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

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