\u2235<\/strong> Baggage deliver 3 tonnes in 5 min.<\/p>\n\n\n\n\u2234 Baggage deliver in 1 min = 3\/6 tonnes.<\/p>\n\n\n\n
For second belt, <\/p>\n\n\n\n
\u2235<\/strong> Baggage deliver 1 tonne in 2 min.<\/p>\n\n\n\n\u2234 Baggage deliver in 1 min = 1\/2 tonne<\/p>\n\n\n\n
\u2234 Total baggage deliver in 1 min = 3\/5 + 1\/2 = 11\/10 tonnes.<\/p>\n\n\n\n
\u2234 11\/10 tonnes delivered in 10\/11 min.<\/p>\n\n\n\n
Hence, time taken to deliver 33 tonnes = 10\/11 X 33 = 30 min.<\/p>\n\n\n\n
<\/span>Q6. Three bells toll at intervals of 9, 12 and 15 min respectively. All the three begin to toll at 8 am. At what time will they toll together again? (2003)<\/strong><\/span><\/h2>\n\n\n\n(a) 8:45 am<\/p>\n\n\n\n
(b) 10:30 am<\/p>\n\n\n\n
(c) 11:00 am<\/p>\n\n\n\n
(d) 1:30 pm<\/p>\n\n\n\n
<\/span>Ans. (c)<\/strong><\/span><\/h3>\n\n\n\nThe bells will ring simultaneously again after a duration equal to the least common multiple (LCM) of their individual tolling intervals, which is the LCM of 9, 12, and 15, resulting in 180 minutes. Thus, they will ring together again after 180 minutes, or 3 hours. Adding 3 hours to the initial time of 8:00 am gives us 11:00 am.<\/p>\n\n\n\n
<\/span>Q7. There are 6 persons; A, B, C, D, E and F. (2005)<\/strong><\/span><\/h2>\n\n\n\nA has 3 items more than C.<\/p>\n\n\n\n
D has 4 items less than B. <\/p>\n\n\n\n
E has 6 items less than F. <\/p>\n\n\n\n
C has 2 items more than E. <\/p>\n\n\n\n
F has 3 items more than D. <\/p>\n\n\n\n
Which one of the following figure cannot be equal to the total number of items possessed by all the 6 persons?<\/em><\/p>\n\n\n\n(a) 41 <\/p>\n\n\n\n
(b) 47 <\/p>\n\n\n\n
(c) 53 <\/p>\n\n\n\n
(d) 59<\/p>\n\n\n\n
<\/span>Ans. (d)<\/strong><\/span><\/h3>\n\n\n\nAccording to the given equations:<\/p>\n\n\n\n
[ A = C + 3 ]
[ D = B – 4 ]
[ E = F – 6 ]
[ C = E + 2 ]
[ F = D + 3 ]<\/p>\n\n\n\n
Adding all the equations together, we get:<\/p>\n\n\n\n
[ A = B – 2 ]<\/p>\n\n\n\n
Substituting this into the expression for the total number of items, we have:<\/p>\n\n\n\n
[ Total number of items= A + B + C + D + E + F = A + (A + 2) + (A – 3) + (A – 2) + (A – 5) + (A + 1) = 6A – 7 ]<\/p>\n\n\n\n
If ( A = 8 ), then the total number of items is ( 8 \\times 6 – 7 = 41 ).<\/p>\n\n\n\n
If ( A = 9 ), then the total number of items is ( 9 \\times 6 – 7 = 47 ).<\/p>\n\n\n\n
If ( A = 10 ), then the total number of items is ( 10 \\times 6 – 7 = 53 ).<\/p>\n\n\n\n
If ( A = 11 ), then the total number of items is ( 11 \\times 6 – 7 = 59 ).<\/p>\n\n\n\n
Thus, option (d) is the correct answer.<\/p>\n\n\n\n
<\/span>Q8. Each of the five persons A, B, C, D and E possesses unequal number of similar items. A, B and C possesses twenty-one items in all, while C, D and E possess seven items in all. How many items do A and B possess in all? (2006)<\/strong><\/span><\/h2>\n\n\n\n(a) 15<\/p>\n\n\n\n
(b) 17<\/p>\n\n\n\n
(c) 18<\/p>\n\n\n\n
(d) Data is insufficient<\/p>\n\n\n\n
<\/span>Ans. (b)<\/strong><\/span><\/h3>\n\n\n\nAccording to the question,<\/p>\n\n\n\n
A + B + C = 21 …(i)<\/p>\n\n\n\n
C + D + E = 7 ….(ii)<\/p>\n\n\n\n
for eq. (ii) C can take values 1, 2 and 4 as 1 + 2 + 4 = 7<\/p>\n\n\n\n
