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- Question
- In a garrison, there was food for 1000 soldiers for one month. After 10 days, 1000 more soldiers joined the garrison. How long would the soldiers be able to carry on with the remaining food?
Options- A. 25 days
- B. 20 days
- C. 15 days
- D. 10 days
- Correct Answer
- 10 days
ExplanationLet us assume that each soldiers eats one unit of food per day. Thus, total units of food at the beginning will be 1000 x 30 = 30000. After 10 days the 100 soldiers would have eaten 1000 x 10 = 10000 units of food. Thus, food left after 10 days equals 20000 units. Now, there are a total of 2000 soldiers who eat one unit of food everyday.
Now, 30 x 1000 = 1000 x 10 + 2000 x D (here, D is number of days food will last)
? 30000 - 10000 = 2000 x D
? D = 20000/2000 = 10 days - 1. The area of circle is 38.5 sq. cm. Its circumference is?
Options- A. 6.20 cm
- B. 11 cm
- C. 22 cm
- D. 121 cm Discuss
- 2. A man buys a watch for Rs1950 in cash and sells it for Rs 2200 at a credit of 1 yr. The rate of interest is 10% per annum. Then he gains or loose how much amount?
Options- A. 50
- B. 55
- C. 15
- D. 45 Discuss
- 3. Find the present worth of a bill of Rs.28000 due 2 years at 12% compound interest. Also find the true discount.
Options- A. 300
- B. 350
- C. 250
- D. 275 Discuss
- 4. The true discount on Rs.1600 due to after a certain of 5% per annum is Rs.160. The time after which it is due is
Options- A. 27months
- B. 23months
- C. 20months
- D. 12months Discuss
- 5. In a triangle ABC, BC = 5 cm AC = 12 cm and AB = 13 cm. The length of a altitude drawn from B on AC is?
Options- A. 4 cm
- B. 5 cm
- C. 6 cm
- D. 7 cm Discuss
- 6. The perimeter of an isosceles triangle is equal to 14 cm . The lateral side is to the base in the ratio 5 : 4 the area of the triangle is?
Options- A. 1/2?21 cm2
- B. 3/2?21 cm2
- C. ?212
- D. 2?21 cm2 Discuss
- 7. The sides of a triangular board are 13 meters, 14 meters and 15 meters. The cost of paining it at the rate of Rs. 8.75 per m 2 is?
Options- A. Rs. 688.80
- B. Rs. 735
- C. Rs. 730.80
- D. Rs. 722.50 Discuss
- 8. A circular disc of area 0.49 ? square meters, rolls down a length of 1.76 km. The number of revolutions it makes is?
Options- A. 300
- B. 400
- C. 600
- D. 4000 Discuss
- 9. The diameter of a wheel is 63 cms. Distance travelled by the wheel in 100 revolutions is?
Options- A. 99 meters
- B. 198 meters
- C. 63 meters
- D. 136 meters Discuss
- 10. The inner circumference of a circular race track, 14 m wide is 440 m . Then the radius of the outer circle is?
Options- A. 70 m
- B. 56 m
- C. 77 m
- D. 84 m Discuss
More questions
Correct Answer: 22 cm
Explanation:
? ( 22 x r2) / 7 = 38.5
? r2 =(38.5 x 7) /22
? r = 3.5 cm
? Circumference = 2 x (22/7) x 3.5 cm
= 22 cm
Correct Answer: 50
Explanation:
S.P = P.W of Rs. 2200 due 1yr hence = Rs. [(2200 x 100)/100+(10 x 1)] = Rs.2000
Gain = Rs.(2000-1950) = Rs. 50
Correct Answer: 300
Explanation:
PW = = Rs.2500
TD = Amount - PW = 2800 - 2500 = Rs.300
Correct Answer: 27months
Explanation:
P.W. = Rs. (1600 - 160) = Rs. 1440
? S.I. on Rs.1440 at 5% is Rs. 160.
? Time = [100 * 160 / 1440 * 5] = 20/9 years = [20/9 * 12] months = 27 months.
Correct Answer: 5 cm
Explanation:
? s= (13 + 5 + 12) / 2 cm
= 15cm
s-a = 2 cm,
s-b = 10 cm and
s-c = 3 cm
? Area = ? (15 x 2 x 10 x 3 ) cm2
= 30 cm2
? (12 x h) / 2 = 30
? h = 5 cm
Correct Answer: 2?21 cm2
Explanation:
Let lateral side = (5y) cm and base = (4y) cm
? perimeter = 5y + 5y + 4y = 14
?y = 1
So, the sides are 5 cm , 5 cm and 4 cm
Now s= 1/2 (5 + 5 + 4) cm = 7 cm
(s-a) = 2 cm
(s-b) = 2 cm and
(s-c) = 3 cm
? Required Area = ? (7 x 2 x 2 x 3) cm2
=2?21 cm2
Correct Answer: Rs. 735
Explanation:
s = (13 + 14 + 15 ) / 2 = 21,
s-a = 8 ,
s-b = 7,
s-c= 6
? Area to be painted = ? [s(s-a) (s-b) (s-c)]
=? [21 x 8 x 7 x 6] m2
= 84 m2
? Cost of painting = Rs. (84 x 8.75) = Rs. 735
Correct Answer: 400
Explanation:
Area = ?r2 = 0.49?
? r = 0.7 m
Number of revolutions = total distance / circumference
= (1.76 x 1000) / (2 x 22/7 x 0.7)
= 400
Correct Answer: 198 meters
Explanation:
Distance travelled in 1 revolution = circumference of the wheel
= ?d
= ( 22/7 ) x 63
= 198 cm
So the distance travelled in 100 revolutions = 100 x distance travelled in 1 revolution
= 100 x 198 cm
= 19800 cm
= 198 m
Correct Answer: 84 m
Explanation:
Circumference of a circular = 2?r
? 2 x (22 / 7) x r = 440
? r = [440 x (7/22) x (1/2)] = 70 m
? Radius of outer circle = (70 + 14) m
= 84 m
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