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- Question
Goods were bought for Rs. 600 and sold the same for Rs. 688.50 at a credit of 9 months and thus gaining 2% The rate of interest per annum is:
Options- A.
16 2 % 3 - B.
14 1 % 2 - C.
13 1 % 3 - D. 15%
- Correct Answer
-
16 2 % 3
ExplanationS.P. = 102% of Rs. 600 = ❨ 102 x 600 ❩ = Rs. 612. 100 Now, P.W. = Rs. 612 and sum = Rs. 688.50.
∴ T.D. = Rs. (688.50 - 612) = Rs. 76.50.
Thus, S.I. on Rs. 612 for 9 months is Rs. 76.50.
∴ Rate = ❨ 100 x 76.50 ❩% = 16⅔% 612 x 3 4
More questions
- 1. The last day of century cannot be either
I. Tuesday
II. Thursday
III. Saturday
IV. Sunday
Options- A. I and II
- B. I and IV
- C. I, II and III
- D. I,III and IV Discuss
Correct Answer: I, II and III
Explanation:
The last day of century cannot be either Tuesday, Thursday or Saturday.
- 2. In a ?ABC, if 2?A = 3?B = 6?C, Then ?A is equal to:
Options- A. 60°
- B. 30°
- C. 90°
- D. 120° Discuss
Correct Answer: 90°
Explanation:
- 3. How many times in a day (24 Hrs), are the hands of a clock in straight line but opposite in direction?
Options- A. 20
- B. 22
- C. 24
- D. 48 Discuss
Correct Answer: 22
Explanation:
The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours.
(Because between 5 and 7 they point in opposite direction at 6 'o clock only). So, in a day, the hands point in the opposite direction 22 times.- 4. Which of the following two months in a particular year will have same calendar.
Options- A. January and August
- B. January and October
- C. March and November
- D. None of these Discuss
Correct Answer: March and November
Explanation:
January and August or October depends on leap year or non leap year. But if we find the number of odd days between March and November we will get number of odd days is 0 hence they will have same calendar.
- 5. In a circle of radius 21 cm an arc subtends an angle of 72° at the centre. The length of the arc is?
Options- A. 13.2 cm
- B. 19.8 cm
- C. 21.6 cm
- D. 26.4 cm Discuss
Correct Answer: 26.4 cm
Explanation:
Arc length = 2?r (?° / 360°)
= 2 x (22/7) x 21 x (72° / 360°) cm
= 26.4 cm- 6. If in a particular year 'X' there are 53 Sundays then how many Sundays will be there in a period of four years X to X + 3 year.
Options- A. 208
- B. 209
- C. 208 or 209
- D. None of these Discuss
Correct Answer: 209
Explanation:
Let us take two cases
Case (i) : When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays
Case (ii) : When year start with Sunday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays.- 7. In a game of 80 points, A can give B 5 points and C 15 points. Then how many points B can give C in a game of 60?
Options- A. 12
- B. 10
- C. 8
- D. 6 Discuss
Correct Answer: 8
Explanation:
? B scores 75 points while A 80
? B scores 60 points while A = 80 x 60 / 75 = 64
Again, A scores 80 points while C = 65
? A scores 64 points while C = 65 x 64 / 80 = 52 points
Hence, in a game of 60, B can give C (60 ? 52) = 8 points- 8. From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus, in three attempts the ratio of wine and water became 343 : 169. The initial amount of wine in the container was:
Options- A. 75 litres
- B. 100 litres
- C. 150 litres
- D. 120 litres Discuss
Correct Answer: 120 litres
Explanation:
It means (since 343 + 169 = 512)
Thus,
Thus the initial amount of wine was 120 liters.
- 9. If a - b = 3 and a2 + b2 = 29, find the value of ab.
Options- A. 10
- B. 12
- C. 15
- D. 18 Discuss
Correct Answer: 10
Explanation:
2 ab = ( a 2 + b 2 ) - ( a - b ) 2= 29 - 9 = 20
⟹ab = 10.
- 10. Milk and water are mixed in a vessel A as 4:1 and in vessel B as 3:2. For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C.
Options- A. 1:9
- B. 9:1
- C. 3:7
- D. 7:3 Discuss
Correct Answer: 7:3
Explanation:
Ratio of Milk and water in a vessel A is 4 : 1
Ratio of Milk and water in a vessel B is 3 : 2
Ratio of only milk in vessel A = 4 : 5
Ratio of only milk in vessel B = 3 : 5
Let 'x' be the quantity of milk in vessel C
Now as equal quantities are taken out from both vessels A & B
=> 4/5 : 3/5
x
3/5-x x - 4/5=> = (equal quantities)
=> x = 7/10
Therefore, quantity of milk in vessel C = 7
=> Water quantity = 10 - 7 = 3
Hence the ratio of milk & water in vessel 3 is 7 : 3
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