For C, = 1, A + B + 1 = 21, A + B = 20<\/p>\n\n\n\n
Similarly, for C = 2, A + B = 21 – 2 = 19<\/p>\n\n\n\n
and for C = 4, A + B = 21 – 4 = 17<\/p>\n\n\n\n
<\/span>Q9. If all the numbers from 501 to 700 are written, what is the total number of times does the digit 6 appear? (2007)<\/strong><\/span><\/h2>\n\n\n\n(a) 138 <\/p>\n\n\n\n
(b) 139 <\/p>\n\n\n\n
(c) 140<\/p>\n\n\n\n
(d) 141<\/p>\n\n\n\n
<\/span>Ans. (c)<\/strong><\/span><\/h3>\n\n\n\nFor numbers between 600 and 700:
Number of occurrences of 6 at the unit’s place = 10
Number of occurrences of 6 at the ten’s place = 10
Number of occurrences of 6 at the hundredth place = 100<\/p>\n\n\n\n
For numbers between 501 and 599:
Number of occurrences of 6 at the unit’s place = 10
Number of occurrences of 6 at the ten’s place = 10<\/p>\n\n\n\n
Hence, the total number of occurrences of 6 between 501 and 700 is:<\/p>\n\n\n\n
10 + 10 + 100 + 10 + 10 = 140<\/p>\n\n\n\n
<\/span>Q10. A person has to completely put each of three liquids: 403 L of petrol, 465 L of diesel and 496 L of mobile oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required? (2007)<\/strong><\/span><\/h2>\n\n\n\n(a) 34<\/p>\n\n\n\n
(b) 44<\/p>\n\n\n\n
(c) 46<\/p>\n\n\n\n
(d) None of these<\/p>\n\n\n\n
<\/span>Ans. (b)<\/strong><\/span><\/h3>\n\n\n\nMaximum capacity of each bottle can be found by taking the HCF of the three given liquids.<\/p>\n\n\n\n
Maximum capacity of each bottle = HCF of 403, 465 and 496 = 31<\/p>\n\n\n\n
Number of bottles for 403 L petrol = 403\/32 = 13<\/p>\n\n\n\n
Number of bottles for 465 L of diesel = 465 \/ 31 = 15<\/p>\n\n\n\n
Number of bottles for 496 L of mobile oil = 496 \/ 31 = 16<\/p>\n\n\n\n
Hence, total number of bottles = 13 + 15 + 16 = 44.<\/p>\n\n\n\n
<\/span>Q11. What is the number of terms in the series 117, 120, 123, 126, …, 333? (2008)<\/strong><\/span><\/h2>\n\n\n\n(a) 72 <\/p>\n\n\n\n
(b) 73 <\/p>\n\n\n\n
(c) 76 <\/p>\n\n\n\n
(d) 79<\/p>\n\n\n\n
<\/span>Ans. (b)<\/strong><\/span><\/h3>\n\n\n\nGiven series is 117, 120, 123, 126, …. 333. <\/p>\n\n\n\n
Given series is an AP series with first term, a = 117 last term, 1333 and common difference, d = 3<\/p>\n\n\n\n
Last term, l = a + (n – 1) * d <\/p>\n\n\n\n
where, n = number of terms. <\/p>\n\n\n\n
\u21d2 117 + (n – 1) * 3 = 333 <\/p>\n\n\n\n
\u21d2 (n – 1) * 3 = 216 <\/p>\n\n\n\n
\u21d2 n – 1 = 72 <\/p>\n\n\n\n
\u2234 n = 72 + 1 = 73<\/p>\n\n\n\n
<\/span>Q12. Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut? (2009)<\/strong><\/span><\/h2>\n\n\n\n(a) 27 <\/p>\n\n\n\n
(b) 36 <\/p>\n\n\n\n
(c) 43 <\/p>\n\n\n\n
(d) 400<\/p>\n\n\n\n
<\/span>Ans. (b)<\/strong><\/span><\/h3>\n\n\n\nSince each rod must be cut into parts of equal length, and each part must be as long as possible, the highest common factor (HCF) should be considered.<\/p>\n\n\n\n
HCF of 78, 104, 117 and 169=13<\/p>\n\n\n\n
Number of parts from 78 cm rod = 78\/13 = 6<\/p>\n\n\n\n
Number of parts from 104 cm rod = 104\/13 = 8<\/p>\n\n\n\n
Number of parts from 117 cm rod = 117\/13 = 9<\/p>\n\n\n\n
Number of parts from 169 cm rod = 169\/13 = 13<\/p>\n\n\n\n
\u2234 Maximum number of pieces=6+8+9+13=36<\/p>\n\n\n\n
