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- Question
If p and q are the roots of the equation x2 ?15x + r = 0 and p ?q = 1, then what is the value of r ?
Options- A. 55
- B. 56
- C. 60
- D. 64
- Correct Answer
- 56
ExplanationIn the below equation,
x2?15x + r = 0
sum of roots = p + q = -(-15)/1 = 15 (sum of roots for equation ax2+ bx + c is -b/a)
product of roots = pq = r/1 = r (product of roots for equation ax2+ bx + c is c/a)
given p ?q = 1
also we know that p+q = 15
subtracting the squares of both
(p+q)2 + (p-q)2 = 15^2?1
p2 + q2 + 2pq ?p2 ?q2 +2pq = 225 -1
4pq = 224
4r = 224
r = 56
- 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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