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

Rate of the stream = 1/2 (8 – 6) km/hr = 1 km/hr.

**Q10. The speed of a boat in still water is 10 km/hr. If it can travel 26 km downstream and 14 km upstream in the same time, the speed of the stream is:**

(a) 2 km/hr

(b) 2.5 km/hr

(c) 3 km/hr

(d) 4 km/hr

Ans.(c)

Sol.

Let the speed of the stream be x km/hr. Then,

Speed downstream = (10 + x) km/hr, Speed upstream = (10 – x) km/hr.

∴ 26/((10 + x))=14/((10 – x)) ⇔ 260 – 26x = 140 + 14x ⇔ 40x = 120 ⇔ x = 3 km/hr.