Nepal Telecom Assistant Level 4 Examination Paper 2075 with Solution of Numericals

Nepal Telecom announced various post and out of them the recent question paper of Assistant Post of level 4 held on on 2075-05-29 is attached below. Also the solutions of Numerical of Mathematics are also done below:

Solution of Question 7:

--->

A can do one work in 20 days.

A can do 1/20 work in 1 days. 

A can do 10/20=1/2 work in 10 days -------- 1

B can do one work in 30 days.

B can do 1/30 work in 1 days.

B can do 10/30=1/3 work in 10 days --------- 2

As from 1 and 2, we have

A+B can do 1/2+1/3 = 5/6 work in 10 days.

A+B can do 1 work in 10*6/5=60/5=12 days

Now Remaing work to do is 1-5/6=1/6

Thus, A+B can do 1/6 work in 12*1/6 = 2 days

Ans: 2 Days.

Solution of Question No. 8:

--->

As given from question:

Cost Price (C.P.) For Annapurna Computer = 67800

Profit % = 12

Delivery Charge = 1500

Selling Price (Without VAT) = ?

Selling Price (With VAT) = ?

we know that,

Proft % = (S.P. - C.P.)*100/C.P

12 = (S.P. - 67800)*100/67800

Net S.P. = 678*12+67800 = 75936

VAT = Net S.P.*13% = 75936*13/100 = 9871.68

Thus,

Selling Price (Without VAT) = Net S.P. + Delivery Charge = 75936 + 1500 = 77436 Ans.

Selling Price (With VAT) = Net S.P. + VAT + Delivery Charge = 75936 + 9871.68 + 1500 = 87307.68 Ans.

Solution of Question No. 9

--->

As per question,

15 chair + 2 table = rs. 4000

3 table + 12 chair = rs. ?

Now,

We have,

2 table + 15 chair  = rs. 4000

1 table + 15 chair = rs. 4000/2 = 2000

3 table + 15 chair = rs. 2000*3 = 6000

3 table + 1 chair = rs. 6000/15 = 400

3 table + 12 chair = rs. 400*12 = 4800 Ans.

Thus the price of 3 table and 12 chairs is Rs. 4800.