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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:
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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:
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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
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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.