1. A clock is set right at 6 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?
- 09 p.m
- 10 p.m
- 11 p.m
- 12 p.m
3
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Solution:
Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
356/15 hrs of this clock = 24 hours of correct clock.
89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
= 90 hrs of correct clock.
So, the correct time is 11 p.m.
2. One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field. 120 m2
- 80 m2 .
- 100 m2 .
- 120 m2
- 140 m2
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Solution:
Other side = ((17)2- (15)2)(1/2) = (289- 225)(1/2) = (64)(1/2) = 8 m.
Area = (15 x 8) m2 = 120 m2.
3. In measuring the sides of a rectangle, one side is taken 5% in excess, and the other 4% in deficit. Find the error percent in the area calculated from these measurements.
- 0.5%
- 0.6%
- 0.7%
- 0.8%
12
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Solution:
Let x and y be the sides of the rectangle. Then, Correct area = xy.
Calculated area = (105/100)*x * (96/100)*y = (504/500 )(xy)
Error In measurement = (504/500)xy- xy = (4/500)xy
Error % = [(4/500)xy *(1/xy) *100] % = (4/5) % = 0.8%.
4. If each side of a square is increased by 25%, find the percentage change in its area.
- 56.25%
- 58.25%
- 60.25%
- 62.25%
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Solution:
Let each side of the square be a. Then, area = a2.
New side =(125a/100) =(5a/4).
New area = (5a/4)2 =(25a2)/16.
Increase in area = ((25 a2)/16)-a2 =(9a2)/16.
Increase% = [((9a2)/16)*(1/a2)*100] % = 56.25%.
5.Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?
- 45000
- 46000
- 47000
- 49000
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Solution:
Volume of the wall = (2400 x 800 x 60) cu. cm.
Volume of bricks = 90% of the volume of the wall =((90/100)*2400 *800 * 60)cu.cm.
Volume of 1 brick = (24 x 12 x 8) cu. cm.
Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.
6. Water flows into a tank 200 m x 160 m through a rectangular pipe of 1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2 meters?
- 92min
- 94min
- 96min
- 98min
23
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Solution:
Volume required in the tank = (200 x 150 x 2) m3 = 60000 m3.
Length of water column flown in1 min =(20*1000)/60 m =1000/3 m
Volume flown per minute = 1.5 * 1.25 * (1000/3) m3 = 625 m3.
Required time = (60000/625)min = 96min
7. Tbe dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.
- 8.02 kg
- 8.03 kg
- 8.04 kg
- 8.05 kg
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Solution:
Volume of the metal used in the box = External Volume - Internal Volume
= [(50 * 40 * 23) - (44 * 34 * 20)]cm3
= 16080 cm3
Weight of the metal =((16080*0.5)/1000) kg = 8.04 kg
8. The surface area of a cube is 1734 sq. cm. Find its volume?
- 4905 cm3
- 4907 cm3
- 4909 cm3
- 4913 cm3
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Solution:
Let the edge of the cube bea. Then,
6a2 = 1734 => a2 = 289 => a = 17 cm.
Volume = a3 = (17)3 cm3 = 4913 cm3.
9.A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.
- 38
- 40
- 42
- 44
34
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View Answer
Solution:
Volume of the block = (6 x 12 x 15) cm3 = 1080 cm3.
Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.
Volume of this cube = (3 x 3 x 3) cm3 = 27 cm3.
Number of cubes = 1080/27 = 40.
10. A cube of edge 15 cm is immersed completely in a rectangular vessel containing water . If the dimensions of the base of vessel are 20 cm x 15 cm, find the rise in water level.
- 11.25 cm
- 11.75 cm
- 13.25 cm
- 13.75 cm
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Solution:
Increase in volume = Volume of the cube = (15 x 15 x 15) cm3.
Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.
11. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas.
- 2/9
- 2/7
- 1/9
- 1/7
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Solution:
Let their edges be a and b. Then,
a3/b3 = 1/27 (or) (a/b)3 = (1/3)3 (or) (a/b) = (1/3).
Ratio of their surface area = 6a2/6b2 = a2/b2 = (a/b)2 = 1/9, i.e. 1:9
.
12. How many iron rods, each of length 7 m and diameter 2 cm can be made out of 0.88 cubic meter of iron?
- 600
- 500
- 400
- 300
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Solution:
Volume of 1 rod = (( 22/7) x (1/100) x (1/100) x 7 ) cu.m = 11/5000 cu.m
Volume of iron = 0.88 cu. m.
Number of rods = (0.88 x 5000/11) = 400.
14.How many words can be formed by using all letters of the word "BIHAR"?
- 74
- 96
- 120
- 240
55
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Solution:
The word BIHAR contains 5 different letters.
Required number of words = 5p5 = 5!
= (5x4x3x2x1) = 120.
15. In how many ways can a cricket eleven be chosen out of a batch of 15 players ?
- 91
- 273
- 1365
- 5460
59
16.In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?
- 100
- 200
- 300
- 500
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Solution:
(3 men out 6) and (2 ladies out of 5) are to be chosen.
Required number of ways = (6c3x5c2)
= [6x5x4/3x2x1] x [5x4/2x1] = 200.
17. Find the present worth of Rs. 930 due 3 years hence at 8% per annum. Also find the discount.
- 150
- 180
- 210
- 230
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Solution:
P.W=100 x Amount /[100 + (R x T)]
=Rs.100 x 930/100+ (8x3)
= (100x930)/124
= Rs. 750,
T.D. = (Amount) - (P.W.) = Rs. (930 - 750) = Rs. 180.
18. The true discount on a certain sum of money due 3 years hence is Rb. 250 and the simple interest on the same sum for the same time and at the same rate is Rs. 375. Find the sum and the rate percent.
- 13 2/3%
- 16 2/3%
- 20 2/3%
- 24 2/3%
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Solution:
T.D. = Rs. 250 and S.I. = Rs. 375.
Sum due =S.I. xT.D./ S.I. -T.D.
=375x250/375- 250
=Rs.750.
Rate=[100*375/750*3]%=16 2/3%
19. If the true discount on a certain sum due 6 months hence at 15% is Rs. 120, what is the banker's discount on the same sum for the same time and at the same rate?
- 127
- 129
- 132
- 135
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Solution:
aB.G. = S.I. on T.D.
= Rs.(120 x 15 x 1/2 x 1/100)
= Rs. 9. (B.D.) - (T.D.) = Rs. 9.
B.D. = Rs. (120 + 9) = Rs. 129.
20. 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.
- 2 months
- 4 months
- 6 months
- 8 months
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Solution:
S.I. on Rs. 1800 = T.D. on Rs. 1872.
P.W. of Rs. 1872 is Rs. 1800.
Rs. 72 is S.I. on Rs. 1800 at 12%.
Time =[(100 x 72)/ (12x1800)]year 1/3year = 4 months.
