Discussion
Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M See What Others Are Saying!
- Question
- What is the um of the digits of the least number which when divided by 54 leaves 35 as remainder, when divided by 80 leaves 61 as remainder and when divided by 119 leaves 100 leaves 100 as remainder?
Options- A. 21
- B. 19
- C. 23
- D. 28
- Correct Answer
- 28
Explanation(54 - 35) = 19, (80 - 61) = 19 and(119 - 100) = 19
LCM of 54, 80 and 119 is 257040.
Required least number = 257040 + 19 = 257059
Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28 - 1. The angle of elevation of the top of a tower standing on a horizontal plane from a point A is ?. After walking a distance a towards the foot of the tower, the angle of elevation is found to be ?. The height of the tower is :
Options- A. a Sin ? Sin ?/ Sin(? - ?)
- B. a Sin ? Sin ?/Sin(? - ?)
- C. a Sin( ? - ? ) Sin ? Sin ?
- D. a Sin(? - ?)/Sin ?Sin ? Discuss
- 2. The number of different permutations of the word 'BANANA' is?
Options- A. 270
- B. 60
- C. 120
- D. 360 Discuss
- 3. A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:
Options- A. gains Rs. 55
- B. gains Rs. 50
- C. loses Rs. 30
- D. gains Rs. 30 Discuss
- 4. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?
Options- A. 10
- B. 13
- C. 18
- D. 136 Discuss
- 5. In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:
Options- A. 8 points
- B. 10 points
- C. 14 points
- D. 40 points Discuss
- 6. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
Options- A. 5 kmph
- B. 6 kmph
- C. 6.25 kmph
- D. 7.5 kmph Discuss
- 7. If 45 m of a uniform rod weight 171 kg, then what will be the weight of 12 m of the same rod?
Options- A. 49 kg
- B. 42.5 kg
- C. 55 kg
- D. 45.6 kg Discuss
- 8. A company has issued 10000 preferred shares and 50000 common shares both of par value $ 100 each. The dividend on a preferred share and a common share is 12% and 17.6%, respectively. The company had a total profit of $ 15 Millions, out of which some amount was kept in reserve fund and the remaining distributed as dividend. Find out the amount kept in reserve fund.
Options- A. $ 5 Million
- B. $ 6 Million
- C. $ 6.5 Million
- D. $ 5.5 Million Discuss
- 9. If a certain sum is doubled in 8 yr on simple interest, in how many year will it be four times?
Options- A. 24 yr
- B. 16 yr
- C. 32 yr
- D. 12 yr Discuss
- 10. The sum of three consecutive odd numbers is 20 more than the first of these numbers.What is the middle number?
Options- A. 7
- B. 9
- C. 11
- D. Data inadequate Discuss
More questions
Correct Answer: a Sin ? Sin ?/ Sin(? - ?)
Explanation:
Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that
?OAP = ?, ?OBP = ? and OB = x
In ?OBP, Use the trigonometry formula
Tan? = P/B = Perpendicular distance / Base distance
Tan? = OP/OB
? OB = OP/Tan?
? OB = OP Cot?
Put the value of OB and OP , We will get
x = h Cot ?...............(1)
In ?OAP, Similarly
Tan? = OP/OA
? OA = OP/ Tan?
? OA = OP Cot ?
Put the value of OA and OP
? a + x = h Cot ?
? x = h Cot ? - a ............(2)
From equation (1) and (2)
? h Cot ? = h Cot ? - a
? a = h Cot ? - h Cot ?
? a = h (Cot ? - Cot ?)
? a = h (Cos ?/ Sin ? - Cos ? / Sin ? )
? a = h( (Cos ? Sin ? - Cos ? Sin ? ) /Sin ? Sin ? )
? a = h( Sin(? - ?) / Sin ? Sin ?)
? h = a Sin ? Sin ?/ Sin(? - ?)
Correct Answer: 60
Explanation:
There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60
Correct Answer: gains Rs. 50
Explanation:
| S.P. | = P.W. of Rs. 2200 due 1 year hence | |||||
|
||||||
| = Rs. 2000. |
∴ Gain = Rs. (2000 - 1950) = Rs. 50.
Correct Answer: 13
Explanation:
Since 39 persons can repair a road by working 5 hours in a day in 12 days.
? 1 persons can repair a road by working 5 hours in a day in 39 x 12 days.
? 1 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 days.
? 30 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 / 30 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 12 x 5 / 30 x 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 2 x 1 / 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 1 x 1 / 3 days.
? 30 persons can repair a road by working 6 hours in a day in 13 days.
Correct Answer: 10 points
Explanation:
A : C = 100 : 72.
| ∴ | B | = | ❨ | B | x | A | ❩ | = | ❨ | 80 | x | 100 | ❩ | = | 10 | = | 100 | = 100 : 90. |
| C | A | C | 100 | 72 | 9 | 90 |
∴ B can give C 10 points.
Correct Answer: 5 kmph
Explanation:
| Then, | 30 | - | 30 | = 3 |
| x | 2x |
⟹ 6x = 30
⟹ x = 5 km/hr.
Correct Answer: 45.6 kg
Explanation:
Let the required weight be w kg.
Less length, Less weight (Direct proportion)
45 : 12 :: 171 : w
? w = (12 x 171)/45 = 45.6 kg
Correct Answer: $ 5 Million
Explanation:
Correct Answer: 24 yr
Explanation:
Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T) = (100 x y) / (y x 8) = 12.5 %
where SI= Simple Interest
P = Principal
T = Time
now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x SI)/(P x R)
= (100 x 3y )/(y x 125)= 24 yr
Correct Answer: 9
Explanation:
Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
? a + a + 2 + a + 4 - a = 20
? 2a + 6 = 20
? 2a + 6 = 20
? 2a = 20 - 6
? 2a = 14
? a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.