**Find solutions for all the questions at the end of each set:**

### Set A

2.The inner and outer radii of a 7 m long hollow iron right circular cylindrical pipe are 2 cm and 4 cm respectively. If 1000 cm^3 of iron weights 5 kg, what is the weight of the pipe?

**ANSWERS AND SOLUTION :**

1.(b) ; Required surface area = πr^2+ 2πrh

= πr(r+2h)

= 22/7 * 7/2 (7/2 + 2 * 4)

= 11* (7 + 16)/2 = 11 * 23/2

= 126.5 sq.mt

2.(b) ; Required weight of pipe = π(R^2 – r^2)h

= 22/7 (4^2 – 2^2) 700

= 26400 cm^2

= 26400 * 5/1000 kg

132 kg

3.(c) Total surface area of two halves = 3πr^2 + 3πr^2

= 6πr^2

Required cost = 8 * 6πr^2 = Rs. 48πr^2

4.(c) Required average price

= (2 × 5.5 + 3 × 3.5 + 3 × 5.5 + 5 × 1.5)/((2+3+3+5))

= (11.0 + 10.5 + 16.5 + 7.5)/13

= 45.5/13

= Rs. 3.50/-

5.(a) Required time = L.C.M of (200, 300, 360 and 450) second

= 1800 seconds

6.(a); Quantity of water flows in 1 hour = πr^2h

= (22 ×7 ×7 ×5000)/(7 ×100 ×100) = 77 cm^3

Volume of water in pipe = (50 ×44 ×7)/100 154 cm^3

so Required time = 154/77 = 2 hours

7.(d) ; Let quantity of acid in mixture = x

And quantity of water in mixture = 3x

According to question,

x + 5/3x = 1/2

=> 3x = 2x + 10

=> x = 10 litres

So, Total quantity of new mixture

= x +3x + 5 = 4x + 5

= 4 * 10 + 5 = 45 litres

8.(a); Let price of a banana be = Rs. 100/-

then, price of an apple and a guava respectively will be Rs. 200/- and Rs. 250/-

Sum of prices of 4 bananas, 2 apples and 3 guavas

= 400 + 400 + 750

= Rs. 1550/-

Sum of prices of 4 bananas, 2 apples and 3 guavas

= 110/100 * 400 * 110/100 * 400 + 110/100 * 750

= 440 + 440 + 825 = Rs. 1705/-

So, increased percentage = (1705 – 1550)/1550 * 100

= 155/1550 * 100 = 10%

9.(a) Let distance between house and station be x km

According to question

x/5 – x/6 = (7+5)/60

(6x – 5x)/30 = 12/60

7x = (30* 12)/60 = 6 km

10.(c) Total increment in age = 11 * 2 = 22 months

so Total ages of two new players

= (18 + 20) years + 22 months

= 38 years + 22 months

Set B

**ANSWERS AND SOLUTION :**

1.(4) Sum of the present ages of four boys

= 9 * 4 + 20 = 56 years

Sum of the present ages of five boys

= 15 * 5 = 75 years

so Present age of new boy

= 75 – 56 = 19 years

2.(4) If the weight of a piece of diamond be 6 x units, then Original price α(6x)^2 = 36kx^2

so 36.kx^2 = 5184 ………(i)

Again,

New price = k(x^2 + 4x^2 + 9x^2)

= 14kx^2

= (14 * 5184)/36 = Rs. 2016

so Loss = 5184 – 2016

= Rs. 3168

3.(1) 30% = Rs. 30

so 100% = Rs. 100

New S.P. = 100 -30 = Rs. 70

4.(4) Part of the tank filled by both pipes in two hours

= 2 (1/8 + 1/6) = 2(3 + 4)/24 = 7/12

Remaining part = 1 – 7/12 = 5/12

Time taken by B in filling the remaining part

= 5/12 * 6 = 5/2 = 2 1/2 hours

5.(4) If the number of females be x, then number of males = 15000 – x

so x * 10/100 + (15000 –x) * 8/100

= 16300 – 15000

= 10x + 120000 – 8x

= 1300 * 100

=> 2x = 130000 – 120000 = 10000

=>x = 5000

6.(2) Interest = 5700 – 5000

= Rs. 700

so Rate = (700 * 100)/(5000 * 1) = 14%

Case II,

Interest = Principal * Time * Rate/100

= (7000 * 5 * 14)/100 = Rs. 4900

Amount = 7000 + 4900

= Rs. 11900

7.(3) Relative speed = 11 – 10 = 1 kmph

Distance covered in 6 minutes

= 1000/60 * 6 metre = 100 metre

so Remaining distance =200-100=100

8.(4) Effective profit percent

= (20 + 25 + (20 * 25)/100) = 50%

so Original cost price = 100/150 * 1200 = Rs.800

9.(1)

10.(4) According to the question.

n/2 + n/4 + n/5 + 7 = n

=> (10n + 5n + 4n)/20 + 7 = n

=> 19n/20 + 7 = n

=> n – 19n/20 = 7

=> n/20 = 7

=> n = 20 * 7 = 140