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.
