Question

A road roller (sometimes called a roller-compactor, or just roller) is a compactor-type
engineering vehicle used to compact soil, gravel, concrete, or asphalt in the construction of

roads and foundations. Similar rollers are used also at landfills or in agriculture. Road rollers

are frequently referred to as steamrollers, regardless of their method of propulsion. RCB

Machine Pvt Ltd started making road roller 10 year ago. Company increased its production

uniformly by fixed number every year. The company produces 800 roller in the 6th year and

1130 roller in the 9th year. On the basis of the above information, answer any four of the

following questions :

( I) What was the company’s production in first year ?

(a) 150 (b) 200 (c) 250 (d) 290

( II) What was the company’s production in the 8th year ?

( a ) 760 (b) 820 (c) 880 (d) 1020

(I II) What roller the company’s total production of the first 6 years?

( a ) 3150 (b) 1775 (c) 2250 (d) 1725

( I V ) What was the increase in the company’s production every year ?

(a) 160 (b) 130 (c) 90 (d) 110

( V) In which year the company’s production was 1350 rollers ?

(a) 5th (b) 6th (c) 11th (d) 9th​

Answer 1

Answer:

( 1 ) (c) 250

( 2 ) (d) 1020

( 3 ) ( a ) 3150

( 4 ) (d) 110

( 5 ) (c) 11th

Step-by-step explanation:

Note : In AP

• Tn = a + ( n - 1 ) d
• Sn = n [2a + ( n - 1 ) d ]

2

As the sales increses uniformly it is in the form of AP

Production of 6th year = 800

=> T6 = a + ( 6 - 1 ) d

=> T6 = a + 5d = 800............. eq 1

Production of 9th year = 1130

=> T9 = a + ( 9 - 1 ) d

=> T9 = a + 8d = 1130 ............. eq 2

eq 2 - eq 1

=> a + 8d = 1130

-( a + 5d ) = - 800

=> a + 8d = 1130

-a - 5d = -800

=> 3d = 330

=> d = 330/3

=> d = 110

=> Company increase production by 110 every year

Substitute d = 110 in eq 1

=> a + 5d = 800

=> a + 5 (110) = 800

=> a + 550 = 800

=> a = 800 - 550

=> a = 250

=> 1st year production will be 250

8th year production

Tn = a + ( n - 1 ) d

T8 = 250 + ( 8 - 1 ) 110

T8 = 250 + 7 (110)

T8 = 250 + 770

T8 = 1020

Sum total production of 1st 6 years

Sn = n [ 2a + ( n - 1 ) d ]

2

=> S6 = 6 [ 2 (250) + ( 6-1 ) 110 ]

2

=> S6 = 3 [ 500 + 5 (110) ]

=> S6 = 3 [ 500 + 550 ]

=> S6 = 3 ( 1050 )

=> S6 = 3,150

1350 Production in the year ( n = ? )

Tn = a + ( n - 1 ) d

=> 1350 = 250 + ( n - 1 ) 110

=> 1350 - 250 = ( n - 1 ) 110

=> 1100 = ( n - 1 ) 110

=> 1100 / 110 = ( n - 1 )

=> 10 = n - 1

=> 10 + 1 = n

=> n = 11