1. Question: The least number which when divided by 18, 27 and 36 leaves the remainders 5, 14 and 23 respectively, is :

    A
    95

    B
    113

    C
    149

    D
    77

    Note: solution: Here `(18 -5 )= 13, (27 -14 ) = 13 & (36 - 23) = 13`. `:.`Required number = `(1.c.m. of 18, 27, 36) - 13 = 95`.
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  2. Question: The least number which when divided by 20, 25, 35 and 40 leaves the remainder 14, 19, 29 and 34 respectively, is :

    A
    1664

    B
    1406

    C
    1404

    D
    1394

    Note: solution: Here `(20 - 14) = 6, (25 - 19) = 6, (35 - 29) = 6 & (40 - 34) = 6`. `:.` Required number = `(1.c.m. of 20, 25, 35, 40)- 6 = 1394`.
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  3. Question: The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is :

    A
    1677

    B
    1683

    C
    2523

    D
    3363

    Note: solution: L.C.M. of `5, 6, 7, 8 = 840`. `:.` Required number is of the from` 840k + 3`. Least value of k for which `(840k +3)` is divisible by 9 is k = 2. `:.` Required number = `(840 xx 2+3) = 1683`.
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  4. Question: The least number which when divided by 5, 6, 8, 9 and 12 leaves a remainder 1 in each case, but when divided by 13 leaves no remainder, is :

    A
    361

    B
    721

    C
    1801

    D
    3601

    Note: solution: L.C.M. of `5, 6, 8, 9, 12 = 360`. `:.` Required number is of the form 360k + 1. Least value of k for which `360k + 1` is divisible by 13 is k = 10. `:.` Required number = `(360 xx 10 +1) = 3601`.
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  5. Question: An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is :

    A
    10.30 a.m.

    B
    10.31 a.m.

    C
    10.59 a.m.

    D
    11 a.m.

    Note: solution: L.C.M. of 60 and 62 seconds is 1860 sec = 31 min. `:.` They will beep together at 10.31 a.m.
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  6. Question: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 second respectively. In 30 minutes, how many times do they toll together ?

    A
    4

    B
    10

    C
    15

    D
    16

    Note: solution: L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds i.e. 2 minutes. In 30 minutes, they will toll together in `(30/2)` + 1 =16 times.
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  7. Question: The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8 :20 :00 hours, then they will again change simultaneously at :

    A
    8 :27 : 12 hrs

    B
    8 : 27 : 24 hrs

    C
    8 : 27 : 36 hrs

    D
    8 : 27 : 48 hrs

    Note: solution: Interval of change = `(1.c.m. of 48, 72, 108) sec. = 432` sec. `:.`The lights will change simultaneously after every 432 seconds, i.e. 7 min. 12 sec. `:.` Next simultaneous change will take place at` 8 : 27 : 12` hrs.
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  8. Question: A can do a piece of work in 30 days while B alone can do it in 40 days. In how many days can A and B working together do it ?

    A
    17`1/7`

    B
    27`1/7`

    C
    42`3/4`

    D
    70

    Note: solution: A's 1 day's work= `1/30` & B's 1 day's work = `1/40`. `:.`(A+B)' s 1 day's work =`(1/30 + 1/40) = 7/120`. `:.`Both together will finish the work in `120/7 =17 1/7` days.
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  9. Question: A dealer professing to sell his at sell his goods at cost price, uses a 900 gm weight for a kilogram . His gain percent is :

    A
    9

    B
    10

    C
    11

    D
    `11 1/2`

    Note: Gain%= `((100)/900 XX 100)` % =`11 1/9` %
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  10. Question: A dishonest dealer professions to sell his goods at cost price. But his uses a 900 gm weight for a kilogram. His gain percent is :

    A
    953 gms

    B
    940 gms

    C
    960 gms

    D
    947 gms

    Note: Let error = x gms Then, `x/(1000-x) xx 100` = ` 6 (18)/47` or, `100x/(1000-x) ` = `300/47` or, `47 xx x` = ` 3(1000 - x)` or, ` 50xx x` =3000 or, x = 60 `:.` weight used =`(1000- 60) =940 gms
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