1. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1 / 3
B. 1 / 4
C. 1 / 5
D. 1 / 7
Ex:Suppose the vessel initially contains 8 litres of liquid. Let x littres of this liquid be replaced with water.
Quantity of water in new mixture = (3 – 3x/8 + x) litres.
Quantity of syrup in new mixture = (5 – 5x/8) litres.
(3 – 3x/8 + x) = (5 – 5x/8) = 5x + 24 = 40 – 5x
=› 10x = 16 =› x = 8/5
So, part of the mixture replaced = (8/5 x 1/8) = 1/5.
2. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?
A. 18750 sq.m
B. 37500 sq.m
C. 40000 sq.m
D. 48000 sq.m
Ex:So length =250 m;breadth=150 m
Area = (250 x 150)m²
= 37500 m²
3. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3.Find his average after 17th inning?
A. 19
B. 29
C. 39
D. 49
Ex:Let the average after 17th inning = x. Then, average after 16th inning = (x – 3)
Average=16 (x-3)+87
= 17x or x=(87-48)
= 39.
4. The banker’s discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs.1872 for the same time at the same rate. Find the time?
A. 3 months
B. 4 months
C. 5 months
D. 6 months
EX:S.I on Rs.1800 = T.D on Rs.1872.
P.W on Rs.1872 is Rs.1800.
Rs.72 is S.I on Rs. 1800 at 12%.
Time =(100×72 / 12×1800)
= 1/3 year
= 4 months.
5. A boat can travel with a speed of 13 km / hr in still water. If the speed of the stream is 4 km / hr. find the time taken by the boat to go 68 km downstream?
A. 2 hours
B. 3 hours
C. 4 hours
D. 5 hours
Ex:Speed Downstream= (13 + 4) km/hr
= 17 km/hr.
Time taken to travel 68 km downstream =(68 / 17)hrs
= 4 hrs.
6. An accurate clock shows 8 o’clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o’clock in the afternoon?
A. 144°
B. 150°
C. 168°
D. 180°
Ex:Angle traced by hour hand in
5 hrs 10 min. = (360/12 x 6)°
= 180°.
7. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is
A. Rs. 1550
B. Rs. 1650
C. Rs. 1750
D. Rs. 2000
Ex:C.I. =Rs[4000x(1+10/100)²-4000]
Rs.(4000×11/10×11/10-4000)= Rs.940.
Sum=Rs. [420×100 /3×8]
= Rs.1750.
8. (7.5×7.5+3.75+2.5×2.5) is equal to
A. 30
B. 60
C. 80
D. 100
Ex:Given expression= (7.5×7.5+2×7.5×2.5+2.5×2.5)²
= (a²+2ab+b²)
=(a+b)²
= (7.5+2.5)²
= 10²
= 100.
9. The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree
A. 30°
B. 45°
C. 60°
D. 90°
EX:Let AB be the tree and AC be its shadow.
Then, < ABC= θ.
Then, AC/AB= √3
cotθ= √3
θ=30°
10. If log 2 = 0.30103, the number of digits in 520 is
A. 14
B. 16
C. 18
D. 25
Ex:Log 520=20 log 5
=20 ×[log(10/2)]
=20 (log 10 – log 2)
=20 (1 – 0.3010)
=20×0.6990
=13.9800.
Characteristics = 13.
Hence, the number of digits in Log 520 is 14.