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- Question
If in the following set of numbers, the first and the third digits are interchanged in each number, which number will be second from the end if arranged in ascending order after interchanging the digits ? 585 546 514 404 206 369
Options- A. 404
- B. 585
- C. 415
- D. 602
- Correct Answer
- 415
ExplanationGiven numbers are 585 546 514 404 206 369
Then interchanging the 1st and 3rd digits of the given numbers, we get
585 645 415 404 602 963
Now, arranging them in ascending order
963 645 602 585 415 404
Then the last second number is 415. - 1. A, B, C, D, E, F are the only six families in Indira Nagar. A, B, C, D, E and F has 7, 8, 10, 13, 6, and 10 member in their families respectively. If 1 member from all the six families left their respective families to accommodate themselves in the hostel of IIM Lucknow, then the average number of member now in each family of Indira nagar is
Options- A. 8
- B. 9
- C. 10
- D. 13 Discuss
- 2. Find the number of diagonals formed in hexagon .
Options- A. 12
- B. 10
- C. 6
- D. 9 Discuss
- 3. If 5a = 3125, then the value of 5(a - 3) is:
Options- A. 25
- B. 125
- C. 625
- D. 1625 Discuss
- 4. A parallelogram has sides 60 m and 40 m and one of the diagonal is 80 m long. Then its area is?
Options- A. 480 sq. m
- B. 320 sq. m
- C. 600? 15 sq. m
- D. 450 ? 15 sq. m Discuss
- 5. (9.95) 2 x (2.01) 3 = 2 x (?) 2
Options- A. 12
- B. 15
- C. 25
- D. 20 Discuss
- 6. In how many ways, a cricket team of 11 players can be made from 15 players, if a particulars player is always chosen?
Options- A. 1835
- B. 1001
- C. 1635
- D. 1365 Discuss
- 7. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
Options- A. 10080
- B. 4989600
- C. 120960
- D. None of these Discuss
- 8. A department had 8 male and female employees each. A project team involving 3 male and 3 female members needs to be chosen from the department employees. How many different projects teams can be chosen?
Options- A. 112896
- B. 3136
- C. 720
- D. 112 Discuss
- 9. In how many different ways, 5 boys and 5 girls can sit on a circular table, so that the boys and girls are alternate?
Options- A. 2880
- B. 2800
- C. 2680
- D. 2280 Discuss
- 10. If ABCD is a parallelogram and AC and BD be its diagonals, then:
Options- A. AB2 + BC2+ CD2+ DA2 = AC2 ? BD2
- B. AB2 + BC2+ CD2 + DA2 = AC2 + BD2
- C. 4AD2 = 2AC2 + 2BD2
- D. 4AB2 = 2AC2 ? 2BC2 Discuss
More questions
Correct Answer: 8
Explanation:
Required average
= (( 7 - 1 ) + ( 8 - 1 ) + ( 10 - 1 ) + ( 13 - 1 ) + (6 - 1) +( 10 - 1 ) )/ 6
= ( 7 + 8 +10 + 13 + 6 + 10 ) / 6 - ( 6 x 1 ) / 6
=9 - 1 = 8
Correct Answer: 9
Explanation:
Hexagon has 6 sides .
? n = 6
? Required number of diagonals = 10C2 - n
= 6C2 - 6
= 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6
= (6 x 5 x 4!) / (2 x 4!) - 6
= 15 - 6
= 9
Correct Answer: 25
Explanation:
5 a = 3125 ⟺ 5 a = 5 5
⟹ a = 5.
∴ 5(a - 3) = 5(5 - 3) = 52 = 25.
Correct Answer: 600? 15 sq. m
Explanation:
AB = 60 m, BC = 40 m and AC = 80 m
? s = (60 + 40 + 80 ) / 2 m = 90 m
(s-a) = 90 - 60 = 30 m,
(s-b) = 90 - 40 = 50 m and
(s-c) = 90 - 80 = 10 m
? Area of ? ABC =
?s(s-a)(s-b)(s-c)
= ?90 x 30 x 50 x 10 m2
= 300?15 m2
? Area of parallelogram ABCD = 2 x area of ? ABC
= 600?15 m2
Correct Answer: 20
Explanation:
(9.95)2 x (2.01)3 = 2 x (?)2
Here, 9.95 is Approximated to 10, 2.01 is approximate to 2 because calculating the square or cubes of whole number is much easier than to calculate the square or cube of decimal numbers
2 x (?) 2 = ( 10)2 x (2)3 = 100 x 8 = 800
? (?)2 = 400
? ? =20
Correct Answer: 1001
Explanation:
Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players.
? Required number of ways = 14C10
= 14 ! / (10! x 4!)
= (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1)
= 7 x 13 x 11
= 91 x 11
= 1001
Correct Answer: 120960
Explanation:
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
| ∴ Number of ways of arranging these letters = | 8! | = 10080. |
| (2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
| Number of ways of arranging these letters = | 4! | = 12. |
| 2! |
∴ Required number of words = (10080 x 12) = 120960.
Correct Answer: 3136
Explanation:
Total ways = 8C3 x 8C3
= (8 x 7 x 6)/(3 x 2) x (8 x 7 x 6)/(3 x 2)
= 56 x 56 = 3136
Correct Answer: 2880
Explanation:
After fixing up one boy on the table, the remaining can be arranged in 4 ! ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 place can be filled by 5 girls in 5 ! ways .
Hence, by the principle of multiplication, the required number of ways = 4 ! x 5 ! = 2880
Correct Answer: AB2 + BC2+ CD2 + DA2 = AC2 + BD2
Explanation:
In ||gm ABCD , we have
Since diagonals of parallelogram(||gm ) bisect each other,
? M will be the mid-point of each of the diagonal AC and BD
? In ?ABC AB2 + BC2 = 2(AM2 + MB2) ......... ( 1 ) [Appolonius Theorem]
In ?ADC AD2 + CD2 = 2(AM2 + DM2) ......... ( 2 )
= 2(AM2 + MB2) [? DM = BM]
Adding equations ( 1 ) and ( 2 )
AB2 + BC2 + CD2 + DA2 = 2(AM2 + MB2 + 2(AM2 + MB2 = 4AM2 + 4MB2
= (2AM)2 + (2MB)2= AC2+ BD2. { ? AM = MC , MB = MD }
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