ML Aggarwal Solution Class 10 Chapter 1 Value Added Tax MCQs

MCQs

A retailer purchases a fan for ₹ 1200 from a wholesaler and sells it to a consumer at 15% profit. If the rate of sales tax (under VAT) at every stage is 8%, then choose the correct answer from the given four options for questions 1 to 5 :

Question 1

The selling price of the fan by the retailer (excluding tax) is
(a) ₹ 1200
(b) ₹ 1380
(c) ₹ 1490.40
(d) ₹ 11296
Sol :

Cost price of a fan = ₹ 1200

Profit = 15%

∴ Sale price $=\frac{\text { C.P. } \times(100+\text { Profit } \%)}{100}$

$=₹ \frac{1200 \times(100+15)}{100}$

$=₹ \frac{1200 \times 115}{100}=₹ 1380$

Ans : (b)


Question 2

VAT paid by the wholesaler is
(a) ₹ 96
(b) ₹ 14.40
(c) ₹ 110.40
(d) ₹ 180

Sol :

Rate of VAT =8%

∴ VAT paid by wholesaler $=₹ 1200 \times \frac{8}{100}$

= ₹ 96 

Ans : (a)


Question 3

VAT paid by the retailer
(a) ₹ 180
(b) ₹ 110.40
(c) ₹ 96
(d) ₹ 14.40
Sol :

VAT deducted by the retailer $= ₹ \frac{1380 \times 8}{100}$

VAT paid by wholesalers = ₹ 96 .

Net VAT paid by his = ₹ 110.40 – 96.00

= ₹14.40

Ans : (d)


Question 4

VAT collected by the Government on the sale of fan is
(a) ₹14.40
(b) ₹96
(c) ₹110.40
(d) ₹180
Sol :

VAT collected by the govt, on the sale of fan$=₹ \frac{11040}{100}$
= ₹110.40
Ans : (c)

Question 5

The cost of the fan to the consumer inclusive of tax is
(a) ₹1296
(b) ₹1380
(c) ₹1310.40
(d) ₹1490.40
Sol :

Cost of fan to the consumer inclusive tax

= ₹1380 + 110.40

= ₹ 1490.40 (d)


A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of ₹ 32000. The shopkeeper sells the TV to a consumer at the listed price. If the sales tax (under VAT) is 6% at every stage, then choose the correct answer from the given four options for questions 6 to 8 :


Question 6

VAT paid by the distributor is
(a) ₹1920
(b) ₹1400
(c) ₹480
(d) ₹8000

Sol :

List price of T.V. set = ₹32000

Discount = 25%

Rate of VAT = 6%

Sale price $=₹ 32000 \times\left(\frac{100-25}{100}\right)$

$=₹ 32000 \times \frac{75}{100}=₹ 24000$

VAT paid by the distributor $=₹ 24000 \times \frac{6}{100}$

= ₹1440

Ans : (b)


Question 7

VAT paid by the shopkeeper is
(a) ₹480
(b) ₹1440
(c) ₹1920
(d) ₹8000

Sol :

Total VAT charged by the shopkeeper

$=₹ 32000 \times \frac{6}{100}$

= ₹1920

VAT already paid by distributor = ₹ 1440

Net VAT paid by shopkeeper

= ₹1920 – ₹1440

= ₹480 

Ans : (a)


Question 8

The cost of the TV to the consumer inclusive of tax is
(a) ₹8000
(b) ₹32000
(c) ₹33920
(d) none of these
Sol :

Cost of T.V. to the consumer inclusive of VAT = ₹32000 + 1920

= ₹33920

Ans : (c)


A wholesaler buys a computer from a manufacturer for ₹ 40000. He marks the price of the computer 20% above his cost price and sells it to a retailer at a discount of 10% on the marked price. The retailer sells the computer to a consumer at the marked price. If the rate of sales tax (under VAT) is 10% at every stage, then choose the correct answer from the given four options for questions 9 to 15 :


Question 9

The marked price of the computer is
(a) ₹40000
(b) ₹48000
(c) ₹50000
(d) none of these

Sol : 

C.R of computer for manufacturer = ₹40000

After marking 20% above the C.R, the price

$=₹ 40000 \times \frac{100+20}{100}$

$=₹ 40000 \times \frac{120}{101}$

= ₹48000 
Ans :(b)


Question 10

Cost of the computer to the retailer (excluding tax) is
(a) ₹36000
(b) ₹40000
(c) ₹43200
(d) ₹47520

Sol :

Rate of discount = 10%

∴ Sales price after discount

$=₹ 48000 \times \frac{100-10}{100}$

$=₹ 48000 \times \frac{90}{100}$

= ₹43200 (c)


Question 11

Cost of the computer to the retailer inclusive of tax is
(a) ₹47520
(b) 43200
(c) 44000
(d) none of these

Sol :

Rate of sales tax (VAT) = 10%

Sales tax charged $=₹ 43200 \times \frac{10}{100}$

= ₹4320

Cost price of T.V. including S.T.

= ₹43200 + ₹4320

= ₹47520

Ans : (a)


Question 12

VAT paid by the manufacturer is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) none of these

Sol :

VAT paid by the manufacturer

$=₹ 40000 \times \frac{10}{100}$

= ₹4000 

Ans : (a)


Question 13

VAT paid by the wholesaler is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) ₹480
Sol :

VAT paid by the wholesaler $=₹ 43200 \times \frac{10}{10}$

= ₹4320

VAT already paid = ₹4000

Net VAT paid by = ₹4320 – ₹4000

= ₹320

Ans : (c)


Question 14

VAT paid by the retailer is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) ₹480

Sol :

VAT paid by retailer $=48000 \times \frac{10}{100}=₹ 4800$

VAT already paid = ₹4320

Net VAT to be paid = ₹4800 – 4320

= ₹480

Ans : (d)


Question 15

Consumer’s cost price inclusive of VAT is
(a) ₹47520
(b) ₹48000
(c) ₹52800
(d) ₹44000

Sol :

Sale price to consumer = ₹4800

VAT paid by the consumer $=₹ 48000 \times \frac{10}{100}$

= ₹4800

Consumers cost price inclusive of VAT = ₹48000 + ₹4800

= ₹52800 

Ans : (c)

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more