Q1. Gorang worked 4 1/2 hours on Monday, 190 minutes Tuesday, from 5 : 20 a.m. to 9 : 10 a.m. on Wednesday, and 220 minutes on Friday. He is paid Rs. 42 per hour. How much did he earn from Monday to Friday
(a) Rs. 560
(b) Rs. 580
(c) Rs. 540
(d) Rs. 637
Sol.
Work done by Gorang on Monday
=4 1/2 h=9/2 h
Work done by Gorang on Tuesday =190/60 h=19/6 h
Work done by Gorang on Wednesday
= (9:10 – 5:20)
=3h+5/6 h
=23/6 h
Work done by Gorang on Friday
=220/60=11/3 h
Total working hours by Gorang
=9/2 h+19/6 h+23/6 h+11/3 h
=((27 + 19 +23 +22))/6 h=91/6 h
Paid@ Rs. 42/h
So, total amount earn by Gorang from Monday to Friday
=91/6×42 = Rs. 637.
Q2. The sum of the greatest 4-digit number and the smallest 3-digit number is
(a) 7000
(b) 9899
(c) 10099
(d) 10999
Sol.
Greatest four digit number = 9999
Smallest three digit number = 100
Required sum = 9999 + 100 = 10099.
Q3. Twenty-six and twenty-six hundredths is written as
(a) 2626
(b) 26.26
(c) 262.6
(d) 2.626
Sol.
Twenty six = 26
Twenty-six hundredth =26/100=0.26
So, adding two, we get =26+26/100 = 26.26
Q4. What number should be subtracted from the product 1109 × 505 so as the get 505050?
(a) 49495
(b) 55005
(c) 54995
(d) 59495
Sol.
Product = 1109 × 505 = 560045
According to question
⇒ 560045 – 505050 = 54995
Q5. A tank contains 240 litres (L) 128 millilitres (mL) of milk, which can be filled completely in 16 jars of the same size. How much milk will be their in 22 such jars?
(a) 330 L, 176 mL
(b) 331 L, 760 mL
(c) 331 L, 176 mL
(d) 332 L, 650 mL
Sol.
As, 1L = 1000 mL
240 litres 128 millilitres
= 240 L + 128 mL
= 240 × 1000 + 128
= 240000 + 128
= 240128 mL
Now,
Milk in 1 Jar =240128/16 mL
Milk in 22 Jars =240128/16×22 mL
= 330176 mL
= 330 L, 176 mL
Q6. Number of degrees in four and two-third right-angles is
(a) 310
(b) 420
(c) 330
(d) 400
Sol.
Right angle = 90 degrees
Four and two third angle =(4+2/3)×90°
=14/3×90°=420°
Q7. A water tank is 11 m long, 10 m wide and 9 m high. It is filled with water to a level of 6 m. What part of the tank is empty?
(a) 1/4
(b) 1/3
(c) 1/6
(d) 2/3
Sol.
Volume of empty space = 11 × 10 × (9 – 6)
= 11 × 10 × 3
Volume of entire tank = 11 × 10 × 9
Hence, the part that is empty of the
entire tank =(11 × 10 × 3)/(11 × 10 × 9)=1/3
Q8. Perimeters of a rectangle and a square are equal. Perimeter of the square is 96 cm and breadth of the rectangle is 4 cm less than the side of the square. Then two times the area (in square cm) of the rectangle is
(a) 560
(b) 960
(c) 1040
(d) 1120
Sol.
Let length of rectangle be x cm and breadth by y cm.
Perimeter of rectangle = 2(x + y)
Let side of a square = a cm
Perimeter of a square = 4a
Now, perimeter of square = 96 cm.
⇒ 4a = 96
∴ a = 24 cm
Breadth of a rectangle (y) = 24 – 4 = 20 cm
According to question
⇒ 2(x + 20) = 96
⇒ x + 20 = 48
∴ x = 48 – 20 = 28 cm
Area of rectangle = 28 × 20 = 560 cm2
Two times of the area of rectangle
= 2 × 560 cm2
= 1120 cm2
Q9. 10290 books are to be packed in wooden boxes. If 98 books can be packed in one box and the cost of one box is Rs. 518, then the total cost of boxes needed for this purpose is:
(a) Rs. 54,390
(b) Rs. 54,292
(c) Rs. 54,908
(d) Rs. 53,872
Sol.
Total number of books = 10290
One box contains = 98 books
Total number of box =10290/98=105
The cost price of one box = Rs. 518
∴ Total cost price of 105 boxes
= 518 × 105 = Rs. 54,390
Q10. The sum of remainders obtained on dividing 12112 by 11 and 13223 by 13 is:
(a) 3
(b) 4
(c) 5
(d) 2
Sol.
When 12112 is divided by 11, it gives remainder 1 and when 13223 is divided by 13, it gives remainder ‘2’.
Hence, sum of the remainders obtained = 1 + 2 = 3
