**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. 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

**Q3. 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

**Q4. 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

**Q5. If each side of a rectangle is increased by 50%, its area will increase by**

**(a) 125%**

(b) 75%

(c) 100%

(d) 225%

Sol.

Original area = l × b

New area =3/2 l×3/2 b=9/4 lb

Increase % =(9/4 lb – lb)/lb×100%

= 125%

**Q6. A man had a certain amount with him. He spent 20% of that to buy an article and 5% of the remaining on transport. Then he gifted Rs. 120. If he is left with Rs. 1,400, the amount he spent on transport is:**

(a) Rs. 76

(b) Rs. 61

(c) Rs. 95

**(d) Rs. 80**

Sol.

Let, the total amount be Rs. x

Now, according to the question,

∴ x – x/5-4x/5×5/100 – 120 = 1400

⇒ x – x/5-x/25 = 1520

⇒ (25x – 5x – x)/25 = 1520

⇒ 19x/25 = 1520

⇒ x = (1520 × 25)/19 = Rs. 2000

∴ Expenditure on transport

=4x/5×5/100=x/25=1/25×2000

= Rs. 80

**Q7. A shopkeeper marks his goods at 40% above their cost price. He is able to sell 3/4th of his goods at this price, and the remaining at 40% discount. Assuming that the shopkeeper is able to sell the goods he buys, find his loss or gain as % of the whole transaction.**

(a) 20% loss

(b) 23% loss

**(c) 26% gain**

(d) 30% gain

Sol.

Total C.P. = Rs. 100 (100 articles)

Total S.P. = 75 × 140/100 + 25 × 60/100 × 1.4

= 105 + 21 = Rs. 126

∴ Gain percent = 26

**Q8. A shopkeeper buys 144 items at 90 paisa each. On the way 20 items are broken. He sells the remainder at Rs. 1.20 each. His gain percent correct to one place of decimal is:**

(a) 13.8%

(b) 14.6%

**(c) 14.8%**

(d) 15.8%

Sol.

20 items are broken out of 144 items.

∴ C.P. of 124 items

= Rs. ((144 × 90)/100) = Rs. 129.60

Total S.P. = Rs. (1.20 × 124) = Rs. 148.8

∴ Gain = Rs. (148.80 – 129.60) = Rs. 19.20

∴ Gain percent = 19.20/129.60 × 100 = 14.8%

**Q9. By selling an article for Rs. 144, a person gained such that the percentage gain equals the cost price of the article. The cost price of the article is:**

(a) Rs. 90

**(b) Rs. 80**

(c) Rs. 75

(d) Rs. 60

Sol.

Let, the C.P. of the article be Rs. x.

Then,

(144 – x)/x × 100 = x

⇒ (144 – x) × 100 = x^2

⇒ x^2 + 100x – 14400 = 0

⇒ x^2 + 180x + 80x – 14400 = 0

⇒ x (x + 180) – 80 (x + 180) = 0

⇒ (x – 80) (x + 180) = 0

Therefore, x = Rs. 80

**Q10. A businessman marks his goods in such a way that even after allowing 12.5% discount on cash purchase, he gains 20%. If the cost price of the goods is Rs. 140, the marked price is:**

(a) Rs. 162

(b) Rs. 172

**(c) Rs. 192**

(d) Rs. 198

Sol.

Suppose marked price = Rs. K

∴ K – 12.5% of K = 140 + 20% of 140

⇒ 87.5% of K = 140 + 28 = 168

∴ K = 16800/87.50 = 192.