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Exercise 1

1.         Answer:
30 days.
            Explanation:
            Before:
            One day work                          =          1 / 20
            One man’s one day work          =          1 / ( 20 * 75)
Now:
                        No. Of workers                        =          50 
                        One day work                          =          50 * 1 /  ( 20 * 75)

                        The total no. of days required to complete the work = (75 * 20) / 50  = 30

2.         Answer:
 0 %
            Explanation:
            Since 3x / 2  = x / (2 / 3)

3.         Answer:
5.3 %
            Explanation:
                        He sells 950 grams of pulses and gains 50 grams.
            If he sells 100 grams of pulses then he will gain (50 / 950) *100  =  5.26

4.         Answer:
250 lines of codes

5.         Answer:
7 days
Explanation:
The equation portraying the given problem is:
                         10 *  x – 2 * (30 – x) =  216   where x is the number of working days.
            Solving this we get x = 23
            Number of days he was absent was 7 (30-23) days.

6.         Answer:
150 men.
Explanation:
            One day’s work                       =          2 / (7 * 90)
            One hour’s work                      =          2 / (7 * 90 * 8)
            One man’s work                       =          2 / (7 * 90 * 8 * 75)

The remaining work (5/7) has to be completed within 60 days, because the total number of days allotted for the project is 150 days.

            So we get the equation
                       
                        (2 * 10 * x * 60) / (7 * 90 *  8 * 75)   =  5/7  where x is the number of men working after the 90th day.

            We get x = 225
            Since we have 75 men already, it is enough to add only 150 men.

7.         Answer:
(c) 1
Explanation:
            a percent of b : (a/100) * b
            b percent of a : (b/100) * a
            a percent of b divided by b percent of a : ((a / 100 )*b) /  (b/100) * a )) = 1

8.         Answer:
Cost price of horse =  Rs. 400 & the cost price of cart = 200.
Explanation:-
            Let x be the cost price of the horse and y be the cost price of the cart.
            In the first sale there is no loss or profit. (i.e.) The loss obtained is equal to the gain.

                        Therefore         (10/100) * x     =  (20/100) * y

                                                            X         =  2 * y     -----------------(1)
            In the second sale, he lost Rs. 10. (i.e.) The loss is greater than the profit by Rs. 10.

                        Therefore         (5 / 100) * x     =  (5 / 100) * y + 10 -------(2)
                        Substituting (1) in (2) we get
                                    (10 / 100) * y   =  (5 / 100) * y + 10
                                    (5 / 100) * y     =  10
                                    y = 200           
From (1) 2 * 200 = x = 400

9.         Answer:
 3.
            Explanation:
            Since inclusion of any male player will reject a female from the team. Since there should be four member in the team and only three males are available, the girl, n should included in the team always irrespective of others selection.

10.       Answer:
5

11.       Answer:
1,2,3 & 4

12.       Answer:
B

13.       Answer:
11 & 9 apples per tree.
            Explanation:
            Let a, b, c, d & e be the total number of apples bored per year in A, B, C, D & E ‘s orchard. Given that             a + 1 = b + 3 = c – 1 = d + 3 = e – 6 
But the question is to find the number of apples bored per tree in C and D ‘s orchard. If is enough to consider c – 1 = d + 3.
            Since the number of trees in C’s orchard is 11 and that of D’s orchard is 13. Let x and y be the number of apples bored per tree in C & d ‘s orchard respectively.
            Therefore 11 x – 1 = 13 y + 3
By trial and error method, we get the value for x and y as 11 and 9
           
14.       Answer:
G.
            Explanation:

            The order in which they are climbing is R – G – K – H – J