<\/span>Q13. While adding the first few continuous natural numbers, a candidate missed one of the numbers and wrote the answer as 177. What was the number missed? (2009)<\/strong><\/span><\/h2>\n\n\n\n(a) 11 <\/p>\n\n\n\n
(b) 12<\/p>\n\n\n\n
(c) 13 <\/p>\n\n\n\n
(d) 14<\/p>\n\n\n\n
<\/span>Ans. (c)<\/strong><\/span><\/h3>\n\n\n\nLet there be n natural numbers and ‘x’ be the number missed out.<\/p>\n\n\n\n
Now, (n(n + 1)\/2) – x = 177<\/p>\n\n\n\n
\u21d2 n(n+1)- 2x = 354<\/p>\n\n\n\n
\u21d2 n(n+1)=354+2x<\/p>\n\n\n\n
Out of the given options only x 13 satisfies it, as n = 19 and x = 13<\/p>\n\n\n\n
19(19+1)=354 + 2(13) = 380.<\/p>\n\n\n\n
\u21d2 380 = 380<\/p>\n\n\n\n
hence, x = 13<\/p>\n\n\n\n
<\/span>Q14. There are five hobby clubs in a college viz, photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day. How many times do all the five groups meet on the same day within 180 days? (2013)<\/strong><\/span><\/h2>\n\n\n\n(a) 3 <\/p>\n\n\n\n
(b) 5 <\/p>\n\n\n\n
(c) 10 <\/p>\n\n\n\n
(d) 18<\/p>\n\n\n\n
<\/span>Ans. (a)<\/strong><\/span><\/h3>\n\n\n\nThe gardening group meets once every 2 days, the electronics group meets once every 3 days, the chess group meets once every 4 days, the yachting group meets once every 5 days, and the photography group meets once every 6 days.<\/p>\n\n\n\n
If they meet on the same day once, the next time they will meet on the same day again will be the least common multiple (LCM) of 2, 3, 4, 5, and 6, which equals 60 days.<\/p>\n\n\n\n
Hence, within 180 days, all five groups will meet on the same day 180\/60 = 3 times.<\/p>\n\n\n\n
<\/span>Q15. The letters L, M, N, O, P, Q, R, S and T in their order are substituted by nine integers 1 to 9 but not in that order. 4 is assigned to P. The difference between P and Tis 5. The difference between Nand Tis 3. What is the integer assigned to N? (2014)<\/strong><\/span><\/h2>\n\n\n\n(a) 7<\/p>\n\n\n\n
(b) 5<\/p>\n\n\n\n
(c) 4<\/p>\n\n\n\n
(d) 6<\/p>\n\n\n\n
<\/span>Ans. (d)<\/strong><\/span><\/h3>\n\n\n\nAccording to the question,<\/p>\n\n\n\n
P=4<\/p>\n\n\n\n
Now, difference between P and T = 5<\/p>\n\n\n\n
\u21d2 T-P=5<\/p>\n\n\n\n
\u21d2 T-4=5<\/p>\n\n\n\n
\u21d2 T=9<\/p>\n\n\n\n
Now, again difference between N and T = 3<\/p>\n\n\n\n
\u21d2 T-N=3<\/p>\n\n\n\n
\u21d2 9-N=3<\/p>\n\n\n\n
\u21d2 N=6<\/p>\n\n\n\n
Hence, option (d) is correct.<\/p>\n\n\n\n
<\/span>Q16. For a charity show, the total tickets sold were 420 Half of these tickets were sold at the rate of \u20b95 each, one-third at the rate of \u20b93 each and the rest for \u20b9 2 each. What was the total amount received? (2014)<\/strong><\/span><\/h2>\n\n\n\n(a) \u20b9900 <\/p>\n\n\n\n
(b) \u20b91540 <\/p>\n\n\n\n
(c) \u20b91610 <\/p>\n\n\n\n
(d) \u20b92000<\/p>\n\n\n\n
<\/span>Ans. (c)<\/strong><\/span><\/h3>\n\n\n\nGiven, total number of tickets = 420<\/p>\n\n\n\n
Number of tickets sold at \u20b95= 420\/2 = 210<\/p>\n\n\n\n
Number of tickets sold at \u20b93 = 420\/3 = 140<\/p>\n\n\n\n
Remaining number of tickets sold at \u20b92=420-(210+140)<\/p>\n\n\n\n
=420-350=70 tickets<\/p>\n\n\n\n
(oros) Now, total amount received<\/p>\n\n\n\n
= (210 x5)+(140 x 3) + (70×2)<\/p>\n\n\n\n
=1050+420 + 140<\/p>\n\n\n\n
= \u20b91610<\/p>\n\n\n\n
Hence, option (c) is correct.<\/p>\n\n\n\n
<\/span>