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- Question
The radius of two circles is 3 cm and 4 cm. The distance between the centres of the circles is 10 cm. What is the ratio of the length of direct common tangent to the length of the transverse common tangent?
Options- A. ?51 : ?68
- B. ?33 : ?17
- C. ?66 : ?51
- D. ?28 : ?17
- Correct Answer
- ?33 : ?17
- 1. AB and AC are the two tangents to a circle whose radius is 6 cm. If ?BAC = 60 deg, then what is the value (in cm) of ?(AB2 + AC2) ?
Options- A. 6?6
- B. 4?6
- C. 9?3
- D. 8?3 Discuss
- 2. Triangle ABC is similar to triangle PQR and AB : PQ = 2 : 3. AD is the median to the side BC in triangle ABC and PS is the median to the side QR in triangle PQR. What is the value of (BD/QS)2 ?
Options- A. 3/5
- B. 4/9
- C. 2/3
- D. 4/7 Discuss
- 3. The area of a trapezium is 21 sq cm, the distance between the parallel sides is 3 cm and one of the parallel sides is 8.3 cm. What is the length of the other parallel side?
Options- A. 12.7 cm
- B. 7.7 cm
- C. 10.7 cm
- D. 5.7 cm Discuss
- 4. An angle is (2/3)rd of its supplementary angle. What is the measure of the angle?
Options- A. 108°
- B. 72°
- C. 54°
- D. 36° Discuss
- 5. The area of a rectangle is 120 sq cm and its perimeter 140 cm. What is the length of its diagonal?
Options- A. 50 cm
- B. 30 cm
- C. 40 cm
- D. 60 cm Discuss
- 6. ABC is a triangle. AB = 5 cm, AC = ?41 cm and BC = 8 cm. AD is perpendicular to BC. What is the area (in cm2) of triangle ABD?
Options- A. 12
- B. 6
- C. 10
- D. 20 Discuss
- 7. Length and breadth of a rectangle are 8 cm and 6 cm respectively. The rectangle is cut on its four vertices such that the resulting figure is a regular octagon. What is the side (in cm) of the octagon?
Options- A. 3(?11) ? 7
- B. 5(?13) ? 8
- C. 4(?7) ? 11
- D. 6(?11) ? 9 Discuss
- 8. The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ?
Options- A. 76
- B. 79
- C. 85
- D. 87 Discuss
- 9. A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ?
Options- A. 4991
- B. 5467
- C. 5987
- D. 6453 Discuss
- 10. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero ?
Options- A. 0
- B. -1
- C. 1
- D. none of these Discuss
Area problems
Search Results
Correct Answer: 6?6
Correct Answer: 4/9
Correct Answer: 5.7 cm
Correct Answer: 72°
Correct Answer: 50 cm
Correct Answer: 6
Correct Answer: 3(?11) ? 7
Correct Answer: 76
Explanation:
Average = total runs / no.of innings = 32
So, total = Average x no.of innings = 32 x 10 = 320.
Now increase in avg = 4runs. So, new avg = 32+4 = 36runs
Total runs = new avg x new no. of innings = 36 x 11 = 396
Runs made in the 11th inning = 396 - 320 = 76
Correct Answer: 4991
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs.[(6500 x 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991.
Correct Answer: 0
Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers = (0 * 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).
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