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Numerical Applications Test 29

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Numerical Applications Test 29
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  • Question 1
    1 / -0
    Find the day of the week on 15 August, 1947. 
    Solution
    $$1600$$ years have $$0$$ odd days ...................(1) 
    $$300$$ years have $$1$$ odd day......................... (2) 
    $$46$$ years have $$11$$ leap years and $$35$$ ordinary years. 
    $$\displaystyle \therefore \quad \left( 11\times 2+35 \right) =57$$ odd days =$$1$$ odd day.
    Till $$30$$ June there are $$6$$ odd days ............(3) 
    From $$1$$ July to $$15$$ August there are $$46$$ days i.e. $$4$$ odd days. 
    $$\displaystyle \therefore$$  $$0 + 1 + 1 + 3 = 5$$ odd days. 
    $$\displaystyle \therefore$$ The day will be Friday. 
  • Question 2
    1 / -0
    At what time are the hands of a clock together between 5 and 6 ?
    Solution
    If the time is 5 hours and $$x$$ minutes after 5, when the two hands of the clock are same, then the angle formed by minute hand is: $$\cfrac{x}{60} \times 360 = 6x$$
    The hour hand moves $$\cfrac{1}{2}$$ degree each minute, then angle traced by hour hand is = $$150 + \cfrac{x}{2}$$
    Hence, $$6x = 150 + \cfrac{x}{2}$$
    $$150 = \cfrac{11x}{2}$$
    $$x = \cfrac{300}{11}$$
    $$x = 27 \cfrac{3}{11}$$ min. past 5
  • Question 3
    1 / -0
    A does half as much work as B in three-fourth of the time. If together they take $$18$$  days to complete the work, how much time shall B take to do it?
    Solution
    Let B takes x days to do the work.
    According to the question 
    A takes $$2\times \dfrac{3x}{4}=\dfrac{3x}{2}$$
    (A+B)s 1 days work=$$\dfrac{1}{18}$$
    then$$\dfrac{1}{x}+\dfrac{2}{3x}=\dfrac{1}{18}$$
    $$\dfrac{5}{3x}=\dfrac{1}{18}$$
    $$ x=\dfrac{18\times 5}{3}=30$$
    Hence B takes 30 days to do work
  • Question 4
    1 / -0
    I carried $$1000$$ kg of watermelon in summer by train. In the beginning, the water content was $$99\%$$. By the time I reached the destination, the water content had dropped to $$98\%$$. The reduction in the weight of the watermelon was
    Solution
    Water + solid = 1000
    Water is  $$\displaystyle  \frac{99}{100} \times 1000 = 990$$
    water evaporated is x.
    so $$\displaystyle \left ( \frac{990 - x}{1000 - x} \right ) 100 = 98$$
    $$99000 - 100 x = 96000 - 98 x$$
    $$1000 = 2x              \Rightarrow x = 500$$
  • Question 5
    1 / -0
    Find the day of the week on 26 January, 1950.
    Solution
    Total number of odd days in first 1600 yeas = 0
    Number of odd days in next 300 years = 1
    Number of odd days in 49 years = $$12 \times 2 + 37 \times 1 = 61$$ days or  5 odd days
    Thus total number of odd days in 1949 = 1 + 5 = 6 days
    Number of days in 190 = 26 = 3 weeks + 5 days.
    Thus, number of odd days in 1950 = 5

    Hence, total number of odd days = 5 + 6 = 11 = 1 week + 4 days
    Now, 0 odd days then Sunday, 1 odd day then Monday and so on
    Since, 4 odd days hence it is Thursday.
  • Question 6
    1 / -0
    How many times are the hands of a clocks coincide in a day ?
    Solution
    The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o'clock).
    AM
    12:00,1:05,2:11,3:16,4:22,5:27,6:33,7:38,8:44,9:49,10:55
    PM
    12:00,1:05,2:11,3:16,4:22,5:27,6:33,7:38,8:44,9:49,10:55
    The hands overlap about every 65 minutes, not every 60 minutes. The hands coincide 22 times in a day.
  • Question 7
    1 / -0
    Find the day of the week on 15 August, 1947.
    Solution
    15th August 1947 has 1946 years and period from 1st January 1947 to 15th August 1947.
    Now, first 1600 years have 0 days.
    Next 300 years have 1 odd day.
    46 years = 11 leap years + 35 non leap years = $$11\times 2 + 35 = 57$$days = 1 odd day
    Number of days from 1st Jan 1947 to 15th August 1947 = 227 days = 3 odd days
    Hence, total number of odd days = $$ 0 + 1 + 1 + 3 = 5$$
    Hence, if 0 odd day it is Sunday, 1 odd day it is a Monday and so on.
    Hence 5 odd days is a Friday.
  • Question 8
    1 / -0
    If February 1, 1966 is Wednesday, what day is March 10, 1966 ?
    Solution
    February 1, 1966 is a Wednesday. Number of days between 1st February and 10th March is = $$27 + 10 = 37$$
    Now, $$\cfrac{37}{7} = 5 + \dfrac{2}{7}$$
    Thus, there are 2 odd days between the two given dates. Hence, the 10th March will be: Friday (2 days after Wednesday)
  • Question 9
    1 / -0
    How many times do the hands of a clock coincide in a day?
    Solution
    The hands of a clock coincide $$11$$ times in every $$12$$ hours . The hands overlap about every $$65$$ minutes, not every $$60$$ minutes. Thus the minute hand and the hour hand coincide $$22$$ times in a day
  • Question 10
    1 / -0
    A fort had provision of food for $$150$$ men for $$45$$ days. After $$10$$ days $$25$$ men left the fort . The number of days for which the remaining food will last is__________.
    Solution
     fort had provision of food for 150 men for 45 days,
    After 10 days 25 men left the fort, so 125 men are there in the fort.
    So, now fort is having food for 150 men for 35 days.

    Suppose 125 men had food for x days.

    Now, Less men, More days (Indirect Proportion)

    $$125\times x=150\times 35$$

    $$x=42days$$

    Answer (C) 42days

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