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- Question
Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
Options- A. 16 years
- B. 18 years
- C. 20 years
- D. Cannot be determined
- E. None of these
- Correct Answer
- 16 years
ExplanationLet the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.Then, (6x + 6) + 4 = 11 (5x + 6) + 4 10 ⟹ 10(6x + 10) = 11(5x + 10)
⟹ 5x = 10
⟹ x = 2.
∴ Sagar's present age = (5x + 6) = 16 years.
More questions
- 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
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- 2. Find the number of diagonals formed in hexagon .
Options- A. 12
- B. 10
- C. 6
- D. 9 Discuss
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- 3. If 5a = 3125, then the value of 5(a - 3) is:
Options- A. 25
- B. 125
- C. 625
- D. 1625 Discuss
Correct Answer: 25
Explanation:
5 a = 3125 ⟺ 5 a = 5 5⟹ a = 5.
∴ 5(a - 3) = 5(5 - 3) = 52 = 25.
- 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
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- 5. (9.95) 2 x (2.01) 3 = 2 x (?) 2
Options- A. 12
- B. 15
- C. 25
- D. 20 Discuss
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- 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
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- 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
Correct Answer: 120960
Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.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.
- 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
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- 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
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- 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
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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