EXERCISE 1.2

1). A book exhibition was held for four days in school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Ans.  Number of tickets sold on 1st day= 1094
Number of tickets sold on 2nd day= 1812
Number of tickets sold on 3rd day= 2050
Number of tickets sold on 4th day= 2751
Therefore total number of tickets sold = 1094 + 1812 + 2050 + 2751
= 7707 tickets

2). Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many  more runs does he need?
Ans
Number of runs scored = 6980 runs
He wishes to complete = 10,000 runs
Runs needed to score more = 10,000 – 6980
= 3020
Shekhar needs 3020 more runs to score 10,000 runs

3). In an election, the successful candidate registered  5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Ans. No of votes of the successful candidate = 5,77,500
No of votes of the rival candidate = 3,48,700
Margin by which election was won = 5,77,500 – 3,48,700
= 2,28,800
Therefore, The successful candidate won by 2,28,800 votes.

4). Kirti bookstore sold books worth Rs 2,85,891 in the first week of June. The bookstore sold books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Ans.
Price of books sold in 1st week = Rs 2,85,891
Price of books sold in 2nd week = Rs 4,00,768
Total sale of books = Rs 2,85,891 + Rs 4,00,768
= Rs 6,86,659
The sale was highest in the second week.
Difference in sale of both weeks = Rs 4,00,768 – Rs 2,85,891
= Rs 1,14,877
Therefore, The sale was greater in the 2nd week by Rs 1,14,877.

5). Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Ans. Greatest number = 76432
Least number = 23467
Difference between the two numbers = 76,432 – 23,467
= 52,965
Therefore, The difference between the two numbers is 52,965.

6). A machine on an average, manufactures 2,625 screws a day. How many screws did it produce in the month of January 2006?
Ans.
No of screws manufactured on a day = 2,825
No of days in January = 31 days
Therefore, No of screws produced in January = 31 x 2,825
= 87,575
Therefore, 87,575 screws were manufactured in the month of January 2006.

7). A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Ans.
Money with the merchant = Rs 78,592
Cost of one radio set = Rs 1,200
No of radio sets to purchase = 40
Therefore, Cost of 40 radio sets = Rs 1,200 x 40
= Rs 48,000
Money left with the merchant = Money had before – Money spent
= Rs 78,592 – Rs 48,000
= Rs 30,592
Therefore, The amount remaining is Rs 30,592.

8). A student multiplied 7236 by 65 instead of multiplying by 56. How much was his answer greater than the correct answer? (Hint : You do not have to do both the multiplication).
Ans. Difference between 65 and 56                                                                                                                                                 = 65 – 56
=  9
The difference between the correct and incorrect = 7236 x 9
= 65124

9). To stitch a shirt 2 m 15 cm cloth is needed. Out of 40m cloth, how many shirts can be stitched and how much cloth will remain?
Ans. i) Total length of cloth = 40m
Total length in cm = 40 x 100 cm
= 4000 cm
Cloth required for 1 shirt = 2 m 15 cm
= 2 x 100 cm + 15 cm
= 200 + 15
= 215 cm
No of shirts that can be stitched = 4000 cm ÷ 215 cm
= 18 shirts
ii) length of remaining cloth
total length of cloth required for shirts = 215 cm x 18
= 3,870 cm.
Length of cloth remaining  = cloth available – cloth used
= 4000 cm – 3870 cm
= 130 cm
= 1 m 30 cm
Therefore, 18 shirts can be stitched out of 40 m and 1 m 30 cm cloth will be left out.

10). Medicine is packed in boxes, each weighting 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Ans.
Weight of one box = 4 kg 500 g
= 4 x 1000 g + 500 g
=4,500 g.
Maximum weight carried by the van = 800 kg
= 800 x 100 g
= 8,00,000 g
No of boxes that can be loaded in the van = 8,00,000 g ÷ 4500 g
= 177 boxes
Therefore, 177 boxes can be loaded in the van.

11). The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways between her school and home. Find the total distance covered by her in six days.
Ans. Distance between the school = 1 km 875 m
And the house               = 1000 m + 875 m
= 1875 m
Since she walks both ways in a day
Distance travelled in one day = 1875 x 2
= 3,750 m
Distance travelled by the student is = 3750 x 6
= 22,500 m
= 22 km 500 m
Therefore, The total distance covered by the student is 22 km 500 m.

12). A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity can it be filled?
Ans.
Quantity of curd in the vessel = 4 litres 500 ml
= 4 x 1000 ml + 500 ml
= 4000 ml + 500 ml
= 4500 ml
Capacity of one glass = 25 ml
No of glasses that can be filled with curd = 4500 ml ÷ 25 ml
= 180 glasses
Therefore, 180 glasses can be filled with curd.