Q1. At a game of billiards, A can give B 15 points in 60 and A can give C 20 points in 60. How many points can B give C in a game of 90?
(a) 30 points
(b) 20 points
(c) 10 points
(d) 12 points
Q2. A solid piece of iron of dimensions 49 × 33 × 24 cm is moulded into a sphere. The radius of the sphere is:
(a) 21 cm
(b) 28 cm
(c) 35 cm
(d) None of these
Q3. The volume of a sphere is 4851 cu. cm. Its curved surface area is:
(a) 1386 cm^2
(b) 1625 cm^2
(c) 1716 cm^2
(d) 3087 cm^2
Q4. The cost of painting the whole surface area of a cube at the rate of 13 paise per sq. cm is Rs. 343.98 Then the volume of the cube is:
(a) 8500 cm^3
(b) 9000 cm^3
(c) 9250 cm^3
(d) 9261 cm^3
Q5. The volume of a rectangular block of stone is 10368 dm^3. Its dimensions are in the ratio of 3 : 2 : 1. If its entire surface is polished at 2 paise per dm^3, then the total cost will be:
(a) Rs. 31.50
(b) Rs. 31.68
(c) Rs. 63
(d) Rs. 63.36
Q6. A sum of money amounts to Rs. 6690 after 3 years and to Rs. 10,035 after 6 years on compound interest. Find the sum.
(a) 4360
(b) 4460
(c) 4260
(d) 4560
Q7. The principal that amounts to Rs. 4913 in 3 years at 61/4% per annum compound interest compounded annually, is:
(a) Rs. 3096
(b) Rs. 4076
(c) Rs. 4085
(d) Rs. 4096
Q8. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
(a) 4 1/2 hrs
(b) 9 hrs
(c) 18 hrs
(d) 36 hrs
Q9. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
(a) 8
(b) 15
(c) 16
(d) 18
Q10. A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
(a) 6 min. to empty
(b) 6 min. to fill
(c) 9 min. to empty
(d) 9 min. to fill
Solutions
S1. Ans.(c)
Sol. A : B = 60 : 45 and A : C = 60 : 40.
∴ (B/C)=(B/A) × (A/C)=(45/60) × (60/40)=45/40=90/80 = 90 : 80
∴ B can give C 10 points in a game of 90.
S2. Ans.(a)
Sol. Volume of the solid = (49 ×33 ×24) cm^3.
Let the radius of the sphere be r.
Then, 4/3 πr^3 = (49 × 33 × 24) ⇔ r^3 = (49 × 33 × 24 × 3 × 7)/(4 × 22)=(21)^3 ⇔ r = 21.
S3. Ans.(a)
Sol. (4/3)×(22/7)×R^3=4851⇒R^3= 4851×(3/4)×(7/22)=(21/2)^3 ⇒ R = 21/2.
∴ Curved surface area = 4×(22/7)×(21/2)×(21/2) cm^2 = 1386 cm^2.
S4. Ans.(d)
Sol. Surface area = (34398/13) = 2646 cm^2.
∴ 6a2 = 2646 ⇒ a2 = 441 ⇒ a = 21.
So, Volume = (21 × 21 × 21) cm^3 = 9261 cm^3.
S5. Ans.(d)
Sol. Let the dimensions be 3x, 2x and x respectively. Then,
3x × 2x × x = 10368 ⇔ x^3=(10368/6) = 1728 ⇔ x = 12.
So, the dimensions of the block are 36 dm, 24 dm and 12 dm.
Surface area [2 (36 × 24 + 24 × 12 + 36 × 12)] dm^2
= [2 × 144 (6 + 2 + 3)] dm^2 = 3168 dm^2
∴ Cost of polishing = Rs. ((2 × 3168)/100) = Rs. 63.36.
S6. Ans.(b)
Sol. Let the sum be Rs. P. Then,
P(1+(R/100))^3 = 6690 …(i)
and P(1+(R/100))^6 = 10035 …(ii)
On dividing, we get (1+R/100)^3=10035/6690=3/2.
Substituting this value in (i), we get:
P × 3/2 = 6690 or P = (6690×2/3) = 4460.
Hence, the sum is Rs. 4460.
S7. Ans.(d)
Sol. Principal = Rs. [4913/(1+(25/(4 × 100))^3 ] = Rs. 4913×(16/17)×(16/17)×(16/17) = Rs. 4096.
S8. Ans.(d)
Sol. Part filled by (A + B) in 1 hour = (1/5)+(1/20)=1/4.
So, A and B together can fill the tank in 4 hours.
Work done by the leak in 1 hour = (1/4)-(2/9)=1/36.
∴ Leak will empty the tank in 36 hrs.
S9. Ans.(d)
Sol. Capacity of the tank = (12 × 13.5) litres = 162 litres.
Capacity of each bucket = 9 litres.
Number of buckets needed = (162/9) = 18.
S10. Ans.(a)
Sol. Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 2/5.
Part emptied by (A + B) in 1 minute = (1/6)-(1/10)=1/15.
∴ 1/15:2/5∷1∶x or x=(2/5)×(1×15) = 6 min.
So, the tank will be emptied in 6 min.
