Exams > Cat > Quantitaitve Aptitude
NUMBERS SET II MCQs
Total Questions : 90
| Page 2 of 9 pages
Answer: Option C. -> 34
:
C
Go from answer options. The numbers are 30, 32, 34, 36 and 38. Only 34 satisfies these conditions.
:
C
Go from answer options. The numbers are 30, 32, 34, 36 and 38. Only 34 satisfies these conditions.
Answer: Option B. -> 1840
:
B
Tn=n(n+5)=n2+5n
Sn=∑Tn=∑n2+5∑n
=n(n+1)(2n+1)6+5×n(n+1)2; Put n = 15
Sn= 1840
:
B
Tn=n(n+5)=n2+5n
Sn=∑Tn=∑n2+5∑n
=n(n+1)(2n+1)6+5×n(n+1)2; Put n = 15
Sn= 1840
Answer: Option D. -> 729
:
D
5y−5y5=100;5y(1−15)=52×4;5y(45)=52×4
⟹5(y−1)×4=52×4;y−1=2;y=3. Answer is (3y)y=(9)3=729.
:
D
5y−5y5=100;5y(1−15)=52×4;5y(45)=52×4
⟹5(y−1)×4=52×4;y−1=2;y=3. Answer is (3y)y=(9)3=729.
Answer: Option C. -> 15
:
C
Conventional Solution: -Let the number be N
N = 247K + 91
247 is divisible by 19.
If N is divide by 19. Remainder will be 91 – 76 = 15.
Alternate solution :
247 is divisible by 19. The first number ,N would be 247+91 = 338. 338 when divided by 19 gives remainder 15.
:
C
Conventional Solution: -Let the number be N
N = 247K + 91
247 is divisible by 19.
If N is divide by 19. Remainder will be 91 – 76 = 15.
Alternate solution :
247 is divisible by 19. The first number ,N would be 247+91 = 338. 338 when divided by 19 gives remainder 15.
Question 15. Parliament of India has 250 MPs. Each of them drink 25 cans of cold drinks per day. The cupboard in the canteen inside the parliament has “a” rows and “a” columns for storing cans of cold drinks. If in each row, we can put only 1000 cans of cold drinks. The cans will last for (Maximum) :-
Answer: Option A. -> 160
:
A
Solution:
Total number of cans =1000×1000=1,00,0000
Daily consumption of cans =250×25=6250
The number of days, that the cans will last =1,00,00006250=160.
:
A
Solution:
Total number of cans =1000×1000=1,00,0000
Daily consumption of cans =250×25=6250
The number of days, that the cans will last =1,00,00006250=160.
Answer: Option B. -> 2500
:
B
2500(1+2×5501Odd). Hence, 2500 is the answer.
:
B
2500(1+2×5501Odd). Hence, 2500 is the answer.
Answer: Option B. -> 118
:
B
1st Method: - 71×47=3337 and 17×47=799, so ab+cd=118 (Option b).
2nd Method: - Odd × Odd = Odd and Odd + Odd = Even.
From the question we can conclude that all the four numbers are odd numbers. So the required answer has to be even, only option (b) is even.
:
B
1st Method: - 71×47=3337 and 17×47=799, so ab+cd=118 (Option b).
2nd Method: - Odd × Odd = Odd and Odd + Odd = Even.
From the question we can conclude that all the four numbers are odd numbers. So the required answer has to be even, only option (b) is even.
Answer: Option D. -> Can't be determined
:
D
Option D :
The value of F = 5 or 3; so, we can’t determine exact value of F. Answer is can’t be determined
:
D
Option D :
The value of F = 5 or 3; so, we can’t determine exact value of F. Answer is can’t be determined
Answer: Option A. -> 0
:
A
(11111)4+5111114=34+(4+1)111114=34+14=44. Thus the required remainder is 0.
:
A
(11111)4+5111114=34+(4+1)111114=34+14=44. Thus the required remainder is 0.
Answer: Option A. -> 8
:
A
Option A.
Given that the combined average is M+F, hence
(M+F)=(2MF+42){(M+F)+1}
Or (M+F)2+(M+F)=2MF+42
Or M2+F2+M+F=42
Or M(M+1)+F(F+1)=42, since M and F are integers, there is only one possibility that M(M+1) =12 or 30 and F(F+1) =30 or 12. Hence M can be 3 or 5 and F can be 5 or 3. So the sum of M and F is 8. Thus option (a).
:
A
Option A.
Given that the combined average is M+F, hence
(M+F)=(2MF+42){(M+F)+1}
Or (M+F)2+(M+F)=2MF+42
Or M2+F2+M+F=42
Or M(M+1)+F(F+1)=42, since M and F are integers, there is only one possibility that M(M+1) =12 or 30 and F(F+1) =30 or 12. Hence M can be 3 or 5 and F can be 5 or 3. So the sum of M and F is 8. Thus option (a).
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